Monsieur Ali Ridha Mahjoub

Je Voudrais Remercier Madame

Professeur Hande Yaman

Monsieur Eduardo Uchoa

Nancy Madame

Perrot

Keywords:

Cette thèse m'a permis de consolider mes connaissances et de confirmer ma passion pour le domaine de la Recherche Opérationnelle en général et l'Optimisation Combinatoire en particulier. Cette expérience fut très enrichissante et constituera un atout important pour mon futur parcours professionnel. J'ai eu la chance d'être accompagné, conseillé et aidé par plusieurs personnes, de près ou de loin, qui se sont vraiment engagées pour m'épauler durant quatre années de travail. Je suis très heureux d'avoir ici l'occasion pour leur témoigner toute ma gratitude.

Je voudrais en premier lieu remercier tout particulièrement

Je remercie aussi tous mes enseignants de maternelle au lycée qui ont, avec mes parents, participé à mon éducation et ont fait de moi ce que je suis aujourd'hui. Je garderai toujours les souvenirs indélébiles de tous ces professeurs et enseignants.

Je voudrais exprimer ma reconnaissance envers mes amis et collègues qui m'ont apporté leur soutien moral et intellectuel tout au long de ces années.

Enfin, je ne saurais pas clore ces remerciements sans dédier ce travail à toute la famille HADHBI qui m'a doté d'une éducation digne et son amour inconditionnel a fait de moi ce que je suis aujourd'hui. C'est grâce à eux que je me dis que rien n'est impossible car je ne serai jamais seul. Remerciements spéciaux à mes très chers parents, pour leurs efforts, leurs sacrifices, leurs peines, leurs bénédictions et pour m'avoir appris à surmonter mes peurs et d'être là quand cela est nécessaire. Ils sont une source inépuisable de tendresse. L'amour qu'ils ont porté à moi, la dignité, l'éducation et le sens de l'honneur me servent de modèle. Ils m'ont toujours dit de prioriser mes études même ils n'avaient rien dans leurs proches. Ce travail est le fruit de leurs sacrifices qu'ils ont consenti pour mon éducation et ma formation le long de ces années. Du plus profond de mon coeur, un grand merci à tous mes frères, et à ma soeur, pour leur soutien moral et matériel qui m'ont permis d'aboutir à ce résultat.

Ils ont pris en charge tous les frais de mes études (inscription, hébergement, vie privée,...) pendant tout mon cursus universitaire. Il est donc impossible d'exprimer à quel point je suis reconnaissant pour tout cela. Ce travail soit témoignage de mon amour sincère et fidèle à eux. Cette thèse est dédiée aussi à leurs épouses (époux), et à leurs enfants, pour l'amour qu'ils m'ont toujours donné. Un grand merci aussi à mes beaux parents et beaux frères, pour leur amour inconditionnel qu'ils me portent.

Tous les mots ne sauraient exprimer la gratitude, l'amour infaillible, le respect, la reconnaissance à ma chère épouse qui a su me soutenir, m'épauler, me supporter, m'encourager ... tout au long de ces années et plus particulièrement durant les moments difficiles et d'inquiétude qui n'ont pas toujours été des plus agréables. Les mots sont peu de choses pour la remercier pour tout cela. Merci donc d'avoir fait de notre vie de couple tout ce dont je rêvais.

Je ne pourrais pas terminer ces remerciements sans dédier ce travail à mes futurs enfants qu'ils vont me donner le plus beau rôle de ma vie : celui d'être papa. Je leur promets mon amour infaillible, mon attention sans limite, mon soutien en toutes circonstances. Je promets surtout de faire mon mieux, même si parfois j'échoue.

facet-defining for the associated polytope under some necessary and sufficient conditions. In addition, we develop separation algorithms for these inequalities. Using these results, we devise a Branch-and-Cut (B&C) algorithm for the problem, and discuss experimental results. A second part of the sis is devoted to an extended formulation for the C-RSA. A column generation algorithm is developed to solve its linear relaxation. We prove that the related pricing problem is equivalent to the so-called resource constrained shortest path problem, which is well known to be NPhard. For this, we propose a pseudo-polynomial time based dynamic programming algorithm. Using this, we devise Branch-and-Price (B&P) and Branch-and-Cut-and-Price (B&C&P) algorithms to solve the problem. An extensive experimental study with comparisons between the different B&C, B&P, and B&C&P algorithms is also presented.

Finally, we turn our attention to the Spectrum Assignment (SA) sub-problem. This has been shown to be equivalent of wavelength assignment, interval coloring, and dynamic storage allocation problems that are well known to be NP-hard. To the best of our knowledge, a polyhedral approach to the SA problem has not been considered before, even to its equivalent problems. For this, first, we propose an integer linear programming compact formulation and investigate the facial structure of the associated polytope. Moreover, we identify several classes of valid inequalities for the polytope and prove that these inequalities are facet-defining. We further discuss the related separation problems. Using this, we devise a Branch-and-Cut (B&C) algorithm for the SA problem, along with some computational results are presented.

Keywords: optical network, network design, integer programming, polyhedron, facet, separation, branch-and-cut, branch-and-price, branch-and-cut-and-price, dynamic programming.

Long Résumé

Pour faire face à une croissance continue de la demande de trafic liée à l'augmentation de la bande passante, les opérateurs de réseaux ont dû faire évoluer l'architecture de leurs réseaux. En conséquence, une nouvelle génération de réseau de transport optique flexible appelée "Spectrally Flexible Optical Networks" (SFONs) a été introduite en 2008 comme une technologie prometteuse en raison de sa flexibilité et de son efficacité par rapport à l'ancienne technologie connue sous le nom "Optical Wavelength Division Multiplexing (WDM)". Les SFONs ont suscité un intérêt intense de la part des laboratoires de recherche, ainsi que dans l'industrie.

Nous étudions dans cette thèse l'un des problèmes clés lors de dimensionnement et planification des SFONs, le problème du routage contraint et assignation spectrale, connue sous le nom " Constrained-Routing and Spectrum Assignment " (CRSA) selon la terminologie anglaise. Il se compose de deux parties: le routage contraint (sélectionner pour chaque demande en trafic un chemin optique physique qui connecte sa source avec sa destination à travers le réseau sans dépasser une longueur maximale de chemin (en km) fixée pour chaque demande en trafic), et l'assignation d'un spectre (assigner à chaque demande en trafic un seul intervalle de "slot" consécutifs (contrainte de contiguïté) au long de son chemin du routage de sorte que le même intervalle de slots consécutifs doit être utilisé sur tous les liens qui appartiennent à son chemin optique physique (contrainte de continuité), et les intervalles de slots consécutifs alloués par un ensemble de demandes dont les chemins ne sont pas des liens disjoints dans le réseau ne peuvent pas partager aucun slot sur les liens partagés (contrainte de non-chevauchement), tout en optimisant une ou plusieurs fonctions objectives linéaires. Le problème CRSA est bien connu comme un problème NP-difficile et très difficile en pratique aussi que de nombreuses études de recherche ont été menées dans ce contexte depuis sa première apparition en 2010.

Certains des algorithmes de résolution proposés dans la littérature sont basés sur des formulations mathématiques utilisant la programmation linéaire (mixte) en nombres entiers qui n'ont pas pu résoudre des instances de grande taille, ainsi que des heuris-tiques et métaheuristiques qui ne peuvent pas garantir l'optimalité de solutions. Il a été jugé approprié de proposer des nouveaux modèles mathématiques plus souples et efficaces en se basant sur la programmation linéaire en nombres entiers, de concevoir et de développer des algorithmes exacts qui pourraient offrir des améliorations prometteuses par rapport aux méthodes existantes. À notre connaissance, l'étude polyédrale n'a pas encore fait l'objet de recherches récentes pour ce problème. Mots clés : réseaux optiques flexibles, polytope, inégalité valide, facette, separation, algorithme de coupes et branchements, algorithme de génération de colonnes et branchements.

Contents

List of Tables

Introduction

The global Internet Protocol (IP) traffic is expected to reach 396 exabytes per month by 2022, up from 194. [START_REF] Balas | On the Maximum Weight Clique Problem[END_REF] Exabytes per month in 2020 [START_REF]The Network Cisco's Technology News Site: Cisco Predicts More IP Traffic in the Next Five Years Than in the History of the Internet[END_REF]. Optical transport networks are then facing a serious challenge related to continuous growth in bandwidth capacity due to the growth of global communication services and networking: mobile internet network (e.g., 5th generation mobile network), cloud computing (e.g., data centers), Full High-definition (HD) interactive video (e.g., TV channel, social networks) [START_REF] Cheng | Routing and Spectrum Assignment Algorithm based on Spectrum Fragment Assessment of Arriving Services[END_REF], etc... as shown in Figure 1. To sustain the network operators face this trend of increase in bandwidth, a new generation of optical transport network architecture called Spectrally Flexible Optical Networks (SFONs) (called also FlexGrid Optical Networks) has been introduced as promising technology because of their flexibility, scalability, efficiency, reliability, and survivability [START_REF] Chatterjee | Fragmentation Problems and Management Approaches in Elastic Optical Networks: A Survey[END_REF][19] compared with the traditional FixedGrid Optical Wavelength Division Multiplexing (WDM) [START_REF] Ramaswami | Optical Networks: A Practical Perspective[END_REF] [START_REF] Ramaswami | Multiwavelength lightwave networks for computer communication[END_REF]. In SFONs the optical spectrum is divided into small spectral units, called frequency slots [START_REF] Santos | Heuristics for Routing and Spectrum Allocation in Elastic Optical Path Networks[END_REF]. They have the same frequency of 12.5 GHz where WDM uses 50 GHz [START_REF] Stern | Multiwavelength Optical Networks: Architectures, Design and Control[END_REF] as recommended by ITU-T [START_REF] Amar | Performance assessment and modeling of flexible optical networks[END_REF]. This can be seen as an improvement in resource utilization.

The concept of slots was proposed initially by Masahiko Jinno et al. in 2008 [57],

and later explored by the same authors in 2010 [START_REF] Walkowiak | Elastic optical networks -a new approach for effective provisioning of cloud computing and content-oriented services[END_REF]. We refer the reader to [START_REF] Lopez | Elastic Optical Networks: Architectures, Technologies, and Control[END_REF] for more information about the architectures, technologies, and control of SFONs.

The Routing and Spectrum Assignment (RSA) problem plays a primary role when dimensioning and designing of SFONs which is the main task for the development of this next generation of optical networks. It consists of assigning for each traffic demand, a physical optical path, and an interval of contiguous slots (called also channels) while optimizing some linear objective(s) and satisfying the following constraints [START_REF] Hadhbi | A novel integer linear programming model for routing and spectrum assignment in optical networks[END_REF]: a) spectrum contiguity: an interval of contiguous slots should be allocated to each demand k with a width equal to the number of slots requested by demand k; Numerous research studies have been conducted on the RSA problem since its first appearance. The RSA is known to be NP-hard [START_REF] Shirazipourazad | On routing and spectrum allocation in spectrum-sliced optical networks[END_REF][109], and more complex than the historical Routing and Wavelength Assignment (RWA) problem [START_REF] Hai | Combining heuristic and exact approaches for solving the routing and spectrum assignment problem[END_REF]. Various (mixed) integer linear programming (ILP) formulations and algorithms have been proposed to solve it. A detailed survey of spectrum management techniques for SFONs is presented in [START_REF] Talebi | Spectrum management techniques for elastic optical networks: A survey[END_REF] where the authors classified variants of the RSA problem into: offline RSA which has been initiated in [START_REF] Klinkowski | An overview of routing methods in optical burst switching networks[END_REF], and online or dynamic RSA which has been initiated in [START_REF] Wan | Dynamic Routing and Spectrum Assignment in Spectrum-Flexible Transparent Optical Networks[END_REF] and recently developed in [START_REF] Patel | On Efficient Candidate Path Selection for Dynamic Routing in Elastic Optical Networks[END_REF] [START_REF] Zhou | Link State Aware Dynamic Routing and Spectrum Allocation Strategy in Elastic Optical Networks[END_REF]. Numerous aspects are investigated in the tutorial [START_REF] Chatterjee | Routing and Spectrum Allocation in Elastic Optical Networks: A Tutorial[END_REF]. This work focuses on the offline RSA problem. There exist two classes of ILP formulations used to solve the RSA problem, called edge-path and edge-node formulations. The ILP edge-path formulation is majorly used in the literature where variables are associated with all possible physical optical paths inducing an explosion of a number of variables and constraints which grow exponentially and in parallel with the growth of the instance size: number of demands, the total number of slots, and topology size: number of links and nodes [START_REF] Hadhbi | A novel integer linear programming model for routing and spectrum assignment in optical networks[END_REF]. We observe that several papers which use the edge-path formulation as an ILP formulation to solve the RSA problem, use a set of precomputed-paths without guaranty of optimality e.g. in [START_REF] Christodoulopoulos | Elastic Bandwidth Allocation in Flexible OFDM-Based Optical Networks[END_REF], [START_REF] Klinkowski | An overview of routing methods in optical burst switching networks[END_REF], [START_REF] Klinkowski | Routing and Spectrum Assignment in Spectrum Sliced Elastic Optical Path Network[END_REF], [START_REF] Klinkowski | A routing and spectrum assignment problem in optical OFDM networks[END_REF], [START_REF] Velasco | Modeling the routing and spectrum allocation problem for flexgrid optical networks[END_REF], [START_REF] Zotkiewicz | Optimization models for flexgrid elastic optical networks[END_REF], [START_REF] Salameh | Routing With Intelligent Spectrum Assignment in Full-Duplex Cognitive Networks Under Varying Channel Conditions[END_REF]. On the other hand, column generation techniques have been used by Klinkowski et al. [START_REF] Ruiz | Column generation algorithm for RSA problems in flexgrid optical networks[END_REF], Jaumard et al. [START_REF] Jaumard | Scalable elastic optical path networking models[END_REF], and recently by Enoch [START_REF] Enoch | Nested Column Generation decomposition for solving the Routing and Spectrum Allocation problem in Elastic Optical Networks[END_REF] to solve the relaxation of the RSA taking into account all the possible paths for each traffic demand. To improve the LP bounds of the RSA relaxation, Klinkowsky et al. proposed a class of valid inequalities induced by cliques separable using a branch-and-bound algorithm [START_REF] Klinkowski | Valid inequalities for the routing and spectrum allocation problem in elastic optical networks[END_REF]. On the other hand, Klinkowski et al. [START_REF] Klinkowski | A Simulated Annealing Heuristic for a Branch and Price-Based Routing and Spectrum Allocation Algorithm in Elastic Optical Networks[END_REF] propose a branch-and-cut-and-price method based on an edge-path formulation for the RSA problem. Recently, Fayez et al. [START_REF] Fayez | Recursive algorithm for selecting optimum routing tables to solve offline routing and spectrum assignment problem[END_REF], and Xuan et al. [START_REF] Xuan | New bi-level programming model for routing and spectrum assignment in elastic optical network[END_REF], proposed a decomposition approach to solve the RSA separately (i.e., R+SA) based on a recursive algorithm and an ILP edge-path formulation.

To overcome the drawbacks of the edge-path formulation usage, a compact edgenode formulation has been introduced as an alternative for it. It holds a polynomial number of variables and constraints that grow only polynomially with the size of the instance. We found just a few works in the literature that use the edge-node formulation to solve the RSA problem e.g., [START_REF] Cai | Novel Node-Arc Model and Multiiteration Heuristics for Static Routing and Spectrum Assignment in Elastic Optical Networks[END_REF], [START_REF] Velasco | Modeling the routing and spectrum allocation problem for flexgrid optical networks[END_REF], [START_REF] Zotkiewicz | Optimization models for flexgrid elastic optical networks[END_REF]. Bertero et al. [START_REF] Bertero | Integer programming models for the routing and spectrum allocation problem[END_REF] present a comparative study between several edge-node formulations and introduce new ILP ones.

On the other hand, and due to the NP-hardness of the C-RSA problem, several heuristics [START_REF] Ding | Hybrid routing and spectrum assignment algorithms based on distance-adaptation combined coevolution and heuristics in elastic optical networks[END_REF], [START_REF] Mesquita | A Routing and Spectrum Assignment Heuristic for Elastic Optical Networks under Incremental Traffic[END_REF], [START_REF] Santos | Heuristics for Routing and Spectrum Allocation in Elastic Optical Path Networks[END_REF], and recently in [START_REF] He | Invalid-Resource-Aware Spectrum Assignment for Advanced-Reservation Traffic in Elastic Optical Network[END_REF], greedy algorithms [START_REF] Mahala | Spectrum assignment technique with firstrandom fit in elastic optical networks[END_REF], metaheuristics as tabu search [START_REF] Goscien | Tabu search algorithm, Routing, Modulation and spectrum allocation, Anycast traffic, Elastic optical networks[END_REF], simulated annealing [START_REF] Klinkowski | A Simulated Annealing Heuristic for a Branch and Price-Based Routing and Spectrum Allocation Algorithm in Elastic Optical Networks[END_REF], genetic algorithms [START_REF] Gong | A Two-Population Based Evolutionary Approach for Optimizing Routing, Modulation and Spectrum Assignments (RMSA) in O-OFDM Networks[END_REF], [START_REF] Hai | An efficient genetic algorithm approach for solving routing and spectrum assignment problem[END_REF], [START_REF] Hai | Combining heuristic and exact approaches for solving the routing and spectrum assignment problem[END_REF], [START_REF] Dinarte | Routing and spectrum assignment: A metaheuristic for hybrid ordering selection in elastic optical networks[END_REF],

ant colony algorithms [START_REF] Lezama | Solving routing and spectrum allocation problems in flexgrid optical networks using pre-computing strategies[END_REF], and a hybrid meta-heuristic approach [START_REF] Ruiz | A hybrid meta-heuristic approach for optimization of routing and spectrum assignment in Elastic Optical Network (EON)[END_REF], have been used

to approach large scale instances of the RSA problem. Furthermore, recent works start using artificial intelligence [START_REF] Reihani | Artificial neural network-based adaptive modulation for elastic optical networks[END_REF], see for example [START_REF] Liu | A Monte Carlo Based Routing and Spectrum Assignment Agent for Elastic Optical Networks[END_REF][62], and deep-learning [START_REF] Chen | Deep-RMSA: A Deep-Reinforcement-Learning Routing, Modulation and Spectrum Assignment Agent for Elastic Optical Networks[END_REF],

and machine-learning [START_REF] Salani | Routing and Spectrum Assignment Integrating Machine-Learning-Based QoT Estimation in Elastic Optical Networks[END_REF][120] [117][48] to get more perefermonce. Selvakumar et al. give a survey [START_REF] Selvakumar | The Recent Contributions of Routing and Spectrum Assignment Algorithms in Elastic Optical Network (EON)[END_REF] in which they summarise the most contributions done for the RSA problem before 2019.

In this thesis, we are interested in the resolution of a complex variant of the RSA problem, called the Constrained-Routing and Spectrum Assignment (C-RSA) problem. Here we suppose that the network should also satisfy the transmission-reach constraint for each traffic demand according to the actual service requirements. To the best of our knowledge a few related works on the RSA, take into account this additional constraint so that the length of the chosen path for each traffic demand should not exceed a certain length (in kms). Recently, Hadhbi et al. [START_REF] Hadhbi | A novel integer linear programming model for routing and spectrum assignment in optical networks[END_REF][51] introduced a novel tractable ILP based on the cut formulation for the C-RSA problem with a polynomial number of variables and an exponential number of constraints that are separable in polynomial time using network flow algorithms. Computational results show that their cut formulation solves larger instances compared with those of Velasco et al. [START_REF] Velasco | Modeling the routing and spectrum allocation problem for flexgrid optical networks[END_REF] and Cai et al. [START_REF] Cai | Novel Node-Arc Model and Multiiteration Heuristics for Static Routing and Spectrum Assignment in Elastic Optical Networks[END_REF]. It has also been used as a basic formulation in the study of Colares et al. [START_REF] Colares | An extended formulation for the Constraint Routing and Spectrum Assignment Problem in Elastic Optical Networks[END_REF], and also by Chouman et al. [START_REF] Chouman | Impact of RSA Optimization Objectives on Optical Network State[END_REF] [START_REF] Chouman | Assessing the Health of Flexgrid Optical Networks[END_REF] to show the impact of several objective functions on the optical networks state. Note that Velasco et al. [START_REF] Velasco | Modeling the routing and spectrum allocation problem for flexgrid optical networks[END_REF], Cai et al. [START_REF] Cai | Novel Node-Arc Model and Multiiteration Heuristics for Static Routing and Spectrum Assignment in Elastic Optical Networks[END_REF], and Bertero et al. [START_REF] Bertero | Integer programming models for the routing and spectrum allocation problem[END_REF], did not take into account the transmission-reach constraint.

However, so far the exact algorithms proposed in the literature could not solve largescale instances. We believe that a cutting-plane-based approach could be powerful for the problem. To the best of our knowledge, such an approach has not been yet considered except the works done by Bianchetti et al. [START_REF] Bianchetti | Valid inequalities and a branch-and-cut algorithm for the routing and spectrum allocation problem[END_REF] for the RSA problem. For this, the main aim of this work is to investigate thoroughly theoretical properties of the C-RSA problem. To this end, we aim at providing a deep polyhedral analysis of the C-RSA problem, and based on this, devise branch-and-cut and branch-and-cut-and-price algorithms for solving large-scale instances of the problem.

So we will introduce a new ILP formulation called cut formulation for the C-RSA problem which can be seen as an improved formulation for the one introduced by Hadhbi et al. [START_REF] Hadhbi | A novel integer linear programming model for routing and spectrum assignment in optical networks[END_REF] [START_REF] Hadhbi | Routage et Affectation Spectrale Optimaux dans des Réseaux Optiques Élastiques FlexGrid[END_REF]. We investigate the facial structure of the associated polytope. We further identify several classes of valid inequalities to obtain tighter LP bounds. Some of these inequalities are obtained by using conflict graphs related to the problem. We then devise separation procedures and give sufficient conditions under which these inequalities are facet defining. Using this, we develop a Branchand-Cut (B&C) algorithm, along with computational results are presented using large-scale instances. On the other hand, we introduce an extended ILP formulation, called path formulation. A column generation algorithm is proposed to solve its linear relaxation. We further adapt the valid inequalities proposed for the cut formulation to obtain also tighter bounds for the path formulation. Based on this, we develop a Branch-and-Cut-and-Price (B&C&P) algorithm to solve the problem.

Computational results are presented using this algorithm. We finally provide a comparative study between the B&C and B&C&P algorithms is presented by using two types of instances: random and realistic ones. The results show that the B&C&P algorithm is more efficient. Furthermore, we have studied the influence of the valid inequalities. The results show that some of them, in particular, clique and cover inequalities are quite efficient. Several enhancements are further investigated and used to speed up and increase the efficiency of our approaches. They are based on a primal heuristic used to produce feasible solutions from fractional solutions given at each node of the branching tree. It allows obtaining good primal bounds and prune some uninteresting nodes of the branching tree. We have also introduced some symmetry-breaking inequalities to manage the equivalent sub-problems in the branching tree.

Several concepts are exploited throughout this dissertation. We start this dissertation by presenting the basic notions of combinatorial optimization, complexity, graph theory, and further give some notations that are used through this manuscript.

In Chapter 2, we present the C-RSA problem. We introduce an integer linear programming formulation namely cut formulation. We then carry out an investigation of the related polytope, the convex hull of all its solutions. Moreover, we describe the classes of valid inequalities and study their facial structure. In particular, we introduce symmetry-breaking inequalities.

In Chapter 3, we discuss the separation procedures for the valid inequalities and describe a Branch-and-Cut algorithm. The comparative study is presented in this chapter, it shows the impact of the additional valid inequalities using several mixedinteger linear program solvers.

In Chapter 4, we give the extended ILP formulation. We present the column generation algorithm to solve its linear relaxation, and the Branch-and-Cut-and-Price (B&C&P) algorithm, along with some computational results are presented. In this chapter, we also provide the comparative analysis of performance between the different algorithms.

Chapter 5 is devoted to the Spectrum Assignment (SA) sub-problem. First, we propose an integer linear programming compact formulation, and investigate the facial structure of the associated polytope. Fuerthremore, we describe several valid inequalities, some of them come from those that are already proposed for the C-RSA.

We also give sufficient conditions under which these inequalities are facet defining.

Based on these results, we develop a B&C algorithm to solve the problem. Furthermore, we describe symmetry-breaking inequalities for the SA, and provide some lower bounds. Finally, we present an extensive experimental study while showing the impact of the valid inequalities and symmetry-breaking inequalities on the effectiveness of the B&C algorithm.

Chapter 1

Preliminary Notions

In this chapter, we present some basic notions of combinatorial optimization, and polyhedra approaches.

Combinatorial Optimization

Operations research is a discipline related to computer science and applied mathematics. In this dissertation, we are interested in one of its branches, called combinatorial optimization. The optimization problems related to combinatorial optimization can be formulated as follows. Let E = {e 1 , ..., e n } be a finite set, namely basic set. Suppose that each element e i , it is associated a weight c(e i ) ∈ R with i ∈ {1, ..., n}. Let F denote a family of subsets of E. The problem aims to identify one subset F from F with the smallest or largest weight given by the sum e i ∈F c(e i ). Such a problem is called combinatorial optimization problem where the set F represents the set of all feasible solutions of the problem. In general, the set F contains an exponential number of solutions. For this, it's known to be very hard to solve combinatorial optimization problems by enumerating all its feasible solutions. To deal with this, various approaches have been developed to approach combinatorial optimization problems. They use different tools, complexity theory, combinatorial optimization, graph theory, linear and non-linear programming, integer programming, mixed integer programming. In the next section, we discuss some concepts from complexity theory.

Complexity Theory

Several researchers in computer science and mathematics are interested in working on the classification of problems into easy or hard problems, and further on the algorithmic complexity whose objective is to find the most efficient algorithm. This has been initiated by Cook [25], Edmonds [START_REF] Edmonds | Covers and packings in a family of sets[END_REF] and Karp [START_REF] Karp | Reducibility among Combinatorial Problems[END_REF].

Theory of complexity [START_REF] Garey | Computers and Intractability: A Guide to the ory of Np-completeness[END_REF] [START_REF] Garey | Computers and Intractability: A Guide to the ory of Np-completeness[END_REF] classifies problems into two essential classes: the class P (polynomial time) class, and the class NP (Nondeterministic polynomial time). In addition, the problems of the NP class are shared into two subclasses: the class of NP-complete problems, and the class of NP-hard problems.

Before defining each class, we first give a general definition of a problem. In general, a problem is a question having parameters given in input such that an answer is needed for it, called solution. A problem is described by giving: a general description of all its parameters, and certain constraints. An instance of a problem is obtained by specifying the value of each input parameter of the problem. For this, one can propose an algorithm to solve the problem. An algorithm for solving a given problem is a procedure that is decomposable into a sequence of finite operations. It allows giving a solution for each instance of the problem. In general, the complexity of an algorithm depends on the size of the problem that reflects the number of parameters needed to describe an instance. The algorithm is said to be polynomial if the maximum number of its operations necessary to solve an instance of size n is bounded by a polynomial function f in n (i.e., f (n)). This means that there exists a scalar c such that the number of its operations necessary is equal to c.f (n). As a result, the notation big O is appeared to express the complexity of an algorithm.

There exists two types of problems: optimization problems and decision problems.

In optimization problems, we want to minimize (or maximize) a function while satisfying a set of constraints. On the other hand, in the a decision problem, the solution is binary like yes / no or 0/1.

An easy problem that can be solved by a polynomial algorithm with respect to its size, is called a problem of class P. A problem is NP if one can verify in polynomial time that a given solution is feasible. A problem is called NP-complete if it belongs NP, and every other problem in NP can be reduced to it in polynomial time [START_REF] Garey | Computers and Intractability: A Guide to the ory of Np-completeness[END_REF].

The Satisfiability Problem (SAT) is the first problem that has been shown to be NP-complete. This was proved in 1971 by Stephen Cook [START_REF] Cook | The complexity of theorem-proving procedures[END_REF] [START_REF] Gherboudj | Méthodes de résolution de problémes difficiles académiques[END_REF].

NP-hard problems are difficult as the NP-complete ones. If a decision problem associated with a optimization problem P is NP-complete then P is said to be NP-hard [START_REF] Gherboudj | Méthodes de résolution de problémes difficiles académiques[END_REF]. Furthermore, note that every problem of the class P is in NP (P ⊆ N P ).

However, the converse is still open. It constitutes a well-known mathematical problem which is part of the 7 problems of the millennium prize. The question P = N P ? is one of the most important questions that has not yet been solved. The answer to this question by "yes" is to prove that all the problems of the NP class are in the P class. Cook has proved in [START_REF] Cook | The complexity of theorem-proving procedures[END_REF]] that all the problems of the NP class are reducible to the SAT problem, which means that if someone finds a polynomial algorithm for this problem, the question P = N P ? is then solved ! [START_REF] Gherboudj | Méthodes de résolution de problémes difficiles académiques[END_REF], i.e. we will be able to solve all NP-complete problems in polynomial time.

Polyhedral Approach and Branch-and-Cut Algorithm 1.3.1 Elements of the Polyhedral Theory

In this section, we will introduce some definitions and properties of polyhedraltheory.

Schrijver [START_REF] Schrijver | Theory of Linear and Integer Programming[END_REF], Nemhauser and Wolsey [START_REF] Nemhauser | Integer and Combinatorial Optimization[END_REF], Wolsey [START_REF] Wolsey | Integer programming[END_REF] and Schrijver [START_REF] Schrijver | Combinatorial Optimization -Polyhedra and Efficiency[END_REF] are the most useful references [START_REF] Zhao | Maximum Bounded Rooted-Tree Problem : Algorithms and Polyhedra[END_REF].

Let x be a vector in R n , with n a positive integer. x is said to be a linear combination of vectors x 1 , x 2 , .., x k ∈ R n if there exist k scaler λ 1 , λ 2 .., λ k such that x = k ı=1 λ i x i . Furthermore, if k ı=1 λ i = 1, then x is said to be affine combination of x 1 , x 2 , .., x k . We say that x is convex combination of x 1 , x 2 , .., x k if x is affine combination of x 1 , x 2 , .., x k and λ i ∈ R + . The vectors x 1 , x 2 , .., x k are affinely independent if λ i = 0 for each i ∈ {1, ..., k} , is the unique solution of the system

k ı=1 λ i x i = 0, k ı=1 λ i = 1,
Given a set S = {x 1 , ..., x k }, the convex hull denoted by conv(S), is the set of all the convex combinations of solutions of S that is

conv(S) = {x ∈ R n | k i=1 λ i x i , ∀λ i ≥ 0 and i λ i = 1}.
This definition ensures that S ⊂ conv(S). A polyhedron P is the set of solutions of a linear system Ax ≤ b. That is P = {x ∈ R n |Ax ≤ b}. A bounded polyhedron is called a polytope.

The dimension of polyhedron P is one less than the maximum number of solution in P that are affinely independent.

An inequality ax ≤ α is valid for a polyhedron P if and only if for every solution

x ∈ P , ax ≤ α. It is said to be violated by a solution x if ax > α. A set F ⊂ P is called face if there exists a valid inequality ax ≤ α for the polyhedron P such that F = {x ∈ P, ax = α}.

We say that the valid inequality ax ≤ α supports a face F if F ̸ = ∅.

A face F is said to be proper face if F ̸ = ∅ and F ̸ = P . If F is a proper face of P , and dim(F ) = dim(P ) -1, then F is called a facet.

A face F of P is a facet if there doesn't exist any proper face F ′ of P containing F .

If P is full-dimensional polyhedron, then ax ≤ α defines a facet P if and only if F is a proper face and there exists a facet defining inequalitybx ≤ β and a scalarρ ̸ = 0 such that F ⊂ {x ∈ P |bx = β} and b = ρa. If P is not full dimensional polyhedron, then ax ≤ α defines a facet of polyhedron P if and only if F is a proper face and there exists a facet of P induced by an inequality bx ≤ β, a scalar ρ ̸ = 0 and a vector λ such that F ⊂ {x ∈ P |bx = β} and b = ρa + λA = .

A solution x ∈ P is an extreme point of P if x is a face of P of dimension 0.

Furthermore, it cannot be written as a convex combination of other points in P .

Cutting Plane Method

Let P be a combinatorial optimization problem and S the set of its feasible solutions.

The problem P can be written as min{cx|x ∈ S}, where c denotes the weight vector associated with the variables x of the problem. Consider the convex hull conv(S) of the feasible solutions of P. The problem P is then equivalent to the linear program min{cx|x ∈ conv(S)}.

The polyhedral approach, introduced by Edmonds [START_REF] Edmonds | Covers and packings in a family of sets[END_REF] consists in describing the polyhedron conv(S) by a set of linear inequalities. This reduces the problem P to solving a linear program. However, a complete description of the polyhedron may contain an exponential number of linear inequalities. The optimization problem on the polyhedron conv(S) can be solved having all its linear inequalities. However, one can have a partial characterization of the polyhedron conv(S). This may be sufficient to solve the problem in polynomial time using the so-called cutting-plane method. This method is based on the so-called separation problem defined as follows.

Let C be a class of valid inequalities for the polyhedron conv(S). The separation problem associated with C consists in deciding whether a given solution x satisfies all inequalities of C, and if not to find an inequality of C violated by x. Grötschel, Losvàsz, and Schrijver [START_REF] Grötschel | Geometric Algorithms and Combinatorial Optimization[END_REF] have shown that an optimization problem over C can be solved in polynomial time if and only if the separation problem associated with C can be solved in polynomial time. This may permit to solve the optimization problem in polynomial time as a sequence of linear programs. Each program is obtained by adding new valid inequalities obtained by solving the related separation problem. For this, we start by solving a linear program containing a small set of valid inequalities. Let us denote by x the obtained optimal solution. We solve the separation problem for C. If x satisfies all the constraints of C, then x i optimal.

Otherwise, at least one constraint violated by x is identified. These should be added to the current linear program. This process is repeated until an optimal solution is found.

Branch-and-Cut Algorithm

The cutting-plane method provides only an optimal solution for the linear relaxation of the problem. This solution may not be integer which means that it is not feasible for the original problem. In this case, we pass to the branching step which consists in dividing the problem into two Sub-problems by choosing a fractional variable x i and setting x i to x i = 1 and x i = 0. We then apply the cutting-plane method for each of the sub-problem. We continue this process until an optimal solution is obtained for the problem. This method is known as Branch-and-Cut method since it combines a branching method with a cutting plane method at each node of the tree.

1.4 Column Generation and Branch-and-Cut-and-Price

Algorithms

Sometimes mathematical formulations of a problem contain a huge number of variables that can be exponential. These are known as "extended formulation". To solve such problems, we use column generation based algorithm. We begin the optimization with a restricted linear program containning a feasible basis. At each iteration, the column generation algorithm checks if there exists a missing variable having a negative reduced cost, and add it to the current restricted linear program. This is the "Pricing Problem". In fact, this consists in determining variables with negative reduced cost. This procedure is repeated until no new variable with negative reduced cost is found. The final solution is optimal for the linear relaxation of the underlaying problem. Furthermore, if it is integral, then it is optimal for the problem. If not, we create two subproblems called children by branching on some fractional variables (variable branching rule) or on some constraints using the Ryan & Foster branching rule [START_REF] Ryan | An integer programming approach to scheduling[END_REF] (constraint branching rule). Such an algorithm is called a Branch-and-Price. A Branch-and-Cut-and-Price algorithm combines a Branch-and-Price algorithm with a cutting-plane procedure.

Graph Theory

In this section, we introduce some elementary definitions in graph theory that are very useful throughout the dessertation, Diestel [START_REF] Diestel | Graph Theory (Graduate Texts in Mathematics)[END_REF], and Golumbic [START_REF] Golumbic | Algorithmic Graph Theory and Perfect Graphs[END_REF] are the most useful references on graph theory [START_REF] Zhao | Maximum Bounded Rooted-Tree Problem : Algorithms and Polyhedra[END_REF].

A graph is a pair G = (V, E), where V is a finite set of nodes (called also vertices) linked by a set of edges (called also links) E which can be oriented or not.

A path p in graph G = (V, E) from node a to node b, is a sequence of nodes such that for each pair of successive nodes v i ,v i+1 , there exist an edge e (v i , v i+1 ) ∈ E.

For any subset of nodes X ⊆ V with X ̸ = ∅, let δ(X) denote the set of edges having one extremity in X and the other one in X = V \ X which is called a cut. When X is a singleton (i.e., X = {v}), we use δ(v) instead of δ({v}) to denote the set of edges incidents with a node v ∈ V . The cardinality of a set K is denoted by |K|.

A vertex coloring of G is an assignment of colors to the vertices of G so that two adjacent vertices v and v ′ cannot get the same color. Same rule for edges, an edge coloring of G is an assignment of colors to the edges of G so that two adjacent edges e and e ′ cannot get the same color. We say that graph G is t-colorable if no more than t different colors assigned in G.

G ′ is called a weighted graph if each node in G ′ is associated with weight.

An interval t-coloring of a weighted graph

G ′ = (V, E, w) is a function c : V -> {1, 2, ..., t} such that c(v)+w(v)-1 ≤ t. We assign an interval [c(v), ..., c(v)+w(v)-1]
of consecutive integers satisfying w(v) of each vertex v that the intervals of colors assigned to two adjacent vertices do not overlap. If interval t-coloring is feasible for a graph G ′ then G ′ is said to be interval t-colorable [START_REF] Shirazipourazad | On routing and spectrum allocation in spectrum-sliced optical networks[END_REF]. The interval chromatic number of G ′ , denoted by χ is the least integer number t such that G ′ has a interval t-coloring [START_REF] Shirazipourazad | On routing and spectrum allocation in spectrum-sliced optical networks[END_REF].

Flexible Optical Networks

The two last decades have seen a big developement in telecommunication networks with a continuous growth in demands. To face this trend of increase in bandwidth, network operators have had to make their network architectures and management evolve. Two significant changes appeared recently in the optical network architec-ture. First the bandwidth-greedy FixedGrid architecture for Optical Wavelength Division Multiplexing (WDM) (called also wavelength routed networks) [START_REF] Ramaswami | Optical Networks: A Practical Perspective[END_REF][93] based on fixed spectrum grid is being replaced by the FlexGrid architecture that is capable of supporting variable data rate (in Gb/s) through flexible spectrum. In this concept, the optical spectrum is divided into slots having the same frequency of The concept of slot was proposed initially by Masahiko Jinno et al. [START_REF] Jinno | Demonstration of novel spectrum-efficient elastic optical path network with perchannel variable capacity of 40 Gb/s to over 400 Gb/s[END_REF], and later explored and more developed by the same authors in 2010 [START_REF] Walkowiak | Elastic optical networks -a new approach for effective provisioning of cloud computing and content-oriented services[END_REF]. In SFONs any optical path can elastically span as many contiguous slots as needed. This technology provides a more efficient use of the spectral domain than the traditional Fixed Grid WDM. Furthermore, a new generation of transponders is becoming available namely, bandwidth-variable transponders (BV-Ts) and bandwidth variable wavelength cross-connects (BV-WXCs) [START_REF] Walkowiak | Elastic optical networks -a new approach for effective provisioning of cloud computing and content-oriented services[END_REF]. These can manage data rates up to 400

Gb/s which cannot be accommodated by a 50 GHz wavelength, and restores the signal which is necessary to re-amplify, re-shape and re-time the passive optical signal (which is called (3R) signal regeneration rule) when the transmission-reach of signals is limited which represents the maximum length (in kms) for the routing of each traffic demand.

The network operators have confronted several optimization problems, in particular some variants of routing and resource allocation problems that arise when designing or planning optical networks. The classical Routing and Wavelength Assignment (RWA) problem is the key issue for design FixedGrid WDM networks. In this problem, we are given an optical network and a set of demands where each demand has an origin and destination. The task is to find a path for each demand and a wavelength such that a single 50 GHz wavelength is assigned to each demand. It was extended by Chlamtac et al. [START_REF] Chlamtac | Lightpath communications: an approach to high bandwidth optical WAN's[END_REF]. It is known to be a NP-hard problem [START_REF] Chlamtac | Lightpath communications: an approach to high bandwidth optical WAN's[END_REF]. It is equivalent to the n-graph-coloring problem where the number n of the colors corresponds to the number of wavelengths and finding the minimum number of wavelengths to route all the traffic demands is equivalent to finding the chromatic number of the conflict graph (where the demands are represented by the nodes and two nodes are linked by an edge if the path of the associated demands share an edge) when the paths are already established. It has been considered as a special case of the integer multicommodity flow (MCF) problem where some specific constraints [START_REF] Brun | Routing and Wavelength Assignment in Optical Networks[END_REF] are added and should be respected. Several models and algorithms have been proposed

to solve the RWA problem. However, in SFONs, RWA cannot handle the changes from wavelength to contiguous slots. As a result, a new problem is appeared to deal with this, called Routing and Spectrum Assignment (RSA) problem. It can be stated as follows. Given an optical network G and a multiset of traffic demands K, it aims at determining for each traffic demand k ∈ K a path and an interval of contiguous slots while optimizing some linear objective(s) and satisfying the following constraints [START_REF] Hadhbi | A novel integer linear programming model for routing and spectrum assignment in optical networks[END_REF]:

1. spectrum contiguity: an interval of contiguous slots should be allocated to each demand k with width equals to the number of slots requested by demand k;

2. spectrum continuity: the interval of contiguous slots allocated to each traffic demand stills the same along the chosen path;

3. non-overlapping spectrum: the intervals of contiguous slots of demands whose paths are not edge-disjoints in the network cannot share any slot over the shared-edge.

The RSA problem is harder than the RWA problem because of the continuity constraint that has not been taken into account when defining the RWA problem. In

such that S k ∩S k ′ = ∅ for each pair of demands k, k ′ ∈ K (k ̸ = k ′ ) with E(p k ) ∩ E(p k ′ ) ̸ = ∅ (

Cut Formulation

Here we introduce an integer linear programming formulation for the C-RSA problem, called formulation. It can be seen as a reformulation of the one introduced by Hadhbi et al. [START_REF] Hadhbi | A novel integer linear programming model for routing and spectrum assignment in optical networks[END_REF]. For k ∈ K and e ∈ E, let x k e be a variable which takes 1 if demand k goes through edge e and 0 if not, and for k ∈ K and s ∈ S, let z k s be a variable which takes 1 if slot s is the last slot allocated for the routing of demand k and 0 if not. The contiguous slots s ′ ∈ {s -w k + 1, ..., s} should be assigned to demand k whenever z k s = 1. The C-RSA problem is equivalent to the following linear integer program min

k∈K e∈E c e x k e , (2.1)
subject to e∈δ(X)

x k e ≥ 1, ∀k ∈ K, ∀X ⊆ V , |X ∩ {o k , d k }| = 1, (2.2) e∈E l e x k e ≤ lk , ∀k ∈ K, (2.3)
z k s = 0, ∀k ∈ K, ∀s ∈ {1, ..., w k -1}, (2.4)
s s=w k z k s ≥ 1, ∀k ∈ K, (2.5)
x k e + x k ′ e + min(s+w k -1,s) s ′ =s z k s ′ + min(s+w k ′ -1,s) s ′ =s z k ′ s ′ ≤ 3, ∀(e, k, k ′ , s) ∈ Q e,s , (2.6
)

0 ≤ x k e ≤ 1, ∀k ∈ K, ∀e ∈ E, (2.7) 0 ≤ z k s ≤ 1, ∀k ∈ K, ∀s ∈ S, (2.8)
x

k e ∈ {0, 1}, ∀k ∈ K, ∀e ∈ E, (2.9)
z k s ∈ {0, 1}, ∀k ∈ K, ∀s ∈ S. (2.10)
where Q e,s denotes the set of all the quadruples (e, k, k Note that the linear relaxation of the C-RSA can be solved in polynomial time given that inequalities (2.2) can be separated in polynomial time using network flows.

Associated Polytope

Let P(G, K, S) be the polytope, convex hull of the solutions of (2.1)-(2.10).

In this section, we discuss the facial structure of the polytope P(G, K, S). First, we describe some structural properties. These will be used for determining the dimension of P(G, K, S). (resp. (o k , j) and (i, d k )) greater that lk then edge ij cannot be in a path routing of demand k, and we then say that ij is a forbidden edge for demand k due to the transmission-reach constraint. Let E k t denote the set of forbidden edges due to the transmission-reach constraint for demand k ∈ K. Note that using Dijkstra's algorithm, one can identify in polynomial time the forbidden edges E k t for each demand k ∈ K. Table 2.1 shows the set of forbidden edges E k t and forbidden nodes V k 0 for each demand k in K Fig. 2.1(b).

k o k ⇒ d k w k lk V k 0 E k t 1 a ⇒ c 2 4 {e, d, g} {cg, dg, de, df, cd, ef } 2 a ⇒ d 1, 00 4 {g} {cg, dg, df } 3 b ⇒ f 2 4 {e, d, g} {cg, dg, de, df, cd, ef } 4 b ⇒ e 1, 00 4 {g} {cg, dg, df } Table 2.1: Sets of V k 0 and E k 0 of the example of Fig. 2.1(b). Consider a subset of nodes W in V \ V k 0 with o k ∈ W and d k ∈ V \ W .
Let f be an edge in a cut δ(W ) such that all the edges e ∈ δ(W ) \ {f } are forbidden for demand k. As a consequence, edge f is an essential edge for demand k. As the forbidden edges, the essential edges can be determined in polynomial time using network flows.

Let E k

1 denote the set of essential edges of demand k, and K e denote the subset of demands in K having e as essential edge. Therefore,

x k e = 1, for all k ∈ K and e ∈ E k 1 .

(2.11)

In addition to the forbidden edges thus obtained due to the transmission-reach constraints, there may exist edges that may be forbidden because of lack of resources for demand k. This is the case when, for instance, the residual capacity of the edge in question does not allow a demand to use this edge, i.e.,

w k > s -k ′ ∈Ke w k ′ .
Let E k c denote the set of forbidden edges for demand k, k ∈ K, due to the resource constraints. Let

E k 0 = E k t ∪ E k c for k ∈ K. Hence,
x k e = 0, for all k ∈ K and e ∈ E k 0 .

(2.12)

As a result of the pre-processing stage, a non-compatibility between the demands may appear due to a lack of resources. For an edge e, two demands k and k ′ with

e / ∈ E k 0 ∪ E k 1 ∪ E k ′ 0 ∪ E k ′ 1
, are said to be non-compatible if the residual capacity of edge e does not allow to route the two demands k, k ′ together through e, i.e.,

w k + w k ′ > s -k"∈Ke w k" . Let K e
c denote the set of pairs of demands (k, k ′ ) in K that are non-compatible for edge e.

On the other hand, a non-compatibility between the edges for a demand may appear due to the transmission-reach constraint. Consider a demand k. Two edges e = ij / ∈

E k 0 ∩E k 1 , e ′ = lm / ∈ E k 0 ∩E k
1 are said non-compatible edges if the length of all (o k , d k )paths formed by e = ij and e ′ = lm together are greater than lk . Note that we are able to determine the non-compatible edges for each demand k in polynomial time using shortest-path algorithms.

Dimension

We first describe some properties that are useful to determine the dimension of P(G, K, S). First the following is easily seen to be true.

Proposition 2.3.1. The follows equation system (2.13) is of full rank         
x k e = 0, for all k ∈ K and e ∈ E k 0 , x k e = 1, for all k ∈ K and e ∈ E k 1 , z k s = 0, for all k ∈ K and s ∈ {1, ..., w k -1}.

(2.13)

The rank of system (2.13) is given by

r = k∈K (|E k 0 | + |E k 1 | + (w k -1)).
Let Q denote a matrix associated with the system (2.13) which contains r lines linear independents.

A solution of the C-RSA problem is given by two sets E k and S k for each demand k ∈ K where E k is a set of edges used for the routing of demand k, and S k is a set of slots assigned to demand k. For the sake of presentaion, we will denote by E k a feasible path, and by S k the last slots assigned to demand k.

Below are some assumptions that will be considered

Let S i = (E i , S i ) be a solution of the C-RSA problem such that E i = (E i 1 , E i 2 , ..., E i |K|-1 , E i |K| ) and S i = (S i 1 , S i 2 , ..., S i |K|-1 , S i |K| )
, and let (x S , z S ) be its incidence vector. Note that when the routing of demands is trivial or already established, one can find a feasible spectrum assignment S i in polynomial time using some heuristics and greedy algorithm as the well-known First-Fit algorithm [START_REF] Amar | Performance assessment and modeling of flexible optical networks[END_REF]. This will be used throughout each proof of polyhedron dimension or facial structure of some valid inequalities such that the set of demands K is considered as an ordered set of demands, i.e., K = {k 1 , k 2 , ..., k |K| }.

Proposition 2.3.2. System (2.13) defines a minimal equation system for P(G, K, S).

Proof. Consider an equation µx + σz = λ of P(G, K, S). To prove that µx + σz is a linear combination of equations system (2.13), it is sufficient to prove that

there exists γ = (γ 1 , γ 2 , γ 3 ) ( with γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = γQ.
We will show that σ k s = 0 for all k ∈ K and s ∈ {w k , ..., s}. Consider a demand k in K and a slot s in {w k , ..., s}. Consider the solution S 0 = (E 0 , S 0 ) given by a) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E 0 k i be the set of edges involved in a shortest path between o k i and d k i , b) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 0 i given by

I 0 i = [ kj ∈D 0 i {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]∩[{w ki , ..., s-w k }∪{s+w ki , ..., s}] if E 0 ki ∩ E 0 k ̸ = ∅ or I 0 i = kj ∈D 0 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,
where

D 0 i = {k j ∈ {k 1 , ..., k i-1 } : E 0 k i ∩ E 0 k j ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 0 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s -w k + 1, ..., s} = ∅ if E 0 k i ∩ E 0 k ̸ = ∅ (
we take into account the possibility of adding slot s in the set of last slots S 0 k assigned to demand k in solution S 0 ).

We let S 0 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S 0 is feasible for the problem, and its incidence vector (x S 0 , z S 0 ) belongs to P(G, K, S).

Let S 1 = (E 1 , S 1 ) be the solution obtained from S 0 by adding slot s as last slot to demand k. Solution S 1 is feasible for the problem. The corresponding incidence vector (x S 1 , z S 1 ) belongs to P(G, K, S). Hence, solutions S 0 and S 0 satisfy equation µx + σz = λ. We then obtain that

µx S 0 + σz S 0 = µx S 1 + σz S 1 = µx S 0 + σz S 0 + σ k s .
It follows that σ k s = 0. In a similar way, we can show that σ k s = 0, for all k and s ∈ {w k , ..., s} Next, we will show that µ k e = 0 for all k ∈ K and e

∈ E \ (E k 0 ∪ E k 1 ). Consider a demand k ∈ K and an edge e ∈ E \ (E k 0 ∪ E k 1 )
. Consider the solution S ′0 = (E ′0 , S ′0 ) such that a) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E ′0 k i be the set of edges involved in a shortest path between o k i and d k i , b) we select slot s k = w k as last slot for demand k in solution S ′0 , c) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′0 i given by

I ′0 i = [ kj ∈D ′0 i {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]∩[{w ki , ..., s k -w k }∪{s k +w ki , ..., s}] if E ′0 ki ∩ (E ′0 k ∪ {e}) ̸ = ∅ or I ′0 i = kj ∈D ′0 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not.
where

D ′0 i = {k j ∈ {k 1 , ..., k i-1 } \ {k} : E ′0 k i ∩ E ′0 k j ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D ′0 i , • {s k i -w k i + 1, ..., s k i } ∩ {s k -w k + 1, ..., s k } = ∅ if E ′0 k i ∩ (E ′0 k ∪ {e}) ̸ = ∅ (
we take into account the possibility of adding edge e in the set of edges

E ′0 k to route demand k).
We let S ′0 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′0 is clearly feasible for the problem, , and its incidence vector (x S ′0 , z S ′0 ) belongs to P(G, K, S). Let S 2 = (E 2 , S 2 ) be the solution obtained from S ′0 by adding edge

e ∈ E \ (E k 0 ∪ E k 1 )
for the routing of demand k in solution S ′0 that E 2 k = E ′0 k ∪ {e}, and remove slot s already selected for demand k as last slot in S ′0 and replaced it by a new slot s ′ such that s ′ is the smallest slot index in {w k , ..., s} such that

{s ′ -w k + 1, ..., s ′ } ∩ {s" -w k ′ + 1, ..., s"} = ∅ for each demand k ′ with E ′0 k ∩ E ′0 k ′ ̸ = ∅. S 2 is
clearly feasible for the problem. The corresponding incidence vector (x S 2 , z S 2 ) belongs to P(G, K, S). Hence, solutions S ′0 and S 2 satisfy equation µx + σz = λ. It follows that

µx S ′0 + σz S ′0 = µx S 2 + σz S 2 = µx S ′0 + µ k e + σz S ′0 -σ k s + σ k s ′ .
Since σ k s = 0 for all k ∈ K and s ∈ {w k , ..., s}, µ k e = 0 for demand k and edge e. In a similar way, we can show that

µ k e = 0, for all k ∈ K and e ∈ E \ (E k 0 ∪ E k 1 ).
Therefore all the equations of the polytope P(G, K, S) are given only in terms of the variables x k e with e ∈ E k 0 ∪ E k 1 and z k s with s ∈ {1, ..., w k }. We distinguish 3 blocks of lines in the matrix Q associated with the system (2.13) a) block Q 1 corresponds to the equations x k e = 0 for all k ∈ K and e

∈ E k 0 such that rang(Q 1 ) = k∈K |E k 0 |, b) block Q 2 corresponds to the equations x k e = 1 for all k ∈ K and e ∈ E k 1 such that rang(Q 2 ) = k∈K |E k 1 |, c) block Q 3 corresponds to the equations z k s = 0 for all k ∈ K and s ∈ {1, ..., w k - 1} such that rang(Q 3 ) = k∈K w k -1.
Note that the 3 blocks of the matrix

Q are independents. Let Q k =     Q 1 k Q 2 k Q 3 k   
 be the submatrix of matrix Q associated to the equations (2.12) and (2.11) and involving variables x k e for all e ∈ E k 0 ∪ E k 1 , and variables z k s with s ∈ {1, ..., w k } for demand k. Note that a forbidden edge can never be an essential edge at the same time.

Otherwise, the problem is infeasible. Furthermore, there is no dependency between essential edges for each demand k and also for different demands in K. Same thing for the forbidden edges. We want to show that

µ k = γ k 1 Q 1 k + γ k 2 Q 2 k and σ k = γ k 3 Q k 3 .
The only solution of these two systems is given by

µ k e = γ k,e 1 , for all k ∈ K and e ∈ E k 0 , (2.14)
µ k e = γ k,e 2 , for all k ∈ K and e ∈ E k 1 , (2.15)
σ k s = γ k,s 3 , for all k ∈ K and s ∈ {1, ..., w k -1}. (2.16)
We conclude at the end that for each k ∈ K and e ∈ E

µ k e =          γ k,e 1 if e ∈ E k 0 γ k,e 2 if e ∈ E k 1 0 otherwise, (2.17)
yielding

µ k = γ k 1 Q 1 k + γ k 2 Q 2 k for each k ∈ K.
Moreover, for each k ∈ K and s ∈ S

σ k s =    γ k,s 3 if s ∈ {1, ..., w k -1} 0 otherwise, (2.18)
i.e., σ k = γ k 3 Q 3 k . As a consequence, (µ, σ) = γQ as desired.

As a consequence, we have the following result.

Theorem 2.3.1. The dimension of P(G, K, S) is given by

dim(P(G, K, S)) = |K| * (|E| + |S|) -r.

Facial Investigation

In this section, we describe facets defining inequalities for the polytope P(G, K, S) from the cut formulation (2.2)-(2.10), and the ones from the valid inequalities. First, we characterize when the basic inequalities (2.2)-(2.10) define facets.

Theorem 2.3.2. Consider a demand k ∈ K, and an edge e ∈ E \ (E k 0 ∪ E k 1 ). Then, inequality x k e ≥ 0 is facet defining for P(G, K, S).

Proof. Let F k e be the face induced by inequality x k e ≥ 0, that is

F k e = {(x, z) ∈ P(G, K, S) : x k e = 0}.
Denote inequality x k e ≥ 0 by αx + βz ≤ λ. Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) : µx + σz = τ }. Suppose that F k e ⊆ F . To prove that F k e is a facet of P(G, K, S), we need to show that there exist ρ ∈ R and γ

= (γ 1 , γ 2 , γ 3 ) ( with γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α, β) + γQ.
First, we will show that σ k ′ s = 0 for all k ′ ∈ K and s ∈ {w k ′ , ..., s}. Consider a slot s in {w k , ..., s}, and solution S 3 = (E 3 , S 3 ) such that a) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we let E 3 k i be the set of edges involved in a shortest path between o k i and d k i , b) for demand k, we let E 3 k be the set of edges involved in a shortest path between o k and d k which does not use edge e, c) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 3 i given by where

I 3 i = [
D 3 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E 3 k i ∩ E 3 k j ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 3 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s -w k + 1, ..., s} = ∅ if E 3 k i ∩ E 3 k ̸ = ∅ (
we take into account the possibility of adding slot s in the set of last slots

S 3
k assigned to demand k in solution S 3 ).

We let S 3 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S 3 is clearly feasible for the problem, and its incidence vector (x S 3 , z S 3 ) belongs to F k e . Now, let S 4 = (E 4 , S 4 ) a solution obtained from S 3 by adding slot s as last slot to demand k. Solution S 4 is feasible for the problem. The corresponding incidence vector (x S 4 , z S 4 ) belongs to F k e . Hence, solutions S 3 and S 4 satisfy equation µx + σz = τ . As a consequence, we have

µx S 3 + σz S 3 = µx S 4 + σz S 4 = µx S 3 + σz S 3 + σ k s .
Hence, σ k s = 0. In a similar way, we can show that

σ k ′ s ′ = 0, for all k ′ ∈ K and s ′ ∈ {w k ′ , ..., s}.
Next, we will show that µ k ′ e ′ = 0 for all demand k ′ ∈ K \ {k} and edge

e ′ ∈ E \ (E k ′ 0 ∪ E k ′ 1 )
, and

µ k e ′ = 0 for edge e ′ ∈ E \ (E k 0 ∪ E k 1 ∪ {e}). Consider an edge e ′ ∈ E \ (E k 0 ∪ E k 1 ∪ {e}) chosen arbitrarily. Let S ′3 = (E ′3 , S ′3
) be the solution given by a) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we let E ′3 k i be the set of edges involved in a shortest path between o k i and

d k i , b) for demand k, we let E ′3
k be the set of edges involved in a shortest path between o k and d k which does not use edge e ′ , c) we select slot s k = w k as last slot for demand k in solution S ′3 , d) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′3 i given by

I ′3 i = [ kj ∈D ′3 i {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]∩[{w ki , ..., s k -w k }∪{s k +w ki , ..., s}] if E ′3 ki ∩ (E ′3 k ∪ {e ′ }) ̸ = ∅ or I ′3 i = kj ∈D ′3 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,
where

D ′3 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E ′3 k i ∩ E ′3 k j ̸ = ∅}. This ensures that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D ′3 i , • {s k i -w k i + 1, ..., s k i } ∩ {s k -w k + 1, ..., s k } = ∅ if E ′3 k i ∩ (E ′3 k ∪ {e ′ }) ̸ = ∅ (
we take into account the possibility of adding edge e in the set of edges

E ′3 k to route demand k).
We let S ′3 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′3 is clearly feasible for the problem, and its incidence vector (x S ′3 , z S ′3 ) belongs to F k e . Let S 5 = (E 5 , S 5 ) be the solution obtained from S ′3 by adding edge e ′ ∈ E\(E k 0 ∪ E k 1 ) for the routing of demand k in solution S ′3 that E 5 k = E ′3 k ∪ {e ′ }, and removing slot s selected for demand k in S ′3 and replaced it by a new slot s ′ ∈ {w k , ..., S} (i.e., 5 is clearly feasible for the problem.

S 5 k = (S ′3 k \ {s}) ∪ {s ′ } such that {s ′ -w k + 1, ..., s ′ } ∩ {s" -w k ′ + 1, ..., s"} = ∅ for each k ′ ∈ K and s" ∈ S ′3 k ′ with E 5 k ∩ E ′3 k ′ ̸ = ∅). S
The corresponding incidence vector (x S 5 , z S 5 ) belongs to F k e . Hence, solutions S ′3 and S 5 satisfy equation µx + σz = τ . As a consequence, we have

µx S ′3 + σz S ′3 = µx S 5 + σz S 5 = µx S ′3 + µ k e ′ + σz S ′3 -σ k s + σ k s ′ .
Since σ k s = 0, it follows that µ k e ′ = 0. As e ′ is chosen arbitrarily, we have that

µ k e ′ = 0, for all e ′ ∈ E \ (E k 0 ∪ E k 1 ∪ {e}).
Using similar argument as above, we can show that

µ k ′ e ′ = 0, for all k ′ ∈ K \ {k} and e ′ ∈ E \ (E k ′ 0 ∪ E k ′ 1 )
.

By (2.17) and (2.18), we know that

         µ k ′ e ′ = γ k ′ ,e ′ 1 for all k ′ ∈ K and e ′ ∈ E k ′ 0 , µ k ′ e ′ = γ k ′ ,e ′ 2 for all k ′ ∈ K and e ′ ∈ E k ′ 1 , σ k ′ s ′ = γ k ′ ,s ′ 3 for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.
Overall, we obtain that

µ k ′ e ′ =                  γ k ′ ,e ′ 1 if e ′ ∈ E k ′ 0 , γ k ′ ,e ′ 2 if e ′ ∈ E k ′ 1 , ρ if k ′ = k and e ′ = e, 0 otherwise,
for each k ′ ∈ K and e ′ ∈ E, and

σ k ′ s ′ =    γ k ′ ,s ′ 3 if s ′ ∈ {1, ..., w k ′ -1}. 0 otherwise, for each k ′ ∈ K and s ′ ∈ S.
Clearly, we have (µ, σ) = ρ(α, β) + γQ.

Theorem 2.3.3. Consider a demand k ∈ K, and a slot s ∈ {w k , .., s}. Then, inequality z k s ≥ 0 is facet defining for P(G, K, S).

Proof. Let F k s denote the face induced by inequality z k s ≥ 0, that is

F k s = {(x, z) ∈ P(G, K, S) : z k s = 0}.
Denote inequality z k s ≥ 0 by αx + βz ≤ λ. Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) : µx + σz = τ }. Suppose that F k s ⊆ F . To prove that F k s is a facet of P(G, K, S), it suffices to show that there exist ρ ∈ R and γ

= (γ 1 , γ 2 , γ 3 ) ( with γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α, β) + γQ.
First, we will show that µ k ′ e ′ = 0 for all demand k ′ ∈ K and edge

e ′ ∈ E \ (E k ′ 0 ∪ E k ′ 1 ). Consider an edge e ∈ E \ (E k 0 ∪ E k 1 ). Let S 6 = (E
I 6 i = [ kj ∈D 6 i {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]∩[{w ki , ..., s k -w k }∪{s k +w ki , ..., s}] if E 6 ki ∩ (E 6 k ∪ {e}) ̸ = ∅ or I 6 i = kj ∈D 6 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,
where

D 6 i = {k j ∈ {k 1 , ..., k i-1 } : E 6 k i ∩ E 6 k j ̸ = ∅}. This verifies that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 6 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s k -w k + 1, ..., s k } = ∅ if E 6 k i ∩ (E 6 k ∪ {e}) ̸ = ∅ (
we take into account the possibility of adding edge e in the set of edges

E 6
k to route demand k).

We let S 6 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S 6 is clearly feasible for the problem, and its incidence vector (x S 6 , z S 6 ) belongs to

F k s .
Based on this, we consider a solution S 7 = (E 7 , S 7 ) obtained from S 6 by adding edge e ∈ E \(E k 0 ∪E k 1 ) for the routing of demand k in solution S 6 that E 7 k = E 6 k ∪{e}. S 7 is clearly feasible for the problem. The corresponding incidence vector (x S 7 , z S 7 ) belongs to F k s . Hence, solutions S 6 and S 7 satisfy equation µx + σz = τ . As a consequence, we have

µx S 6 + σz S 6 = µx S 7 + σz S 7 = µx S 6 + µ k e + σz S 6 .
As a result, µ k e = 0 for demand k and edge e. In a similar way, we can show that

µ k ′ e = 0, for all k ′ ∈ K and e ∈ E \ (E k ′ 0 ∪ E k ′ 1 ).
Next, we will show that, σ k ′ s ′ = 0 for all k ′ ∈ K \ {k} and s ′ ∈ {w k ′ , ..., s}, and σ k s ′ = 0 for all slots s ′ ∈ {w k , ..., s} \ {s}. Consider a slot s ′ in {w k , ..., s} \ {s}. Let S ′6 = (E ′6 , S ′6 ) be the solution given by a) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E ′6 k i be the set of edges involved in a shortest path between o k i and d k i , b) we select the smallest slot index s k in {w k , ..., s}\{s, s ′ } as last slot for demand k, c) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′6 i given by

I ′6 i = [ kj ∈D ′6 i {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]∩[{w ki , ..., s ′ -w k }∪{s ′ +w ki , ..., s}] if E ′6 ki ∩ E ′6 k ̸ = ∅ or I ′6 i = kj ∈D ′6 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,
where

D ′6 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E ′6 k i ∩ E ′6 k j ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s} ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D ′6 i , • and {s k i -w k i + 1, ..., s} ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E ′6 k i ∩ E ′6 k ̸ = ∅ (
we take into account the possibility of adding slot s ′ in the set of last slots s′6

k assigned to demand k in solution S ′6 ).

We let S ′6 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′6 is clearly feasible for the problem, its incidence vector (x S ′6 , z S ′6 ) belongs to F k s . Then consider the solution S 8 obtained from S ′6 by adding slot s ′ as last slot to demand k. Solution S 8 is clearly feasible for the problem. The corresponding incidence vector (x S 8 , z S 8 ) belongs to F k s . Hence, solutions S ′6 and S 8 satisfy equation µx + σz = τ . We have so

µx S ′6 + σz S ′6 = µx S 8 + σz S 8 = µx S ′6 + σz S ′6 + σ k s ′ .
Hence, σ k s ′ = 0. In a similar way, we can show that

σ k ′ s ′ = 0, for all k ′ ∈ K and s ′ ∈ {w k ′ , ..., s} with s ̸ = s ′ if k = k ′ .          µ k ′ e ′ = γ k ′ ,e ′ 1 for all k ′ ∈ K and e ′ ∈ E k ′ 0 , µ k ′ e ′ = γ k ′ ,e ′ 2 for all k ′ ∈ K and e ′ ∈ E k ′ 1 , σ k ′ s ′ = γ k ′ ,s ′ 3 for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.
We conclude that for each k ′ ∈ K and e ∈ E

µ k ′ e =          γ k ′ ,e 1 if e ∈ E k ′ 0 , γ k ′ ,e 2 if e ′ ∈ E k ′ 1 , 0 otherwise,
and for each k ′ ∈ K and s ′ ∈ S

σ k ′ s ′ =                  γ k ′ ,s ′ 3 if s ′ ∈ {1, ..., w k ′ -1}, 0 if s ′ ∈ {w k ′ , ..., s} and k ′ ̸ = k, 0 if s ′ ∈ {w k ′ , ..., s} \ {s} and k ′ = k, ρ if s ′ = s and k ′ = k.
Clearly, we have (µ, σ) = ρ(α, β) + γQ.

Proposition 2.3.3. Consider a demand k ∈ K. Let (e, e ′ ) be a pair of noncompatible edges for demand k. Then, the inequality

x k e + x k e ′ ≤ 1, (2.19)
is valid for P(G, K, S).

Proof. It is trivial due to the transmission-reach constraint and given the definition of non-compatible edges for demand k.

Based on the definition of a non-compatible demands for an edge e, we introduce the following inequality.

∪ E k 1 ∪ E k ′ 0 ∪ E k ′ 1 .
Then, the inequality

x k e + x k ′ e ≤ 1, (2.20)
is valid for P(G, K, S).

Proof. It is trivial given the definition of non-compatible demands for edge e. As a result, inequality x k e ≤ 1 is not facet defining for P(G, K, S). Sufficiency.

Let F ′k

e denote the face induced by inequality x k e ≤ 1, that is

F ′k e = {(x, z) ∈ P(G, K, S) : x k e = 1}.
Denote inequality x k e ≤ 1 by αx + βz ≤ λ. Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) : µx + σz = τ }. Suppose that F ′k e ⊆ F . To prove that F ′k e is a facet of P(G, K, S), we need to show that there exist ρ ∈ R and γ

= (γ 1 , γ 2 , γ 3 ) ( with γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α, β) + γQ.
First, we will show that σ k ′ s = 0 for all k ′ ∈ K and s ∈ {w k ′ , ..., s}. Consider a slot s in {w k , ..., s}. Let S 9 = (E 9 , S 9 ) be the solution given by a) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we let E where

D 9 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E 9 k i ∩ E 9 k j ̸ = ∅}. This ensures that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 9 i ,
• and {s k i -w k i +1, ..., s k i }∩{s-w k +1, ..., s} = ∅ ( we take into account the possibility of adding slot s in the set of last slots S 9 k assigned to demand k in solution S 9 ). We let S 9 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S 9 is clearly feasible for the problem, and its incidence vector (x S 9 , z S 9 ) belongs to F ′k e . Then consider the solution S 10 = (E 10 , S 10 ) obtained from S 9 by adding slot s as last slot to demand k. Solution S 10 is feasible for the problem. The corresponding incidence vector (x S 10 , z S 10 ) belongs to F ′k e . Hence, solutions S 9 and S 10 satisfy equation µx + σz = τ . We then obtain that µx S 9 + σz S 9 = µx S 10 + σz S 10 = µx S 9 + σz S 9 + σ k s .

Hence, σ k s = 0. In a similar way, we can show that

σ k ′ s = 0, for all k ′ ∈ K and s ∈ {w k ′ , ..., s} Next, we will show that µ k ′ e ′ = 0 for all demand k ′ ∈ K \ {k} and e ′ ∈ E \ (E k ′ 0 ∪ E k ′ 1 ), and µ k e ′ = 0 for demand k and e ′ ∈ E \ (E k 0 ∪ E k 1 ∪ {e}). Consider an edge e ′ ∈ E \ (E k 0 ∪ E k 1 ∪ {e}) chosen arbitrarily. Let S ′9 = (E ′9 , S ′9
) be the solution given by a) for each demand k i ∈ K \ {k} with i ∈ {1, .. where Hence, µ k e ′ = 0. In a similar way, we can show that

D ′9 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E ′9 k i ∩ E ′9 k j ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D ′9 i , • and {s k i -w k i +1, ..., s k i }∩{s k -w k +1, ..., s k } = ∅ if E ′9 k i ∩(E ′9 k ∪{e ′ }) ̸ = ∅ (
µ k e ′ = 0, for all e ′ ∈ E \ (E k 0 ∪ E k 1 ∪ {e}).
Moreover, we re-do the same procedure for all k ′ ∈ K \ {k} and e ′ ∈ E \ (E k 0 ∪ E k 1 ). We conclude at the end that

µ k ′ e ′ = 0, for all k ′ ∈ K \ {k} and e ′ ∈ E \ (E k ′ 0 ∪ E k ′ 1 ), µ k e ′ = 0, for all e ′ ∈ E \ (E k 0 ∪ E k 1 ∪ {e}). 50
We know from (2.17) and (2.18) that

         µ k ′ e ′ = γ k ′ ,e ′ 1 for all k ′ ∈ K and e ′ ∈ E k ′ 0 , µ k ′ e ′ = γ k ′ ,e ′ 2 for all k ′ ∈ K and e ′ ∈ E k ′ 1 , σ k ′ s ′ = γ k ′ ,s ′ 3
for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.

We conclude that for each k ′ ∈ K and e ′ ∈ E

µ k ′ e ′ =                  γ k ′ ,e ′ 1 if e ′ ∈ E k ′ 0 , γ k ′ ,e ′ 2 if e ′ ∈ E k ′ 1 , ρ if k ′ = k and e ′ = e, 0 otherwise,
and for each k ′ ∈ K and s ∈ S

σ k ′ s =    γ k ′ ,s 3 if s ∈ {1, ..., w k ′ -1}, 0 otherwise.
Consequently, (µ, σ) = ρ(α, β) + γQ which ends the proof.

Theorem 2.3.5. Consider a demand k ∈ K, and a slot s ∈ {w k , .., s}. Then, inequality z k s ≤ 1 is facet defining for P(G, K, S) if and only if there does not exist a demand

k ′ ∈ K \ {k} with E k 1 ∩ E k ′ 1 ̸ = ∅.
Proof. Neccessity.

For a demand k ∈ K and a slot s ∈ {w k , .., s}, if there exists a demand k ′ ∈ K \ {k}

with E k 1 ∩ E k ′ 1 ̸ = ∅.
Then, the inequality z k s ≤ 1 is domined by the non-overlapping inequality (2.6) for each edge e ∈ E k 1 ∩ E k ′ 1 . As a result, the inequality z k s ≤ 1 is not facet defining for P(G, K, S).

Sufficiency.

Let F ′k s be the face induced by inequality z k s ≤ 1, that is

F ′k s = {(x, z) ∈ P(G, K, S) : z k s = 1}.
We denote inequality z k s ≤ 1 by αx + βz ≤ λ. Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) : µx + σz = τ }. Suppose that F ′k s ⊆ F . To prove that F ′k s is a facet of P(G, K, S), we need to show there exist ρ ∈ R and γ

= (γ 1 , γ 2 , γ 3 ) ( with γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α, β) + γQ.
First, we will show that µ k ′ e = 0 for all demand k ′ ∈ K and edge e where

∈ E \ (E k ′ 0 ∪ E k ′ 1 ). Consider an edge e ∈ E \ (E k 0 ∪ E k 1 ). Let S 12 = (E
D 12 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E 12 k i ∩ E 12 k j ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 12 i , • and {s k i -w k i +1, ..., s k i }∩{s k -w k +1, ..., s k } = ∅ if E 12 k i ∩(E 12
k ∪{e}) ̸ = ∅ ( we take into account the possibility of adding edge e in the selected path

E 12
k to route demand k in solution S 12 ).

We let S 12 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S 12 is clearly feasible for the problem, its incidence vector (x S 12 , z S 12 ) belongs to F ′k s . Then consider the solution S 13 = (E 13 , S 13 ) obtained from S 12 by adding edge e ∈ E \ (E k 0 ∪ E k 1 ) for the routing of demand k in solution S 12 which means that E 13 k = E 12 k ∪ {e}. The last slots assigned to the demands K, and paths assigned the set of demands K \ {k} in S 12 remain the same in solution S 13 , i.e., S 13 k = S 12 k for each k ∈ K, and 13 is clearly feasible for the problem. The corresponding incidence vector (x S 13 , z S 13 ) belongs to F ′k s . Hence, solutions S 12 and S 13 satisfy equation µx + σz = τ . It follows that µx S 12 + σz S 12 = µx S 13 + σz S 13 = µx S 12 + µ k e + σz S 12 .

E 13 k ′ = E 12 k ′ for each k ′ ∈ K \ {k}. S
As a result, µ k e = 0. In a similar way, we can show that

µ k ′ e = 0, for all k ′ ∈ K and e ∈ E \ (E k ′ 0 ∪ E k ′ 1 ).
Next, we will show that, σ k ′ s ′ = 0 for all k ′ ∈ K \ {k} and s ′ ∈ {w k ′ , ..., s}, and σ k s ′ = 0 for all slots s ′ ∈ {w k , ..., s} \ {s}. Consider a slot s ′ in {w k , ..., s} \ {s}. Let S ′12 = (E ′12 , S ′12 ) be the solution given by a) for each demand where

k i ∈ K with i ∈ {1, ...,
D ′12 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E ′12 k i ∩ E ′12 k j ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D ′12 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E ′12 k i ∩ E ′12
k ̸ = ∅ ( we take into account the possibility of adding slot s ′ in the selected last slots S ′12 k to route demand k in solution S ′12 ).

We let S ′12 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′12 is clearly feasible for the problem, its incidence vector (x S ′12 , z S ′12 ) belongs to F ′k s . Then we derive solution S 14 from S ′12 by adding slot s ′ as last slot to demand k in S ′12 . Solution S 14 is clearly feasible for the problem. The corresponding incidence vector (x S 14 , z S 14 ) belongs to F ′k s . Hence, solutions S ′12 and S 14 satisfy equation µx + σz = τ . We have so

µx S ′12 + σz S ′12 = µx S 14 + σz S 14 = µx S ′12 + σz S ′12 + σ k s ′ .
Hence, σ k s ′ = 0. In a similar way, we can show that

σ k ′ s ′ = 0, for all k ′ ∈ K and s ′ ∈ {w k ′ , ..., s} with s ̸ = s ′ if k = k ′ .          µ k ′ e ′ = γ k ′ ,e ′ 1 for all k ′ ∈ K and e ′ ∈ E k ′ 0 , µ k ′ e ′ = γ k ′ ,e ′ 2 for all k ′ ∈ K and e ′ ∈ E k ′ 1 , σ k ′ s ′ = γ k ′ ,s ′ 3
for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.

Overall, we obtain that

µ k ′ e =          γ k ′ ,e 1 if e ∈ E k ′ 0 , γ k ′ ,e 2 if e ′ ∈ E k ′ 1 , 0 otherwise,
for each k ′ ∈ K and e ∈ E, and

σ k ′ s ′ =                  γ k ′ ,s ′ 3 if s ′ ∈ {1, ..., w k ′ -1}, 0 if s ′ ∈ {w k ′ , ..., s} and k ′ ̸ = k, 0 if s ′ ∈ {w k ′ , ..., s} \ {s} and k ′ = k, ρ if s ′ = s and k ′ = k, for each k ′ ∈ K and s ′ ∈ S.
As a consequence, (µ, σ) = ρ(α, β) + γQ.

Theorem 2.3.6. Consider a demand k ∈ K. Then, inequality (2.5), s s=w k z k s ≥ 1, is facet defining for P(G, K, S).

Proof. Let F k S be the face induced by inequality s s=w k z k s ≥ 1, that is

F k S = {(x, z) ∈ P(G, K, S) : s s=w k z k s = 1}.
Denote inequality

s s=w k z k s ≥ 1 by αx + βz ≤ λ.
Let µx + σz ≤ τ be a valid inequality that defines a facet F of P(G, K, S). Suppose that F k S ⊆ F . To prove that F k S is a facet of P(G, K, S), we show that there exist ρ ∈ R and γ = (γ 1 , γ 2 , γ 3 ) ( with

γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1 ) such that (µ, σ) = ρ(α, β) + γQ.
First, we will show that µ k ′ e = 0 for all demand k ′ ∈ K and edge e where As a result, µ k e = 0. In a similar way, we can show that

∈ E \ (E k ′ 0 ∪ E k ′ 1 ). Consider an edge e ∈ E \ (E k 0 ∪ E k 1 ). Let S 15 = (E
D 15 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E 15 k i ∩ E 15 k j ̸ = ∅}. This ensures that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 15 i , • {s k i -w k i + 1, ..., s k i } ∩ {s -w k + 1, ..., s} = ∅ if E 15 k i ∩ (E
µ k ′ e = 0, for all k ′ ∈ K and e ∈ E \ (E k ′ 0 ∪ E k ′ 1 ).
Next, we will show that, σ k ′ s ′ = 0 for all k ′ ∈ K \ {k} and s ′ ∈ {w k ′ , ..., s}. Consider a demand k ′ in K \ {k} and a slot s ′ in {w k ′ , ..., s} \ {s}. Let S ′15 = (E ′15 , S ′15 ) be the solution given by a) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E ′15 k i be the set of edges involved in a shortest path between o k i and d k i , b) we select slot s k = w k as last slot for demand k, and let S ′15 k = {s k }, c) we select the smallest slot index s k ′ from the set of slots I ′15 k ′ given by where

I ′15 k ′ = {w ki , ..., s k -w k }∩{s k +w ki , ..., s}\{s ′ } if E ′15 k ′ ∩E ′15 k ̸ = ∅ or I ′15 k ′ = {w k ′ , ..., s}\{s ′ } if not. d) for each demand k i ∈ K \ {k, k ′ } with i ∈ {1, ...,
D ′15 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k, k ′ } : E ′15 k i ∩ E ′15 k j ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D ′15 i , • {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k ′ + 1, ..., s} = ∅ if E ′15 k i ∩ E ′15 k ′ ̸ = ∅ (
we take into account the possibility of adding slot s ′ in the set of last slots

S ′15

k ′ to route demand k ′ in solution S ′15 ).

We let S ′15 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′15 is feasible for the problem. The corresponding incidence vector (x S ′15 , z S ′15 ) belongs to F k S . Then we derive a solution S 17 from S ′15 by adding slot s ′ as last slot to demand k ′ . Solution S 17 is clearly feasible for the problem. The corresponding incidence vector (x S 17 , z S 17 ) belongs to F k S . Hence, solutions S ′15 and S 17 satisfy equation µx + σz = τ . We have so

µx S ′15 + σz S ′15 = µx S 17 + σz S 17 = µx S ′15 + σz S ′15 + σ k ′ s ′ . Hence, σ k ′ s ′ = 0.
In a similar way, we can show that

σ k ′ s ′ = 0,
i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : Ẽ15 k i ∩ Ẽ15 k j ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D15 i , • {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if Ẽ15 k i ∩ Ẽ15 k ̸ = ∅ (
we take into account the possibility of adding slot s ′ in the set of last slots S15 k to route demand k in solution S15 ).

We let S15 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}.

S15 is clearly feasible for the problem, and its incidence vector (x S15 , z S15 ) belongs to F k S . Then consider the solution S 18 obtained from S15 by adding slot s ′ as last slot to demand k ′ in S 18 k and removing the last slot s assigned to k in S15 k (i.e., S 18 k = ( S15 k \ {s}) ∪ {s ′ } for demand k). Solution S 18 is feasible for the problem. The corresponding incidence vector (x S 18 , z S 18 ) belongs to F k S . Hence, solutions S15 and S 18 satisfy equation µx + σz = τ . We have so

µx S15 + σz S15 = µx S 18 + σz S 18 = µx S15 + σz S15 + σ k s ′ -σ k s .
As a result, σ k s ′ = σ k s . In a similar way, we can show that

σ k s ′ = σ k s , for all slots s, s ′ ∈ {w k , ..., s}.
Consequently, we obtain that σ k s = ρ for demand k and slot s in {w k , ..., s}. We know from (2.17) and (2.18) that

         µ k ′ e ′ = γ k ′ ,e ′ 1 for all k ′ ∈ K and e ′ ∈ E k ′ 0 , µ k ′ e ′ = γ k ′ ,e ′ 2 for all k ′ ∈ K and e ′ ∈ E k ′ 1 , σ k ′ s ′ = γ k ′ ,s ′
We conclude that for each k ′ ∈ K and e ∈ E

µ k ′ e =          γ k ′ ,e 1 if e ∈ E k ′ 0 , γ k ′ ,e 2 if e ∈ E k ′ 1 , 0 otherwise,
and for each k ′ ∈ K and s ∈ S x k e ≥ 1, that is

σ k ′ s =          γ k ′ ,s 3 if s ∈ {1, ..., w k ′ -1}, ρ if k ′ = k
F k X = {(x, z) ∈ P(G, K, S) : e∈(δ(X)\E k 0 )
x k e = 1}.

Let X = {o k }. Denote inequality e∈(δ(X)\E k 0 )
x k e ≥ 1 by αx + βz ≤ λ. Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) : µx + σz = τ }. Suppose that F k X ⊆ F . To prove that F k X is a facet of P(G, K, S), we need to show that there exist ρ ∈ R and γ = (γ 1 , γ 2 , γ 3 ) ( with

γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α, β) + γQ.
First, we will show that σ k ′ s = 0 for all k ′ ∈ K and s ∈ {w k ′ , ..., s}. Consider a slot s in {w k , ..., s}. Let S 19 = (E 19 , S 19 ) be the solution given by a) for each demand where Hence, σ k s = 0. In a similar way, we can show that

k i ∈ K \ {k} with i ∈ {1, ...,
D 19 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E 19 k i ∩ E 19 k j ̸ = ∅}. This ensures that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 19 i , • {s k i -w k i + 1, ..., s k i } ∩ {s -w k + 1, ..., s} = ∅ if E 19 k i ∩ E 19 k ̸ = ∅ (
σ k ′ s ′ = 0, for all k ′ ∈ K and s ∈ {w k ′ , ..., s}.
Next, we will show that µ k e ′ = 0 for all demand where

k ′ ∈ K \ {k} and e ′ ∈ E \ (E k ′ 0 ∪ E k ′ 1 ), and µ k e ′ = 0 for demand k and e ′ ∈ E \ (E k 0 ∪ E k 1 ∪ δ(X)). Consider an edge e ′ ∈ E \ (E k 0 ∪ E k 1 ∪ δ(X)) chosen
D ′19 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E ′19 k i ∩ E ′19 k j ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D ′19 i , • {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E ′19 k i ∩ (E ′19 k ∪ {e}) ̸ = ∅ (
we take into account the possibility of adding edge e ′ in the selected path E ′19 k to route demand k in solution S ′19 ). We let S ′19 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′19 is feasible for the problem. its incidence vector (x S ′19 , z S ′19 ) belongs to F k X . Let S 21 = (E 21 , S 21 ) be the solution obtained from S ′19 by adding edge e ′ ∈ E \(E k 0 ∪E k 1 ) for the routing of demand k in solution S ′19 which means that E 21 k = E ′19 k ∪{e ′ }. S 21 is clearly feasible for the problem. The corresponding incidence vector (x S 21 , z S 21 ) belongs to F k X . Hence, solutions S ′19 and S 21 satisfy equation µx + σz = τ . It follows that

µx S ′19 + σz S ′19 = µx S 21 + σz S 21 = µx S ′19 + µ k e ′ + σz S ′19 .
Hence, µ k e ′ = 0. In a similar way, we can show that

µ k ′ e ′ = 0, for all k ′ ∈ K \ {k} and e ′ ∈ E \ (E k ′ 0 ∪ E k ′ 1 ), µ k e ′ = 0, for all e ′ ∈ E \ (E k 0 ∪ E k 1 ∪ δ(X)).
Next, we will prove that the µ k e for all edge e ∈ (δ(X) \ E k 0 ) are equivalent. Consider an edge e ′ ∈ (δ(X) \ E k 0 ) such that e ′ / ∈ E 19 k . Let S19 = ( Ẽ19 , S19 ) be the solution given by a) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we let Ẽ19 k i be the set of edges involved in a shortest path between o k i and d k i , b) for demand k, we let Ẽ19 k be the set of edges involved in a shortest path between o k and d k . This guarantees that one edge e from (δ(X) where

\ E k 0 ) is chosen to route demand k, i.e., |(δ(X) \ E k 0 ) ∩ Ẽ19 k | = 1, c)
D19 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : Ẽ19 k i ∩ Ẽ19 k j ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D19 i , • {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if Ẽ19 k i ∩ ( Ẽ19 k ∪ {e}) ̸ = ∅ (
we take into account the possibility of using edge e ′ in the selected path Ẽ19 k to route demand k in solution S19 ).

We let S19

k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}.

S19 is clearly feasible for the problem, and corresponding incidence vector (x S19 , z S19 ) belongs to F k X . Then consider the solution S 22 obtained from S19 by modifying the path assigned to demand k in S19 from Ẽ19 k to a path E 22 k passed through edge e ′ with |(δ(X) \ E k 0 ) ∩ E 22 k | = 1, and modifying the last slots assigned to some demands K ⊂ K from S19 k to S 22 k for each k ∈ K while satisfying non-overlapping constraint. The paths assigned to the demands K \ {k} in S19 remain the same in S 22 (i.e., E 22 k" = Ẽ19 k" for each k" ∈ K \ {k}), and also without modifying the last slots assigned to the demands K\ K in S19 , i.e., S19 k = S 22 k for each demand k ∈ K\ K. Solution S 22 is feasible for the problem. The corresponding incidence vector (x S 22 , z S 22 ) belongs to F k X . Hence, solutions S19 and S 22 satisfy equation µx + σz = τ . We have so

µx S19 + σz S19 = µx S 22 + σz S 22 = µx S19 + σz S19 + µ k e ′ -µ k e + k∈ K s ′ ∈S 22 k σ k s ′ - s∈ S19 k σ k s + e"∈E 22 k \{e ′ } µ k e" - e"∈ Ẽ19 k \{e} µ k e" .
Since µ k e" = 0 for all e" ∈ E \ (E k 0 ∪ E k 1 ∪ δ(X)), and σ k ′ s = 0 for all k ′ ∈ K and s ∈ {w k ′ , ..., s}, it follows that µ k e ′ = µ k e . In a similar way, we can show that µ k e = µ k e ′ , for all pairs e, e ′ ∈ (δ(X) \ E k 0 ).

Consequently, we obtain that µ k e = ρ for all e ∈ (δ(X) \ E k 0 ). By (2.17) and (2.18), we know that

         µ k ′ e ′ = γ k ′ ,e ′ 1 for all k ′ ∈ K and e ′ ∈ E k ′ 0 , µ k ′ e ′ = γ k ′ ,e ′ 2 for all k ′ ∈ K and e ′ ∈ E k ′ 1 , σ k ′ s ′ = γ k ′ ,s ′ 3 for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.
We conclude that for each k ′ ∈ K and e ∈ E

µ k ′ e =                  γ k ′ ,e 1 if e ∈ E k ′ 0 , γ k ′ ,e 2 if e ∈ E k ′ 1 , ρ if k = k ′ and e ∈ (δ(X) \ E k 0 ), 0 otherwise,
and for each k ′ ∈ K and s ∈ S

σ k ′ s =    γ k ′ ,s 3 if s ∈ {1, ..., w k ′ -1}, 0 otherwise.
As a consequence, (µ, σ) = ρ(α, β) + γQ.

In what follows, we present several valid inequalities for P(G, K, S), and study their facial structure.

Valid Inequalities and Facets

We start this section by introducing some classes of valid inequalities that can be defined using Chvàtal-Gomory procedure. Then, the inequality k∈Ke min(s+w k -1,s)

Edge-Slot-Assignment Inequalities

s ′ =s z k s ′ ≤ 1, (2.21)
is valid for P(G, K, S).

Proof. Inequality (2.21) ensures that the set of demands K e cannot share slot s over edge e, which means that slot s is assigned to at most one demand k from K e over edge e. We know from the inequality (2.6) that for each pair of demands k, k

′ ∈ K e with k ̸ = k ′ min(s+w k -1,s) s ′ =s z k s ′ + min(s+w k ′ -1,s) s ′ =s z k ′ s ′ ≤ 1, given that x k e = x k ′ e = 1.
After that, we use the Chvàtal-Gomory and recurrence procedures to prove that (2.21) is valid for P(G, K, S). For any subset of demands K ⊆ K e , by using a recurrence procedure, we get that for all demands K ′ ⊆ K with

|K ′ | = | K| -1 k∈K ′ min(s+w k -1,s) s ′ =s z k s ′ ≤ 1.
By adding the previous inequalities for all subset of demands

K ′ ⊆ K with |K ′ | = | K| -1 K ′ ⊆ K |K ′ |=| K|-1 k∈K ′ min(s+w k -1,s) s ′ =s z k s ′ ≤ K ′ ⊆ K |K ′ |=| K|-1 1.
Note that for each k ∈ K, the sum

min(s+w k -1,s) s ′ =s z k s ′ appears ( | K| | K|-1 -1) = | K|-1 times in the previous sum. This implies that k∈ K min(s+w k -1,s) s ′ =s (| K| -1)z k s ′ ≤ | K|.
By dividing the two sides of the previous sum by | K| -1, we have

k∈ K min(s+w k -1,s) s ′ =s z k s ′ ≤ | K| | K| -1 ⇒ k∈ K min(s+w k -1,s) s ′ =s z k s ′ ≤ | K| | K| -1 .
As a result,

k∈ K min(s+w k -1,s) s ′ =s z k s ′ ≤ 1 given that | K| | K| -1 = 1.
We conclude at the end that inequality (2.21) is valid for P(G, K, S).

Inspiring
∩ E k ′ 0 ∩ E k" 0 , (k, k ′ ) / ∈ K e c , (k, k") / ∈ K e c
, and (k ′ , k") /

∈ K e c . Then, the inequality

x k e + x k ′ e + x k" e + min(s+w k -1,s) s ′ =s z k s ′ + min(s+w k ′ -1,s) s ′ =s z k ′ s ′ + min(s+w k" -1,s) s"=s z k" s" ≤ 4, (2.22)
is valid for P(G, K, S).

Proof. Consider an edge e ∈ E. Let s be a slot in S. Inequality (2.22) ensures that if the three demands k, k ′ , k" pass through edge e, they cannot share slot s.

Let show that inequality (2.22) can be seen as Chvàtal-Gomory cuts using Chvàtal-Gomory procedure. We know from (2.24) that

x k e + x k ′ e + min(s+w k -1,s) s ′ =s z k s ′ + min(s+w k ′ -1,s) s ′ =s z k ′ s ′ ≤ 3, x k e + x k" e + min(s+w k -1,s) s ′ =s z k s ′ + min(s+w k" -1,s) s"=s z k" s" ≤ 3, x k ′ e + x k" e + min(s+w k ′ -1,s) s ′ =s z k ′ s ′ + min(s+w k" -1,s) s"=s z k" s" ≤ 3.
By adding the three previous inequalities, we get the following inequality

2x k e + 2x k ′ e + 2x k" e + 2 min(s+w k -1,s) s ′ =s z k s ′ + 2 min(s+w k ′ -1,s) s ′ =s z k ′ s ′ + 2 min(s+w k" -1,s) s"=s z k" s" ≤ 9.
By dividing the two sides of the previous inequality by 2, we obtain that

x k e + x k ′ e + x k" e + min(s+w k -1,s) s ′ =s z k s ′ + min(s+w k ′ -1,s) s ′ =s z k ′ s ′ + min(s+w k" -1,s) s"=s z k" s" ≤ 9 2 .
As a result,

x k e + x k ′ e + x k" e + min(s+w k -1,s) s ′ =s z k s ′ + min(s+w k ′ -1,s) s ′ =s z k ′ s ′ + min(s+w k" -1,s) s"=s z k" s" ≤ 4.
We conclude at the end that inequality (2.22) is valid for P(G, K, S).

Inequality (2.22) can then be generalized for any subset of demand K ⊆ K under certain conditions.

Proposition 2.4.3. Consider an edge e ∈ E, and a slot s in S. Let K be a subset of demands of K with e / ∈ E k 0 for each demand k ∈ K, (k, k ′ ) / ∈ K e c for each pair of demands (k, k ′ ) in K, and k∈ K w k ≤ sk"∈Ke\ K w k" . Then, the inequality

k∈ K x k e + k ′ ∈ K min(s+w k ′ -1,s) s ′ =s z k ′ s ′ ≤ | K| + 1, (2.23)
is valid for P(G, K, S).

Let n k denote the total number of possibilities to choose a k element in a set of n elements.

Proof. Inequality (2.23) ensures that if the demands k ∈ K pass through edge e, they cannot share slot s. For this, we use the Chvàtal-Gomory and recurrence procedures to prove that (2.23) is valid for P(G, K, S). For any subset of demands K ⊆ K with e / ∈ E k 0 for each demand k ∈ K, by recurrence procedures we get that for all demands

K ′ ⊆ K with |K ′ | = | K| -1 k∈K ′ x k e + k∈K ′ min(s+w k -1,s) s ′ =s z k s ′ ≤ |K ′ | + 1.
By adding the previous inequalities for all subset of demands

K ′ ⊆ K with |K ′ | = | K| -1 K ′ ⊆ K |K ′ |=| K|-1 k∈K ′ x k e + K ′ ⊆ K |K ′ |=| K|-1 k∈K ′ min(s+w k -1,s) s ′ =s z k s ′ ≤ K ′ ⊆ K |K ′ |=| K|-1 (|K ′ | + 1).
Note that for each k ∈ K, the variable x k e and the sum

min(s+w k -1,s) s ′ =s z k s ′ appear ( | K| | K|-1 -1) times in the previous sum. This implies that k∈ K( | K| | K| -1 -1)x k e + k∈ K min(s+w k -1,s) s ′ =s ( | K| | K| -1 -1)z k s ′ ≤ | K| | K| -1 (|K ′ |+1) Given that |K ′ | = | K| -1, this is equivalent to say that k∈ K( | K| | K| -1 -1)x k e + k∈ K min(s+w k -1,s) s ′ =s ( | K| | K| -1 -1)z k s ′ ≤ | K| | K| -1 | K|
Moreover, and taking into account that (

| K| | K|-1 -1) = | K| -1, we found that k∈ K(| K| -1)x k e + k∈ K min(s+w k -1,s) s ′ =s (| K| -1)z k s ′ ≤ | K| 2
By dividing the two sides of the previous sum by | K| -1, we have

k∈ K x k e + k∈ K min(s+w k -1,s) s ′ =s z k s ′ ≤ | K| 2 | K| -1 ⇒ k∈ K x k e + k∈ K min(s+w k -1,s) s ′ =s z k s ′ ≤ | K| | K| | K| -1 .
After some simplifications, we obtain that k∈

K x k e + k∈ K min(s+w k -1,s) s ′ =s z k s ′ ≤ | K| + | K| | K| -1 .
As a result,

k∈ K x k e + k∈ K min(s+w k -1,s) s ′ =s z k s ′ ≤ | K| + 1 given that | K| | K| -1 = 1.
We conclude at the end that inequality (2.23) is valid for P(G, K, S).

Inequality
∈ K with e / ∈ E k 0 ∩ E k ′ 0 and (k, k ′ ) / ∈ K e c .
Then, the inequality

x k e + x k ′ e + min(s+w k -1,s) s ′ =s z k s ′ + min(s+w k ′ -1,s) s ′ =s z k ′ s ′ + k"∈Ke\{k,k ′ } min(s+w k" -1,s) s ′ =s z k" s ′ ≤ 3, (2.24)
is valid for P(G, K, S).

Proof. Consider an edge e ∈ E, and a pair of demands k, k ′ ∈ K. Let s be a slot in S. Inequality (2.24) ensures that if the two demands k, k ′ pass through edge e, they cannot share slot s with the set of demands in K e \ {k, k ′ }. This can be seen as a particular case for inequality (2.21) induced by subset of demands

K = {k, k ′ } ∪ K e .
Let generalize inequality (2.24) for any subset of demand K ⊆ K under certain conditions.

Proposition 2.4.5. Consider an edge e ∈ E, and a slot s in S. Let K be a subset of demands of K with e / ∈ E k 0 for each demand k ∈ K, (k, k ′ ) / ∈ K e c for each pair of demands (k, k ′ ) in K, and k∈ K w k ≤ sk"∈Ke\ K w k" . Then, the inequality

k∈ K x k e + k∈ K min(s+w k -1,s) s ′ =s z k s ′ + k ′ ∈Ke\ K min(s+w k ′ -1,s) s"=s z k ′ s" ≤ | K| + 1, (2.25)
is valid for P(G, K, S).

This can be seen as a strengthened version of inequality (2.24).

Proof. Inequality (2.25) ensures that if the demands k ∈ K pass through edge e, they cannot share slot s with the set of demands in K e \ K. This can be seen be a particular case inequality (2.23) induced by K ∪ K e for slot s over edge e.

Definition 2.4.1. An interval I = [s i , s j ] represents an ordered set of contiguous slots situated between the two slots s i and s j with j ≥ i + 1 and s j ≤ s (e.g., interval I = [START_REF] Accorsi | Guidelines for the Computational Testing of Machine Learning approaches to Vehicle Routing Problems[END_REF][START_REF] Balcan | Learning to Branch[END_REF] contains all slots situated between the slots s i = 1 and s j = 6).

Theorem

k ′ ∈ K w k -1, ..., s j -max k∈ K w k + 1}, c) and w k + w k ′ ≥ |I| + 1 for each k, k ′ ∈ K, d) and 2w k ≥ |I| + 1 for each k ∈ K.
Proof. Neccessity.

• if K e \ K ̸ = ∅,

Sufficiency.

Let F e,s K be the face induced by inequality (2.23), that is

F e,s K = {(x, z) ∈ P(G, K, S) : k∈ K x k e + min(s+w k -1,s) s ′ =s z k s ′ = | K| + 1}. Let denote by αx + βz ≤ λ inequality k∈ K x k e + min(s+w k -1,s) s ′ =s z k s ′ ≤ | K| + 1.
Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) : µx + σz = τ }. Suppose that F e,s K ⊆ F . To prove that F e,s K is a facet of P(G, K, S), we need to show that there exists ρ ∈ R and γ = (γ 1 , γ 2 , γ 3 ) (such that such that s / ∈ {s

γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α, β) + γQ. Let first show that µ k e ′ = 0 for each edge e ′ ∈ E \ (E k 0 ∪ E k 1 ) for each demand k ∈ K with e ̸ = e ′ if k ∈ K. Consider a demand k ∈ K and an edge e ′ ∈ E \ (E k 0 ∪ E k 1 ) with e ̸ = e ′ if k ∈ K. Let S 38 = (E
k i -w k i +1, ..., s k i } if k i ∈ K, where D 38 i = {k j ∈ {k 1 , ..., k i-1 }∪{k ′ } : E 38 k i ∩ E 38 k j ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 38 i ,
• and

{s k i -w k i + 1, ..., s k i } ∩ {s k -w k + 1, ..., s k } = ∅ if E 38 k i ∩ (E 38 k ∪ {e ′ }) ̸ = ∅ (
we take into account the possibility of using edge e ′ in the selected path E ′38 k to route demand k in solution S ′38 ). We let S 38 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S 38 is feasible for the problem. its incidence vector (x S 38 , z S 38 ) belongs to F e,s K . Then consider the solution S 39 = (E 39 , S 39 ) obtained from S 38 by adding edge As a result, µ k e ′ = 0. In a similar way, we can show that

e ′ ∈ E \ (E k 0 ∪ E k 1 )
µ k e ′ = 0, for all k ∈ K and e ′ ∈ E \ (E k 0 ∪ E k 1 ) with e ̸ = e ′ if k ∈ K.
Let show that σ k s ′ = 0 for all k ∈ K and s ′ ∈ {w k , ..., s} with s ′ / ∈ {s, ..., s

+ w k -1} if k ∈ K.
Consider a demand k in K and a slot s ′ in {w k , ..., s} with s ′ / ∈ {s, ..., s

+ w k -1} if k ∈ K. Let S ′38 = (E ′38 , S ′38
) be the solution given by a) for each demand

k i ∈ K \ K with i ∈ {1, ..., |K|}, we let E ′38 k i be the set of edges involved in a shortest path between o k i and d k i , b) for demand k, we let E ′38
k be the set of edges involved in a shortest path between o k and d k which uses edge e, c) for each demand k ′ ∈ K \ {k}, we let E ′38 k ′ be the set of edges involved in a shortest path between o k ′ and d k ′ which use edge e, d) for one demand k ∈ K, we select the smallest slot index s k in {w k , ..., s} as last slot such that s ∈ {s k -w k + 1, ..., s k }, e) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′38 i given by

I ′38 i = [ kj ∈D ′38 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}] if E ′38 ki ∩ E ′38 k ̸ = ∅ or I ′38 i = kj ∈D ′38 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not.
where

D ′38 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E ′38 k i ∩ E ′38 k j ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D ′38 i , • s / ∈ {s k i -w k i + 1, ..., s k i } if k i ∈ K,
• and

{s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E ′38 k i ∩ E ′38 k ̸ = ∅ (
we take into account the possibility of adding slot s ′ in the selected set of last slots

S ′38

k to route demand k in solution S ′38 ).

We let S ′38 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′38 is clearly feasible for the problem. The corresponding incidence vector (x S ′38 , z S ′38 ) belongs to F e,s K . Then consider the solution S 40 obtained from S ′38 by adding slot s ′ as last slot to demand k. Solution S 40 is feasible for the problem. The corresponding incidence vector (x S 40 , z S 40 ) belongs to F e,s K . Hence, solutions S ′38 and S 40 satisfy equation µx + σz = τ . We have so

µx S ′38 + σz S ′38 = µx S 40 + σz S 40 = µx S ′38 + σz S ′38 + σ k s ′ .
It follows that σ k s ′ = 0. In a similar way, we can show that

σ k s ′ = 0, for all k ∈ K and s ′ ∈ {w k , ..., s} with s ′ / ∈ {s, ..., s + w k -1} if k ∈ K. Let prove that σ k s ′ for all k ∈ K and s ′ ∈ {s, ..., s+w k -1} are equivalent. Consider a demand k ′ ∈ K and a slot s ′ ∈ {s, ..., s + w k ′ -1} with k ′ ∈ K. Let S 41 = (E 41 , S 41 )
be a solution obtained from S 38 by adding slot s ′ as last slot to demand k ′ with modifying the paths assigned to a subset of demands K ⊂ K in S 38

(i.e., E 41 k = E 38 k for each k ∈ K \ K, and E 41 k ̸ = E 38 k for each k ∈ K)
, and also the last slots assigned to the demands K \ {k, k ′ } in S 38 remain the same in S 41 , i.e., S 38 k" = S 41 k" for each demand k" ∈ K \ {k, k ′ }, and S 41 k ′ = S 38 k ′ ∪ {s ′ } for demand k ′ , and modifying the last slots assigned to demand k by adding a new last slot s and removing the last slot s ′ ∈ S 38 k with s ′ ∈ {s i +w k +1, ..., s j } and s / ∈ {s i +w k +1, ..., s j } for demand

k with k ∈ K such that S 41 k = (S 38 k \{s})∪{s} such that {s-w k +1, ..., s}∩{s ′ -w k ′ +1, ..., s ′ } = ∅ for each k ′ ∈ K and s ′ ∈ S 41 k ′ with E 41 k ∩ E 41 k ′ ̸ = ∅. Solution S 41
is feasible for the problem. The corresponding incidence vector (x S 41 , z S 41 ) belongs to F e,s K . Hence, solutions S 38 and S 41 satisfy equation µx + σz = τ . We have so

µx S 38 + σz S 38 = µx S 41 + σz S 41 = µx S 38 + σz S 38 + σ k ′ s" -σ k s ′ + σ k s - k∈ K e ′ ∈E 38 k µ k e ′ + k∈ K e ′ ∈E 41 k µ k e ′ . Since σ k s = 0 for s / ∈ {s, ..., s + w k -1} with k ∈ K, and µ k e ′ = 0 for all k ∈ K and e ′ ∈ E \ (E k 0 ∪ E k 1 ) with e ′ ̸ = e if k ∈ K, it follows that σ k ′ s" = σ k s ′ .
In a similar way, we can show that

σ k s ′ = σ k ′ s" , for all pairs (k, k ′ ) ∈ K
with s ′ ∈ {s, ..., s+w k -1} and s ′ ∈ {s, ..., s+w k ′ -1}. We re-do the same procedure for each two slots s, s ′ ∈ {s, ..., s

+ w k -1} for each demand k ∈ K with k ∈ K such that σ k s ′ = σ k s"
, for all k ∈ K and s, s ′ ∈ {s, ..., s + w k -1}.

We will prove that µ k e for all k ∈ K are equivalent. Let S 42 = (E 42 , S 42 ) be the solution given by a) for each demand where

k i ∈ K \ K with i ∈ {1, ...,
D 42 i = {k j ∈ {k 1 , ..., k i-1 } ∪ K : E 42 k i ∩ E 42 k j ̸ = ∅}. This ensures that {s k i - w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 42
i . We let S 42 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}.

Obviously, S 42 is feasible for the problem. Moreover, the corresponding incidence vector (x S 42 , z S 42 ) belongs to F e,s K . Consider now a demand k ′ in K such that e / ∈ E 42 k ′ . We derive a feasible solution S 43 = (E 43 , S 43 ) for the problem from S 42 by a) the paths assigned to the demands K \ {k, k ′ } in S 42 remain the same in S 43 (i.e.,

E 43 k" = E 42 k" for each k" ∈ K \ {k, k ′ }),
b) without modifying the last slots assigned to the demands K in S 42 , i.e., S 42

k = S 43 k for each demand k ∈ K, c) modifying the path assigned to demand k ′ in S 42 from E 42 k ′ to a path E 43 k ′ passed through edge e (i.e., e ∈ E 43 k ′ ) with k ′ ∈ K such that {s -w k + 1, ..., s} ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ for each k ∈ K and s ′ ∈ S 42 k ′ and s ′ ∈ S 42 k with E 42 k ∩ E 43 k ′ ̸ = ∅, d)
w k + 1, ..., s} ∩ {s ′ -w k" + 1, ..., s ′ } = ∅ for each k" ∈ K \ {k, k ′ } and s ′ ∈ S 42 k and s ′ ∈ S 42 k" with E 42 k" ∩ E 43 k ̸ = ∅, and {s -w k + 1, ..., s} ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ for each s ′ ∈ S 42 k and s ′ ∈ S 42 k ′ with E 43 k" ∩ E 43 k ̸ = ∅.
The corresponding incidence vector (x S 43 , z S 43 ) belongs to F e,s K . Hence, solutions S 42 and S 43 satisfy equation µx + σz = τ . We then obtain that

µx S 42 + σz S 42 = µx S 43 + σz S 43 = µx S 42 + σz S 42 + µ k ′ e -µ k e + e"∈E 43 k ′ \{e} µ k ′ e" - e"∈E 42 k ′ µ k ′ e" + e"∈E 43 k µ k e" - e"∈E 42 k \{e} µ k e" .
Since

µ k e" = 0 for all k ∈ K and e" ∈ E \ (E k 0 ∪ E k 1 ) with k ∈ K, it follows that µ k ′ e = µ k e .
In a similar way, we can show that

µ k e = µ k ′ e , for all pairs (k, k ′ ) ∈ K.
Furthermore, let prove that all σ k s ′ and µ k e are equivalent for all k ∈ K and s ′ ∈ {s, ..., s + w k -1}. Now let us consider for each demand k ′ with k ′ ∈ K, a solution S 44 = (E 44 , S 44 ) obtained from S 42 as below a) the paths assigned to the demands K \ {k ′ } in S 42 remain the same in S 44 (i.e.,

E 44 k" = E 42 k" for each k" ∈ K \ {k ′ }),
b) without modifying the last slots assigned to the demands K \ {k} in S 42 , i.e., S 42 k" = S 44 k" for each demand k" ∈ K \ {k}, c) modifying the set of last slots assigned to demand

k ′ in S 42 from S 42 k ′ to S 44 k ′ such that S 44 k ′ ∩ {s, ..., s + w k ′ -1} = ∅.
Hence, there are | K| -1 demands from K that share slot s over edge e (i.e., all the demands in K \ {k ′ }), and two demands {k, k ′ } from K that use edge e in solution S 44 . Solution S 44 is then feasible for the problem. The corresponding incidence vector (x S 44 , z S 44 ) belongs to F e,s K . Hence, solutions S 42 and S 44 satisfy equation µx + σz = τ . We then obtain that

µx S 42 + σz S 42 = µx S 44 + σz S 44 = µx S 42 + σz S 42 + µ k ′ e -σ k ′ s ′ + e"∈E 44 k ′ \{e} µ k ′ e" - e"∈E 42 k ′ µ k ′ e" .
Since µ k e" = 0 for all k ∈ K and e"

∈ E \ (E k 0 ∪ E k 1 ) with e ̸ = e" if k ∈ K, it follows that µ k ′ e = σ k ′ s ′ .
In a similar way, we can show that

µ k e = σ k s ′ , for all k ∈ K and s ′ ∈ {s, ..., s + w k -1}.
Based on this, and given that all µ k e are equivalent for all k ∈ K, and that σ k s ′ are equivalent for all k ∈ K and s ′ ∈ {s, ..., s + w k ′ -1}, we obtain that

µ k e = σ k ′ s ′ , for all k, k ′ ∈ K and s ′ ∈ {s, ..., s + w k ′ -1}.
Consequently,

µ k e = σ k ′ s ′ = ρ, for all k, k ′ ∈ K and s ′ ∈ {s, ..., s + w k ′ -1}.
We know from (2.17) and (2.18) that

         µ k ′ e ′ = γ k ′ ,e ′ 1 for all k ′ ∈ K and e ′ ∈ E k ′ 0 , µ k ′ e ′ = γ k ′ ,e ′ 2 for all k ′ ∈ K and e ′ ∈ E k ′ 1 , σ k ′ s ′ = γ k ′ ,s ′ 3 for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.
Overall, we obtain that

µ k ′ e ′ =                  γ k ′ ,e ′ 1 if e ′ ∈ E k ′ 0 , γ k ′ ,e ′ 2 if e ′ ∈ E k ′ 1 , ρ if k ′ ∈ K and e ′ = e, 0 otherwise,
for each k ′ ∈ K and e ′ ∈ E, and

σ k s ′ =          γ k,s ′ 3 if s ′ ∈ {1, ..., w k -1} ρ if k ∈ K and s ′ ∈ {s, ..., s + w k -1}, 0 otherwise.
for each k ∈ K and s ′ ∈ S.

As a consequence, we have (µ, σ) = ρ(α, β) + γQ.

Theorem 2.4.2. Consider an edge e ∈ E, and a slot s ∈ S. Let K be a subset of

demands in K with | K| ≥ 3, and k∈ K w k ≤ s -k ′ ∈Ke\ K w k ′ . Then, inequality
(2.25) is facet defining for P(G, K, S) if and only if there does not exist an interval

of contiguous slots I = [s i , s j ] such that a) |{s i + w k -1, ..., s j }| ≥ w k for each demand k ∈ K, b) and s ∈ {s i + max k ′ ∈ K w k -1, ..., s j -max k∈ K w k + 1}, c) and w k + w k ′ ≥ |I| + 1 for each k, k ′ ∈ K, d) and w k + w k ′ ≥ |I| + 1 for each k ∈ K and k ′ ∈ K e \ K, e) and 2w k ≥ |I| + 1 for each k ∈ K, f ) and 2w k ′ ≥ |I| + 1 for each k ′ ∈ K e \ K.
Proof. Neccessity.

Suppose that there exists an interval of contiguous slots I = [s i , s j ] such that all the conditions a) -f ) are verified. Then inequality (2.25) is dominated by another valid inequality which will be presented later. As a result, inequality (2.25) is not facet defining for P(G, K, S).

Sufficiency.

Let denote F ′e,s K the face induced by inequality (2.25), that is

F ′e,s K = {(x, z) ∈ P(G, K, S) : k∈ K x k e + k∈ K min(s+w k -1,s) s ′ =s z k s ′ + Ke\ K min(s+w k ′ -1,s) s ′ =s z k ′ s ′ = | K| + 1}. We denote inequality k∈ K x k e + k∈ K min(s+w k -1,s) s ′ =s z k s ′ + Ke\ K min(s+w k ′ -1,s) s ′ =s z k ′ s ′ ≤ | K|+1 by αx+βz ≤ λ.
Let µx+σz ≤ τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) : µx + σz = τ }. Suppose that F ′e,s K ⊆ F . We show that there exist ρ ∈ R and γ

= (γ 1 , γ 2 , γ 3 ) (such that γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ - 1}) such that (µ, σ) = ρ(α, β) + γQ.
We re-do the same technique of proof already detailed to prove that inequality (2.23) is facet defining for P(G, K, S) such that the solutions S 38 -S 44 still feasible for F ′e,s K given that their incidence vector are composed by k∈

K x k e + k∈ K min(s+w k -1,s) s ′ =s z k s ′ + Ke\ K min(s+w k ′ -1,s) s ′ =s z k ′ s ′ ≤ | K|+1. We conclude at the end that for each k ′ ∈ K and e ′ ∈ E µ k ′ e ′ =                  γ k ′ ,e ′ 1 if e ′ ∈ E k ′ 0 , γ k ′ ,e ′ 2 if e ′ ∈ E k ′ 1 , ρ if k ′ ∈ K and e ′ = e, 0 otherwise,
and for each k ∈ K and s ′ ∈ S

σ k s ′ =          γ k,s ′ 3 if s ′ ∈ {1, ..., w k -1}, ρ if k ∈ K ∪ K e and s ′ ∈ {s, ..., s + w k -1}, 0 otherwise.
As a result, we have (µ, σ) = ρ(α, β) + γQ.

Edge-Interval-Capacity-Cover Inequalities

Let now introduce some valid inequalities which can be seen as cover inequalities using some notions of cover related to the problem.

Definition 2.4.2. For an interval of contiguous slots

I = [s i , s j ], a subset of de- mands K ′ ⊆ K is said a cover for the interval I = [s i , s j ] if and only if k∈ K w k > |I| and w k < |I| for each k ∈ K. Moreover, it is said to be a minimal cover if k ′ ∈ K\{k} w k ′ ≤ |I| for each demand k ∈ K.
Based on these definitions, we introduce the following inequalities. over edge e. Then, the inequality

k∈K ′ s j s=s i +w k -1 z k s ≤ |K ′ | -1, (2.26)
is valid for P(G, K, S).

Proof. The interval I = [s i , s j ] can cover at most |K ′ | -1 demands given that K ′ is a minimal cover for interval I = [s i , s j ] over edge e. Otherwise, the nonoverlapping constraint is violated given that there exists at least one slot s ∈ I such

that k∈K ′ s+w k -1 s ′ =s z k s > 1.
Inequality (2.26) can be lifted using a sequential lifting procedure [START_REF] Balas | Facets of the Knapsack Polytope From Minimal Covers[END_REF] to be facet defining and generate lifted facets for a sub-polytope of P(G, K, S).

Theorem 2.4.3. Let I = [s i , s j ] be an interval of contiguous slots in [1, s]. Let K ⊆ K e be a minimal cover for interval I = [s i , s j ] over edge e. Let K e \ K = {k 1 , ..., k q } be arbitrarily ordred with q = |K e \ K|. Consider the following sequence of knapsack problems defined as

                     z i = max j∈ K a j + i-1 j=1 β j a j , j∈ K w j a j + i-1 j=1 w k j a j ≤ |I| -w k i , a j ∈ {0, 1}, ∀j ∈ K ∪ {1, ..., i -1}, (2.27)
for all i ∈ {1, ..., q} with

β j = | K| -1 -z j for all j ∈ {1, ..., i -1}. Then, the inequality k∈ K s j s=s i +w k -1 z k s + q j=1 s j s ′ =s i +w k j -1 β j z k j s ′ ≤ | K| -1, (2.28)
is valid for P(G, K, S). Moreover, inequality (5.13) defines facet of

P(G, K, S, K, e, E) = {(x, z) ∈ P(G, K, S) : k ′ ∈K E k ′ 1 ∩E k 1 ̸ =∅ for all k∈ K s j s ′ =s i +w k ′ -1 z k ′ s ′ = 0}.
if there does not exist an interval of contiguous slots

I ′ = [s ′ i , s ′ j ] in [1, s] with I ⊂ I ′ such that K defines a minimal cover for the interval I ′ .
Proof. It is trivial given that inequality (5.13) can never be dominated in P(G, K, S, K, e, E) if there does not exist an interval of contiguous slots [1, s] with I ⊂ I ′ such that K defines a minimal cover for the interval I ′ . Inequality (2.26) can then be generalized over all edges e ∈ E. Moreover, it should be lifted to be facet definig for the polytope P(G, K, S) as follows.

I ′ = [s ′ i , s ′ j ] in
Proposition 2.4.7. Let I = [s i , s j ] be an interval of contiguous slots in [1, s]. Let

K ′ ⊂ K be a minimal cover for interval I = [s i , s j ] such that E k 1 ∩ E k ′ 1 ̸ = ∅ for each pair (k, k ′ ) ∈ K ′ . Then, the inequality k∈K ′ s j s=s i +w k -1 z k s ≤ |K ′ | -1, (2.29)
is valid for P(G, K, S).

Proof. The interval I can cover at most |K ′ | -1 demands given that K ′ is a minimal cover for interval I.

Inequality (2.29) can then be lifted using a sequential lifting procedure [START_REF] Balas | Facets of the Knapsack Polytope From Minimal Covers[END_REF] to generate several facets for the polytope P(G, K, S).

Theorem 2.4.4. Let I = [s i , s j ] be an interval of contiguous slots. Let K ⊆ K be a minimal cover for interval

I = [s i , s j ] such that E k 1 ∩ E k ′ 1 ̸ = ∅ for each pair (k, k ′ ) ∈ K. Let K ′ ⊆ K \ K = {k 1 , ..., k q } such that E k 1 ∩ E k ′ 1 ̸ = ∅ for each pair (k, k ′ ) ∈ K ∪ K ′ .
Consider the following sequence of knapsack problems defined as

                     z i = max j∈ K a j + i-1 j=1 β j a j , j∈ K w j a j + i-1 j=1 w k j a j ≤ |I| -w k i , a j ∈ {0, 1}, ∀j ∈ K ∪ {1, ..., i -1}, (2.30)
for all i ∈ {1, ..., q} with β j = | K| -1 -z j for all j ∈ {1, ..., i -1}. Then, the inequality k∈ K

s j s=s i +w k -1 z k s + q j=1 s j s ′ =s i +w k j -1 β j z k j s ′ ≤ | K| -1, (2.31)
is valid for P(G, K, S). Moreover, inequality (2.31) defines facet of P(G, K, S) if there does not exist an interval of contiguous slots

I ′ = [s ′ i , s ′ j ] in [1, s] with I ⊂ I ′ such that K defines a minimal cover for the interval I ′ .
Proof. It is trivial given that inequality (2.31) can never be dominated in P(G, K, S) if there does not exist an interval of contiguous slots [1, s] with I ⊂ I ′ such that K defines a minimal cover for the interval I ′ .

I ′ = [s ′ i , s ′ j ] in
Inspiring from inequalities (2.26) and (2.29), we define another valid inequality induced by any subset of demands K defining a minimal cover for any interval I as follows.

Definition 2.4.3. Consider an inequality αx T ≤ β which is not valid for a polyhedron P(G, K, S). It is said to be optimality cut for P(G, K, S) if it is valid for a semi-polytope of P(G, K, S) which covers at least one optimal solution for the problem.

Let Q(G, K, S) = {(x, z) ∈ P(G, K, S) : s s=w k z k s = 1, ∀k ∈ K} be a semi-polytope of P(G, K, S)
. Note that each valid inequality of Q(G, K, S) which is not valid for P(G, K, S), it defines an optimality cut for P(G, K, S).

Proposition 2.4.8. Consider an edge e ∈ E. Let I = [s i , s j ] be an interval of contiguous slots in [1, s]. Let K be a minimal cover for the interval I such that a)

k∈ K w k ≤ s - k ′ ∈Ke\ K w k ′ , b) e / ∈ E k 0 for each demand k ∈ K, c) (k, k ′ ) / ∈ K e c for each pair of demands (k, k ′ ) in K.
Then, the inequality

k∈ K x k e + k∈ K s j s=s i +w k -1 z k s ≤ 2| K| -1, (2.32
)

is valid for Q(G, K, S).
Proof. The interval I = [s i , s j ] can cover at most | K| -1 demands given that K is a minimal cover for interval I = [s i , s j ] over edge e. It follows that if demands K pass together through edge e (i.e., k∈ K x k e = | K|), there are at most | K| -1 demands that can share the interval I over edge e. We ensure that inequalities (2.32) are verified by any feasible solution having an incidence vector in Q(G, K, S).

Otherwise, the non-overlapping constraint is violated such if there exists a solution S that violates inequality (2.32), this will certainly prove that there exists a slot s ∈ I over edge e such that

k∈ K s+w k -1 s ′ =s z k s ′ > 1 given that k∈ K x k e ≤ | K| and k∈ K s j s=s i +w k -1 z k
s ≤ |K| for any feasible solution S with incidence vector in Q(G, K, S). Inequality (2.32) can also be lifted using a sequential lifting procedure [START_REF] Balas | Facets of the Knapsack Polytope From Minimal Covers[END_REF] to be facet defining and generate lifted facets for the polytope Q(G, K, S).

Theorem 2.4.5. Let I = [s i , s j ] be an interval of contiguous slots in [1, s]. Let K be a minimal cover for the interval I such that K does not define a minimal cover for an edge e, where e / ∈ E k 0 for each demand k ∈ K. Let K e \ K = {k 1 , ..., k q } be arbitrarily ordred with q = |K e \ K|. Consider the following sequence of knapsack problems defined as

                     z i = max j∈ K a j + i-1 j=1 β j a j , j∈ K w j a j + i-1 j=1 w k j a j ≤ |I| -w k i , a j ∈ {0, 1}, ∀j ∈ K ∪ {1, ..., i -1}, (2.33)
for all i ∈ {1, ..., q} with β j = | K| -1 -z j for all j ∈ {1, ..., i -1}. Then, the inequality k∈

K x k e + k∈ K s j s=s i +w k -1 z k s + q j=1 s j s ′ =s i +w k j -1 β j z k j s ′ ≤ 2| K| -1, (2.34
)

is valid for Q(G, K, S). Moreover, inequality (2.34) defines facet of Q(G, K, S) if
there does not exist an interval of contiguous slots

I ′ = [s ′ i , s ′ j ] in [1, s] with I ⊂ I ′ such that K defines a minimal cover for the interval I ′ .
Proof. It is trivial given that inequality (2.34) can never be dominated in Q(G, K, S) if there does not exist an interval of contiguous slots [1, s] with I ⊂ I ′ such that K defines a minimal cover for the interval I ′ .

I ′ = [s ′ i , s ′ j ] in

Edge-Interval-Clique Inequalities

Using inequalities (2.32), and based on the set of minimal cover K with cardinality | K| = 2, we introduce the following inequality.

Proposition 2.4.9. Consider an edge e ∈ E. Let I = [s i , s j ] be an interval of contiguous slots. Let {k, k ′ } be a minimal cover for the interval I over edge e such that e / ∈ E k 0 ∪ E k ′ 0 . Then, the inequality

x k e + x k ′ e + s j s=s i +w k -1 z k s + s j s=s i +w k ′ -1 z k ′ s ≤ 3, (2.35
)

is valid for Q(G, K, S). Moreover, it is valid for P(G, K, S) if 2w k > |I| and 2w k ′ > |I|.
Proof. Inequality (2.35) is a particular case of inequality (2.36) for a minimal cover K = {k, k ′ }.

Using this, we introduce the following conflict graph.

H e I if w k + w k ′ > |I| and (k, k ′ ) / ∈ K e c
. This is equivalent to say that two linked nodes v k and v k ′ means that the two demands k, k ′ define a minimal cover for the interval I over edge e.

For an edge e ∈ E, the conflict graph H e I is a threshold graph with threshold value equals to t = |I| such that for each node v k with e / ∈ E k 0 ∪ E k 1 , we associate a positive weight wv k = w k such that all two nodes v k and v k ′ are linked by an edge if and only if wv k + wv k ′ > t which is equivalent to the conflict graph H e I .

Proposition 2.4.10. Consider an edge e ∈ E. Let I = [s i , s j ] be an interval of contiguous slots. Let C be a clique in the conflict graph H e I with |C| ≥ 3, and

v k ∈C w k ≤ s -k ′ ∈Ke\C w k ′ .
Then, the inequality

v k ∈C x k e + s j s=s i +w k -1 z k s ≤ |C| + 1, (2.36
)

is valid for Q(G, K, S). Moreover, It is valid for P(G, K, S) if 2w k > |I| for each v k ∈ C.
Proof. For each edge e ∈ E and interval of contiguous slots I ⊆ S, inequality (2.36) ensures that if the set of demands in clique C pass through edge e, they cannot share the interval I = [s i , s j ] over edge e. This means that there are at most one demand from the demands in C that can be totally covered by the interval I over edge e (i.e., all the slots assigned to the demand are in I). Inequality (2.36) can be shown as Chvàtal-Gomory cuts using Chvàtal-Gomory and recurrence procedures.

For all two linked node v k and v k ′ in H e I , we know from inequality (2.35)

x k e + x k ′ e + s j s=s i +w k -1 z k s + s j s ′ =s i +w k ′ -1 z k ′ s ′ ≤ 3.
By adding the previous inequalities for all two linked node v k and v k ′ in the clique set C, and by recurrence procedure we obtain that for all

K ′ ⊆ C with |K ′ | = |C|-1 v k ∈C ′ x k e + v k ∈C ′ s j s=s i +w k -1 z k s ≤ |K ′ | + 1.
By adding the previous inequalities for all

K ′ ⊆ C with |K ′ | = |C| -1, we get K ′ ⊆C |K ′ |=|C|-1 v k ∈C ′ x k e + K ′ ⊆C |K ′ |=|C|-1 v k ∈C ′ s j s=s i +w k -1 z k s ≤ K ′ ⊆C |K ′ |=|C|-1 (|K ′ | + 1).
Note that for each demand k with v k ∈ C, the variable x k e and the sum

s j s=s i +w k -1 z k s appear ( |C| |C|-1 -1) times in the previous sum. It follows that v k ∈C ( |C| |C| -1 -1)x k e + v k ∈C s j s=s i +w k -1 .( |C| |C| -1 -1)z k s ≤ |C| |C| -1 |C|. Given that ( |C| |C|-1 -1) = |C| -1, we obtain that v k ∈C (|C| -1)x k e + v k ∈C s j s=s i +w k -1 (|C| -1)z k s ≤ |C| 2 .
By dividing the two sides of the previous sum by |C| -1, we have

v k ∈C x k e + v k ∈C s j s=s i +w k -1 z k s ≤ |C| 2 |C| -1 ⇒ v k ∈C x k e + v k ∈C s j s=s i +w k -1 z k s ≤ |C| |C| |C| -1 ⇒ v k ∈C x k e + v k ∈C s j s=s i +w k -1 z k s ≤ |C| |C| -1 + 1 |C| -1 .
By doing the following simplification

v k ∈C x k e + v k ∈C s j s=s i +w k -1 z k s ≤ |C| |C| -1 |C| -1 + |C| |C| -1 ⇒ v k ∈C x k e + v k ∈C s j s=s i +w k -1 z k s ≤ |C| + |C| |C| -1 .
As a result,

v k ∈C x k e + v k ∈C s j s=s i +w k -1 z k s ≤ |C|+ |C| |C| -1 ⇒ v k ∈C x k e + v k ∈C s j s=s i +w k -1 z k s ≤ |C|+1 given that |C| |C| -1 = 1.
We conclude at the end that inequality (2.36) is valid for

Q(G, K, S). Moreover, it is valid for P(G, K, S) if 2w k > |I| for each v k ∈ C.
Moreover, inequality (2.36) can be strengthened as follows.

Proposition 2.4.11. Consider an edge e ∈ E. Let I = [s i , s j ] be an interval of contiguous slots. Let C be a clique in the conflict graph H e I with |C| ≥ 3, and

v k ∈C w k ≤ s -k ′ ∈Ke\C w k ′ . Let C e ⊆ K e \ C be a clique in the conflict graph H e I such that w k + w k ′ ≥ |I| + 1 for each v k ∈ C and v k ′ ∈ C e .
Then, the inequality

v k ∈C x k e + v k ∈C s j s=s i +w k -1 z k s + v k ′ ∈Ce s j s ′ =s i +w k ′ -1 z k ′ s ′ ≤ |C| + 1, (2.37) is valid for Q(G, K, S). Moreover, it is valid for P(G, K, S) if 2w k > |I| for each v k ∈ C ∪ C e .
Proof. For each edge e ∈ E and interval of contiguous slots I ⊆ S, inequality (2.37) ensures that if the set of demands in clique C pass through edge e, they cannot share the interval I = [s i , s j ] over edge e with a subset of demands in C e . On the other hand, inequality (2.37) can be seen as a particular case of inequality (2.36) induced by a clique

C ′ = C ∪ C e given that x k e = 1 for all v k ∈ C e .
Theorem 2.4.6. Consider an edge e ∈ E. Let I = [s i , s j ] be an interval of contiguous slots. Let C be a clique in the conflict graph

H e I with |C| ≥ 3, k∈C w k ≤ s -k ′ ∈Ke\C w k ′ , and |{s i + w k -1, ..., s j }| ≥ w k for each demand k with v k ∈ C.
Then, inequality (2.36) is facet defining for P(G, K, S) if and only if a) there does not exist a demand

k ′ ∈ K e \ C with w k + w k ′ > |I| and w k ′ ≤ |I|,
b) and there does not exist an interval I ′ of contiguous slots with I ⊂ I ′ such that C defines also a clique in the associated conflict graph H e I ′ .

Proof. Neccessity.

It is trivial given that

• if there does not exist a demand k ′ ∈ K e \ C with w k + w k ′ > |I| and w k ′ ≤ |I|,
and • if there exists an interval I ′ of contiguous slots with I ⊂ I ′ such that C defines also a clique in the associated conflict graph H e I ′ . This implies that inequality (2.36) induced by clique C for the interval I is dominated by inequality (2.36) induced by the same clique C for the interval I ′ given that {s i + w k -1, ..., s j } ⊂ I ′ for each k ∈ C. As a result, inequality (2.36) is not facet defining for P(G, K, S).

|{s i + w k -1, ..., s j }| ≥ w k for each demand k with v k ∈ C.

Sufficiency.

Let F H e I C denote the face induced by inequality (2.36), that is

F H e I C = {(x, z) ∈ P(G, K, S) : v k ∈C x k e + s j s=s i +w k -1 z k s = |C| + 1}. Let denote inequality v k ∈C x k e + s j s=s i +w k -1 z k s ≤ |C| + 1 by αx + βz ≤ λ. Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) : µx+σz = τ }. Suppose that F H e I C ⊆ F . In order to prove that inequality v k ∈C x k e + s j s=s i +w k -1 z k s ≤ |C|+1 is facet defining for P(G, K, S), we need to show that there exists ρ ∈ R and γ = (γ 1 , γ 2 , γ 3 ) (such that γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ - 1}) such that (µ, σ) = ρ(α, β) + γQ.
We first show that µ k e ′ = 0 for each edge 53 ) be the solution given by a) for each demand

e ′ ∈ E \ (E k 0 ∪ E k 1 ) for each demand k ∈ K with e ̸ = e ′ if k ∈ C. Consider a demand k ∈ K and an edge e ′ ∈ E \ (E k 0 ∪ E k 1 ) with e ̸ = e ′ if k ∈ C. Let S 53 = (E 53 , S
k i ∈ K \ C with i ∈ {1, ..., |K|}, we let E 53 k i be the set of edges involved in a shortest path between o k i and d k i , b) for each demand k ∈ C, we let E 53
k be the set of edges involved in a shortest path between o k and d k which uses edge e, c) for one demand k ′ from C, we select the slot

s k ′ = s i + w k ′ -1 as last slot, d) for each demand k i ∈ C \ {k ′ } with i ∈ {1, ...,
I 53 i = [ kj ∈D 53 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] \ {s i , ..., s j } if E 53 ki ∩ E 53 k ′ ̸ = ∅,
where

D 53 i = {k j ∈ {k 1 , ..., k i-1 } ∩ C : E 53 ki ∩ E 53 kj ̸ = ∅}.
e) for each demand k i ∈ K \ C with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 53 i given by

I 53 i = [ kj ∈R 53 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s k -w k } ∪ {s k + w ki , ..., s}] if E 53 ki ∩ (E 53 k ∪ {e ′ }) ̸ = ∅ or I 53 i = kj ∈R 53 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,
where

R 53 i = {k j ∈ {k 1 , ..., k i-1 } ∪ C such that E 53 k i ∩ E 53 k j ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ R 53 i ,
• and

{s k i -w k i + 1, ..., s k i } ∩ {s k -w k j + 1, ..., s k } = ∅ if E 53 k i ∩ (E 53 k ∪ {e ′ }) ̸ = ∅ (
µx S 53 + σz S 53 = µx S 54 + σz S 54 = µx S 53 + µ k e ′ + σz S 53 .
As a result, µ k e ′ = 0. In a similar way, we can show that

µ k e ′ = 0, for all k ∈ K and e ′ ∈ E \ (E k 0 ∪ E k 1 ) with e ̸ = e ′ if k ∈ C. Let show that σ k s = 0 for all k ∈ K and s ∈ {w k , ..., s} with s / ∈ {s i + w k -1, ..., s j } if v k ∈ C. Consider a demand k in K and a slot s ′ in {w k , ..., s} with s ′ / ∈ {s i + w k -1, ..., s j } if v k / ∈ C. Let S ′53 = (E ′53 , S ′53
) be the solution given by a) for each demand k i ∈ K \ C with i ∈ {1, ..., |K|}, we let E ′53 k i be the set of edges involved in a shortest path between o k i and

d k i , b) for each demand k ∈ C, we let E ′53
k be the set of edges involved in a shortest path between o k and d k , c) for one demand k ′ from C, we select the slot

s k ′ = s i + w k ′ -1 as last slot, d) for each demand k i ∈ C \ {k ′ } with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′53 i
given by

I ′53 i = [ kj ∈D ′53 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] \ {s i , ..., s j } if E ′53 ki ∩ E ′53 k ′ ̸ = ∅,
where

D ′53 i = {k j ∈ {k 1 , ..., k i-1 } ∩ C : E ′53 ki ∩ E ′53 kj ̸ = ∅}.
e) for each demand k i ∈ K \ C with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′53 i given by

I ′53 i = [ kj ∈R ′53 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}] if E ′53 ki ∩ E ′53 k ̸ = ∅ or I ′53 i = kj ∈R ′53 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,
where

R ′53 i = {k j ∈ {k 1 , ..., k i-1 } ∪ C : E ′53 k i ∩ E ′53 k j ̸ = ∅}. We let S ′53 k i = {s k i }
µx S ′53 + σz S ′53 = µx S 55 + σz S 55 = µx S ′53 + σz S ′53 + σ k s ′ .
Hence, σ k s ′ = 0. In a similar way, we can show that

σ k s = 0, for all k ∈ K and s ∈ {w k , ..., s} with s / ∈ {s i + w k -1, ..., s j } if v k ∈ C. Let prove that σ k s for all v k ∈ C and s ∈ {s i + w k -1, ..., s j } are equivalent. Consider a demand k ′ ∈ K and a slot s ′ ∈ {s i + w k ′ -1, ..., s j } with v k ′ ∈ C. Let S53 = ( Ẽ53 , S53 ) be the solution given by a) for each demand k i ∈ K \ C with i ∈ {1, ...

, |K|}, we let Ẽ53

k i be the set of edges involved in a shortest path between o k i and d k i , b) for each demand k ∈ C, we let Ẽ53 k be the set of edges involved in a shortest path between o k and d k which uses edge e, c) for one demand k" from C, we select the slot s k" = s i + w k" + 1 as last slot, d) for each demand k i ∈ C \ {k"} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots Ĩ53 i given by

Ĩ53 i = [ kj ∈ D53 i {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]∩[{w ki , ..., s ′ -w k ′ }∪{s ′ +w ki , ..., s}]\{s i , ..., s j } if Ẽ53 ki ∩ Ẽ53 k ′ ̸ = ∅ or Ĩ53 i = [ kj ∈ D53 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] \ {s i , ..., s j } if not, where D53 i = {k j ∈ {k 1 , ..., k i-1 } ∩ C : Ẽ53 k i ∩ Ẽ53 k j ̸ = ∅}, e) for each demand k i ∈ K \ C with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots Ĩ53 i given by Ĩ53 i = [ kj ∈ R53 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k ′ } ∪ {s ′ + w ki , ..., s}] if Ẽ53 ki ∩ Ẽ53 k ′ ̸ = ∅ or Ĩ53 i = kj ∈ R53 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not, where R53 i = {k j ∈ {k 1 , ..., k i-1 } ∪ C : Ẽ53 k i ∩ Ẽ53 k j ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ R53 i ,
• and

{s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ if Ẽ53 k i ∩ Ẽ53 k ′ ̸ = ∅ (
we take into account the possibility of adding slot s ′ as a last slot in the selected last slots S53

k ′ to route demand k ′ in solution S53 ).

We let S53 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}.

S53 is clearly feasible for the problem given that it satisfies all the constraints of cut formulation (2.2)-(2.10). Hence, the corresponding incidence vector (x

S53 , z S53 ) belongs to F H e I C
. Let S 56 = (E 56 , S 56 ) be a solution obtained from S53 by adding slot s ′ as last slot to demand k ′ in S53 , and modifying the last slots assigned to demand k by adding a new last slot s in S53 and removing the last slot s ∈ S53 k from S53 with s ∈ {s i + w k + 1, ..., s j } and s /

∈ {s i + w k + 1, ..., s j } for demand k ∈ C such that {s -w k + 1, ..., s} ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ for each k ′ ∈ K and s ′ ∈ S 56 k ′ with E 56 k ∩ E 56 k ′ ̸ = ∅. Solution S 56
is feasible for the problem. The corresponding incidence vector (x S 56 , z S 56 ) belongs to F H e I C . Hence, solutions S 53 and S 56 satisfy equation µx + σz = τ . We have so

µx S53 + σz S53 = µx S 56 + σz S 56 = µx S53 + σz S53 + σ k ′ s ′ -σ k s + σ k s . Since σ k s = 0 for s / ∈ {s i + w k -1, ..., s j } with v k ∈ C, it follows that σ k ′ s ′ = σ k s .
In a similar way, we can show that

σ k s = σ k ′ s ′ , for all pairs (v k , v k ′ ) ∈ C
with s ∈ {s i + w k -1, ..., s j } and s ′ ∈ {s i + w k ′ -1, ..., s j }. We re-do the same procedure for each two slots s, s

′ ∈ {s i + w k -1, ..., s j } for each demand k ∈ K with v k ∈ C such that σ k s = σ k s ′ , for all v k ∈ C and s, s ′ ∈ {s i + w k -1, ..., s j }. Let prove now that µ k e for all k ∈ K with v k ∈ C are equivalent. Consider a demand k ′ ∈ K with v k ′ in C such that e / ∈ E 57 k ′ .
For this, we derive a solution S" 58 = (E" 58 , S" 58 ) from S 53 by we derive a solution S 58 = (E 58 , S 58 ) from S 53 by a) the paths assigned to the demands K \ {k, k ′ } in S 53 remain the same in S 58 (i.e.,

E 58 k" = E 53 k" for each k" ∈ K \ {k, k ′ }),
b) without modifying the last slots assigned to the demands K in S 53 , i.e., S 53 k = S 58

k for each demand k ∈ K, c) modifying the path assigned to demand k ′ in S 53 from E 53 k ′ to a path E 58 k ′ passed through edge e (i.e., e ∈ E 58 k ′ ) with v k ′ ∈ C such that {s -w k + 1, ..., s} ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ for each k ∈ K and s ′ ∈ S 53 k ′ and s ∈ S 53 k with E 53 k ∩ E 58 k ′ ̸ = ∅, d)
w k + 1, ..., s} ∩ {s ′ -w k" + 1, ..., s ′ } = ∅ for each k" ∈ K \ {k, k ′ } and s ∈ S 53 k and s ′ ∈ S 53 k" with E 53 k" ∩ E 58 k ̸ = ∅, and {s -w k + 1, ..., s} ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ for each s ∈ S 53 k and s ′ ∈ S 53 k ′ with E 58 k" ∩ E 58 k ̸ = ∅.
Solution S" 58 is feasible for the problem. The corresponding incidence vector (x S" 58 , z S" 58 ) belongs to F

H e I C
. Hence, solutions S 53 and S" 58 satisfy equation µx + σz = τ . We then obtain that

µx S 53 + σz S 53 = µx S 58 + σz S 58 = µx S 53 + σz S 53 + µ k ′ e -µ k e + e"∈E" 58 k ′ \{e} µ k ′ e" - e"∈E 53 k ′ µ k ′ e" + e"∈E" 58 k µ k e" - e"∈E 53 k \{e} µ k e" .
Since

µ k e" = 0 for all k ∈ K and e" ∈ E \ (E k 0 ∪ E k 1 ) with v k / ∈ C, it follows that µ k ′ e = µ k e .
In a similar way, we can show that

µ k e = µ k ′ e , for all pairs (v k , v k ′ ) ∈ C.
Furthermore, let prove that all σ k s and µ k e are equivalent for all k ∈ C and s ∈ {s i + w k -1, ..., s j }.

Now let us consider a demand k

′ ∈ K with v k ′ ∈ C, a solution S 59 = (E 59 , S 59 )
obtained from S 53 as below a) the paths assigned to the demands K \ {k ′ } in S 53 remain the same in S 59 (i.e.,

E 59 k" = E 53 k" for each k" ∈ K \ {k ′ }), b) without modifying the last slots assigned to the demands K \ {k} in S 53 , i.e., S 53 k" = S 59 k" for each demand k" ∈ K \ {k}, c) modifying the set of last slots assigned to demand k ′ in S 53 from S 53 k ′ to S 59 k ′ such that S 59 k ′ ∩ {s i + w k ′ -1, ..., s j } = ∅.
Hence, there are |C| -1 demands from C that are covered by the interval I (i.e., all the demands in C \ {k ′ }), and two demands {k, k ′ } from C that use edge e in solution S 59 . Solution S 59 is then feasible for the problem. The corresponding incidence vector (x S 59 , z S 59 ) belongs to F H e I C . Hence, solutions S57 and S 59 satisfy equation µx + σz = τ . We then obtain that

µx S57 + σz S57 = µx S 59 + σz S 59 = µx S57 + σz S57 + µ k ′ e -σ k ′ s + e"∈E 59 k ′ \{e} µ k ′ e" - e"∈ Ẽ57 k ′ µ k ′ e" .
Since µ k e" = 0 for all k ∈ K and e"

∈ E \ (E k 0 ∪ E k 1 ) with e ̸ = e" if v k ∈ C, it follows that µ k ′ e = σ k ′ s .
In a similar way, we can show that

µ k e = σ k s , for all v k ∈ C and s ∈ {s i + w k -1, ..., s j }.
Based on this, and given that all µ k e are equivalent for all v k ∈ C, and that σ k s are equivalent for all v k ∈ C and s ∈ {s i + w k ′ -1, ..., s j }, we obtain that

µ k e = σ k ′ s , for all k, k ′ ∈ C and s ∈ {s i + w k ′ -1, ..., s j }.
Consequently,

µ k e = σ k ′ s = ρ, for all k, k ′ ∈ C and s ∈ {s i + w k ′ -1, ..., s j }.
By (2.17) and (2.18), we know that

         µ k ′ e ′ = γ k ′ ,e ′ 1 for all k ′ ∈ K and e ′ ∈ E k ′ 0 , µ k ′ e ′ = γ k ′ ,e ′ 2 for all k ′ ∈ K and e ′ ∈ E k ′ 1 , σ k ′ s ′ = γ k ′ ,s ′ 3 for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.
We conclude that for each

k ′ ∈ K and e ′ ∈ E µ k ′ e ′ =                  γ k ′ ,e ′ 1 if e ′ ∈ E k ′ 0 , γ k ′ ,e ′ 2 if e ′ ∈ E k ′ 1 , ρ if k ′ ∈ C and e ′ = e, 0 otherwise,
and for each k ∈ K and s ∈ S

σ k s =          γ k,s 3 if s ∈ {1, ..., w k -1} ρ if v k ∈ C and s ∈ {s i + w k -1, ..., s j }, 0 otherwise.
As a consequence, (µ, σ) = ρ(α, β) + γQ. Proof. Neccessity.

• if there exists a demand k" ∈ K e \ C e with w k + w k" ≥ |I| + 1 for each v k ∈ C, and

w k ′ + w k" ≥ |I| + 1 for each v k ′ ∈ C e . Then, inequality (2.37) is dominated by its lifted with C ′ e = C e ∪ {k"}. Moreover, if |{s i + w k -1, ..., s j }| < w k for each demand k with v k ∈ C ∪ C e ,
then inequality (2.37) is then dominated by inequality (2.25) for a set of demands K = {k ∈ K such that v k ∈ C} and slot s = s i + min k∈C∪Ce w k + 1 over edge e. As a result, inequality (2.37) is not facet defining for P(G, K, S).

• if there exists an interval I ′ of contiguous slots with I ⊂ I ′ such that C ∪ C e defines also a clique in the associated conflict graph H e I ′ . This implies that inequality (2.37) induced by clique C ∪ C e for the interval I is dominated by inequality (2.37) induced by the same clique C ∪ C e for the interval I ′ given that {s i + w k -1, ..., s j } ⊂ I ′ for each k ∈ C ∪ C e . As a result, inequality (2.37) is not facet defining for P(G, K, S).

Sufficiency.

Let F ′H e I C be the face induced by inequality (2.37), that is

F ′H e I C = {(x, z) ∈ P(G, K, S) : v k ∈C x k e + s j s=s i +w k -1 z k s + v k ′ ∈Ce s j s ′ =s i +w k ′ -1 z k ′ s ′ = |C| + 1}. We denote inequality v k ∈C x k e + s j s=s i +w k -1 z k s + v k ′ ∈Ce s j s ′ =s i +w k ′ -1 z k ′ s ′ ≤ |C|+1 by αx + βz ≤ λ.
Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and

F = {(x, z) ∈ P(G, K, S) : µx + σz = τ }. Suppose that F ′H e I C
⊆ F . We use the same proof of the facial structure done for inequality (2.36) in the proof of theorem 2.4.6 to prove that inequality

v k ∈C x k e + s j s=s i +w k -1 z k s + v k ′ ∈Ce s j s ′ =s i +w k ′ -1 z k ′ s ′ ≤ |C|+1 is facet
= (γ 1 , γ 2 , γ 3 ) (such that γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α, β) + γQ.
For this, we show that a) σ k s = 0 for all demand k ∈ K and slot s ∈ {w k , ..., s} with s /

∈ {s i + w k -1, ..., s j } if v k ∈ C ∪ C e , b)
i + w k ′ - 1, ..., s j }.
At the end, we obtain that for each

k ′ ∈ K and e ′ ∈ E µ k ′ e ′ =                  γ k ′ ,e ′ 1 if e ′ ∈ E k ′ 0 , γ k ′ ,e ′ 2 if e ′ ∈ E k ′ 1 , ρ if k ′ ∈
C and e ′ = e, 0 otherwise, and for each k ∈ K and s ∈ S

σ k s =          γ k,s 3 if s ∈ {1, ..., w k -1} ρ if v k ∈ C ∪ C e and s ∈ {s i + w k -1, ..., s j }, 0 otherwise.
As a result, we have (µ, σ) = ρ(α, β) + γQ.

Interval-Clique Inequalities

We have looked at the definition of inequality (2.36), we detected that there may exist some cases that we can face which are not covered by inequality (2.36). For this, we provide the following inequality and its generalization.

Proposition 2.4.12. Consider an interval of contiguous slots I = [s i , s j ] in S with

s i ≤ s j -1. Let k, k ′ be a pair of demands in K with E k 1 ∩ E k ′ 1 ̸ = ∅, and w k ≤ |I|.
Then, the inequality

s j s=s i +w k -1 z k s + s j s ′ =s i +w k ′ -1 z k ′ s ′ ≤ 1, (2.38) is valid for Q(G, K, S). Moreover, it is valid for P(G, K, S) if 2w k > |I| and 2w k ′ > |I|.
Proof. It is trivial given that the interval I = [s i , s j ] cannot cover the two demands k, k ′ shared an essential edge with total sum of number of slots exceeds |I|. Furthermore, inequality (2.38) is a particular case of inequality (2.36

) for K = {k, k ′ } over each edge e ∈ E k 1 ∩ E k ′ 1 .
However, it will be used for a generalized inequality using the following conflict graph. Definition 2.4.5. Let I = [s i , s j ] be an interval of contiguous slots in [1, s] with

s i ≤ s j -1. Consider the conflict graph H E I defined as follows. For each demand k ∈ K with w k ≤ |I|, consider a node v k in H E I . Two nodes v k and v k ′ are linked by an edge in H E I if w k + w k ′ > |I| and E k 1 ∩ E k ′ 1 ̸ = ∅.
Proposition 2.4.13. Let I = [s i , s j ] be an interval of contiguous slots in [1, s] with

s i ≤ s j -1,
and C be a clique in the conflict graph H E I with |C| ≥ 3. Then, the inequality

v k ∈C s j s=s i +w k -1 z k s ≤ 1, (2.39
)

is valid for Q(G, K, S). Moreover, it is valid for P(G, K, S) if 2w k > |I| for each v k ∈ C.
Proof. It is trivial given the definition of clique set in the conflict graph H E I such that for all two linked node v k and v k ′ in H E I , we know from inequality (2.38)

s j s=s i +w k -1 z k s + s j s ′ =s i +w k ′ -1 z k ′ s ′ ≤ 1.
By adding the previous inequalities for all two linked node v k and v k ′ in the clique set C, and by recurrence procedure we obtain that for all

C ′ ⊆ C with |C ′ | = |C| -1 v k ∈C ′ s j s=s i +w k -1 z k s ≤ 1.
By adding the previous inequalities for all

C ′ ⊆ C with |C ′ | = |C| -1, we get C ′ ⊆C |C ′ |=|C|-1 v k ∈C ′ s j s=s i +w k -1 z k s ≤ C ′ ⊆C |C ′ |=|C|-1 1.

91

Note that for each demand k with v k ∈ C, the sum

s j s=s i +w k -1 z k s appears ( |C| |C|-1 - 1) = |C| -1 times in the previous sum. It follows that v k ∈C s j s=s i +w k -1 (|C| -1)z k s ≤ |C|.
By dividing the two sides of the previous sum by |C| -1, we have so

v k ∈C s j s=s i +w k -1 z k s ≤ |C| |C| -1 ⇒ v k ∈C s j s=s i +w k -1 z k s ≤ 1 given that |C| |C| -1 = 1.
We conclude at the end that inequality (2.39) is valid for Q(G, K, S). Moreover, it

is valid for P(G, K, S) if 2w k > |I| for each v k ∈ C.
I ′ in [1, s] such that I ⊂ I ′ with • w k + w k ′ ≥ |I ′ | for each k, k ′ ∈ C, • w k ≤ |I ′ | for each k ∈ C.
Proof. Neccessity.

We distinguish two cases a) if there exists a clique C ′ that contains all the demands k ∈ C. Then, inequality (2.39) induced by clique C is dominated by another inequality (2.39) induced by clique C ′ . Hence, inequality (2.39) cannot be facet defining for P(G, K, S).

b) if there exists an interval of contiguous slots

I ′ in [1, s] such that I ⊂ I ′ with • w k + w k ′ ≥ |I ′ | for each k, k ′ ∈ C, • w k ≤ |I ′ | for each k ∈ C.
This means that inequality (2.39) induced by clique C for the interval I is dominated by inequality (2.39) induced by clique C for the interval I ′ . Hence, inequality (2.39) cannot be facet defining for P(G, K, S).

Let F

H E I C
be the face induced by inequality (2.39), that is

F H E I C = {(x, z) ∈ P(G, K, S) : v k ∈C s j s=s i +w k -1 z k s = 1}. Denote inequality v k ∈C s j s=s i +w k -1 z k s ≤ 1 by αx + βz ≤ λ.
Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) : µx+σz = τ }.

Suppose that

F H E I C ⊆ F . In order to prove that inequality v k ∈C s j s=s i +w k -1 z k
s ≤ 1 is facet defining for P(G, K, S), we need to show that there exist ρ ∈ R and

γ = (γ 1 , γ 2 , γ 3 ) (such that γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α, β) + γQ.
We first show that µ k e = 0 for each edge e

∈ E \ (E k 0 ∪ E k 1 ) for each demand k ∈ K. Consider a demand k ∈ K and an edge e ∈ E \ (E k 0 ∪ E k 1 )
. Let S 60 = (E 60 , S 60 ) be the solution given by a) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E 60 k i be the set of edges involved in a shortest path between o k i and where

d k i , b) for one demand k ′ from C, we select the slot s k ′ = s i + w k ′ -1, c) for each demand k i ∈ C \ {k ′ } with i ∈ {1, ...,
D 60 i = {k j ∈ {k 1 , ..., k i-1 } ∩ C : E 60 k i ∩ E 60 k j ̸ = ∅}, d) for each demand k i ∈ K \ C with i ∈ {1, ...,
I 60 i = [ kj ∈R 60 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s k -w k } ∪ {s k + w ki , ..., s}] if E 60 ki ∩ (E 60 k ∪ {e}) ̸ = ∅ or I 60 i = kj ∈R 60 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,
where As a result, µ k e = 0. In a similar way, we can show that

R 60 i = {k j ∈ {k 1 , ..., k i-1 } ∪ C such that E 60 k i ∩ E 60 k j ̸ = ∅}. We let S 60 k i = {s k i }
µ k e = 0, for all k ∈ K and e ∈ E \ (E k 0 ∪ E k 1 ).
Let show that σ k s = 0 for all k ∈ K and s ∈ {w k , ..., s} with s /

∈ {s i + w k -1, ..., s j } if v k ∈ C. Consider a demand k ∈ K and a slot s ′ in {w k , ..., s} with s ′ / ∈ {s i + w k -1, ..., s j } if v k ∈ C. Let S ′60 = (E ′60 , S ′60
) be the solution given by a) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E ′60 k i be the set of edges involved in a shortest path between o k i and where

d k i , b) for one demand k ′ from C, we select the slot s k ′ = s i + w k -1, c) for each demand k i ∈ C \ {k ′ } with i ∈ {1, ...,
D ′60 i = {k j ∈ {k 1 , ..., k i-1 } ∪ C : E ′60 k i ∩ E ′60 k j ̸ = ∅}, d) for each demand k i ∈ K \ C with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′60 i
given by

I ′60 i = [ kj ∈R ′60 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}] if E ′60 ki ∩ E ′60 k ̸ = ∅ or I ′60 i = kj ∈R ′60 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,
where

R ′60 i = {k j ∈ {k 1 , ..., k i-1 } ∪ C such that E 60 k i ∩ E 60 k j ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ R ′60 i ,
• and

{s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E ′60 k i ∩ E ′60 k ̸ = ∅ (
we take into account the possibility of adding slot s ′ as a last slot in the selected last slots S ′60 k to route demand k in solution S ′60 ).

We let S ′60 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′60 is feasible for the problem. Hence, the corresponding incidence vector (x S ′60 , z S ′60 ) belongs to F H E I C . Then consider the solution S 62 obtained from S ′60 by adding slot s ′ as last slot to demand k in S ′60 . Solution S 62 is feasible for the problem. The corresponding incidence vector (x S 62 , z S 62 ) belongs to F H E I C . Hence, solutions S ′60 and S 62 satisfy equation µx + σz = τ . We have so

µx S ′60 + σz S ′60 = µx S 62 + σz S 62 = µx S ′60 + σz S ′60 + σ k s ′ .
Hence, σ k s ′ = 0. In a similar way, we can show that

σ k s = 0, for all k ∈ K and s ∈ {w k , ..., s} with s / ∈ {s i + w k -1, ..., s j } if v k ∈ C.
Let prove that σ k s for all v k ∈ C and s ∈ {s i + w k -1, ..., s j } are equivalent. Consider a demand k ′ ∈ K and a slot s ′ ∈ {s i + w k ′ -1, ..., s j } with v k ′ ∈ C, and a solution S60 = ( Ẽ60 , S60 ) given by a) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let Ẽ60 k i be the set of edges involved in a shortest path between o k i and d k i , b) for one demand k from C, we select the slot s k = s i + w k -1, c) for each demand k i ∈ C \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots Ĩ60 i given by

Ĩ60 i = [ kj ∈ D60 i {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]∩[{w ki , ..., s ′ -w k ′ }∪{s ′ +w ki , ..., s}]\{s i , ..., s j } if Ẽ60 ki ∩ Ẽ60 k ′ ̸ = ∅ or Ĩ60 i = [ kj ∈ D60 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] \ {s i , ..., s j } if not, where D60 i = {k j ∈ {k 1 , ..., k i-1 } ∩ C : D60 ki ∩ D60 kj ̸ = ∅}, d) for each demand k i ∈ K \ C with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots Ĩ60 i given by Ĩ60 i = [ kj ∈ R60 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k ′ } ∪ {s ′ + w ki , ..., s}] if Ẽ60 ki ∩ Ẽ60 k ′ ̸ = ∅ or Ĩ60 i = kj ∈ R60 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not, where R60 i = {k j ∈ {k 1 , ..., k i-1 } ∪ C such that D60 k i ∩ D60 k j ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ R60 i ,

95

• and

{s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ if Ẽ60 k i ∩ Ẽ60 k ′ ̸ = ∅ (
we take into account the possibility of adding slot s ′ as a last slot in the selected last slots S60 k ′ to route demand k ′ in solution S60 ).

We let S60 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}.

S60 is feasible for the problem. Hence, the corresponding incidence vector (x S60 , z S60 ) belongs to F

H E I C .
Then consider the solution S 63 obtained from S60 by adding slot s ′ as last slot to demand k ′ , and modifying the last slots assigned to demand k by adding a new last slot s and removing the last slot s ∈ S60 k with s ∈ {s i + w k + 1, ..., s j } and s / 63 is feasible for the problem. The corresponding incidence vector (x S 63 , z S 63 ) belongs to F H E I C . Hence, solutions S60 and S 63 satisfy equation µx + σz = τ . We have so

∈ {s i + w k + 1, ..., s j } for demand k ∈ K with v k ∈ C such that S 63 k = ( S60 k \{s})∪{s} such that {s-w k +1, ..., s}∩{s ′ -w k ′ +1, ..., s ′ } = ∅ for each k ′ ∈ K and s ′ ∈ S 63 k ′ with E 63 k ∩ E 63 k ′ ̸ = ∅. Solution S
µx S60 + σz S60 = µx S 63 + σz S 63 = µx S60 + σz S60 + σ k ′ s ′ -σ k s + σ k s . Since σ k s = 0 for s / ∈ {s i + w k -1, ..., s j } with v k ∈ C, it follows that σ k ′ s ′ = σ k s .
In a similar way, we can show that

σ k s = σ k ′ s ′ , for all pairs (v k , v k ′ ) ∈ C,
with s ∈ {s i + w k -1, ..., s j } and s ′ ∈ {s i + w k ′ -1, ..., s j }. We re-do the same procedure for each two slots s, s ′ ∈ {s i + w k -1, ..., s j } for each demand k ∈ K with

v k ∈ C such that σ k s = σ k s ′ , for all v k ∈ C and s, s ′ ∈ {s i + w k -1, ..., s j }.
Consequently, we obtain that σ k s = ρ for all v k ∈ C and s ∈ {s i + w k -1, ..., s j }. By (2.17) and (2.18), we know that

         µ k ′ e ′ = γ k ′ ,e ′ 1 for all k ′ ∈ K and e ′ ∈ E k ′ 0 , µ k ′ e ′ = γ k ′ ,e ′ 2 for all k ′ ∈ K and e ′ ∈ E k ′ 1 , σ k ′ s ′ = γ k ′ ,s ′ 3 for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.
We conclude that for each k ∈ K and e ∈ E

µ k e =          γ k,e 1 if e ∈ E k 0 , γ k,e 2 if e ∈ E k 1 , 0 otherwise,
and for each k ∈ K and s ∈ S

σ k s =          γ k,s 3 if s ∈ {1, ..., w k -1} ρ if v k ∈ C and s ∈ {s i + w k -1, ..., s j }, 0 otherwise.
As a consequence, (µ, σ) = ρ(α, β) + γQ.

Let N (v) denote the set of neighbors of node v in a given graph.

Theorem 2.4.9. Consider an interval of contiguous slots I = [s i , s j ], and a pair of

demands k, k ′ ∈ K with (v k , v k ′ ) in G E I . Then, inequality (2.38) is facet defining for P(G, K, S) if and only if a) N (v k ) ∩ N (v k ′ ) = ∅ in the conflict graph H E I , b
) and there does not exist an interval of contiguous slots

I ′ in [1, s] such that I ⊂ I ′ with w k + w k ′ ≥ |I ′ |, w k ≤ |I ′ |, and w k ′ ≤ |I ′ |.
Proof. Neccessity.

We distinguish two cases:

a) if N (v k ) ∩ N (v k ′ ) ̸ = ∅ in the conflict graph H E I ,

Sufficiency.

We use the same proof of theorem 2.4.8 for a clique

C = {v k , v k ′ } in the conflict graph H E I .

Interval-Odd-Hole Inequalities

Proposition 2.4.14. Let I = [s i , s j ] be an interval of contiguous slots in [1, s] with s i ≤ s j -1, and H be an odd-hole H in the conflict graph H E I with |H| ≥ 5. Then, the inequality

v k ∈H s j s=s i +w k -1 z k s ≤ |H| -1 2 , (2.40) is valid for Q(G, K, S). Moreover, it is valid for P(G, K, S) if 2w k > |I| for each v k ∈ H.
Proof. It is trivial given the definition of odd-hole set in the conflict graph H E I . We strengthen the proof as belows. For each pair of nodes (v k , v k ′ ) linked in H by an edge, we know that

s j s=s i +w k -1 z k s + s j s ′ =s i +w k ′ -1 z k ′ s ′ ≤ 1.
Given that H is an odd-hole which means that we have |H| -1 pair of nodes (v k , v k ′ ) linked in H, and by doing a sum for all pairs of nodes (v k , v k ′ ) linked in H, it follows that

(v k ,v k ′ )∈E(H) s j s=s i +w k -1 z k s + s j s ′ =s i +w k ′ -1 z k ′ s ′ ≤ |H| -1.
where

(v k ,v k ′ )∈E(H) s j s=s i +w k -1 z k s + s j s ′ =s i +w k ′ -1 z k ′ s ′ = v k ∈H 2 s j s=s i +w k -1 z k s ≤ |H| -1.
By dividing the two sides of the previous sum by 2, it follows that

v k ∈H s j s=s i +w k -1 z k s ≤ |H| -1 2 = |H| -
H ∩ C = ∅, c) and the nodes (v k , v k ′ ) are linked in H E I for all v k ∈ H and v k ′ ∈ C.
Then, the inequality

v k ∈H s j s=s i +w k -1 z k s + |H| -1 2 v k ′ ∈C s j s ′ =s i +w k ′ -1 z k ′ s ′ ≤ |H| -1 2 , ( 2.41
)

is valid for Q(G, K, S). Moreover, it is valid for P(G, K, S) if 2w k > |I| for each v k ∈ C ∪ H.
Proof. It is trivial given the definition of odd-hole set and clique set in the conflict graph H E I such that if

s j s ′ =s i +w k ′ -1 z k ′ s ′ = 1 for v k ′ ∈ C, it forces the quantity v k ∈H s j
s=s i +w k -1 z k s to be equal to 0. Otherwise, we know from inequality (2.40) that the sum v k ∈H s j s=s i +w k -1 z k s is always smaller than |H|-1 2 . We strengthen the proof as belows. For each pair of nodes (v k , v k ′ ) linked in H by an edge, we know that

s j s=s i +w k -1 z k s + s j s ′ =s i +w k ′ -1 z k ′ s ′ + v k" ∈C s j
s"=s i +w k" -1 z k" s" ≤ 1 given that all the nodes v k" ∈ C are linked with the nodes v k and v k ′ . Given that H is an odd-hole which means that we have |H| -1 pair of nodes (v k , v k ′ ) linked in H, and by doing a sum for all pairs of nodes (v k , v k ′ ) linked in H, it follows that

(v k ,v k ′ )∈E(H) s j s=s i +w k -1 z k s + s j s ′ =s i +w k ′ -1 z k ′ s ′ + v k" ∈C s j s"=s i +w k" -1 z k" s" ≤ |H| -1.
Taking into account that each node v k in H has two neighbors in H, this implies that

s j s=s i +w k -1 z k s appears twice in the previous inequality. The sum v k" ∈C s j s"=s i +w k" -1 z k" s"
appears |H| -1 times in in the previous inequality. As a result,

(v k ,v k ′ )∈E(H) s j s=s i +w k -1 z k s + s j s ′ =s i +w k ′ -1 z k ′ s ′ + (|H| -1) v k" ∈C s j s"=s i +w k" -1 z k" s" ≤ |H| -1 ⇒ v k ∈H 2 s j s=s i +w k -1 z k s + (|H| -1) v k" ∈C s j s"=s i +w k" -1 z k" s" ≤ |H| -1.
By dividing the two sides of the previous sum by 2, and since |H| is an odd number, it follows that

v k ∈H s j s=s i +w k -1 z k s + |H| -1 2 v k" ∈C s j s"=s i +w k" -1 z k" s" ≤ |H| -1 2 = |H| -1 2 .
We conclude at the end that inequality (2.41) is valid for Q(G, K, S). Moreover, it c) and there does not exist an interval I ′ of contiguous slots with I ⊂ I ′ such that H defines also an odd-hole in the associated conflict graph H E I ′ .

is valid for P(G, K, S) if 2w k > |I| for each v k ∈ C ∪ H.
H E I ((H \ {v k }) ∪ {v k ′ }) does not contain an odd-hole H ′ = (H \ {v k }) ∪ {v k ′ }, b) and there does not exist a node v k ′ / ∈ H in H E I such that v k ′ is linked with all nodes v k ∈ H,
Proof. Neccessity.

We distinguish the following cases:

a) if for a node v k ′ / ∈ H in H E I , there exists a node v k ∈ H such that the induced graph H E I ((H \ {v k }) ∪ {v k ′ }) contains an odd-hole H ′ = (H \ {v k }) ∪ {v k ′ }.
This implies that inequality (2.40) can be dominated by doing some lifting procedures using the following valid inequalities

v k ∈H s j s ′ =s i +w k -1 z k s ′ ≤ |H| -1 2
, and

v k ′ ∈H ′ s j s ′ =s i +w k ′ -1 z k ′ s ′ ≤ |H| -1 2 ,
as follows

s j s ′ =s i +w k -1 z k s ′ + s j s ′ =s i +w k ′ -1 z k ′ s ′ + 2 v k" ∈H\{k,k ′ } s j s"=s i +w k" -1 z k" s" ≤ |H| -1.
By adding the sum

s j s ′ =s i +w k ′ -1 z k ′ s ′
to the previous inequality, we obtain

s j s ′ =s i +w k -1 z k s ′ + 2 s j s ′ =s i +w k ′ -1 z k ′ s ′ + 2 v k" ∈H\{k,k ′ } s j s"=s i +w k" -1 z k" s" ≤ |H| -1 + s j s ′ =s i +w k ′ -1 z k ′ s ′ . Since s j s ′ =s i +w k ′ -1 z k ′ s ′ ≤ 1, it follows that s j s ′ =s i +w k -1 z k s ′ + 2 s j s ′ =s i +w k ′ -1 z k ′ s ′ + 2 v k" ∈H\{k,k ′ } s j s"=s i +w k" -1 z k" s" ≤ |H|.
By dividing the last inequality by 2, we obtain that

s j s ′ =s i +w k -1 1 2 z k s ′ + s j s ′ =s i +w k ′ -1 z k ′ s ′ + v k" ∈H\{k,k ′ } s j s"=s i +w k" -1 z k" s" ≤ |H| 2 .
Given that

H ′ = (H \ {k}) ∪ {k ′ } such that |H ′ | = |H|, and |H| is an odd number which implies that |H| 2 = |H|-1 2 . As a result s j s ′ =s i +w k -1 1 2 z k s ′ + v k ′ ∈H ′ s j s"=s i +w k ′ -1 z k ′ s" ≤ |H ′ | -1 2 .
That which was to be demonstrated. b) if there exists a node

v k ′ ∈ H in H E I such that v k ′ is linked with all nodes v k ∈ H. As a result, inequality (2.

40) is dominated by the following inequality

v k ∈H s j s ′ =s i +w k -1 z k s ′ + |H| -1 2 s j s ′ =s i +w k ′ -1 z k ′ s ′ ≤ |H| -1 2 .
c) if there exists an interval I ′ of contiguous slots with I ⊂ I ′ such that H defines also an odd-hole in the associated conflict graph H E I ′ . This implies that inequality (2.40) induced by odd-hole H for the interval I is dominated by inequality (2.40) induced by the same odd-hole H for the interval I ′ given that {s i + w k -1, ..., s j } ⊂ I ′ for each k ∈ H. As a result, inequality (2.40) is not facet defining for P(G, K, S).

If no one of these two cases is verified, inequality (2.40) can never be dominated by another inequality without changing its right-hand side.

Sufficiency.

Let F

H E I H
be the face induced by inequality (2.40), that is

F H E I H = {(x, z) ∈ P(G, K, S) : v k ∈H s j s=s i +w k -1 z k s = |H| -1 2 }.
We denote inequality v k ∈H

s j s=s i +w k -1 z k s ≤ |H|-1 2 by αx + βz ≤ λ. Let µx + σz ≤
τ be a facet defining inequality for P(G, K, S) and

F = {(x, z) ∈ P(G, K, S) : µx + σz = τ }. Suppose that F H E I H ⊆ F . In order to prove that inequality v k ∈H s j s=s i +w k -1 z k s ≤ |H|-1 2
is facet defining for P(G, K, S), we will show that there exist ρ ∈ R and

γ = (γ 1 , γ 2 , γ 3 ) (such that γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α, β) + γQ.
We first show that µ k e = 0 for each edge e ∈ E \ (E k 0 ∪ E k 1 ) for each demand k ∈ K. Consider a demand k ∈ K and an edge e ∈ E \ (E k 0 ∪ E k 1 ). Let S 64 = (E 64 , S 64 ) be the solution given by a) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E 64 k i be the set of edges involved in a shortest path between o k i and

d k i , b) select a subset of demands H from H with | H| = |H|-1
2 , c) for each demand k i ∈ H with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 64 i given by

I 64 i = [ kj ∈L 64 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ {s i + w ki -1, ..., s j }.
where where As a result, µ k e = 0. In a similar way, we can show that

L 64 i = {k j ∈ {k 1 , ..., k i-1 } ∪ H : E 64 k i ∩ E 64 k j ̸ = ∅}, d) for each demand k i ∈ H \ H with i ∈ {1, ...,
R 64 i = {k j ∈ {k 1 , ..., k i-1 } ∪ H such that E 64 k i ∩ E 64 k j ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ R 64 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s k -w k j + 1, ..., s k } = ∅ if E 64 k i ∩ (E 64 k ∪ {e ′ }) ̸ = ∅ (
µ k e = 0, for all k ∈ K and e ∈ E \ (E k 0 ∪ E k 1 ).
Let show that σ k s = 0 for all k ∈ K and s ∈ {w k , ..., s} with s /

∈ {s i + w k -1, ..., s j } if v k ∈ H.
Consider a demand k in K and a slot s ′ in {w k , ..., s} with

s ′ / ∈ {s i + w k -1, ..., s j } if v k ∈ H. Let S ′64 = (E ′64 , S ′64
) be the solution given by a) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E ′64 k i be the set of edges involved in a shortest path between o k i and d k i , b) select a subset of demands H from H with | H| = |H|-1 2 , c) for each demand k i from H with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′64 i given by

I ′64 i = [ kj ∈D ′64 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ {s i + w ki -1, ..., s j },
where

D ′64 i = {k j ∈ {k 1 , ..., k i-1 } ∩ H : E ′64 k i ∩ E ′64 k j ̸ = ∅}, d) for each demand k i ∈ H \ H with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′64 i
given by

I ′64 i = kj ∈D" 64 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} \ {s i + w ki -1, ..., s j },
where D"

64 i = {k j ∈ {k 1 , ..., k i-1 } ∩ H : E" 64 k i ∩ E" 64 k j ̸ = ∅}, e) for each demand k i ∈ K \ H with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′64 i
given by

I ′64 i = [ kj ∈R ′64 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}] if E ′64 ki ∩ E ′64 k ̸ = ∅ or I ′64 i = kj ∈R ′64 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,
where

R ′64 i = {k j ∈ {k 1 , ..., k i-1 } ∪ H such that E ′64 k i ∩ E ′64 k j ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ R ′64 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E ′64 k i ∩ E ′64 k ̸ = ∅ (
we take into account the possibility of adding slot s ′ as a last slot in the selected last slots S ′64 k to route demand k in solution S ′64 ).

We let S ′64

k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′64 is feasible for the problem. Hence, the corresponding incidence vector (x S ′64 , z S ′64 ) belongs to F H E I H . Then consider the solution S 66 obtained from S ′64 by adding slot s ′ as last slot to demand k in S ′64 . Solution S 66 is clearly feasible for the problem.

The corresponding incidence vector (x S 66 , z S 66 ) belongs to F H E I H . Hence, solutions S ′64 and S 66 satisfy equation µx + σz = τ . We have so

µx S ′64 + σz S ′64 = µx S 66 + σz S 66 = µx S ′64 + σz S ′64 + σ k s ′ .
Hence, σ k s ′ = 0. In a similar way, we can show that σ k s = 0, for all k ∈ K and s ∈ {w k , ..., s} with s / where where

∈ {s i + w k -1, ..., s j } if v k ∈ H. Let prove that σ k ′ s ′ for all v k ′ ∈ H and s ′ ∈ {s i + w k ′ -1, ..., s j } are equivalent. Consider a demand k ′ ∈ K with v k ′ ∈ H and a slot s ′ ∈ {s i + w k ′ -1, ...,
L 66 i = {k j ∈ {k 1 , ..., k i-1 } ∩ H : E 66 k i ∩ E 66 k j ̸ = ∅}, d) for each demand k i ∈ H \ H with i ∈ {1, ...,
R 66 i = {k j ∈ {k 1 , ..., k i-1 } ∪ H such that E 66 k i ∩ E 66 k j ̸ = ∅}. Hence, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ R 66 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ if E 66 k i ∩ E 66 k ′ ̸ = ∅ (
k ′ s ′ + σ k i s′ -σ k i s .
Since σ k s = 0 for all demand k ∈ K and slot s ∈ {w k , ..., s} with s / ∈ {s i +w k +1, ..., s j }

if v k ∈ H, it follows that σ k i s = σ k ′ s ′ .
In a similar way, we can show that

σ k s = σ k ′ s ′ , for all pairs (v k , v k ′ ) ∈ H.
Consequently, we obtain that σ k s = ρ for all v k ∈ H and s ∈ {s i + w k -1, ..., s j }. Overall, and using the results (2.17) and (2.18), we obtain that

µ k e =          γ k,e 1 if e ∈ E k 0 , γ k,e 2 if e ∈ E k 1 , 0 otherwise,
for each k ∈ K and e ∈ E, and

σ k s =          γ k,s 3 if s ∈ {1, ..., w k -1}, ρ if v k ∈ H and s ∈ {s i + w k -1, ..., s j }, 0 otherwise.
for each k ∈ K and s ∈ S.

As a consequence, (µ, σ) = ρ(α, β) + γQ.

Theorem 2.4.11. Let H be an odd-hole, and C be a clique in the conflict graph

H E I with a) |H| ≥ 5, b) and H ∩ C = ∅, c) 2w k > |I| for each v k ∈ C ∪ H, d) and the nodes (v k , v k ′ ) are linked in H E I for all v k ∈ H and v k ′ ∈ C.
v k ∈H s j s=s i +w k -1 z k s + |H| -1 2 v k ′ ∈C s j s ′ =s i +w k ′ -1 z k ′ s ′ + |H| -1 2 s j s"=s i +w k" -1 z k" s ′ ≤ |H| -1 2 .
b) if there exists an interval I ′ of contiguous slots with I ⊂ I ′ such that H and C define also an odd-hole and its connected clique in the associated conflict graph H E I ′ . This implies that inequality (2.41) induced by odd-hole H and clique C for the interval I is dominated by inequality (2.41) induced by the same odd-hole H and clique C for the interval I ′ given that {s i + w k -1, ..., s j } ⊂ I ′ for each k ∈ H.

If these cases are not verified, we ensure that inequality (2.41) can never be dominated by another inequality without modifying its right-hand side. Otherwise, inequality (2.41) is not facet defining for P(G, K, S).

Sufficiency.

Let F H E I H,C be the face induced by inequality (2.41), that is

F H E I H,C = {(x, z) ∈ P(G, K, S) : v k ∈H s j s=s i +w k -1 z k s + |H| -1 2 v k ′ ∈C s j s ′ =s i +w k ′ -1 z k ′ s ′ = |H| -1 2 }. Let denote inequality v k ∈H s j s=s i +w k -1 z k s ≤ |H|-1 2 by αx+βz ≤ λ. Let µx+σz ≤
τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) :

µx + σz = τ }. Suppose that F H E I H,C ⊆ F . To prove that F H E I H,C
is a facet of P(G, K, S), we need to show that there exist ρ ∈ R and γ = (γ 1 , γ 2 , γ 3 ) (such that where where

γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α,
D 70 i = {k j ∈ {k 1 , ..., k i-1 } ∩ H : E 70 k i ∩ E 70 k j ̸ = ∅}, e) for each demand k i ∈ K \ H with i ∈ {1, ...,
R 70 i = {k j ∈ {k 1 , ..., k i-1 } ∪ H such that E 70 k i ∩ E 70 k j ̸ = ∅}. Hence, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ R 70 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k′ + 1, ..., s′ } = ∅ if E 70 k i ∩ E 70 k′ ̸ = ∅ (
k i + 1, ..., s j }) for each k i ∈ H, b) and select a new last slot s′ i / ∈ {s i + w k i + 1, ..., s j } for each k i ∈ H i.e., S 71
k i = (S 70 k i \ {s i }) ∪ {s ′ i } such that {s ′ i -w k i -1, ..., s′ i } ∩ {s -w k + 1, ..., s} = ∅ for each k ∈ K and s ∈ S 70 k with E 71 k ∩ E 71 k i ̸ = ∅ for each k i ∈ H,
+ k i ∈ H σ k i s′ i - k i ∈ H σ k i si . Since σ k s = 0 for all demand k ∈ K and slot s ∈ {w k , ..., s} with s / ∈ {s i +w k +1, ..., s j } if v k ∈ H ∪ C, it follows that k i ∈ H σ k i si = σ k′ s′ for v k′ ∈ C.
In a similar way, we can show that

σ k ′ s ′ = ρ |H| -1 2 , for all v k ′ ∈ C and s ′ ∈ {s i + w k ′ + 1, ..., s j }.
As a result,

σ k s = σ k ′ s ′ , for all (v k , v k ′ ) ∈ C and s ∈ {s i + w k + 1, ..., s j } and s ′ ∈ {s i + w k ′ + 1, ..., s j }.
Consequently, we obtain that

σ k ′ s ′ = ρ |H|-1 2 for all v k ′ ∈ C and s ′ ∈ {s i + w k ′ - 1, ..., s j }.
By (2.17) and (2.18), we know that

         µ k ′ e ′ = γ k ′ ,e ′ 1 for all k ′ ∈ K and e ′ ∈ E k ′ 0 , µ k ′ e ′ = γ k ′ ,e ′ 2 for all k ′ ∈ K and e ′ ∈ E k ′ 1 , σ k ′ s ′ = γ k ′ ,s ′ 3
for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.

We conclude that for each k ∈ K and e ∈ E

µ k e =          γ k,e 1 if e ∈ E k 0 , γ k,e 2 if e ∈ E k 1 , 0 otherwise,
and for each k ∈ K and s ∈ S

σ k s =                  γ k,s 3 if s ∈ {1, ..., w k -1}, ρ if v k ∈ H and s ∈ {s i + w k -1, ..., s j }, ρ |H|-1 2 if v k ∈ C and s ∈ {s i + w k -1, ..., s j }, 0 otherwise.
As a result, we have (µ, σ) = ρ(α, β) + γQ.

Edge-Slot-Assignment-Clique Inequalities

Here, we introduce another conflict graph totally different compared with the conflict graphs presented previously.

Definition 2.4.6. Let H e S be a conflict graph defined as follows. For each slot s ∈ {w k , ..., s} and demand k ∈ K with e / ∈ E k 0 , consider a node v k,s in H e S . Two nodes v k,s and v k ′ ,s ′ are linked by an edge in

H e S if a) k = k ′ , b) or {s -w k + 1, ..., s} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ if k ̸ = k ′ and (k, k ′ ) / ∈ K e c .
Based on this definition, we introduce the following inequalities.

v k,s ∈C (x k e + z k s ) ≤ |C| + 1, (2.42
)

is valid for Q(G, K, S). Moreover, it is valid for P(G, K, S) if {s -w k + 1, ..., 1} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ for each (v k,s , v k ′ ,s ′ ) ∈ C.
Proof. It is trivial given the definition of a clique set in the conflict graph H e S such that for each two linked nodes v k,s and v k ′ ,s ′ in H e S , we have

x k e + x k ′ e + z k s + z k ′ s ′ ≤ 3.
This can be generalized for a triplet of linked nodes v k,s and v k ′ ,s ′ and v k ′′ ,s ′′ with

w k + w k ′ + w k" ≤ s -k∈Ke\{k,k ′ ,k"} w k, such that for each linked nodes (v k,s , v k ′ ,s ′ )
and (v k,s , v k ′′ ,s ′′ ) and (v k ′ ,s ′ , v k ′′ ,s ′′ ), we have

x k e + x k ′ e + z k s + z k ′ s ′ ≤ 3, x k e + x k" e + z k s + z k" s" ≤ 3, x k ′ e + x k" e + z k ′ s ′ + z k" s" ≤ 3.
By adding the three previous inequalities, we get the following inequality using the chvatal gomory procedure

2x k e + 2x k ′ e + 2x k" e + 2z k s + 2z k ′ s ′ + 2z k" s" ≤ 9 ⇒ x k e + x k ′ e + x k" e + z k s + z k ′ s ′ + z k" s" ≤ 4 given that 9 2 = 4.
This can be generalized for each clique C with |C| ≥ 4 while showing that inequality (2.42) can be seen as Chvàtal-Gomory cuts. Using the Chvàtal-Gomory and recurrence procedures, we obtain that

v k,s ∈C ′ x k e + z k s ≤ |C ′ | + 1, for all C ′ ⊂ C with |C ′ | = |C| -1 and |C ′ | ≥ 3.
By adding the previous inequalities for all C ′ ⊂ C with |C ′ | = |C| -1, and doing then some simplification, we get at the end that

v k,s ∈C x k e + z k s ≤ |C| + |C| |C| -1 ⇒ v k,s ∈C x k e + z k s ≤ |C| + 1,
given that |C| |C| -1 = 1. We conclude at the end that inequality (2.42) is valid for

Q(G, K, S). Moreover, it is valid for P(G, K, S) if {s -w k + 1, ..., 1} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ for each (v k,s , v k ′ ,s ′ ) ∈ C.
Theorem 2.4.12. Consider an edge e ∈ E, and a clique C in the conflict graph H e S with {s -

w k + 1, ..., 1} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ for each (v k,s , v k ′ ,s ′ ) ∈ C.
I = [s i , s j ] ⊂ [1, s] with a) [ min v k,s ∈C (s -w k + 1), max v k,s ∈C s] ⊂ I, b) and w k + w k ′ ≥ |I| + 1 for each (v k , v k ′ ) ∈ C, c) and 2w k ≥ |I| + 1 and w k ≤ |I| for each v k ∈ C. Proof. Neccessity.
If C is not maximal clique in the conflict graph H e S , this means that inequality (2.42) can be dominated by another inequality associated with a clique C ′ such that C ⊂ C ′ without changing its right-hand side. Moreover, if there exists an interval

of contiguous slots I = [s i , s j ] ⊂ [1, s] with a) [ min v k,s ∈C (s -w k + 1), max v k,s ∈C s] ⊂ I, b) and w k + w k ′ ≥ |I| + 1 for each (v k , v k ′ ) ∈ C, c) and 2w k ≥ |I| + 1 and w k ≤ |I| for each v k ∈ C.
Then, inequality (2.42) is dominated by inequality (2.36). As a result, inequality (2.42) cannot be facet defining for P(G, K, S).

Sufficiency.

Let F H e S C be the face induced by inequality (2.42), that is

F H e S C = {(x, z) ∈ P(G, K, S) : v k,s ∈C x k e + z k s = 1}.
Let denote inequality v k,s ∈C x k e + z k s ≤ 1 by αx + βz ≤ λ. Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) : µx+σz = τ }.

Suppose that F

H e S C ⊆ F . In order to prove that inequality v k,s ∈C x k e +z k s ≤ 1 is facet defining for P(G, K, S), we need to show that there exist ρ ∈ R and γ = (γ 1 , γ 2 , γ 3 )

(such that γ k,e ′ 1 ∈ R for all k ′ ∈ K and e ′ ∈ E k ′ 0 , γ k,e ′ 2 ∈ R for all k ′ ∈ K and e ′ ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α, β)+γQ.
In a similar way with the proof of theorem 2.4.1, we obtain that

µ k ′ e ′ =                  γ k ′ ,e ′ 1 if e ′ ∈ E k ′ 0 , γ k ′ ,e ′ 2 if e ′ ∈ E k ′ 1 , ρ if k ′ ∈ K(C
) and e ′ = e, 0 otherwise, for each k ′ ∈ K and e ′ ∈ E, and

σ k s ′ =          γ k,s ′ 3 if s ′ ∈ {1, ..., w k -1} ρ if v k,s ∈ C, 0 otherwise.
for each k ∈ K and s ′ ∈ S, where K(C) = {k ∈ K : ∃s ∈ {w k , ..., s} with v k,s ∈ C}.

As a consequence, (µ, σ) = ρ(α, β) + γQ.

Slot-Assignment-Clique Inequalities

On the other hand, we detected that there may exist some cases that are not covered by inequality (2.42) and (2.25) previously introduced. For this, we provide the following definition of a conflict graph and its associated inequality.

• k = k ′ , • or E k 1 ∩ E k ′ 1 ̸ = ∅ and {s -w k + 1, ..., s} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ if k ̸ = k ′ .
Proposition 2.4.17. Let C be a clique in the conflict graph H E S with |C| ≥ 3. Then, the inequality

v k,s ∈C z k s ≤ 1, (2.43)
is valid for Q(G, K, S). Moreover, it is valid for P(G, K, S) if {s -

w k + 1, ..., 1} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ for each (v k,s , v k ′ ,s ′ ) ∈ C.
Proof. It is trivial given the definition of a clique set in the conflict graph H E S such that for each two linked nodes v k,s and v k ′ ,s ′ in H E S , we know from the inequality (2.6) that

z k s + z k ′ s ′ ≤ 1, given that x k e = x k ′ e = 1 for all e ∈ E k 1 ∩ E k ′ 1 and {s -w k + 1, ..., s} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅.
By adding the previous inequalities for all two nodes v k,s and v k ′ ,s ′ in C, and by recurrence procedure we obtain that for all

C ′ ⊆ C with |C ′ | = |C| -1 v k,s ∈C ′ z k s ≤ 1.
By adding the previous inequalities for all

C ′ ⊆ C with |C ′ | = |C| -1, we get C ′ ⊆C |C ′ |=|C|-1 v k,s ∈C ′ z k s ≤ C ′ ⊆C |C ′ |=|C|-1 1.
112 Note that for each demand k and slot s with v k,s ∈ C, the variable z k s appears ( |C| |C|-1 -1) = |C| -1 times in the previous sum. It follows that

v k,s ∈C (|C| -1)z k s ≤ |C|.
By dividing the two sides of the previous sum by |C| -1, we have so

v k,s ∈C z k s ≤ |C| |C| -1 ⇒ v k,s ∈C z k s ≤ 1 given that |C| |C| -1 = 1.
We conclude at the end that inequality (2.43) is valid for Q(G, K, S). Moreover,

it is valid for P(G, K, S) if {s -w k + 1, ..., 1} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ for each (v k,s , v k ′ ,s ′ ) ∈ C.
Theorem 2.4.13. Consider a clique C in the conflict graph H E S with {s -

w k + 1, ..., 1} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ for each (v k,s , v k ′ ,s ′ ) ∈ C. Then, inequality (2.43)
is facet defining for P(G, K, S) if and only if C is a maximal clique in the conflict graph H E S , and there does not exist an interval of contiguous slots

I = [s i , s j ] ⊂ [1, s] with a) [ min v k,s ∈C (s -w k + 1), max v k,s ∈C s] ⊂ I, b) and w k + w k ′ ≥ |I| + 1 for each (v k , v k ′ ) ∈ C,
= [s i , s j ] ⊂ [1, s] with a) [ min v k,s ∈C (s -w k + 1), max v k,s ∈C s] ⊂ I, b) and w k + w k ′ ≥ |I| + 1 for each (v k , v k ′ ) ∈ C, c) and 2w k ≥ |I| + 1 and w k ≤ |I| for each v k ∈ C.
Then, inequality (2.43) is dominated by inequality (2.39). As a result, inequality (2.43) cannot be facet defining for P(G, K, S).

Sufficiency.

Let F

H E S C
be the face induced by inequality (2.43), that is

F H E S C = {(x, z) ∈ P(G, K, S) : v k,s ∈C z k s = 1}.
Let denote inequality v k,s ∈C z k s ≤ 1 by αx + βz ≤ λ. Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) : µx + σz = τ }.

Suppose that

F H E S C
⊆ F . In order to prove that inequality v k,s ∈C z k s ≤ 1 is facet defining for P(G, K, S), we need to show that there exist ρ ∈ R and γ = (γ 1 , γ 2 , γ 3 )

(such that γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α, β)+γQ.
We first show that µ k e = 0 for each edge e where where

∈ E \ (E k 0 ∪ E k 1 ) for each demand k ∈ K. Consider a demand k ∈ K and an edge e ∈ E \ (E k 0 ∪ E k 1 ). Let S 72 = (E
D 72 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k ′ } : E 72 k i ∩ E 72 k j ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 72 i , • {s k i -w k i + 1, ..., s k i } ∩ {s k -w k j + 1, ..., s k } = ∅ if E 72 k i ∩ E 72 k ̸ = ∅ (
D ′72 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k ′ } : E ′72 k i ∩ E ′72 k j ̸ = ∅}. This satisfies that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D ′72 i , • {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E ′72 k i ∩ E ′72 k ̸ = ∅ (
we take into account the possibility of adding slot s ′ in the selected set of last slots S ′72 k to route demand k in solution S ′72 ).

We let S ′72

k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′72 is feasible for the problem. Hence, the corresponding incidence vector (x S ′72 , z S ′72 ) belongs to F H E S C . Then consider the solution S 74 obtained from S ′72 by adding slot s ′ as last slot to demand k in S ′72 . Solution S 74 is clearly feasible for the problem.

The corresponding incidence vector (x S 74 , z S 74 ) belongs to F H E S C . Hence, solutions S ′72 and S 74 satisfy equation µx + σz = τ . We have so

µx S ′72 + σz S ′72 = µx S 74 + σz S 74 = µx S ′72 + σz S ′72 + σ k s ′ .
Hence, σ k s ′ = 0. In a similar way, we can show that σ k s = 0, for all k ∈ K and s ∈ {w k , ..., s} with v k,s / ∈ C.

Let prove that σ k s for all v k,s ∈ C are equivalent. Consider a node v k ′ ,s ′ in C such that s ′ / ∈ S 72 k ′ . Let S72 = ( Ẽ72 , S72 ) be the solution given by 1. for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we let Ẽ72 k i be the set of edges involved in a shortest path between o k i and d k i , 2. select a pair of demand k and slot s from clique C (i.e., v k,s ∈ C) such that slot s k = s will be used as last slot for demand k, 3. for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots Ĩ72 i given by

Ĩ72 i = [ kj ∈ D72 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k ′ } ∪ {s ′ + w ki , ..., s}] if Ẽ72 ki ∩ Ẽ72 k ′ ̸ = ∅ or Ĩ72 i = kj ∈ D72 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not, where D72 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : Ẽ72 k i ∩ Ẽ72 k j ̸ = ∅}. This ensures that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D72 i , • {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ if Ẽ72 k i ∩ Ẽ72 k ′ ̸ = ∅ (
we take into account the possibility of adding slot s ′ in the selected set of last slots S72 k ′ to route demand k ′ in solution S72 ).

We let S72

k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}. S72 is feasible for the problem. Hence, the corresponding incidence vector (x

S72 , z S72 ) belongs to F H E S C
. Then consider the solution S 75 obtained from S72 by adding slot s ′ as last slot to demand k ′ in S72 , and modifying the last slots assigned to demand k by adding a new last slot s and removing the last slot 75 is clearly feasible for the problem. The corresponding incidence vector (x S 75 , z S 75 ) belongs to F

s ∈ S72 k with v k,s ∈ C and v k,s / ∈ C such that S 75 k = ( S72 k \ {s}) ∪ {s} and {s -w k + 1, ..., s} ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ for each k ′ ∈ K and s ′ ∈ S 75 k ′ with E 75 k ∩ E 75 k ′ ̸ = ∅. Solution S
H E S C .
Hence, solutions S72 and S 75 satisfy equation µx + σz = τ . We have so

µx S72 + σz S72 = µx S 75 + σz S 75 = µx S72 + σz S72 + σ k ′ s ′ -σ k s + σ k s . Since σ k s = 0 for v k,s / ∈ C, and µ k e = 0 for all k ∈ K and e ∈ E \ (E k 0 ∪ E k 1 ), it follows that σ k ′ s ′ = σ k s .
In a similar way, we can show that

σ k s = σ k ′ s ′ , for all pairs (v k,s , v k ′ ,s ′ ) ∈ C.
Consequently, we obtain that σ k s = ρ for all pairs v k,s ∈ C. We know from (2.17) and (2.18) that

         µ k ′ e ′ = γ k ′ ,e ′ 1 for all k ′ ∈ K and e ′ ∈ E k ′ 0 , µ k ′ e ′ = γ k ′ ,e ′ 2 for all k ′ ∈ K and e ′ ∈ E k ′ 1 , σ k ′ s ′ = γ k ′ ,s ′ 3 for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.
As a result, we obtain that for each k ∈ K and e ∈ E

µ k e =          γ k,e 1 if e ∈ E k 0 γ k,e 2 if e ∈ E k 1 0 otherwise
and for each k ∈ K and s ∈ S

σ k s =          γ k,s 3 if s ∈ {1, ..., w k -1} ρ if v k,s ∈ C, 0 if v k,s / ∈ C.
As a consequence, (µ, σ) = ρ(α, β) + γQ.

Slot-Assignment-Odd-Hole Inequalities

One can strengthen inequality (2.43) by introducing the following inequalities based on the so-called odd-hole inequalities.

Proposition 2.4.18. Let H be an odd-hole in the conflict graph H E S with |H| ≥ 5. Then, the inequality

v k,s ∈H z k s ≤ |H| -1 2 , ( 2.44
)

is valid for Q(G, K, S). Moreover, it is valid for P(G, K, S) if {s -w k + 1, ..., 1} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ for each pair of nodes (v k,s , v k ′ ,s ′ ) that are linked in H.
Proof. It is trivial given the definition of the odd-hole in the conflict graph H E S such that for each pair of nodes (v k,s , v k ′ ,s ′ ) linked in H by an edge, we know that

z k s + z k ′ s ′ ≤ 1.
Given that H is an odd-hole which means that we have |H| -1 pair of nodes (v k,s , v k ′ ,s ′ ) linked in H, and by doing a sum over all pairs of nodes (v k,s , v k ′ ,s ′ ) linked in H, it follows that

(v k,s ,v k ′ ,s ′ )∈E(H) z k s + z k ′ s ′ ≤ |H| -1.
Taking into account that each node v k in H has two neighbors in H, this implies that z k s appears twice in the previous inequality. As a result,

(v k,s ,v k ′ ,s ′ )∈E(H) z k s + z k ′ s ′ = v k,s ∈H 2z k s =⇒ v k,s ∈H 2z k s ≤ |H| -1.
As a result,

v k,s ∈H z k s ≤ |H| -1 2 = |H| -1 2 since |H| is an odd number.
We conclude at the end that inequality (2.44) is valid for P(G, K, S).

Note that inequality (2.44) can be strengthened without modifying its right-hand side by combining inequality (2.44) and (2.43).

Proposition 2.4.19. Let H be an odd-hole, and C be a clique in the conflict graph

H E S with a) |H| ≥ 5, b) and H ∩ C = ∅, c) and the nodes (v k,s , v k ′ ,s ′ ) are linked in H E S for all v k,s ∈ H and v k ′ ,s ′ ∈ C.
Then, the inequality

v k,s ∈H z k s + |H| -1 2 v k ′ ,s ′ ∈C z k ′ s ′ ≤ |H| -1 2 , ( 2.45
)

is valid for Q(G, K, S). Moreover, it is valid for P(G, K, S) if {s -w k + 1, ..., 1} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ for each (v k,s , v k ′ ,s ′ ) ∈ C and pair of nodes (v k,s , v k ′ ,s ′ ) linked in H.
Proof. It is trivial given the definition of the odd-hole and clique in

H E S such that if v k ′ ,s ′ ∈C z k ′ s ′ = 1 for a v k ′ ,s ′ ∈ C ∈ C which implies that the quantity v k,s ∈H z k s
is forced to be equal to 0. Otherwise, we know from inequality (2.44) that the sum v k,s ∈H z k s is always smaller than |H|-1 2 . We strengthen the proof as belows. For each pair of nodes (v k,s , v k ′ ,s ′ ) linked in H by an edge, we know that z k s + z k ′ s ′ + v k",s" ∈C z k" s" ≤ 1 given that all the nodes v k",s" ∈ C are linked with the nodes v k,s and v k ′ ,s ′ . Given that H is an odd-hole which means that we have |H| -1 pair of nodes (v k,s , v k ′ ,s ′ ) linked in H, and by doing a sum for all pairs of nodes (v k,s , v k ′ ,s ′ ) linked in H, it follows that

(v k,s ,v k ′ ,s ′ )∈E(H) z k s + z k ′ s ′ + v k",s" ∈C z k" s" ≤ |H| -1.
Taking into account that each node v k,s has two neighbors in H, this implies that z k s appears twice in the previous inequality. The sum v k",s" ∈C z k" s" appears |H| -1 times in in the previous inequality. As a result,

(v k,s ,v k ′ ,s ′ )∈E(H) z k s + z k ′ s ′ + (|H| -1) v k",s" ∈C z k" s" ≤ |H| -1 ⇒ v k,s ∈H 2z k s + (|H| -1) v k",s" ∈C z k" s" ≤ |H| -1.
By dividing the two sides of the previous sum by 2, and since |H| is an odd number, it follows that

v k,s ∈H z k s + |H| -1 2 v k",s" ∈C z k" s" ≤ |H| -1 2 = |H| -1 2 .
We conclude at the end that inequality (2.45) is valid for P(G, K, S).

Theorem 2.4.14. Let H be an odd-hole in the conflict graph H E S with |H| ≥ 5, and {s - c) and there does not exist an interval of contiguous slots

w k + 1, ..., 1} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ for each pair of nodes (v k,s , v k ′ ,s ′ ) linked in H.
v k ′ ,s ′ / ∈ H in H E S such that v k ′ ,s ′ is linked with all nodes v k,s ∈ H,
I = [s i , s j ] ⊂ [1, s] with • [ min v k,s ∈H (s -w k + 1), max v k,s ∈H ] ⊂ I, • and w k + w k ′ ≥ |I| + 1 for each (v k , v k ′ ) linked in H,
• and 2w k ≥ |I| + 1 and w k ≤ |I| for each v k ∈ H.

Proof. Neccessity.

We distinguish the following cases:

a) if for a node v k ′ ,s ′ / ∈ H in H E S , there exists a node v k,s ∈ H such that the induced graph H E S (H \{v k,s }∪{v k ′ ,s ′ }) contains an odd-hole H ′ = (H \{v k,s })∪{v k ′ ,s ′ }.
This implies that inequality (2.44) can be dominated using some technics of lifting based on the following two inequalities v k,s ∈H z k s ≤ |H|-1 2 , and

v k ′ ,s ′ ∈H ′ z k ′ s ′ ≤ |H ′ |-1 2 . b) if there exists a node v k ′ ,s ′ / ∈ H in H E S such that v k ′ ,s ′ is
linked with all nodes v k,s ∈ H. This implies that inequality (2.44) can be dominated by the following valid inequality

v k,s ∈H z k s + |H| -1 2 z k ′ s ′ ≤ |H| -1 2 .
c) if there exists an interval of contiguous slots

I = [s i , s j ] ⊂ [1, s] with • [ min v k,s ∈H (s -w k + 1), max v k,s ∈H ] ⊂ I,
• and

w k + w k ′ ≥ |I| + 1 for each (v k , v k ′ ) linked in H,
• and 2w k ≥ |I| + 1 and

w k ≤ |I| for each v k ∈ H.
This implies that inequality (2.44) is dominated by inequality (2.40).

If no one of these cases is verified, inequality (2.44) can never be dominated by another inequality without changing its right-hand side. Otherwise, inequality (2.44) cannot be facet defining for P(G, K, S).

Sufficiency.

Let F

H E S H
be the face induced by inequality (2.44), that is

F H E S H = {(x, z) ∈ P(G, K, S) : v k,s ∈H z k s = |H| -1 2 }. Denote inequality v k,s ∈H z k s ≤ |H|-1 2 by αx + βz ≤ λ. Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) : µx + σz = τ }. Suppose that F H E S H ⊆ F . In order to prove that inequality v k,s ∈H z k s ≤ |H|-1 2 is
facet defining for P(G, K, S), we will show that there exist ρ ∈ R and γ = (γ 1 , γ 2 , γ 3 )

(such that γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α, β)+γQ.
We first show that µ k e = 0 for each edge e where given by

∈ E \ (E k 0 ∪ E k 1 ) for each demand k ∈ K. Consider a demand k ∈ K and an edge e ∈ E \ (E k 0 ∪ E k 1 ). Let S 76 = (E
D 76 i = {k j ∈ {k 1 , ..., k i-1 } ∪ H76 : E 76 k i ∩ E 76 k j ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 76 i , • {s k i -w k i + 1, ..., s k i } ∩ {s k -w k j + 1, ..., s k } = ∅ if E 76 k i ∩ E 76 k ̸ = ∅ (
I ′76 i = [ kj ∈D ′76 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}] if E ′76 ki ∩ E ′76 k ̸ = ∅ or I ′76 i = kj ∈D ′76 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,
where

D ′76 i = {k j ∈ {k 1 , ..., k i-1 } ∪ H′76 : E ′76 k i ∩ E ′76 k j ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D ′76 i , • {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E ′76 k i ∩ E ′76
k ̸ = ∅ ( we take into account the possibility of adding slot s ′ in the selected set of last slots S ′76 k to route demand k in solution S ′76 ).

We let S ′76

k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′76 is feasible for the problem. Hence, the corresponding incidence vector (x S ′76 , z S ′76 ) belongs to F H E S H . Then consider the solution S 78 obtained from S ′76 by adding slot s ′ as last slot to demand k in S ′76 . Solution S 78 is feasible for the problem. The corresponding incidence vector (x S 78 , z S 78 ) belongs to F H E S H . Hence, solutions S ′76 and S 78 satisfy equation µx + σz = τ . We have so

µx S ′76 + σz S ′76 = µx S 78 + σz S 78 = µx S ′76 + σz S ′76 + σ k s ′ .
Hence, σ k s ′ = 0. In a similar way, we can show that where

σ k s = 0,
D 80 i = {k j ∈ {k 1 , ..., k i-1 } ∪ H80 : E 80 k i ∩ E 80 k j ̸ = ∅}. Hence, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 80 i , • {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ if E 80 k i ∩ E 80 k ′ ̸ = ∅ (
k with v k,s ∈ H and v k,s / ∈ H such that S ′80 k = (S 80 k \ {s}) ∪ {s} for demand k such that {s -w k + 1, ..., s} ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ for each k ′ ∈ K and s ′ ∈ S ′80 k ′ with E ′80 k ∩ E ′80 k ′ ̸ = ∅.
The corresponding incidence vector (x S ′80 , z S ′80 ) belongs to F H E S H . Hence, solutions S 80 and S ′80 satisfy equation µx + σz = τ . We have so

µx S 80 + σz S 80 = µx S ′80 + σz S ′80 = µx S 80 + σz S 80 + σ k ′ s ′ -σ k s + σ k s . It follows that σ k ′ s ′ = σ k s for demand k ′ and a slot s ′ ∈ {w k ′ , ..., s} with v k ′ ,s ′ ∈ H given that σ k s = 0 for v k,s / ∈ H.
123 By (2.17) and (2.18), we know that

         µ k ′ e ′ = γ k ′ ,e ′ 1 for all k ′ ∈ K and e ′ ∈ E k ′ 0 , µ k ′ e ′ = γ k ′ ,e ′ 2 for all k ′ ∈ K and e ′ ∈ E k ′ 1 , σ k ′ s ′ = γ k ′ ,s ′ 3
for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.

We conclude that for each k ∈ K and e ∈ E

µ k e =          γ k,e 1 if e ∈ E k 0 γ k,e 2 if e ∈ E k 1 0 otherwise,
and for each k ∈ K and s ∈ S

σ k s =          γ k,s 3 if s ∈ {1, ..., w k -1} ρ if v k,s ∈ H, 0 if v k,s / ∈ H.
As a result, we have (µ, σ) = ρ(α, β) + γQ.

Theorem 2.4.15. Let H be an odd-hole, and C be a clique in the conflict graph

H E S with a) |H| ≥ 5, b) and H ∩ C = ∅, c) and the nodes (v k,s , v k ′ ,s ′ ) are linked in H E S for all v k,s ∈ H and v k ′ ,s ′ ∈ C, d) {s -w k + 1, ..., 1} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ for each (v k,s , v k ′ ,s ′ ) ∈ C and pair of nodes (v k,s , v k ′ ,s ′ ) linked in H.
Then, inequality (2.45) is facet defining for P(G, K, S) if and only if

a) for each node v k",s" in H E S with v k",s" / ∈ H ∪ C and C ∪ {v k",s" } is a clique in H E S , there exists a subset of nodes H ⊆ H of size |H|-1 2 such that H ∪ {v k",s" } is stable in H E S ,
b) and there does not exist an interval of contiguous slots

I = [s i , s j ] ⊂ [1, s] with • [ min v k,s ∈H∪C (s -w k + 1), max v k,s ∈H∪C ] ⊂ I,
• and

w k + w k ′ ≥ |I| + 1 for each (v k , v k ′ ) linked in H, • and w k + w k ′ ≥ |I| + 1 for each (v k , v k ′ ) linked in C,
• and

w k + w k ′ ≥ |I| + 1 for each v k ∈ H and v k ′ ∈ C,
• and 2w k ≥ |I| + 1 and w k ≤ |I| for each v k ∈ H,

• and

2w k ′ ≥ |I| + 1 and w k ′ ≤ |I| for each v k ′ ∈ C.
Proof. Neccessity.

We distinguish the following cases:

a) if there exists a node v k",s" / ∈ H ∪ C in H E S such that v k",s"
is linked with all nodes v k,s ∈ H and also with all nodes v k ′ ,s ′ ∈ C. This implies that inequality (2.45) can be dominated by the following valid inequality

v k,s ∈H z k s + |H| -1 2 v k ′ ,s ′ ∈C z k ′ s ′ + |H| -1 2 z k" s" ≤ |H| -1 2 . b) if
F H E S H,C = {(x, z) ∈ P(G, K, S) : v k,s ∈H z k s + |H| -1 2 v k ′ ,s ′ ∈C z k ′ s ′ = |H| -1 2 }.

Let denote inequality

v k,s ∈H z k s + |H|-1 2 v k ′ ,s ′ ∈C z k ′ s ′ ≤ |H|-1 2 by αx + βz ≤ λ.
Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and

F = {(x, z) ∈ P(G, K, S) : µx + σz = τ }. Suppose that F H E S H,C ⊆ F . To prove that F H E S
H,C is a facet of P(G, K, S), it suffices to show that there exist ρ ∈ R and γ = (γ 1 , γ 2 , γ 3 ) (such that where

γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α, β) + γQ.
D 82 i = {k j ∈ {k 1 , ..., k i-1 } ∪ H82 : E 82 k i ∩ E 82 k j ̸ = ∅}. Hence, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 82 i , • {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ if E 82 k i ∩ E 82 k ′ ̸ = ∅ (
we take into account the possibility of adding slot s ′ in the selected set of last slots S 82 k ′ to route demand k ′ in solution S 82 ). We let S 82 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S 82 is feasible for the problem. Hence, the corresponding incidence vector (x S 82 , z

k ∈ S 82 k with v k,s k ∈ H and v k,s k / ∈ H ∪ C such that S 83 k = (S 82 k \ {s k }) ∪ {s k } for each demand k ∈ { k ∈ K with v k,s ∈ H82 } such that {s -w k + 1, ..., s} ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ for each k ′ ∈ K and s ′ ∈ S 83 k ′ with E 83 k ∩ E 83 k ′ ̸ = ∅.
Solution S 83 is feasible for the problem. The corresponding incidence vector (x S

S 82 + σ k ′ s ′ - (k,s k )∈ H82 σ k s k + k∈K H σ k sk .
where

K H = { k ∈ K with v k,s ∈ H82 }. Since σ k sk = 0 for v k,s k / ∈ H ∪ C, it follows that σ k ′ s ′ = (k,s k )∈ H82 σ k s k . As a result, σ k ′ s ′ = ρ |H|-1 2
given that σ k s are equivalent for all v k,s ∈ H. Given that the pair v k ′ ,s ′ is chosen arbitrarily in clique C, we re-do the same procedure for all v k ′ ,s ′ ∈ C. Consequently, we obtain that

σ k ′ s ′ = ρ |H|-1 2 for all v k ′ ,s ′ ∈ C.
Overall, and using the results (2.17) and (2.18), we obtain that

µ k e =          γ k,e 1 if e ∈ E k 0 , γ k,e 2 if e ∈ E k 1 , 0 otherwise,
for each k ∈ K and e ∈ E, and

σ k s =                  γ k,s 3 if s ∈ {1, ..., w k -1}, ρ if v k,s ∈ H, ρ |H|-1 2 if v k,s ∈ C, 0 otherwise,
for each k ∈ K and s ∈ S.

As a consequence, we obtain that (µ, σ) = ρ(α, β) + γQ.

Let us now introduce some valid inequalities that are related to the routing sub-problem issus from the transmission-reach constraint.

Incompatibility-Clique Inequalities

C ′ ⊆ C with |C ′ | = |C| -1 v k,e ∈C ′
x k e ≤ 1.

By adding the previous inequalities for all

C ′ ⊆ C with |C ′ | = |C| -1, we get C ′ ⊆C |C ′ |=|C|-1 v k,e ∈C ′ x k e ≤ C ′ ⊆C |C ′ |=|C|-1 1.
Note that for each demand k and edge e with v k,e ∈ C, the variable x k e appears ( |C| |C|-1 -1) = |C| -1 times in the previous sum. It follows that

v k,e ∈C (|C| -1)x k e ≤ |C|.
By dividing the two sides of the previous sum by |C| -1, we have so

v k,e ∈C x k e ≤ |C| |C| -1 ⇒ v k,e ∈C x k e ≤ 1 given that |C| |C| -1 = 1.
This ends the proof.

F H K E C = {(x, z) ∈ P(G, K, S) : v k,e ∈C
x k e = 1}.

Let denote inequality v k,e ∈C x k e ≤ 1 by αx + βz ≤ λ. Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) : µx + σz = τ }.

Suppose that

F H K E C
⊆ F . In order to prove that inequality v k,e ∈C x k e ≤ 1 is facet defining for P(G, K, S), we show that there exist ρ ∈ R and γ = (γ 1 , γ 2 , γ 3 ) (such that where As a result, µ k e = 0. In a similar way, we can show that

γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α, β) + γQ. We first show that µ k e = 0 for each edge e ∈ E \ (E k 0 ∪ E k 1 ) for each demand k ∈ K with v k,e / ∈ C
D 84 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E 84 k i ∩ E 84 k j ̸ = ∅}. Hence, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 84 i , • {s k i -w k i + 1, ..., s k i } ∩ {s k -w k + 1, ..., s k } = ∅ if E 84 k i ∩ (E 84 k ∪ {e}) ̸ = ∅ (
µ k e = 0, for all k ∈ K and e ∈ E \ (E k 0 ∪ E k 1 ) with v k,e / ∈ C.
Let show that σ k s = 0 for all k ∈ K and s ∈ {w k , ..., s}. Consider a demand k in K and a slot s ′ in {w k , ..., s}, and a solution S ′84 = (E ′84 , S ′84 ) such that a) select one pair of demand k ′ and edge e ′ from clique C (i.e., v k ′ ,e ′ ∈ C), we let E ′84 k ′ be the set of edges involved in a shortest path between o k ′ and d k ′ which uses edge e ′ , b) for each pair of demand k" and edge e" with v k",e" ∈ C \{v k,e }, we let E ′84 k" be the set of edges involved in a shortest path between o k" and d k" which uses edge e" which does not use edge e", c) for each demand k i ∈ K \ C with i ∈ {1, ..., |K|} \ {k}, we let E ′84 k i be the set of edges involved in a shortest path between o k i and d k i , d) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′84 i given by

I ′84 i = [ kj ∈D ′84 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}] if E ′84 ki ∩ E ′84 k ̸ = ∅ or I ′84 i = kj ∈D ′84 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not.
where

D ′84 i = {k j ∈ {k 1 , ..., k i-1 } : E ′84 k i ∩ E ′84 k j ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D ′84 i , • {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E ′84 k i ∩ E ′84 k ̸ = ∅ (
we take into account the possibility of adding slot s ′ as a last slot in the set of last slots S ′84 k to route demand k in solution S ′84 ). We let S ′84 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′84 is feasible for the problem. Hence, the corresponding incidence vector (x S ′84 , z ) such that v k ′ ,e ′ and v k,e" linked in C, c) modifying the last slots assigned to some demands K ⊂ K from S 84 k to S 87 k for each k ∈ K while satisfying non-overlapping constraint.

The paths assigned to the demands K \ {k, k ′ } in S 84 remain the same in S 87 (i.e., E 87 k" = E 84 k" for each k" ∈ K \ {k, k ′ }), and also without modifying the last slots assigned to the demands K \ K in S 84 µ k e" .

Since µ k e" = 0 for all k ∈ K and e" ∈ E \ (E k 0 ∪ E k 1 ) with v k,e" / ∈ C, and σ k s = 0 for all k ∈ K and s ∈ {w k , ..., s}, it follows that µ k ′ e ′ = µ k e . Given that the pair (v k,e , v k ′ ,e ′ ) are chosen arbitrarily in clique C, we re-do the same procedure for all pairs (v k,e , v k ′ ,e ′ ) such that we find

µ k e = µ k ′ e ′ , for all pairs (v k,e , v k ′ ,e ′ ) ∈ C.
Consequently, we obtain that µ k e = ρ for all v k,e ∈ C. By (2.17) and (2.18), we know that

         µ k ′ e ′ = γ k ′ ,e ′ 1 for all k ′ ∈ K and e ′ ∈ E k ′ 0 , µ k ′ e ′ = γ k ′ ,e ′ 2 for all k ′ ∈ K and e ′ ∈ E k ′ 1 , σ k ′ s ′ = γ k ′ ,s ′ 3 for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.
We conclude that for each k ∈ K and e ∈ E As a consequence, (µ, σ) = ρ(α, β) + γQ.

µ k e =                  γ k,e 1 if e ∈ E k 0 , γ k,e 2 if e ∈ E k 1 , ρ if v k,e ∈ C,

Incompatibility-Odd-Hole Inequalities

(v k e ,v k ′ e ′ )∈E(H) x k e + x k ′ e ′ ≤ |H| -1.
Taking into account that each node v k e in H has two neighbors in H, this implies that x k e appears twice in the previous inequality. As a result,

(v k e ,v k ′ e ′ )∈E(H) x k e + x k ′ e ′ = v k e ∈H 2x k e =⇒ v k e ∈H 2x k e ≤ |H| -1 =⇒ v k e ∈H x k e ≤ |H| -1 2 = |H| -1 2 since |H| is an odd number.
We conclude at the end that inequality (2.47) is valid for P(G, K, S). Then, the inequality

Inequality
v k e ∈H
x k e + |H| -1 2

v k ′ e ′ ∈C x k ′ e ′ ≤ |H| -1 2 , (2.48)
is valid for P(G, K, S).

Proof. It is trivial given the definition of the odd-hole and clique in H

K E such that if v k ′ e ′ ∈C x k ′ e ′ = 1 for a v k ′ e ′ ∈ C, which implies that the quantity v k e ∈H
x k e is forced to be equal to 0. Otherwise, we know from inequality (2.47) that the sum v k e ∈H x k e should be smaller than |H|-1 2 . We strengthen the proof as belows. For each pair of nodes (v k,e , v k ′ ,e ′ ) linked in H by an edge, we know that x k e + x k ′ e ′ + v k",e" ∈C x k" e" ≤ 1 given that all the nodes v k",e" ∈ C are linked with the nodes v k,e and v k ′ ,e ′ . Given that H is an odd-hole which means that we have |H| -1 pair of nodes (v k,e , v k ′ ,e ′ ) linked in H, and by doing a sum for all pairs of nodes (v k,e , v k ′ ,e ′ ) linked in H, it follows that

(v k,e ,v k ′ ,e ′ )∈E(H) x k e + x k ′ e ′ + v k",e" ∈C
x k" e" ≤ |H| -1.

Taking into account that each node v k,e has two neighbors in H, this implies that

x k e appears twice in the previous inequality. The sum v k",e" ∈C x k" e" appears |H| -1 times in in the previous inequality. As a result,

(v k,e ,v k ′ ,e ′ )∈E(H) x k e + x k ′ e ′ + (|H| -1) v k",e" ∈C x k" e" ≤ |H| -1 ⇒ v k,e ∈H 2x k e + (|H| -1) v k",e" ∈C
x k" e" ≤ |H| -1.

By dividing the two sides of the previous sum by 2, and since |H| is an odd number, it follows that

v k,e ∈H x k e + |H| -1 2 v k",e" ∈C x k" e" ≤ |H| -1 2 = |H| -1 2 .
We conclude at the end that inequality (2.48) is valid for P(G, K, S).

v k ′ ,e ′ ∈H ′ x k ′ e ′ ≤ |H ′ |-1 2 . b) if there exists a node v k ′ ,e ′ / ∈ H in H K E such that v k ′ ,e ′ is
linked with all nodes v k,e ∈ H. This implies that inequality (2.47) can be dominated by the following valid inequality

v k,e ∈H x k e + |H| -1 2 x k ′ e ′ ≤ |H| -1 2 .
If no one of these cases is verified, inequality (2.47) can never be dominated by another inequality without changing its right-hand side. Otherwise, inequality (2.47)

is not facet defining for P(G, K, S).

Sufficiency.

Let F

H K E H
denote the face induced by inequality (2.47), that is

F H K E H = {(x, z) ∈ P(G, K, S) : v k,e ∈H
x k e = |H| -1 2 }. Suppose that F

Denote inequality v

H K E H ⊆ F .
In order to prove that inequality v k,e ∈H x k e = |H|-1 2 is facet defining for P(G, K, S), we show that there exist ρ ∈ R and γ = (γ 1 , γ 2 , γ 3 )

(such that γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α, β)+γQ.
Let first show that µ k e = 0 for each edge e where As a result, µ k e = 0. In a similar way, we can show that

∈ E \ (E k 0 ∪ E k 1 ) for each demand k ∈ K with v k,e / ∈ H
D 88 i = {k j ∈ {k 1 , ..., k i-1 } : E 88 k i ∩ E 88 k j ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 88 i , • {s k i -w k i + 1, ..., s k i } ∩ {s k -w k + 1, ..., s k } = ∅ if E 88 k i ∩ (E 88 k ∪ {e}) ̸ = ∅ (
µ k e = 0, for all k ∈ K and e ∈ E \ (E k 0 ∪ E k 1 ) with v k,e / ∈ H.
Let show that σ k s = 0 for all k ∈ K and s ∈ {w k , ..., s}. Consider a demand k in K and a slot s ′ in {w k , ..., s}. Let S ′88 = (E ′88 , S ′88 ) be the solution given by a) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E ′88 k i be the set of edges involved in a shortest path between o k i and where

d k i , b) select a subset of nodes H′88 from H with | H′88 | = |H|-1 2 ,
D ′88 i = {k j ∈ {k 1 , ..., k i-1 } : E ′88 k i ∩ E ′88 k j ̸ = ∅}. This ensures that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D ′88 i , • {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E ′88 k i ∩ E ′88 k ̸ = ∅ (
we take into account the possibility of adding slot s ′ as a last slot in the set of last slots S ′88 k to route demand k in solution S ′88 ).

We let S ′88

k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′88 is feasible for the problem. Hence, the corresponding incidence vector (x S ′88 , z S ′88 ) belongs to F H K E H . After that, we derive solution S 90 obtained from S ′88 by adding slot s ′ as last slot to demand k in S ′88 . Solution S 90 is clearly feasible for the problem.

The corresponding incidence vector (x S 90 , z S 90 ) belongs to F H K E H . Hence, solutions S ′88 and S 90 satisfy equation µx + σz = τ . We have so

µx S ′88 + σz S ′88 = µx S 90 + σz S 90 = µx S ′88 + σz S ′88 + σ k s ′ .
Hence, σ k s ′ = 0. In a similar way, we can show that µ k e" .

σ k s = 0,
Since µ k e" = 0 for all k ∈ K and e" ∈ E \ (E k 0 ∪ E k 1 ) with v k,e" / ∈ H, and σ k s = 0 for all k ∈ K and s ∈ {w k , ..., s}, it follows that µ k ′ e ′ = µ k e . Given that the pair (v k,e , v k ′ ,e ′ ) are chosen arbitrarily in odd-hole H, we re-do the same procedure for all pairs (v k,e , v k ′ ,e ′ ) such that we find

µ k e = µ k ′ e ′ , for all pairs (v k,e , v k ′ ,e ′ ) ∈ H.
Consequently, we obtain that µ k e = ρ for all v k,e ∈ H. We know from (2.17) and (2.18) that

         µ k ′ e ′ = γ k ′ ,e ′ 1 for all k ′ ∈ K and e ′ ∈ E k ′ 0 , µ k ′ e ′ = γ k ′ ,e ′ 2 for all k ′ ∈ K and e ′ ∈ E k ′ 1 , σ k ′ s ′ = γ k ′ ,s ′
We then conclude that for each k ∈ K and e ∈ E

µ k e =                  γ k,e 1 if e ∈ E k 0 , γ k,e 2 if e ∈ E k 1 , ρ if v k,e ∈ H, 0 otherwise,
and for each k ∈ K and s ∈ S

σ k s =    γ k,s 3 if s ∈ {1, ..., w k -1}, 0 otherwise.
As a consequence, (µ, σ) = ρ(α, β) + γQ.

Theorem 2.4.18. Let H be an odd-hole, and C be a clique in the conflict graph Proof. Neccessity.

H K E with a) |H| ≥ 5, b) and H ∩ C = ∅, c)
If there exists a node v k",e" / ∈ H ∪ C in H K E such that v k",e" is linked with all nodes v k,e ∈ H and also with all nodes v k ′ ,e ′ ∈ C. This implies that inequality (2.48) can be dominated by the following valid inequality

v k,e ∈H x k e + |H| -1 2 v k ′ ,e ′ ∈C x k ′ e ′ + |H| -1 2 x k" e" ≤ |H| -1 2 .
As a result, inequality (2.48) is not facet defining for P(G, K, S).

Sufficiency.

Let F

H K E H,C be the face induced by inequality (2.48), that is

F H K E H,C = {(x, z) ∈ P(G, K, S) : v k,e ∈H x k e + |H| -1 2 v k ′ ,e ′ ∈C x k ′ e ′ = |H| -1 2 }. Let denote inequality v k,e ∈H x k e + |H|-1 2 v k ′ ,e ′ ∈C x k ′ e ′ ≤ |H|-1 2 by αx + βz ≤ λ.
Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) : µx + σz = τ }. Suppose that F

H K E H,C ⊆ F . To prove that F H K E H,C
is a facet of P(G, K, S), we need to show that there exists ρ ∈ R and γ = (γ 1 , γ 2 , γ 3 ) (such that c) modifying the last slots assigned to some demands K ⊂ K from S 92 k to S ′93 k for each k ∈ K while satisfying non-overlapping constraint.

γ k,e 1 ∈ R for all k ′ ∈ K and e ∈ E k ′ 0 , γ k,e 2 ∈ R for all k ′ ∈ K and e ∈ E k ′ 1 , γ k ′ ,s ′ 3 ∈ R for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}) such that (µ, σ) = ρ(α,
The paths assigned to the demands K \ (K(H 92 ) ∪ {k ′ }) in S 92 remain the same in S ′93 (i.e., E ′93 k" = E 92 k" for each k" ∈ K \ {k, k ′ }), and also without modifying the last slots assigned to the demands K \ K in S 92 , i.e., S 92 k = S ′93 k for each demand k ∈ K \ K. Solution S ′93 is feasible for the problem. The corresponding incidence vector (x S ′93 , z S ′93 ) belongs to F H K E H,C . Hence, solutions S 92 and S ′93 satisfy equation µx + σz = τ . We have so

µx S 92 + σz S 92 = µx S ′93 + σz S ′93 = µx S 92 + σz S 92 + µ k ′ e ′ - v k,e ∈H 92 µ k e + k∈ K s ′ ∈S ′93 k σ k s ′ - s∈S 92 k σ k s + e"∈E ′93 k ′ \{e ′ } µ k ′ e" - e"∈E 92 k ′ µ k ′ e" + e"∈E ′93 k µ k e" - k∈K(H 92 ) e"∈E 92 k µ k e" .
Since µ k e" = 0 for all k ∈ K and e" ∈ E \ (E k 0 ∪ E k 1 ) with v k,e" / ∈ H ∪ C, and σ k s = 0 for all k ∈ K and s ∈ {w k , ..., s}, it follows that µ k ′ e ′ = v k,e ∈H 92 µ k e . As a result, µ k ′ e ′ = ρ |H|-1 2 . Given that the pair v k ′ ,e ′ is chosen arbitrarily in clique C, we re-do the same procedure for all pairs v k ′ ,e ′ ∈ C such that we find

µ k ′ e ′ = ρ |H| -1 2 , for all pairs v k ′ ,e ′ ∈ C.
As a result, all µ k ′ e ′ ∈ C are equivalent such that

µ k ′ e ′ = µ k" e" = ρ |H| -1 2
, for all pairs v k ′ ,e ′ , v k",e" ∈ C.

By (2.17) and (2.18), we know that

         µ k ′ e ′ = γ k ′ ,e ′ 1 for all k ′ ∈ K and e ′ ∈ E k ′ 0 , µ k ′ e ′ = γ k ′ ,e ′ 2 for all k ′ ∈ K and e ′ ∈ E k ′ 1 , σ k ′ s ′ = γ k ′ ,s ′ 3 for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.
As a result, we obtain that for each k ∈ K and e ∈ E

µ k e =                        γ k,e 1 if e ∈ E k 0 , γ k,e 2 if e ∈ E k 1 , ρ if v k,e ∈ H, ρ |H|-1 2 if v k,e ∈ C, 0 otherwise,
and for each k ∈ K and s ∈ S

σ k s =    γ k,s 3 if s ∈ {1, ..., w k -1}, 0 otherwise.
As a result, we have (µ, σ) = ρ(α, β) + γQ.

Tranmission

P(G, K, S, C, k) = {(x, z) ∈ P(G, K, S) : e ′ ∈E\(E k 1 ∪E k 0 )
x k e ′ = 0}.

Proof. It is trivial given that inequality (2.49) can never be dominated in P(G, K, S, C, k).

On the other hand, one can use sequential lifting procedure [START_REF] Balas | Facets of the Knapsack Polytope From Minimal Covers[END_REF] to sequentially lift the inequality (2.49) and generate lifted valid inequalities that are facet defining for the polytope P(G, K, S) as follows.

Theorem 2.4.20. Let C be a minimal cover for a demand

k ∈ K. Let E \ (E k 1 ∪ C ∪ E k 0 ) = {e 1 , ..., e q } be arbitrarily ordred with q = |E \ (E k 1 ∪ C ∪ E k 0 )|.
Consider a sequence of knapsack problems defined as

                     z i = max j∈C u j + i-1 j=1 α j u j , j∈C l j u j + i-1 j=1 l j u j ≤ lk - e ′ ∈E k 1 ℓ e ′ -l e i , u j ∈ {0, 1}, ∀j ∈ C ∪ {1, ..., i -1}, (2.50)
for all i ∈ {1, ..., q} with α j = |C| -1 -z j for all j ∈ {1, ..., i -1}. Then, the inequality e∈C

x k e + q j=1 α j x k e j ≤ |C| -1, (2.51)
is valid for P(G, K, S). Moreover, it's facet defining for P(G, K, S).

Proof. It's trivial given that inequality (2.51) can never be dominated in P(G, K, S).

Definition 2.4.10. Consider a demand k ∈ K. Let p be a sub-path in G such that each pair of edges

(v k,e , v k,e ′ ) ∈ E \(E k 0 ∪E k 1 )
are not linked by an edge in the conflict graph HK E . We say that the path p is infeasible for the demand k if it does not occur as a subpath in any feasible routing for the demand k, i.e., there does not exist a feasible path for demand k containning p due to the transmission-reach constraint.

Moreover, it is said to be minimal infeasible if each sub-path p ′ of p of cardinality

|E(p ′ )| = |E(p)| -1,
can be used in a feasible routing for the demand k.

Note that one can verify in polynomial time using Dijkstra algorithm if a subpath p in G if it is infeasible or not for a demand k ∈ K. is valid for P(G, K, S).

Proof. It is trivial given that p is minimal infeasible sub-path for demand k this means that there are at most |E(p)| -1 edges from the set of edges in E(p) that can be used to route demand k.

Edge-Capacity-Cover Inequalities

Let provide now some inequalities related to the capacity constraint over edges.

Proposition 2.4.25. Consider an edge e in E. Then, the inequality

k∈K\Ke w k x k e ≤ s - k ′ ∈Ke w k ′ , (2.53)
is valid for P(G, K, S).

Proof. The number of slots allocated in edge e ∈ E should be less than the residual capacity of edge e which is equal to s -

k ′ ∈Ke w k ′ .
Based on this, we introduce the following definitions.

Definition 2.4.11. For an edge e ∈ E, a subset of demands

C ⊆ K with e / ∈ E k 0 ∩E k 1 For each demand k ∈ C, is said a cover for edge e if k∈C w k > s - k ′ ∈Ke w k ′ .
Moreover, it is said to be a minimal cover if

k ′ ∈C\{k} w k ′ ≤ s - k"∈Ke w k" . Proposition
k ′ ∈K\(C∪Ke) x k ′ e = 0}.
Proof. It is trivial given that inequality (2.54) can never be dominated in P(G, K, S, C, e).

One can use the sequential lifting procedure [START_REF] Balas | Facets of the Knapsack Polytope From Minimal Covers[END_REF] to sequentially lift the inequality (2.54) and generate lifted facets for the polytope P(G, K, S) as follows.

Theorem 2.4.22. Let C be a minimal cover for an edge e ∈ E. Let K \ (K e ∪ C ∪ Ke ) = {k 1 , ..., k q } be arbitrarily ordred with q = |K \ (K e ∪ C ∪ Ke )|. Consider a sequence of knapsack problems defined as

                     z i = max j∈C u j + i-1 j=1 α j u j , j∈C w j u j + i-1 j=1 w k j u j ≤ s - k ′ ∈Ke w k ′ -w k i , u j ∈ {0, 1}, ∀j ∈ C ∪ {1, ..., i -1}, (2.55)
for all i ∈ {1, ..., q} with α j = |C| -1 -z j for all j ∈ {1, ..., i -1}. Then, the inequality k∈C

x k e + q j=1 α j x k j e ≤ |C| -1, (2.56)
is valid for P(G, K, S). Moreover, it's facet defining for P(G, K, S).

Proof. It's trivial given that inequality (2.56) can never be dominated in P(G, K, S).

Symmetry-Breaking Inequalities

We have noticed that several symmetrical solutions may appear given that there exist several feasible solutions that have the same value of the solution (called equivalents solutions), and they can be found by doing some permutations between the slots assigned to some demands without changing the selected paths (routing) while satisfying the C-RSA constraints. There exists several methods to break the symmetry. See, for example, perturbation method proposed by Margot [START_REF] Margot | Symmetry in integer linear programming[END_REF], isomorphism pruning method by Margot et al. [START_REF] Margot | Pruning by isomorphism in branch-and-cut[END_REF][68], orbital branching method by Ostrowski et al. [START_REF] Ostrowski | Symmetry in scheduling problems[END_REF][76], orbital fixing method by Kaibel et al. [START_REF] Kaibel | Orbitopal fixing[END_REF], and symmetry-breaking constraints by Kaibel and Pfetsch [START_REF] Kaibel | Packing and partitioning orbitopes[END_REF] which is applied in our study. We aim to introduce breaking-symmetry inequalities to remove the sub-problems in the enumeration tree that are equivalent due to the equivalency of their associated solutions.

For this, we derive the following inequalities.

Proposition 2.5.1. Consider a demand k, slot s ∈ {1, ..., s -1}. Let s ′ be a slot in {s, ..., s}

min(s ′ +w k -1,s) s"=s ′ z k s" - k ′ ∈K min(s+w k ′ -1,s) s"=s z k ′ s" ≤ 0. (2.57)
This ensures that slot s ′ can be assigned to demand k if and only if slot s (which precedes slot s ′ ) is already assigned to at least one demand k ′ in K. A similar idea was proposed by Mendez-Diaz and Zabala [START_REF] Méndez-Díaz | A Branch-and-Cut algorithm for graph coloring[END_REF] to break the symmetry for the vertex coloring problem. Note that inequalities (5.17) are not valid for the polytope P(G, K, S) given that they cut off some feasible regions in the polytope P(G, K, S). In any case, we ensure that there exists at least one optimal solution from our original problem that remains feasible and belongs to the convex hull of non-symmetric solutions of the C-RSA problem.

Lower Bounds

In this section, we derive some lower bounds for the C-RSA. Let p * k denote the minimum-cost path between origin node o k and destination node d k for demand k with total length smaller than the transmission-reach lk . We know in advance that the optimal path chosen for each demand k ∈ K in the optimal solution, its total cost is at least equal to the total cost of the minimum-cost path p * k . Based on this, we introduce the following inequalities. is valid for P(G, K, S).

Proof. It's trivial given that in any feasible solution S in P(G, K, S), the total cost of the path selected to demand k is greatest than or equal to the total cost of the minimum-cost path p * k .

Inequality (2.58) is then used to derive a lower bounds for the C-RSA as follows. Proof. It's trivial given that the optimal value is at least equal to the sum of the total cost of minimum-cost path over all the demands in K.

The separation problem associated with inequality (2.59) is equivalent to solving the Resource Constrained Shortest Path (RCSP) Problem for each demand k.

The RCSP is well known to be a NP-hard problem [START_REF] Dror | Note on the Complexity of the Shortest Path Models for Column Generation in VRPTW[END_REF]. For this, we propose a pseudo-polynomial time algorithm using dynamic programming [START_REF] Dumitrescu | Algorithms for the weight constrained shortest path problem[END_REF] to compute the minimum-cost path for each demand k while satisfying the transmission-reach constraint. For each demand k ∈ K, we associate to each node v ∈ V \ V k 0 in the graph G a set of labels L v such that each label corresponds to differents paths from th origin node o k to the node v, and each label p is specified by a cost equals to e∈E(p) c e , and a weight equals to e∈E(p) ℓ e . We denote by T v the set of labels on node v ∈ V . For each demand k and slot s ∈ {w k , ..., s}, the complexity of the algorithm is bounded by O(|E \ E k 0 | * lk ) [START_REF] Dumitrescu | Algorithms for the weight constrained shortest path problem[END_REF]. Algorithm 1 summarizes the different steps of the dynamic programming algorithm.

Chapter 3

Branch-and-Cut Algorithm for the C-RSA Problem

Based on theoretical results presented in chapter (2), we devise a Branch-and-Cut algorithm to solve the C-RSA problem. Our aim is to study the effectiveness of the algorithm, and assess the impact of the valid inequalities on the effectiveness of the Branch-and-Cut algorithm. First, we give an overview of the algorithm. Then, we describe the separation procedure used for each valid inequality based on exact algorithms, greedy-algorithms, and heuristics. At the end, we provide a detailed behavior study of the Branch-and-Cut algorithm.

Branch-and-Cut Algorithm

Description

In what follows, we describe the Branch-and-Cut algorithm. Consider an undirected, loopless, and connected graph G = (V, E), which is specified by a set of nodes V , and a multiset E of links. Each link e = ij ∈ E is associated with a length ℓ e ∈ R + (in kms), a cost c e ∈ R + such that each link e ∈ E is divided into s ∈ N + slots. Let S = {1, . . . , s} be an optical spectrum of available frequency slots with s ≤ 320, and K be a multiset of demands such that each demand k ∈ K is specified by an origin node

o k ∈ V , a destination node d k ∈ V \ {o k }, a slot-width w k ∈ Z + ,
x k e = 0, ∀k ∈ K, ∀e ∈ E k 0 , x k e = 1, ∀k ∈ K, ∀e ∈ E k 1 , z k s = 0, ∀k ∈ K, ∀s ∈ {1, ..., w k -1}, s s=w k z k s = 1, ∀k ∈ K, 0 ≤ x k e ≤ 1, ∀k ∈ K, ∀e ∈ E, 0 ≤ z k s ≤ 1, ∀k ∈ K, ∀s ∈ S.

Test of Feasibility

Given an optimal solution (x, z) for the relaxation of LP 0 . It is feasible for the C-RSA problem if and only if (x, z) is integral and it satisfies the cut inequalities (2.2)

and non-overlapping inequalities (2.6). Usually, (x, z) does not satisfy inequalities (2.2) and (2.6). As a result, (x, z) is not feasible for the C-RSA problem. For this, we generate several valid inequalities violated by a solution (x, z) at each iteration of the Branch-and-Cut algorithm. This is known under the name of Separation Procedure.

It consists in identifying for a given class of valid inequalities the existence of one

or more inequalities of this class that are violated by the current solution. We repeat this procedure in each iteration of the algorithm until no violated inequality is identified. As a result, the final solution is optimal for the linear relaxation of the cut formulation. Furthermore, if it is integral, then it is optimal for the C-RSA problem. Otherwise, we create two subproblems called childs by branching on a fractional variable (variable branching rule) or on some constraints using the Ryan & Foster branching rule (constraint branching rule). Based on this, we devise a basic Branch-and-Cut algorithm by combining cutting-plane algorithm based on the separation of the cut inequalities (2.2) and non-overlapping inequalities (2.6), and a Branch-and-Bound algorithm.

On the other hand, to make more efficient the Branch-and-Cut algorithm, we already introduced several classes of valid inequalities used to obtain tighter LP bounds.

Based on this, and at each iteration in each node of the Branch-and-Cut tree, one can identify one or more than one violated inequality by the current fractional solution for a given class of valid inequalities. Algorithm 2 summarizes the different steps of Branch-and-Cut algorithm taking into account additional valid inequalities for a given class of valid inequalities.

For this, we study the separation problem of each class of valid inequality introduced in this dissertation as follows. Consider a fractional solution (x, z).

Separation of Non-Overlapping Inequalities

Let e be an edge in E and s a slot in S.

s ′ =s zk s ′ + min(s+w k ′ -1,s) s"=s zk ′ s" > 3.
For this, we propose an exact algorithm in O(|E| * s * |K| * log(|K|)) which works as follows. We select each pair of demands k, k ′ ∈ K with x k e > 0,

min(s+w k -1,s) s ′ =s z k s ′ > 0, xk ′ e > 0 and min(s+w k ′ -1,s) s"=s zk ′ s" > 0.
We then add the following inequality induced by each selected pair of demands k, k ′ for slot s over edge e to the current LP if it is violated

x k e + x k ′ e + min(s+w k -1,s) s ′ =s z k s ′ + min(s+w k ′ -1,s) s"=s z k ′ s" ≤ 3.
Otherwise, we conclude that such inequality does not exist for the current solution (x, z). On the other hand, given that inequalities (2.5) are taken in format of equalities when implementing the B&C algorithm (i.e., s s=w k z k s = 1 for all k ∈ K). Based on this, and taking into account the non-overlapping inequalities (2.6), we propose a new non-overlapping inequality (3.1) more efficient compared to the ones of (2.6). Proposition 3.1.1. Consider an edge e, and a pair of demands k, k

′ ∈ K with e / ∈ E k 0 ∪ E k ′ 0 .
Let s be a slot in {w k , ..., s}. Then, the inequality

x k e + x k ′ e + z k s + min(s+w k ′ -1,s) s"=s-w k +1 z k ′ s" ≤ 3, (3.1) is valid for Q(G, K, S).
The separation problem associated with inequality (3. zk ′ s" > 0. We then add the following inequality to the current LP if it is violated

x k e + x k ′ e + z k s + min(s+w k ′ -1,s) s"=s-w k +1 z k ′ s" ≤ 3.
Otherwise, we conclude that there does not exist an inequality from the non-overlapping inequalities (3.1) violated by the current solution (x, z). Note that, from an efficiency point of view, inequalities (3.1) replace inequalities (2.6) in the B&C algorithm.

Separation of Cut Inequalities

In this section we discuss the separation problem of the cut inequalities (2.2). Its associated separation problem consists in identifying a cut inequalities (2.2) that is violated by (x, z). For each demand k ∈ K, this can be done in polynomial time [START_REF] Ford | Maximal flow through a network[END_REF] as shown in theorem of Ford and Fulkerson by finding a minimum cut separating the origin-node o k and destination-node d k . As a result, this can be done exactly [START_REF] Ford | Maximal flow through a network[END_REF] and very effectively in

O(|V \ V k 0 | 2 * |E \ E k 0 |
) for each demand k using an efficient implementation of minimum cut algorithm based on the so-called preflow push-relabel algorithm of Goldberg and Tarjan [START_REF] Goldberg | A New Approach to the Maximum Flow Problem[END_REF]

Separation of Edge-Slot-Assignment Inequalities

Consider an edge e ∈ E and a slot s ∈ S. The separation problem associated with inequality (2.23) consists in identifying a subset of demands K * ⊂ K such that

k∈ K * xk e + min(s+w k -1,s) s ′ =s zk s ′ > | K * | + 1.
For this, we propose an exact algorithm in O(|K| * |E| * s) which works as follows.

The main idea is to iteratively add each demand k ∈ K to K * if and only if x k e > 0 and

min(s+w k -1,s) s ′ =s z k s ′ > 0.
We then add the following inequality induced by K * to the current LP if it is violated and satisfies some conditions about validity of inequality (2.23) k∈ K *

x k e + min(s+w k -1,s)

s ′ =s z k s ′ ≤ | K * | + 1.
Otherwise, we conclude that such inequality does not exist for the current solution (x, z). Moreover, if such violated inequality is identified, it can be easily lifted by introducing inequality (2.25) induced by K * and a subset of demands K e \ K * as follows k∈ K *

x k e + min(s+w k -1,s)

s ′ =s z k s ′ + k ′ ∈Ke\ K * min(s+w k ′ -1,s) s ′ =s ≤ | K * | + 1.
Remark 3.1.1. Inequality (3.1) is a particular case of inequality (2.42) for a clique

C = {v k,s } ∪ {v k ′ ,s ′ ∈ H e S such that {s ′ -w ′ k + 1, ..., s ′ } ∩ {s -w k + 1, ..., s} ̸ = ∅}.

Separation of Edge-Slot-Assignment-Clique Inequalities

Consider an edge e ∈ E. The separation algorithm for inequality (2.42) To do this, we use the greedy algorithm introduced by Nemhauser and Sigismondi [START_REF] Nemhauser | A Strong Cutting Plane/Branch-and-Bound Algorithm for Node Packing[END_REF] to identify a maximal clique C * in the conflict graph H e S given that computing a maximal clique in such a graph is also NP-hard problem [START_REF] Karp | Reducibility among Combinatorial Problems[END_REF]. Based on this, we first assign a positive weight zk s * xk e to each node v k,s in the conflict graph H e S . We then select a node v k,s in the conflict graph H e S having the largest weight compared with the other nodes in H e S , and set C * = {v k,s }. After that, we iteratively add each node v k ′ ,s ′ to the current C * if it is linked with all the nodes v k,s already assigned to the current clique C * and zk ′ s ′ > 0 and xk ′ e > 0. At the end, we add inequality (2.42) induced by clique C * for edge e to the current LP if it is violated and satisfies some conditions about validity of inequality (2.42). Hence, we add the following inequality

v k,s ∈C * x k e + z k s ≤ |C| + 1.
Furthermore, it can be lifted by identifying a maximal clique N * such that each

v k ′ ,s ′ ∈ N * is linked with all the nodes v k,s ∈ C * ∪ (N * \ {v k ′ ,s ′ }) in H e S .
For this, we also use the greedy algorithm introduced by Nemhauser and Sigismondi [START_REF] Nemhauser | A Strong Cutting Plane/Branch-and-Bound Algorithm for Node Packing[END_REF] to identify clique N * as follows. We first set

N * = {v k ′ ,s ′ } with v k ′ ,s ′ / ∈ C *
) + v k ′ ,s ′ ∈N * z k ′ s ′ ≤ 1.

Separation of Edge-Interval-Clique Inequalities

Let discuss the separation problem of inequality (2.32). Consider an edge e ∈ E.

We first construct a set of intervals of contiguous slots I ∈ I e such that each interval of contiguous slots I e is identified by generating two slots s i and s j randomly in S with s j ≥ s i + 2 max k∈K\ Ke w k . Consider now an interval of contiguous slots I = [s i , s j ] ∈ I e over edge e. The separation problem associated with inequality (2.32) is NP-hard [START_REF] Klabjan | The complexity of cover inequality separation[END_REF] given that it consists in identifying a cover K * for the interval

I = [s i , s j ] over edge e, such that k∈ K * xk e + s j s ′ =s i +w k -1 zk s ′ > 2| K * | -1.
For this, we use a greedy algorithm introduced by Nemhauser and Sigismondi [START_REF] Nemhauser | A Strong Cutting Plane/Branch-and-Bound Algorithm for Node Packing[END_REF] as follows. We first select a demand k ∈ K having the largest number of requested slot w k with xk e > 0 and

s j s ′ =s i +w k -1
zk s ′ > 0, and then set K * to K * = {k}. After that, we iteratively add each demand k ′ ∈ K \ K * to K * with xk ′ e > 0 and

s j s ′ =s i +w k ′ -1
zk ′ s ′ > 0, until a cover K * is obtained for the interval I over edge e with k∈ K * w k > |I|. We further derive a minimal cover from the cover K * by deleting each demand k ∈ K * if k ′ ∈ K * \{k} w k ′ ≤ |I|. We then add inequality (2.32) induced by the minimal cover K * for the interval I and edge e if it is violated and satisfies some conditions about validity of inequality (2.32). The following valid inequality is then added to the current LP

k∈ K * x k e + s j s ′ =s i +w k -1 z k s ′ ≤ 2| K * | -1.

Separation of Edge-Interval-Clique Inequalities

The separation problem related to inequality (2.36) is NP-hard [77][81] given that it consists in identifying a maximal clique C * in the conflict graph H e I for a given edge e and a given interval I = [s i , s j ] such that

k∈C * xk e + s j s ′ =s i +w k -1 zk s ′ > |C * | + 1.
We start our procedure of separation by constructing a set of intervals of contiguous slots I = [s i , s j ] ∈ I e for a given edge e ∈ E such that each interval of con- x k e +

tiguous slots I = [s i , s j ] ∈ I e is
s j s ′ =s i +w k -1 z k s ′ + k ′ ∈C * e s j s ′ =s i +w k ′ -1 z k ′ s ′ ≤ |C * | + 1, such that a) w k ′ + w k ≥ |I| + 1 for each k ∈ C * and k ′ ∈ C * e , b) w k ′ + w k" ≥ |I| + 1 for each k ′ ∈ C * e and k" ∈ C * e , c) w k ′ ≤ |I| for each k ′ ∈ C * e .

Separation of Interval-Clique Inequalities

Given an interval of contiguous slots I = [s i , s j ]. Our separation algorithm for inequality (2.39) consists in identifying a maximal clique C * in the conflict graph

H E I such that k∈C * s j s ′ =s i +w k -1 zk s ′ > 1.
As a result, its associated separation problem is NP-hard given that computing a maximal clique in a given graph is known to be a NP-hard problem [START_REF] Karp | Reducibility among Combinatorial Problems[END_REF]. For this, we also use the greedy algorithm introduced by Nemhauser and Sigismondi [START_REF] Nemhauser | A Strong Cutting Plane/Branch-and-Bound Algorithm for Node Packing[END_REF] to identify a maximal clique in the conflict graph H E I as follows. We first generate a set of intervals of contiguous slots denoted by I E such that each interval of contiguous slots I = [s i , s j ] ∈ I E is given for each slot s i ∈ S and slot s j with

s j ∈ {s i + max k∈K, |E k 1 |≥1
w k , ..., min(s,

s i + 2 max k∈K, |E k 1 |≥1
w k )}. We then consider an interval of contiguous slots I = [s i , s j ] ∈ I E and its associated conflict graph H E I . We associate a positive weight

s j s ′ =s i +w k -1 zk s ′ for each node v k in H E I .
We select a demand k having the largest number of slots w k and weight

s j s ′ =s i +w k -1 zk
s ′ , and then set C * = {k}. After that, we iteratively add each demand k ′ having

s j s ′ =s i +w k ′ -1 zk ′ s ′ > 0 such that its corresponding node v k ′ is linked with all the nodes v k with k ∈ C * .
At the end, we consider inequality (2.39) induced by the maximal clique C * if it is violated, i.e., by adding the following inequality to the current LP

k∈C * s j s ′ =s i +w k -1 z k s ′ ≤ 1.
Moreover, this additional inequality can be strengthened as follows

k∈C * s j s ′ =s i +w k -1 z k s ′ + k ′ ∈C * e s j s ′ =s i +w k ′ -1 z k ′ s ′ ≤ 1,
where

C * E ⊂ K \ C * such that a) w k ′ + w k ≥ |I| + 1 and E k 1 ∩ E k ′ 1 ̸ = ∅ for each k ∈ C * and k ′ ∈ C * E , b) w k ′ + w k" ≥ |I| + 1 and E k ′ 1 ∩ E k" 1 ̸ = ∅ for each k ′ ∈ C * E and k" ∈ C * E , c) w k ′ ≤ |I| for each k ′ ∈ C * E .

Separation of Interval-Odd-Hole Inequalities

For inequality (2.40), we propose a separation algorithm that consists in identifying an odd-hole H * in the conflict graph H E I for a given Interval I and a fractional solution (x, z) such that

k∈H * s j s ′ =s i +w k -1 zk s ′ > |H * | -1 2 .
This can be done in polynomial time as shown by Rebennack et al. [94][95]. Based on this, we use the exact algorithm proposed by the same authors which consists of finding a minimum weighted odd-cycle in a graph. For this, we should first generate a set of intervals of contiguous slots I E as we did before in the section 3.1.9. We then consider a conflict graph H E I associated with a given interval of contiguous slots I ∈ I E . We construct an auxiliary conflict graph H ′E I which can be seen as a bipartite graph by duplicating each node v k in H E I (i.e., v k and v ′ k ) and two nodes are linked in H ′E I if their original nodes are linked in H E I . We assign to each link

(v a , v b ) in H ′E I a weight equals to 1- s j s ′ =s i +wa-1 za s ′ - s j s ′ =s i +w b -1 zb s ′ 2
. We then compute for each node v k in H E I , the shortest path between v k and its copy in the auxiliary conflict graph

H ′E I denoted by p v k ,v ′ k .
After that, we check if the total sum of weight over edges belong this path is smallest than 1 2 ,

(va,v b )∈E(p v k ,v ′ k ) 1 - s j s ′ =s i +wa-1 za s ′ - s j s ′ =s i +w b -1 zb s ′ 2 < 1 2 .
If so, odd-hole H * is composed by all the original nodes of nodes belong the computed

shortest path p v k ,v ′ k , i.e., V (p v k ,v ′ k ) \ {v ′ k }.
We then add inequality (2.40) induced by odd-hole H * to the current LP, i.e.,

k∈H * s j s ′ =s i +w k -1 z k s ′ ≤ |H * | -1 2 .
It can be lifted using the greedy algorithm introduced by Nemhauser and Sigismondi [START_REF] Nemhauser | A Strong Cutting Plane/Branch-and-Bound Algorithm for Node Packing[END_REF] to identify a maximal clique C * in the conflict graph

H E I such that a) w k ′ + w k ≥ |I| + 1 and E k 1 ∩ E k ′ 1 ̸ = ∅ for each k ∈ H * and k ′ ∈ C * , b) w k ′ + w k" ≥ |I| + 1 and E k ′ 1 ∩ E k" 1 ̸ = ∅ for each k ′ ∈ C * and k" ∈ C * , c) w k ′ ≤ |I| for each k ′ ∈ C * .
For this, we first assign a positive weight equals to the number of slots request w k ′ by demand k ′ for each node v k ′ linked with all the nodes v k ∈ H * in the conflict graph H E I . We then select the node v k ′ linked with all the nodes v k ∈ H * in the conflict graph H E I having the largest weight, and set C * to {k ′ }. After that, we iteratively add each demand k" to the current clique C * if its associated node v k" is linked with all the nodes v k ∈ H * and nodes v k ′ ∈ C * . As a result, we add inequality (2.41) induced by odd-hole H * and clique C * to the current LP, i.e.,

k∈H * s j s ′ =s i +w k -1 z k s ′ + |H * | -1 2 k ′ ∈C * s j s"=s i +w k ′ -1 z k ′ s" ≤ |H * | -1 2 .

Separation of Slot-Assignment-Clique Inequalities

Now, we describe the separation algorithm for inequality (2.43). It consists in identifying a maximal clique C * in the conflict graph

H E S such that v k,s ∈C * zk s > 1,
for a given fractional solution (x, z) of the current LP.

For this, we use the greedy algorithm introduced by Nemhauser and Sigismondi [START_REF] Nemhauser | A Strong Cutting Plane/Branch-and-Bound Algorithm for Node Packing[END_REF] to identify a maximal clique C * in the conflict graph H E S given that computing a maximal clique in such a graph is also NP-hard problem [START_REF] Karp | Reducibility among Combinatorial Problems[END_REF]. Based on this, we first assign a positive weight zk s to each node v k,s in the conflict graph H E S . We then select a node v k,s in the conflict graph H E S having the largest weight compared with the other nodes in H E S , and set C * = {v k,s }. After that, we iteratively add each node v k ′ ,s ′ to the current C * if it is linked with all the nodes v k,s already assigned to the current clique C * and zk ′ s ′ > 0. At the end, we add inequality (2.43) induced by clique C * to the current LP if it is violated, i.e., we add the following inequality

v k,s ∈C * z k s ≤ 1.
Furthermore, it can be lifted by identifying a maximal clique N * such that each

v k ′ ,s ′ ∈ N * is linked with all the nodes v k,s ∈ C * ∪ (N * \ {v k ′ ,s ′ }) in H E S .
For this, we also use the greedy algorithm introduced by Nemhauser and Sigismondi [START_REF] Nemhauser | A Strong Cutting Plane/Branch-and-Bound Algorithm for Node Packing[END_REF] to identify clique N * as follows. We first set

N * = {v k ′ ,s ′ } with v k ′ ,s ′ / ∈ C *
C * ∪ N * to the current LP, i.e., v k,s ∈C * z k s + v k ′ ,s ′ ∈N * z k ′ s ′ ≤ 1.

Separation of Slot-Assignment-Odd-Hole Inequalities

The separation algorithm of inequality (2.44) can be performed by identifying an odd-hole H * in the conflict graph H E S for a given fractional solution (x, z) such that

v k,s ∈H * zk s > |H * | -1 2 .
This can be done in polynomial time as shown by Rebennack et al. [START_REF] Rebennack | Stable Set Problem: Branch & Cut Algorithms[END_REF][95] by finding a minimum weighted odd-cycle in the conflict graph H E S . For this, we first construct an auxiliary conflict graph H ′E S which can be seen also as a bipartite graph by duplicating each node v k,s in H E S (i.e., v k,s and v ′ k,s ) such that each two nodes are linked in H ′E S if their original nodes are linked in H E S . We assign to each link (ṽ k,s , ṽk ′ ,s ′ ) in H ′E S a weight equals to

1-z k s -z k ′ s ′

2

. We then compute for each node

v k,s in H E S ,
C * v k,s ∈H * z k s + |H * | -1 2 v k ′ ,s ′ ∈C * z k ′ s ′ ≤ |H * | -1 2 ,

Separation of Incompatibility-Clique Inequalities

Consider now inequality (2.46). Its associated separation algorithm consists in identifying a maximal clique C * in the conflict graph

H K E such that v k,e ∈C * xk e > 1.
The separation problem related to this inequality is NP-hard given that computing a maximal clique in the conflict graph H K E is NP-hard problem [START_REF] Karp | Reducibility among Combinatorial Problems[END_REF]. For this, we also use the greedy algorithm introduced by Nemhauser and Sigismondi [START_REF] Nemhauser | A Strong Cutting Plane/Branch-and-Bound Algorithm for Node Packing[END_REF]

v k ′ ,e ′ ∈ N * is linked with all the nodes v k,e ∈ C * ∪ (N * \ {v k ′ ,e ′ }) in H K E .
For this, we also use the greedy algorithm introduced by Nemhauser and Sigismondi [START_REF] Nemhauser | A Strong Cutting Plane/Branch-and-Bound Algorithm for Node Packing[END_REF] to identify clique N * as follows. We first set

N * = {v k ′ ,e ′ } with v k ′ ,e ′ / ∈ C * a node in H K E having the largest degree |δ(v k ′ ,e ′ )| in H K E
and should be also linked with all the nodes v k,e ∈ C * in H K E . We then iteratively add each node x k ′ e ′ ≤ 1.

v k ′ ,e ′ / ∈ C * ∪ N * to the current N * if it is linked in H K E with

Separation of Incompatibility-Odd-Hole Inequalities

The separation algorithm related to inequality (2.47) can be done in polynomial time by finding a minimum weighted odd-cycle in the conflict graph H K E as shown by Rebennack et al. [94][95]. For this, our aims is to identify an odd-hole H * in the conflict graph

H K E such that v k,e ∈H * xk e > |H * | -1 2 ,
for a given fractional solution (x, z) of the current LP.

We start its procedure of separation by constructing an auxiliary conflict graph

v k ′ ,e ′ ∈C * x k ′ e ′ ≤ |H * | -1 2 .

Separation of Transmission-Reach-Cover Inequalities

In this section, we study the separation problem of inequality (2.49). Consider a demand k ∈ K. The separation problem associated with inequality (2.49) is NP-hard [START_REF] Klabjan | The complexity of cover inequality separation[END_REF] given that it consists in identifying a cover C * related to the transmission-reach constraint of demand k, such that

e∈c * xk e > |C * | -1.
For this, we propose a separation algorithm based on a greedy algorithm introduced by Nemhauser and Sigismondi [START_REF] Nemhauser | A Strong Cutting Plane/Branch-and-Bound Algorithm for Node Packing[END_REF]. We first select an edge e ∈ E \ (E k 0 ∪ E k 1 ) having the largest length ℓ e with xk e > 0, and set C * to C * = {e}. After that, we iteratively add each edge e ′ ∈ E \(E k 0 ∪E k 1 ∪C * ) to C * while e∈C * ℓ e ≤ lk and e ′ is compatible with the edges already added to the cover C * , i.e., until a cover C * is obtained for the demand k with e∈C * ℓ e > lk . We further derive a minimal cover from the cover C * by deleting each edge e ∈ C * if e ′ ∈C * \{e} ℓ e ′ ≤ lk . We then add inequality (2.49) induced by the minimal cover C * for demand k to the current LP if it is violated, i.e., e∈C *

x k e ≤ |C * | -1.

Furthermore, inequality (2.49) induced by the minimal cover C * can be lifted by introducing an extended coverinequality as follows

e∈C * x k e + e ′ ∈E(C * ) x k e ′ ≤ |C * | -1,
where ℓ e ′ ≥ ℓ e for each e ∈ C * and e ′ ∈ E(C * ) with e ′ / ∈ E k 0 ∪E k 1 and e ′ is compatible with each edge e ∈ C * .

Separation of Edge-Capacity-Cover Inequalities

Let now study the separation problem of inequality (2.54). Given an edge e ∈ E.

The separation problem associated with inequality (2.54) is NP-hard [START_REF] Klabjan | The complexity of cover inequality separation[END_REF] given that it consists in identifying a cover K * edge e, such that

k∈ K * xk e > | K * | -1.
For this, we propose a separation algorithm based on a greedy algorithm introduced by Nemhauser and Sigismondi [START_REF] Nemhauser | A Strong Cutting Plane/Branch-and-Bound Algorithm for Node Packing[END_REF]. We first select a demand k ∈ K \ K e having largest number of requested slot w k with xk e > 0, and set K * to K * = {k}. After that, we iteratively add each demand k ′ ∈ K \ (K e ∪ K * ) to K * while k∈ K * w k ≤ s -k∈Ke w k, i.e., until a cover K * is obtained for the edge e with k∈ K * w k > s -k∈Ke w k. We further derive a minimal cover from the cover K * by deleting

each demand k ∈ K * if k ′ ∈ K * \{k} w k ′ ≤ s -k∈Ke w k.
We then add inequality (2.54) induced by the minimal cover K * for edge e to the current LP if it is violated, i.e.,

k∈ K * x k e ≤ | K * | -1.

Primal Heuristic

Here, we propose a primal heuristic to boost the performance of the Branch-and-Cut algorithm. It is based on a hybrid method between a local search algorithm and a greedy-algorithm. Given an optimal fractional solution (x, z) in a certain node of the B&C tree, our primal heuristic consists in constructing an integral "feasible"

ℓ e ≤ lk .
This can be done in polynomial time using network flow algorithms.

Afterwards, we use a local search algorithm. It consists in generating at each iteration a sequence of numeroted demands L (order) with

L = 1 ′ , 2 ′ , ..., |K| ′ -1, |K| ′ .
Based on this sequence of demands, we use a greedy algorithm to select a path p from R k ′ and a slot s ∈ {w k ′ , ..., s} for each demand k ′ ∈ L with zk ′ s ̸ = 0 and xk ′ e ̸ = 0 for each e ∈ E(p), while respecting the non-overlapping constraint with the set of demands that precede demand k ′ in the list L (i.e., the demands 1 ′ , 2, ..., k ′ -1).

However, if there does not exist such pair of path p and slot s for demand k ′ , we then select a path p and a slot s ∈ {w k ′ , ..., s} for demand k ′ ∈ L with zk After that, we compute the associated total cost of the paths selected for the set of demands K in the final solution S given by the greedy-algorithm (i.e., k∈K e∈E k c e ).

Our local search algorithm generates a new sequence by doing some permutation of demands in the last sequence of demands, if the value of the solution given by greedy-algorithm is smaller than the value of the best solution found until the current iteration. Otherwise, we stop the algorithm, and we give in output the best solution found during the primal heuristic induced by the best sequence of demands having the smallest value of total cost of the selected path compared with the others generated sequences.

Computational Study

Implementation's Feature

We have used C++ programming language to implement the B&C algorithm under Linux using three framworks, Cplex 12.9 [START_REF] Cplex | V12. 9: User's Manual for Cplex[END_REF], Gurobi 9.0 [START_REF] Gurobi Optimization | Gurobi Optimizer Reference Manual[END_REF], and "Solving Constraint Integer Programs" (Scip 7.0) [START_REF] Gamrath | The Scip Optimization Suite 7.0[END_REF] framework using Cplex 12.9 as LP solver.

It has been tested on LIMOS high performance server with a memory size limited to 64 gb while benefiting from parallelism by activating 8 threads when using Gurobi or Scip (which is not possible when using cutting-plane based method under Cplex).

We limit the CPU time to 5 hours (18000 s).

Description of Instances

We further present computational results using two types of graphs: real, and other realistics from SND-Lib [START_REF] Orlowski | SNDlib 1.0-Survivable Network Design Library[END_REF] with total number of nodes |V | up to 161 and total number of edges |E| up to 166 as shown in Table 3

Computational Results

We consider 4 criteria in our computational study, average number of nodes in the B&C tree (Nb Nd), average gap (Gap) which represents the relative error between the lower bound gotten at the end of the resolution and best upper bound, average number of violated inequalities added during the algorithm (Nbr Cuts), and average Cpu time computation (TT).

Based on preliminary results, the cover-based inequalities (2.54) and (2.32) are

shown to be efficient than the clique-based inequalities (2.43), (2.42) and (2.36).

In fact, the B&C algorithm performs very well when adding the cover-based inequalities (2.54) and (2.32) under Scip and Gurobi. We notice that adding these valid inequalities allows solving to optimality some instances that are not solved to optimality when using Cplex, Gurobi and Scip. Furthermore, they allow reducing the average gap, average number of nodes, and the average cpu time. On the other hand, we observed that the valid inequalities do not work well when using Cplex.

This is due to deactivating the Cplex's cut generation such that Cplex does not work well without its proper cut generation even if the valid inequalities are shown to be efficient when using Gurobi and Scip for the instances tested. The results also show that several inequalities of the cover-based inequalities (2. Using this, we provide a comparative study between Cplex, Gurobi and Scip. For this, we aim to evaluate the impact of the valid inequalities used within the B&C algorithm. Our first series of computational results presented in Tables 3.2, it concerns a comparaison between the results obtained for the B&C algorithm using Cplex and Scip (without or with additional valid inequalities). On the other hand, in the second series of computational results shown in Table 3.3, we present the results found for the B&C algorithm using Gurobi and Scip (without or with additional valid inequalities). In the third series shown in Table 3.4, we compare the results found by the B&C algorithm using Cplex (without or with additional valid inequalities) with those that are found when using Scip (without or with additional valid inequalities).

For each instance, we run

• Cplex with benefiting of its automatic cut generation and without our additional valid inequalities (denoted by B&C CPX in the different tables),

• Cplex using our valid inequalities and disabling its proper cut generation (denoted by Own B&C CPX),

• Gurobi with benefiting of its automatic cut generation and without our additional valid inequalities (denoted by B&C GRB),

• Gurobi using our valid inequalities and disabling its proper cut generation (denoted by Own B&C GRB),

• Scip with benefiting of its automatic cut generation and without our additional valid inequalities (denoted by B&C SCIP),

• Scip using our valid inequalities and disabling its proper cut generation (denoted by Own B&C SCIP).

To make the results and the comparison more readable, we just present some computational results using a subset of instances based on 2 real topologies: German, Nsfnet, and 2 realistic topologies: India35 and Pioro40.

We first notice that our valid inequalities allows solving several instances to optimality that are not solved to optimality when using B&C CPX, B&C GRB and B&C SCIP. Furthermore, they enabled reducing the average number of nodes in the B&C tree, and also the average Cpu time for several instances. On the other hand, and when the optimality is not proven, adding valid inequalities decreases the average gap for several instances. However, there exists a few instances in which adding valid inequalities does not improve the results of B&C algorithm. We further observe that Own B&C SCIP is shown to be very efficient compared with Cplex and Gurobi (see for example Table 3.2 and 3.3). However, and looking at the instances that are solved to optimality by Own B&C GRB and Own B&C SCIP, we notice that we have less number of nodes and time cpu when using Own B&C SCIP compared with Own B&C GRB (see for example Table 3.3). Furthermore, Own B&C SCIP works much betther than SCIP, Cplex and Gurobi even when using their proper cuts such that Own B&C SCIP is able to solve several instances to optimality that are not solved when using B&C CPX, B&C GRB and B&C SCIP. This means that we are able to beat Cplex, Gurobi and Scip using Own B&C SCIP. On the other hand, and considering large-scale instances with |K| ≥ 200, we noticed that adding valid inequalities does not improve the effectiveness of the B&C algorithm such that there exist some instances that are solved to optimality using B&C CPX and B&C GRB that are not solved to optimality when using Own B&C CPX, Own B&C GRB and Own B&C SCIP. Based on these results, we conclude that using the valid inequalities allows obtaining tighter LP bounds. They significantly improve the results yielded by the B&C CPX, B&C GRB, and B&C SCIP for several instances with number of demands up to 150.

Given that Own B&C SCIP is able to beat Cplex, Gurobi and Scip, we turn our attention to the numerical results found when using SCIP. They are shown in the following Table 3.5.

We can see from Table 3.5 that our B&C algorithm (Own B&C) is able to solve to optimality more instances than B&C SCIP. Indeed, 137 instances are solved to optimality when our inequalities are used (Own B&C) while 101 instances are solved to optimality in run B&C SCIP. Also, when our inequalities are used, the number of nodes in the B&C tree is decreased in most cases compared to the case where they are not used. Moreover, the CPU time is, in general, smaller when our inequalities are used. Finally, when comparing the instances which are not solved to optimality, we can see that the optimality gap is smaller, for most of the instances, when our inequalities are used.

Concluding Remarks

In this chapter, we have devised a B&C algorithm, and conducted some computational experiments. Our study shows that the valid inequalities are effective for solving real and realistic instances of the problem. It could be interesting to study the impact of the symmetry breaking inequalities and the precomputed lower bounds on the performance of the Branch-and-Cut algorithm. not, such that s represents the last slot of the interval of contiguous slots of width w k allocated by demand k ∈ K, with s ∈ S and p ∈ P k . Note that all the slots s ′ ∈ {s -w k + 1, ..., s} should be assigned to demand k along the path p whenever y k p,s = 1. Let P k (e) denote set of all admissible (o k ,d k ) paths going through edge e in G for demand k.

In this case, the C-RSA is also equivalent to the following integer linear program min

k∈K p∈P k e∈E(p) s s=w k c e y k p,s , (4.1)
subject to

p∈P k w k -1 s=1 y k p,s = 0, ∀k ∈ K, (4.2)
p∈P k s s=w k y k p,s = 1, ∀k ∈ K, (4.3)
k∈K p∈P k (e) s+w k -1 s ′ =s y k p,s ′ ≤ 1, ∀e ∈ E, ∀s ∈ S, (4.4)
y k p,s ≥ 0, ∀k ∈ K, ∀p ∈ P k , ∀s ∈ S, (4.5)
y k p,s ∈ {0, 1}, ∀k ∈ K, ∀p ∈ P k , ∀s ∈ S. (4.6)
Inequalities (4.2) express the fact that a demand k ∈ K cannot occupy a slot s as the last slot before her slot-width w k . Inequalities (4.3) express the routing and spectrum constraints at the same time such that they ensure that exactly one slot s ∈ {w k , . . . , s} is assigned as last slot for the routing of demand k, and exactly one single path from P k is allocated by each demand k ∈ K. Note that a slot s ∈ S is said an allocated slot by demand k if and only if p∈P k s+w k -1 s ′ =s y k p,s ′ = 1 which means that s is covered by the interval of contiguous slots allocated by demand k.

Inequalities (4.4) ensure that a slot s over edge e cannot be allocated to at most by one demand k ∈ K. Inequalities (4.5) are trivial inequalities, and constraints (4.6) are the integrality constraints.

To benefit from some theoretical results done in chapter 2, we introduce the two variables x k e and z k s used in the cut formulation already presented in chapter 2. As a result, all the valid inequalities for the polytope associated with the cut formulation, they are still valid for the polytope associated with the path formulation following the addition of these two variables and the two following constraints

x k e - p∈B k (e) s s=w k y k p,s = 0, ∀k ∈ K, ∀e ∈ E, (4.7)
and

z k s - p∈B k y k p,s = 0, ∀k ∈ K, ∀s ∈ S. (4.8)
Therefore, the C-RSA is then equivalent to the extended formulation based on the following integer linear program min

k∈K e∈E c e x k e , (4.9
)

p∈P k w k -1 s=1 y k p,s = 0, ∀k ∈ K, (4.10)
p∈P k s s=w k y k p,s = 1, ∀k ∈ K, (4.11)
x k e -

p∈P k (e) s s=w k y k p,s = 0, ∀k ∈ K, ∀e ∈ E, (4.12
)

z k s - p∈P k y k p,s = 0, ∀k ∈ K, ∀s ∈ S, (4.13)
k∈K p∈P k (e)

s+w k -1 s ′ =s y k p,s ′ ≤ 1, ∀e ∈ E, ∀s ∈ S, (4.14)
y k p,s ≥ 0, ∀k ∈ K, ∀p ∈ P k , ∀s ∈ S, (4.15)
x k e ≥ 0, ∀k ∈ K, ∀e ∈ E, (4.16)

z k s ≥ 0, ∀k ∈ K, ∀s ∈ S, (4.17)
y k p,s ∈ {0, 1}, ∀k ∈ K, ∀p ∈ P k , ∀s ∈ S. (4.18)

Column Generation Algorithm

As it has been mentioned previously, our path formulation contains a huge number of variables which can be exponentiel in the worst case due to the number of all feasible paths for each traffic demand. To deal with this, we use a column generation algorithm to solve its linear relaxation. For this, we begin the algorithm with a restricted linear program of our path formulation by considering a feasible subset of variables (columns). For this, we first generate a subset of feasible paths for each demand k ∈ K denoted by B k ⊂ P k such that the variables y k p,s for each k ∈ K, p ∈ B k and s ∈ S induce a feasible basis for the restricted linear program. This means that there exists at least one feasible solution for the restricted linear program. Based on this, we derive the so-called restricted master problem (RMP) as follows min k∈K e∈E c e x k e , subject to

p∈B k w k -1 s=1 y k p,s = 0, ∀k ∈ K, p∈B k s s=w k y k p,s = 1, ∀k ∈ K, x k e - p∈B k (e) s s=w k y k p,s = 0, ∀k ∈ K, ∀e ∈ E, z k s - p∈B k y k p,s = 0, ∀k ∈ K, ∀s ∈ S, k∈K p∈B k (e) s+w k -1 s ′ =s y k p,s ′ ≤ 1, ∀e ∈ E, ∀s ∈ S, y k p,s ≥ 0, ∀k ∈ K, ∀p ∈ B k , ∀s ∈ S, x k e ≥ 0, ∀k ∈ K, ∀e ∈ E, z k s ≥ 0, ∀k ∈ K, ∀s ∈ S.
At each iteration, the column generation algorithm checks if there exists a variable y k p,s with p / ∈ B k for a demand k and slot s having a negative reduced cost using the solution of the dual problem associated with the constraints of the linear relaxation (4.1)-(4.5), and add it to B k . This can be achieved by solving the so-called pricing problem (PP).

The Pricing Problem

As noted later, we consider an initial restricted master problem denoted by RM P 0 which is based on an initial subset of variables induced by a subset of feasible path

B k ⊂ P k for each demand k ∈ K.
The pricing problem consists in finding a feasible path p for a demand k and slot s having a negative reduced cost using the optimal solution of the dual problem. For this, we consider the following dual variables a) α associated with the equations (4.10) such that α k ∈ R for all k ∈ K, b) β associated with the equations (4.11) such that β k ∈ R for all k ∈ K, c) µ associated with inequalities (4.14) such that µ e s ≤ 0 for all e ∈ E and s ∈ S, d) λ associated with the equations (4.12) such that λ k e ∈ R for all k ∈ K and e ∈ E, e) ρ associated with the equations (4.13) such that ρ k s ∈ R for all k ∈ K and s ∈ S,

The dual problem of the linear relaxation (4.9)-(4.17) is equivalent to max -

k∈K β k + e∈E s∈S µ e s , (4.19)
subject to

β k - e∈E(p) (λ k e + s s ′ =s-w k +1 µ e s ′ ) -ρ k s ≥ 0, ∀k ∈ K, ∀p ∈ P k , ∀s ∈ {w k , ..., s}, (4.20)
α k - e∈E(p) s s ′ =max(1,s-w k +1) µ e s ′ ≥ 0, ∀k ∈ K, ∀p ∈ P k , ∀s ∈ {1, ..., w k -1}, (4.21
)

c e + λ k e ≥ 0, ∀k ∈ K, ∀e ∈ E, (4.22)
α k + ρ k s ≥ 0, ∀k ∈ K, ∀s ∈ S, (4.23)
µ e s ≤ 0, ∀e ∈ E, ∀s ∈ S. (4.24)
As a result, the so-called reduced-cost rc k s (p) related to each demand k ∈ K, path p ∈ P k and slot s ∈ {w k , ..., s}, is given by

rc k s = β k -ρ k s + min p∈P k [ e∈E(p) -λ k e - s s ′ =s-w k +1 µ e s ′ ], (4.25)
Therefore, for each demand k ∈ K and slot s ∈ {w k , ..., s}, the pricing problem aims at finding a path p * of P k such that

rc k s (p * ) = β k -ρ k s + min p∈P k [ e∈E(p) -λ k e - s s ′ =s-w k +1 µ e s ′ ], (4.26)
Finding such path p * can be seen as a separation procedure for the dual constraint (4.20) which consists in identifying a path p * for each demand k ∈ K and slot s ∈ {w k , ..., s} such that

β k -ρ k s + e∈E(p * ) (-λ k e - s s ′ =s-w k +1 µ e s ′ ) < 0 and e∈E(p * ) ℓ e ≤ lk .
As a result, the pricing problem consists in solving the Resource Constrained Shortest Path (RCSP) problem. The RCSP problem is well known to be weakly NP-hard [START_REF] Dror | Note on the Complexity of the Shortest Path Models for Column Generation in VRPTW[END_REF]. Several algorithms have been proposed in the literature to solve this problem based on dynamic programming algorithms, heuristics and some techniques related to the lagrangian decomposition. As background references we mention [START_REF] Carlyle | Lagrangian relaxation and enumeration for solving constrained shortest-path problems[END_REF][START_REF] Dumitrescu | Algorithms for the weight constrained shortest path problem[END_REF][START_REF] Eppstein | Finding the k shortest paths[END_REF][START_REF] Joksch | The shortest route problem with constraints[END_REF][START_REF] Lozano | On an exact method for the constrained shortest path problem[END_REF].

Dynamic Programming Algorithm for the Pricing Problem

In this work, we propose a pseudo-polynomial time based dynamic programming algorithm [START_REF] Dumitrescu | Algorithms for the weight constrained shortest path problem[END_REF]. It consists in finding the minimum-cost path for each demand k and slot s while satisfying the transmission-reach constraint. It is based on the dynamic programming algorithm proposed by Dumitrescu et al. [START_REF] Dumitrescu | Algorithms for the weight constrained shortest path problem[END_REF] to solve the RCSP problem. For each demand k ∈ K and slot s, we associate to each node v ∈ V in the graph G a set of labels L v such that each label corresponds to differents paths from th origin node o k to the node v, and each label p is specified by a cost equals

to e∈E(p) (-λ k e -s s ′ =s-w k +1 µ e s ′ )
, and a weight equals to e∈E(p) ℓ e . We denote by T v the set of labels on node v ∈ V . For each demand k and slot s ∈ {w k , ..., s}, the complexity of the algorithm is bounded by O(|E \ E k 0 | * lk ) [START_REF] Dumitrescu | Algorithms for the weight constrained shortest path problem[END_REF]. Algorithm 3 summarizes the different steps of the dynamic programming algorithm.

Initial Columns

The basic subset of paths used to define the restricted master problem, they are generated using a brute-force search algorithm which creates a search tree that covers all the feasible paths P k for each demand k. It is then used to pre-compute an initial subset B k of feasible paths for each demand k ∈ K taking into account the transmission-reach constraint to prune some non intersecting nodes in our search tree of this algorithm.

Branch-and-Price and Branch-and-Cut-and-Price Algorithms

Based on these features, we derive a Branch-and-Cut-and-Price algorithm for solving the C-RSA problem.

Description

The main purpose of this algorithm is to solve a sequence of linear programs using the column generation algorithm at each node of a Branch-and-Bound algorithm.

At each iteration of the algorithm, we solve our pricing problem by identifying one or more than one new column by solving a RCSP problem for each demand k and slot s ∈ {w k , ..., s} using the dynamic programming algorithm. We repeat this pro-cedure in each iteration of the column generation until no new column is found (i.e., rc k s ≥ 0 for all k ∈ K and s ∈ {w k , ..., s}. As a result, the final solution is optimal for the linear relaxation of the path formulation. Furthermore, if it is integral, then it is optimal for the C-RSA problem. Otherwise, we create two subproblems by branching on fractional variables (variable branching rule) or on some constraints using the Ryan & Foster branching rule [START_REF] Ryan | An integer programming approach to scheduling[END_REF] (constraint branching rule). Algorithm 4 summarizes the different steps of the Branch-and-Price algorithm.

By combining the Branch-and-Price algorithm with a cutting-plane based algorithm, we devise a Branch-and-Cut-and-Price which works as follows. Consider a fractional solution (ȳ, x, z). At each iteration of the Branch-and-Price algorithm, and for a given class of valid inequalities, our aim is to identify the existence of one or more than one inequalities of this class that are violated by the current solution. We repeat this procedure in each iteration of the algorithm until no violated inequality is identified.

As mentioned before, the Branch-and-Cut-and-Price algorithm also uses the different classes of valid inequalities presented in chapter [START_REF] Amar | Performance assessment and modeling of flexible optical networks[END_REF]. They are performed in the order (2.54), (2.32), (2.42), (2.36), (2.43).

Algorithm 5 summarizes the different steps of the Branch-and-Cut-and-Price algorithm for a given class of valid inequalities.

Primal Heuristic

Here, we propose a primal heuristic based on a hybrid method between local search algorithm and a greedy-algorithm. Given a feasible fractional solution (ȳ, x, z), our primal heuristic consists in constructing an integral "feasible" solution from this fractional solution. For this, we propose a local search algorithm which consists in generating at each iteration a sequence of demands

L = 1 ′ , 2 ′ , ..., |K| ′ -1, |K| ′ .
Based on this sequence of demands, our greedy algorithm selects a path p and a slot s for each demand k ′ ∈ L with y k ′ p,s ̸ = 0 while respecting the non-overlapping constraint with the set of demands that precede demand k ′ in the list L (i.e., the demands 1 ′ , 2, ..., k ′ -1). However, if there does not exist such pair of path p and slot s for demand k ′ , we then select a path p and a slot s for demand k ′ ∈ L with y k ′ p,s = 0 and s ∈ {w k ′ , ..., s} while respecting the non-overlapping constraint with the set of demands that precede demand k ′ in the list L. After that, we compute the associated total length of the paths selected for the set of demands K in the final solution S given by the greedy-algorithm. Our local search algorithm generates a new sequence by doing some permutation of demands in the last sequence of demands, if the value of the solution given by greedy-algorithm is smaller than the value of the best solution found until the current iteration. Otherwise, we stop the algorithm, and we give in output the best solution found during our primal heuristic induced by the best sequence of demands having the smallest value of total length of the selected path compared with the others generated sequences.

Computational Study

Implementation's Feature

The B&P and B&C&P algorithms described in the current chapter have been implemented in C++ under Linux using the "Solving Constraint Integer Programs" framework (Scip 6.0.2), and Cplex 12.9 as LP solver. These have been tested on LIMOS high-performance server with a memory size limited to 64 Gb while benefiting from parallelism by activating 8 threads, and with a CPU time limited to 5 hours (18000 s).

Computational Results

Throughout this section, we present the performance results of the B&C&P algorithm. Our main goal is to show the effectiveness of the valid inequalities used within the B&C&P algorithm. Table 4.1 reports the experiment results for both the Branch-and-Price (B&P) (i.e., B&C&P without using our additional valid inequalities) and the B&C&P algorithms.

Each line corresponds to the average results of 4 tested instances. Note that we deactivate the SCIP's proper cut generation for both the B&P and B&C&P algorithms given that they may change the dual problem, as well as the calculation of the reduced-cost. In order to evaluate the impact of the additional valid inequalities used within the B&C&P algorithm, we consider 5 criteria, the average number of nodes in the branching tree (Nb Nd), the average optimality gap (Gap) which represents the relative error between the lower bound and the best upper bound obtained at the end of the resolution, the average number of generated columns (Nbr Cols), the average number of violated inequalities added (Nbr Cuts), and the average CPU time in seconds (TT).

The results show that the B&C&P is able to solve 187 instances to optimality while 147 instances are solved to optimality when using the B&P. Hence, our valid inequalities allow solving several instances to optimality within a reasonable amount Based on the reported results, we notice that the B&C&P algorithm seems to be very efficient compared with B&C algorithm such that it is able to provide optimal solutions for several instances, which is not the case for the B&C algorithm (without or with additional valid inequalities) within the CPU time limit (5 hours).

Furthermore, several instances are solved to optimality by the B&C algorithm using Cplex, Gurobi, and Scip could also be solved to optimality within the B&C&P algorithm. The average number of explored nodes using the B&C&P algorithm is greatly reduced for several instances compared with the B&C algorithm. Moreover, the average CPU time is significantly reduced using the B&C&P algorithm compared with the B&C algorithm. On the other hand, and when using the B&P algorithm, we notice that we are able to beat Own B&C SCIP such that B&P is able to provide optimal solutions for several instances that are not solved to optimality by the B&C algorithm using Cplex (see Table 4.2), and Gurobi (see Table 4.3). Furthermore, we noticed that the average number of explored nodes and the average CPU time using the B&P algorithm are greatly reduced for several instances compared with the B&C algorithm using Cplex and Gurobi. However, Own B&C SCIP is able to beat the B&P algorithm. The results in Table 4.4 show that Own B&C SCIP provide optimal solutions for several instances that are not solved to optimality by the B&P algorithm. But when the optimality is verified by these two algorithms, we found that using the B&P algorithm reduces the average number of explored nodes and the average CPU time for several instances compared with Own B&C SCIP.

Concluding Remarks

In this chapter, we first have given an extended formulation for the problem, and solve its linear relaxation using a column generation algorithm. We have discussed the associated pricing problem. Moreover, we have investigated the polytope associated with our formulation, and introduced several classes of valid inequalities.

Their separation procedures are further presented. Using this, we have devised the B&C&P algorithm. Computational experiments have convincingly shown the strength of the valid inequalities. They significantly improve the results yielded by the B&P algorithm. Hence, the B&C&P algorithm performs very well compared with the B&P algorithm. Furthermore, the B&C&P algorithm is shown to be able to beat the B&C algorithm. A computational analysis is conducted to show the effectiveness of our approach for solving large-scale instances. Chapter 5

Compact Formulation and Polyhedra for the Spectrum Assignment Sub-problem

In this chapter, we focus on the Spectrum Assignment (SA) sub-problem. First, we propose an integer linear programming compact formulation, and further investigate the facial structure of the associated polytope. Moreover, we identify several classes of valid inequalities for the polytope such that some of them come from those that are already proposed for the C-RSA. We further prove that these inequalities are facet-defining, and discuss their separation problems. Based on these results, we devise a Branch-and-Cut (B&C) algorithm for the SA problem.

The Spectrum Assignment Sub-problem

The SA problem can be stated as follows. We consider an optical spectrum of s ∈ Z + The SA is well known to be NP-hard problem [START_REF] Bermond | On spectrum assignment in elastic optical treenetworks[END_REF]. It is equivalent to the problems of wavelength assignment, interval coloring, and dynamic storage allocation [START_REF] Bermond | On spectrum assignment in elastic optical treenetworks[END_REF] that are well known to be NP-hard.

Compact Formulation

Here we introduce an integer linear programming compact formulation for the SA problem. For s ∈ S, let u s be a variable which takes 1 if slot s is used and 0 if not, and for k ∈ K and s ∈ S, let z k s be a variable which takes 1 if slot s is the last slot allocated for the routing of demand k and 0 if not. The contiguous slots s ′ ∈ {s -w k + 1, ..., s} should be assigned to demand k whenever z k s = 1. The SA is equivalent to the following integer linear program

min s∈S u s , (5.1)
subject to z k s = 0, for all k ∈ K and s ∈ {1, ..., w k -1}, (5.2)

s s=w k z k s ≥ 1, for all k ∈ K, (5.3)
k∈ Ke min(s,s+w k -1)

s ′ =s z k s -u s ≤ 0, for all e ∈ E, and s ∈ S,

z k s ≥ 0, for all k ∈ K and s ∈ S, (5.5)

u s ≤ 1, for all s ∈ S, (5.6)
z k s ∈ {0, 1}, for all k ∈ K and s ∈ S, (5.7) where

u s ∈ {0,
D 105 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E(p k i ) ∩ E(p k j ) ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j {k 1 , ..., k i-1 } ∪ {k} with E(p k i ) ∩ E(p k j ) ̸ = ∅,
• and s / ∈ {s k i - for each demand k ∈ K. S 106 is feasible for the SA problem. Hence, the corresponding incidence vector (u S 106 , z S 106 ) belongs to P sa (G, K, S). We then obtain that µu S 105 + σz S 105 = µu S 106 + σz S 106 = µu S 105 + σz S 105 + µ s. Hence, µ s = 0.

w k i + 1, ..., s k i } (slot
In a similar way, we can show that µ s = 0, for all slots s ∈ S.

Let show now that σ k s = 0 for all k ∈ K and s ∈ {w k , ..., s}. Consider a demand k in K and a slot s in {w k , ..., s}. Let S 107 = (U where Since µ s = 0 for each s ∈ S, it follows that σ k s = 0. In a similar way, we can show that σ k s = 0, for all k ∈ K and s ∈ {w k , ..., s}.

D 107 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E(p k i ) ∩ E(p k j ) ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 107 i , • and {s -w k + 1, ..., s} ∩ {s k i -w k i + 1, ..., s k i } = ∅ if E(p k ) ∩ E(p k i ) ̸ = ∅ (
Therefore, we obtain that all the equations of the polytope P sa (G, K, S) are given only in terms of the variables z k s with s ∈ {1, ..., w k } for each demand k ∈ K. We distinguish |K| blocks of lines in the matrix M associated with the system (5.2)

• block M k corresponds to the equations z k s = 0 for all s ∈ {1, ..., w k -1} such that rang(M k ) = w k -1.

Note that the |K| blocks of the matrix M are independents. Furthermore, there is no dependency between slots such that for each demand k, the slots s ∈ {1, ..., w k -1} are independants such that

w k -1 s=1 σ k s = w k -1 s=1 γ k,s ⇒ w k -1 s=1 (σ k s -γ k,s ) = 0,
for each demand k ∈ K. The only solution of this system is σ k s = γ k,s for each s ∈ {1, ..., w k -1} for demand k. As k is chosen arbitrarily in K, we re-do the same procedure for all k ′ ∈ K \ {k}. We then get that

σ k s = γ k,s
, for all k ∈ K and s ∈ {1, ..., w k -1}.

(5.9)

As a result, we have (µ, σ) = γM which ends the proof. Proof. Given the rank of the matrix M which equals to r ′ and the results of proposition (5.3.1).

Facial Investigation

Here we study the facial structure of the basic constraints of the compact formulation (5.1)-(5.8) that are facets defining for the polyhedron P sa (G, K, S) under certain conditions.

Theorem 5.3.2. Consider a demand k ∈ K and a slot s ∈ {w k , .., s}. Then, inequality z k s ≥ 0 is facet defining for P sa (G, K, S).

Proof. Let us denote F k s the face induced by inequality z k s ≥ 0, that is

F k s = {(u, z) ∈ P sa (G, K, S) : z k s = 0}.
We denote inequality z k s ≥ 0 by αu + βz ≤ λ. Let µu + σz ≤ τ be a valid inequality that defines a facet F of P sa (G, K, S). Suppose that F k s ⊂ F = {(u, z) ∈ P sa (G, K, S) : µu + σz = τ }. To prove that F k s is facet defining for P sa (G, K, S), it sufficient to show that there exist ρ ∈ R and γ ∈ R k∈K (w k -1) ) such that (µ, σ) = ρ(α, β) + γM . where

D 109 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E(p k i ) ∩ E(p k j ) ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j {k 1 , ..., k i-1 } ∪ {k} with E(p k i ) ∩ E(p k j ) ̸ = ∅,
• and s / ∈ {s k i - for each demand k ∈ K. Solution S ′109 is feasible for the SA problem. Hence, the corresponding incidence vector (u S ′109 , z S ′109 ) belongs to F k s . Solutions S 109 and S ′109 satisfy equation µu + σz = τ . We then obtain that µu S 109 + σz S 109 = µu S ′109 + σz S ′109 = µu S 109 + σz S 109 + µ s. Hence, µ s = 0.

w k i + 1, ..., s k i } (slot
In a similar way, we can show that µ s = 0, for all slots s ∈ S.

Next, we will show that, σ k s ′ = 0 for all s ′ ∈ {w k , . where Given that µ s = 0 for all s ∈ S, it follows that σ k s ′ = 0. In a similar way, we can show that σ k s ′ = 0, for all slots s ′ ∈ {w k , ..., s} \ {s}, σ k ′ s ′ = 0, for all k ′ ∈ K \ {k} and s ′ ∈ {w k ′ , ..., s}.

D 110 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E(p k i ) ∩ E(p k j ) ̸ = ∅}. This ensures that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 110 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E(p k ) ∩ E(p k i ) ̸ = ∅ (
It follows that σ k s = ρ for demand k and slot s in {w k , ..., s}. By (5.9), we know that

σ k ′ s ′ = γ k ′ ,s ′ , for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.
Overall, we obtain that µ s = 0 for each slot s ∈ S , and for each k ′ ∈ K and s ′ ∈ S

σ k ′ s ′ =          γ k ′ ,s ′ if s ′ ∈ {1, ..., w k ′ -1}, ρ if k ′ = k and s ′ = s, 0 otherwise.
As a consequence, we have (µ, σ) = ρ(α, β) + γM .

Theorem 5.3.3. Consider a slot s ∈ S. Then, inequality u s ≤ 1 is facet defining for P sa (G, K, S).

Proof. Let us denote F s the face induced by inequality u s ≤ 1 given by

F s = {(u, z) ∈ P sa (G, K, S) : u s = 1}.
We denote inequality u s ≤ 1 by αu + βz ≤ λ. Let µu + σz ≤ τ be a valid inequality that defines a facet F of P sa (G, K, S). Suppose that F s ⊂ F = {(u, z) ∈ P sa (G, K, S) : µu + σz = τ }. To prove that F s is facet defining for P sa (G, K, S), it sufficient to show that there exist ρ ∈ R and γ ∈ R k∈K (w k -1) ) such that (µ, σ) = ρ(α, β) + γM . where This implies that µ s = 0.

D 113 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k ′ } : E(p k i ) ∩ E(p k j ) ̸ = ∅}. This ensures that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D
In a similar way, we can show that µ s = 0, for all slots s ∈ S \ {s}.

Next, we will show that, σ k s ′ = 0 for all k ∈ K and s ′ ∈ {w k , ..., s}. Consider a demand k ∈ K and a slot s ′ in {w k , ..., s}. Let S 114 = (U 114 , S 114 ) be the solution given by a) for one demand k ′ ∈ K \ {k}, we select the smallest slot index s k ′ ∈ {w k ′ , ..., s ′ } ∩ {s, ..., s + w k ′ -1} as last slot, b) we select the slot s k in {w k , ..., s} \ {s} \ {s ′ } as last slot for demand k with {s k - where Since µ s = 0 for all s ∈ S \ {s}, it follows that σ k s ′ = 0. In a similar way, we can show that σ k ′ s ′ = 0, for all k ′ ∈ K \ {k} and s ′ ∈ {w k ′ , ..., s}.

w k + 1, ..., s k } ∩ {s k ′ -w k ′ + 1, ..., s k ′ } = ∅ if E(p k ) ∩ E(p k ′ ) ̸ = ∅, c) for each demand k i ∈ K \ {k, k ′ } with i ∈ {1, ...,
D 114 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k, k ′ } : E(p k i ) ∩ E(p k j ) ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 114 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E(p k ) ∩ E(p k i ) ̸ = ∅ (
It follows that µ s = ρ for slot s in S.

We know from (5.9) that

σ k ′ s ′ = γ k ′ ,s ′ , for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.
We conclude that

µ s ′ =    ρ if s ′ = s, 0 otherwise,
and for each k ′ ∈ K and s ′ ∈ S

σ k ′ s ′ =    γ k ′ ,s if s ∈ {1, ..., w k ′ -1}, 0 otherwise.
As a consequence, we have (µ, σ) = ρ(α, β) + γM as desired.

Theorem 5.3.4. For a demand k ∈ K, inequality s s=w k z k s ≥ 1 is facet defining for P sa (G, K, S).

Proof. Let F k

S be the face induced by inequality

s s=w k z k s ≥ 1, that is F k S = {(x, z) ∈ P sa (G, K, S) : s s=w k z k s = 1}.

200

We denote inequality s s=w k z k s ≥ 1 by αu + βz ≤ λ. Let µu + σz ≤ τ be a valid inequality that defines a facet F of P sa (G, K, S). Suppose that F k S ⊂ F = {(u, z) ∈ P sa (G, K, S) : µu + σz = τ }. To prove that F k S is facet defining for P sa (G, K, S), it sufficient to show that there exist ρ ∈ R and γ ∈ R k∈K (w k -1) ) such that (µ, σ) = ρ(α, β) + γM . where Hence, µ s = 0.

D 116 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E(p k i ) ∩ E(p k j ) ̸ = ∅}. This verifies that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D
In a similar way, we can show that µ s = 0, for all slots s ∈ S.

Next, we will show that, σ k ′ s ′ = 0 for all k ′ ∈ K \ {k} and s ′ ∈ {w k ′ , ..., s}. Consider a demand k ′ in K \ {k} and a slot s ′ in {w k ′ , ..., s}. Let S 117 = (U 117 , S 117 ) be the solution given by a) we select slot s k = w k as last slot for demand k, b) we select the smallest slot index s k ′ from the set of slots I 117 k ′ given by where Since µ s = 0 for all s ∈ S, it follows that σ k ′ s ′ = 0. In a similar way, we can show that

I 117 k ′ = {w ki , ..., s k -w k } ∩ {s k + w ki , ..., s} \ {s ′ } if E(p k ′ ) ∩ E(p k ) ̸ = ∅ or I 117 k ′ = {w k ′ , ..., s} \ {s ′ } if not. c) for each demand k i ∈ K \ {k, k ′ } with i ∈ {1, ...,
D 117 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k, k ′ } : E(p k i ) ∩ E(p k j ) ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 117 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ if E(p k ′ ) ∩ E(p k i ) ̸ = ∅ (
σ k ′ s ′ = 0, for all k ′ ∈ K \ {k}
= {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E(p k i ) ∩ E(p k j ) ̸ = ∅}. Hence, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D117 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ if E(p k ′ ) ∩ E(p k i ) ̸ = ∅ (
= ( S118 k \ {s}) ∪ {s} for demand k such that {s ′ -w k + 1, ..., s ′ } ∩ {s" -w k ′ + 1, ..., s"} = ∅ for each k ′ ∈ K and s" ∈ S 119 k ′ with E 119 k ∩ E 119 k ′ ̸ = ∅.
The last slots assigned to the demands K \ {k} in S118 remain the same, i.e., S118

k" = S 119 k"

for each demand k" ∈ K \ {k}. Solution S 119 is feasible for the SA problem. The corresponding incidence vector (u S 119 , z S 119 ) belongs to F k S . Hence, solutions S 118 and S 119 satisfy equation µu + σz = τ . We then obtain that

µu S118 + σz S118 = µu S ′119 + σz S ′119 = µu S118 + σz S118 -σ k s + σ k s ′ - s∈U 118 \U 119 µ s + s′ ∈U 119 \U 118 µ s′ .
Since µ s = 0 for all s ∈ S, it follows that σ k s ′ = σ k s . In a similar way, we can show that σ k s ′ = σ k s , for all slots s, s ′ ∈ {w k , ..., s}.

Consequently, we obtain that σ k s = ρ for demand k and slot s in {w k , ..., s}. By (5.9), we know that

σ k ′ s ′ = γ k ′ ,s ′ , for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.
We then conclude that µ s = 0 for each slot s ∈ S , and for each k ′ ∈ K and s ∈ S

σ k ′ s =          γ k ′ ,s if s ∈ {1, ..., w k ′ -1}, ρ if k ′ = k and s ∈ {w k ′ , ..., s}, 0 otherwise.
As a result, we have (µ, σ) = ρ(α, β) + γM as desired.

Valid Inequalities and Facets

In what follows, we present several valid inequalities for P sa (G, K, S), and prove that they are facet-defining under certain conditions.

Interval-Capacity-Cover Inequalities

We start this section by introducing some classes of valid inequalities related to the knapsack constraints. Let us introduce the following conflict graph.

i +w k -1 z k s ≤ |K ′ | -1, (5.10)
is valid for P sa (G, K, S).

Proof. The interval I can cover at most |K ′ | -1 demands given that K ′ is a minimal cover for interval I.

Inequality (5.10) can be strengthened by extending each minimal cover K ′ ⊂ K for an interval I as follows.

Proposition 5.4.2. Let I = [s i , s j ] be an interval of contiguous slots in [1, s]. Let K ′ ⊆ K be a minimal cover for interval I = [s i , s j ] such that K ′ defines a clique in H sa , and Ξ(K ′ ) be a subset of demands in

K \ K ′ such that Ξ(K ′ ) = {k ∈ K \K ′ such that w k ≥ w k ′ and E(p k )∩E(p k ′ ) ̸ = ∅ ∀k ′ ∈ K ′ }. Then, the inequality k∈K ′ s j s=s i +w k -1 z k s + k ′ ∈Ξ(K ′ ) s j s ′ =s i +w k ′ -1 z k ′ s ′ ≤ |K ′ | -1, (5.11)
is valid for P sa (G, K, S).

Proof. The interval I = [s i , s j ] can cover at most |K ′ |-1 demands from the demands in K ′ ∪ Ξ(K ′ ) given that K ′ is a minimal cover for interval I = [s i , s j ] and the definition of the set Ξ(K ′ ) such that for each pair (k, k ′ ) with k ∈ K ′ and k ′ ∈ Ξ(K ′ ), the set (K ′ \ {k}) ∪ {k ′ } stills defining minimal cover for the interval I over edge e.

Furthermore, for each quadruplet (k, k ′ , k, k′ ) with k, k ′ ∈ K ′ and k, k′ ∈ Ξ(K ′ ), the set (K ′ \ {k, k ′ }) ∪ { k, k′ } stills defining minimal cover for the interval I given that

w k + w k ′ ≤ w k + w k′ . Theorem
P sa (G, K, S, K, I) = {(u, z) ∈ P sa (G, K, S) : k ′ ∈K\ K (v k ,v k ′ )∈H r ∀k∈ K s j s ′ =s i +w k ′ -1 z k ′ s ′ = 0}.

Proof. Necessity

If there exists an interval of contiguous slots I ′ in [1, s] with I ⊂ I ′ such that K defines a minimal cover for the interval I ′ . This means that {s i + w k -1, ..., s j } ⊂ I ′ .

As a result, inequality (5.10) induced by the minimal cover K for the interval I, it is dominated by another inequality (5.10) induced by the same minimal cover K for the interval I ′ . Hence, inequality (5.10) cannot be facet defining for the polytope P sa (G, K, S, I).

Sufficiency.

Let F I K denote the face induced by inequality (5.10), that is

F I K = {(u, z) ∈ P sa (G, K, S, K, I) : k∈ K s j s=s i +w k -1 z k s = | K| -1}. Denote inequality k∈ K s j s=s i +w k -1 z k s ≤ | K| -1 by αu + βz ≤ λ. Let µu + σz ≤ τ be a valid inequality that is facet defining F of P sa (G, K, S, I). Suppose that F I K ⊂ F = {(u, z) ∈ P sa (G, K, S, I) : µu + σz = τ }. In order to prove that inequality k∈ K s j s=s i +w k -1 z k s ≤ | K| - 1
is facet defining for P sa (G, K, S, I), we show that there exist ρ ∈ R and γ ∈ R k∈K (w k -1) ) such that (µ, σ) = ρ(α, β) + γM .

First, we show that µ s = 0 for all s ∈ S. Consider a slot s ∈ S. Let S 120 = (U 120 , S 120 ) be the solution given by a) for one demand k ′ from K, we select the smallest slot index where As a result, µ s = 0.

s k ′ in [{w k ′ , ..., s}\[{s i + w k ′ -1, ..., s j } ∪ {s, ..., s + w k ′ -1}]] \ {s, ..., s + w k i -1} (slot
D 120 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E(p k i ) ∩ E(p k j ) ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 120 i , • and s / ∈ {s k i -w k i + 1, ..., s k i } (slot
In a similar way, we can show that µ s = 0, for all slots s ∈ S. where Since µ s = 0 for all slots s ∈ S, it follows that σ k s ′ = 0. In a similar way, we can show that σ k s = 0, for all k ∈ K and s ∈ {w k , ..., s} with s /

D 122 i = {k j ∈ {k 1 , ..., k i-1 } ∩ K : E(p k i ) ∩ E(p k j ) ̸ = ∅}, c) for each demand k i ∈ K \ K with i ∈ {1, ...,
= {k j ∈ {k 1 , ..., k i-1 } ∪ K : E(p k i ) ∩ E(p k j ) ̸ = ∅}. This ensures that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 122 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E(p k i ) ∩ E(p k ) ̸ = ∅ (
∈ {s i + w k -1, ..., s j } if k / ∈ K.
Let prove that σ k s for all k ∈ K and s ∈ {s i + w k -1, ..., s j } are equivalent. Consider a demand k ′ ∈ K and a slot s ′ ∈ {s i + w k ′ -1, ..., s j } with k ′ ∈ K. Let S 124 = (U 124 , S 124 ) be the solution given by a) for one demand k" from K, we select the smallest slot index s k" in {w k" , ..., s} \ {s i + w k" -1, ..., s j } as last slot, b) for each demand k i ∈ K \ {k"} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 124 i given by

I 124 i = [ kj ∈D 124 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ {s i + w ki -1, ..., s j },
where

D 124 i = {k j ∈ {k 1 , ..., k i-1 } ∩ K : E(p k i ) ∩ E(p k j ) ̸ = ∅}, c) for each demand k i ∈ K \ K with i ∈ {1, ...,
= {k j ∈ {k 1 , ..., k i-1 } ∪ K : E(p k i ) ∩ E(p k j ) ̸ = ∅}. Hence, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 124 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ if E(p k i ) ∩ E(p k ′ ) ̸ = ∅ (
= (S 125 k \{s})∪{s} such that {s-w k +1, ..., s}∩{s ′ -w k ′ +1, ..., s ′ } = ∅ for each k ′ ∈ K and s ′ ∈ S 125 k ′ with E 125 k ∩ E 125 k ′ ̸ = ∅.
Solution S 125 is feasible for the SA problem. The corresponding incidence vector (u S 125 , z S 125 ) belongs to F I K . Hence, solutions S 124 and S 125 satisfy equation µu + σz = τ . We then obtain that µu S 124 + σz S 124 = µu S 125 + σz S 125 = µu S 124 + σz

S 124 + σ k ′ s ′ -σ k s + σ k s + s"∈U 125 \U 124 µ s" - s"∈U 124 \U 125 µ s" .
Since σ k s = 0 for s / ∈ {s i + w k -1, ..., s j } with k ∈ K, and µ s" = 0 for all s" ∈ S, it follows that σ k ′ s ′ = σ k s . The pair (k, k ′ ) are chosen arbitrarily in the minimal cover K, we then re-do the same procedure for all pairs (k, k ′ ) such that we find

σ k s = σ k ′ s ′ , for all pairs (k, k ′ ) ∈ K,
with s ∈ {s i + w k -1, ..., s j } and s ′ ∈ {s i + w k ′ -1, ..., s j }. We re-do the same procedure for each two slots s, s ′ ∈ {s i + w k -1, ..., s j } for each demand k ∈ K with k ∈ K such that σ k s = σ k s ′ , for all k ∈ K and s, s ′ ∈ {s i + w k -1, ..., s j }.

By (5.9), we know that

σ k ′ s ′ = γ k ′ ,s ′ , for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.
Overall, we obtain that µ s = 0 for each slot s ∈ S, and

σ k ′ s =          γ k ′ ,s if s ∈ {1, ..., w k ′ -1}, ρ if k ′ ∈ K and s ∈ {s i + w k ′ -1, , ..., s j }, 0 otherwise, for each k ′ ∈ K and s ∈ S.
As a consequence, we have (µ, σ) = ρ(α, β) + γM as desired.

Inequality (5.10) can then be lifted using a sequential lifting procedure [START_REF] Balas | Facets of the Knapsack Polytope From Minimal Covers[END_REF] to be facet defining and generate several facets for the polytope P sa (G, K, S).

Theorem 5.4.2. Let I = [s i , s j ] be an interval of contiguous slots. Let K ⊂ K be a minimal cover for interval I = [s i , s j ] such that K defines a clique in H sa . Let

K ′ ⊂ K \ K = {k 1 , ..., k q } such that K ∪{k 1 , ..., k q } defines a clique in H sa . Consider
the following sequence of knapsack problems defined as

                     z i = max j∈ K a j + i-1 j=1 β j a j , j∈ K w j a j + i-1 j=1 w k j a j ≤ |I| -w k i , a j ∈ {0, 1}, ∀j ∈ K ∪ {1, ..., i -1}, (5.12)
for all i ∈ {1, ..., q} with β j = | K| -1 -z j for all j ∈ {1, ..., i -1}. Then, the inequality k∈ K

s j s=s i +w k -1 z k s + q j=1 s j s ′ =s i +w k j -1 β j z k j s ′ ≤ | K| -1, (5.13)
is valid for P sa (G, K, S). Moreover, inequality (5.13) defines facet of P sa (G, K, S) if there does not exist an interval of contiguous slots

I ′ = [s ′ i , s ′ j ] in [1, s] with I ⊂ I ′ such that K defines a minimal cover for the interval I ′ .
Proof. It is trivial given that inequality (5.13) can never be dominated in P sa (G, K, S) if there does not exist an interval of contiguous slots [1, s] with I ⊂ I ′ such that K defines a minimal cover for the interval I ′ .

I ′ = [s ′ i , s ′ j ] in

Interval-Clique Inequalities

if w k + w k ′ > |I| and E(p k ) ∩ E(p k ′ ) ̸ = ∅. Let Q sa (G, K, S) = {(x, z) ∈ P sa (G, K, S) : k∈K s s=w k z k s = 1} be a semi- polytope of P sa (G, K, S).
Proposition 5.4.3. Let I = [s i , s j ] be an interval of contiguous slots in [1, s] with s i ≤ s j -1, and C be a clique in the conflict graph H ′E I with |C| ≥ 3. Then, inequality (2.39) is also valid for Q sa (G, K, S). Moreover, it is valid for P sa (G, K, S)

if 2w k > |I| for each v k ∈ C.
Proof. We use the same proof of proposition (2.4.13).

′ in [1, s] such that I ⊂ I ′ with • w k + w k ′ ≥ |I ′ | for each k, k ′ ∈ C, • 2w k ≥ |I ′ | + 1 and w k ≤ |I ′ | for each k ∈ C.
c) and there does not exist a slot s ∈ I such that s ∈ {s ′ -w k + 1, .., s ′ } for each k ∈ C and s ′ ∈ {s i + w k -1, .., s j }.

Proof. Neccessity.

We distinguish three cases a) if there exists a clique C ′ that contains all the demands k ∈ C.

F H ′E I C = {(u, z) ∈ P sa (G, K, S) : v k ∈C s j s=s i +w k -1 z k s = 1}. We denote inequality v k ∈C s j s=s i +w k -1 z k s ≤ 1 by αu + βz ≤ λ. Let µu + σz ≤ τ be a valid inequality that is facet defining F of P sa (G, K, S). Suppose that F H ′E I C ⊂ F = {(u, z) ∈ P sa (G, K, S) : µu + σz = τ }.
In order to prove that inequality v k ∈C s j s=s i +w k -1 z k s ≤ 1 is facet defining for P sa (G, K, S), we need to show that there exist ρ ∈ R and γ ∈ R k∈K (w k -1) ) such that (µ, σ) = ρ(α, β) + γM .

Let first show that µ s = 0 for all s ∈ S. Consider a slot s ∈ S. Let S 127 = (U 127 , S 127 ) be the solution given by a) for one demand k ′ from C, we select the smallest slot index s k ′ = {s i +w k ′ -1, ..., s j }\ {s, ..., s -w k ′ -1} as last slot (slot assignment constraint taking into account the possibility of adding slot s in the set of used slots U 127 ), b) for each demand k i ∈ C \ {k ′ } with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 127 i given by

I 127 i = [ kj ∈D 127 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}][∩{s i + w ki -1, ..., s j }] \ {s, ..., s + w ki -1},
where where Since µ s = 0 for all slots s ∈ S, it follows that σ k s ′ = 0. In a similar way, we can show that σ k s = 0, for all k ∈ K and s ∈ {w k , ..., s} with s /

D 127 i = {k j ∈ {k 1 , ..., k i-1 } ∩ C : E(p k i ) ∩ E(p k j ) ̸ = ∅}, c) for each demand k i ∈ K \ C with i ∈ {1, ...,
D 129 i = {k j ∈ {k 1 , ..., k i-1 } ∩ C : E(p k i ) ∩ E(p k j ) ̸ = ∅}, c) for each demand k i ∈ K \ C with i ∈ {1, ...,
= {k j ∈ {k 1 , ..., k i-1 } ∪ C : E(p k i ) ∩ E(p k j ) ̸ = ∅}. This ensures that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 129 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E(p k i ) ∩ E(p k ) ̸ = ∅ (
∈ {s i + w k -1, ..., s j } if v k / ∈ C.
Let prove that σ k s for all v k ∈ C and s ∈ {s i + w k -1, ..., s j } are equivalent. Consider a demand k ′ ∈ K and a slot s ′ ∈ {s i + w k ′ -1, ..., s j } with v k ′ ∈ C, and a solution S 131 = (U 131 , S 131 ) given by a) for one demand k from C, we select theslot Since σ k s = 0 for s / ∈ {s i + w k -1, ..., s j } with v k ∈ C, and µ s" = 0 for all s" ∈ S, it follows that σ k ′ s ′ = σ k s . Given that the pair (v k , v k ′ ) are chosen arbitrarily in clique C, we re-do the same procedure for all pairs (v k , v k ′ ) such that we find

s k = s i + w k -1 as last slot, b) for each demand k i ∈ C \ {k},
= {k j ∈ {k 1 , ..., k i-1 } ∪ C such that E(p k i ) ∩ E(p k j ) ̸ = ∅}. This ensures that • {s k i -w k i + 1, ..., s k i } ∩ {s -w k j + 1, ..., s} = ∅ for each k j ∈ D 131 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ if E(p k i ) ∩ E(p k ′ ) ̸ = ∅ (
σ k s = σ k ′ s ′ , for all pairs (v k , v k ′ ) ∈ C,
with s ∈ {s i + w k -1, ..., s j } and s ′ ∈ {s i + w k ′ -1, ..., s j }. We re-do the same procedure for each two slots s, s ′ ∈ {s i + w k -1, ..., s j } for each demand k ∈ K with

v k ∈ C such that σ k s = σ k s ′ , for all v k ∈ C and s, s ′ ∈ {s i + w k -1, ..., s j }, σ k s = σ k ′ s ′ , for all v k , v k ′ ∈ C, s ∈ {s i + w k -1, ..., s j } and s ′ ∈ {s i + w k ′ -1, ..., s j }.
Consequently, we obtain that σ k s = ρ for all v k ∈ C and s ∈ {s i + w k -1, ..., s j }. We know from (5.9) that

σ k ′ s ′ = γ k ′ ,s ′ , for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.
As a consequence, we obtain that µ s = 0 for each slot s ∈ S , and

σ k ′ s =          γ k ′ ,s if s ∈ {1, ..., w k ′ -1}, ρ if v k ′ ∈ C and s ∈ {s i + w k ′ -1, , ..., s j }, 0 otherwise,
for each k ′ ∈ K and s ∈ S. As a result, we have (µ, σ) = ρ(α, β) + γM as desired.

Interval-Odd-Hole Inequalities

F H ′E I H = {(u, z) ∈ P sa (G, K, S) : v k ∈H s j s=s i +w k -1 z k s = |H| -1 2 }.
We denote inequality v k ∈H where where

s j s=s i +w k -1 z k s ≤ |H|-
D 134 i = {k j ∈ {k 1 , ..., k i-1 } ∩ H : E(p k i ) ∩ E(p k j ) ̸ = ∅}. We let S 134 k i = {s k i } be the set of last slots assigned to demand k i , d) for each demand k i ∈ K \ H with i ∈ {1, ...,
R 134 i = {k j ∈ {k 1 , ..., k i-1 } ∪ H such that E(p k i ) ∩ E(p k j ) ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ R 134 i ,
• where where Since µ s = 0 for all slots s ∈ S, it follows that σ k s ′ = 0. In a similar way, we can show that

L 136 i = {k j ∈ {k 1 , ..., k i-1 } ∩ H : E(p k i ) ∩ E(p k j ) ̸ = ∅}. c) for each demand k i ∈ H \ H with i ∈ {1, ...,
D 136 i = {k j ∈ {k 1 , ..., k i-1 } ∪ H : E(p k i ) ∩ E(p k j ) ̸ = ∅}. This ensures that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 136 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E(p k i ) ∩ E(p k ) ̸ = ∅ (
σ k s ′ = 0, for demand k and s ′ ∈ {w k , ..., s} with s ′ / ∈ {s i + w k -1, ..., s j } if v k ∈ H.
We re-do the same procedure for all demand k ′ in K \ {k} such that where where if v k ∈ H, and µ s" = 0 for all s" ∈ S, it follows that σ k i s = σ k ′ s ′ . Given that the pair (v k , v k ′ ) are chosen arbitrarily in odd-hole H, we re-do the same procedure for all pairs (v k , v k ′ ) such that we find σ k s = σ k ′ s ′ , for all pairs (v k , v k ′ ) ∈ H, s ∈ {s i + w k -1, ..., s j } and {s i + w k ′ -1, ..., s j }.

σ k ′ s = 0, for all k ′ ∈ K \ {k} and s ∈ {w k ′ , ..., s} with s / ∈ {s i + w k ′ -1, ..., s j } if v k ′ ∈ H. Let prove that σ k ′ s ′ for all v k ′ ∈ H and s ′ ∈ {s i + w k ′ -1, ..., s j } are equivalent. Consider a demand k ′ ∈ K with v k ′ ∈ H and a slot s ′ ∈ {s i + w k ′ -1, ...,
L 138 i = {k j ∈ {k 1 , ..., k i-1 } ∩ H : E(p k i ) ∩ E(p k j ) ̸ = ∅}, c) for each demand k i ∈ H \ H with i ∈ {1, ...,
D 138 i = {k j ∈ {k 1 , ..., k i-1 } ∩ H : E(p k i ) ∩ E(p k j ) ̸ = ∅}, d) for each demand k i ∈ K \ H with i ∈ {1, ...,
Consequently, we obtain that σ k s = ρ for all v k ∈ H and s ∈ {s i + w k -1, ..., s j }. Overall, and using the result (5.9), we obtain that µ s = 0 for each slot s ∈ S , and for each k ′ ∈ K and s ∈ S. As a result, we have (µ, σ) = ρ(α, β) + γM as desired.

σ k ′ s =          γ k ′ ,

Slot-Assignment-Clique Inequalities

On the other hand, we also noticed that there may exist some cases that are not covered by inequality (2.25). For this, we provide an adapted definition of a conflict graph H E S for the SA problem and its associated inequality.

Definition 5.4.3. Let H ′E S be a conflict graph defined as follows. For all slot s ∈ {w k , ..., s} and demand k ∈ K, consider a node v k,s in H ′E S . Two nodes v k,s and v k ′ ,s ′ are linked by an edge in H ′E S if and only if

• k = k ′ ,
• or E k 1 ∩ E k ′ 1 ̸ = ∅ and {s -w k + 1, ..., s} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ when k ̸ = k ′ .

Based on the conflict graph H ′E S , we introduced the following inequalities. Proof. We use the same proof of proposition (2.4.17).

Theorem On the other hand, if there exists a slot s ′ ∈ S such that s ′ ∈ {s-w k +1, .., s} for each v k,s ∈ C, then inequality (2.43) is dominated by the non-overlapping inequality (5.4).

Hence, inequality (2.43) cannot be facet defining for P sa (G, K, S). Sufficiency. We denote inequality v k,s ∈C z k s ≤ 1 by αu + βz ≤ λ. Let µu + σz ≤ τ be a valid inequality that is facet defining F of P sa (G, K, S). Suppose that F H ′E S C ⊂ F = {(u, z) ∈ P sa (G, K, S) : µu + σz = τ }. In order to prove that inequality v k,s ∈C z k s ≤ 1 is facet defining for P sa (G, K, S), we show that there exist ρ ∈ R and γ ∈ R k∈K (w k -1) ) such that (µ, σ) = ρ(α, β) + γM . Since µ s = 0 for all slots s ∈ S, it follows that σ k s ′ = 0. In a similar way, we can show that σ k s = 0, for all k ∈ K and s ∈ {w k , ..., s} with v k,s / ∈ C.

Let prove that σ k s for all v k,s ∈ C are equivalent. Consider a demand k ′ ∈ K and a slot s ′ ∈ {w k ′ , ..., s} with v k ′ ,s ′ ∈ C, and a solution S 145 = (U 145 , S 145 ) given by a) select a pair of demand k and slot s from clique C (i.e., v k,s ∈ C) such that slot s k = s will be used as last slot for demand k, b) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 145 i given by Since σ k s = 0 for v k,s / ∈ C, and µ s" = 0 for all s" ∈ S, it follows that σ k ′ s ′ = σ k s . In a similar way, we can show that σ k s = σ k ′ s ′ , for all pairs (v k,s , v k ′ ,s ′ ) ∈ C, Consequently, we obtain that σ k s = ρ for all v k,s ∈ C. Overall, and using the result (5.9), we obtain that µ s = 0 for each slot s ∈ S, and Since µ s = 0 for all slots s ∈ S, it follows that σ k s ′ = 0. In a similar way, we can show that σ k s ′ = 0, for demand k and s ′ ∈ {w k , ..., s} with v k,s ′ / ∈ H.

I 145 i = [
σ k s =          γ k,
We re-do the same procedure for all demand k ′ in K \ {k} such that Since σ k s = 0 for v k,s / ∈ H, and µ s" = 0 for all s" ∈ S, it follows that σ k ′ s ′ = σ k s . Consequently, we obtain that σ k s = ρ for all v k,s ∈ H. By (5.9), we know that σ k ′ s ′ = γ k ′ ,s ′ , for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.

σ k ′ s = 0,
We then conclude that µ s = 0 for each slot s ∈ S , and

σ k ′ s =          γ k ′ ,s if s ∈ {1, ..., w k ′ -1}, ρ if v k ′ ,s ∈ H, 0 otherwise,
As a consequence, we have (µ, σ) = ρ(α, β) + γM as desired.

In the next section, we will derive some symmetry breaking inequalities for the SA subproblem in which some symmetrical solutions may appeared.

Symmetry-Breaking Inequalities

In this section, we address some symmetry issues that can appear when solving the SA problem.

Proposition 5.5.1. We ensure that for all slot s ∈ {1, ..., s -1} u s -u s+1 ≥ 0, (5.14) which means that a slot s + 1 can be used if and only if slot s is used.

Similar idea was proposed by Mendez-Diaz et al. [69][70] to break the symmetry for the vertex coloring problem.

To strengthen inequality (5.14), we propose the following inequalities. Similar idea was proposed by Friedman [START_REF] Friedman | Fundamental Domains for Integer Programs with Symmetries[END_REF]. However, the coefficient 2 |K|-k can provoques some numerical intractabilities for the computer machine [START_REF] Bendotti | Sub-symmetry-breaking inequalities and application to the Unit Commitment Problem[END_REF]. For this, we introduce the following inequality.

Inequality (5.20) is not valid for P sa (G, K, S) given that there exist some feasible solutions in P sa (G, K, S) which violate inequality (5.20) when for example a slot s ∈ S is used (i.e., u s = 1) but there is no demand k ∈ K which use slot s (i.e., k∈K min(s+w k -1,s) s ′ =s z k s ′ ). On the other hand, we ensure that all the optimization algorithms developed to solve the MWC problem can be used to compute the upper bound based on the conflict graph H r w . Based on inequalities (5.19) and (5.20), we conclude that the minimum number of slots to be used by the set of demands K while satisfying the SA constraints, it's equal to the total weight of the maximum weighted clique in the conflict graph H r w . Based on theoretical results presented in this chapter, we devise a Branch-and-Bound (B&B) and Branch-and-Cut algorithms to solve the SA problem. Moreover, we study the effectiveness of these algorithms and assess the impact of the valid inequalities on the effectiveness of the Branch-and-Cut algorithm.

Branch-and-Cut Algorithm

Description

Here we describe the Branch-and-Cut algorithm. We consider the following linear problem which can be seen as a strenghtned formulation for the compact formulation Note that the separation procedures of the valid inequalities presented in this chapter are still the same as those presented in chapter (2) for the C-RSA. However, we need to present the separation procedure for the interval-capacity-cover inequalities (5.10) as follows. Given a fractional solution (ū, z). We first consider an interval of contiguous slots I = [s i , s j ] which is identified by generating two slots s i and s j randomly in S with s j ≥ s i + 2 max k∈K w k . The separation problem associated with inequality (5.10) is NP-hard [START_REF] Klabjan | The complexity of cover inequality separation[END_REF] given that it consists in identifying a cover K * for the interval I = [s i , s j ], such that k∈ K *

s j s ′ =s i +w k -1 zk s ′ > | K * | -1.
For this, we use a greedy algorithm introduced by Nemhauser and Sigismondi [START_REF] Nemhauser | A Strong Cutting Plane/Branch-and-Bound Algorithm for Node Packing[END_REF] as follows. We first select a demand k ∈ K having the largest number of requested slot w k with s j s ′ =s i +w k -1 zk s ′ > 0, and then set K * to K * = {k}. After that, we iteratively add each demand k ′ ∈ K \ K * to K * with s j s ′ =s i +w k ′ -1 zk ′ s ′ > 0 and demand k ′ share an edge with all the demands already added K * , until a cover K * is obtained for the interval I over edge e with k∈ K * w k > |I|. We further derive a minimal cover from the cover K * by deleting each demand k ∈ K * if k ′ ∈ K * \{k} w k ′ ≤ |I|. We then add inequality (5.10) induced by the minimal cover K * for the interval I if it is violated, i.e., we add the following valid inequality to the current LP k∈ K *

s j s ′ =s i +w k -1 z k s ′ ≤ | K * | -1.

Primal Heuristic

Let us present now a primal heuristic useful to boost the performance of the Branchand-Cut algorithm. It is based on a hybrid method between a local search algorithm and a greedy-algorithm. Given an optimal fractional solution (ū, z) in a certain node of the B&C tree, it consists in constructing an integral solution and "feasi- We also consider the valid inequalities (5. [START_REF] Cheng | Routing and Spectrum Assignment Algorithm based on Spectrum Fragment Assessment of Arriving Services[END_REF]) introduced previously that are shown to be as a precomputed lower bounds for the SA problem. They can be separated as follows. For each demand k ∈ K, we use a greedy algorithm introduced by Nemhauser and Sigismondi [START_REF] Nemhauser | A Strong Cutting Plane/Branch-and-Bound Algorithm for Node Packing[END_REF] to generate a maximum clique in H sa containning demand k. We first set Ck to Ck = {k}. After that, we iteratively add each demand k ′ ∈ K \ Ck to Ck such that demand k ′ must share an edge with all the demands already generated in Ck . We further add inequality (5.19) induced by clique Ck for demand k to the compact formulation (5.1)-(5.8)

s∈S u s ≥ k ′ ∈ Ck w k ′ .
Based on this, we provide a comparative study between the B&B (without additional valid inequalities) and the B&C (with additional valid inequalities) algorithms. Our objective in this study is to show the efficiency of the inequalities we have introduced for solving the SA problem. We present some computational results using several instances with a number of demand ranges in {10, 20, 30, 40, 50, 100, 150, 200, 250, 300} and s up to 320 slots. We use two types of topologies: real, and realistic ones from SND-LIB already described in Table 3.1. We first run our B&C algorithm with SCIP in which our valid inequalities are used, and all the Scip's internal cuts are deactivated. We call this run Own B&C SCIP. Then, we run the B&C algorithm with SCIP, and activating all the internal cuts we had deactivated prior in run 1.

We call this run B&B SCIP. Tables 5.1 and 5.2 below report the results obtained for the two runs. For each run and each instance, we report the number of nodes in B&C tree (Nbr Nd), the optimality gap (Gap), the number of violated inequalities added during the algorithm (Nbr Cuts), and the total CPU Time (TT) in seconds.

Finally, notice that each line of each table 5.1 and 5.2, corresponds to the average results of 4 instances.

The results show that Own B&C SCIP is able to solve several instances to optimality that are not solved to optimality when using the B&B SCIP even if Scip uses its proper cuts. Furthermore, we noticed that our valid inequalties allow solving to optimality more instances than B&B SCIP. Also, they enable reducing the average number of nodes in the B&C tree for several instances such that there exist some cases that we are able to solve some instances in the root of the B&C tree which is not the case when using the B&B. On the other hand, and looking at the instances that are not solved to optimality (i.e., gap > 0, 00), adding valid inequalities decreases the average gap for several instances and much more for the large instances with a number of demands |K| ≥ 150. However, there exist a few instances very rare, for example the triplet (German, 300, 320), in which adding valid inequalities does not improve the results of the B&B algorithm. Based on these results, we ensure that using the valid inequalities allows obtaining tighter LP bounds and improve the effectiveness of the B&B algorithm such that the B&C algorithm is able to beat the B&B algorithm even if Scip use its proper cuts that are shown to be very efficient for another optimization problems studied in the literature.

Concluding Remarks

In this chapter, we have studied the Spectrum Assignment sub-problem. We have introduced an integer linear programming compact formulation, and further investigated the facial structure of the associated polyhedron. Moreover, we have derived several valid inequalities that are facet-defining under sufficient conditions. Using the polyhedral results and the separation procedures, we have devised a Branchand-Cut (BC) algorithm to solve the problem. We have also presented experimental results. The results have shown the effectiveness of the valid inequalities such that the B&C algorithm is shown to be very performant for solving large-scale instances of the problem. It could be very interesting to study the impact of the symmetry breaking inequalities on the performance of the Branch-and-Cut algorithm. valid inequalities allowed improving the effectiveness of the B&C&P algorithm. On the other, we have presented a comparative study between the B&C, B&P, and B&C&P algorithms. The results have shown that the B&C&P algorithm is able to provide optimal solutions for several instances, which is not the case for the B&C algorithm within the CPU time limit (5 hours). Moreover, both B&C and B&P algorithms perform well. However, some instances are still difficult to solve with both B&C, B&P and B&C&P algorithms. For this, some enhancements are further investigated and integrated into our algorithms. They are based on a warm-start algorithm using some metaheuristics, and a primal heuristic using a hybrid method between a greedy algorithm and local search algorithm that is shown to be very useful to obtain good primal bounds. Moreover, we introduce some symmetry-breaking inequalities that allow avoiding the equivalents sub-problems in the different enumeration trees of B&C, B&P, and B&C&P algorithms.

Instances

Afterward, we have studied the Spectrum Assignment (SA) sub-problem when the routing is trivial or a routing path is pre-selected for each demand. First, we have presented a compact formulation for the SA problem. We have carried out an investigation of the associated polytope. Moreover, we have identified several valid inequalities for the polytope, some of them come from those that are already proposed for the C-RSA. We have proved that they are facet defining under certain necessary and sufficient conditions. They were further incorporated within a Branch-and-Cut e) and provide a deeper comparative study between the algorithms [START_REF] Accorsi | Guidelines for the Computational Testing of Machine Learning approaches to Vehicle Routing Problems[END_REF].

  istrative au LIMOS, pour sa sympathie, son aide précieuse et son efficacité dans l'organisation et la résolution des problèmes administratifs. J'associe à ces remerciements Madame Fatiha Bendali, Enseignante-Chercheuse à l'Université de Clermont Auvergne, et Monsieur Jean Mailfert, Maitre de conférences à l'Université de Clermont Auvergne, pour leur gentillesse, leur écoute, leurs conseils et le temps conséquent qu'ils m'ont accordés. Je les remercie pour tout cela. Ces remerciements seraient incomplets sans une mention particulière pour le groupe POC (Polyèdre et Optimisation Combinatoire). Un énorme merci pour les discussions intéressantes que nous avons pu avoir autour de l'Optimisation Combinatoire et les approches polyédrales en particulier. Ils ont si généreusement contribué à la réussites des différents évènements organisés pour les doctorants en France (JPOC, séminaires, ROADEF,...), et une conférence internationale ISCO qui donne l'opportunité d'échanger avec des chercheurs très connus à l'échelle internationale et de profiter de leurs expériences scientifiques et professionnelles. Chaque jour, je suis reconnaissant à tous mes anciens professeurs de Master AN-DROIDE à la Sorbonne Université (ex UPMC-Université Pierre et Marie Curie, Paris), mes encadrants de stage du Master à EDF (Paris) et ceux de Licence Recherche Opérationnelle à l'USTHB (Alger), qui par leurs paroles, leurs écrits, leurs conseils et leurs commentaires m'ont aidé à exceller tout au long de ces années.

  Nous fournissons donc une analyse théorique approfondie et concevons des algorithmes exacts de type coupes, branchements et génération de colonnes pour résoudre le problème CRSA en considérant des réseaux de taille réaliste. Pour ce faire, notre contribution consiste à introduire un programme linéaire en nombres entiers basée sur des coupes, où le nombre de variables n'augmente que de manière polynomiale avec la taille de l'instance traitée. En outre, nous étudions la structure polyédrale du polyèdre associé, et dérivons plusieurs classes d'inégalités valides. Nous donnons quelques conditions nécessaires et suffisantes pour que certaines inégalités valides soient des facettes pour le polyèdre associé. Nous élaborons ensuite des procédures de séparation pour ces inégalités valides. Ces inégalités sont ensuite utilisées dans la relaxation linéaire afin d'obtenir des bornes duales plus serrées. En se basant sur ça, nous développons un algorithme de coupes et branchements pour le problème CRSA. D'autre part, nous avons proposé une nouvelle formulation étendue basée sur des chemins, où les variables sont associées à tous les chemins possibles pour chaque demande en trafic induisant donc une explosion de nombre de variables qui croissent de manière exponentielle et en parallèle avec la croissance de la taille de l'instance traitée. Nous développons également un algorithme de génération de colonnes pour la résolution de sa relaxation linéaire. Les inégalités valides de la formulation de coupes restent aussi valides pour le polyèdre associé à cette formulation étendue. Nous développons ensuite un algorithme exact qui combine un algorithme de coupes et branchements avec un algorithme de génération de colonnes pour résoudre le problème CRSA. D'autre part, vu la complexité du problème, le problème CRSA peut être décomposé en deux sous-problèmes de telle sorte que le routage contraint précède l'assignation du spectre (CR+SA). Nous analysons la structure polyédrale du sous-problème d'assignation du spectre (SA) lorsque le routage est déjà établi. Tout d'abord, nous proposons une formulation compacte pour le problème SA. Nous étudions ensuite la structure du polyèdre associé. Nous définissons quelques classes supplémentaires d'inégalités valides et introduire quelques inégalités pour bien gérer la symétrie afin de supprimer les solutions symétriques obtenues lors de la résolution du problème. Nous donnons également quelques conditions nécessaires et suffisantes pour que certaines inégalités valides définissent des facettes pour le polyèdre. Des procédures de séparation sont ensuite proposées pour certaines de ces inégalités valides et qui seront utilisées par la suite pour obtenir des bornes plus étroites dans la relaxation linéaire. Nous élaborons ensuite un algorithme de coupes et branchements pour le sous problème SA. A la fin de chaque étape, nous examinons plus en profondeur l'efficacité et le comportement de nos algorithmes, et augmentons leurs efficacités grâce à plusieurs améliorations basant sur des heuristiques primales et aussi quelques techniques de branchement qui pourraient offrir une promesse d'amélioration par rapport aux méthodes existantes compte tenu des réseaux de taille réaliste de SndLib, et d'autres de taille réelle. Nous menons aussi une étude comparative d'efficacité entre les différents algorithmes proposés dans cette thèse.

Figure 1 :

 1 Figure 1: Historical Evolution of Optical Transport Networks.

Figure 1 . 1 :

 11 Figure 1.1: Relation between P, NP, NP-complete and NP-hard problems.

Figure 1 . 2 :

 12 Figure 1.2: conv(S) vs S.

Figure 1 .

 1 Figure 1.3 illustrates the polyhedron P , valid inequality, face, facet and extreme point.

Figure 1 . 3 :

 13 Figure 1.3: Geometric interpretation for the polyhedron P , valid inequality, face, facet and extreme point.

12. 5

 5 GHz (where FixedGrid networks use 50 GHz, the width of a wavelength) as recommended by ITU-T [2]. See for example Figure 1.4 which shows a fixed-grid with 4 wavelengths of 50 GHz to serve 4 demandes of two of 10 Gb/s, one of 40 Gb/s, one of 100 Gb/s. However, in the flex-grid we use just 9 slots of frequency 12.5 GHz to serve these demands.

Figure 1 . 4 :

 14 Figure 1.4: FixedGrid Vs FlexGrid.

2 Cut

 2 our work, we focus on a variant of the RSA problem, called Constrained-Routing and Spectrum Assignment Problem (C-RSA). Chapter Formulation and Polyhedra for the C-RSA Problem 2.1 The Constrained-Routing and Spectrum Assignment Problem The Constrained-Routing and Spectrum Assignment Problem can be stated as follows. We consider an optical spectrum of s ∈ Z + available contiguous frequency slots, denoted by S = {1, . . . , s}. A SFON topology can be represented by an undirected, loopless, and connected graph G = (V, E), with V is the set of vertices representing the optical nodes (data centers, users, stations,...), and E the set of links representing optical-fibers. A length ℓ e ∈ R + (in kms), a cost c e ∈ R + , and a set of s of contiguous frequency slots are associated with each edge e. Let K be a set of non-splittable traffic demands. Each demand k ∈ K has an origin nodeo k ∈ V , a destination node d k ∈ V \ {o k }, a slot-width w k ∈ Z + ,and a transmissionreach lk ∈ R + (in kms). The C-RSA consists in determining for each demand k ∈ K, a (o k ,d k )-path p k in G (non-splittable demands) such that e∈E(p k ) l e ≤ lk (tranmission-reach constraint), and an interval of contiguous frequency slots S k ⊆ S of width equal to w k (continuity and contiguity constraint)

Fig. 2 .

 2 Fig. 2.1 provides a feasible solution for an instance of the C-RSA problem containing 4 demands routed in a graph G consisting of 7 nodes and 10 edges. Each edge e is specified by a triplet [l e , c e , s] with s = 9.

Figure 2 . 1 :

 21 Figure 2.1: Set of established paths and spectrums in graph G (Fig. 2(a)) for the set of demands {k 1 , k 2 , k 3 , k 4 } defined in Table 2(b).

For each demand k and

  node v, one can compute a shortest path between each of the pairs of nodes (o k , v), (v, d k ). If the length of the (o k , d k )-paths formed by the concatenation of the shortest paths (o k , v) and (v, d k ) is greater that lk then node v cannot be in a path routing demand k, and we then say that v is a forbidden node for demand k. Let V k 0 denote the set of forbidden nodes for demand k ∈ K. Note that using Dijkstra's algorithm, one can identify in polynomial time the forbidden nodes V k 0 for each demand k ∈ K. On the other hand and regarding the edges, for each demand k and edge e = ij, one can compute a shortest path between each of the pairs of nodes (o k , i), and (j, d k ), and (o k , j), and (i, d k ). If the length of the (o k , d k )-path formed by e together with the shortest paths (o k , i) and (j, d k )

  a) graph G contains at least one feasible path between o k and d k , for all k ∈ K, b) the number of slots s is largely sufficient to route all the demands, c) for each demand k ∈ K and e ∈ E \ (E k 0 ∪ E k 1 ), there exists at least a feasible route E k between o k and d k such that e ′ ∈E k ℓ e ′ +ℓ e ≤ lk , and for each e ′ ∈ E k , the edges (e, e ′ ) are compatible edges for demand k.

Proposition 2 . 4 . 1 .

 241 Consider an edge e ∈ E with K e ̸ = ∅. Let s be a slot in S.

Proposition 2 . 4 . 6 .

 246 Consider an edge e ∈ E. Let I = [s i , s j ] be an interval of contiguous slots in[1, s]. Let K ′ ⊆ K e be a minimal cover for interval I = [s i , s j ]

Definition 2 . 4 . 4 .

 244 Consider an edge e ∈ E. Let I = [s i , s j ] be an interval of contiguous slots in [1, s] with s i ≤ s j -1. Consider the conflict graph H e I defined as follows. For each demand k ∈ K with w k ≤ |I| and e / ∈ E k 0 , consider a node v k in H e I . Two nodes v k and v k ′ are linked by an edge in

  we take into account the possibility of using edge e ′ in the selected path E 53 k to route demand k in solution S 53 ). We let S 53 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S 53 is feasible for the problem. The corresponding incidence vector (x S 53 , z S 53 ) belongs to F H e I C . Then we derive a solution S 54 = (E 54 , S 54 ) obtained from S 53 by adding edge e ′ ∈ E \ (E k 0 ∪ E k 1 ) for the routing of demand k in solution S 53 which means that E 54 k = E 53 k ∪ {e ′ }. The last slots assigned to the demands K, and paths assigned the set of demands K \ {k} in S 53 remain the same in solution S 54 , i.e., S 54 k = S 53 k for each k ∈ K, and E 54 k ′ = E 53 k ′ for each k ′ ∈ K \ {k}. S 54 is clearly feasible for the problem. The corresponding incidence vector (x S 54 , z S 54 ) belongs to F H e I C . Hence, solutions S 53 and S 54 satisfy equation µx + σz = τ . It follows that

  be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′53 is clearly feasible for the problem. Hence, the corresponding incidence vector (x S ′53 , z S ′53 ) belongs to F H e I C . Then we derive solution S 55 from S ′53 by adding slot s ′ as last slot to demand k in solution S ′53 , i.e., S 55 k = S ′53 k ∪ {s ′ }. Solution S 55 is feasible for the problem. The corresponding incidence vector (x S 55 , z S 55 ) belongs to F H e I C . Hence, solutions S ′53 and S 55 satisfy equation µx + σz = τ . We have so

Theorem 2 . 4 . 7 .

 247 Consider an edge e ∈ E. Let I = [s i , s j ] be an interval of contiguous slots. Let C be a clique in the conflict graph H e I with |C| ≥ 3, v k ∈C w k ≤ sk ′ ∈Ke\C w k ′ , and |{s i +w k -1, ..., s j }| ≥ w k for each demand k with v k ∈ C ∪C e . Let C e ⊆ K e \ C be a clique in the conflict graph H e I such that w k + w k ′ ≥ |I| + 1 for each v k ∈ C and v k ′ ∈ C e . Then, inequality (2.37) is facet defining for P(G, K, S) if and only if a) there does not exist a demand k" ∈ K e \ C e with w k + w k" ≥ |I| + 1 for each v k ∈ C, and w k ′ + w k" ≥ |I| + 1 for each v k ′ ∈ C e , b) and there does not exist an interval I ′ of contiguous slots with I ⊂ I ′ such that C ∪ C e defines also a clique in the associated conflict graph H e I ′ .

  this means that there exists a clique C in the conflict graph H E I of cardinality equals to |C| ≥ 3 with k, k ′ ∈ C. As a result, inequality (2.38) is dominated by inequality (2.39) induced by clique C. Hence, inequality (2.38) is not facet defining for P(G, K, S). b) if there exists an interval of contiguous slots I ′ in [1, s] such that I ⊂ I ′ with w k + w k ′ ≥ |I ′ |, w k ≤ |I ′ |, and w k ′ ≤ |I ′ |. This means that inequality (2.38) induced by the two demands k, k ′ for the interval I is dominated by inequality (2.38) induced by the same demands for the interval I ′ .

70 i 70 i= kj ∈D 70 i

 707070 Consider a demand k′ with v k′ ∈ C and a slot s′ ∈ {s i + w k′ -1, ..., s j }. Let S 70 = (E 70 , S 70 ) be the solution given by a) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E 70 k i be the set of edges involved in a shortest path between o k i and d k i , b) select a subset of demands H from H with | H| = |H|-1 2 , c) for each demand k i ∈ H with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 70 i given by {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ {s i + w ki -1, ..., s j },whereL 70 i = {k j ∈ {k 1 , ..., k i-1 } ∩ H : E 70 k i ∩ E 70 k j ̸ = ∅}, d) for each demand k i ∈ H \ H with i ∈ {1, ...,|K|}, we select the smallest slot index s k i in the set of slots I 70 i given by I {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} \ {s i + w ki -1, ..., s j },

  we take into account the possibility of adding slot s′ as a last slot in the selected last slots S70 k′ to route demand k′ in solution S 70 ). We let S 70 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S 70 is feasible for the problem. Hence, the corresponding incidence vector (x S 70 , z S 70 ) belongs to F H E I H,C . Then consider the solution S 71 obtained from S 70 as belows a) remove all the last slots si totally covered by the interval I and which has been selected by each demand k i ∈ H in solution S 70 (i.e., s ∈ S 70 k i and s ∈ {s i + w

S 70 k

 70 c) and add slot s′ to the set of last slots S 70 k′ assigned to demand k′ in solution S 70 , i.e., S 71 k′ = S 70 k′ ∪ {s ′ }, d) without changing the set of last slots assigned to the demands K \ H, i.e., S 71 k = for each demand K \ H. Solution S 71 is feasible for the problem. The corresponding incidence vector (x S 71 , z S 71 ) belongs to F H E I H,C . Hence, solutions S 70 and S 71 satisfy equation µx + σz = τ . We have so µx S 70 + σz S 70 = µx S 71 + σz S 71 = µx S 70 + σz S 70 + σ k′ k′

Proposition 2 . 4 . 16 .

 2416 Consider an edge e ∈ E. Let C be a clique in the conflict graph H e S with |C| ≥ 3, and k∈C w k ≤ sk ′ ∈Ke\C w k ′ . Then, the inequality

  c) and 2w k ≥ |I| + 1 and w k ≤ |I| for each v k ∈ C.Proof. Neccessity.If C is not maximal clique in the conflict graph H E S , this means that inequality (2.43) can be dominated by another inequality associated with a clique C ′ such that C ⊂ C ′ without changing its right-hand side. Moreover, if there exists an interval of contiguous slots I

  we take into account the possibility of adding slot s ′ in the selected set of last slots S 80 k ′ to route demand k ′ in solution S 80 ). We let S 80 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S 80 is feasible for the problem. Hence, the corresponding incidence vector (x S 80 , z S 80 ) belongs to F H E S H . After that, we derive the solution S ′80 = (E ′80 , S ′80 ) from S 80 by a) and adding slot s ′ as last slot to demand k ′ , i.e., S ′80 k ′ = S 80 k ′ ∪ {s ′ } for demand k ′ , b) and modifying the last slots assigned to demand k by adding a new last slot s and removing the last slot s ∈ S 80

  . Then we derive solution S 83 obtained from S 82 by a) adding slot s ′ as last slot to demand k ′ , i.e., S 83 k ′ = S 82 k ′ ∪ {s ′ } with v k ′ ,s ′ ∈ C, b) and modifying the last slots assigned to each demand k ∈ { k ∈ K with v k,s ∈ H82 } by adding a new last slot sk and removing the last slot s

1 2

 1 k,e ∈H x k e ≤ |H|-by αx + βz ≤ λ. Let µx + σz ≤ τ be a facet defining inequality for P(G, K, S) and F = {(x, z) ∈ P(G, K, S) : µx + σz = τ }.

Proposition 2 . 4 . 24 .

 2424 Consider a demand k ∈ K. Let p be a minimal infeasible sub-path for demand k in G. Then, the inequality e∈E(p) x k e ≤ |E(p)| -1.(2.52)

Theorem 2 . 4 . 21 .

 2421 Consider an edge e in E. Let C be a minimal cover in K for edge e. Then, inequality (2.54) is facet defining for the polytope P(G, K, S, C, e) where P(G, K, S, C, e) = {(x, z) ∈ P(G, K, S) :

Proposition 2 . 6 . 1 .

 261 Consider a demand k ∈ K. Then, the inequality e∈E c

  is valid for P(G, K, S).

  . It consists in computing a maximum flow/minimum cut in G of demand k by assigning a positif weight xk e for each edge e in G. For this, we use a C++ library proposed by the LEMON GRAPH library [59] which calls the algorithm of Goldberg and Tarjan for the minimum cut computation. Based on this, we conclude that the separation of the cut inequalities (2.2) can be done in O(|V | 2 * |E| * |K|) in the worst case.

  all the nodes already assigned to C * and N * . At the end, we add inequality (2.46) induced by clique C * ∪ N * to the current LP, i.e., v k,e ∈C * x k e + v k ′ ,e ′ ∈N *

  ′ s = 0 and xk e ̸ = 0 for each e ∈ E(p) while respecting the non-overlapping constraint with the set of demands that precede demand k ′ in the list L. The complexity of this algorithm is bounded by O(|K| * |S| * |P | * log(|K|)) where |P | = max k∈K R k .

  54)(2.32), and clique-based inequalities (2.43), (2.42) and (2.36), are generated along the B&C algorithm. However, the number of clique-based inequalities (2.43) generated is very less compared with other inequalities. Based on these results, we conclude that the valid inequalities are very useful to obtain tighter LP bounds using Gurobi and Scip. On the other hand, the clique-based inequalities (2.46), cover-based inequalities (2.49), and the different families of odd-hole inequalities, are shown to be not efficient for the instances tested such that the number of their violated inequalities generated is very less and equal to 0 for several instances. However, they are still very interesting from a theoretical point of view. Based on this, the separation of our valid inequalities, is performed along the B&C algorithm (using Cplex, Gurobi and Scip) in the following order a) edge-capacity-cover inequalities (2.54), b) edge-Interval-Capacity-Cover inequalities (2.32), c) edge-slot-assignment-clique inequalities (2.42), d) edge-interval-clique inequalities (2.36), e) slot-assignment-clique inequalities (2.43).

  available contiguous frequency slots, denoted by S = {1, . . . , s}. A spectrally flexible optical network can be represented by an undirected, loopless, and connected graph G = (V, E), with V is the set of vertices representing the optical nodes (data centers, users, stations,...), and E the set of links representing the optical-fibers. A length ℓ e ∈ R + (in kms), a cost c e ∈ R + , and a set of s of contiguous frequency slots are associated with each edge e. Let K be a multiset of demands such that each demand k is specified by an origin node o k ∈ V , a destination node d k ∈ V \ {o k }, a slot-width w k ∈ Z + , and a routing path p k from its origin o k to its destination d k through G. The SA consists of determining for each demand k ∈ K an interval of contiguous frequency slots S k ⊂ S of width equal to w k (continuity and contiguity constraints) such that S k ∩ S k ′ = ∅ for each pair of demands k, k ′ ∈ K (k ̸ = k ′ ) with paths sharing an edge , i.e., E(p k ) ∩ E(p k ′ ) ̸ = ∅ , while optimizing the number of slots allocated in S.

  assignment constraint taking into account the possibility of adding slot s in the set of used slots U 105 ), We let S 105 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. c) we let U 105 be the set of slots used in S such that for each demand k and last slot s k ∈ S 105 k and s ′ ∈ {s k -w k + 1, ..., s k }, we have s ′ ∈ U 105 . S 105 is feasible for the SA problem. Hence, the corresponding incidence vector (u S 105 , z S 105 ) belongs to P sa (G, K, S). Then we derive a solution S 106 = (U 106 , S 106 ) obtained from S 105 by adding slot s as an used slot in U 106 without modifying the last slots assigned to the demands K in S 105 which remain the same in solution S 106 i.e., S 105 k = S 106 k

Theorem 5 . 3 . 1 .

 531 The dimension of P sa (G, K, S) is given bydim(P sa (G, K, S)) = |K| * |S| + |S| -r ′ = |K| * |S| + |S| -k∈K (w k -1).

First 109 i

 109 , let show that µ s = 0 for all s ∈ S. Consider a slot s ∈ S, and a solution S 109 = (U 109 , S 109 ) given by a) we select the smallest slot index s k in {w k , ..., s} \ [{s, ..., s + w k -1} ∪ {s}] as last slot for demand k (slot assignment constraint taking into account the possibility of adding slot s in the set of used slots U 109 ), b) for each demand k i ∈ K with i ∈ {1, ..., |K|} \ {k}, we select the smallest slot index s k i in the set of slots I 109 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] \ {s, ..., s + w ki -1}

  assignment constraint taking into account the possibility of adding slot s in the set of used slots U 109 ), Let S 109 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. c) we let U 109 be the set of slots used in S such that for each demand k ′ ∈ K and last slot s k ′ ∈ S 109 k and s ′ ∈ {s k ′ -w k ′ + 1, ..., s k ′ }, we have s ′ ∈ U 109 . S 109 is clearly feasible for the SA problem. Hence, the corresponding incidence vector (u S 109 , z S 109 ) belongs to F k s . Then consider the solution S ′109 = (U ′109 , S ′109 ) obtained from S 109 by adding slot s as an used slot in U ′109 without modifying the last slots assigned to the demands K in S 109 which remain the same in solution S ′109 i.e., S 109 k = S ′109 k

First, let show 113 i

 113 that µ s ′ = 0 for all s ′ ∈ S \ {s}. Consider a slot s ∈ S \ {s}, and a solution S 113 = (U 113 , S 113 ) given by a) for one demand k ′ ∈ K, we select the smallest slot index s k ′ ∈ [{w k ′ , ..., s ′ }∩{s, ..., s+ w k ′ -1}] \ {s, ..., s + w k ′ -1} as last slot, b) for each demand k i ∈ K \ {k ′ } with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 113 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] \ {s, ..., s + w ki -1},

  we take into account the possibility of adding slot s ′ in the set of last slots S 114 k assigned to demand k in solution S 114 ), Let S 114 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. d) we let U 114 be the set of slots used in S.

S

  114 is clearly feasible for the SA problem. The corresponding incidence vector (u S 114 , z S 114 ) belongs to F s . Then consider the solution S 115 = (U 115 , S115 ) obtained from S 114 by adding slot s ′ as last slot to demand k without modifying the last slots assigned to the demands K \{k} in S 114 k remain the same in solution S 115 i.e., S

First, let show 116 i

 116 that µ s = 0 for all s ∈ S. Consider a slot s ∈ S, and a solutionS 116 = (U116 , S 116 ) given by a) we select the smallest slot index s k in {w k , ..., s} \ {s, ..., s + w k -1} as last slot for demand k (slot assignment constraint taking into account the possibility of adding slot s in the set of used slots U 116 ), b) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 116 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] \ {s, ..., s + w ki -1}

120 i 120 i= 120 i

 120120120 assignment constraint taking into account the possibility of adding slot s in the set of used slots U 120 ), b) for each demand k i ∈ K \ {k ′ } with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 120 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}][∩{s i + w ki -1, ..., s j }] \ {s, ..., s + w ki -1},whereD {k j ∈ {k 1 , ..., k i-1 } ∩ K : E(p k i ) ∩ E(p k j ) ̸ = ∅}, c) for each demand k i ∈ K \ K with i ∈ {1, ...,|K|}, we select the smallest slot index s k i in the set of slots I 120 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] \ {s, ..., s + w ki -1}

120 k = S 121 k

 120121 assignment constraint taking into account the possibility of adding slot s in the set of used slots U 120 ). We let S 120 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}.d) let U 120 be the set of slots used in S such that for each demand k ∈ K and last slots k ∈ S 120 k and s ′ ∈ {s k -w k + 1, ..., s k }, we have s ′ ∈ U 120 .206S 120 is clearly feasible for the SA problem. Hence, the corresponding incidence vector (u S 120 , z S 120 ) belongs to F I K . Then consider the solution S 121 = (U 121 , S 121 ) obtained from S 120 by adding slot s as an used slot in U 121 without modifying the last slots assigned to the demands K in S 120 which remain the same in solution S 121 i.e.,Sfor each demand k ∈ K. S 121 is feasible for the SA problem. Hence, the corresponding incidence vector (u S 121 , z S 121 ) belongs to F I K . Hence, solutions S 120 and S 121 satisfy equation µu + σz = τ . We then obtain that µu S 120 + σz S 120 = µu S 121 + σz S 121 = µu S 120 + σz S 120 + µ s.

Theorem 5 . 4 . 3 .

 543 Let I = [s i , s j ] be an interval of contiguous slots in [1, s] with s i ≤ s j -1, and C be a clique in the conflict graph H ′E I with |C| ≥ 3, and 2w k > |I| for each v k ∈ C. Then, inequality (2.39) is facet defining for P sa (G, K, S) if and only if a) C is a maximal clique in the conflict graph H ′E I , b) and there does not exist an interval of contiguous slots I

Proposition 5 . 4 . 5 .

 545 Let C be a clique in the conflict graph H ′E S with |C| ≥ 3. Then, inequality (2.43) is valid for Q sa (G, K, S). Moreover, it is valid for P sa (G, K, S) if{s -w k + 1, ..., 1} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ for each (v k,s , v k ′ ,s ′ ) ∈ C.

v

  k,s ∈C s] ⊂ I, • and w k + w k ′ ≥ |I| + 1 for each (v k , v k ′ ) ∈ C,• and 2w k ≥ |I| + 1 and w k ≤ |I| for each v k ∈ C.

3 .

 3 and there does not exist a slot s ′ ∈ S such that s ′ ∈ {s -w k + 1, .., s} for each v k,s ∈ C.Proof. Neccessity.If C is not maximal clique in the conflict graph H ′E S , this means that inequality (2.43) can be dominated by another inequality associated with a clique C ′ such that C ⊂ C ′ without changing its right-hand side. Moreover, if there exists an interval of contiguous slots I = [s i , s j ] ⊂ [1, s] satisfying the conditions of the condition 2 of the theorem. Then, inequality (2.43) is dominated by inequality (2.39). As a result, inequality (2.43) cannot be facet defining for P sa (G, K, S).

  u, z) ∈ P sa (G, K, S) : v k,s ∈C z k s = 1}.

First, we showii

  that µ s = 0 for all s ∈ S. Consider a slot s ∈ S, and a solution S 141 = (U 141 , S 141 ) given by a) select one pair of demand k ′ and slot s ′ from clique C (i.e., v k ′ ,s ′ ∈ C), and use slots k ′ =s ′ as last slot with s / ∈ {s ′ -w k ′ + 1, ..., s ′ } (slot assignment constraint taking into account the possibility of adding slot s in the set of used slots U 141 ), b) for each demand k i ∈ K \ {k ′ } with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 141 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] \ {s, ..., s + w ki -1},whereD 141 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k ′ } : E(p k i ) ∩ E(p k j ) ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 141 i , • and s / ∈ {s k i -w k i + 1, ...,s k i } (slot assignment constraint taking into account the possibility of adding slot s in the set of used slots U 141 ). We let S 141 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}. c) Let U 141 be the set of slots used in S such that for each demand k and last slot s k ∈ S 141 k and s ′ ∈ {s k -w k + 1, ..., s k }, we have s ′ ∈ U 141 . S 141 is clearly feasible for the SA problem. Hence, the corresponding incidence vector (u S 141 , z S 141 ) belongs to F H ′E S C . Then we derive a solution S 142 = (U 142 , S 142 ) from S 141 by adding slot s as an used slot in U 142 without modifying the last slots assigned to the demands K in S 141 which remain the same in solution S 142 i.e., k ∈ K. S 142 is feasible for the SA problem. Hence, the corresponding incidence vector (u S 142 , z S 142 ) belongs to F H ′E S C . Hence, solutions S 141 and S 142 satisfy equation µu + σz = τ . We then obtain that µu S 141 + σz S 141 = µu S 142 + σz S 142 = µu S 141 + σz S 141 + µ s. Hence, µ s = 0. In a similar way, we can show that µ s = 0, for all slots s ∈ S. Let show that σ k s = 0 for all k ∈ K and s ∈ {w k , ..., s} with v k,s / ∈ C. Consider a demand k in K and a slot s ′ in {w k , ..., s} with v k,s ′ / ∈ C. Let S 143 = (U 143 , S 143 ) be the solution given by a) select one pair of demand k ′ and slot s ′ from clique C (i.e., v k ′ ,s ′ ∈ C), and use slot s k ′ = s ′ as last slot with {s ′-w k ′ + 1, ..., s ′ } ∩ {s ′ -w k + 1, ..., s} = ∅ if E(p k ) ∩ E(p k ′ ) ̸ = ∅, b) for each demand k i ∈ K \ {k ′ } with i ∈ {1, ...,|K|}, we select the smallest slot index s k i in the set of slots I 143 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}] if E(p ki ) ∩ E(p k ) ̸ = ∅ or I 143 i = kj ∈D 143 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,whereD 143 i = {k j ∈ {k 1 , ..., k i-1 } ∪ {k ′ } : E(p k i ) ∩ E(p k j ) ̸ = ∅}. This ensures that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 143 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E(p k i ) ∩ E(p k ) ̸ = ∅ (we take into account the possibility of adding slot s ′ as a last slot in the selected last slots S 143 k to route demand k in solution S 143 ). We let S 143 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}. c) a set of slots U 143 are then used in S such that for each demand k ′ ∈ K and last slot s" ∈ S 143 k ′ and s" ∈ {s k ′ -w k ′ + 1, ..., s k ′ }, we have s" ∈ U 143 . S 143 is clearly feasible for the problem. Hence, the corresponding incidence vector (u S 143 , z S 143 ) belongs to F H ′E S C . Then we derive a solution S 144 = (U 144 , S 144 ) from S 143 by adding slot s ′ as last slot to demand k without modifying the last slots assigned to the demands K \ {k} in S 143 , i.e., S 143 k ′ = S 144 k ′ for each demand k ′ ∈ K \ {k}, and S 144 k = S 143 k ∪ {s ′ } for demand k. Solution S 144 is feasible for the SA problem. The corresponding incidence vector (u S 144 , z S 144 ) belongs to F H ′E S C . Hence, solutions S 143 and S 144 satisfy equation µu + σz = τ . We then obtain that µu S 143 + σz S 143 = µu S 144 + σz S 144 = µu S 143 + σz S 143 + σ k s ′ + s∈U 144 \U 143 µ s -s∈U 143 \U 144 µ s.

SH= 1 2⊂Fii

 1 {(u, z) ∈ P sa (G, K, S) : Denote inequality v k,s ∈H z k s ≤ |H|-by αu + βz ≤ λ. Let µu + σz ≤ τ be a valid inequality that is facet defining F of P sa (G, K, S). Suppose that F H ′E S H = {(u, z) ∈ P sa (G, K, S) : µu + σz = τ }. To prove that F H ′E S H is a facet of P sa (G, K, S), we need to show that there exist ρ ∈ R and γ ∈ R k∈K (w k -1) ) such that (µ, σ) = ρ(α, β) + γM .We first show that µ s = 0 for all s ∈ S. Consider a slot s ∈ S, and a solutionS 148 = (U 148 , S 148 ) such that a) select a subset of nodes H148 from H with | H148 | = |H|-1 2 , and each pair of nodes (v k,s , v k ′ ,s ′ ) ∈ H148 are not linked in the conflict graph H ′E S , and s / ∈ {s wk + 1, ..., s} for each v k,s ∈ H148 , b) for each pair of demand k and slot s with v k,s ∈ H148 , we select slot s k = s as last slot for demand k, c) for each demand k i ∈ K \ H148 with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 148 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] \ {s, ..., s + w ki -1},whereD 148 i = {k j ∈ {k 1 , ..., k i-1 } ∪ H148 : E(p k i ) ∩ E(p k j ) ̸ = ∅}. This guarantees that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 148 i , • and s / ∈ {s k i -w k i + 1, ...,s k i } (slot assignment constraint taking into account the possibility of adding slot s in the set of used slots U 148 ). We let S 148 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}. d) let U 148 be the set of slots used in S such that for each demand k and last slot s k ∈ S 148 k and s ′ ∈ {s k -w k + 1, ..., s k }, we have s ′ ∈ U 148 . S 148 is clearly feasible for the SA problem. Hence, the corresponding incidence vector (u S 148 , z S 148 ) belongs to F H ′E S H . Then consider the solution S 149 = (U 149 , S 149 ) obtained from S 148 by adding slot s as an used slot in U 149 without modifying the last slots assigned to the demands K in S 148 which remain the same in solution S 149 i.e., S 148 k = S 149 k for each demand k ∈ K. S 149 is clearly feasible for the SA problem. Hence, the corresponding incidence vector (u S 149 , z S 149 ) belongs to F H ′E S H . Hence, solutions S 148 and S 149 satisfy equation µu + σz = τ . We then obtain that µu S 148 + σz S 148 = µu S 149 + σz S 149 = µu S 148 + σz S 148 + µ s.It follows that µ s = 0.In a similar way, we can show that µ s = 0, for all slots s ∈ S.Let show thatσ k s = 0 for all k ∈ K and s ∈ {w k , ..., s} with v k,s / ∈ H. Consider a demand k in K and a slot s ′ in {w k , ..., s} with v k,s ′ / ∈ H. Let S 150 = (U 150 , S 150 ) be the solution given by a) select a subset of nodes H150 from H with | H150 | = |H|-1 2 , and each pair of nodes (v k,s , v k ′ ,s ′ ) ∈ H150 are not linked in the conflict graph H ′E S , , b) for each pair of demand k and slot s with v k,s ∈ H150 , we select slot s k = s as last slot for demand k, c) for each demand k i ∈ K \ H150 with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 150 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}] if E(p ki ) ∩ E(p k ) ̸ = ∅ or I 150 i = kj ∈D 150 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,whereD 150 i = {k j ∈ {k 1 , ..., k i-1 } ∪ H150 : E(p k i ) ∩ E(p k j ) ̸ = ∅}. This ensures that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 150 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k + 1, ..., s ′ } = ∅ if E(p k i ) ∩ E(p k ) ̸ = ∅ (we take into account the possibility of adding slot s ′ as a last slot in the selected last slots S 150 k to route demand k in solution S 150 ). We let S 150 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}. d) a set of slots U 150 are used in S such that for each demand k" and last slot s k" ∈ S 150 k" and s" ∈ {s k" -w k" + 1, ..., s k" }, we have s" ∈ U 150 . S 150 is clearly feasible for the problem. Hence, the corresponding incidence vector (u S 150 , z S 150 ) belongs to F H ′E S H . Then consider the solution S 151 = (U 151 , S 151 ) obtained from S 150 by adding slot s ′ as last slot to demand k without modifying the last slots assigned to the demands K \ {k} in S 150 , i.e., S 150 k ′ = S 151 k ′ for each demand k ′ ∈ K \ {k}, and S 151 k = S 150 k ∪ {s ′ } for demand k. Solution S 151 is feasible for the SA problem. The corresponding incidence vector (u S 151 , z S 151 ) belongs to F H ′E S H . Hence, solutions S 150 and S 151 satisfy equation µu + σz = τ . We then obtain that µu S 150 + σz S 150 = µu S 151 + σz S 151 = µu S 150 + σz S 150 + σ k s ′ + s∈U 151 \U 150 µ s -s∈U 150 \U 151 µ s.

SH.

  Then consider the solution S 153 = (U 153 , S 153 ) obtained from S 152 such that a) the last slots assigned to the demands K \{k, k ′ } in S 152 remain the same in S 153 , i.e., S 152 k" = S 153 k" for each demand k" ∈ K \ {k, k ′ }, where k is a demand with v k,s ∈ H152 and s ∈ S 152 k such that v k ′ ,s ′ is not linked with any node v k",s" ∈ H152 \ {v k,s }, b) and adding slot s ′ as last slot to demand k ′ , i.e., S 153 k ′ = S 152 k ′ ∪ {s ′ } for demand k ′ , c) and modifying the last slots assigned to demand k by adding a new last slot s and removing the last slot s ∈ S 152 k with v k,s ∈ H and v k,s / ∈ H such that S 153 k = (S 152 k \ {s}) ∪ {s} for demand k such that {s -w k + 1, ..., s} ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ for each k ′ ∈ K and s ′ ∈ S 153 k ′ with E(p k ) ∩ E(p k ′ ) ̸ = ∅.

Proposition 5 . 5 . 2 .

 552 Consider a slot s ∈ {1, ..., s -1}. Then, k∈K min(s+w k -1,s)

  for all k ∈ K and s ∈ {1, ..., w k -1},(5.22) s s=w k z k s = 1, for all k ∈ K,(5.23) k∈ Ke min(s,s+w k -1)s ′ =s z k s -u s ≤ 0,for all e ∈ E, and s ∈ S, (5.24)u s -k∈K min(s+w k -1,s) s ′ =s z k s ′ ≤ 0, for all s ∈ S,(5.25)z k s ≥ 0, for all k ∈ K and s ∈ S, (5.26)0 ≤ u s ≤ 1, for all s ∈ S,(5.27)z k s ∈ {0, 1}, for all k ∈ K and s ∈ S,(5.28)u s ∈ {0, 1}, for all s ∈ S.(5.29) Inequality (5.25) ensures that if slot s is not used by at least one demand, its associated variable u s is forced to be equal to zero.On the other hand, and to boost the performance of the B&B algorithm, we already introduced several classes of valid inequalities to obtain tighter LP bounds. Based on this, and at each iteration in a certain level of the B&B algorithm, one can identify one or more than one violated inequality by the current fractional solution for a given class of valid inequalities. Algorithm 6 summarizes the different steps of the Branch-and-Cut algorithm taking into account additional valid inequalities for a given class of valid inequalities.

  ble" if possible from this fractional solution. For this, we first use a local search algorithm to generate at each iteration a sequence of demands L numeroted with a) interval-odd-hole inequalities (2.40), b) slot-assignment-odd-hole inequalities (2.44), c) interval-clique inequalities (2.43), d) slot-assignment-clique inequalities (2.43), e) interval-capacity-cover inequalities (5.10).

  algorithm. The results have shown the efficiency of the valid inequalities allowed enhancing the resolution of the SA problem. Hence, the Branch-and-Cut is shown to be very performant compared with the Branch-and-Bound algorithm. Finally, it would be interesting to further investigate a combination of the different algorithms with some machine learning and reinforcement learning algorithms to well manage the B&C, B&P, and B&C&P trees and particularly for a) the node selection [27][36], b) variable selection and branching rule [6][36], c) column selection [39][111], d) cut selection [54][110],

  2.1 Sets of V k0 and E k 0 of the example of Fig. 2.1(b). . . . . . . . . . . . 3.1 Characteristics of Different Topologies Used for our Experiments. . . Table of Comparison Between B&C, B&P and B&C&P Algorithms: Scip (Without or With Additional Valid Inequalities) Vs Scip (Without or With Additional Valid Inequalities). . . . . . . . . . . . . . . 5.1 Table of Comparison Between: B&B SCIP Vs Own B&C SCIP Using Real Graphs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Table of Comparison Between: B&B SCIP Vs Own B&C SCIP Using Realistic Graphs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 Relation between P, NP, NP-complete and NP-hard problems. . . . .

	List of Figures
	1	Historical Evolution of Optical Transport Networks. . . . . . . . . .
	1.1

3.2

Table of Comparison for the B&C Algorithm: Cplex (Without or With Additional Valid Inequalities) Vs Scip (Without or With Additional Valid Inequalities). . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Table of Comparison for the B&C Algorithm: Gurobi (Without or With Additional Valid Inequalities) Vs Scip (Without or With Additional Valid Inequalities). . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Table of Comparison for the B&C Algorithm: Cplex (Without or With Additional Valid Inequalities) Vs Gurobi (Without or With Additional Valid Inequalities). . . . . . . . . . . . . . . . . . . . . . . . 3.5 The Impact of Valid Inequalities in the Own B&C SCIP Performance Using Realistic Graphs. . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Influence of the Valid Inequalities: B&P Vs B&C&P. . . . . . . . . . 4.2 Table of Comparison Between B&C, B&P and B&C&P Algorithms: Cplex (Without or With Additional Valid Inequalities) Vs Scip (Without or With Additional Valid Inequalities). . . . . . . . . . . . . . . 4.3 Table of Comparison Between B&C, B&P and B&C&P Algorithms:

Gurobi (Without or With Additional Valid Inequalities) Vs Scip (Without or With Additional Valid Inequalities). . . . . . . . . . . . . . . 4.4 1.2 conv(S) vs S. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Geometric interpretation for the polyhedron P , valid inequality, face, facet and extreme point. . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 FixedGrid Vs FlexGrid. . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Set of established paths and spectrums in graph G (Fig. 2(a)) for the set of demands {k 1 , k 2 , k 3 , k 4 } defined in Table 2(b). . . . . . . . . .

  6 , S6 ) be the solution given by a) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E 6 k i be the set of edges involved in a shortest path between o k i and d k i , b) we select the smallest slot index s k in {w k , ..., s} \ {s} as last slot for demand

	k,
	c) for each demand k

i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 6 i given by

  Theorem 2.3.4. Consider a demand k ∈ K, and an edge e ∈ E \ (E k 0 ∪ E k 1 ). Then, inequality x k e ≤ 1 is facet defining for P(G, K, S) if and only if a) there does not exist a demand k ′ ∈ K \ {k} such that the two demands k and k ′ are non-compatible demands for edge e, b) and there does not exist an edge e ′ ∈ E \ (E k 1 ∪ E k 0 ∪ {e}) such that the two edges e and e ′ are non-comptible edges for demand k.

	Proof. Neccessity.
	For demand k and edge e ∈ E \ (E k 0 ∪ E k 1 ), if
	a) there exists a demand k

′ ∈ K \ {k} such that the two demands k and k ′ are non-compatible demands for edge e. Then, inequality x k e ≤ 1 is dominated by inequality (2.20). b) there exists an edge e ′ ∈ E \ (E k 1 ∪ E k 0 ∪ {e}) such that the two edges e and e ′ are non-comptible edges for demand k. Then, inequality x k e ≤ 1 is dominated by inequality (2.19).

  9 k i be the set of edges involved in a shortest path between o k i and d k i , b) for demand k, we let E 9 k be the set of edges involved in a shortest path between o k and d k which uses edge e, c) we select the smallest slot index s k in {w k , ..., s} \ {s} as last slot for demand k in solution S 9 , d) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 9 i given by {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]∩[{w ki , ..., s-w k }∪{s+w ki , ..., s}]

	I 9 i = [
	kj ∈D 9 i
	if E 9 ki ∩ E 9 k ̸ = ∅ or I 9 i =
	kj ∈D 9

i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not.

  ., |K|}, we let E ′9 k i be the set of edges involved in a shortest path between o k i and d k i , b) for demand k, we let E ′9 k be the set of edges involved in a shortest path between o ) we select the slot s k = w k as last slot for demand k, d) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′9 i given by kj -w kj } ∪ {s kj + w ki , ..., s} if not,

	I ′9 i =	{w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if E ′9 ki ∩ (E ′9 k ∪ {e ′ }) ̸ = ∅
	kj ∈D ′9 i
		or I ′9 i =	{w ki , ..., s
		kj ∈D ′9 i

k and d k which uses edge e, and edge e ′ is compatible-edge with all the selected edges e" ∈ E ′9 k in solution S ′9 , i.e., e"∈E ′9 k l e" + ℓ e ′ ≤ lk . c

corresponding incidence vector (x S 11 , z S 11 ) belongs to F ′k e . Hence, solutions S ′9 and S 11 satisfy equation µx + σz = τ . It follows that µx S ′9 + σz S ′9 = µx S 11 + σz S 11 = µx S ′9 + µ k e ′ + σz S ′9 .

  we take into account the possibility of adding edge e ′ in the selected path E ′9 k to route demand k in solution S ′9 ).We let S ′9 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′9 is clearly feasible for the problem, and its incidence vector (x S ′9 , z S ′9 ) belongs to F ′k e . Let S 11 = (E 11 , S 11 ) be the solution obtained from solution S ′9 by adding edge e ′ ∈ E \ (E k

	0 ∪ E k 1 ) for the routing of demand k in solution S ′9 which means
	that E 11 k = E ′9 k ∪ {e}. The last slots assigned to the demands K, and paths assigned
	the set of demands K \ {k} in S ′9 remain the same in solution S 11 , i.e., S 11 k = S ′9 k
	for each k ∈ K, and E 11 k ′ = E ′9 k ′ for each k ′ ∈ K \ {k}. S 11 is clearly feasible for the
	problem. The

  12 , S12 ) be the solution given by a) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we let E 12 k i be the set of edges involved in a shortest path between o k i and d k i , b) for demand k, we let E12 k be the set of edges involved in a shortest path between o k and d k which does not use edge e, and edge e ′ is compatible-edge with all {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]∩[{w ki , ..., s k -w k }∪{s k +w ki , ..., s}] {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s} if not,

	the selected edges e" ∈ E 12 k , i.e., e"∈E 12 k	l e" + ℓ e ′ ≤ lk .
	c) we select slot s k = s as last slot for demand k, and we let S 12 k = {s},
	d) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot
	index s k i in the set of slots I 12 i given by
	I 12 i = [
	kj ∈D 12 i
	if E 12 ki ∩(E 12 k ∪{e}) ̸ = ∅ or I 12 i =
	kj ∈D 12 i

  |K|}, we let E ′12 k i be the set of edges involved in a shortest path between o k i and d k i , b) we select slot s k = s as last slot for demand k, and we let S ′12 k = {s}, c) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′12 {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]∩[{w ki , ..., s ′ -w k }∪{s ′ +w ki , ..., s}] {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not.

			i	given by
	I ′12 i	= [
		kj ∈D ′12 i
	if E ′12 ki ∩ E ′12 k ̸ = ∅ or I ′12 i	=
			kj ∈D ′12 i

  15 , S15 ) be the solution given by a) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we let E 15 k i be the set of edges involved in a shortest path between o k i and d k i , b) for demand k, we let E 15 k be the set of edges involved in a shortest path between o k and d k such that edge e is compatible-edge with all the selected {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]∩[{w ki , ..., s k -w k }∪{s k +w ki , ..., s}] {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s} if not,

	edges e" ∈ E 15 k , i.e., e"∈E 15 k	l e" + l e ≤ lk ,
	c) we select slot s k = w k as last slot for demand k, and let S 15 k = {s k },
	d) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot
	index s k i in the set of slots I 15 i given by
	I 15 i = [
	kj ∈D 15 i
	if E 15 ki ∩(E 15 k ∩{e}) ̸ = ∅ or I 15 i =
		kj ∈D 15 i

assigned to the demands K, and paths assigned the set of demands K \ {k} in S 15 remain the same in solution S 16

  15 k ∪ {e}) ̸ = ∅ ( we take into account the possibility of adding edge e in the selected pathE 15 k to route demand k in solution S 15 ).We let S 15 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}.S15 is feasible for the problem, its incidence vector (x S 15 , z S 15 ) belongs to F k S . Then we derive a solution S 16 = (E 15 , S 15 ) obtained from S 15 by adding edge e ∈ E \(E k , i.e., S 16 k = S 15 k for each k ∈ K, and E 16 k ′ = E 15 k ′ for each k ′ ∈ K \ {k}. S 16 is clearly feasible for the problem. The corresponding incidence vector (x S 16 , z S 16 ) belongs to F k S . Hence, solutions S 15 and S 16 satisfy equation µx + σz = τ . It follows that µx S 15 + σz S 15 = µx S 16 + σz S 16 = µx S 15 + µ k e + σz S 15 .

	0 ∪
	E k 1 ) for the routing of demand k in solution S 15 which means that E 16 k = E 15 k ∪ {e}.
	The last slots

  |K|}, we select the smallest slot index s k i in the set of slots I ′15 {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]∩[{w ki , ..., s ′ -w k ′ }∪{s ′ +w ki , ..., s}] {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

			i	given by
	I ′15 i	= [
		kj ∈D ′15 i
	if E ′15 ki ∩ E ′15 k ′ ̸ = ∅ or I ′15 i	=
			kj ∈D ′15 i

  for all k ′ ∈ K and s ′ ∈ {w k ′ , ..., s}.Let prove now that σ k s for demand k and slot s in {w k , ..., s} are equivalent. Consider a slot s ′ ∈ {w k , ..., s}. Let S15 = ( Ẽ15 , S15 ) be the solution given by a) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let Ẽ15 k i be the set of edges involved in a shortest path between o k i and d k i , b) we select the smallest slot index s k from {w k , ..., s}\{s ′ } as last slot for demand {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]∩[{w ki , ..., s ′ -w k }∪{s ′ +w ki , ..., s}] {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

	k,
	c) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot
	index s k i in the set of slots Ĩ15 i given by
	Ĩ15 i = [
	kj ∈ D15
	if Ẽ15 ki ∩ Ẽ15 k ̸ = ∅ or Ĩ15 i =
	kj ∈ D15
	where D15

i i

  and s ∈ {w k ′ , ..., s},

	0	otherwise.
	As a consequence, (µ, σ) = ρ(α, β) + γQ.
	Theorem 2.3.7. Consider a demand k in K and a subset of node X ⊂ V, with |X ∩
	{o k , d k }| = 1 and δ(X) ∩ E k 1 = ∅. Then, inequality (2.2), e∈δ(X) x k e ≥ 1, is facet
	defining for P(G, K, S).
	Proof. Let F k X denote the face induced by inequality
		e∈(δ(X)\E k 0 )

  |K|}, we let E 19 k i be the set of edges involved in a shortest path between o k i and d k i , we select the smallest slot index s k in {w k , ..., s} \ {s} as last slot for demand k, d) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 19 i given by {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s -w k } ∪ {s + w ki , ..., s}] {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

	b) for demand k, we let E 19 k be the set of edges involved in a shortest path between o k and d k . This guarantees that one edge e from (δ(X) \ E k 0 ) is chosen to route demand k, i.e., |(δ(X) \ E k 0 ) ∩ E 19 k | = 1, i = [ kj ∈D 19 i if E 19 ki ∩ E 19 k ̸ = ∅ or I 19 i = c) for demand k, I 19 kj ∈D 19 i

19 and S 20 satisfy equation µx + σz = τ . We then obtain that µx S 19 + σz S 19 = µx S 20 + σz S 20 = µx S 19 + σz S 19 + σ k s .

  we take into account the possibility of adding slot s in the set of last slots S19 k to route demand k in solution S19 ).We let S 19k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}.S19 is clearly feasible for the problem, and its incidence vector (x S 19 , z S 19 ) belongs toF k X .Then consider the solution S 20 = (E 20 , S 20 ) obtained from S 19 by adding slot s as last slot to demand k without modifying the paths assigned to the demands K

	in S 19 (i.e., E 20 k = E 19 1 for each k ∈ K), and the last slots assigned to the demands
	K \ {k} in S 19 remain the same in solution S 20 i.e., S 19 k ′ = S 20 k ′ for each demand
	k ′ ∈ K \ {k}, and S 20 k = S 19 k ∪ {s} for demand k. Solution S 20 is feasible for the
	problem. The corresponding incidence vector (x S 20 , z S 20 ) belongs to F k X . Hence,
	solutions S

  arbitrarily. Let S ′19 = (E ′19 , S ′19 ) be the solution given by a) for each demandk i ∈ K \ {k} with i ∈ {1, ...,|K|}, we let E ′19 k i be the set of edges involved in a shortest path between o k i and d k i , b) for demand k, we let E ′19 k be the set of edges involved in a shortest path between o k and d k such that edge e ′ is compatible-edge with the selected edges e" ∈ E ′19 k , i.e., {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s k -w k } ∪ {s k + w ki , ..., s}] {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

	e"∈E ′19 k	l e" + ℓ e ′ ≤ lk .
	c) we select the slot s k = w k as last slot for demand k, and let S ′19 k = {s k },
	d) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index
	s k i in the set of slots I ′19 i	given by
	I ′19 i	= [
		kj ∈D ′19 i
	if E ′19 ki ∩ (E ′19 k ∪ {e ′ }) ̸ = ∅ or I ′19 i	=
			kj ∈D ′19 i

  we select the slot s k = w k as last slot for demand k , and let S19k = {s k }, d) for each demand k i ∈ K \ {k} with i ∈ {1, ...,|K|}, we select the smallest slot index s k i in the set of slots Ĩ19 , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s k -w k } ∪ {s k + w ki , ..., s}] , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not.

	i given by
	Ĩ19 i = [
	kj ∈ D19
	if Ẽ19 ki ∩ ( Ẽ19 k ∪ {e ′ }) ̸ = ∅ or Ĩ19 i =
	kj ∈ D19

i {w ki i {w ki

  2.4.1. Consider an edge e ∈ E, and a slot s ∈ S. Let K be a subset of demands in K, and k∈ K w k ≤ sk ′ ∈Ke\ K w k ′ . Then, inequality (2.23) is facet defining for P(G, K, S) if and only if K e \ K = ∅, and there does not exist an interval of contiguous slots I = [s i , s j ] such that a) |{s i + w k -1, ..., s j }| ≥ w k for each demand k ∈ K,

b) and s ∈ {s i + max

  38 , S38 ) be the solution given by a) for each demand k i ∈ K \ K with i ∈ {1, ..., |K|}, we let E 38 k i be the set of edges involved in a shortest path between o k i and d k i , b) for demand k, we let E 38 k be the set of edges involved in a shortest path between o k and d k which uses edge e such that edge e ′ is compatible with all the selected edges e" ∈ E 38 k of demand k in solution S 38 , i.e., e"∈E 38 k l e" + ℓ e ′ ≤ lk , c) for each demand k ′ ∈ K \ {k}, we let E 38 k ′ be the set of edges involved in a shortest path between o k ′ and d k ′ which does uses edge e, d) for one demand k ′ ∈ K, we select the smallest slot index s k ′ in {w k ′ , ..., s} as last slot such that s ∈ {s k ′ -w k ′ + 1, ..., s k ′ }, e) for each demand k i ∈ K \ {k ′ } with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 38 i given by kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s k -w k } ∪ {s k + w ki , ..., s}] kj -w kj } ∪ {s kj + w ki , ..., s} if not.

	I 38 i = [ kj ∈D 38 i {w ki , ..., s if E 38 ki ∩ (E 38 k ∪ {e ′ }) ̸ = ∅ or I 38 i =	{w ki , ..., s
	kj ∈D 38 i

  for the routing of demand k in solution S38 which means thatE 39 k = E 38 k ∪ {e ′ }.The last slots assigned to the demands K, and paths assigned the set of demands K \ {k} in S 38 remain the same in solution S 39 , i.e., S 39 k = S 38 k for each k ∈ K, and E 39 k ′ = E 38 k ′ for each k ′ ∈ K \ {k}. S 39 is clearly feasible for the problem. The corresponding incidence vector (x S 39 , z S 39 ) belongs to F e,s K . Hence, solutions S 38 and S 39 satisfy equation µx + σz = τ . It follows that µx S 38 + σz S 38 = µx S 39 + σz S 39 = µx S 38 + µ k e ′ + σz S 38 .

  |K|}, we let E 42 k i be the set of edges involved in a shortest path between o k i and d k i , b) for one demand k from K, we let E 42 k i be the set of edges involved in a shortest path between o k and d k which uses edge e, c) for each demand k ′ ∈ K \ {k}, we let E 42 k ′ be the set of edges involved in a shortest path between o k ′ and d k ′ which does not use edge e,

d) for each demand k ∈ K, we select the smallest slot index s k in {w k , ..., s} ∩ {s, ..., s + w k -1} as last slot, e) for each demand

k i ∈ K \ K with i ∈ {1, ...,

|K|}, we select the smallest slot index s k i in the set of slots I 42 i given by I 42 i = [ kj ∈D 42 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s -w k } ∪ {s + w ki , ..., s}] if E 42 ki ∩ E 42 k ̸ = ∅, or I 42 i = [ kj ∈D 42 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] if not,

  modifying the path assigned to demand k in S 42 with e ∈ E 42 k and k ∈ K from E42

k to a path E

43

k without passing through edge e (i.e., e / ∈ E 43 k ) and {s -

  modifying the path assigned to demand k in S 53 with e ∈ E 53 k and v k ∈ C from E 53

k to a path E 58 k without passing through edge e (i.e., e / ∈ E 58 k ) and {s -

  53 defined in the proof of theorem 2.4.6 which stills feasible such that its incidence vector (x S 53 , z S 53 ) belongs to F C . Furthermore, and based on the solutions S 53 to S 59 with corresponding incidence vector (x S 53 , z S 53 ) to (x S 59 , z S 59 )

	I solution S ′H e belong to F ′H e I	′H e I C	is a proper face based on

defining for P(G, K, S). We first prove that F C , we show that there exist ρ ∈ R and γ

  and σ k s are equivalent for all v k ∈ C ∪ C e and s ∈ {s i + w k -1, ..., s j }, c) and µ k e ′ = 0 for all demand k ∈ K and edge e ∈ E \ (E k 0∪ E k 1 ) with e ̸ = e ′ if v k ∈ C,

d) and µ k e are equivalent for the set of demands in C, e) and σ k ′ s and µ k e are equivalent for all v k ∈ C and v k ′ ∈ C ∪ C e and s ∈ {s

  |K|}, we select the smallest slot index s k i in the set of slots I 60 i given by

	I 60 i =
	kj ∈D 60

i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} \ {s i , ..., s j },

  |K|}, we select the smallest slot index s k i in the set of slots I 60 i given by

  be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S 60 is feasible for the problem, and its incidence vector (x S 60 , z S 60 ) belongs to F The last slots assigned to the demands K, and paths assigned the set of demands K \ {k} in S60 remain the same in solution S 61 , i.e., S 61 k = S 60 k for each k ∈ K, and E 61 k ′ = E 60 k ′ for each k ′ ∈ K \ {k}. S 61 is clearly feasible for the problem. The corresponding incidence vector (x S 61 , z S 61 ) belongs to F 60 and S 61 satisfy equation µx + σz = τ . It follows that µx S 60 + σz S 60 = µx S 61 + σz S 61 = µx S 60 + µ k e + σz S 60 .

	H E I C . Hence,
	solutions S
	H E I

C . Then we derive a solution S 61 = (E 61 , S 61 ) obtained from S 60 by adding edge

e ∈ E \ (E k 0 ∪ E k 1 )

for the routing of demand k in solution S 60 which means that

E 61 k = E 60 k ∪ {e}.

  |K|}, we select the smallest slot index s k i in the set of slots I ′60

		i	given by
	I ′60 i	=
		kj ∈D ′60

i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} \ {s i , ..., s j },

  E(H) denotes the set of edges in the sub-graph of the conflict graph H E

I

induced by H. Taking into account that each node v k in H has two neighbors in H, this implies that s j s=s i +w k -1 z k s appears twice in the previous inequality. As a result,

  Theorem 2.4.10. Let H be an odd-hole in the conflict graph H E I with |H| ≥ 5 and 2w k > |I| for each v k ∈ H. Then, inequality (2.40) is facet defining for P(G, K, S)

	if and only if
	a) for each node v k ′ / ∈ H in H E

I , there exists a node v k ∈ H such that the induced graph

  |K|}, we select the smallest slot index s k i in the set of slots I 64 i given by 64 i = {k j ∈ {k 1 , ..., k i-1 } ∩ H : E 64 k i ∩ E 64 k j ̸ = ∅}. We let S 64 k i = {s k i } be the set of last slots assigned to demand k i , e) for each demand k i ∈ K \ H with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 64 i given by {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s k -w k } ∪ {s k + w ki , ..., s}] {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

	I 64 i = [
		kj ∈R 64 i
	if E 64 ki ∩ (E 64 k ∪ {e}) ̸ = ∅ or I 64 i =
		kj ∈R 64 i
	I 64 i	=
		kj ∈D 64

i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} \ {s i + w ki -1, ..., s j }, where D

  we take into account the possibility of using edge e ′ in the selected path E 64 k to route demand k in solution S 64 ).64 is feasible for the problem. Hence, the corresponding incidence vector (x S 64 , z S 64 ) The last slots assigned to the demands K, and paths assigned the set of demands K \ {k} in S 64 remain the same in solution S 65 , i.e.,65 is clearly feasible for the problem. The corresponding incidence vector (x S 65 , z S 65 ) belongs to H . Hence, solutions S 64 and S 65 satisfy equation µx + σz = τ . It follows that µx S 64 + σz S 64 = µx S 65 + σz S 65 = µx S 64 + µ k e + σz S 64 .

	We let S 64 k belongs to F	H E
	S 65 k = S 64
	F	H E I

i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S I H . Then we derive a solution S 65 = (E 65 , S 65 ) obtained from S 64 by adding edge e ∈ E \ (E k 0 ∪ E k 1 ) for the routing of demand k in solution S 64 which means that E 65 k = E 64 k ∪ {e}. k for each k ∈ K, and E 65 k ′ = E 64 k ′ for each k ′ ∈ K \ {k}. S

  s j }. Let S 66 = (E 66 , S 66 ) be the solution given by a) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E 66 k i be the set of edges involved in a shortest path between o k i andd k i , b) select a subset of demands H from H with | H| = |H|-12 , c) for each demand k i from H with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 66 i given by

	I 66 i	= [
		kj ∈L 66

i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ {s i + w ki -1, ..., s j }.

  |K|}, we select the smallest slot index s k i in the set of slots I 66 i given by

	I 66 i	=
		kj ∈D 66
	I 66 i = [
		kj ∈R 66

i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} \ {s i + w ki -1, ..., s j }, where

D 66 i = {k j ∈ {k 1 , ..., k i-1 } ∩ H : E 66 k i ∩ E 66 k j ̸ = ∅}, e) for each demand k i ∈ K \ H with i ∈ {1, ...,

|K|}, we select the smallest slot index s k i in the set of slots I 66 i given by i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k ′ } ∪ {s ′ + w ki , ..., s}] if E 66 ki ∩ E 66 k ′ ̸ = ∅ or I 66 i = kj ∈R 66 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

  we take into account the possibility of adding slot s ′ as a last slot in the selected last slots S 66 k ′ to route demand k ′ in solution S 66 ).66 is feasible for the problem. Hence, the corresponding incidence vector (x S 66 , z S 66 ) H . Based on this, we construct a feasible solution S 67 = (E 67 , S 67 ) obtained from S 66 as belows a) without changing the established paths for the demands K \ K in solution S 66 , i.e.,E 67 k = E 66 k for each demand k ∈ K \ K, b)remove the last slot s totally covered by the interval I and which has been selected by a demand k i ∈ {v k 1 , ..., v kq } in solution S 66 (i.e., s ∈ S 66 k i and s′ ∈ {s i +w k i +1, ..., s j }) such that each pair of nodes (v k ′ , v k j ) are not linked in odd-hole H with j ̸ = i, c) and select a new last slot s′ / ∈ {s i + w k i + 1, ..., s j } for demand k i i.e., S 67 k i = (S 66 k i \ {s}) ∪ {s ′ } such that {s ′ -w k i -1, ..., s′ } ∩ {s -w k + 1, ..., s} = ∅ for each k ∈ K and s ∈ S 66 k with E 67 k ∩ E 67 k i ̸ = ∅, d) and add slot s ′ to the set of last slots S 66 k ′ assigned to demand k ′ in solution S 66 , i.e., S 67 k ′ = S 66 k It follows that µx S 66 + σz S 66 = µx S 67 + σz S 67 = µx S 66 + σz S 66 + σ

	We let S 66 k belongs to F	H E I

i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S ′ ∪ {s ′ }. Solution S 67 is feasible for the problem. The corresponding incidence vector (x S 67 , z S 67 ) belongs to F H E I H . Hence, solutions S 66 and S 67 satisfy equation µx + σz = τ .

  Then, inequality (2.41) is facet defining for P(G, K, S) if and only ifa) for each node v k" in H E I with v k" / ∈ H ∪ C and C ∪ {v k" } is a clique in H E I , there exists a subset of nodes H ⊆ H of size |H|-1 2 such that H ∪ {v k" } is stable in H E E I such that v k" is linked with all nodes v k ∈ H and nodes v k ′ ∈ C.This implies that inequality (2.41) is dominated by the following inequality

I , b) and there does not exist an interval I ′ of contiguous slots with I ⊂ I ′ such that H and C define also an odd-hole and its connected clique in the associated conflict graph H E I ′ . Proof. Neccessity. a) Note that if there exists a node v k" / ∈ H ∪ C in H

  β) + γQ. For this, we show that a) σ k s = 0 for all demand k ∈ K and slot s ∈ {w k , ..., s} with s / ∈ {s i + w k -1, ..., s j } if v k ∈ H ∪ C as we did in the proof of theorem 2.4.14, b) and µ k e = 0 for all demand k ∈ K and edge e ∈ E \ (E k 0 ∪ E k 1 ) as we did in the proof of theorem 2.4.14, c) and σ k s are equivalent for all v k ∈ H and s ∈ {s i + w k -1, ..., s j } as we did in the proof of theorem 2.4.14. Solutions S 49 -S 69 still feasible for F . We should prove now that σ k s are equivalent for all v k ∈ C and s ∈ {s i + w k -1, ..., s j }.

	H E I
	H,C

  |K|}, we select the smallest slot index s k i in the set of slots I 70 i given by {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s′ -w k′ } ∪ {s ′ + w ki , ..., s}]

	I 70 i = [
	kj ∈R 70 i

if E

70

ki ∩ E 70 k ′ ̸ = ∅ or I 70 i = kj ∈R 70 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not, 107

  Definition 2.4.7. Let H E S be a conflict graph defined as follows. For all slot s ∈ {w k , ..., s} and demand k ∈ K, consider a node v k,s in H E S . Two nodes v k,s and v k ′ ,s ′ are linked by an edge in H E S if and only if

  72 , S72 ) be the solution given by 1. for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we let E 72 k i be the set of edges involved in a shortest path between o k i and d k i , 2. for demand k, we let E 72 k be the set of edges involved in a shortest path between o k and d k such that edge e is compatible with all the selected edges e ∈ E 72 k , i.e.,

	I 72 i = [
	kj ∈D 72

e ′ ∈E

72

k ℓ e ′ + ℓ e ≤ lk ,

[START_REF] Balas | Facets of the knapsack polytope[END_REF]

. select one pair of demand k ′ and slot s ′ from clique C (i.e., v k ′ ,s ′ ∈ C), and use slot

s k ′ = s ′ as last slot, 4. for each demand k i ∈ K \ {k ′ } with i ∈ {1, ...,

|K|}, we select the smallest slot index s k i in the set of slots I 72 i given by i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s k -w k } ∪ {s k + w ki , ..., s}] if E 72 ki ∩ (E 72 k ∪ {e}) ̸ = ∅ or I 72 i = kj ∈D 72 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

  we take into account the possibility of using edge e in the selected path E72 k to route demand k in solution S72 ). We let S 72 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}.S 72 is feasible for the problem. Hence, the corresponding incidence vector (x S 72 , z S 72 )C . Then we derive a solution S 73 = (E 73 , S 73 ) obtained from S 72 by adding edge e ∈ E \ (E k 0 ∪ E k 1 ) for the routing of demand k in solution S 72 which means that E73 k = E 72 k ∪ {e}. The last slots assigned to the demands K, and paths assigned the set of demands K \ {k} in S 72 remain the same in solution S 73 , i.e.,73 is clearly feasible for the problem. The corresponding incidence vector (x S 73 , z S 73 ) belongs to C . Hence, solutions S 72 and S 73 satisfy equation µx + σz = τ . It follows that µx S 72 + σz S 72 = µx S 73 + σz S 73 = µx S 72 + µ k e + σz S 72 . Let show that σ k s = 0 for all k ∈ K and s ∈ {w k , ..., s} with v k,s / ∈ C. Consider a demand k in K and a slot s ′ in {w k , ..., s} with v k,s ′ / ∈ C. Let S ′72 = (E ′72 , S ′72 ) be the solution given by 1. for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E ′72 k i be the set of edges involved in a shortest path between o k i and d k i , 2. select one pair of demand k ′ and slot s ′ from clique C (i.e., v k ′ ,s ′ ∈ C), and use slot s k ′ = s ′ as last slot, 3. for each demand k i ∈ K \ {k ′ } with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′72

	F	H E S
	As a result, µ k e = 0.
	In a similar way, we can show that
			µ k e = 0, for all k ∈ K and e ∈ E \ (E k 0 ∪ E k 1 ).
			i	given by
		I ′72 i	= [
			kj ∈D ′72
	belongs to F	H E S
	S 73

k = S 72 k for each k ∈ K, and E 73 k ′ = E 72 k ′ for each k ′ ∈ K \ {k}. S i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}] if E ′72 ki ∩ E ′72 k ̸ = ∅ or I ′72 i = kj ∈D ′72 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

  76 , S76 ) be the solution given by 1. for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we let E 76 k i be the set of edges involved in a shortest path between o k i and d k i , 2. for demand k, we let E 76 k be the set of edges involved in a shortest path between o k and d k such that edge e is compatible with all the selected edges e ∈ E 76 k , 3. select a subset of nodes H76 from H with | H76 | = |H|-1 2 , and each pair of nodes (v k,s , v k ′ ,s ′ ) ∈ H76 are not linked in the conflict graph H E S , 4. for each pair of demand k and slot s with v k,s ∈ H76 , we select slot s k = s as last slot for demand k, 5. for each demand k i ∈ K \ H76 with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 76 i given by

	I 76 i = [
	kj ∈D 76

i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s k -w k } ∪ {s k + w ki , ..., s}] if E 76 ki ∩ (E 76 k ∪ {e}) ̸ = ∅ or I 76 i = kj ∈D 76 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

  we take into account the possibility of using edge e in the selected path E76 k to route demand k in solution S76 ). ′77 , S ′77 ) obtained from S 76 by adding edge e ∈ E \ (E k 0 ∪ E k 1 ) for the routing of demand k in solution S 76 which means that E ′77 k = E 76 k ∪ {e}. The last slots assigned to the demands K, and paths assigned the set of demands K \ {k} in S 76 remain the same in solution S ′77 , i.e.,S ′77 k = S 76 k for each k ∈ K, and E ′77 k ′ = E 76 k ′ for each k ′ ∈ K \ {k}. S ′77is clearly feasible for the problem. The corresponding incidence vector (x S ′77 , z S ′77 ) belongs H . Hence, solutions S 76 and S ′77 satisfy equation µx + σz = τ . It follows that µx S 76 + σz S 76 = µx S ′77 + σz S ′77 = µx S 76 + µ k e + σz S 76 . Let show that σ k s = 0 for all k ∈ K and s ∈ {w k , ..., s} with v k,s / ∈ H. Consider a demand k in K and a slot s ′ in {w k , ..., s} with v k,s ′ / ∈ H. Let S ′76 = (E ′76 , S ′76 ) be the solution given by 1. for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E ′76 k i be the set of edges involved in a shortest path between o k i and d k i , 2. select a subset of nodes H′76 from H with | H′76 | = |H|-1 2 , and each pair of nodes (v k,s , v k ′ ,s ′ ) ∈ H′76 are not linked in the conflict graph H E S , 3. for each pair of demand k and slot s with v k,s ∈ H′76 , we select slot s k = s as last slot for demand k, 4. for each demand k i ∈ K \ H′76 with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′76 i

	We let S 76 k to F H E S
	As a result, µ k e = 0.
	In a similar way, we can show that
	µ k

i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}. S 76 is feasible for the problem. Hence, the corresponding incidence vector (x S 76 , z S 76 ) belongs to F H E S H . Then we derive a solution S ′77 = (E e = 0, for all k ∈ K and e ∈ E \ (E k 0 ∪ E k 1 ).

  for all k ∈ K and s ∈ {w k , ..., s} with v k,s / ∈ H. we consider the solution S 80 = (E 80 , S 80 ) defined as follows 1. for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E 80 k i be the set of edges involved in a shortest path between o k i and d k i , 2. select a subset of nodes H80 from H with | H80 | = |H|-1 2 , and each pair of nodes (v k,s , v k",s" ) ∈ H80 are not linked in the conflict graph H E S , and each v k,s ∈ H80 is not linked with node v k ′ ,s ′ in H E S , 3. for each pair of demand k and slot s with v k,s ∈ H80 , we select slot s k = s as last slot for demand k, 4. for each demand k i ∈ K \ H80 with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 80 i given by

	I 80 i = [
	kj ∈D 80

Let prove that σ k s for all v k,s ∈ H are equivalent. Consider a node v k ′ ,s ′ in H. i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k ′ } ∪ {s ′ + w ki , ..., s}] if E 80 ki ∩ E 80 k ̸ = ∅ or I 80 i = kj ∈D 80 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not.

  . In what follows, we prove that σ k ′ s ′ are equivalent for all v k ′ ,s ′ ∈ C. For this, we consider a node v k ′ s ′ ∈ C, and a solution S 82 = (E 82 , S 82 ) given by a) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we let E 82 k i be the set of edges involved in a shortest path between o k i and d k i , b) select a subset of nodes H82 from H with | H82 | = |H|-1 2 , and each pair of nodes (v k,s , v k",s" ) ∈ H82 are not linked in the conflict graph H E S , c) for each pair of demand k and slot s with v k,s ∈ H82 , we select slot s k = s as last slot for demand k, d) for each demand k i ∈ K \ H82 with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 82 i given by

	H E S
	H,C I 82 i = [
	kj ∈D 82
	For this,
	we need to show that
	a) σ k s = 0 for all demand k ∈ K and slot s ∈ {w k , ..., s} with v k,s / ∈ H ∪ C as done in
	the proof of theorem 2.4.14,
	b) and µ k e = 0 for all demand k ∈ K and edge e ∈ E \ (E k 0 ∪ E k 1 ) as done in the proof
	of theorem 2.4.14,
	c) and σ k

s are equivalent for all v k,s ∈ H as done in the proof of theorem 2.4.14, given that the solutions S 65 -S 80 still feasible such that their corresponding incidence vectors belong to F i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k ′ } ∪ {s ′ + w ki , ..., s}] if E 82 ki ∩ E 82 k ̸ = ∅ or I 82 i = kj ∈D 82 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

  83 , z S 83 ) . Hence, solutions S 82 and S 83 satisfy equation µx + σz = τ . We have so µx S 82 + σz S 82 = µx S 83 + σz S 83 = µx S 82 + σz

	H E S H,C belongs to F

  ) if k ̸ = k ′ : k and k ′ are non compatible demands for edge e. Proposition 2.4.20. Let C be a clique in H K E . Then, the inequality

	x k e ≤ 1,	(2.46)
	v k e ∈C
	is valid for P(G, K, S).
	Proof. It is trivial given the definition of a clique set in the conflict graph H K E . We
	know from inequalities (2.19) or (2.20) that for all pairs of nodes (v k e , v k ′ e ′ ) in a clique
	C in H K E
	x k e + x k ′ e
	Based on inequalities (2.19) and (2.20), we introduce the following conflict graph.
	Definition 2.4.8. Let H K E be a conflict graph defined as follows. For each demand
	k and edge e / ∈ E k 0 ∪ E k 1 , consider a node v k e in H K E . Two nodes v k e and v k ′ e ′ are linked
	by an edge in H K

E a) if k = k ′ : e and e ′ are non compatible edges for demand k.

b′ ≤ 1, By adding the previous inequalities for all two nodes v k,e and v k ′ ,e ′ in C, and by recurrence procedure we obtain that for all

  |K|} \ {k}, we let E 84 k i be the set of edges involved in a shortest path between o k i and d k i , d) for demand k, we select the slot s k = w k as last slot, e) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 84 i given by

	I 84 i = [
	kj ∈D 84
	if E 84 ki ∩ (E 84 k ∪ {e}) ̸ = ∅ or I 84 i =
	kj ∈D 84

. Consider a demand k ∈ K and an edge e ∈ E \ (E k 0 ∪ E k 1 ) with v k,e / ∈ C. Let S 84 = (E 84 , S 84 ) be the solution given by a) select one pair of demand k ′ and edge e ′ from clique C (i.e., v k ′ ,e ′ ∈ C), we let E 84 k ′ be the set of edges involved in a shortest path between o k ′ and d k ′ which uses edge e ′ , b) for each pair of demand k" and edge e" with v k",e" ∈ C \ {v k,e }, we let E 84 k" be the set of edges involved in a shortest path between o k" and d k" which uses edge e" which does not use edge e", c) for each demand k i ∈ K \ C with i ∈ {1, ..., i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s k -w k } ∪ {s k + w ki , ..., s}] i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

  we take into account the possibility of using edge e in the selected path E84 k to route demand k in solution S 84 ). We let S 84 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S 84 is feasible for the problem. Hence, the corresponding incidence vector (x S 84 , z S 84 ) C . Then we derive a solution S 85 = (E 85 , S 85 ) obtained from S 84 by adding edge e ∈ E \ (E k 0 ∪ E k 1 ) for the routing of demand k in solution S 84 which means that E 85 k = E 84 k ∪ {e}. The last slots assigned to the demands K, and paths assigned the set of demands K \ {k} in S 84 remain the same in solution S 85 , i.e., S 85 k = S 84 k for each k ∈ K, and E 85 k ′ = E 84 k ′ for each k ′ ∈ K \ {k}. S 85 is clearly feasible for the problem. The corresponding incidence vector (x S 85 , z S 85 ) belongs to C . Hence, solutions S 84 and S 85 satisfy equation µx + σz = τ . It follows that µx S 84 + σz S 84 = µx S 85 + σz S 85 = µx S 84 + µ k e + σz S 84 .

	belongs to F	H K E
	F	H K E

  Then consider the solution S 86 obtained from S ′84 by adding slot s ′ as last slot to demand k in S ′84 . Solution S 86 is clearly feasible for the problem.The corresponding incidence vector (x S 86 , z S 86 ) belongs to F ′84 and S 86 satisfy equation µx + σz = τ . We have so µx S ′84 + σz S ′84 = µx S 86 + σz S 86 = µx S ′84 + σz S ′84 + σ k s ′ .

	S ′84 )
	belongs to F C . H K H K E E C . Hence, solutions
	S Hence, σ k s ′ = 0.
	In a similar way, we can show that
	σ k

s = 0, for all k ∈ K and s ∈ {w k , ..., s} with v k,s / ∈ C. Let prove that µ k e for all v k,e are equivalent. Consider a node v k ′ ,e ′ in C such that e ′ / ∈ E 84 k ′ . For this, we derive solution S 87 from S 84 by a) modifying the path assigned to demand k ′ in S 84 from E 84 k ′ to a path E 87 k ′ passed through edge e ′ with v k ′ ,e ′ ∈ C, b) modifying the path assigned to demand k in S 84 with e ∈ E 84 k and v k,e ∈ C from E 84 k to a path E 87 k without passing through any edge e" ∈ E \ (E k 0 ∪ E k 1

  , i.e., S 84 k = S87 k for each demand k ∈ K \ K. Solution S 87 is feasible for the problem. The corresponding incidence vector (x S 87 , z S 87 ) belongs to F C . Hence, solutions S 84 and S 87 satisfy equation µx + σz = τ . We have so µx S 84 + σz S 84 = µx S 87 + σz S 87 = µx S 84 + σz S 84 + µ k ′

	H K E
		e ′ -µ k e +	k∈ K s ′ ∈S 87 k σ k s ′ -	s∈S 84 k σ k s
	+	µ k ′ e" -	µ k ′ e" +	µ k e" -
	e"∈E 87 k ′ \{e ′ }	e"∈E 84 k ′	e"∈E 87 k

e"∈E

84

k \{e}

  pair of nodes (v k e , v k ′ e ′ ) linked in H, and by doing a sum for all pairs of nodes (v k e , v k ′ e ′ ) linked in H, it follows that

	Proposition 2.4.21. Let H be an odd-hole in the conflict graph H K E with |H| ≥ 5.
	Then, the inequality
	v k e ∈H	x k e ≤	|H| -1 2	,	(2.47)
	is valid for P(G, K, S).
	Proof. It is trivial given the definition of the odd-hole in the conflict graph H K E . We
	strengthen the proof as belows. For each pair of nodes (v k e , v k ′ e

′ ) linked in H by an edge, we know that x k e + x k ′ e ′ ≤ 1. Given that H is an odd-hole which means that we have |H| -1

  \ {v k,e } ∪ {v k ′ ,e ′ }) does not contain an odd-holeH ′ = H \ {v k,e } ∪ {v k ′ ,e ′ }, b) and there does not exist a node v k ′ ,e ′ / ∈ H in H K E such that v k ′ ,e ′ is linked with all nodes v k,e ∈ H.We distinguish the following cases:a) if for a node v k ′ ,e ′ / ∈ H in H K E , there exists a node v k,e ∈ H such that the induced graph H K E (H \{v k,e }∪{v k ′ ,e ′ }) contains an odd-hole H ′ = (H \{v k,e })∪{v k ′ ,e ′ }.This implies that inequality (2.47) can be dominated using some technics of lifting based on the following two inequalities v k,e ∈H x k e ≤ |H|-1 2 , and

	Proof. Neccessity.

Theorem 2.4.17. Let H be an odd-hole in the conflict graph H K E with |H| ≥ 5. Then, inequality (2.47) is facet defining for P(G, K, S) if and only if a) for each v k ′ ,e ′ / ∈ H, there exists a node v k,e ∈ H such that the induced graph H K E (H

  |K|}, we let E 88 k i be the set of edges involved in a shortest path between o k i and d k i , b) for demand k, we let E 88 k be the set of edges involved in a shortest path between o k and d k such that edge e is compatible with all the selected edges e ∈ E 88 |H|-1 2 , and each pair of nodes (v k ′ ,e ′ , v k",e" ) ∈ H88 are not linked in the conflict graph H K E , d) for each pair of demand k ′ and edge e ′ with v k ′ ,e ′ ∈ H88 , we consider a new set of edges E 88 k ′ involved in a shortest path between o k ′ and d k ′ if edge e ′ is not compatible with all the selected edges e" ∈ E 88 k ′ , or we add edge e ′ in E 88 k ′ if not, i.e., E 88 k ′ = E 88 k ′ ∪ {e ′ }, e) for each demand k ′ and edge e ′ with v k ′ ,e ′ ∈ H \ H88 , we modify the set of edges E 88 k ′ if E 88 k ′ contains some edges e ′ that are non compatible with the selected edges E 88 k" with v k",e" ∈ H88 . This can be done by selecting a new set of edges E 88 k ′ which contains all edges involved in a shortest path between o k ′ and d k ′ such that edge e ′ is compatible with each edge e" and demand k" with v k",e" ∈ H88 , f) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 88 i given by {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s k -w k } ∪ {s k + w ki , ..., s}]

	I 88 i = [
	kj ∈D 88 i
	if E 88 ki ∩ (E 88 k ∪ {e}) ̸ = ∅ or I 88 i =
	kj ∈D 88

. Consider a demand k ∈ K and an edge e ∈ E \ (E k 0 ∪ E k 1 ). Let S 88 = (E 88 , S 88 ) be the solution given by a) for each demand k i ∈ K \ {k} with i ∈ {1, ..., k , i.e., e ′ ∈E 88 k ℓ e ′ + ℓ e ≤ lk , c) select a subset of nodes H88 from H with | H88 | = i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

  we take into account the possibility of using edge e in the selected path E88 k to route demand k in solution S 88 ). We let S 88 k i = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. S 88 is feasible for the problem. Hence, the corresponding incidence vector (x S 88 , z S 88 )H . Then we derive a solution S 89 = (E 89 , S 89 ) obtained from S 88 by adding edge e ∈ E \ (E k 0 ∪ E k 1 ) for the routing of demand k in solution S 88 which means that E 89 k = E 88 k ∪ {e}. The last slots assigned to the demands K, and paths assigned the set of demands K \ {k} in S 88 remain the same in solution S 89 , i.e.,89 is clearly feasible for the problem. The corresponding incidence vector (x S 89 , z S 89 ) belongs to H . Hence, solutions S 88 and S 89 satisfy equation µx + σz = τ . It follows that µx S 88 + σz S 88 = µx S 89 + σz S 89 = µx S 88 + µ k e + σz S 88 .

	belongs to F	H K E
	S 89 k = S 88
	F	H K E

k for each k ∈ K, and

E 89 k ′ = E 88 k ′ for each k ′ ∈ K \ {k}. S

  and each pair of nodes (v k ′ ,e ′ , v k",e" ) ∈ H′88 are not linked in the conflict graph H K E , c) for each pair of demand k ′ and edge e ′ with v k ′ ,e ′ ∈ H′88 , we consider a new set of edges E ′88 k ′ involved in a shortest path between o k ′ and d k ′ if edge e ′ is not compatible with all the selected edges e" ∈ E ′88 k ′ , or we add edge e ′ in E ′88 k ′ if not, i.e., E ′88 k ′= E ′88 k ′ ∪ {e ′ }, d)for each demand k ′ and edge e ′ with v k ′ ,e ′ ∈ H \ H′88 , we modify the set of edgesE ′88 k ′ if E ′88k ′ contains some edges e ′ that are non compatible with the selected edges E ′88k" with v k",e" ∈ H′88 . This can be done by selecting a new set of edges E ′88 k ′ which contains all edges involved in a shortest path between o k ′ and d k ′ such that edge e ′ is compatible with each edge e" and demand k" with v k",e" ∈ H′88 , e) for each demand k i ∈ K with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I ′88

		i	given by
	I ′88 i	= [
		kj ∈D ′88

i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∩ {s ′ + w ki , ..., s}] if E ′88 ki ∩ E ′88 k ̸ = ∅ or I ′88 i = kj ∈D ′88 i {w ki , ...,

s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

  the path assigned to demand k ′ in S 88 from E 88 k ′ to a path E91 k ′ passed through edge e ′ with v k ′ ,e ′ ∈ H, b) and selecting a pair of demand-edge (k, e) from the set of pairs of demand-edge inH 88 such that v k ′ ,e ′ is not linked with any node v k",e" in H 88 \ {v k,e }, c) modifying the path assigned to demand k in S 88 with e ∈ E 88 k and v k,e ∈ H from E 88 k to a path E 91 k without passing through any edge e" ∈ E \ (E k 0 ∪ E k 1 ) such that v k ′ ,e ′ and v k,e" linked in H, d) modifying the last slots assigned to some demands K ⊂ K from S 88 k to S 91 k for each k ∈ K while satisfying non-overlapping constraint. The paths assigned to the demands K \ {k, k ′ } in S 88 remain the same in S 91 (i.e., E 91 k" = E 88 k" for each k" ∈ K \ {k, k ′ }), and also without modifying the last slots assigned to the demands K \ K in S 88 , i.e., S 88 k = S 91 k for each demand k ∈ K \ K. Solution S 91 is feasible for the problem. The corresponding incidence vector (x S 91 , z S 91 ) belongs to F H . Hence, solutions S 88 and S 91 satisfy equation µx + σz = τ . We have so µx S 88 + σz S 88 = µx S 91 + σz S 91 = µx S 88 + σz S 88 + µ k ′ e

	H K E
				s∈S 88 k σ k s
	+	µ k ′ e" -	µ k ′ e" +	µ k e" -
	e"∈E 91 k ′ \{e ′ }	e"∈E 88 k ′	e"∈E 91 k	e"∈E 88 k \{e}

for all k ∈ K and s ∈ {w k , ..., s} with v k,s / ∈ H. Let prove now that µ k e for all v k,e are equivalent. Consider a node v k ′ ,e ′ in H such that e ′ / ∈ E 88 k ′ . For this, we derive solution S 91 from S 88 by a) modifying ′ -µ k e + k∈ K s ′ ∈S 91 k σ k s ′ -

  Now let us consider a node v k ′ ,e ′ in C such that e ′ / ∈ E 92 k ′ . For this, we derive solution S ′93 from S 92 by a) modifying the path assigned to demand k ′ in S 92 from E 92 k ′ to a path E ′93 k ′ passed through edge e ′ with v k ′ ,e ′ ∈ C, b) and modifying the path assigned to each demand k with v k,e ∈ H 92 in S 92 with

	β) + γQ. For this,
	we show that
	a) σ k s = 0 for all demand k ∈ K and slot s ∈ {w k , ..., s} as done in the proof of theorem
	2.4.17,
	b) and µ k e = 0 for all demand k ∈ K and edge e ∈ E \ (E k 0 ∪ E k 1 ) with v k,e / ∈ H ∪ C as
	done in the proof of theorem 2.4.17,
	c) and µ k

e are equivalent for all v k,e ∈ H as done in the proof of theorem 2.4.17, given that the solutions defined in the proof of theorem 2.4.17, their corresponding incidence vector belong to F

H K E H,C . Let prove now that µ k ′ e ′ are equivalent for all v k ′ ,e ′ ∈ C.

e ∈ E 92 k and v k,e ∈ H from E 92 k to a path E ′93 k without passing through any edge e" ∈ E \ (E k 0 ∪ E k 1 ),

  ℓ e > lke ′ ∈E k 1 ℓ e ′ , and each pair of edges (e, e ′ ) ∈ C are compatible edges for demand k. Furthermore, it is said minimal cover for demand k if and only if for each e ∈ C we have e ′ ∈C\{e} ℓ e ′ ≤ lke"∈E k

	1	l e" .
	Based on this, we introduce the following inequalities.
	Proposition 2.4.23. Consider a demand k ∈ K. Let C be a minimal cover related
	to the tranmission-reach constraint for demand k. Then, the inequality
	x k e ≤ |C| -1,	(2.49)
	e∈C
	is valid for P(G, K, S).
	-Reach-Cover Inequalities
	Inequalities (2.46), (2.47) and (2.48) can be strengthened by defining a minimal
	cover related to the transmission-reach constraint.

Definition 2.4.9. Consider a demand k ∈ K. A cover C for demand k related to the transmission-reach constraint is a subset of edges in E \ (E k 0 ∪ E k 1 ) such that e∈C Proof. It is trivial given that C is minimal cover for demand k this means that there are at most |C| -1 edges from the set of edges in C that can be used to route demand k. Theorem 2.4.19. Consider a demand k in K. Let C be a minimal cover related to the tranmission-reach constraint for demand k. Then, inequality (2.49) is facet defining for the polytope P(G, K, S, C, k) where

  2.4.26. Consider an edge e in E. Let C be a minimal cover in K for edge e. Then, the inequality

	x k e ≤ |C| -1,	(2.54)
	k∈C

is valid for P(G, K, S).

Proof. If C is minimal cover for edge e ∈ E this means that there are at most |C| -1 demands from the set of demands in C that can use edge e.

  For this, we propose an exact algorithm in O(|E| * s * |K| * (|K| -1)) which works as follows. For each demand k and slot s ∈ {w k , ..., s} over edge e with x k e > 0, z k s > 0, we select each demand k ′ ∈ K \ {k} with xk ′ e > 0 and

	min(s+w k ′ -1,s)
	s"=s-w k +1

1) consists in identifying for each edge e, demand k ∈ K, and slot s ∈ {w k , ..., s}, a demands k ′ ∈ K such that xk e + xk ′ e + zk s + min(s+w k ′ -1,s) s"=s-w k +1

zk ′ s" > 3.

  a node in H e S having the largest value of node-degree (i.e., |δ(v k ′ ,s ′ )|) in H e S and v k ′ ,s ′ is linked with all the nodes v k,s ∈ C * in H e S and k ′ ∈ K e . Afterwards, we iteratively add each node v k",s" / ∈ C * ∪ N * to the current N * if it is linked in H e S with all the nodes already assigned to C * and N * and k" ∈ K e . At the end, we add inequality (2.42) induced by clique C * ∪ N * to the current LP, i.e.,

	(x k e + z k s
	v k,s ∈C *

  identified for each slot s i ∈ S and slot s j with s j ∈ {s i + max k∈K\ Ke w k , ..., min(s, s i + 2 max k∈K\ Ke w k )}. Consider now an interval of contiguous slots I = [s i , s j ] ∈ I e over an edge e, and its associated conflict graph H e I . We then use a greedy algorithm introduced by Nemhauser and Sigismondi[START_REF] Nemhauser | A Strong Cutting Plane/Branch-and-Bound Algorithm for Node Packing[END_REF] to identify a maximal clique in the conflict graph H e I as follows. We first associate a positive weight for each node v k in H e I equals to xk e * s j s ′ =s i +w k -1 zk s ′ . We then set C * = {k} such that k is a demand in K having the largest number of slots w k and weight xk e * s j s ′ =s i +w k -1 zk s ′ . After that, we iteratively add each demand k ′ having xk ′e > 0 ands j s ′ =s i +w k ′ -1zk ′ s ′ > 0 such that its corresponding node v k ′ is linked with all the nodes v k with k already assigned to the current C * . After that, we check if inequality (2.36) induced by the maximal clique C * for the interval I and edge e is violated or not. If so, we add inequality (2.36) induced by the maximal clique C * to the current LP, i.e.,

		s j
	x k e +	z k s ′ ≤ |C
	k∈C *	s ′ =s i +w k -1

* | + 1. One can strengthen this additional inequality by adding inequality (2.37) induced by the maximal clique C * and C * e ⊂ K e \ C * , i.e., k∈C *

  E S and v k ′ ,s ′ is linked with all the nodes v k,s ∈ C * in H E S . Afterwards, we iteratively add each node v k ′ ,s ′ / ∈ C * ∪ N * to the current N * if it is linked in H E S with all the nodes already assigned to C * and N * . At the end, we add inequality (2.43) induced by clique

a node in H E S having the largest value of node-degree (i.e., |δ(v k ′ ,s ′ )|) in H

  the shortest path between v k,s and its copy v ′ k,s in the auxiliary conflict graph H ′E S denoted by p v k,s ,v ′ k,s . After that, we check if the total sum of weight over edges belonging to this path is smaller than 1 2 . If so, odd-hole H * is composed by all the original nodes of nodes belong the computed shortest path p v k,s ,v ′ k,s , i.e., V (p v k,s ,v ′ k,s ) \ {v ′ k,s }. As a result, the following inequality (2.44) induced by odd-hole H * in the conflict graph H E S such that each node v k ′ ,s ′ ∈ C * should have a link with all the nodes v k,s ∈ H * , and the nodes v k",s" ∈ C * \ {v k ′ ,s ′ } in the conflict graph H E S . For this, we first assign a node v k ′ ,s ′ / ∈ H

	z k s ≤	|H
	v k,s ∈H *

* | -1 2 , should be added to the current LP. Moreover, one can strengthen inequality (2.44) induced by odd-hole H * using the greedy algorithm introduced by Nemhauser and Sigismondi [73] to identify a maximal clique C * * to clique C * (i.e., C * = {v k ′ ,s ′ }) such that v k ′ ,s ′ has the largest value of node-degree (i.e., |δ(v k ′ ,s ′ )|) in H E S and v k ′ ,s ′ is linked with all the nodes v k,s ∈ H * in H E S . After that, we iteratively add each node v k ′ ,s ′ / ∈ H * ∪ C * to the current clique C * if it is linked in H E S with all the nodes already assigned to odd-hole H * and clique C * . We then add inequality (2.45) induced by odd-hole H * and clique

  to identify a maximal clique in the conflict graph H K E taking into account the fractional solution (x, z) as follows. We first assign a positive weight xk e to each node v k,e in the conflict graph H K E . We then select a node v k,e in the conflict graph H K E having the largest weight xk e , and set C * = {v k,e }. After that, we iteratively add each node v k ′ ,e ′ to the current C * if it is linked with all the nodes v k,e ∈ C * and xk ′ e ′ > 0. At the end, the following inequality (2.46) induced by clique C *

	x k e ≤ 1,
	v k,e ∈C *
	should be added to the current LP if it is violated. Furthermore, one can strengthen
	the additional inequality (2.46) by identifying a maximal clique N

* such that each

  v k,e ,v ′ k,e . Note that if the total sum of weight over edges belonging to this path is smaller than 1 2 , this means that there exists odd-hole H * composed by all the original nodes of nodes belong the computed shortest path p v k,e ,v ′ k,s , i.e., V (p v k,e ,v ′ Moreover, inequality (2.47) induced by odd-hole H * can be lifted using the greedy algorithm introduced by Nemhauser and Sigismondi [73] by identifying a maximal clique C * in the conflict graph H K E such that each node v k ′ ,e ′ ∈ C * should have a link with all the nodes v k,e ∈ H * , and the nodes v k ′ ",e" ∈ C * \{v k ′ ,e ′ } in the conflict graph H K E . For this, we first assign a node v k ′ ,e ′ / ∈ H * to clique C * (i.e., C * = {v k ′ ,e ′ }) having the largest degree |δ(v k ′ ,e ′ )| in H K E , and v k ′ ,e ′ should be linked with all the nodes v k,e ∈ H . We then add inequality (2.48) induced by odd-hole H * and clique C *

					k,s ) \ {v ′ k,s },
	such that its associated inequality (2.47) is violated by the current fractional solution
	(x, z) to the current LP. As a result, we add following inequality (2.47) induced by
	odd-hole H *
	v k,e ∈H *	x k e ≤	|H * | -1 2	.
	x k e +	|H
	v k,e ∈H *
					H ′K E
	by duplicating each node v k,e in H K E (i.e., v k,e and v ′ k,e ) such that each two nodes
	are linked in H ′K E if their original nodes are linked in H K E . We assign to each link (ṽ k,e , ṽk ′ ,e ′ ) in H ′K E a weight 1-x k e -x k ′ e ′

2

. After that, we compute for each node v k,e in H K E , the shortest path between v k,e and its copy v ′ k,e . We denote this shortest path by p * in H K E . After that, we iteratively add each node v k ′ ,e ′ / ∈ H * ∪ C * to the current clique C * if it is linked in H K E with all the nodes already assigned to H * ∪ C * * | -1 2

  solution from (x, z). For this, we first construct several paths R k for each demand k ∈ K based on the fractional values xk e such that for each p ∈ R k

	xk e ≥ 1, ∀X ⊂ V s.t. |X ∩ {o k , d k }| = 1 and
	e∈δ(X)∩E(p)	e∈E(p)

  .1. The demands K are randomly generated with |K| ∈ {10, 20, 30, 40, 50, 100, 150}, and s up to 320 slots. Note that we tested 4 instances for each triplet (G, K, s) with |K| ∈ {10, 20, 30, 40, 50, 100, 150, 200, 250, 300}, and s up to 320 slots.

	Graphs	Number	Number	Max Node	Min Node	Average Node
			of Nodes	of Links	Degree	Degree	Degree
		German	17	25	5	2	2.94
	Real Topology	Nsfnet Spain Conus75	14 30 75	21 56 99	4 6 5	2 2 2	3 3.73 2.64
		Coronet100	100	136	5	2	2.72
		Europe	28	41	5	2	2.92
		France	25	45	10	2	3.6
		German50	50	88	5	2	3.52
	Realistic Topology	Brain161 Giul39 India35	161 39 35	166 86 80	37 8 9	1 3 2	2.06 4.41 4.57
		Pioro40	40	89	5	4	4.45
		Ta65	65	108	10	1	3.32
		Zib54	54	80	10	1	2.96

Table 3 .

 3

1: Characteristics of Different Topologies Used for our Experiments.

Table 3 .

 3 2: Table of Comparison for the B&C Algorithm: Cplex (Without or With Additional Valid Inequalities) Vs Scip (Without or With Additional Valid Inequalities).

	Own B&C SCIP	Nbr Nd Gap Nbr Cuts TT	59 0,00 429,75 0,83	141 0,00 2403,50 3,89	160376,50 1,46 129867,25 8334,95	383058,66 3,70 224642,33 16624,87	251152,50 13,73 305309,75 17074,95	3014,25 0,00 17668,50 617,80	3609 0,00 24782,25 3057,79	1 0,00 95,75 0,15	21586 0,00 24587 192,27	281569,66 3,29 340177 11048,71	119841,66 1,17 163519,33 5673,46	148476,50 5,91 340399,25 17405,09	1 0,00 464,25 40,87
	B&C SCIP	Nbr Nd Gap TT	1310,25 0,00 14,35	185956 0,27 3895,50	401335,75 1,60 11740,04	315993,66 8,33 16206,36	246146,50 9,62 16675,88	1158,50 0,00 340,10	12759 0,01 7329,06	13462 0,00 113,64	699646 9,51 18000	272065 40,99 18000	225696,67 46,74 18000	247873,25 43,09 18000	56598,50 57,19 18000
	Own B&C GRB	Nbr Nd Gap Nbr Cuts TT	3460,25 0,00 1866,50 149,13	28093,75 0,27 33413 4962,48	3614,50 0,00 19141,50 882,68	9920 1,82 216792,33 15453,54	6679 17,55 263086,50 18000	1746,50 0,00 200920 15425,53	979,75 73,06 44188,25 18000	4385 0,00 4095,25 216,83	181730 2,58 99411 18000	19702,67 13,20 263905,66 18000	6239,33 24,31 307366 18000	5265,75 47,22 347095,75 18000	2253,25 41,60 326605 18000
	B&C GRB	Nbr Nd Gap TT	4906,25 0,00 981,48	20752 0,27 4889,90	24321,50 0,39 9279,54	35451,67 5,52 18000	18901,50 6,15 18000	1 0,00 1634,37	64,75 0,00 3184,56	15222,75 0,00 2087,33	51525 12,77 18000	27735 22,41 18000	12631 34,18 18000	8733,50 29,35 18000	7790,50 29,60 18000
	Instances	Instances |K| |S|	10 15	20 45	30 45	German 40 45	50 55	100 140	150 210	10 15	20 20	30 30	40 35	50 50	100 120

Table 3 .

 3 3: Table of Comparison for the B&C Algorithm: Gurobi (Without or With Additional Valid Inequalities) Vs Scip (Without or With Additional Valid Inequalities).

Table 3 .

 3 4: Table of Comparison for the B&C Algorithm: Cplex (Without or With Additional Valid Inequalities) Vs Gurobi (Without or With Additional Valid Inequalities).

	Instances		B&C SCIP	Own B&C SCIP
	Graph	|K|	s	Nbr Nd	Gap	TT	Nbr Nd	Gap	TT
		10	15	1310,25	0,00	14,35	59	0,00	0,83
		20	45	185956	0,27	3895,5	141	0,00	3,89
		30	45	401335,75	1,60	11740,04	160376,50	1,46	8334,95
	German	40 50	45 55	315993,66 246146,50	8,33 9,62	16206,36 16675,88	383058,66 251152,50 13,73 17074,95 3,70 16624,87
		100 140	1158,50	0,00	340,10	3014,25	0,00	617,80
		150 210	12759	0,01	7329,06	3609	0,00	3057,79
		200 260	5099,33	0,78	10095,88	3067	0,00	6770,75
		10	320	1	0,00	10,37	1	0,00	462,15
		20	320	10,50	0,00	19,21	15	0,00	832,22
		30	40	66534,75	16,40	18000	11304,25	6,08	5006,31
	Coronet100	40 50	40 80	81051 11385,25	3,96 0,01	18000 4496,92	2127 19,75	0,00 0,00	707,54 139,55
		100 120	12787,50	13,36 14228,34	8390,25	7,66	10920,70
		150 200	4454,50	27,12 13692,63	3165,75	29,13 15527,10
		200 280	3579,25	33,35	18000	1	38,97	18000
		10	15	13462	0,00	113,64	1	0,00	0,15
		20	20	699646	9,51	18000	21586	0,00	192,27
		30	30	272065	40,99	18000	281569,66	3,29	11048,71
	Nsfnet	40 50	35 50	225696,67 46,74 247873,25 43,09	18000 18000	119841,66 148476,50	1,17 5,91	5673,46 17405,09
		100 120	56598,50	57,19	18000	1	0,00	40,87
		150 160	12663	58,50	18000	1	0,00	136,02
		200 210	7726,50	54,85	18000	710	0,28	9121,79
		10	40	1907,25	0,00	87,60	1	0,00	1,80
		20	40	9	0,00	4	7	0,00	5,92
		30	40	91798	0,00	7821,5	32156,75	0,00	2309,66
	India35	40 50	40 80	161514 34	2,42 0,00	17486,08 22,13	191812 69,25	0,18 0,00	17333,53 112,19
		100 120	24797	0,32	9137,26	23403,75	0,44	9494,52
		150 200	16809	0,21	13739,65	1026	0,00	4101,80
		200 280	11197	0,37	13930,35	2027,75	3,69	14516,65
		10	40	1	0,00	1,49	1	0,00	1,69
		20	40	1,50	0,00	3,44	1	0,00	4,88
		30	40	1,50	0,00	5,72	6,25	0,00	10,54
	Pioro40	40 50	40 80	83597 14	0,20 0,00	8692,5 15,93	67151 4	0,12 0,00	8711,30 54,39
		100	80	21281,75	0,04	9087,52	23785,75	0,04	9916,63
		150 160	823,50	0,00	816,89	124,50	0,00	1509,87
		200 280	1503,75	0,00	3772,9	423,50	0,00	7424,98
		10	40	1	0,00	1,58	1	0,00	1,83
		20	40	1,50	0,00	2,92	1	0,00	3,71
		30	40	4	0,00	4,50	1	0,00	6,10
	Giul39	40 50	40 40	4,50 54420	0,00 0,00	7,17 4376,98	1 52156,75	0,00 0,00	10,15 4361,26
		100	40	55472,50	6,88	17781,71	54675,50	8,38	17802,83
		150 120	836	0,00	1050,13	11655,50	0,00	9411,30
		200 120	10191,25	0,24	13794,32	6518	0,01	9914,02

Table 3 .

 3

5: The Impact of Valid Inequalities in the Own B&C SCIP Performance Using Realistic Graphs.

Table 4 .

 4 1: Influence of the Valid Inequalities: B&P Vs B&C&P.

Table 4 .

 4 2: Table of Comparison Between B&C, B&P and B&C&P Algorithms: Cplex (Without or With Additional Valid Inequalities) Vs Scip (Without or With Additional Valid Inequalities).

Table 4 .

 4 3: Table of Comparison Between B&C, B&P and B&C&P Algorithms: Gurobi (Without or With Additional Valid Inequalities) Vs Scip (Without or With Additional Valid Inequalities).

Table 4 .

 4 4: Table of Comparison Between B&C, B&P and B&C&P Algorithms: Scip (Without or With Additional Valid Inequalities) Vs Scip (Without or With Additional Valid Inequalities).

  1}, for all s ∈ S.(5.8) where Ke denotes the set of demands in K passing through edge e (i.e., Ke = {k ∈ K, e ∈ E(p k )}. Equations (5.2) ensure that demand k cannot occupy a slot s as last slot before her slot-width w k . Inequalities(5.3) ensure than more than one interval of contiguous slots can be assigned to each demand k ∈ K. It should normally be an equation form ensuring that exactly one slot s ∈ {w k , . . . , s} (one interval of contiguous slots) must be assigned to demand k as last-slot. Here we relax this constraint. Optimizing the spectrum-usage objective function, the equality is a) we select the smallest slot index s k in {w k , ..., s} \ {s, ..., s + w k -1} as last slot for demand k (slot assignment constraint taking into account the possibility of adding slot s in the set of used slots U 105 ), b) for each demand k i ∈ K with i ∈ {1, ..., |K|} \ {k}, we select the smallest slot index s k i in the set of slots I 105 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] \ {s, ..., s + w ki -1}

		i	given by
	I 105 i	= [
		kj ∈D 105 i

  107 , S 107 ) be the solution given by a) we select the smallest slot index s k in {w k , ..., s} \ {s} as last slot for demand k, b) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 107 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s -w k } ∪ {s + w ki , ..., s}] if E(p ki ) ∩ E(p k ) ̸ = ∅ or I 107 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

		i	given by
	I 107 i	= [
		kj ∈D 107 i
			i	=
			kj ∈D 107 i

  we take into account the possibility of adding slot s in the set of last slots S107 We let S 107 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}.c) we let U 107 be the set of slots used in S such that for each demand k and last slots k ∈ S 107 k and s ′ ∈ {s k -w k + 1, ..., s k }, we have s ′ ∈ U 107 .S 107 is clearly feasible for the problem given that it satisfies all the SA constraints of the compact formulation (5.1)-(5.8). Hence, the corresponding incidence vector (u S 107 , z S 107 ) belongs to P sa (G, K, S). Then consider the solution S 108 = (E 108 , S 108 ) obtained from S 107 by adding slot s as last slot to demand k without modifying the last slots assigned to the demands K \ {k} in S 107 remain the same in solution S 108 i.e., S 107 k ′ = S 108 k ′ for each demand k ′ ∈ K \ {k}, and S 108 Solution S 108 is feasible for the SA problem. The corresponding incidence vector (u S 108 , z S 108 ) belongs to P sa (G, K, S). We then obtain that µu S 107 + σz S 107 = µu S 108 + σz S 108 = µu S 107 + σz S 107 + σ k s +

			k	assigned to
	demand k in solution S 107 ).
	k	= S 107 k	∪ {s} for demand
	k. s∈{s,...,s-w k +1}\U 107	µ s.

  .., s} \ {s}. Consider a slot s ′ in {w k , ..., s} \ {s}. Let S 110 = (U 110 , S 110 ) be the solution given by a) we select the smallest slot index s k in {w k , ..., s} \ {s, s ′ } as last slot for demand k, b) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 110 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}] if E(p ki ) ∩ E(p k ) ̸ = ∅ or I 110 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

		i	given by
	I 110 i	= [
		kj ∈D 110 i
			i	=
			kj ∈D 110 i

  we take into account the possibility of adding slot s ′ in the set of last slots S 110 = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}. c) we let U 110 be the set of slots used in S such that for each demand k ′ ∈ K and lastslot s k ′ ∈ S 10 k ′ and s" ∈ {s k ′ -w k ′ + 1, ..., s k ′ }, we have s" ∈ U 110 .S 110 is clearly feasible for the SA problem. Hence, the corresponding incidence vector (u S 110 , z S 110 ) belongs to F k s . Then we derive a solution S 112 = (U 112 , S 112 ) obtained from S 110 by adding slot s ′ as last slot to demand k without modifying the last slots assigned to the demands K \ {k} in S 110 Solution S 112 is feasible for the SA problem. The corresponding incidence vector (u S 112 , z S 112 ) belongs to F k s . Hence, solutions S 110 and S 112 satisfy equation µu + σz = τ . We the obtain that µu S 110 + σz S 110 = µu S 112 + σz S 112 = µu S 110 + σz S 110 + σ k s ′ + s∈{s ′ -w k +1,...,s ′ }\U 110

			k	assigned to
	demand k in solution S 110 ),
	Let S 110 k i k	remain the same in solution
	S 112 i.e., S 110 k ′ = S 112 k ′ for each demand k ′ ∈ K \ {k}, and S 112 k	= S 110
			µ s.

′

∪ {s ′ } for demand k.

  113 i , • and s / ∈ {s k i -w k i + 1, ..., s k i } (slot assignment constraint taking into account the possibility of adding slot s in the set of used slots U 113 ), = {s k i } be the set of last slots assigned to each demand k i with i ∈ {1, ..., |K|}. c) we let U 113 be the of slots used in S such that for each demand k and last slot s ∈ S 113 k and s ′ ∈ {s k -w k + 1, ..., s k }, we have s ′ ∈ U 113 . S 113 is clearly feasible for the SA problem, and its incidence vector (u S 113 , z S 113 ) belongs to F s . After that, we derive a solution S ′113 = (U ′113 , S ′113 ) obtained from S 113 by adding slot s as an used slot in U ′113 without modifying the last slots assigned to the demands K in S 113 which remain the same in solution S ′113 i.e., Solution S ′113 is feasible for the SA problem. The corresponding incidence vector (u S ′113 , z S ′113 ) belongs to F s . Hence, solutions S 113 and S ′113 satisfy equation µu + σz = τ . We then obtain that µu S 113 + σz S 113 = µu S ′113 + σz S ′113 = µu S 113 + σz S 113 + µ s.

	S 113 k	= S ′113 k	for each demand k ∈ K.
	Let S 113 k i

  |K|}, we select the smallest slot index s k i in the set of slots I 114 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}] if E(p ki ) ∩ E(p k ) ̸ = ∅ or I 114 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

		i	given by
	I 114 i	= [
		kj ∈D 114 i
		i	=
			kj ∈D 114 i

  } for demand k. Solution S 115 is feasible for the SA problem. The corresponding incidence vector (u S 115 , z S 115 ) belongs to F s . Hence, solutions S 114 and S 115 satisfy equation µu + σz = τ . We the obtain that µu S 114 + σz S 114 = µu S 115 + σz S 115 = µu S 114 + σz S 114 + σ k s ′ + s∈{s ′ -w k +1,...,s ′ }\U 114

		114 k ′ =
	S 115 k ′ for each demand k ′ ∈ K \ {k}, and S 115 k	= S 114
		µ s.

′

∪ {s ′

  116 i ,• and s / ∈ {s k i -w k i + 1, ..., s k i } (slot assignment constraint taking into account the possibility of adding slot s in the set of used slots U 116 ).We let S 116 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}. c) a set of slots U 116 is then used in S such that for each demand k ′ ∈ K and last slots ∈ S 116 k ′ and s ′ ∈ {s k ′ -w k ′ + 1, ..., s k ′ }, we have s ′ ∈ U 116 ,S116 is clearly feasible for the SA problem. The corresponding incidence vector (u S 116 , z S 116 ) belongs to F k S . Next, we derive a solution S ′116 = (U ′116 , S ′116 ) obtained from S 116 by adding slot s as an used slot in U ′116 without modifying the last slots assigned to the demands K in S 116 which remain the same in solution ∈ K. Solution S ′116 is feasible for the SA problem, and its incidence vector (u S ′116 , z S ′116 ) belongs to F k S . Hence, solutions S 116 and S ′116 satisfy equation µu + σz = τ . We then obtain that µu S 116 + σz S 116 = µu S ′116 + σz S ′116 = µu S 116 + σz S 116 + µ s.

	S ′116 i.e., S 116 k	= S ′116 k	for each demand k

  |K|}, we select the smallest slot index s k i in the set of slots I 117 {w ki , ..., s kj -w kj } ∪{s kj +w ki , ..., s}]∩[{w ki , ..., s ′ -w k ′ } ∪{s ′ +w ki , ..., s}] if E(p ki ) ∩ E(p k ′ ) ̸ = ∅ or I 117 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

		i	given by
	I 117 i	= [
		kj ∈D 117 i
		i	=
			kj ∈D 117 i

  we take into account the possibility of adding slot s ′ in the set of last slots S117 k ′ assigned to demand k ′ in solution S 117), We let S 110 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}. d) a set of slots U 117 are then used in S such that for each demand k" ∈ K and last slot s ∈ S 117 k" and s" ∈ {s k" -w k" + 1, ..., s k" }, we have s" ∈ U 117 . S 117 is clearly feasible for the SA problem. The corresponding incidence vector (u S 117 , z S 117 ) belongs to F k S . Then consider the solution S 118 = (U 118 , S 118 ) obtained from S 117 by adding slot s ′ as last slot to demand k ′ without modifying the last slots assigned to the demands K \ {k ′ } in S 117 ∈ K \ {k ′ }, and S 118 k ′ = S 117 k ′ ∪ {s ′ } for demand k ′ . Solution S 118 is feasible for the SA problem. The corresponding incidence vector (u S 118 , z S 118 ) belongs to F k S . Hence, solutions S 117 and S 118 satisfy equation µu + σz = τ . We the obtain that µu S 117 + σz S 117 = µu S 118 + σz S 118 = µu S 117 + σz S 117 + σ k ′ s ′ + s∈{s ′ -w k ′ +1,...,s ′ }\U 117

			k	remain the same in solution S 118
	i.e., S 117 k	= S 118 k	for each demand k µ s.

  and s ′ ∈ {w k ′ , ..., s}.Let prove now that σ k s for demand k and slot s in {w k , ..., s} are equivalent. Consider a slot s ′ ∈ {w k , ..., s} such that s ′ / ∈ S 119 k . Let S119 = ( Ũ 119 , S119 ) be the solution given by a) we select the smallest slot index s k from {w k , ..., s} \ {s ′ } as last slot for demand k, b) for each demand k i ∈ K \ {k} with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots Ĩ117

		i	given by
	Ĩ117 i	= [
		kj ∈ D117

i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}] if E(p ki ) ∩ E(p k ) ̸ = ∅ or Ĩ117 i = kj ∈ D117 i {w ki , ...,

s kj -w kj } ∪ {s kj + w ki , ..., s} if not, where D117 i

  we take into account the possibility of adding slot s ′ in the set of last slots S117 k ′ assigned to demand k ′ in solution S117 ). = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}. c) we let Ũ 117 be the set of slots used in S such that for each demand k ′ ∈ K and last slot s k ′ ∈ S117 k ′ and s" ∈ {s k ′ -w k ′ + 1, ..., s k ′ }, we have s" ∈ Ũ 117 . S118 is feasible for the SA problem. Hence, the corresponding incidence vector S . Based on this, we derive a solution S 119 = (U 119 , S 119 ) from S118 by adding slot s ′ as last slot to demand k and removing the last slot

	Let S117 k i (u S118 , z	S118 ) belongs to F k
	s ∈ S 118 k , i.e., S 119 k

  This is equivalent to say that two linked nodes v k and v k ′ means that the routing paths of the demands k, k ′ share an edge in G.Based on the conflict graph H sa , we introduce the following inequalities. Proposition 5.4.1. Let I = [s i , s j ] be an interval of contiguous slots in[1, s]. Let K ′ ⊂ K be a minimal cover for interval I = [s i , s j ] such that K ′ defines a clique in H sa . Then, the inequality

		s j
	k∈K ′	s=s

Definition 5.4.1. Consider the conflict graph H sa defined as follows. For each demand k ∈ K, consider a node v k in H sa . Two nodes v k and v k ′ are linked by an edge in H sa if and only if E(p k ) ∩ E(p k ′ ) ̸ = ∅.

  Let show that σ k s = 0 for all k ∈ K and s ∈ {w k , ..., s} with s / ∈ {s i + w

		i	given by
	I 122 i	= [
		kj ∈D 122

k -1, ..., s j } if k ∈ K. Consider a demand k in K and a slot s ′ in {w k , ..., s} with s ′ / ∈ {s i + w k -1, ..., s j } if k ∈ K. Let S 122 = (U 122 ,

S 122 ) be the solution given by a) for one demand k ′ from K, we select the smallest slot index s k ′ in {w k ′ , ..., s} \ {s i + w k ′ -1, ..., s j } as last slot, b) for each demand k i ∈ K \ {k ′ } with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 122 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ {s i + w ki -1, ..., s j },

  |K|}, we select the smallest slot index s k i in the set of slotsI 122 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}] if E(p ki ) ∩ E(p k ) ̸ = ∅ or I 122

		i	given by
	I 122 i	= [
		kj ∈D 122 i
			i	=
			kj ∈D 122

i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not, where D 122 i

  we take into account the possibility of adding slot s ′ as a last slot in the selected last slots let S 122 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}.d) a set of slots U 122 are then used in S such that for each demand k ′ ∈ K and lastslot s k ′ ∈ S 120 k ′ and s" ∈ {s k ′ -w k ′ + 1, ..., s k ′ }, we have s" ∈ U 120 .S 122 is clearly feasible for the problem. Hence, the corresponding incidence vector (u S 122 , z S 122 ) belongs to F I K . Then we derive a solution S 123 = (U 123 , S 123 ) obtained from S 122 by adding slot s ′ as last slot to demand k without modifying the last slots assigned to the demands K \ {k} in S 122 , i.e., S 122k ′ = S 123 k ′ for each demand k ′ ∈ K \ {k},and S 123 } for demand k. Solution S 123 is feasible for the SA problem. The corresponding incidence vector (u S 123 , z S 123 ) belongs to F I K . Hence, solutions S 122 and S 123 satisfy equation µu + σz = τ . We then obtain that µu S 122 + σz S 122 = µu S 123 + σz S 123 = µu S 122 + σz S 122 + σ k s ′ +

	We k	= S 122 k	∪ {s ′ s∈U 123 \U 122	µ s
			-	µ s.
			s∈U 122 \U 123
	S 122

k to route demand k in solution S 122 ).

  |K|}, we select the smallest slot index s k i in the set of slots I 124 {w ki , ..., s kj -w kj } ∪{s kj +w ki , ..., s}]∩[{w ki , ..., s ′ -w k ′ } ∪{s ′ +w ki , ..., s}] if E(p ki ) ∩ E(p k ) ̸ = ∅ or I 124

		i	given by
	I 124 i	= [
		kj ∈D 124 i

i = kj ∈D 124 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not, where D 124 i

  we take into account the possibility of adding slot s ′ as a last slot in the selected last slotsS 124 k ′ to route demand k ′ in solution S 124 ).We let S 124 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}.d) let U 124 be the set of slots used in S such that for each demand k" ∈ K and last slot s k" ∈ S 124 k" and s" ∈ {s k" -w k" + 1, ..., s k" }, we have s" ∈ U 124 .S 124 is clearly feasible for the problem given that it satisfies all the constraints of cut formulation (2.2)-(2.10). Hence, the corresponding incidence vector (u S 124 , z S 124 ) belongs to F I K . Then consider the solution S 125 = (E 125 , S 125 ) obtained from S 125 by adding slot s ′ as last slot to demand k such that the last slots assigned to the demands K \ {k, k ′ } in S 125 remain the same in S 125 , i.e., S 125 k" = S 125 k" for each demand k" ∈ K \ {k, k ′ }, and S 125 k ′ = S 125 k ′ ∪ {s ′ } for demand k ′ , and modifying the last slots assigned to demand k by adding a new last slot s and removing the last slot s ∈ S 125

k with s ∈ {s i + w k + 1, ..., s j } and s / ∈ {s i + w k + 1, ..., s j } for demand k with k ∈ K such that S 125 k

  Based on the definition of the conflict graph H E I , we define a new conflict graph adapted to the SA problem. Definition 5.4.2. Let I = [s i , s j ] be an interval of contiguous slots in [1, s] with s i ≤ s j -1. Consider the conflict graph H ′E I defined as follows. For each demand k ∈ K with w k ≤ |I|, consider a node v k in H ′E I . Two nodes v k and v k ′ are linked by an edge in H ′E I

  + w k ′ ≥ |I ′ | for each k, k ′ ∈ C, • 2w k ≥ |I ′ | + 1 and w k ≤ |I ′ | for each k ∈ C.This means that inequality (2.39) induced by clique C for the interval I is dominated by inequality (2.39) induced by clique C for the interval I ′ . Hence, inequality (2.39) cannot be facet defining for P sa (G, K, S). c) if there exists a slot s ∈ I such that s ∈ {s ′ -w k + 1, .., s ′ } for each k ∈ C and s ′ ∈ {s i + w k -1, .., s j }, this implies that inequality (2.39) is dominated by the the non-overlapping inequality (5.4). Hence, inequality (2.39) cannot be facet defining for P sa (G, K, S).

	Then, inequality (2.39) induced by clique C is dominated by another inequality (2.39) induced by clique C Sufficiency. Let F H ′E I C be the face induced by inequality (2.39), that is

′

. Hence, inequality (2.39) cannot be facet defining for P sa (G, K, S). b) if there exists an interval of contiguous slots I ′ in

[1, s]

such that I ⊂ I ′ with • w k

  |K|}, we select the smallest slot index s k i in the set of slots I 127 This guarantees that• {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ...s k j } = ∅ for each k j ∈ D 127 i , • and s / ∈ {s k i -w k i + 1, ...,s k i } (slot assignment constraint taking into account the possibility of adding slot s in the set of used slots U 127 ), We let S 127 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}. d) a set of slots U 127 are used in S such that for each demand k ∈ K and last slot s k ∈ S 127 k and s ′ ∈ {s k -w k + 1, ..., s k }, we have s ′ ∈ U 127 . S 127 is feasible for the SA problem. Hence, the corresponding incidence vector (u S 127 , z S 127 ) belongs to F H ′E I C . Then we derive a solution S 128 = (U 128 , S 128 ) obtained from S 127 by adding slot s as an used slot in U 128 without modifying the last slots assigned to the demands K in S 127 which remain the same in solution S 128 i.e., Solution S 128 is feasible for the SA problem. The corresponding incidence vector (u S 128 , z S 128 ) belongs to F H ′E I C . Hence, solutions S 127 and S 128 satisfy equation µu + σz = τ . We then obtain that µu S 127 + σz S 127 = µu S 128 + σz S 128 = µu S 127 + σz S 127 + µ s. |K|}, we select the smallest slot index s k i in the set of slots I 129

	S 127 k	= S 128
	Hence, µ s = 0.
	In a similar way, we can show that
			µ s = 0, for all slots s ∈ S.
	Let show that σ k
			i	given by
			I 129 i	= [
			kj ∈D 129
			i	given by
	I 127 i	= [
			kj ∈D 127

i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] \ {s, ..., s + w ki -1}

where

D 127 i = {k j ∈ {k 1 , ..., k i-1 } ∩ C : E(p k i ) ∩ E(p k j ) ̸ = ∅}. k for each demand k ∈ K. s = 0 for all k ∈ K and s ∈ {w k , ..., s} with s / ∈ {s i + w k -1, ..., s j } if v k ∈ C.

Consider a demand k in K and a slot s ′ in {w k , ..., s} with s ′ / ∈ {s i + w k -1, ..., s j } if k ∈ C. be the solution given by S 129 = (U 129 , S 129 ) be the solution given by a) for one demand k ′ from C, we select the slot s k ′ = s i + w k ′ -1 as last slot, b) for each demand k i ∈ C \ {k ′ } with i ∈ {1, ..., i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ {s i + w ki -1, ..., s j },

  |K|}, we select the smallest slot index s k i in the set of slots I 129 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}]

		i	given by
	I 129 i	= [
		kj ∈D 129 i

if E(p ki ) ∩ E(p k ) ̸ = ∅ or I 129 i = kj ∈D 129 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not, where D 129 i

  we take into account the possibility of adding slot s ′ as a last slot in the selected last slots We let S 122 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}.d) let U 129 be the slot used in S such that for each demand k ′ ∈ K and last slots k ′ ∈ S 129 k ′ and s" ∈ {s k ′ -w k ′ + 1, ..., s k ′ }, we have s" ∈ U 129 .S 129 is clearly feasible for the problem, and its incidence vector (u S 129 , z S 129 ) belongs to F H ′E I C . Then consider the solution S 130 = (U 130 , S 130 ) obtained from S 129 by adding slot s ′ as last slot to demand k without modifying the last slots assigned to the demands K \ {k} in S 129 , i.e., S 129 k ′ = S 130 k ′ for each demand k ′ ∈ K \ {k}, and S 130 k = S 129 k ∪ {s ′ } for demand k. Solution S 130 is feasible for the SA problem. The corresponding incidence vector (u S 130 , z S 130 ) belongs to F H ′E I C . Hence, solutions S 129 and S 130 satisfy equation µu + σz = τ . We then obtain that µu S 129 + σz S 129 = µu S 130 + σz S 130 = µu S 129 + σz S 129 + σ k s ′ +

	S 129 k	to route demand k in solution S 129 ),
		µ s -
		s∈U 130 \U 129

s∈U 129 \U 130 µ s.

  we select the smallest slot index s k i in the set of slotsI 131 {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]∩[{w ki , ..., s ′ -w k ′ }∪{s ′ +w ki , ..., s}]\{s i , ..., s j } if E(p ki )∩E(p k ′ ) ̸ = ∅ or I 131{w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]\{s i , ..., s j } if not, ) for each demand k i ∈ K \ C with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 131 {w ki , ..., s kj -w kj } ∪{s kj +w ki , ..., s}]∩[{w ki , ..., s ′ -w k ′ } ∪{s ′ +w ki , ..., s}] if E(p ki ) ∩ E(p k ′ ) ̸ = ∅ or I 131 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

			i	given by
		I 131 i	= [
			kj ∈D 131 i
			i	=
			kj ∈D 131 i
	where D 131 i
	i	given by
		I 131 i	= [
			kj ∈D 131 i
			i	= [
			kj ∈D 131 i
	where D 131 i	= {k j ∈ {k 1 , ..., k i-1 } ∪ C such that D 131 ki ∩ D 131 kj ̸ = ∅},

c

  we take into account the possibility of adding slot s ′ as a last slot in the selected last slotsS 131 k ′ to route demand k ′ in solution S 131 ).We let S 131 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}. d) Let U 131 be the set of slots used in S such that for each demand k and last slots k ∈ S 131 k and s" ∈ {s k -w k + 1, ..., s k }, we have s" ∈ U 131 .S 131 is clearly feasible for the problem. Hence, the corresponding incidence vector (u S 131 , z S 131 ) belongs to F H ′E I C . Then consider the solution S 132 = (U 132 , S 132 ) from S 131 by adding slot s ′ as last slot to demand k without modifying the last slots assigned to the demands K \ {k, k ′ } in S 131 , i.e., S 131 k" = S 132 k" for each demand k" ∈ K \ {k, k ′ }, and S 132 k ′ = S 131 k ′ ∪ {s ′ } for demand k ′ , and with modifying the last slots assigned to demand k by adding a new last slot s and removing the last slot s ∈ S 131 k with s ∈ {s i +w k +1, ..., s j } and s / ∈ {s i +w k +1, ..., s Solution S 132 is feasible for the SA problem. The corresponding incidence vector (u S 132 , z S 132 ) belongs to F H ′E I C . Hence, solutions S 131 and S 132 satisfy equation µu + σz = τ . We then obtain that µu S 131 + σz S 131 = µu S 132 + σz S 132 = µu S 131 + σz S 131 + σ k ′

	s ′ -σ k s + σ k s +	µ s"
	s"∈U 132 \U 131
	-	µ s" .
	s"∈U 131 \U 132

j } for demand k ∈ K with v k ∈ C such that S 132 k = (S 131 k \{s})∪{s} such that {s-w k +1, ..., s}∩{s ′ -w k ′ +1, ..., s ′ } = ∅ for each k ′ ∈ K and s ′ ∈ S 132 k ′ with E(p k ) ∩ E(p k ′ ) ̸ = ∅.

  Then, inequality (2.40) is valid for Q sa (G, K, S). Moreover, it is valid for P(G, K, S) if2w k > |I| for each v k ∈ H.Proof. We use the same proof of proposition(5.4.4). 5.4.4. Let H be an odd-hole in the conflict graph H ′E I with |H| ≥ 5, and 2w k > |I| for each v k ∈ H. Then, inequality (2.40) is facet defining for P sa (G, K, S) if and only if a) for each node v k ′ / ∈ H in H ′E I , there exists a node v k ∈ H such that the induced graph H ′E I((H \ {v k }) ∪ {v k ′ }) does not contain an odd-hole H ′ = (H \ {v k }) ∪ {v k ′ }, b) and there does not exist a node v k ′ / ∈ H in H ′E I such that v k ′ is linked with all nodes v k ∈ H,c) and there does not exist an interval I ′ of contiguous slots with I ⊂ I ′ such that H defines also an odd-hole in the associated conflict graph H E I ′ .We use the same proof presented in the proof of theorem (2.4.10).

	Sufficiency. Theorem Proof. Neccessity. Let F H ′E I H be the face induced by inequality (2.40), that is

Proposition 5.4.4. Let I = [s i , s j ] be an interval of contiguous slots in [1, s] with s i ≤ s j -1, and H be an odd-hole H in the conflict graph H ′E I with |H| ≥ 5.

1 2

 1 by αu + βz ≤ λ. Let µu + σz ≤ τ be a valid inequality that is facet definingF of P sa (G, K, S). |H|-12 , b) for each demand k i from H with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 134

				Suppose that
	F H ′E I H	⊂ F = {(u, z) ∈ P sa (G, K, S) : µu + σz = τ }. To prove that	F H ′E I H	is a facet
	of P i	given by
	I 134 i	= [
		kj ∈L 134
		i	given by
	I 134 i	=
		kj ∈D 134

sa (G, K, S), we need to show that there exist ρ ∈ R and γ ∈ R k∈K (w k -1) ) such that (µ, σ) = ρ(α, β) + γM .

Let first show that µ s = 0 for all s ∈ S. Consider a slot s ∈ S. Let S 134 = (U 134 , S 134 ) be the solution given by a) select a subset of demands H from H with | H| = i {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}]∩[{s i +w ki -1, ..., s j }]\{s, ..., s+w ki -1},

where

L 134 i = {k j ∈ {k 1 , ..., k i-1 } ∩ H : E(p k i ) ∩ E(p k j ) ̸ = ∅}. c) for each demand k i ∈ H \ H with i ∈ {1, ...,

|K|}, we select the smallest slot index s k i in the set of slots I 134 i {w ki , ..., s kj -w kj }∪{s kj +w ki , ..., s}\[{s i +w ki -1, ..., s j }∪{s, ..., s+w ki -1}],

  |K|}, we select the smallest slot index s k i in the set of slots I 134 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] \ {s, ..., s + w ki -1}

		i	given by
	I 134 i	= [
		kj ∈R 134 i

  and s / ∈ {s k i -w k i + 1, ..., s k i } (slot assignment constraint taking into account the possibility of adding slot s in the set of used slots U 134 ). We let S 134 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}.e) let U 134 be the set of slots used in S such that for each demand k and last slots k ∈ S 134 k and s ′ ∈ {s k -w k + 1, ...,s k }, we have s ′ ∈ U 134 , and and s / ∈ U 134 (slot assignment constraint taking into account the possibility of adding slot s in the set of used slots U 134 ). S 134 is clearly feasible for the SA problem. Hence, the corresponding incidence vector (u S 134 , z S 134 ) belongs to F H ′E I H . Then we derive a solution S 135 = (U 135 , S 135 ) obtained from S 134 by adding slot s as an used slot in U 135 without modifying the last slots assigned to the demands K in S 134 which remain the same in solution S 135 i.e., S 134 Solution S 135 is feasible for the SA problem. Hence, the corresponding incidence vector (u S 135 , z S 135 ) belongs to F H ′E I H . Hence, solutions S 134 and S 135 satisfy equation µu + σz = τ . We then obtain that µu S 134 + σz S 134 = µu S 135 + σz S 135 = µu S 134 + σz S 134 + µ s. |H|-1 2 , b) for each demand k i from H with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 136

		k	= S 135
	Hence, µ s = 0.
	In a similar way, we can show that
			µ s = 0, for all slots s ∈ S.
	Let show that σ k
			i	given by
	I 136 i	= [
		kj ∈L 136

k for each demand k ∈ K. s = 0 for all k ∈ K and s ∈ {w k , ..., s} with s / ∈ {s i + w k -1, ..., s j } if v k ∈ H. Consider a demand k in K and a slot s ′ in {w k , ..., s} with s ′ / ∈ {s i + w k -1, ..., s j } if v k ∈ H. Let S 136 = (U 136 , S 136 ) be the solution given by a) select a subset of demands H from H with | H| = i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{s i + w ki -1, ..., s j }],

  |K|}, we select the smallest slot index s k i in the set of slotsI 136 ∈ {k 1 , ..., k i-1 } ∩ H : E(p k i ) ∩ E(p k j ) ̸ = ∅}. We let S 136 k i = {s k i } be the set of last slots assigned to demand k i , d) for each demand k i ∈ K \ H with i ∈ {1, ...,|K|}, we select the smallest slot index s k i in the set of slots I 136 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}] if E(p ki ) ∩ E(p k ) ̸ = ∅ or I 136 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

		i	given by
	I 136 i	= [
		kj ∈D 136 i
			i	=
			kj ∈D 136 i
		i	given by
	I 136 i	=
		kj ∈D 136

i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} \ {s i + w ki -1, ..., s j }, where D 136 i = {k j

  we take into account the possibility of adding slot s ′ as a last slot in the selected last slots S 136 k to route demand k in solution S 136 ). We let S 136 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}, e) let U 136 be the set of slots used in S such that for each demand k ′ ∈ K and last slot s" ∈ S 136 k 137 , S 137 ) from S 136 by adding slot s ′ as last slot to demand k without modifying the last slots assigned to the demands K \ {k} in S 136 , i.e., S 136 k ′ = S 137 k ′ for each demand k ′ ∈ K \ {k}, and S 137 } for demand k. Solution S 137 is feasible for the SA problem. The corresponding incidence vector (u S 137 , z S 137 ) belongs to F H ′E I H . Hence, solutions S 136 and S 137 satisfy equation µu + σz = τ . We then obtain that µu S 136 + σz S 136 = µu S 137 + σz S 137 = µu S 136 + σz S 136 + σ k s ′ +

	k	= S 136 k	∪ {s ′ s∈U 137 \U 136	µ s
			-	µ s.
			s∈U 136 \U 137

′ and s" ∈ {s k ′ -w k ′ + 1, ..., s k ′ }, we have s" ∈ U 136 .

S 136 is clearly feasible for the problem. Hence, the corresponding incidence vector (u S 136 , z S 136 ) belongs to F H ′E I H . After that, we derive a solution S 137 = (U

  s j }. Let S 138 = (U 138 , S 138 ) be the solution given by a) select a subset of demands H from H with | H| = |H|-1 2 , b) for each demand k i from H with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 138 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ {s i + w ki -1, ..., s j }.

		i	given by
	I 138 i	= [
		kj ∈L 138 i

  |K|}, we select the smallest slot index s k i in the set of slots I 138

		i	given by
	I 138 i	=
		kj ∈D 138

i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} \ {s i + w ki -1, ..., s j },

  |K|}, we select the smallest slot index s k i in the set of slots I 138• {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for eachk j ∈ R 138 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ if E 138 k i ∩ E 138 k ′ ̸ = ∅ (we take into account the possibility of adding slot s ′ as a last slot in the selected last slots S 138 k ′ to route demand k ′ in solution S 138 ). e) a set of slots U 138 are used in S such that for each demand k and last slot s k ∈ S 138 k and s ′ ∈ {s k -w k + 1, ..., s k }, we have s ′ ∈ U 138 . S 138 is clearly feasible for the problem. Hence, the corresponding incidence vector (u S 138 , z S 138 ) belongs to F H ′E I H . Then we derive a solution S 139 from S 138 as belows a) remove the last slot s totally covered by the interval I and which has been selected by a demand k i ∈ {v k 1 , ..., v kq } in solution S 139 (i.e., s ∈ S 139 k i and s′ ∈ {s i + w k i + 1, ..., s j }) such that each pair of nodes (v k ′ , v k j ) are not linked in odd-hole H with j ̸ = i, b) and select a new last slot s′ / ∈ {s i + w k i + 1, ..., s j } for demand k i i.e., S 139 k i = (S 138 k i \ {s}) ∪ {s ′ } such that {s ′ -w k i -1, ..., s′ } ∩ {s -w k + 1, ..., s} = ∅ for each k ∈ K and s ∈ S 139 k with E 139 k ∩ E 139 k i ̸ = ∅, c) and add slot s ′ to the set of last slots S 139 k ′ assigned to demand k ′ in solution S 139 , i.e., S 139 k ′ = S 138 k ′ ∪ {s ′ }. solution S 139 is clearly feasible for the SA problem. The corresponding incidence vector (u S 139 , z S 139 ) belongs to F H ′E I H . Hence, solutions S 138 and S 139 satisfy equation µu + σz = τ . We have so µu S 138 + σz S 138 = µu S 139 + σz S 139 = µu S 138 + σz S 138 + σ k ′ s ′ + σ k i s′ -σ k i s + Since σ k s = 0 for all demand k ∈ K and slot s ∈ {w k , ..., s} with s / ∈ {s i +w k +1, ..., s j }

			µ s"
			s"∈U 139 \U 138
			-	µ s" .
			s"∈U 138 \U 139
		i	given by
	I 138 i	= [
		kj ∈R 138

i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k ′ } ∪ {s ′ + w ki , ..., s}] if E(p ki ) ∩ E(p k ′ ) ̸ = ∅ or I 138 i = kj ∈R 138 i {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,

where

R 138 i = {k j ∈ {k 1 , ..., k i-1 } ∪ H : E 138 k i ∩ E 138 k j ̸ = ∅}. Hence,

  s if s ∈ {1, ..., w k ′ -1}, ρ if v k ′ ∈ H and s ∈ {s i + w k ′ -1, , ..., s j },

	0	otherwise,

  5.4.5. Consider a clique C in the conflict graph H ′E S with {s -w k + 1, ..., 1} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ for each (v k,s , v k ′ ,s ′ ) ∈ C.Then, inequality (2.43) is facet defining for P sa (G, K, S) if and only if 1. C is a maximal clique in the conflict graph H ′E S , 2. and there does not exist an interval of contiguous slots I = [s i , s j ] ⊂ [1, s] with

	• [ min

v k,s ∈C (s -w k + 1), max

  kj ∈D 145 i {w ki , ..., s kj -w kj } ∪{s kj +w ki , ..., s}]∩[{w ki , ..., s ′ -w k ′ } ∪{s ′ +w ki , ..., s}] if E(p ki ) ∩ E(p k ′ ) ̸ = ∅ or I 145 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not, where D 145 i= {k j ∈ {k 1 , ..., k i-1 } ∪ {k} : E(p k i ) ∩ E(p k j ) ̸ = ∅}. This ensures that • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 145 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ if E(p k i ) ∩ E(p k ′ ) ̸ = ∅ ( we takeinto account the possibility of adding slot s ′ as a last slot in the selected last slotsS 145 k ′ to route demand k ′ in solution S 145 ).We let S 145k i = {s k i }be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}. c) let U 145 be the slots used in S such that for each demand k" ∈ K and last slot s k" ∈ S 145 k" and s" ∈ {s k" -w k" + 1, ..., s k" }, we have s" ∈ U 145 . S 145 is feasible for the SA problem. Hence, the corresponding incidence vector (u S 145 , z S 145 ) belongs to F H ′E S C . After that, we derive a solution S 146 = (E 146 , S 146 ) from S 145 by adding slot s ′ as last slot to demand k ′ without modifying the last slots assigned to the demands K \ {k, k ′ } in S 145 , i.e., S 145 k" = S 146 k" for each demand k" ∈ K \ {k, k ′ }, and S 146 k ′ = S 145 k ′ ∪ {s ′ } for demand k ′ , and with modifying the last slots assigned to demand k by adding a new last slot s and removing the last slot s ∈ S 145 k with s ∈ {s i + w k + 1, ..., s j } and s ∈ {w k , ..., s} for demand k with v k,s / Solution S 146 is feasible for the SA problem. The corresponding incidence vector (u S 146 , z S 146 ) belongs to F H ′E S C . Hence, solutions S 145 and S 146 satisfy equation µu + σz = τ . We then obtain that µu S 145 + σz S 145 = µu S 146 + σz S 146 = µu S 145 + σz S 145 + σ k ′ s ′ -σ k s + σ k s +

		i	=
			kj ∈D 145 i
			∈ C
	such that S 146 k	= (S 145
			µ s"
			s"∈U 146 \U 145
			-	µ s" .
			s"∈U 145 \U 146

k \ {s}) ∪ {s}.

  s if s ∈ {1, ..., w k -1}, ∈ K and s ∈ S. As a result, we have (µ, σ) = ρ(α, β) + γM .5.4.5 Slot-Assignment-Odd-Hole InequalitiesProposition 5.4.6. Let H be an odd-hole in the conflict graph H ′E S with |H| ≥ 5, and {s -w k + 1, ..., 1} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ for each pair of nodes (v k,s , v k ′ ,s ′ ) linked in H.Then, inequality (2.44) is valid for P sa (G, K, S). Proof. We use the same proof of proposition (2.4.14). Theorem 5.4.6. Let H be an odd-hole in the conflict graph H ′E S with |H| ≥ 5, and {s -w k + 1, ..., 1} ∩ {s ′ -w k ′ + 1, ..., s ′ } ̸ = ∅ for each pair of nodes (v k,s , v k ′ ,s ′ ) linked in H. Then, inequality (2.44) is facet defining for P sa (G, K, S) if and only if a) for each node v k ′ ,s ′ / ∈ H in H ′E S , there exists a node v k,s ∈ H such that the induced graph H ′E S ((H \ {v k,s }) ∪ {v k ′ ,s ′ }) does not contain an odd-hole, b) and there does not exist a node v k ′ ,s ′ / ∈ H in H ′E S such that v k ′ ,s ′ is linked with all nodes v k,s ∈ H, c) and there does not exist an interval of contiguous slots I = [s i , s j ] ⊂ [1, s] with • and w k + w k ′ ≥ |I| + 1 for each (v k , v k ′ ) linked in H, • and 2w k ≥ |I| + 1 and w k ≤ |I| for each v k ∈ H. We distinguish the following cases: a) if for a node v k ′ ,s ′ / ∈ H in H ′E S , there exists a node v k,s ∈ H such that the induced graph H ′E S (H \{v k,s }∪{v k ′ ,s ′ }) contains an odd-hole H ′ = (H \{v k,s })∪{v k ′ ,s ′ }. This implies that inequality (2.44) can be dominated using some technics of lifting based on the following two inequalitiesv k,s ∈H z k s ≤ |H|-1 2 , and v k ′ ,s ′ ∈H ′ z k ′ s ′ ≤ |H ′ |-1 2 . b) if there exists a node v k ′ ,s ′ / ∈ H in H ′E S such that v k ′ ,s ′ islinked with all nodes v k,s ∈ H. This implies that inequality (2.44) can be dominated by the following ) if there exists an interval of contiguous slots I = [s i , s j ] ⊂ [1, s] satisfying the conditions of c). Hence, inequality (2.44) is dominated by inequality (2.40).If no one of these cases is verified, inequality (2.44) can never be dominated by another inequality without changing its right-hand side. Otherwise, inequality (2.44) cannot be facet defining for P sa (G, K, S).

	valid inequality
	v k,s ∈H	ρ z k s +	if v k,s ∈ C, |H| -1 2 z k ′ s ′ ≤	|H| -1 2	.
			0	otherwise,
	v k,s ∈H Sufficiency. (s -w k + 1), max v k,s ∈H for each k • [ min Let F H ′E	] ⊂ I,
	Proof. Neccessity.

cS H denote the face induced by inequality (2.40), that is F H ′E

  for all k ′ ∈ K \ {k} and s ∈ {w k ′ , ..., s} with v k,s ′ / ∈ H.Let prove that σ ks for all v k,s ∈ H are equivalent. Consider a node v k ′ ,s ′ in H. Let S 152 = (U 152 , S 152 ) be the solution given by a) select a subset of nodes H152 fromH with | H152 | = |H|-1 2 , and each pair of nodes (v k,s , v k ′ ,s ′ ) ∈ H152 are not linked in the conflict graph H′E S , b) for each pair of demand k and slot s with v k,s ∈ H152 , we select slot s k = s as last slot for demand k, c) for each demand k i ∈ K \ H152 with i ∈ {1, ..., |K|}, we select the smallest slot index s k i in the set of slots I 152 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s}] ∩ [{w ki , ..., s ′ -w k } ∪ {s ′ + w ki , ..., s}] if E(p ki ) ∩ E(p k ) ̸ = ∅ or I 152 {w ki , ..., s kj -w kj } ∪ {s kj + w ki , ..., s} if not,whereD 152 i = {k j ∈ {k 1 , ..., k i-1 } ∪ H152 : E(p k i ) ∩ E(p k j ) ̸ = ∅}. As a result, • {s k i -w k i + 1, ..., s k i } ∩ {s k j -w k j + 1, ..., s k j } = ∅ for each k j ∈ D 152 i , • and {s k i -w k i + 1, ..., s k i } ∩ {s ′ -w k ′ + 1, ..., s ′ } = ∅ if E(p k i ) ∩ E(p k ′ ) ̸ = ∅ ( we takeinto account the possibility of adding slot s ′ as a last slot in the selected last slotsS 152 k ′ to route demand k ′ in solution S 152 ).We let S 152 k i = {s k i } be the set of last slots assigned to demand k i with i ∈ {1, ..., |K|}.d) let U 152 be the set of slots used in S such that for each demand k" ∈ K and last slot s k" ∈ S 152 k" and s" ∈ {s k" -w k" + 1, ..., s k" }, we have s" ∈ U 152 .S 152 is clearly feasible for the problem. Hence, the corresponding incidence vector (u S 152 , z S 152 ) belongs to F H′E

		i	given by
	I 152 i	= [
		kj ∈D 152 i
			i	=
			kj ∈D 152 i

  Solution S 153 is feasible for the SA problem. The corresponding incidence vector (u S 153 , z S 153 ) belongs to F H ′E S H . Hence, solutions S 152 and S 153 satisfy equation µu + σz = τ . We then obtain that µu S 152 + σz S 152 = µu S 153 + σz S 153 = µu S 152 + σz S 152 + σ k ′ s ′ -σ k s + σ k

			s
	+	µ s" -	µ s" .
	s"∈U 153 \U 152	s"∈U 152 \U 153

Table 5 .

 5 1: Table of Comparison Between: B&B SCIP Vs Own B&C SCIP Using Real Graphs.

	Instances			B&B SCIP		Own B&C SCIP
	Topology |K| |S|	Nbr Nd Gap	TT	Nbr Nd Gap Nbr Cuts	TT
		10	40	1	0,00	0,02	1	0,00	0	0,03
		20	40	1	0,00	0,53	1	0,00	0	0,1
		30	40	1	0,00	3,74	1	0,00	5,5	0,57
		40	40	4	0,00	1,32	3	0,00	12,5	5,84
	Pioro40	50 100	80 80	5 3	0,00 0,00	2,66 44,31	1,25 18,5	0,00 0,00	15,75 77,5	13,52 2769,13
		150 160	56	1,95	9335,82	57	0,00	48,75	9169,93
		200 280	1	0,14	4934,59	1,25	0,00	28,5	3023,14
		250 280	1	0,00 3782,08	1	0,00	73,5	2580
		300 320	4,25	0,18 10548,18	3,25	0,36	96	13502,49
		10	80	1	0,00	0,04	1	0,00	0	0,06
		20	40	1	0,00	0,08	1	0,00	0	0,14
		30	40	2	0,00	3,52	1,25	0,00	12	6,11
		40	80	4,5	0,00	4,43	1	0,00	0	3,82
	India35	50 100 240 160	1 13,5	0,00 1,55 13278,76 7,64	9,25 10,5	0,00 0,20	7 64,50	67,06 10572,62
		150 400	8	4,71	18000	15	5,18	89	18000
		200 280	1	10,58 13577,39	1,25	4,11	0,75	8531,99
		250 280	1	1,45	18000	1	0,72	61	18000
		300 320	1	1,8	16858,2	3	1,97	62,25	18000
		10	40	1	0,00	0,08	1	0,00	0,50	0,17
		20	40	1	0,00	0,04	1	0,00	0	0,09
		30	40	1	0,00	0,36	1	0,00	0	0,47
		40	80	6,75	0,00	11,91	5,50	0,00	26	18,12
	Brain161	50 100 160 120	9 65	0,00 0,00 3297,48 25,23	3 6	0,00 0,00	16,25 35	25,17 1009,43
		150 320	58,5	0,26 10284,04	43,25	0,27	148,25	12232,16
		200 400	8	0,40 12172,23	1,67	0,36	45,67	18000
		250 480	1	0,86 13492,92	1,67	0,33	52	18000
		300 320	1	1,30	18000	1	0,32	11,50	18000

Table 5 .

 5 2: Table of Comparison Between: B&B SCIP Vs Own B&C SCIP Using

	Realistic Graphs.

à Orange Labs, pour avoir bien voulu examiner mes travaux de thèse et accepter de participer au jury. Je la remercie aussi pour les commentaires intéressants qu'elle a formulés et sa clairvoyance sur le sujet. Je voudrais témoigner toute ma reconnaissance à Monsieur Mourad Baïou, directeur de laboratoire LIMOS, et Monsieur Farouk Toumani, ancien directeur de LIMOS et actuellement directeur de l'ISIMA, pour leur confiance, leurs disponibilités, leur écoute et leur accompagnement dont j'ai pu bénéficier pendant l'élaboration de ma thèse. Ils ont été une grande ressource pour le développement personnel et professionnel durant cette thèse. Bien sûr, atteindre ces objectifs n'aurait pas été possible sans le soutien de l'UCA et le LIMOS, qui m'ont permis, grâce à une allocation de recherches de l'ANR et diverses aides financières durant la crise sanitaire, de me consacrer sereinement à l'élaboration de cette thèse. Il est important donc de remercier tout le personnel de l'UCA et le LIMOS pour m'avoir accueilli chaleureusement. Ils ont tout mis en oeuvre pour que ma thèse se déroule dans les meilleures conditions possibles. Merci pour toutes ces années de guidance.Je souhaite remercier spécialement Madame Béatrice Bourdieu, responsable admin-

This work was supported by the French National Research Agency grant ANR-17-CE25-0006, FlexOptim Project.

for all k ′ ∈ K and s ′ ∈ {1, ..., w k ′ -1}.

return constrained minimum-cost path p * k for demand k;

Remerciements

0 and a label p ∈ L i \ T i having the smallest value of e∈E(p) c e ;

for each e = ij ∈ δ(i) \ E k 0 such that e ′ ∈E(p) ℓ e ′ + ℓ e ≤ lk do if j / ∈ V (p) then

Concluding Remarks

In this chapter, we have focused on a complex variant of the Routing and Spectrum Assignment (RSA) problem, called the Constrained-Routing and Spectrum Assignment (C-RSA). We first have proposed a new integer linear programming formulation based on the so-called cut formulation for the C-RSA. We have investigated the facial structure of the associated polytope by showing that some basic inequalities are facet-defining under certain conditions. We have further identified several families of valid inequalities to obtain tighter LP bounds. Moreover, we have studied the facial structure of these valid inequalities, and have shown that they are facet defining for the polytope under certain necessary and sufficient conditions.

We have also introduced some symmetry-breaking inequalities to well manage the so-called equivalents sub-problems.

Problem

In this chapter, we first introduce an extended integer linear programming formulation based on the so-called path formulation. All the different valid inequalities presented in chapter 2, they are still valid for the path formulation. Using this, we derive a Branch-and-Cut-and-Price algorithm to solve the C-RSA problem. In this section, we describe the framework of this algorithm. First, we give an overview of the column generation algorithm. Then, we discuss the pricing problem. We further present a primal heuristic used to boost the performance of the algorithm.

We give at the end some computational results and a comparative study between Branch-and-Cut and Branch-and-Cut-and-Price algorithms. We close our chapter with some concluding remarks.

Path Formulation

Let P k denote the set of all feasible (o k ,d k ) paths in G such that for each demand k ∈ K, we have

ℓ e ≤ lk , for all p k ∈ P k .

We consider for k ∈ K and p ∈ P k and s ∈ S, a variable y k p,s which takes 1 if slot s is the last slot allocated along the path p for the routing of demand k and 0 if Result: Optimal path p * for demand k and slot s

Update the set of label of CPU time. Furthermore, they enable reducing the average number of nodes in the B&C&P tree, and also the average CPU time for several instances. We also notice that several instances have been solved to optimality in the root of the B&C&P tree (i.e., Nb Nd=1) that necessitates a large number of branching nodes when using the B&P algorithm. On the other hand, and when the optimality is not proven, adding valid inequalities decreases the gap for several instances. However, there exist a few instances very rare in which adding valid inequalities have no any impact. Moreover, some instances are still difficult to solve with both the B&P and B&C&P algorithms. Branch-and-Cut-and-Price using SCIP. We denote by B&C CPX the Branch-and-Cut algorithm when using Cplex with benefiting of its automatic cut generation and without our additional valid inequalities, and by Own B&C CPX when using Cplex with our additional valid inequalities and disabling its proper cut generation. On the other hand, in the second series of computational results are shown in Table 4.3, we present the results found for the Branch-and-Cut algorithm using Gurobi (without or with additional valid inequalities) compared with those of Branch-and-Price and Branch-and-Cut-and-Price using SCIP. We denote by B&C GRB when using Gurobi with benefiting of its automatic cut generation and without our additional

Associated Polytope

Let P sa (G, K, S) be the polytope, convex hull of the solutions for the formulation (5.1)-(5.8). Here we study the facial structure of the polytope P sa (G, K, S).

A solution of the SA problem is based on the variables (u, z) is given by two sets S k for each demand k ∈ K and U for the spectrum-usage of S where a) S k denotes the set of index of the last slots selected for demand k such that

b) U denotes the set of slots allocated over the spectrum S such that for each demand k ∈ K and last slot s ∈ S k ⇒ each slot s ′ ∈ {s-w k +1, ..., s} should be in U i.e. s ′ ∈ U.

We suppose that the number of slots s is largely sufficient to route all the demands, and to avoid the existence of some slots s ∈ S such that u s = 1 in any feasible solution S of the SA problem. This means that there does not exist a slot s ∈ S such that u s = 1.

Dimension

Let M denote the matrix associated with the equations (5.2). We ensure that the matrix M is of full rank given that the demands are independants, and the slots in S are independents for each demand k ∈ K. As a result, rank(M ) = k∈K (w k -1).

Let us denote by r ′ the rank of the matrix M .

Proposition 5.3.1. The equation system (5.2) defines a minimal equation system for P sa (G, K, S).

Proof. To prove that σz + µu = λ is a linear combination of equations (5.2), it's sufficient to prove that for each demand k ∈ K, there exists for each demand k ∈ K a γ k ∈ R w k -1 such that (µ, σ) = γM . Let u S and z S denote the incidence vector of a solution S of the SA problem.

Let first show that µ s = 0 for all s ∈ S. Consider a slot s ∈ S, and solution S 105 = (U 105 , S 105 ) given by Proposition 5.5.3. We ensure that for all slot s ∈ {1, ..., s -1} k∈K min(s+w k -1,s) (5.16) which means that the number of intervals of contiguous slots allocated which cover slot s + 1 (cardinality of slot-usage) cannot be greater than the number of channels allocated which cover slot s.

Similar idea was proposed by Mendez-Diaz et al. [START_REF] Méndez-Díaz | A Polyhedral Approach for Graph Coloring 1[END_REF][70] to break the symmetry for the vertex coloring problem. Our inequalities and those of Mendez-Diaz et al.

[69] [START_REF] Méndez-Díaz | A Branch-and-Cut algorithm for graph coloring[END_REF] differ in their right and left hand sides.

Proposition 5.5.4. Due to inequality (5.14), we ensure that for all k ∈ K, and s 0 ∈ {1, ..., s -1} and s ∈ {s 0 , ..., s}

which means that for a slot S 0 ∈ {1, ..., s -1}, a demand k can allocate a slot in the sub-spectrum {S 0 , ..., s} if slot S 0 is used.

Similar idea was proposed by Mendez-Diaz et al. [START_REF] Méndez-Díaz | A Branch-and-Cut algorithm for graph coloring[END_REF] for the vertex coloring problem. Inequalities (5.17)} and those of Mendez-Diaz et al. [START_REF] Méndez-Díaz | A Branch-and-Cut algorithm for graph coloring[END_REF] differ in their left hand sides.

Lower Bounds

Here we propose some lower bounds issus from the conflict graph H sa . They can be seen as a valid inequalities for the polytope P sa (G, K, S). is valid for P sa (G, K, S).

Proof. Inequality (5.18) ensures that the number of slots used in the spectrum S is greater than the flow over all the edges (the flow for an edge e is equal to the number of slots that should be used over edge e).

Inequality (5.18) can be generalized as follows using the conflict graph H sa .

Proposition 5.6.2. Let C be a clique in H sa . Then, the inequality

is valid for P sa (G, K, S).

Proof. It's trivial given the definition of clique C in the conflict graph H sa such that we know in advance that the demands in C share an edge in E which means that they cannot share a slot in S. Hence, the number of allocated slots s∈S u s is at least equal to the number of requested slots of the demands in C.

Upper Bounds

Let us introduce the following weighted conflict graph in which a positive integer called weight is assigned to each node.

Definition 5.7.1. Consider the conflict graph H r w defined as follows. For each demand k ∈ K, consider a node v k in H r w . Two nodes v k and v k ′ are linked by an edge in H r w if and only if

Each node v k is associated with a positive weight which equals to the requested number of slots w k of demand k. Definition 5.7.2. Let C be a clique in H r w . It's known to be the maximum weight clique in H r w if the total weight of the nodes in C ( v k ∈C w k ) defines the maximum total weight over all cliques in H r w , i.e.,

Based on these definitions, we introduce the following inequality and showing that computing the upper bound for the SA is equivalent to solving the Maximum Weighted Clique Problem (MWC) which is well known to be NP-hard problem [START_REF] Balas | On the Maximum Weight Clique Problem[END_REF].

Proposition 5.7.1. Let C be the maximum weighted clique in H r w . Then, the upper bound is defined as follows

Proof. It's trivial given the definition of the maximum weighted clique C in the conflict graph H r w such that the maximum number of allocated slots s∈S u s is at most equal to the number of requested slots of the demands in C. end return the best optimal solution (u * , z * ) for the SA;

Based on this sequence of demands, our greedy algorithm selects a slot s for each demand k ′ ∈ L with zk ′ s ̸ = 0, while respecting the non-overlapping constraint with the set of demands that precede demand k ′ in the list L (i.e., the demands 1 ′ , 2, ..., k ′ -1). However, if there does not exist such slot s for demand k ′ , we then select a slot s for demand k ′ ∈ L with zk ′ s = 0 with s ∈ {w k ′ , ..., s} while respecting the non-overlapping constraint with the set of demands that precede demand k ′ in the list L. The complexity of this algorithm can be bounded by O(|K| * |S| * log(|K|)).

Afterwards, we compute the total number of slots in S used by the set of demands K in the final solution S given by the greedy-algorithm (i.e., s∈S u s ). Our local search algorithm generates a new sequence by doing some permutation of demands in the last sequence of demands, if the value of the solution given by greedy-algorithm is smaller than the value of the best solution found until the current iteration. Otherwise, we stop the algorithm, and we give in output the best solution found during our primal heuristic induced by the best sequence of demands having the smallest value of the total number of slots in S used compared with the others generated sequences.

Computational Study

Implementation's Feature

We use C++ to implement the B&B and B&C algorithms under Linux using the "Solving Constraint Integer Programs" framework (Scip 6.0.2) such that Cplex 12.9 is used as LP solver. These have also been tested on LIMOS high-performance server with a memory size limited to 64 Gb while benefiting from parallelism by activating 8 threads, and with a CPU time limited to 5 hours (18000 s). We use the same graphs presented in Table 3.1, and the same instances used in the section 3.2.2.

Computational Results

Preliminary results show that introducing some families of valid inequalities allows solving several instances to optimality. Moreover, they enable reducing the average number of nodes in the B&C tree, and also the average CPU time for several instances. On the other hand, the results show that the odd-hole inequalities

Conclusion

In this thesis, we have studied the Constrained-Routing and Spectrum Assignment (C-RSA) problem related to the dimensioning and designing of Spectrally Flexible Optical Networks (SFONs). It's well known to be NP-hard. The main aim of this thesis was to provide a deep polyhedral investigation and design a cutting plane method for the problem and handle large-scale instances.

First, we have proposed an integer linear programming formulation namely cut formulation. We have investigated the related polytope defined by the convex hull of all its solutions. Moreover, we have identified several classes of valid inequalities for the polytope and studied their facial structure. We further have discussed their separation problems. We have also proposed a primal heuristic to obtain tighter primal bounds and enhance the resolution of the problem. These results are used to devise a Branch-and-Cut (B&C) algorithm for the C-RSA problem, along with some computational results are presented using two types of instances: random and In the second part of thesis, we have discussed an extended formulation based on the so-called path formulation. It can be seen as a reformulation of the cut formulation using the so-called path variables. We have developed a column generation algorithm to solve its linear relaxation. We have shown that the pricing problem is equivalent to the resource-constrained shortest path problem, which is well known to be NPhard. For this, we have developed a pseudo-polynomial algorithm based on dynamic programming enabled solving the pricing problem in polynomial time. Using this, we have devised Branch-and-Price and Branch-and-Cut-and-Price algorithms. The results show that the Branch-and-Cut-and-Price performs very well compared with the Branch-and-Price. Hence, the significant impact and the power of the introduced