M Chazal
email: chazal@ensae.fr

Ceremade

E Jouini Ceremade

R Tahraoui Ceremade

Production Planning and Inventories Optimization : A Backward Approach in the Convex Storage Cost Case

Keywords: Production planning, inventory management, integro-differential backward equations

, we study the deterministic optimization problem of a profit-maximizing firm which plans its sales/production schedule. The firm knows the revenue associated to a given level of sales, as well as its production and storage costs. The revenue and the production cost are assumed to be respectively concave and convex. Here, we also assume that the storage cost is convex. This allows us to relate the optimal planning problem to the study of an integro-differential backward equation, from which we obtain an explicit construction of the optimal plan.

Introduction

We consider a firm which produces and sells a good which can be stored. The firm acts in continuous time on a finite period in order to maximize dynamically its profit. Here, the instantaneous profit of the firm is the revenue entailed by the instantaneous sales, diminished by the cost of the instantaneous production, and by the cost of storage of the current inventories. Our approach of this production planning and inventory management problem is in the same vein as the one launched in 1958 by K. J. Harrow, S. Karlin and H. Scarf [START_REF] Arrow | Studies in the Mathematical Theory of Inventory and Production[END_REF]. Many contributions to this theory have been brought from the 50's until now, with many different approaches. However, authors generally consider firms that do not have any control on the level of the (possibly stochastic) demand driven sales. AJOUTER DES REF, Rochet, Van der Heyden....

In this paper, we work in a competitive and deterministic context. Following, L. Arvan and L. N. Moses [START_REF] Arvan | Inventory Investment and the Theory of the Firm[END_REF], we assume that the firm controls not only its production rate but also its sales rate. It knows the revenue associated to the selling of x units of goods, the cost of producing y units, and the cost of storing S units of the good. We assume that the marginal revenue is nonincreasing and that the marginal cost of production is nondecreasing, i.e. the revenue function is concave and the production cost function is convex.

The sales/production planning problem of the profit-maximizing firm is formulated as an optimal control problem where the controls, namely the sales and production paths are integrable. In other words, the cumulative production and sales processes are assume to be absolutely continuous.

This problem has been studied in [START_REF] Chazal | Production Planning and Inventories Optimization With a General Storage Cost Function[END_REF] with a general storage cost function. The authors proved an existence result for a relaxed problem in which the cumulative sales process is allowed to have a jump at time 0. This means that if the firm is allowed to make a partial depletion of its eventual initial inventory at time 0. They also derived the first order conditions of optimality for both problems and thus provided a qualitative description of the optimal plans. In particular, they establish the following result : the optimal inventory level must decrease until it reaches 0. It is then kept null by producing for immediate selling.

In the present article, we go further in this analysis. For this purpose, we assume that the storage cost is convex. In this context, we first see that the initial problem has a solution if and only if the relaxed one has a solution without jumps at time 0. Second, the first order conditions of [START_REF] Chazal | Production Planning and Inventories Optimization With a General Storage Cost Function[END_REF] allows us to characterize the (unique) optimal plan for the relaxed problem. Third, using this characterization, we relate the relaxed optimization problem to the study of a class of solutions of an integro-differential backward equation, indexed by a class of admissible terminal conditions. We establish existence and uniqueness of a maximal solution of this integro-differential backward equation, for every terminal condition. We then study the behavior of the solution with respect to the terminal condition.

This allows us to provide a constructive description of the optimal plan. The optimal plan is determined by selecting the greatest terminal condition r such that a certain functional of the solution of the integro-differential backward equation, representing the inventory level at time 0+, S 0 (r), remains lower than the exogenous initial inventory s 0 . The difference α = s 0 -S 0 (r) corresponds to the size of the jump of the cumulative sales process at time 0. After 0, the sales and production rates appear to be functions of the corresponding maximal solution.

We also prove that there exists an exogenous threshold s0 on the initial inventory above which the size of the jump, α, is positive. The economic interpretation is the following. If the initial inventory s 0 entails too high storage costs, i.e. s 0 is greater than s0 , then, it is optimal for the firm to sell out immediately the quantity s 0 -s0 , so as to reduce its initial inventory to s0 . If the initial inventory in lower than s0 , then the firm has no interest in selling out some stock immediately. The level s0 appears as the maximal level that it can afford to hold.

The paper is organized as follows. Section 1 provides a precise description of the model and recalls some basic results of [START_REF] Chazal | Production Planning and Inventories Optimization With a General Storage Cost Function[END_REF]. Section 2 is devoted to the backward characterization of the optimal plan for the relaxed problem. The constructive resolution of the production planning problem is given in Section 3.

The Model Formulation

The firm acts in continuous time on a finite period [0, T ]. It is endowed with an initial inventory of s 0 ∈ R + units of the good.

A sales/production plan is represented by a couple (x, y) of functions in L 1 + [0, T ], the set of nonnegative elements of L 1 [0, T ], where x(t) (resp. y(t)) is the sales (resp. production) rate in units of the good at time t. In other words t 0 x(u)du (resp. t 0 y(u)du) is the cumulative quantity of the good sold out (resp. produced) up to time t. We shall say that (x, y) ∈ L 1 + [0, T ] × L 1 + [0, T ] is a sales/production plan if the induced inventory S (x,y) satisfies

S (x,y) (t) s 0 + t 0 y(u)du - t 0 x(u)du ≥ 0, ∀t ∈ [0, T ].
This means that the company must never be out of stock. We denote by A the set of all sales/production plans :

A {(x, y) ∈ L 1 + [0, T ] × L 1 + [0, T ] | S (x,y) (t) ≥ 0, ∀t ∈ [0, T ]}.
When selling out at the rate x(t) at time t, the firm has a revenue rate of π(x(t)). The cost of producing at the rate y(t) at time t is c(y(t)). Both π and c are continuous, nondecreasing functions on R + . They satisfy π(0) = 0, c(0) = 0 and π(x) > 0, c(x) > 0, for all positive x. The function π (resp. c) is assumed to be concave (resp. convex).

The cost of storing an amount S(t) of goods at time t is denoted by s(S(t)). The function s is assumed to be continuous, nondecreasing on R + and to satisfy s(0) = 0 and s(S) > 0, for all positive S.

Given the discount rate λ > 0, the profit over time induced by (x, y) ∈ A is defined by

J(x, y) T 0 e -λt [π(x(t)) -c(y(t)) -s(S (x,y) (t))]dt.
Observe that by concavity of π and Jensen's inequality

T 0 e -λt π(x(t))dt < ∞ .
The functions c and s being nonnegative, it follows that J is well defined as a map from A into R ∪ {-∞}.

The profit-maximizing company plans its sales/production schedule by solving the following optimization problem sup

(x,y)∈A J(x, y). (1)
It turns out that the function J may fail to have a maximum on A. This is typically the case when s 0 is to high (see section 3). Nevertheless, it was proved in [START_REF] Chazal | Production Planning and Inventories Optimization With a General Storage Cost Function[END_REF] that existence holds for some relaxed problem that we now describe.

The sales rate is no longer described by an integrable function, but by a nonnegative finite Borel measure on [0, T ] which has its singular part positively proportional to the Dirac measure at 0. In this framework, for a sales path equal to αδ 0 + x, where x ∈ L 1 + [0, T ] represents the absolutely continuous part of the considered Borel measure, the cumulative sales process is given by

X(0) = 0 and X(t) = α + t 0 x(u)du , ∀t ∈ (0, T ] .
This means that the firm is allowed to sell out, at time 0, a share α of its initial inventory.

The production path is still assumed to be integrable. A sales/production plan is now a triplet (α, x, y) in

R + × L 1 + [0, T ] × L 1 + [0, T ] which satisfies the inventory constraint S (α,x,y) (t) = s 0 + t 0 y(u)du -α - t 0 x(u)du ≥ 0, ∀t ∈ (0, T ].
For t = 0, we set S (α,x,y) (0) = s 0 . The inventory level S (α,x,y) can have a downward jump (α ≥ 0) at 0+.

We denote by B the set of relaxed sales/production plans :

B {(α, x, y) ∈ R + × L 1 + [0, T ] × L 1 + [0, T ] | S (α,x,y) (t) ≥ 0, ∀t ∈ (0, T ]}.
The relaxed profit is defined on B by

F(α, x, y) = α π(∞) + T 0 e -λt [π(x(t)) -c(y(t)) -s(S (α,x,y) (t))]dt
where, we have set π(∞) lim

x→∞ π(x)/x
which is well defined in R + , by concavity and nonnegativity of π. This is the price at which the firm can sell at an infinite rate. It is also the lowest price accessible for the company. However, since holding inventories has a cost, the firm may take advantage of an immediate depletion, even at this price. Observe that, when π is differentiable, we have

lim x→∞ π(x) x = lim x→∞ π(x).
The relaxed optimization problem is sup

(α,x,y)∈B F(α, x, y). (2)
It was proved in [START_REF] Chazal | Production Planning and Inventories Optimization With a General Storage Cost Function[END_REF] that

Theorem 1 F has a maximum in B and max

(α,x,y)∈B F(α, x, y) = sup (x,y)∈A J(x, y).
Notice that in the context of [START_REF] Chazal | Production Planning and Inventories Optimization With a General Storage Cost Function[END_REF], where the storage cost is not assumed to be convex, uniqueness is not guaranteed in Problem [START_REF] Arvan | Inventory Investment and the Theory of the Firm[END_REF]. In this article, we shall go further in the analysis of the set of solutions of both problems. For this purpose, we require some regularity conditions on π, c and s and we assume that s is convex. In particular, this will allow us to assert that Problem (2) has a unique solution and thus that Problem 1 has a (unique) solution if and only if the solution of Problem (2) is in {0} × A.

To be more precise, we shall work under the following conditions.

Standing assumption (H).

(i) Argmax (π -c) = {a} for some a > 0.

(ii) π is differentiable on (0, ∞) and π is continuous and one to one on [a, ∞). (iii) c is differentiable on R + and ċ is continuous and one to one on [0, a].

(iv) s is continuously differentiable on R + . (v) s is convex.
Assumptions (i) to (iv) were already required in [START_REF] Chazal | Production Planning and Inventories Optimization With a General Storage Cost Function[END_REF] in order to obtain a precise characterization of the set of plans which satisfy the first order conditions of optimality in Problem [START_REF] Arvan | Inventory Investment and the Theory of the Firm[END_REF]. The existence result has been obtained under a weaker assumption : the functions π, c and s are continuous, and π -c admits a, possibly not unique, maximum.

Theorem 2 Problem (2) has a unique solution (α, x, y). Problem (1) has a solution if and only if α = 0. If α = 0 then, (x, y) is the unique solution of (1).

Proof. Existence for Problem (2) is stated in Theorem 1 and was proved in [START_REF] Chazal | Production Planning and Inventories Optimization With a General Storage Cost Function[END_REF]. Moreover, Theorem 7 in [START_REF] Chazal | Production Planning and Inventories Optimization With a General Storage Cost Function[END_REF] states that, if (α, x, y) is a solution then x ≥ a ≥ y a.e. Since π and c are respectively strictly concave on [a, ∞) and strictly convex on [0, a], and since s is convex, uniqueness in Problem (2) is straightforward. By noticing that (0, x, y) ∈ B ⇔ (x, y) ∈ A and F(0, •) |A ≡ J and recalling that by Theorem 1 max

(α,x,y)∈B F(α, x, y) = sup (x,y)∈A J(x, y) the proof of Theorem 2 is easy to complete.
In light of Theorem 2, we see that planning the optimal sales/production schedule can be done by solving the relaxed problem. Indeed, if the initial problem (1) has a solution (x, y) then, (α = 0, x, y) is the solution of the relaxed problem [START_REF] Arvan | Inventory Investment and the Theory of the Firm[END_REF]. Since, we further exhibit situations where the solution of Problem (2) is not regular i.e. α > 0 (see section 3), meaning that Problem (1) has no solution, we must in actual fact consider Problem (2) to solve the planning problem. We shall see that for some revenue and costs functions π, c and s and some initial inventory s 0 , the firm can not act in an optimal way without getting rid at time 0 of a certain share of s 0 . From now on, we focus our attention on the relaxed problem and what we call the optimal plan is the unique solution (α, x, y) of Problem (2).

2 Characterization of the Optimal Plan.

In this section we begin with recalling some characterization and thus some qualitative description of the optimal plan derived from the first order conditions obtained in [START_REF] Chazal | Production Planning and Inventories Optimization With a General Storage Cost Function[END_REF]. We then introduce some more tractable formulation of these optimality conditions in order to reach our main goal here : to provide a constructive resolution of the planning problem.

The first order conditions

It was shown in [START_REF] Chazal | Production Planning and Inventories Optimization With a General Storage Cost Function[END_REF] that the optimal plan must be such that there is no inventory accumulation. In particular, if the firm has no starting inventories (s 0 = 0) then it adopts a static strategy consisting in producing for immediate sales. Moreover, the concavity of the revenue and convexity of the cost of production urges the firm to minimize the variations of its sales and production rates, so that it must produce and sell at the same constant rate a which maximize π -c. To sum up, when s 0 = 0, the optimal plan is (0, a, a). Let us now turn to the case where s 0 is positive.

Standing assumption. The initial inventory s 0 is positive.

As a direct consequence of Theorem 7 in [START_REF] Chazal | Production Planning and Inventories Optimization With a General Storage Cost Function[END_REF], we have the following characterization of the solution of Problem (2). Proposition 3 (α, x, y) ∈ B is the solution of (2) if and only if it satisfies :

(i) x (resp. y) is nonincreasing (resp. nondecreasing) and x ≥ a ≥ y on [0, T ],

(ii) S (α,x,y) (T ) = 0 and T 0 inf{t ∈ (0, T ] | S (α,x,y) (t) = 0} > 0,

(iii) if T 0 < T then, a) S (α,x,y) = 0 on [T 0 , T ], b) x = y = a on (T 0 , T ],
(iv) S (α,x,y) is decreasing on (0, T 0 ], (v) a) x and y are both continuous on (0, T 0 ) and satisfy b)

e -λt π(x(t)) = π(x(0+)) + t 0 e -λu ṡ S (α,x,y) (u) du, ∀t ∈ (0, T 0 ) (3) c) y(t) = g(x(t)), ∀t ∈ (0, T 0 ) (4)
where g(r)

ċ-1 ( π(r)) if π(r) > ċ(0) 0 elsewhere , ∀ r ∈ [a, ∞). (5)
(vi) If T 0 < T then, x(T 0 -) = a.

(vii) If α > 0 then, x(0+) = ∞.

Proof. By Theorem 7 in [START_REF] Chazal | Production Planning and Inventories Optimization With a General Storage Cost Function[END_REF], items (i) to (vii) are equivalent to :

lim sup ε>0 ε→0 F((α, x, y) + ε(β, h, k)) -F(α, x, y) ε ≤ 0, for all (β, h, k) ∈ R × L 1 [0, T ] × L 1 [0, T ] such that (α, x, y) + ε 0 (β, h, k) ∈ B for some ε 0 > 0.
Then, Proposition 3 holds by concavity of F and uniqueness in Problem (2).

Proposition 3 furnishes some qualitative properties of the optimal plan which were obtained in [START_REF] Chazal | Production Planning and Inventories Optimization With a General Storage Cost Function[END_REF] without assuming that the cost of storage is convex. The main property is that the inventory level is nonincreasing and must be null at the end of the period. This phenomenon arises directly from the concavity of the revenu and the convexity of the cost of production and the fact that holding inventories has a cost.

EXPLIQUER ???

Let us now put in light the qualitative description provided by Proposition 3. The optimal way to deplete the initial inventory is in two phases. This leads to a three phases sales/production plan. The first phase is devoted to the selling activity. The firm begins with possibly depleting a share α ≥ 0 of its initial inventory at time 0, selling at an infinite rate. The sales rate is then nonincreasing or equivalently the marginal revenue t -→ π(x(t)) is nondecreasing. If the marginal revenue overtakes the lowest marginal cost of production ċ(0) then, the production activity actually starts (see (4) and ( 5)). During this second phase, the sales rate and the production rate are such that the marginal revenue and the marginal production cost remain equal : ċ(y(t)) = π(x(t)). The production rate is nondecreasing. During this destocking stage the sales rate is greater than a and the production rate lower. If inventories are all cleared before the end of the period (T 0 < T ) then, the third phase starts : production and sales are at the same constant rate, a, maximizing the instantaneous profit.

Remark that by (ii), even if the firm is allowed to get rid of its whole initial inventory at time 0, it does not. This is mainly due to the fact that α is sold out time 0 at an infinite rate and hence at the lowest price π(∞).

The Backward Characterization

We now turn to our first step on the way to the explicit determination of the optimal plan (α, x, y), that is providing a backward procedure to find (α, x, y). We shall extract from Proposition 3 some sufficient conditions of optimality which are expressed by mean of some integro-differential backward equation that must be satisfied by the sales rate x. We first obtain these conditions by arguing in the necessary way. We then check that they are sufficient.

In the sequel we will use the following properties of the function g defined in (5).

Remark 4 Under assumption (H) and by construction the function g is continuous, nonincreasing on [a, ∞). Moreover, it satisfies g(a) = ċ-1 ( π(a)) = a and takes values in [0, a].

First of all, remark that by Proposition 3 the optimal production rate y is an exogenous function of the optimal sales rate x. More precisely, if T 0 = T then, by (4) we have y(t) = g(x(t)) for any t ∈ (0, T ). If T 0 < T then, by (4) again we have y(t) = g(x(t)) for any t ∈ (0, T 0 ) and by (iiib) and since a = g(a), we also have y(t) = a = g(a) = g(x(t)) for any t ∈ (T 0 , T ]. In both cases we have :

y = g(x) a.e. on [0, T ]. (6)
In light of this result, we now seek for some conditions on some sales policy (α, x) for the plan ( α, x, g(x)) to be the optimal one.

Observe that, by (iiia) we have S (α,x,y) (T 0 ) = 0 and therefore, for any t ∈ [0, T ],

S (α,x,y) (t) = S (α,x,g(x)) (t) = S (α,x,g(x)) (t) -S (α,x,g(x)) (T 0 ) = s 0 -α + t 0 [g(x(u)) -x(u)]du -s 0 -α + T 0 0 [g(x(u)) -x(u)]du = T 0 t [x(u) -g(x(u))]du .
Using this backward formulation of the inventory level, it is easy to see that (α, x) satisfies the backward system deduced from (3) and from the equality

S (α,x,g(x)) (T 0 ) = 0 e -λt π(x(t)) = e -λT 0 π(x(T 0 -)) - T 0 t e -λu ṡ T 0 u [x(s) -g(x(s))]ds du, ∀t ∈ (0, T 0 ) α = s 0 - T 0 0 [x(u) -g(x(u))]du
Now, by using appropriated changes of variable, one can check that, in term of the translation of x |(0,T 0 ) on (T -T 0 , T ) which is defined by w(t) = x(t-(T -T 0 )), for any t ∈ (T -T 0 , T ], this system reads :

e -λt π(w(t)) = e -λT π(w(T -)) - T t e -λu ṡ T u [w(s) -g(w(s))]ds du, ∀t ∈ (T -T 0 , T ) α = s 0 - T T -T 0 [w(u) -g(w(u))]du
We finally see that the translation w of x |(0,T 0 ) on (T -T 0 , T ) can be found in the set of solutions of the above integro-differential backward equation indexed by the set [a, ∞) of possible terminal conditions (by (i) x takes values in [a, ∞)). We shall see that w is characterized, among these solutions, by some boundary conditions derived from (iiia), (vi) and (vii), which moreover furnish α. This is Corollary 6 below.

For sake of readability we introduce a notation for the difference between the optimal sales rate and the optimal production rate. We shall put :

δ(r) r -g(r) , ∀ r ∈ [a, ∞).
We have x -y = δ(x) a.e. on [0, T ]. The following remark embodies some useful properties of δ.

Remark 5 By Remark 4, δ is continuous and increasing on [a, ∞), satisfies δ(a) = 0 and hence is positive on (a, ∞). Moreover lim r→∞ δ(r) = ∞.

We can now state our backward procedure to find the optimal plan. Corollary 6 Consider some function w and some couple (τ 0 , r) ∈ [0, T )×[a, ∞) such that :

(i) w(T ) = r, w is continuous and decreasing on (τ 0 , T ], (ii) if r > a then τ 0 = 0, (7) e -λt π(w(t)) = e -λT π(r) - T t e -λu ṡ T u δ(w(s))ds du, ∀t ∈ (τ 0 , T ], (8)
T τ 0 δ(w(u))du ≤ s 0 , (9) if
T τ 0 δ(w(u))du < s 0 then w(τ 0 +) = ∞. (10) Then,    α x y    =    s 0 - T τ 0 δ(w(u))du w(• + τ 0 )1 (0,T -τ 0 ] + a1 (T -τ 0 ,T ] g w(• + τ 0 )1 (0,T -τ 0 ] + a1 (T -τ 0 ,T ]    (11)
Proof. Let w and (τ 0 , r) be as in the statement. We shall prove that the plan ( α, x, ỹ) defined by

α s 0 - T τ 0 δ(w(u))du (12) x w(• + τ 0 )1 (0,T -τ 0 ] + a1 (T -τ 0 ,T ] (13)
ỹ g(x) (14)
is in B and satisfies conditions (i) to (vii) of Proposition 3 which characterize (α, x, y).

Since w is decreasing on (τ 0 , T ] and w(T

) = r ∈ [a, ∞), by (
T τ 0 w(u)du ≤ s 0 + T τ 0 g(w(u))du ≤ s 0 + T a.
Since w ≥ 0, this implies that w ∈ L 1 + [τ 0 , T ] and then, by construction

x ∈ L 1 + [0, T ]. By (9) again, we have α ≥ 0. Thus (α, x, ỹ) ∈ R + × L 1 + [0, T ] × L 1 + [0, T ]

and x and ỹ satisfy (i).

Let us now prove that S ( α,x,ỹ) ≥ 0 on (0, T ] and that (ii), (iii) and (iv) hold.

We shall obtain that T0 inf{t ∈ (0, T ] | S ( α,x,ỹ) (t) = 0} = T -τ 0 . We begin with proving that x = ỹ = a and S ( α,x,ỹ) = 0 on (T -τ 0 , T ]. By (13) and ( 14) and since g(a) = a we have :

x = a = g(a) = g(x) = ỹ on (T -τ 0 , T ], (16)
so that

S ( α,x,ỹ) (t) = S ( α,x,ỹ) (T -τ 0 ), ∀t ∈ (T -τ 0 , T ]. (17)
By ( 14), ( 12), (13) and after a change of variable, for any t ∈ (0, T -τ 0 ] we have :

S ( α,x,ỹ) (t) = s 0 -α + t 0 [g(x(u)) -x(u)]du = T τ 0 δ(w(u))du + t 0 [g(w(u + τ 0 )) -w(u + τ 0 )]du = T τ 0 δ(w(u))du + t+τ 0 τ 0 [g(w(u)) -w(u)]du,
that is by definition of δ,

S ( α,x,ỹ) (t) = T t+τ 0 δ(w(u))du . (18)
Therefore S ( α,x,ỹ) (T -τ 0 ) = 0 and by ( 17)

S ( α,x,ỹ) (t) = 0, ∀t ∈ [T -τ 0 , T ]. ( 19
)
In order to complete the proof of (15) it suffices to check that

S ( α,x,ỹ) is decreasing on (0, T -τ 0 ]. (20)
Indeed from ( 19) and since by assumption τ 0 ∈ [0, T ) we will therefore deduce that S ( α,,x,ỹ) ≥ 0 on (0, T ], T0 inf{t ∈ (0, T ] | S ( α,x,ỹ) (t) = 0} = T -τ 0 and T0 > 0, so that (ii) and (iiia) hold by (19) again, (iiib) holds by ( 16) and (iv) holds by (20).

As for (20), recall that by Remark 5, δ takes positive values on (a, ∞). Moreover, since by (i) w is decreasing on (τ 0 , T ] and satisfies w(T ) = r ≥ a, it takes values in (a, ∞) on (τ 0 , T ). Consequently, δ • w is positive on (τ 0 , T ). It then follows from (18) that S ( α,x,ỹ) is decreasing on (0, T -τ 0 ]. This concludes the proof of (15).

Let us now turn to the proof of (v). Item (vc) holds by construction (see 14). Since w is continuous on (τ 0 , T ], by (13) x is continuous on (0, T -τ 0 ] = (0, T0 ] and so is ỹ = g(x) since g is continuous on [a, ∞). This proves (va). Let us establish (vb), i.e. let us prove that (3) is satisfied. By (13), equation (8) reads :

e -λt π(x(t -τ 0 )) = e -λT π(r) - T t e -λu ṡ T u δ(w(s))ds du, ∀t ∈ (τ 0 , T ],
which, after a change of variable from (τ 0 , T ] onto (0, T -τ 0 ] = (0, T0 ] yields :

e -λ(t+τ 0 ) π(x(t)) = e -λT π(r) - T t+τ 0 e -λu ṡ T u δ(w(s))ds du, ∀t ∈ (0, T0 ],
equivalently,

e -λt π(x(t)) = e -λ(T -τ 0 ) π(r) - T t+τ 0 e -λ(u-τ 0 ) ṡ T u δ(w(s))ds du, ∀t ∈ (0, T0 ].
Using a change of variable in the integral and since T -τ 0 = T0 we obtain :

e -λt π(x(t)) = e -λ T0 π(r) - T0 t e -λu ṡ T u+τ 0 δ(w(s))ds du, ∀t ∈ (0, T0 ].
Plugging (18) in this last equation we get :

e -λt π(x(t)) = e -λ T0 π(r) - T0 t e -λu ṡ S ( α,x,ỹ) (u) du, ∀t ∈ (0, T0 ]. ( 21
)
Observe that since x is nonincreasing and takes values in [a, ∞), since π is continuous and decreasing on [a, ∞) and satisfies lim x→∞ π(x) = π(∞) we have :

lim t 0 π(x(t)) = π(x(0+)) ∈ [ π(∞), π(a)].
Therefore, sending t to 0 in (21) we obtain :

π(x(0+)) = e -λ T0 π(r) - T0 0 e -λu ṡ S ( α,x,ỹ) (u) du . (22)
By computing the difference between ( 21) and ( 22), we obtain (3).

To complete the proof of Corollary 6 it remains to show that (vi) and (vii) are satisfied i.e. to check that :

if T0 < T then x( T0 -) = a , ( 23
) if α > 0 then x(0+) = ∞ . ( 24
)
Observe that, by (i) and ( 13)

x( T0 -) = w(T -) = w(T ) = r ∈ [a, ∞). (25)
Therefore, since by (7) [ r > a ⇒ τ 0 = 0 ] and since [ T0 < T ⇔ τ 0 > 0 ], condition (23) is satisfied. As for the second condition, we have by (13) again : x(0+) = w(τ 0 +) and by (12

) α = s 0 - T τ 0 δ(w(u))du
Condition (24) therefore holds since it is equivalent to condition (10). This ends the proof of Corollary 6.

3 Constructive resolution of the production planning problem.

In the sequel, for r ∈ [a, ∞), BW (r) stand for the integro-differential equation involved in Corollary 6 :

BW (r) : e -λt π(w(t)) = e -λT π(r) - T t e -λu ṡ T u δ(w(s))ds du , t ≤ T
By studying the set of solutions of the equation BW (r) when r varies in [a, ∞) in order to determine the optimal plan (α, x, y) by applying Corollary 6 , we achieve here our main aim : to provide a constructive resolution of the production planning problem.

In addition, we shall give necessary and sufficient conditions for the optimal plan to be regular (α = 0) i.e. without depletion at time 0. Given the revenue and costs functions and the length of the planning period, the regularity of the optimal plan, depends only on the level of the initial inventory s 0 . We shall obtain that there is a positive (possibly infinite) level s0 , depending on π, c, s and T such that : if s 0 ≤ s0 (resp. s 0 > s0 ) then the solution is regular : α = 0 (resp. not regular : α > 0). We shall also exhibit some qualitative distinction between the optimal schedule obtained on a long planning period or on a short one.

In the sequel, we use the following notation : for all interval I and all set E ⊂ R, we denote by C(I; E) the set of functions that are continuous on I with values in E. For E = R, we simply write C(I) for C(I; E). The proofs of the technical results on equation BW that are stated in this section are postponed to Appendix A and B.

We start with the existence and uniqueness of a maximal solution for BW (r), given any terminal condition r ∈ [a, ∞). From this maximal feature we deduce a necessary and sufficient condition for the solution to explode at some given time. This property may be needed when applying Corollary 6 (see 10). We also check that the solution has the continuity and monotony required in Corollary 6. For this purpose we work under the following technical

Standing Assumptions

(Hπ) The inverse of π|[a,∞) is locally Lipschtiz on ( π(∞), π(a)]. (Hc) The inverse of ċ|[0,a] is locally Lipschitz on [ ċ(0), ċ(a)].
Theorem 7 For every r ∈ [a, ∞), there exists a unique couple (τ (r), w) ∈ [-∞, T ) × C((τ (r), T ]), such that w satisfies BW (r) on (τ (r), T ] :

e -λt π(w(t)) = e -λT π(r) - T t e -λu ṡ T u δ(w(s))ds du , ∀t ∈ (τ (r), T ],
and such that either τ (r) = -∞, or τ (r) > -∞ and w(τ (r)+) = ∞.

(26)

In addition, w is decreasing and takes values in [a, ∞).

Notation

Given r ∈ [a, ∞), if (τ (r), w) is associated to r by Theorem 7 we shall write in short w r and say that w r is the maximal solution of BW (r).

We denote by τ the map from [a, ∞) into [-∞, T ) defined by Theorem 7.

In the sequel we shall use the following Theorem 7 is sufficient to initialized the constructive resolution of the planning problem by using Corollary 6. Indeed, one may begin with considering the solution w a of BW (a) and answer the question : Does there exist some τ 0 ∈ [max{0, τ (a)}, T ) such that :

Remark 8 Let r ∈ [a, ∞).
(a) either

T τ 0 δ(w a (u))du = s 0 , (b) or T τ 0 δ(w a (u))du < s 0 and w a (τ 0 +) = ∞ ? (27)
Remark 9 Assume that the answer to (27) is positive. First, by Theorem 7, w a is continuous, decreasing on (τ 0 , T ] and such that w a (T ) = a, so that (i) of Corollary 6 holds for w a with τ 0 and r = a. Second, by definition and by (27), w a satisfies (8), ( 9) and (10) so that (ii) also holds. The optimal plan is then given by (11) in function of w a .

The situations where the answer to (27) is affirmative are described in Theorems 12 and 13 below, obtained by using monotony and continuity of the (translated) backward inventory process associated to w a .

Remark 10 By Remark 8, the function

t ∈ (τ (a), T ] -→ S(t, a) T t δ(w a (s))ds ∈ R +
is well-defined, decreasing and continuous on (τ (a), T ]. It has a limit as t goes to τ (a) which is :

s a sup t∈(τ (a),T ] T t δ(w a (u))du = T τ (a) δ(w a (u))du ∈ R + ∪ {∞} Since τ (a) < T , it is positive.
We claim that there are two cases to keep distinct : either τ (a) > 0 and then the answer to (27) is positive, or τ (a) ≤ 0 and the answer may be negative. If The alternative between τ (a) > 0 and τ (a) ≤ 0 can be economically interpreted as the one between a long planning period and a short one. Indeed, saying that τ (a) > 0 (resp. τ (a) ≤ 0) amounts to saying that the length of the planning period T is greater (resp. lower) than the length T a T -τ (a) of the definition interval of the maximal solution w a of BW (a). It turns out that T a does in actual fact not depend on T . This comes from the time homogeneity of BW (invariance by translation on the terminal time) : by using Theorem 7 and changes of variable, it is easy to prove the following

Remark 11 Let r ∈ [a, ∞) and T 1 , T 2 ∈ [0, ∞). If (w 1 , τ 1 (r)) (resp. (w 2 , τ 2 (r)))
is the maximal solution of BW (r) with terminal time T = T 1 (resp. T 2 ) then we have :

T 1 -τ 1 (r) = T 2 -τ 2 (r) and w 1 (t) = w 2 (t + T 2 -T 1 )
,

T 1 t δ(w r 1 (u))du = T 2 t+T 2 -T 1 δ(w r 2 (u))du for any t ∈ (τ 1 (r), T 1 ].
Therefore the length of the domain of the maximal solution of BW (a), T a = T -τ (a) and the level s a = T τ (a) δ(w a (u))du = Ta 0 δ(w a Ta (u))du depend only on π, c and s and not on the length of the planning period T . In the sequel, we shall say that the planning period is long (resp. short) if τ (a) > 0 (resp. τ (a) ≤ 0).

The Long Planning Period Case

Theorem 12 Assume that τ (a) > 0.

1. If s 0 ≤ s a = T τ (a) δ(w a (u))du then, there exists a unique τ 0 ∈ [0, T ) such that T τ 0 δ(w a (u))du = s 0 and the optimal plan is regular, given by

   α x y    =    0 w a (• + τ 0 )1 (0,T -τ 0 ] + a1 (T -τ 0 ,T ] g w a (• + τ 0 )1 (0,T -τ 0 ] + a1 (T -τ 0 ,T ]    2. If s 0 > s a = T τ (
a) δ(w a (u))du then, the optimal plan is not regular, given by

   α x y    =    s 0 - T τ (a) δ(w a (u))du w a (• + τ (a))1 (0,T -τ (a)] + a1 (T -τ (a),T ] g w a (• + τ (a))1 (0,T -τ (a)] + a1 (T -τ (a),T ]    Proof. Recall that s 0 > 0 = S(T, a). If s 0 ≤ T τ (
a) δ(w a (u))du = S(τ (a)+, a) the existence of a unique τ 0 ∈ [τ (a), T ) such that s 0 = T τ 0 δ(w a (u))du = S(τ 0 , a) is insured by Remark 10 and the Mean-Value Theorem. In addition, since τ (a) > 0, we have τ 0 > 0. Therefore condition (27a) holds for τ 0 and the proof of item 1 is ended by using Remark 9. Since τ (a) > -∞, by (26) we have w a (τ (a)+) = ∞. Therefore, if s 0 > T τ (a) δ(w a (u))du, condition (27b) holds for τ 0 = τ (a). Using Remark 9 again we then obtain item 2.

From Theorem 12, we see that when the planning period is long (τ (a) > 0), the optimal plan is regular if and only if the initial inventory is below the exogenous level

s 0 = s a = T τ (a)
δ(w a (s))ds.

Notice that it may be infinite. In that case, whatever the initial inventory is, the optimal plan is regular.

From an economic point of view, s 0 acts as a threshold. If the initial inventory s 0 is greater than s 0 , then holding such a (high) amount of stock is too expensive. It is then optimal for the firm to sell out immediately the quantity α = s 0 -s 0 , so as to reduce its initial inventory to s 0 . This is the maximal level that it can afford to hold. Recall that the quantity α is sold at an infinite rate and hence at the lowest price π(∞). Therefore the firm has no interest in selling more than necessary. This explain why it sells exactly s 0 -s 0 .

For every s 0 > s 0 , the regular part (x, y) of the optimal plan is the same, it corresponds to the optimal plan obtained for an initial inventory level equal to s 0 . Only the size of the jumps on the sales α = s 0 -s 0 changes with s 0 . If s 0 ≤ s 0 then, the optimal plan is regular and actually depends on s 0 since it is obtain from the translation of w a on the interval (0, T -τ 0 ] where τ 0 solves s 0 = T τ 0 δ(w a (u))du. Remark that the existence of such a threshold and the corresponding properties of the optimal plan will also hold in the case of a short planning period. We shall determine this threshold but not reproduce the discussion.

To the contrary what follows is specific to the long planning periods (T > T a ). First, the threshold s 0 does not depend on T . Second, the plan that the firm adopts is in two (non trivial) phases. The first phase consists in selling out the initial inventory in an optimal way. During the second phase, the firm follows the just-in-time strategy, producing and selling at the same constant rate a, until the end of the period. The depletion phase does not depend on T . It is the one associated to the optimal plan obtained on the period [0, T a ]. In view of item 2, this property is straightforward when s 0 > s 0 = T τ (a) δ(w a (u))du. When s 0 ≤ s 0 , it follows from Remark 11.

We end up this first step of the constructive resolution by stressing the fact that, in the case of a long planning period, nothing more than the computation of w a is necessary to solve the planning problem. Let us now turn to the short planning period case.

The Short Planning Period Case

In this section we assume that τ (a) ≤ 0. By using Remark 10 and the Mean-Value Theorem, we see that, if 0 < s 0 ≤ T 0 δ(w a (s))ds = S(0, a) then, there exists some τ 0 ∈ [0, T ) such that s 0 = T 0 δ(w a (s))ds. Therefore the answer to ( 27) is affirmative and the optimal plan is obtained as previously by using Remark 9. We have :

Theorem 13 Assume that τ (a) ≤ 0.

If s 0 ≤ T 0 δ(w a (u))du then, there exists a unique τ 0 ∈ [0, T ) such that T τ 0 δ(w a (u))du = s 0 and the optimal plan is regular given by :

   α x y    =    0 w a (• + τ 0 )1 (0,T -τ 0 ] + a1 (T -τ 0 ,T ] g w a (• + τ 0 )1 (0,T -τ 0 ] + a1 (T -τ 0 ,T ]   
From Theorem 13 we see that even if the planning period is short, when the initial inventory s 0 is lower than T 0 δ(w a (u))du, the firm has time to clear out its stock before the horizon T , selling out at the rate x = w a (• + τ 0 ) on (0, T -τ 0 ] and it can afford a static management final phase : x = y = a on (τ 0 , T ]. We shall find out that this is not the case when the initial inventory is greater than

T 0 δ(w a (u))du.
Assume that s 0 > T 0 δ(w a (s))ds = S(0, a). Since the function t -→ S(t, a) = T t δ(w a (u))du is decreasing, if there exists some τ ∈ [τ (a), T ) such that T τ δ(w a (s))ds = s 0 then, τ < 0. Therefore the answer to (27) is negative. In that case, the planning problem can not be solved by using w a . One has to consider the solutions w r of BW (r) for r > a and hence, by condition (7) of Corollary 6, which are defined at least on (0, T ], i.e. such that τ (r) ≤ 0.

The planning problem can be solved by finding r ∈ {r ∈ [a, ∞) | τ (r) ≤ 0} such that (a) either T 0 δ(w r(u))du = s 0 , (b) or T 0 δ(w r(u))du < s 0 and w r(0+) = ∞.

(28) Indeed, we have the following

Remark 14 If r ∈ τ -1 (R -) {r ∈ [a, ∞) | τ (r) ≤ 0}
is such that (28) holds then, by Theorem 7, w r satisfies the requirements of Corollary 6 with (τ 0 , r) = (0, r). The optimal plan is then given by (11) in function of w r which reads

(α, x, y) = (0, w r, g(w r)) if (28a) holds, (s 0 - T 0 δ(w r(u))du, w r, g(w r)) if (28b) holds. (29)
We therefore now focus our attention on the function

S 0 : τ -1 (R -) = {r ∈ [a, ∞) | τ (r) ≤ 0} -→ R + ∪ {∞} r -→ S 0 (r) = T 0 δ(w r (s))ds.
It represents the level at time 0+ of the backward cumulative inventory process associated to any w r solution of BW (r) defined at least on (0, T ]. In particular, we shall investigate the monotony and continuity of the function S 0 with respect to r in order to find the greatest value that it can reach when the terminal condition r varies in τ -1 (R -). This will allow us to determine the threshold s 0 below (above) which the optimal plan is (not) regular. For this purpose we work under the following technical

Standing Assumption (Hs) The function s is strictly convex and ṡ is locally Lipschitz on R + .

Proposition 15 The function τ

: r ∈ [a, ∞) -→ τ (r) ∈ [-∞, T ) is nondecreasing, right-continuous and satisfies lim r→∞ τ (r) = T . Therefore, if the set τ -1 (R -) is not empty then, it is a bounded interval with min τ -1 (R -) = a. We set m sup{τ -1 (R -)} < ∞.
Proposition 16 Assume that τ -1 (R -) = ∅.

1. The function S 0 is well-defined, nondecreasing and left-continuous as a map from

τ -1 (R -) into R + ∪ {∞}.
2. The function S 0 is continuous on its domain Dom(S 0 ) {r ∈ τ -1 (R -) | S 0 (r) ∈ R + }.

If τ (r) < 0 then, r ∈ Dom(S 0 ).

If τ (m)

> 0 i.e. τ -1 (R -) = [a, m) then, sup τ -1 (R -) S 0 = ∞.
Corollary 17 1. The function S 0 has at most one discontinuity as a map from τ

-1 (R -) into R + ∪ {∞}. If d ∈ τ -1 (R -) is such a point of discontinuity then, d < m, τ (d) = 0, Dom(S 0 ) = [a, d] and hence, max Dom(S 0 ) S 0 = S 0 (d) = T 0 δ(w d (u))du < ∞ and inf r>d r∈τ -1 (R -) S 0 (r) = ∞ . 2. If sup τ -1 (R -) S 0 < ∞ then, τ (m) = 0, and hence, Dom(S 0 ) = τ -1 (R -) = [a, m] and max Dom(S 0 ) S 0 = max τ -1 (R -) S 0 = S 0 (m) = T 0 δ(w m (u))du < ∞ .
From Corollary 17 we see that, as for continuity and boundness of S 0 , there are exactly three cases to keep distinct. The function S 0 may be continuous and unbounded on τ -1 (R -), continuous and bounded on τ -1 (R -) or it may have a unique discontinuity at some point d ∈ τ -1 (R -). In addition, sup

Dom(S 0 ) S 0 =      ∞ if S 0 is continuous and unbounded, S 0 (m) = T 0 δ(w m (u))du < ∞ if S 0 is continuous and bounded, S 0 (d) = T 0 δ(w d (u))du < ∞ if S 0 has a discontinuity at d ∈ τ -1 (R -) .
By continuity of S 0 on its domain, we shall obtain that the threshold s 0 on the initial inventory below (above) which the optimal plan is (not) regular exists and is given by

s 0 = sup Dom(S 0 ) S 0 .
Theorem 18 Assume that τ (a) ≤ 0 and that S 0 is continuous and unbounded on τ -1 (R -). Consider s 0 > T 0 δ(w a (u))du. There exists some r ∈ τ -1 (R -), r > a, such that T 0 δ(w r(u))du = s 0 and the optimal plan is regular, given by (α, x, y) = (0, w r, g(w r)) .

Proof. Since S 0 is continuous and unbounded on the interval τ -1 (R -) and s 0 > T 0 δ(w a (u))du = S 0 (a) = min τ -1 (R -) S 0 , by the Mean-Value Theorem there exists some r ∈ τ -1 (R -) such that S 0 (r) = T 0 δ(w r(u))du = s 0 (r > a since s 0 > S 0 (a)). This means that condition (28a) holds for r. The proof is completed by using Remark 14.

Theorem 19 Assume that τ (a) ≤ 0 and that S 0 is continuous and bounded on τ -1 (R -). Consider s 0 > T 0 δ(w a (u))du.

1. If s 0 ≤ S 0 (m) = T 0 δ(w m (u))du then, there exists some r ∈ (a, m] such that T 0 δ(w r(u))du = s 0 and the optimal plan is regular, given by (α, x, y) = (0, w r, g(w r)) .

If s 0 > S 0 (m) =

T 0 δ(w m (u))du then, the optimal plan is not regular, given by

(α, x, y) = (s 0 - T 0 δ(w m (u))du, w m , g(w m )) . Proof. By 2 of Corollary 17, if S 0 is bounded on τ -1 (R -) then τ (m) = 0, τ -1 (R -) = [a, m]
and max τ -1 (R -) S 0 = S 0 (m). Therefore, if in addition S 0 is continuous on τ -1 (R -) then, by the Mean-Value Theorem there exists some r ∈ (a, m] such that T 0 δ(w r(u))du = s 0 and then, item 1 holds by Remark 14. Now, assume that s 0 > S 0 (m) = T 0 δ(w m (u))du. Since τ (m) = 0 > -∞, Proposition 7 implies that w m (0+) = w m (τ (m)+) = ∞. Therefore condition (28b) holds with r = m and the proof is concluded by using Remark 14.

Theorem 20 Assume that τ (a) ≤ 0 and that S 0 has a (unique) discontinuity at d ∈ τ -1 (R -). Consider s 0 > T 0 δ(w a (u))du.

1. If s 0 ≤ S 0 (d) =
T 0 δ(w d (u))du then, there exists some r ∈ (a, d] such that T 0 δ(w r(u))du = s 0 and the optimal plan is regular, given by (α, x, y) = (0, w r, g(w r)) .

If s

0 > S 0 (d) =
T 0 δ(w d (u))du then, the optimal plan is not regular, given by

(α, x, y) = (s 0 - T 0 δ(w d (u))du, w d , g(w d )) .
Proof. By 1 of Corollary 17, if S 0 has a discontinuity at some d ∈ τ -1 (R -) then τ (d) = 0, Dom(S 0 ) = [a, d] and max Dom(S 0 ) S 0 = S 0 (d) < ∞. Since by Proposition 16 S 0 is continuous on its domain, item 1 is obtained as above by using the Mean-Value Theorem and Remark 14. Item 2 is also obtained by the same arguments as above with d instead of m.

We have seen from Theorem 13, that when s 0 ≤ T 0 δ(w a (u))du the firm can afford to sell out its inventory at some rate which ranges a part of the path of w a . From Theorems 18, 19 and 20 we see that if the initial inventory is higher than T 0 δ(w a (u))du then, the firm must move out its stock faster : the optimal sales rate is given by x = w r on (0, T ] with r > a so that it is greater than w a on the whole planning period (see Proposition 26 in Appendix B for the increasing feature of w r with respect to r). In this context, the production may never start. Typically, if ċ(0) > π(∞) and r > π-1 ( ċ(0)) then, π(x) = π(w r) ≤ π(r) < ċ(0) on (0, T ] and hence by definition of g, y = 0 on (0, T ]. In any case, the inventory is totally depleted right in T , there is no phase where the firm produces for immediate sales at the constant rate a. This makes qualitatively different the optimal plan obtained on a short planning period and the one obtained on a long period.

From Theorems 13, 18, 19 and 20 we check that there is positive threshold above (below) which the optimal plan is with (without) depletion at time 0. It is given by

s 0 sup Dom(S 0 ) S 0 = sup {r∈[a,∞)|τ (r)≤0 , T 0 δ(w r (u))du<∞} T 0 δ(w r (u))du .
Recall that by time homogeneity of BW (see Remark 11) we have observe that the length T a of the domain of the maximal solution of BW (a) and the threshold s 0 obtained for long planning periods (T > T a ) do not depend on T . To the contrary, when the planning period is short (T < T a ), the threshold may depend on T . Let us denote (w r Ta , τ Ta (r)) the maximal solution of BW (r) with terminal time T a , for any r ∈ [a, ∞). By using Remark 11 we see that, given π, c, s and hence T a , the dependence of s 0 with respect to T is given by :

s 0 (T ) = sup {r∈[a,∞)|τ Ta (r)≤0 , Ta Ta-T δ(w r Ta (u))du<∞} Ta Ta-T δ(w r Ta (u))du
It would be of interest to study the monotony of s 0 with respect to T and it would be economically funded to obtain that the shorter the period is, the lower the threshold is. Also, notice that, while we indeed prove the existence of such a threshold, we do not provide some conditions for deciding whether if it is infinite or finite, with S 0 continuous or not. These are directions for future research.

and such that the boundary condition (30) holds. Since z is continuous on (τ, T ], by Remark 21 and by Assumption (Hs), we see that the function

u -→ λz(u) + ṡ T u ∆(z(s))ds
is continuous on (τ, T ]. We then deduce from (31) that z is in actual fact continuously differentiable on (τ, T ] with

ż(t) = λz(t) + ṡ T t ∆(z(s))ds , ∀t ∈ (τ, T ]. (32)
Then, it is easy to check that z satisfies e -λt z(t) = e -λT π(r) -

T t e -λu ṡ T u ∆(z(s))ds du, ∀t ∈ (τ, T ]. (33)
Under Assumption (H), we can set w π-1 • z and see that, since z is in C((τ, T ]; D), w is in C((τ, T ]; [a, ∞)). From (32) and (30), we deduce that w solves BW (r) on (τ, T ] and satisfies the following boundary condition

either τ = -∞, or, τ > -∞ and w(τ +) lim t τ w(t) = ∞.
Besides, since z is increasing, it follows from assumption (H) again, that w is decreasing. This concludes the proof of Theorem 7.

We now turn to the proof of Theorem 7bis. It is based on the three following propositions and an induction argument. More precisely, some regularity and Lipschitzianity properties of the right-hand side of the equation BW (θ) are given in Proposition 22. Using this properties we will prove that for every terminal condition θ ∈ D, the equation BW (θ) has at least one solution : there exists some continuous function z which satisfies BW (θ) on some non-empty interval (γ, T ]. We will also provide some result of continuous dependence of solutions with respect to the terminal condition. This is Proposition 23. With Proposition 24, we will see that a solution (γ, z) which does not satisfy the boundary condition (30) can be extended in another solution on some larger interval.

In the sequel, for φ ∈ C([a, b]), we denote φ [a,b] sup a≤s≤b |φ(s)|.
The right-hand side of the backward equation is studied in terms of the following operator : for fixed η ∈ [-∞, T ),

F η : (η, T ] × C((η, T ]; D) -→ R (t, v) -→ λv(t) + ṡ T t ∆(v(s))ds Proposition 22 Let η ∈ [-∞, T
) and let L be some compact subset of D.

(o) The operator F η is well defined and takes values in (0, ∞).

(i) For all v ∈ C((η, T ]; D), the function t -→ F η (t, v) is continuous on (η, T ].
(ii) There exists some constant K η,L > 0 such that, for all t ∈ (η, T ],

v , ṽ ∈ C([t, T ]; L) =⇒ |F η (t, v) -F η (t, ṽ)| ≤ K η,L v -ṽ [t,T ] ,
Proof. Let v ∈ C((η, T ]; D). First, since D = ( π(∞), π(a)] ⊂ (0, ∞), v takes values in (0, ∞). Second, from Remark 21, ∆•v is continuous and nonnegative on (η, T ]. This implies that the function t -→ T t ∆(v(s))ds is well-defined, taking values in R + and continuous on (η, T ]. Since ṡ is continuous on R + , (o) and (i) of Proposition 22 hold. We now prove (ii).

Let t ∈ (η, T ]. Notice that, by Remark 21,

0 ≤ ∆(u) ≤ π-1 (u), ∀u ∈ D.
Therefore, since π-1 is decreasing on D, and v, ṽ are taking values in L ⊂ D, we have

T t ∆(v(s))ds , T t ∆(ṽ(s))ds ∈ [0, (T -η) π-1 (min L)].
It then follows from (Hs) that there exists some constant C η,L > 0 such that

|F η (t, v) -F η (t, ṽ)| ≤ λ|v(t) -ṽ(t)| + ṡ T t ∆(v(s))ds -ṡ T t ∆(ṽ(s))ds ≤ λ|v(t) -ṽ(t)| + C η,L T t ∆(v(s))ds - T t
∆(ṽ(s))ds .

By Remark 21, it follows from the last inequality that there exists some constant K L > 0 such that

|F η (t, v) -F η (t, ṽ)| ≤ λ|v(t) -ṽ(t)| + K L C η,L T t |v(s) -ṽ(s)|ds, (34)
and hence

|F η (t, v) -F η (t, ṽ)| ≤ K η,L v -ṽ [t,T ] ,
for some constant K η,L . This ends the proof of Proposition 22.

Observe that if

η 1 ≤ η 2 then, F η 1 |(η 2 ,T ]×C((η 2 ,T ];D) ≡ F η 2 .
Therefore, in the sequel we shall omit the index and always consider F with its maximal set of definition.

Definition 1 Given θ ∈ D, we shall say that (γ, z) is a solution of BW (θ) if : γ ∈ [-∞, T ), z ∈ C((γ, T ]; D) and z satisfies the equation BW (θ) on (γ, T ].
Recall that the equation BW (θ) reads as follows in terms of F :

BW (θ) : z(t) = θ - T t F (u, z)du, t ≤ T.
Proposition 23 1. For all θ ∈ D, the equation BW (θ) has at least one solution. If (γ, z) is such a solution then, z is increasing on (γ, T ].

2. Let η ∈ (-∞, T ] and let L be a compact subset of D. Then, there exists some constant K η,L > 0 such that : if (γ, z) is a solution of BW (θ) and (γ , z ) is a solution of BW (θ ) then, for all t ∈ [η, T ] such that z and z are both defined on [t, T ] and both map [t, T ] into L we have

z -z [t,T ] ≤ K η,L |θ -θ | .
Proof. We begin with the Proof of item 1. Let θ ∈ D. First observe that the last assertion is a direct consequence of the nonnegativity of ṡ and the positivity of elements of D. We now concentrate on the first assertion. We shall use a Picard's approximations argument.

Step 1. We start with the construction of a suitable set in which we will construct our Picard's approximations. For this purpose, let us fix some η ∈ (-∞, T ) and choose some b ≥ 0 such that the compact

L [θ -b, min{(θ + b), π(a)}] is included in D. (35)
We claim that there exists some constant M > 0 for which

∀t ∈ (η, T ] , ∀v ∈ C([t, T ]; D) : v -θ [t,T ] ≤ b =⇒ |F (t, v)| ≤ M. (36)
To see this, first notice that by Proposition 22 (o) and (i), the function t -→ F (t, θ) is continuous on (-∞, T ], with positive values and therefore bounded on [η, T ] by some constant M 0 > 0. Let us now prove that (36) holds for M = M 0 + K η,L where K η,L is given by Proposition 22 (ii). Let t ∈ (η, T ] and v ∈ C([t, T ]; D) with v -θ [t,T ] ≤ b. Then, by definition of L, v is in C([t, T ]; L). Since the taking values in L, we therefore have by Proposition 22 (ii)

|F (t, v) -F (t, θ)| ≤ K η,L v -θ [t,T ] ≤ K η,L b and hence, |F (t, v)| ≤ K η,L b + M 0 M. This provides (36).
We are now in position to construct a suitable set. Recalling that η < T , we may choose some γ ∈ (η, T ) such that

0 < T -γ < min{T -η, b M , 1 K η,L }. ( 37
)
Our Picard's approximations will be constructed on the set

S { v ∈ C([γ, T ]; L) | v(T ) = θ, |v(t) -v(u)| ≤ M |t -u|, ∀t, u ∈ [γ, T ] }.
Observe that S is closed for the pointwise convergence on [γ, T ]. We now define an operator P on S by

P (v)(t) θ - T t F (u, v)du, ∀ t ∈ [γ, T ] .
We shall prove in the following steps that P admits a fixed point in S.

Step 2. We first prove that P maps S into S. Let v ∈ S. First recall that by Proposition 22 (o) we have

F (u, v) > 0 , ∀u ∈ [γ, T ] (38)
and notice that by (36), (37) and by definition of S we have

F (u, v) ≤ M , ∀u ∈ [γ, T ]. ( 39
)
We now check that P (v) takes values in L.

Let t ∈ [γ, T ]. By equations ( 38) and (39) and by definition of γ (see (37)), we have

P (v)(t) = θ - T t F (u, v)du ≤ θ ≤ min{(θ + b), π(a)}, and
P (v)(t) = θ - T t F (u, v)du ≥ θ -M (T -γ) ≥ θ -b which implies that P (v)(t) ∈ L. The proof of
Step 2 is concluded by writing

|P (v)(t) -P (v)(u)| = t u F (s, v)ds ≤ M |t -u|, ∀t, u ∈ [γ, T ],
where the last inequality follows from (39).

Step 3. We shall now construct a Cauchy sequence in S that converges to a fixed point of P . Let (z n ) n be the sequence defined by z 0 = θ on [γ, T ] and z n+1 = P (z n ), ∀n ∈ N.

Since γ ∈ (η, T ], it follows from Proposition 22 (ii) that, for all t ∈ [γ, T ] we have

|z n+1 (t) -z n (t)| = T t [F (u, z n ) -F (u, z n-1 )]du ≤ |T -t|K η,L z n -z n-1 [t,T ] ≤ ((T -γ)K η,L ) n z 1 -z 0 [γ,T ] .
By (37), 0 < (T -γ)K η,L < 1. Therefore, (z n ) is a Cauchy sequence. Since S is closed, its limit z is in S. Arguing as above, we see that

P (z) -P (z n ) [γ,T ] ≤ K η,L (T -γ) z -z n [γ,T ] .
It follows that

P (z) -z [γ,T ] = P (z) -P (z n ) [γ,T ] + z n+1 -z [γ,T ] ≤ K η,L (T -γ) z -z n [γ,T ] + z n+1 -z [γ,T ] ,
which shows, by passing to the limit, that P (z) = z on [γ, T ]. This completes the proof of the first item of Proposition 23. We now turn to the Proof of item 2. Let t ∈ [η, T ] such that z and z both map [t, T ] into L. Then, for all u ∈ [t, T ], z and z both map [u, T ] into L. Therefore, by (34) in the proof of Proposition 22, there exists some C = C η,L > 0 such that

|F (u, z) -F (u, z )| ≤ λ|z(u) -z (u)| + C T u |z(v) -z (v)|dv, ∀u ∈ [t, T ].
This implies that, for all s in [t, T ], we have

|z(s) -z (s)| ≤ |θ -θ | + T s |F (u, z) -F (u, z )|du ≤ |θ -θ | + T s λ|z(u) -z (u)| + C T u |z(v) -z (v)|dv du ≤ |θ -θ | + (λ + C(T -η)) T s |z(u) -z (u)|du ≤ |θ -θ | + C η,L T s |z(u) -z (u)|du.
By applying Gronwall's Lemma to the last inequality and since η ≤ s we obtain

|z(s) -z (s)| ≤ |θ -θ |e C η,L (T -s) ≤ |θ -θ |e C η,L (T -η) ,
which ends the proof of Proposition 23.

We now turn to the extension Proposition.

Proposition 24 Let (γ, z) be a solution of BW (θ) for some θ ∈ D. If γ > -∞ and if z satisfies z(γ+) lim

t γ z(t) > π(∞)
then, BW (θ) has a solution (γ, ẑ) which satisfies : γ < γ and ẑ|(γ,T ] ≡ z.

Proof. Let (γ, z) be a solution of BW (θ) satisfying γ > -∞. Since z is increasing the limit z(γ+) = lim t γ z(t) exists in D = [ π(∞), π(a)]. Let us assume that z(γ+) > π(∞).

We have to prove that there exists some γ < γ and some ẑ ∈ C((γ, T ]; D) such that ẑ satisfies BW (θ) on (γ, T ] i.e. (40) Indeed, the result will then hold for (γ, ẑ) where ẑ is defined by setting ẑ|(γ,γ] ≡ z and ẑ|(γ,T ] ≡ z. Let us establish (40). For ease of notation, let us write I = Since the translation ṡ(. + I) inherits of the increasing and Lipschitzian feature of ṡ, one should be convinced that Fη satisfies the regularity and Lipschitzianity properties gathered in Proposition 22. This properties together with the fact that z(γ+) is in D are sufficient to let the Picard's approximations argument be valid to prove the existence of some γ < γ and some z ∈ C((γ, γ]; D) such that z satisfies (40) on (γ, γ]. This ends the proof of Proposition 24.

We are now in position to give the Proof of Theorem 7bis.

Let us first establish the uniqueness. Let (γ, z) and (γ , z ) satisfying the requirements of Theorem 7bis. Fix some arbitrary t ∈ (max(γ, γ ), T ]. Since z and z are in C((max{γ, γ }, T ]; D), there exists some compact L ⊂ D such that z and z both map [t, T ] into L. It follows from item 2 of Proposition 23 that z ≡ z on [t, T ]. By arbitrariness of t in (max(γ, γ ), T ], we then have z ≡ z on (max(γ, γ ), T ], so that γ, γ ≥ max{γ, γ }. Therefore γ = γ and hence (γ, z) = (γ, z ). This provides uniqueness.

We claim that the right-hand side is negative, so that, for some η > 0, π(w) -π(w ) < 0, on (σ -η, σ), and hence, by the increasing feature of π on [a, ∞), w > w on (σ -η, σ). This will contradict the definition of σ and finally proves Proposition 26.

To see that the above claim holds, recall that δ is increasing on D. Since ṡ is increasing, the claim follows from the definition of σ and the fact the continuous functions w and w satisfy w(T ) = r > r = w (T ).

Proof of Proposition 15.

Step 1. We first prove that τ : r ∈ [a, ∞) -→ τ (r) ∈ [-∞, T ) is nondecreasing. Fix r and r in [a, ∞) such that r < r, and denote w (resp. w) for w r (resp. w r ). By Proposition 26, w ≤ w < ∞ on (max{τ (r ), τ (r)}, T ]. This implies that τ (r ) ≤ max{τ (r ), τ (r)}, i.e. τ (r ) ≤ τ (r).

Step 2. We now prove that τ is right-continuous as a map from [a, ∞) into [-∞, T ). Assume to the contrary that there exists some r

∈ [a, ∞) such that lim r r τ (r) > τ (r ) ∈ [-∞, T ) (42)
For ease of notation, we write w for w r . Since > τ (r ), w ( ) is finite and we can find some real number M such that

M > w ( ). (43)
Since, by Theorem 7, w is decreasing, we have M > w ( ) > r . Combining these inequalities with (42), we deduce that there exists some r ∈ [a, M ) such that

0 < r -r < M -w ( ) K ,[a,M ] , (44)
where K ,[a,M ] given by Corollary 25. Moreover, since τ is nondecreasing, we have

τ (r) ≥ inf z>r τ (z) = > -∞.
It follows that τ (r) > -∞. We deduce from Theorem 7 that w r (τ (r)+) = ∞. Now, since w r is continuous, decreasing, and satisfies w r (T ) = r < M < ∞, it follows from the Mean-Value Theorem that there exists some t M ∈ (τ (r), T ) such that w r (t M ) = M . Observing that

≤ τ (r) < t M , (45)
it follows from the decreasing feature of w together with (44) that

w r (t M ) -w (t M ) > M -w ( ) > K ,[a,M ] (r -r ). ( 46
)
We shall now put in light the required contradiction. Since w r and w are decreasing, it follows from the equality w r (t M ) = M , (45) and ( 43

≤ w r (t M ) -w (t M ) ≤ K ,[a,M ] (r -r ),
which combined with (46) yields the required contradiction.

Step 3. We finally prove that lim r→∞ τ (r) = T . Assume to the contrary that

T lim r→∞ τ (r) < T. (47)
Considering the sequence (w n ) n≥a , we see that

where the inequality holds because T < T and π(∞) ≥ 0. In order to obtain a contradiction, we shall prove that the right-hand side is negative. Observe that, since ṡ is nonnegative,

- T T e -λt ṡ T t δ(w n (s))ds dt ≤ - (T +T )/2 T e -λt ṡ T t δ(w n (s))ds dt. (49)
Recall that, from Theorem 7, w n is decreasing and satisfies w n (T ) = n. By Remark 5, we deduce that there exists some n 0 ≥ a such that for all n ≥ n 0 ,

δ(w n (s)) ≥ δ(w n (T )) ≥ 2 , ∀s ∈ [(T + T )/2, T ].
Therefore, for all n ≥ n 0 and t

∈ [T , (T + T )/2] T t δ(w n (s))ds ≥ T (T +T )/2 δ(w n (s))ds ≥ 2 T - T + T 2 ≥ T -T ,
where we used the nonnegativity of δ. By the increasing feature and nonnegativity of ṡ, this implies that for all n ≥ n 0 ,

- (T +T )/2 T e -λt ṡ T t δ(w n (s))ds dt ≤ -ṡ(T -T ) (T +T )/2 T e -λt dt. (50)
From ( 49) and (50) we deduce that

lim n→∞ - T T e -λt ṡ T t δ(w n (s))ds dt ≤ -ṡ(T -T ) (T +T )/2 T e -λt dt < 0, (51)
where the last inequality holds because T < T and ṡ is increasing, nonnegative and therefore positive on (0, ∞). The required contradiction follows from (48) and (51).

The second assertion Proposition 15 is a direct consequence of the above steps.

Proof of Proposition 16.

Assume that τ -1 (R -) is not empty. is well-defined, decreasing and continuous on (τ (r), T ]. In addition, it has has a limit as t goes to τ (r) which is :

sup t∈(τ (r),T ] T t δ(w r (u))du = T τ (r) δ(w r (u))du ∈ R + ∪ {∞} .
Therefore, for any r ∈ [a, ∞) such that τ (r) ≤ 0 we have : S 0 (r) T 0 δ(w r (u))du ∈ R + ∪ {∞}. The function S 0 is thus well-defined as a map from τ -1 (R -) into R + ∪ {∞}. Besides, from Proposition 26 it follows that it is nondecreasing. Let us prove that it is left-continuous. Since S 0 is nondecreasing, the limit lim z r S 0 (z) exists in R + ∪ {∞} for all r ∈ τ -1 (R -). We shall prove that lim z r S 0 (z) = S 0 (r) in R + ∪ {∞}, ∀r ∈ τ -1 (R -) ∩ (a, ∞).

(52)

Fix r ∈ τ -1 (R -) ∩ (a, ∞). We first consider the case S 0 (r ) ∈ R + . Fix ε > 0. By 2 of Remark 27, there exists some t ε ∈ (0, T ] such that 0 ≤ S 0 (r ) -S(t ε , r ) < ε 3 . This proves (52) when S 0 (r ) < ∞.

We now consider the case where S 0 (r ) = ∞. Assume to the contrary that M lim r r S 0 (r) < ∞. Since S 0 is nondecreasing by (i), we have M = sup r<r S 0 (r).

In order to obtain a contradiction, we shall find some r ∈ τ -1 (R -) such that r < r and S 0 (r) > M.

(56)

First notice that since S 0 (r ) = ∞ we have τ (r ) = 0. Then, by 2 of Remark 27, there exists some t M ∈ (0, T ] such that S(t M , r ) > 2M . Let K K 0,[a,w(t M )] be given by Corollary 25. We set :

r max{r - M K(T -t M )
, a}.

By construction, r < r since by assumption r > a. Then, by the same arguments as in the previous case, we can apply Corollary 25 for η = 0 and L = [a, w r (t M )] on [t M , T ], to obtain S(t M , r) = S(t M , r) -S(t M , r ) + S(t M , r )

≥ T t M
[δ(w r (s)) -δ(w r (s))]ds + S(t M , r ) ≥ (T -t M )K(r -r ) + S(t M , r )

> -M + 2M = M.
This proves (56) and therefore (52) when S 0 (r ) = ∞. The proof of item 1 is completed.

2.

From Remark 27, we see that if τ (r) < 0 then S 0 (r) = S(0, r) < ∞, i.e. r ∈ DomS 0 {r ∈ τ -1 (R -) | S 0 (r) < ∞} We shall prove that S 0 is continuous on its domain DomS 0 .

Observe that, since S 0 is nondecreasing, DomS 0 is an interval. Since S 0 is left-continuous, we only have to prove that if r ∈ Dom and r < sup(Dom) then lim r r S 0 (r) = S 0 (r ).

Since r < sup(Dom), there exists r > r such that S 0 (r) < ∞. Fix ε > 0, then, by Remark 27, there exists some t ε ∈ (0, T ] such that 0 ≤ S 0 (r) -S(t ε , r) < ε 3

Now, for all r ∈ (r , r], we have, by Proposition 26 and Remark 5, 0 ≤ S 0 (r ) -S(t ε , r ) ≤ S 0 (r) -S(t ε , r) ≤ S 0 (r) -S(t ε , r) < ε 3 .

Notice that by Proposition 26, for all r ∈ [r , r], w r maps [t ε , T ] into [a, w r(t ε )]. The proof can therefore be completed by applying Corollary 25, with η = 0 and L = [a, w r(t ε )], to w r and w r on [t ε , T ]. This ends the proof of item 2.

3.

We now prove that if τ (m) > 0 then, sup τ -1 (R -) S 0 = ∞. We claim that lim

r m w r (τ (m)) = ∞, (57)
Then, there exists a sequence (r n ) in τ -1 (R -) increasing to m such that Noticing that, by (61), K(r -m) ≥ -M/2, we obtain w r (t M ) ≥ 2M -M/2 > M.

Since t M > τ (m) and w r is decreasing, this implies that w r (τ (m)) > M, which ends the proof of (59) and therefore proves (57). The proof of Proposition 16 is completed

Proof of Corollary 17.

  13) x is nonincreasing and takes values in [a, ∞) on (0, T ]. By Remark 4, g is nonincreasing, taking values in [0, a] on [a, ∞). Therefore, by (14) ỹ is nondecreasing and takes values in [0, a] on (0, T ]. Besides, by (9), by definition of δ and since τ 0 ≥ 0 and since g takes values in [0, a] we have

  s))ds du , ∀t ∈ (γ, T ], and such that ẑ|(γ,T ] ≡ zTo do this, recall that z satisfiesz(t) = θ -(s))ds du , ∀t ∈ (γ, T ],and hence, by passing to the limit as t goes to γ z(γ+) = θγ < γ and some z ∈ C((γ, γ]; D) such that z(t) = z(γ+) -γ t λz(u) + ṡ γ u ∆(z(s))ds + T γ ∆(z(s))ds du , ∀t ∈ (γ, γ].

Tγ

  ∆(z(s))ds. Now, for all η ∈ [-∞, γ] consider the operator defined by Fη : (η, γ] × C((η, γ]; D) -→ R (t, v) -→ λv(t) + ṡ T t ∆(v(s))ds + I

  and e -λt π(w (t)) = e -λσ π(w (σ)) -(s))ds du.Since w (σ) = w(σ), this implies that, for all t ∈ ( t, σ)., π(w(t)) -π(w (t)) being continuous on ( t, T ], we can let t tend to σ in the above equality to obtain lim t σ[ π(w(t)) -π(w (t))]

  ), that they both map the interval [t M , T ] into [a, M ]. Therefore, applying Corollary 25 with η = and L = [a, M ] we have 0

e

  -λT π(w n (T )) -e -λT π(n) = -n (s))ds dt.Since, by Theorem 7, w n is decreasing, we have w n (T ) ≥ n, ∀n ≥ a. Then letting n tend to ∞ in the previous inequality shows that 0 ≤ (e -λT -e -λT ) π(∞) = lim n→∞ -n (s))ds dt

1 .

 1 We shall use the following Remark 27 It follows from Remark 5 that, for any r ∈ [a, ∞) the function t ∈ (τ (r), T ] -→ S(t, r) T t δ(w r (s))ds ∈ R +

( 53 )

 53 Let r ∈ τ -1 (R -) such that r < r . Since by Proposition 15 τ is nondecreasing , w r and w r are both defined on (τ (r ), T ]. Observe that w r maps [t ε , T ] into [a, w r (t ε )]. Besides, by Proposition 26, a ≤ w r ≤ w r on [t ε , T ]. Therefore, w r and w r map [t ε , T ] into [a, w r (t ε )]. Then, by Corollary 25, for η = τ (r ) and L = [a, w r (t ε )], there exists someK > 0 such that 0 ≤ δ(w r ) -δ(w r ) ≤ K(r -r) on [t ε , T ],(54)where the first inequality follows from Proposition 26 and Remark 5 . Using Proposition 26 and Remark 5 again, we also have0 ≤ S 0 (r) -S(t ε , r) ≤ S 0 (r ) -S(t ε , r) < ε 3 . (55) Now, defining η(ε, t ε (r )) ε 3(T -t ε )K ,and using (53), (54) and (55), we see that for all r ∈ [r -η, r )|S 0 (r ) -S 0 (r)| ≤ |S 0 (r ) -S(t ε , r )| + |S(t ε , r ) -S(t ε , r)| + |S(t ε ,r) -S 0 (r)| w r (s)) -δ(w r (s))]ds ≤ 2ε 3 + (T -t ε )K(r -r) ≤ ε.

  rn (s))ds ≥ τ (m)δ(w rn (τ (m))).

1 . 2 .

 12 Assume that S 0 has a discontinuity at d ∈ τ -1 (R -). Then, since by Proposition 16 S 0 is left-continuous, we see that d < sup{τ -1 (R -)} = m and S 0 (d) = lim r d S 0 (r). Since S 0 is nondecreasing, we must haveS 0 (d) = lim r d S 0 (r) < lim r d S 0 (r),and therefore S 0 (d) ∈ R + . It follows thatS 0 (r) = ∞ , ∀r ∈ τ -1 (R -) , r > d .(62)Otherwise, we would have [a, r] ⊂ Dom(S 0 ) by the increasing feature of S 0 . Since, by Proposition 16 S 0 is continuous on its domain, this would implies that S 0 is continuous on [a, r], a contradiction with the definition of d together with the inequality d < r. This shows that Dom(S 0 ) = [a, d] and that S 0 can not have any other discontinuity point, since it is continuous on its domain. Finally, since by Proposition 16 [ τ (r) < 0 ⇒ S 0 (r) < ∞ ], it follows from (62) that τ (r) = 0 for all r ∈ (d, m]. By right-continuity of τ , this implies that τ (d) = 0. Assume that sup τ -1 (R -) S 0 < ∞. From Proposition 16 we know that if τ (m) > 0 then, sup τ -1 (R -) S 0 = ∞. Therefore, τ (m) ≤ 0. Now, since m = sup τ -1 (R -) = sup{t ∈ [a, ∞) | τ (r) ≤ 0} and since by Proposition 15 τ is right-continuous, we also have τ (m) ≥ 0. This implies that τ -1 (R -) = [a, m]. The sequel of assertion 2 follows from the increasing feature of S 0 .

  Since w r is continuous, decreasing and takes values in [a, ∞) on (τ (r), T ] and since by Remark 5 δ is continuous, increasing and nonnegative on [a, ∞), we see that δ • w

r is continuous, decreasing and nonnegative on (τ (r), T ]. It is therefore positive on (τ (r), T ).

By Assumption (H) we can define ∆ on D by

For later purpose, we make the following Remark 21 By Assumptions (H), (Hπ) and (Hc), ∆ is nonnegative, locally Lipschitz and bounded from above by π-1 on D.

We shall obtain Theorem 7 as a Corollary of

Theorem 7bis

For all θ ∈ D, there exists a unique couple (τ, z) ∈ [-∞, T ) × C((τ, T ]; D), such that z satisfies the following equation on (τ, T ]

and such that,

Moreover, z is increasing.

If Theorem 7bis holds then, for all r ∈ [a, ∞), since π(r) ∈ D, there exists a unique couple (τ,

We now prove that existence holds. Define E {(γ, z) solution of BW (θ)}.

By item 1 of Proposition 23, E is not empty. It is easy to check that E is inductive for the order defined by

Thus, by Zorn's lemma, it admits a maximal element (τ, z). If τ = -∞, then the proof is concluded. We now assume that τ = -∞. We have to prove that z(τ +) = π(∞). Assume to the contrary that z(τ +) > π(∞). Then, by Proposition 24 there exists some (τ , ẑ) in E such that τ < τ and ẑ|(τ,T ] ≡ z. This contradicts the maximal feature of (τ, z) in E. The proof of Theorem 7bis is completed.

Appendix B : Proofs of Propositions 15 and 16 and Corollary

17.

The proofs of Proposition 15 and 16 are based on the following corollary of item 2 of Proposition 23, which obviously holds under Assumption (Hπ) and Remark 5.

Corollary 25 Let η ∈ (-∞, T ] and let L be a compact subset of [a, ∞). Then, there exists some K η,L > 0 such that : if for some r, r ∈ [a, ∞) and some t ∈ [η, T ] the functions w r and w r both map

We will also use the following result on the increasing feature of the maximal solution of BW (r) with respect to the terminal condition r.

Proposition 26 Fix r and r in [a, ∞) such that r < r. Then, for all t ∈ [max{τ (r), τ (r )}, T ), we have : w r ≤ w r on ( t, T ].

Proof. For ease of notation, we shall write w (resp. w ) for w r (resp. w r ). Assume to the contrary that the set {t ∈ ( t, T ] | w(t) < w (t)} is not empty. Then, by continuity of w and w , and since w(T ) = r > r = w (T ), we have

By definition of w and w , we see that, for all t ∈ ( t, σ],

e -λt π(w(t)) = e -λσ π(w(σ)) -

The proof is completed by using (58) and recalling that τ (m) > 0 and that lim r→∞ δ(r) = ∞ by Remark 5. We now prove (57). Observe that, by Proposition 15 and Proposition 26, the function r -→ w r (τ (m)) is nondecreasing on [a, m). We therefore have

We shall prove that M = ∞. Assume to the contrary that

In order to have a contradiction, we shall find some r ∈ τ -1 (R -) such that r < m and w r (τ (m)) > M.

First notice that, since 0 > τ (m) > -∞, we have, by Theorem 7, w m (τ (m)+) = ∞. Therefore, by continuity, there exists some t M ∈ (τ (m), T ] such that

Let K K τ (m),[a,2 max{M,m}] be given by Corollary 25 and define By (60), we therefore have w r (t M ) ≥ 2 max{M, m} + K(r -m).


Elyès Jouini
email: jouini@ceremade.dauphine.fr

Clotilde Napp

Ceremade

A class of models satisfying a dynamical version of the CAPM

Keywords: CAPM, CCAPM, market beta, equilibrium, …nancial markets. JEL numbers D50, G12. y

Under a comonotonicity assumption between aggregate dividends and the market portfolio, the CCAPM formula becomes more tractable and more easily testable. In this paper, we provide theoretical justi…cations for such an assumption.

Introduction

In this paper we provide a dynamical version of the CAPM, i.e., an equilibrium model, analogous to the CCAPM, replacing in the resulting formula aggregate consumption by the market portfolio. Indeed, we show in a continuous time framework that the instantaneous rate of return of any security in excess of the riskless interest rate is a multiple common to all securities of the "instantaneous covariance" of the 1 return with the market portfolio changes. In order to derive such a dynamical version of the CAPM, we need an important assumption: the stocks prices are nondecreasing functions of the dividends. The main contribution of this paper is to prove that for almost all the classical utility functions, there exist equilibria satisfying this monotonicity assumption.

The main advantage of our approach is that in order to estimate the risk premium of any asset we do not need to observe the aggregate consumption process, which, as underlined by Breeden, Gibbons and Litzenberger (1989), is a delicate issue. Furthermore, as underlined by [START_REF] Mankiw | Risk and return : Consumption beta versus market beta[END_REF] "a stock's market beta contains much more information on its return than does its consumption beta".

The model

We …x a …nite-time horizon T > 0 and we set T [0; T ]. As usual, we let ( ; F; P) denote a …xed probability space on which is de…ned a k-dimensional Brownian motion W = W 1 t ; :::; W k t t2T and a P-augmentation of the natural …ltration generated by W denoted by (F t ) t2T .

We consider a complete …nancial market with (k + 1) securities whose prices are denoted by S 0 ; S 1 ; :::; S k .

Each security S i is in total net supply equal to i with 0 = 0 and we denote by D i the dividends process associated to asset S i for i = 1; :::; k; and by D = P k i=1 i D i the aggregate dividends process. We assume that

dS 0 t = S 0 t r t dt, S 0 0 = 1 (2.1)
dS i t = S i t i t i t dt + i t dW t for i = 1; :::; k; (2.2)
where the real valued processes r, i (the dividend yield process); i as well as the (1 k) matrix valued process i are progressively measurable and uniformly bounded. We assume that for all t in T;

( t ) 1 = i j (t) 1
1 i;j k exists and is uniformly bounded. The k -dimensional process can then be de…ned by:

t ( t ) 1 [( t r t 1 k )]
where = 1 ; :::; k and where 1 k denotes the k-dimensional vector whose every component is one.

We let for all t 2 T,

M t E t ( ) exp n R t 0 ( s ) dW s 1=2 R t 0 k s k 2 ds o .
The market is complete, so we can apply the results of [START_REF] Huang | An Intertemporal General Equilibrium Asset Pricing Model: The Case of Di¤usion Information[END_REF] and our economy can be supported by a representative agent, endowed with one unit of the market portfolio N P k i=1 i S i and whose preferences for consumption and terminal wealth are characterized by

U (c; X) = E " Z T 0 u (t; c t ) dt + V (X)

#

where the utility functions u and V satisfy the following assumption.

Assumption (U)

The function u : T R + ! R is continuous. The function V : R + ! R as well as for all t 2 T; u (t; ) are C 3 on (0; 1), stricly increasing, strictly concave and satisfy inf x u c (t; x) = inf x V 0 (x) = 0 . The function

I u : T R + ! R is of class C 1;2
, where I u (t; ) denotes the generalized inverse of u c (t; ).

Recall that an equilibrium consists of a collection S 0 ; :::; S k ; (b ; b c) such that (b ; b c) is an optimal investment-consumption process for the representative agent as de…ned in [START_REF] Merton | Optimum consumption and portfolio rules in a continuous time model[END_REF] and such that markets clear, i.e., b c t = D t and b i t = i S i t for all t 2 T. At the equilibrium the optimal consumption process of the representative agent must equal the aggregate dividends process, and we deduce from e.g. [START_REF] Karatzas | Optimization problems in the theory of continuous trading[END_REF] that a necessary and su¢ cient condition for an equilibrium to be reached is that there exists > 0 such that

t M t = u c (t; D t ) ; t 2 T (2.3) T M T = V 0 (N T ) (2.4)
where t 1=S 0 t and with no additional condition because in this case, the budget constraint, i.e. This means that it su¢ ces to observe e.g. an index (seen as a proxy of the market portfolio) in order to estimate the risk premium of any asset.

E h R T 0 t M t D t dt + T M T N T i = N 0 ,
In the next, we prove the existence of equilibria satisfying Assumption (N) for a large class of utility functions.

Existence of equilibria satisfying Assumption (N)

More precisely, we shall establish that for a given choice of utility functions and of risk level, there exists an equilibrium satisfying Assumption (N).

Proposition 1. Let be given satisfying the regularity assumptions of Section 2. For the following choices of (u; V ), there exists an equilibrium satisfying Assumption (N) :

1. Logarithmic utility function: u (t; c) = exp ( t) log (c + ), V (x) = exp ( T ) log (x + 1), n (t; x) = (1= ) x for some positive constant . 1) x , n (t; x) = (1=a) x for some positive constant a.

Power utility function:

u (t; c) = c for 2 ]0; 1[, V (x) = a (
3. Exponential utility function: u (t; c) = 1 exp ( c), V (x) = (1+x) , n (t; x) = exp fx= (1 )g 1 for some 2 ]0; 1[.

Proof

An equilibrium is characterized by (2:3) and (2:4). We proceed as in Karatzas-Lehoczky-Shreve (1990) and by Itô's Lemma, we obtain that (2:3) and (2:4) as well as Assumption (N) are satis…ed if there exists a function d of class C 1;2 such that for all t, d x (t; ) > 0 and

8 > > > > > > < > > > > > > : N (t) h 1 + fx(t;Nt) f (t;Nt) N t i = ft(t;Nt) f (t;Nt) + fx(t;Nt) f (t;Nt) h d (t; N t ) k N (t)k 2 N t i 1 2 fxx(t;Nt) f (t;Nt) k N (t)k 2 N 2 t : i (t) r t = fx(t;Nt) f (t;Nt) N t i (t) ( N (t)) d (T; N T ) I u [T; V 0 (N T )]
where f (t; x) u c (t; d (t; x)) =u c (0; d (0; N 0 )).

In our examples, solutions are given by

1. i t = + Nt Nt+1 i (t) ( N (t)) ; r t = ; d (t; x) = x 2. i t = r t + (1 ) i (t) ( N (t)) ; r t = 1 2 ( 1) k N (t)k 2 + a( 1) ; d (t; x) = ax 3.

> > <

> > :

N (t) = ( 1) 2 log(1+Nt) 1+ Nt ( 1) Nt 1+ Nt k N (t)k 2 h 1 + 1 2 ( 2) Nt (1+Nt) i r t = 1 2 ( 1) (Nt) 2 (1+Nt)(1+ Nt) k N (t)k 2 ( 1)
2 log(1+Nt)

(1+ Nt) ; d (t; x) = (1 ) log (1 + x)
which completes the proof.

Proposition 2. If is given satisfying the regularity assumptions of Section 2 and if x V 00 (x) V 0 (x) and x 2 V 000 (x)

V 0 (x)
are bounded, if V 0 (x) uc(t;x) is bounded in a neighborhood of 0 and if inf x x V 00 (x) V 0 (x) > 1, then there is an equilibrium satisfying Assumption (N).

Proof Immediate using the proof of Proposition 1 and taking d (t; x) = I u [t; V 0 (x)] and

r t = V t h 1 + V 00 (Nt) V 0 (Nt) N t i V 0 (N t ) V (N t ) N t k N (t)k 2 i t = r t + V 00 (N t ) V 0 (N t ) N t i t ( N (t))
where

V t = V 00 (Nt) V 0 (Nt) N t k N (t)k 2 + V 00 (Nt) V 0 (Nt) I u [t; V 0 (N t )] 1 2 V 000 (Nt) V 0 (Nt) k N (t)k 2 (N t ) 2 .
We have so far obtained existence of an equilibrium satisfying Assumption (N) when is given.

Another way of modelling the introduction of uncertainty is to replace uncertainty on the securities price process by uncertainty on the dividends process.

Proposition 3. Let D ; u and V be given.

1. If D = D for = 1 ; :::; k in R k , if V ( ) =
n (T; x) = I V [u c (T; x)].
This is in turn equivalent to the following equation

Z t (t; x) = Z xx (t; x) + F (t; x) with terminal condition Z (0; x) = H (x) (u c I V u c ) T; exp 1 p 2 k k x where Z (t; x) (nu c ) [T t; exp ( x + t)], F (t; x) exp ( x + t) u c (T t; exp ( x + t)), = 1 p 2 k k and = 2
The solution of this partial di¤erential equation (see [START_REF] Cannon | The one-dimensional heat equation[END_REF]) is given by

Z (t; x) = Z +1 1 K (t; x y) H (y) dy + Z t 0 Z +1 1 K (t ; x y) F ( ; y) dyd (3.2)
where

K (t; x) = 1 p 4 t exp x 2 4t .
By di¤erentiation, we obtain Z x (t; x) = K (t; ) H 0 (x) + Z t 0 K (t ; ) F x ( ; ) (x) d .

We check then that under our conditions we have H 0 (x) > 0 and F x (t; x) 0. Hence Z x (t; x) > 0 which implies that n x (t; ) > 0. This ends the proof of 1.

If we do not impose a particular choice for D ; it su¢ ces to impose n (t; ) = I V (u c (t; )) in the previous partial di¤erential equation and to choose D accordingly.

Conclusion

We proved that there exist equilibria satisfying our comonotonicity condition for a large class of utility functions. In such equilibria, the state price density M t decreases with the market portfolio value and the consumption beta is replaced by the market beta in the CCAPM formula. These two features are consistent with empirical evidence as underlined by Aït-Sahalia and Lo (2000) for the …rst one and by [START_REF] Mankiw | Risk and return : Consumption beta versus market beta[END_REF] for the second one.

  is automatically binding.

	This leads to the following classical CCAPM formula, where the market clearing conditions permit to
	replace aggregate consumption by aggregate dividends

i (t) r t = t i (t) ( D (t)) for all t 2 T and for all i = 1; :::; k (CCAPM) where t = u cc (t; D t ) =u c (t; D t ) and dD t = D (t)dt + D (t)dW t for all t 2 T. The market portfolio value satis…es a stochastic di¤erential equation of the form dN t = N t (( N (t) N (t)) dt + N (t)dW t ) : When the process N can be written in the form N t = fn (t; D t ) ; t 2 Tg where n : T R + ! R + is of class C 1;2 with n x (t; x) > 0 (Assumption (N)), then Itô's Lemma gives us the following Dynamical CAPM formula i (t) r t = t i (t) ( N (t)) for all t 2 T and for all i = 1; :::; k: (Dynamical CAPM)

  u (T; ) and if ũ : x 7 ! xu c (t; x) is increasing, then for all dividend process D = fD t ; t 2 Tg of the form dD t = D t dt + D (t)dW t , for 2 R, there In the general case, for all utility functions u and V; there exists D such that N t = n (t; D t ) for n (t; ) = I V [u c (t; )] and the equilibrium conditions (2:3) and (2:4) are satis…ed.

	2. Proof	We must have n (T; x) = I V [u c (T; x)]. Adopting the same approach as in the proof of
	Proposition 1, it is easy to obtain that there is an equilibrium with N t = n (t; D t ) if n satis…es the
	following partial di¤erential equation
		n t (t; D t ) + n x (t; D t ) D (t) +	1 2	n xx (t; D t ) k D (t)k 2 + D t A t n (t; D t )
		=	u cc (t; D t ) u c (t; D t )	n x (t; D t ) k D (t)k 2 ,	(3.1)
	for A t	1 uc(t;Dt) u
	exist price processes S 0 ; :::; S k such that the equilibrium conditions (2:3) and (2:4) are satis…ed as
	well as Assumption (N).

ct (t; D t ) + u cc (t; D t ) D (t) + 1 2 u ccc (t; D t ) k D (t) k 2 and with terminal condition


Thermal aware scheduling on distributed computing water heaters

Issam Rais, Eddy Caron and Laurent Lefevre Inria -LIP Laboratory, ENS, University of Lyon, France {Issam.Rais, Eddy.Caron, Laurent.Lefevre}@inria.fr

Abstract-Combining computation and water warming is an efficient alternative for recycling dissipated energy. Based on a set of Computing Water Heaters, used as a distributed super computer, we introduce a computing water heater scheduling system that has been validated by simulation on extreme workloads. These simulations lead us to reach an approximate CPU usage of about 60% while minimizing the thermal resistance usage to 10% (only when necessary) on a complex steady state.

I. INTRODUCTION

Energy consumption is a real limiting factor for large scale data centers and supercomputers [START_REF] Koomey | Implications of historical trends in the electrical efficiency of computing[END_REF]. For the cooling of such systems, generated heat must be processed through air, water or free cooling facilities. At the same time, there exists a real need for hot water which represents a huge part of consumed energy by households. The idea here is to combine the need of water warming and the need to compute. The proposed architecture explores potential energy reduction by supporting a set of distributed computing water heaters (CWH), recovering dissipated energy from computation directly into hot water for household usage.

We collaborate with a CWH industrial provider (the deFab company) who needed a scheduling framework. 1 Computing needed by a customer has to be efficiently delivered and scheduled by the "deFab Platform" to one of the machine composing the distributed set of computing water heaters.

The main contributions of this paper consists in:

1) choosing a coherent scheduling architecture compatible with a set of distributed computing entities and with the needs of the multiple users of the system; 2) proposing adapted multi level scheduling policies and heuristics while properly fulfilling constraints imposed by a set of computing water heaters; 3) studying crippling and steady state cases that could happened in such infrastructures through simulation.

II. COMPUTING WATER HEATERS (CWH)

A Computing Water Heater (CWH) differs in many ways from a Traditional Water Heater (TWH) (Figure 1). A TWH is an entity composed of a tank of water built together with a thermal resistance and respecting standards for inputting and outputting water. The first valve represents the arrival of water. Once a water pipe is connected, cold water can be inputted. The second one is for the output pipe. Once the water has been heated, through the usage of the thermal resistance, hot water can be outputted for common usage.

As shown in Figure 1, a CWH differs from a TWH in an architecture level. In fact, a CWH is built with a Computing Unit (noted C.U. on Figure 1) which warms the water. This CU is connected to the Internet to receive computational tasks.

B. Water life cycle

Globally, a TWH has only one goal: heat water to reach a target temperature set at a factory level. When hot water is requested and delivered, cold water is inputed, making the temperature of water in the tank drop and at the same time, leaving the tank at the same volume of water. This newly inputted cold water will be heated by the thermal resistance and outputted when someone will ask for hot water.

A CWH has approximately the same water life cycle as a TWH. It differs in the way of heating water. Instead of using only the thermal resistance, a CWH could use dissipated energy from the CU. This entity is composed of servers connected like a cluster, supervised by a machine delivering work. The CWH usage is specific:

• Availability of a CWH is based on the temperature of water in the tank. So the availability of a CWH is true if and only if temperature of water is below the target temperature.

• When a CWH reaches its target temperature, it stops computing, making the components of the CU unavailable to computation. • Such a CWH is available again if and only if a water usage is witnessed and that the temperature of water decreases. Several crippling cases can be encountered :

• Starvation : This case occurs when there is not enough work to reach the target temperature of every CWH by using only the CU • Saturation : When a CWH reaches its target temperature, it stops requesting work. If no water usage is made, the temperature will remain the same, placing the CWH in a saturation state

III. A SCHEDULING ARCHITECTURE FOR CWH A. Scheduling architecture description

A multiple hierarchical scheduling is proposed. Every scheduling point only knows the level underneath. The pull/push approach and the multiple hierarchical scheduling has been designed to lighten the decision making process of every decision making point in the architecture.

A customer based system implies an online scheduling. We consider here that all the incoming applications will be based on embarrassingly data parallelism with no constraints nor communications between tasks.

A scheduler manages a fixed number of machines. The scheduler asks work through a pull action. It is the one that delivers the work to the CUs. So a scheduler only communicates with the metaScheduler and the CUs under its management. The metaScheduler is the entity that answers the pull and stores work delivered by the customer. Due to the distributed aspect of the set of machines, the life cycle of these tasks has to be described.

B. The computing life cycle

Fig. 2: The deFab hierarchical architecture From the "deFab Platform" to CUs on CWH, Figure 2 presents the proposed architecture applied to the deFab infrastructure.

A user sends a request with all his needed computation. This computation is known and supported by the deFab infrastructure. The "deFab platform" determines the needed information for the metaScheduler concerning this application.

A Computing Water Heater Scheduler (CWHS) is the one to start communication. In function of temperature of water in the tank and in function of the computation power owned by the CWH, a "Pull" request will be delivered. This pull contains a number of FLOP for each core in the CWH. The number of FLOP asked represents the needed work to reach the target temperature of the CWH.

When the CWHS receives a task (arrow 2), and if there is some unused CU under its command, it sends it to the free unused CU as shown on the arrow 3. If there is no free CU, the received task is put on a queue (position of the arriving task in the queue depends on the chosen policy). Thus, the CWHS is also the one to deliver work to the CUs existing on the CWH.

The metaScheduler receives work from customers. Users upload needed files for their computation and applications are already deployed on every machine. It also is the one answering the pull requested by one of the CWHS from the deFab infrastructure. When the metaScheduler receives a pull, it tries to answer it as soon as possible. If there is enough work to answer to this request, metaScheduler starts to send chosen tasks.

The application model is based on the Expected Time To Compute Model [START_REF] Ali | Task execution time modeling for heterogeneous computing systems[END_REF]. We consider that the FLOP estimation is perfect (because the deFab infrastructure allows only known application to run, so we consider that the applications are very well known).

The supported scheduling is non preemptive and a task cannot leave the concerned CWH until it has finished its execution. Solutions like migrating extra tasks could be applied, while implying other problematics like good prediction of water usage.

IV. METASCHEDULER POLICIES

Two heuristics were developed for the metaScheduler : 1) First Requested First Served policy (FRFS): The goal of this heuristic is to maximize correspondence between the pull and the answer to this pull. Thus, maximizing the usage of CU on the CWH to reach the target temperature. This heuristic is also aiming for a maximization of priority so financial gain. When a pull is received, an immediate response is made, considering only available work. This heuristic is not fair when there is starvation. In fact, if there is enough work only for one CWH, all work will be delivered to the first puller while other won't receive any work.

2) Resource and Load Aware Policy (RLA): This heuristic allows the metaScheduler to be fair between all CWH. It also maximizes the usage rate of every core between every CWH. At the same time, it permits minimizing the inactivity of a CWH compared to another one. It relies the hypothetical temperature(the temperature that will reach the water in the tank if all the received task are executed if no water usage is seen) of every CWH to fairly give work to the remaining CWH.

V. RESULTS AND VALIDATION

We study several cases that could append during a life cycle of a set of CWH (steady state, saturation and starvation). The SimGrid framework [START_REF] Casanova | Simgrid: a sustained effort for the versatile simulation of large scale distributed systems[END_REF] was chosen as the base for the simulator.

The green curves of all the following graphs represent the accumulated power used by all components present on the CWH (CPUS and the thermal resistance). The black curves represents the progression of the temperature of the water in the CWH.

In the figures, the red part represents the usage, in percentage, of the CPUs during the simulation for a specific CWH. While the blue part represents time, in percentage, when the resistance is on. Finally, the green part represents the part of the time when the CPUs are waiting for work.

A. Starvation and saturation: complex scenarios

The following section presents how the implemented heuristics behave concerning complex scenarios (Section II).

1) First Requested First Served (FRFS) policy under starvation: The results shown in this section are made with a set of 10 CWH. Each one has 10 CPUs. The distributed platform has to perform a set of computing works.

Two metrics (power and temperature) from selected CWH are exposed. CWH 0 (Figure 3) is not getting any work from the metaScheduler and uses the thermal resistance to warm water, while CWH 2 (Figure 3) is getting work from metaScheduler and warms water through the CPU usage. FRFS properly pulls from CWH 1 and 2. Considering that these CWH are asking for enough FLOP, the metaScheduler is delivering all the pending work. All other CWH will be in a starvation state, and must heat their water using the thermal resistance, like CWH 0 (Figure 3). Figure 4 shows that only 2 out of 10 CWH received work.

2) Resource and Load Aware Policy (RLA) under starvation: This validation considers the same distributed infrastructure of CWH (10 CWH with 10 CPUs).

Figure 5 represents the CWH 2 behavior with RLA policiy, which benefits from CPU usage to warm water. But after second 6000, the thermal resistance is activated due to the lack of work left on the metaScheduler.

C W H _ 0 C W H _ 1 C W H _ 2 C W H _ 3 C W H _ 4 C W H _ 5 C W H _ 6 C W H _ 7 C W H _ 8 C W H _ 9
Percentage Usage Idle Resistance Fig. 6: CWH behavior during starvation (RLA Policy)

Figure 6 shows a balanced and fair distribution of work between Computing Water Heaters. Actually, the usage of the CPU is around 75% for every CWH.

It should be noted that under the same work demand and with the same set of CWH, FRFS finishes the simulation at t = 45000 while Resource and Load Aware Policy finishes the simulation at t = 7000. In other words, a reduction of about 85% of the completion time is observed with RLA under starvation.

3) Saturation: This section presents a simulation of 3 CWH, with a work demand superior to what it is needed to reach the target temperature of every CWH. Resource and Load Aware Policy is the chosen heuristic.

Figure 7 shows the evolution of temperature and power used during the simulation of CWH 0. A peak on the temperature curve shows a water usage on the CWH. The saturation state is reached around t = 70000. Some water usages occur This behavior can be observed on Figure 7, by analyzing the oscillations of power when the target temperature is reached. When all CUs received work, power used is then around 1000W (all CUs at work). A peak is then observed. It is the thermal resistance that is triggered to reach the target temperature.

Even with a CWH receiving enough work to reach the target temperature only with CPU usage, we can observe unequal CPU usage (Figure 8). It shows that the results concerning the usage of CPUs depends not only on the type of workLoad or the chosen heuristic, but also on the water usage.

B. Steady state: experimental validation scenario

The studied steady state presents a work demand representing an online work arrival for a set of CWH over 2 weeks.

Every CWH follows a realistic set of water usage which represents an usual water usage of a French household over a period of two weeks.

Figure 9 represents the evolution of temperature and power consumed by the representative CWH 9.

This figure shows that the heating of the set of CWH is achieved by thermal dissipation from CPUs. Besides the first two reach of the target temperature, the instantaneous consumption is around 1000W (10 × 100W , 100W is the consumption of a CPU at full power). When the CWH almost reaches target temperature, peaks are observed. It corresponds to a water usage not important enough to deploy a task on a CPU without exceeding the target temperature. Figure 10 shows a fair distribution of work. One can also note that a high rate of idle exists. It can be explained by the fact that a CWH doesn't ask work when its target temperature is reached. The part of resistance usage can be interpreted by the fact that the heuristics implemented on the CWHS chooses not to deliver a task when the execution of this task will make the water reach a temperature higher than the target temperature.

VI. CONCLUSION AND PERSPECTIVES

In this work, we established an efficient and adapted scheduling system for a distributed infrastructure of computing water heaters. We implemented the scheduling architecture and relevant heuristics on a SimGrid based simulator.

The results obtained with various extreme and steady workloads, show that the chosen architecture and heuristic are compatible and efficient with the requirements of a set of computing water heaters. Thus, on a steady state, we reached an approximate CPU usage of about 60% and minimized the thermal resistance usage (10%, only when no computation is available on metaScheduler).

Fig. 1 :

 1 Fig. 1: from TWH to CWH

Fig. 3 :

 3 Fig. 3: Warming water of CWH 0 (through thermal resistance) and CWH 2 (through CPU usage)

Fig. 4 :

 4 Fig. 4: CWH behavior during starvation (FRFS policy)

Fig. 8 :

 8 Fig. 7: CWH 0: with water usage

Fig. 9 :Fig. 10 :

 910 Fig. 9: Life cycle of a CWH : temperature and energy consumption

This work is supported by the ELCI FSN project ("Fonds pour la Société Numérique") project that associates academic and industrial partners to design and provide software environment for very high performance computing. We thank B. Laplane from the deFab company for his technical support.