OPTIMAL POLICIES FOR A DETERMINISTIC CONTINUOUS-TIME INVENTORY MODEL WITH SEVERAL SUPPLIERS

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https://doi.org/10.1051/ro/2020037 www.rairo-ro.org

OPTIMAL POLICIES FOR A DETERMINISTIC CONTINUOUS-TIME INVENTORY MODEL WITH SEVERAL SUPPLIERS

Lakdere Benkherouf

^1,∗

and Brian H. Gilding

²

Abstract. This paper is concerned with determining the optimal inventory policy for an infinite- horizon deterministic continuous-time continuous-state inventory model, where, in the absence of intervention, changes in inventory level are governed by a differential evolution equation. The decision maker has the option of ordering from several suppliers, each of which entails differing ordering and purchasing costs. The objective is to select the supplier and the size of the order that minimizes the discounted cost over an infinite planning horizon. The optimal policy is formulated as the solution of a quasi-variational inequality. It is shown that there are three possibilities regarding its solvability: it has a unique solution that corresponds to an (s, S) policy; it does not admit a solution corresponding to an (s, S) policy but does have a unique solution that corresponds to a generalized (s, S) policy; or, it does not admit a solution corresponding to an (s, S) policy or a generalized (s, S) policy. A necessary and sufficient condition for each possibility is obtained. Examples illustrate their occurrence.

Mathematics Subject Classification. 90B05.

Received November 2, 2019. Accepted April 4, 2020.

1. Introduction

Inventory control is concerned with the efficient management of stock of diverse items that are used in a production process or the provision of a service. Such items may include raw materials, components for work in progress, finished products, and consumables. The capital tied up with stocking such items is large enough to call for the application of quantitative methods of management. Holding costs influence the balance between variation in supply of the stock and demand.

A policy in inventory control that has proven to be practical and effective, and has stood the test of time, is the (s, S) policy. This policy sets two levels of stock denoted bysandS > s. The policy recommends ordering to bring the level up toS whenever it drops to s, and not replenishing when it exceedss. Such a policy was introduced in [1]. A policy of this type was shown to be optimal in each period of a multi-period dynamic inventory problem with stochastic demand if the holding and shortage costs are linear functions of the level of stock, and, in a single period problem if the expected cost function is a concave function of the level of stock, in the well-known paper [21]. Assumptions in this paper regarding the differentiability of certain cost

Keywords.Optimal inventory policy, quasi-variational inequality, (s, S) policy, generalized (s, S) policy.

1 Department of Statistics and Operations Research, College of Science, Kuwait University, P.O. Box 5969, Safat 13060, Kuwait.

2 Department of Mathematics, College of Science, Kuwait University, P.O. Box 5969, Safat 13060, Kuwait.

∗Corresponding author:lakdere.benkherouf@ku.edu.kw

Article published by EDP Sciences c EDP Sciences, ROADEF, SMAI 2021

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functions were subsequently relaxed in [28]. An (s, S) policy was further shown to be optimal in an infinite- horizon dynamic inventory model in [13]. All of the aforementioned results were generalized and shown to be valid under alternative conditions in [25]. For a discrete-time stationary inventory model, an (s, S) policy was shown to be optimal using a quasi-variational inequality (QVI) in [3]. Similar results confirming the optimality of an (s, S) policy using a QVI were obtained in [4,7].

A variation on the (s, S) inventory policy is the (s, S, T) periodic review inventory policy, which was introduced in [11]. Incorporating a cost as the consequence of a review, the policy is to bring the level of stock up toS > s when it is sor less, and take no action otherwise, with periodT. For practical applications, particularly those involving a stochastic demand that is expensive to monitor, such a policy has operational benefits. Policies of this type are studied in [14] and references cited therein.

Another variation on the (s, S) inventory policy is what has become known in the inventory control literature as the generalized (s, S) policy. It involves several suppliers, and stock levelssN < sN−1 <· · · < s1 < S1 <

S2 <· · · < SN for some natural number N. If the level of stock is greater than s1 one does not replenish it, if it is between s₂ and s₁ one orders from supplier 1 to complete the stock to level S₁, and so on, down to a level of stock betweens_N ands_N₋₁where one orders from supplierN−1 to bring the stock up to levelS_N₋₁. If the stock level is less than sN one orders from supplier N to complete the stock to level SN. The policy may intentionally exclude further available suppliers. Introductions to generalized (s, S) policies may be found in [5,19].

Although a generalized (s, S) policy seems to be very practical in real life, there are few research articles on the topic. A discrete-time stochastic inventory model with an ordering-cost function that is concave is considered in [17,18]. A dynamic model with two suppliers, one with a high purchasing cost and no setup cost, and the other with a low purchasing cost and a fixed ordering cost, is examined in [9]. With a finite and with an infinite horizon, the existence of a generalized (s, S) policy is shown under the assumption that the probability density function of the demand is log-concave. This assumption is met by many probability distributions, including the normal and gamma distributions. More recently, the optimal policy for the model treated in [17] with a general demand distribution has been shown to be a generalized (s, S) policy for all but a finite interval of inventory levels, in [2].

In the present paper, an inventory manager is faced with the problem of determining an optimal replenishment policy for a continuous-time continuous-state deterministic inventory model with a single item, where, in the absence of intervention, the changes in the inventory level are modelled by a differential equation, and where the control policies are of an impulse control type. Replenishment is offered by a number of suppliers, each of which charges differently. The objective is to find an admissible control that minimizes the overall inventory costs over an infinite planning horizon. For a single supplier model the optimal policy is of (s, S) type. However, when several suppliers are available, the problem is far from straightforward. As it turns out, there are three possibilities. Either there is a unique optimal policy that is an (s, S) policy, there is a unique optimal policy that is a generalized (s, S) policy but not an (s, S) policy, or, an optimal policy is neither of these. The technical machinery employed to solve the problem is based on QVI techniques. A model related to that studied in the present paper has been investigated in [23]. For further information on QVIs, the reader is referred to [6,8].

Our results differ from those in [12,15,16,26,27]. These papers deal with the long-term average cost criterion for continuous-time models, and indicate that under the respective circumstances an (s, S) policy is optimal.

This is not the case, in general, for the infinite-horizon discounted model considered in the present paper.

The next section contains the formal statement of the inventory control problem as well as the QVI formula- tion. The derivation of the optimality of the (s, S) policy for a single supplier is reviewed in Section3as a prelude to the main contribution of the paper. The analysis of the problem with several suppliers is subsequently to be found in Section4. Examples4.22–4.24provide concrete illustrations of the three possible outcomes alluded to above. A conclusion and some general remarks are provided in Section5.

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2. Problem statement

Consider a stock of a single item with J possible suppliers. The level of stock at time t is given by x(t).

A level x≥0 corresponds to the number of items held. A levelx <0 indicates a shortage of −xitems. In the absence of intervention, changes inxare governed by the evolution equation

˙

x(t) =−G(x(t)), (2.1)

whereGis a positive continuous function defined onR. From a modelling point of view, the latter accounts for a stock-dependent demand rate and the deterioration of items. The forms of Gthat are commonly used have been reviewed in [10,24]. A generic expression that amalgamates these forms is

G(x) =

(g0 for x <0

g0+g1x+gβx^β for x≥0, (2.2)

whereg0>0,g1≥0,gβ≥0, and 0< β <1 are prescribed constants.

Placing an order with supplierj∈ J, where

J ={1,2, . . . , J},

entails a fixed cost kj and a cost cj per unit item. Given a choice between two suppliers j and `, for which kj < k` and cj ≤ c`, or for which kj ≤k` and cj < c`, one would obviously opt for supplier j. Also, as far as minimizing costs is involved, given a choice between two suppliers j and ` with kj =k` and cj =c`, it is immaterial which of them one chooses. So, with no loss of generality, it will be supposed that

k1> k2>· · ·> kJ>0 (2.3) and

c1< c2<· · ·< cJ. (2.4)

Under this assumption using a supplier with a higher set-up cost involves a lesser cost per item. For any supplier, this may well be the case with regard to a larger competitor whose warehouse is located further away from the customer. Whereas purchasing from the competitor requires a higher cost for transportation, say, which is then transferred into the set-up cost, the price charged per unit by the competitor could be lower.

The holding cost is given by a continuous nonnegative functionf defined onR. The classical expression is f(x) =

(−px for x <0

qx for x≥0, (2.5)

wherep >0 andq >0 are constants. This is a linear function of the inventory level for items in stock, and for a shortage of items.

Variability of money-value in time is considered by the exponential discount of costs at a constant rate

α >0. (2.6)

This method of discounting costs was proposed in [20], is an integral component of the continuous-time inventory models in [4,5,7,8,22,23], and is possibly the most popular method of discounting applied in economics.

An admissible replenishment strategy consists of a sequence

V_n ={(ti, ξ_i, j_i) :i= 1, . . . , n},

where t_i ≥0 represents the time of the ith control (intervention),ξ_i >0 is the quantity ordered at that time, andj_i∈ J labels the supplier. One may note that should one wish to order a total of Ξ items from a selection

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of suppliers I ⊆ J at the same time, then the minimum of the total cost P

j∈I(k_j+c_jξ_j) subject to the constraintsξj ≥0 for everyj ∈ Iand P

j∈Iξj = Ξ>0 is necessarily given by a combination in which ξj = Ξ for somej ∈ I, andξ`= 0 for every other`∈ I [5]. Thus, at any timeti, an order will be placed with a single supplier.

In this setting, the problem of finding an optimal impulse control policy reduces to that of determining a sequence V = lim

n→∞V_n that solves

u(x(0)) = min

V

Z ∞ 0

f(x(t)) e^−αtdt+

∞

X

i=1

(kj_i+cj_iξi) e^−αtⁱ

. (2.7)

Note that ifxsatisfies equation (2.1) then

˜ x=

Z x 0

dη G(η)

satisfies the same equation with Greplaced by the constant function 1. Concurrently, if usatisfies (2.7) then

˜

u(˜x(0)) = u(x(0)) satisfies (2.7) with f(x(t)) replaced by ˜f(˜x(t)). Conversely, by the reverse transformation, any solution of (2.7) in whichxis governed by (2.1) with G≡1 gives rise to a solution satisfying the original equation (2.1) if

Z x 0

dη

G(η) → ±∞ as x→ ±∞.

This condition is fulfilled by (2.2). Hence, with nominal loss of generality, we shall take G≡1.

With the aforementioned simplification, it will be shown that the optimal impulse control policy of (2.7) corresponds to a solution of a QVI. If no order is made within a time interval from t to t+τ where τ >0, then (2.1) withG≡1 and (2.7) lead to

u(x(t))≤u(x(t+τ)) e^−ατ+ Z t+τ

t

f(x(ρ)) e^{−α(ρ−t)}dρ

=u(x(t)−τ) e^−ατ+ Z τ

0

f(x(t)−υ) e^−αυdυ.

Hence, dropping the dependence ofxont,

u(x)−u(x−τ) +ατ u(x−τ)≤αu(x−τ) Z τ

0

1−e^−αυ dυ+

Z τ 0

f(x−υ) e^−αυdυ.

Dividing byτ, and then lettingτ →0, yields Au

(x)≤f(x), where Au

(x) =u⁰(x) +αu(x). (2.8)

On the other hand, if an order is made with supplier j at timet, then the quantity ordered should minimize kj+u(x(t) +ξ) +cjξwith respect toξ≥0. In other words, one should haveu(x(t)) = Mju

(x(t)), where Mju

(x) =kj+ min{u(x+ξ) +cjξ:ξ≥0}.

We deduce that the optimal impulse control policy corresponds to the solution of the QVI





 Au≤f u≤M u

(Au−f) (u−M u) = 0,

(2.9) in whichAis given by (2.8), and

M u

(x) = min M_ju

(x) :j∈ J . (2.10)

To show the existence of a solution of the QVI (2.9) it will be helpful to first examine the case whereJ= 1.

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3. The (s, S) policy for a single supplier

When there is a single supplier, we may identify M with M₁, and drop the subscript from k₁ and c₁. Subsequently, it can be verified that ifu satisfies the QVI (2.9), then the function ˜ugiven by ˜u(x) = u(x) + cx−c/α satisfies (2.9) with f replaced by ˜f, where ˜f(x) =f(x) +αcx, and c replaced by 0, and vice versa.

Hence, with no loss of generality, we may suppose thatc= 0.

In the expectation that the optimal inventory control policy will involve levels of stockxdefining a stopping interval in which u=M u, and levels of stock xnot necessitating replenishment defining an interval in which u < M uandAu=f, separated by some stock levels, we search for a solution of the QVI based on the following premise.

Ansatz 3.1. The solution of (2.9) is a differentiable real functionusuch thatu=M uon (−∞, s], and,Au=f andu < M uin (s,∞), for some numbers.

Lemma 3.2. Under Ansatz3.1, a solution of the QVI is given by

u(x) =

(y(s) f or x < s

y(x) f or x≥s, (3.1)

wherey is a solution of the differential equation

y⁰+αy=f in R (3.2)

satisfying

y⁰(s) =y⁰(S) = 0 (3.3)

and

y(s) =y(S) +k (3.4)

for some numberS > s.

Proof. Suppose thatuis a solution of the QVI that satisfies the ansatz. Letx < s. Then, since M u

(x) is well defined,umust have a least absolute minimum in [x,∞), atSsay. Because this implies thatu(x) = M u

(x) = k+u(S), necessarilyS > x. In turn this means that u has a least absolute minimum in [η,∞) at S for all x≤η ≤S. Whence, u(η) = M u

(η) =k+u(S) for all x≤η ≤min{s, S}, and, u(min{s, S}) =k+u(S).

It follows thatS > s, thatS is the least absolute minimum ofuon [s,∞), thatu(s) =k+u(S), and, in view of the arbitrariness ofx < s, thatuis constant on (−∞, s]. Given the differentiability ofu, the minimality of S requires u⁰(S) = 0, and the constance of u on (−∞, s] requires u⁰(s) = 0. Moreover, the condition Au=f in (s,∞) says that umust be a solution of (3.2) there. However, under the assumption thatf is a continuous function onR, any solution of the ordinary differential equation (3.2) in (s,∞) can be extended ontoR. Thus

umust be as in (3.1) withy as stated.

The following hypothesis is sufficient for the existence of a function yas indicated in Lemma3.2.

Hypothesis 3.3. The functionf is continuous on R, strictly decreasing on (−∞, γ] and strictly increasing on [γ,∞) for someγ∈R, and is coercive,i.e.f(x)→ ∞as x→ ±∞.

Lemma 3.4. There exists a unique functionϕ: (−∞, γ)→(γ,∞)such that (3.2)admits a solutiony satisfying (3.3) fors < S if and only ifs < γ andS=ϕ(s), in which casey is unique,

y⁰<0 in (s, S), and y⁰ >0 in (−∞, s)∪(S,∞). (3.5) Furthermore,ϕis continuous and strictly decreasing, and such thatϕ(s)→γ ass→γ.

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Proof. Equation (3.2) has a unique solution satisfyingy⁰(s) = 0 for anys. It is given by y(x) = e^−αx

e^αsf(s)

α +

Z x s

e^αηf(η) dη

. Hence, it is such that

y⁰(x) = e^−αx

e^αxf(x)−e^αsf(s)−α Z x

s

e^αηf(η) dη

.

Applying the formula for integration by parts of Riemann–Stieltjes integrals, this yields y⁰(x) = e^−αx

Z x s

e^αηdf(η). (3.6)

The conditiony⁰(S) = 0 consequently requires Z S

s

e^αηdf(η) = 0. (3.7)

In view of the monotonicity off, this necessitatess < γ andS=ϕ(s)> γ, where Z ϕ(s)

γ

e^αηdf(η) =− Z γ

s

e^αηdf(η). (3.8)

Because

Z z γ

e^αηdf(η)>

Z z γ

e^αγdf(η) = e^αγ{f(z)−f(γ)} for z > γ,

andf(z)→ ∞asz→ ∞, such a valueϕ(s) exists for alls < γ. The identity (3.8) leads to the properties ofϕ.

Formulae (3.6) and (3.7) together with the monotonicity off deliver (3.5).

Given (3.2) and (3.3) fors < S, the condition (3.4) is equivalent to

f(s) =f(S) +αk. (3.9)

Consequently, in the light of Lemma3.4, it can be reformulated as

F(s) =k, (3.10)

where

F(s) =f(s)−f(ϕ(s))

α =−1

α Z ϕ(s)

s

df(η) (3.11)

fors < γ. The next lemma leads to the conclusion that (3.10) has a unique solution.

Lemma 3.5. The functionF is continuous, strictly decreasing, and such thatF(s)→0ass→γ, andF(s)→

∞ass→ −∞.

Proof. The continuity ofF and its behaviour ass→γfollow from (3.11) and the properties ofϕprovided by Lemma3.4. To verify the remaining characteristics ofF, takes < r < γ. By (3.11),

F(s)−F(r) = Z r

s

e^{α{η−ϕ(r)}}−1

α df(η) +

Z ϕ(s) ϕ(r)

e^{α{η−ϕ(r)}}−1

α df(η)

+ e^−αϕ(r) α

Z ϕ(r) r

e^αηdf(η)−e^−αϕ(r) α

Z ϕ(s) s

e^αηdf(η).

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The last two terms on the right-hand side of the above equality are zero by (3.8). The first two are positive by the monotonicity of the exponential function and off. In fact, of the very first it can be said that

Z r s

e^{α{η−ϕ(r)}}−1

α df(η)>

Z r s

e^{α{r−ϕ(r)}}−1

α df(η)

=1−e^{α{r−ϕ(r)}}

α {f(s)−f(r)}.

It follows thatF(s)> F(r) for alls < r < γ, andF(s)→ ∞as s→ −∞.

The above analysis yields the following.

Theorem 3.6. Suppose that f satisfies Hypothesis 3.3. Then the QVI (2.9) has a unique solution satisfying Ansatz3.1.

Proof. Let u be given by (3.1), where y is as in Lemma 3.4. Then, u is constant on (−∞, s], and, by (3.5), strictly decreasing on [s, S], and strictly increasing on [S,∞). Forx≤sthis implies that M u

(x) =k+u(S) = k+y(S)−y(s) +u(x), for s < x < S it implies that M u

(x) = k+u(S) > k+y(S)−y(s) +u(x), and for x≥S it implies that M u

(x) = k+u(x) > u(x). Concurrently, Au

(x) =f(s) < f(x) forx < s, and Au

(x) = Ay

(x) =f(x) forx > s. So, if y(s)> y(S) +kthenu > M uin (−∞, s). Ify(s)< y(S) +k then (Au−f) (u−M u)>0 in (−∞, s). Finally, if (3.4) holds thenAu < f and u=M uin (−∞, s), and,Au=f

andu < M uin (s,∞).

The above delivers the (s, S) policy in which the optimal course of action for an inventory manager is to order up to levelS when the inventory level is sor less,i.e.the level of shortage has reached−sor more, and to freely let the inventory level decrease and shortages accrue in response to the demand otherwise.

Example 3.7. Suppose thatf is given by (2.5) withp >0 andq >0. Then Hypothesis3.3holds with γ= 0, ϕ(s) = ln

1 +p(1−e^αs)/q /α, and F(s) =− {ps+qϕ(s)}/α. It follows thatsis the unique solution of αps+qln

1 +p(1−e^αs)/q +α²k= 0 in the interval (−∞,0),

S=−(αk+ps)/q,

and the functiony in the solution (3.1) of (2.9) satisfying Ansatz3.1is given by y(x) =

(p

1−αx−e^α(s−x)

/α² for x≤0 q

αx−1 + e^α(S−x)

/α² for x >0.

The values ofsandSin the (s, S) policy naturally depend on the numberkand the functionf. The ensuing corollaries of Theorem3.6deliver the monotonicity of this dependence.

Corollary 3.8. Let(s^±, S^±)be such that equation (3.2)has a solutiony_±satisfying (3.3)and (3.4)fork=k^± with 0< k⁻< k⁺. Thens⁺< s⁻ andS⁻< S⁺.

Proof. Recalling thats^±is the unique solution of the equationF(s) =k^±fors∈(−∞, γ), and thatF is strictly decreasing on (−∞, γ) by Lemma3.5, necessarilys⁺< s⁻. SinceS^±=ϕ(s^±) whereϕis given by Lemma3.4, andϕis strictly decreasing on (−∞, γ), this implies thatS⁺> S⁻. Corollary 3.9. Let (s^±, S^±) be such that equation (3.2) with f replaced by f^± has a solution y_± satisfying (3.3) and (3.4), for functions f^± satisfying Hypothesis 3.3 with the same number γ. If f⁺ −f⁻ is non-decreasing, then s⁺ ≤ s⁻ and S⁺ ≤ S⁻ with strict inequality in each case if f⁺ −f⁻

(S⁻) >

f⁺−f⁻

(s⁺)and only if f⁺−f⁻

(S⁺)> f⁺−f⁻ (s⁻).

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Proof. Let ϕ^± denote the function given by Lemma 3.4, and F^± the function (3.11), for f^± respectively.

In addition, letS^±_∞= limη→−∞ϕ^±(η), andψ^± with domain (γ, S_∞^±) denote the inverse ofϕ^±. By (3.8), Z ϕ⁻(s)

ϕ⁺(s)

e^αηdf⁺(η) =

Z ϕ⁻(s) s

e^αηd f⁺−f⁻

(η) (3.12)

for all s < γ. So, ϕ⁺ ≤ ϕ⁻ in (−∞, γ). Consequently, S_∞⁺ ≤ S_∞⁻, andψ⁺ ≤ ψ⁻ in (γ, S_∞⁺). Now, by (3.8) and (3.11),

F⁺(s) =

Z ϕ⁺(s) s

e^α(η−X)−1

α df⁺(η) (3.13)

for anys < γ andX. The same holds with the superscript changed from + to−. Thus, takingX =ϕ⁺(s), F⁺(s)−F⁻(s) =−

Z ϕ⁺(s) s

1−e^α{^η−ϕ⁺^(s)}

α d f⁺−f⁻ (η)

− Z ϕ⁻(s)

ϕ⁺(s)

e^α{η−ϕ⁺(s)} −1

α df⁻(η). (3.14)

Sinceγ < ϕ⁺≤ϕ⁻, this implies thatF⁺≤F⁻ in (−∞, γ). In particular,k=F⁺(s⁺)≤F⁻(s⁺). Therefore, F⁻(s⁻) = k ≤F⁻(s⁺). In view of the monotonicity of F⁻, this necessitates s⁺ ≤ s⁻. On the other hand, takings=ψ^±(S) andX =ψ⁻(S) in (3.13) and its counterpart forγ < S < S⁺_∞,

F⁺ ψ⁺(S)

−F⁻ ψ⁻(S)

= Z S

ψ⁻(S)

e^α{^η−ψ⁻^(S)} −1

α d f⁺−f⁻ (η) +

Z ψ⁻(S) ψ⁺(S)

1−e^α{^η−ψ⁻^(S)}

α df⁺(η). (3.15)

Hence, F⁺(ψ⁺) ≥F⁻(ψ⁻) in (γ, S_∞⁺). As a consequence, k = F⁺(ψ⁺(S⁺)) ≥F⁻(ψ⁻(S⁺)). This implies that F⁻(ψ⁻(S⁻)) = k ≥ F⁻(ψ⁻(S⁺)), from which, in view of the monotonicity of F⁻ and ψ⁻, it follows that S⁺ ≤S⁻. It remains to establish the necessary and sufficient conditions for strictness in the inequalities s⁺ ≤s⁻ and S⁺ ≤S⁻. Suppose first that f⁺−f⁻

(S⁺) = f⁺−f⁻

(s⁻). This means that ϕ⁺ =ϕ⁻ in [ψ⁻(S⁺), γ). In particular,ϕ⁺(ψ⁻(S⁺)) =ϕ⁻(ψ⁻(S⁺)) =S⁺. Hence,s⁺=ψ⁺(S⁺) =ψ⁺(ϕ⁺(ψ⁻(S⁺))) = ψ⁻(S⁺) ≥ ψ⁻(S⁻) = s⁻. Therefore, s⁺ = s⁻. Consequently, ϕ⁺ = ϕ⁻ on [s⁺, γ), which in turn gives S⁺ = ϕ⁺(s⁺) = ϕ⁺(s⁻) = ϕ⁻(s⁻) = S⁻. Suppose next that f⁺ −f⁻

(S⁻) > f⁺ −f⁻

(s⁺). This necessitates f⁺−f⁻

(S⁺)> f⁺−f⁻

(s⁻), for if this were not the case, then by what we have just proven we would have s⁺ = s⁻ and S⁺ =S⁻, which would lead to an immediate contradiction of our supposition.

However, if f⁺−f⁻

(S⁺)> f⁺−f⁻

(s⁻), then f⁺−f⁻

(S⁻)> f⁺−f⁻

(s⁻). Hence, (3.12) gives S⁻ = ϕ⁻(s⁻) > ϕ⁺(s⁻). So, by (3.14), F⁺(s⁻) > k. Whence, s⁺ < s⁻. The inequality S⁻ > ϕ⁺(s⁻) further implies thatψ⁺(S⁻)< s⁻. Therefore, by (3.15), F⁺(ψ⁺(S⁻))> k. This gives ψ⁺(S⁻)< s⁺. Thus,

S⁺< ϕ⁺(ψ⁺(S⁻)) =S⁻.

Hypothesis3.3includes the supposition thatf(x)→ ∞asx→ ∞. The only place above where this has been used is in the proof of Lemma3.4. Careful consideration of this usage reveals the following.

Remark 3.10. Theorem3.6remains valid when the supposition that f(x)→ ∞asx→ ∞is replaced by Z ∞

γ

e^αηdf(η)≥ − Z γ

−∞

e^αηdf(η).

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Discounting future costs, the problem of finding an optimal control policy becomes that of finding a strategy to minimize the long-term average cost per unit time of holding and replenishing the inventory.

Remark 3.11. In the limit α → 0, the (s, S) policy given by Theorem 3.6 converges to the classical EOQ model with shortages.

See [5] for further details, and [22,23] for the corresponding result for related problems.

4. The generalized (s, S) policy for several suppliers

The results on the existence and uniqueness of an (s, S) policy for the inventory model with a single supplier provide a motif for the analysis of the problem with several suppliers, notwithstanding that emulation of these results turns out not to be so straightforward. We shall analyse the problem with more than one supplier in three phases. The first is to consider an ansatz that suitably generalizes the one employed for the problem with one supplier. The second is to characterize a solution satisfying this ansatz to such an extent that it becomes apparent that the QVI admits at most one such solution. The third and final phase is to prove that the characterization yields a solution of the problem, thereby obtaining an existence result to complement that of uniqueness.

4.1. The ansatz

As we shall duly demonstrate, the QVI (2.9) with several suppliers need not have a solution fulfilling Ansatz3.1. The scope for finding a solution can be widened by relaxing the assumptions within the ansatz. The following involves just a minor relaxation, with far reaching consequences for the analysis.

Ansatz 4.1. The solution of (2.9) is a continuous real function u such that u= M u on (−∞, s], and, u is differentiable,Au=f, andu < M uin (s,∞), for some numbers.

Ans¨atze3.1 and 4.1are equivalent for a single supplier, as is revealed by the next theorem, which expands on Lemma3.2.

Theorem 4.2. Under Ansatz4.1, a solution of the QVI is given by

u(x) =

(v(x) f or x < s y(x) f or x≥s,

wherey is a solution of the differential equation (3.2) satisfying

y⁰(s) =y⁰(S) =−cj (4.1)

and

y(s) =y(S) +kj+cj(S−s) (4.2)

for somej∈ J andS > s, and

v(x) = min M_`y

(s) +c_`(s−x) :`∈ J . (4.3)

Furthermore,

y(s) = M_jy

(s)< M_`y

(s) f or every `∈ J \ {j}, (4.4)

and

y(x)≤v(x) f or x < s.

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We realize the proof of the above theorem through a sequence of lemmata. Throughout,uwill be assumed to be a given solution of (2.9) that satisfies Ansatz4.1. We define

u`(x) =u(x) +c`x (4.5)

for`∈ J andx∈R, and note that M_`u

(x) =k_`−c_`x+ min{u_`(η) :η ≥x}. (4.6) Lemma 4.3. If u(x) = M_`u

(x) thenu_` has a least absolute minimum on[x,∞)at anX > x such that

u`(x) =u`(X) +k` (4.7)

and

u_`(z)≤u_`(x) f or all z≤X. (4.8)

Proof. Since M`u

(x) is well defined,u`has a least absolute minimum on [x,∞), atX say, whereby M`u (x) = k_`−c_`x+u_`(X). As u_`(x) = M_`u

(x) +c_`x, this gives (4.7). Because k_` >0, the possibility thatX =xis excluded. Finally, for any z≤X, by (2.9), (2.10), (4.6), and (4.7),u(z)≤ M u

(z)≤ M_`u

(z)≤k_`−c_`z+

u`(X) =u`(x)−c`z. Whence, (4.8).

Lemma 4.4. Suppose that u=M uin the proper interval[a, b]. Thenuis concave on [a, b].

Proof. Sinceuis continuous onR, each of the functionsu_`is uniformly continuous on [a, b]. Hence there exists aδ∈(0, b−a] such that|u`(x)−u_`(z)|< k_` for every`∈ J and allxandz in [a, b] with|x−z| ≤δ. Now, let w,x, andz be any three numbers in [a, b] with w≤x≤z≤w+δ, and`∈ J be such thatu(x) = M`u

(x).

By contrivance, u`(x)< u`(η) +k` for all x≤η ≤x+δ. Hence, the number X given by Lemma 4.3is such thatX > x+δ≥w+δ≥z. Consequently, by (4.8),

(z−x)u_`(w) + (x−w)u_`(z)≤(z−x)u_`(x) + (x−w)u_`(x) = (z−w)u_`(x).

Eliminatingu` using (4.5) yields

(z−x)u(w) + (x−w)u(z)≤(z−w)u(x).

In view of the arbitrariness ofw,x, andz, this establishes thatuis concave on any subinterval of [a, b] of length δ. Because (a, b) can be covered by a finite number of such open subintervals, the concavity on the whole of

[a, b] follows.

Lemma 4.5. Suppose that u = M u on [a, x] for some a < x. Let ` be the least number in J such that u(x) = M_`u

(x). Thenu_` is constant and u=M_`uon [w, x]for some w∈[a, x).

Proof. In the first instance, let` ∈ J be such that u(w_i) = M_`u

(w_i) for an increasing sequence {w_i}^∞_i=1 ⊂ [a, x) converging to x. By continuity,u(x) = M_`u

(x). By Lemma 4.4, u_` is concave on [a, x]. Therefore, u_` has a right derivative D⁺u` at every point in [a, x), and a left derivativeD⁻u` at every point in (a, x], with

D⁺u`

(η0) ≥ D⁻u`

(η1)≥ D⁺u`

(η1)≥ D⁻u`

(η2) for all a≤η0 < η1 < η2 ≤x. By (4.8) applied at x=w1, D⁺u`

(w1)≤0. By (4.8) applied as it is, D⁻u`

(x)≥0. Hence, D⁺u`

(η) = D⁻u`

(η) = 0 for all η ∈ (w₁, x). This implies that u_` is constant on [w₁, x]. Consequently, M_`u

(w) = k_` − c_`w + min{u`(η) :η≥w}=k`−c`w+min{u`(η) :η≥x}= M`u

(x)+c`(x−w) =u(x)+c`(x−w) =u`(x)−c`w= u`(w)−c`w=u(w) for allw∈[w1, x]. If now there is anL∈ J \ {`}for whichu(x) = MLu

(x), then by (4.8) with`replaced byL, there holds D⁻uL

(x)≥0. So,cL−c`= D⁻uL−D⁻u`

(x)≥0. This can be the case only ifL≥`. Consequently,`must be the least numberL∈ J for whichu(x) = M_Lu

(x).

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Lemma 4.6. Suppose that u = M u on [x, b] for some b > x. Let ` be the greatest number in J such that u(x) = M`u

(x). Thenu` is constant and u=M`uon [x, z] for somez∈(x, b].

Proof. The proof of this lemma is analogous to that of the previous one. We omit the details.

Lemma 4.7. Let j be the least number in J such that u(s) = Mju

(s). Then u is differentiable at s, u⁰(s) = −c_j, and, u = y in [s,∞), where y is a solution of (3.2) that satisfies (4.1) and (4.2) for some s < S.

Proof. The requirement thatusatisfiesAu=f in (s,∞) is the same as asking that it is a solution of (3.2) there.

Furthermore, sincef is continuous onR, said solution can be extended to one,ysay, onR. By Lemma4.5, there is aw < ssuch thatu⁰ =−cj in (w, s). Hence,y⁰(x) +αy(x) =f(x)≥ Au

(x) =u⁰(x) +αu(x) =αu(x)−c_j for allx∈(w, s). Passage to the limitx→sgivesy⁰(s)≥ −cj. On the other hand, by Lemma4.3,ujhas a least absolute minimumS in [s,∞) such thatS > s, and,uj(x)≤uj(s) for allx∈(s, S). The latter inequality can be reformulated asy(x)−y(s)≤ −cj(x−s). Hence, dividing byx−sand passing to the limitx→s, we obtain y⁰(s)≤ −c_j. Combining the two inequalities comparingy⁰(s) toc_jyieldsy⁰(s) =−c_j, from which it follows that uis differentiable atsandu⁰(s) =−cj. BecauseSis a minimum ofuj, andujis differentiable there,u⁰_j(S) = 0.

Inasmuch uj(x) =y(x) +cjxfor x≥s, this is equivalent toy⁰(S) = −cj. Thus we have obtained (4.1). The identity (4.2) follows from the observation that Lemma4.3additionally says thatuj(S) =uj(s)−kj. Lemma 4.8. Further to Lemma4.7,(4.4)holds.

Proof. By Ansatz3.1, u(s) = M u

(s)≤ M_`u

(s) for every ` ∈ J. Suppose now that u(s) = M_`u (s) for some ` ∈ J \ {j}. Then by (4.8) at x = s, u⁰_`(s) ≤ 0. Hence,cj = −u⁰(s) = c`−u⁰_`(s) ≥ c`. This implies that ` < j, which contradicts the definition of j as the least number`∈ J for whichu(s) = M`u

(s). Hence, necessarily,y(s) =u(s)≤ M`u

(s) = M`y

(s) for every`∈ J with equality if and only if`=j.

Lemma 4.9. There holdsy≤uin(−∞, s].

Proof. By Lemmata4.4–4.6,uis differentiable at all but a finite number of points in (−∞, s). At anη < sat whichuis differentiable,

d

dη{e^αηu(η)}= e^αη Au

(η)≤e^αηf(η) = e^αη Ay

(η) = d

dη {e^αηy(η)}.

Hence, integrating piecewise, from x < s to s, we obtain e^αsu(s)−e^αxu(x) ≤ e^αsy(s)−e^αxy(x). Inasmuch

u(s) =y(s) by Lemma4.7, the result follows.

Lemma 4.10. There holdsu=v in(−∞, s].

Proof. Letx < s. By (4.6), M`u

(x)≤k`−c`x+ min{u`(η) :η≥s}= M`u

(s) +c`(s−x) for every`∈ J. So, u(x) = M u

(x) = min M`u

(x) :`∈ J ≤ v(x). Next, let ` be the greatest number in J such that u(x) = M`u

(x). By Lemma 4.6, there is a z ∈ (x, s] such that u` is constant on [x, z]. From the concavity of u established in Lemma 4.4, it follows that u_` is non-increasing on [x, s]. Recalling (4.6), this means that u(x) =k`−c`x+ min{u`(η) :η≥s}= M`u

(s) +c`(s−x)≥v(x). Thus we have proven that u(x)≤v(x) andu(x)≥v(x). In view of the arbitrariness ofx < s, this ascertains thatu=v in (−∞, s). Lemmata4.7and

4.8giveu=v ats.

Theorem4.2is a product of Lemmata4.7–4.10.

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4.2. Uniqueness

The basic thought behind our further analysis of the inventory model with several suppliers is that there should be an optimal (s, S) policy for each of the suppliers in the absence of competitors. Thus, the functionf_j defined by

fj(x) =f(x) +αcjx for x∈R (4.9)

should satisfy Hypothesis 3.3 for some number γj for every j ∈ J. In the light of (2.4) and (2.6), this necessitates

γ₁≥γ₂≥ · · · ≥γ_J. (4.10) Subsequently, appealing to the theory in Section 3, for every j ∈ J there is a unique pair (s_j, S_j) withs_j <

γ_j< S_j such that (3.2) has a solutiony_j satisfying (4.1) and (4.2) for somes < Sif and only if (s, S) = (s_j, S_j).

Furthermore,yj is unique,

y⁰_j<−cj on (s_j, S_j), and y⁰_j>−cj on (−∞, sj)∪(S_j,∞). (4.11) Theorem4.2tells us that a solutionuof (2.9) satisfying Ansatz4.1is necessarily such thats=sj,u=yj in [s_j,∞), andy_j≤uin (−∞, s_j) for somej∈ J. Consequently, our search for a solution of (2.9) can be reduced to that of finding an appropriatej∈ J.

Note that for everyj∈ J and`∈ J, the functionyj−y`is a solution of (3.2) withf ≡0. Hence, given any ζ∈R, there holds

y_j(x)−y_`(x) ={y_j(ζ)−y_`(ζ)}e^α(ζ−x) for all x∈R, which in turn implies that

y_j⁰(x)−y_`⁰(x) =−α{yj(ζ)−y`(ζ)}e^α(ζ−x) for all x∈R. (4.12) Lemma 4.11. The number j is the greatest minimizer ofy`(ζ)with respect to`∈ J for allζ∈R.

Proof. Pick ζ∈R. Suppose thaty`(ζ)≤yj(ζ) for some`∈ J. Under this supposition, (4.12) givesy_j⁰ ≤y_`⁰ in R, with equality only ify`(ζ) =yj(ζ). This has two consequences. Firstly,

yj(S`) +y`(s`)−y`(S`) =yj(s`) + Z S`

s`

y_j⁰(η)−y⁰_`(η) dη≤yj(s`) (4.13) with equality only ify_`(ζ) =y_j(ζ). Secondly, if ` > j, theny⁰_j(S_`)≤y⁰_`(S_`) =−c` <−cj. Hence, by (4.11), S`> sj. On the other hand, if `≤j, thenS`> γ`≥γj > sj by (4.10). So,S`> sj irrespective of the ordering ofj and`. Therefore, recalling (4.6),

M`u

(s`)≤k`−c`s`+u`(S`) =yj(S`) +k`+c`(S`−s`) =yj(S`) +y`(s`)−y`(S`). (4.14) However, by Theorem4.2,

y_j(s_`)≤u(s_`)≤ M u

(s_`)≤ M_`u

(s_`) (4.15)

with equality in the second inequality only if s` ≤sj. In which case, by (4.11),−cj ≤y⁰_j(s`)≤y⁰_`(s`) =−c`. Hence, by (2.4),`≤j. In combination, (4.13)–(4.15) are compatible only if they hold with equality throughout.

Moreover, this necessitatesy`(ζ) =yj(ζ) and`≤j. We conclude that for every`∈ J either y`(ζ)> yj(ζ), or,

y_`(ζ) =y_j(ζ) and`≤j.

To further characterize a solution of (2.9), we have the next lemma.

Lemma 4.12. Suppose thatj is a minimizer ofy`(ζ) with respect to`∈ J for someζ∈R. Then there exists a sequence of numbers

S_j,j =S_j< S_j,j−1< S_j,j−2<· · ·< S_j,1

such that y⁰_j<−c` in (s_j, S_j,`), andy⁰_j>−c` in (S_j,`,∞) for every`∈ {1,2, . . . , j}.

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Proof. Pick`∈ {1,2, . . . , j}. By (4.12),y_j⁰ ≥y⁰_`inR. Together with (4.11) forj =`, this tells us thaty⁰_j>−c`in (−∞, s`)∪(S`,∞). Hence, in view of (4.11) as it stands, we can definesj,`= min

x≤sj:y⁰_j(x) =−c` ≥s`and Sj,` = max

x≥Sj:y⁰_j(x) =−c` ≤S`. Inasmuchf` satisfies Hypothesis3.3, Lemma3.4 says thaty_j⁰ <−c`

in (sj,`, Sj,`), andy⁰_j>−c`in (−∞, sj,`)∪(Sj,`,∞). The ordering of the sequence{Sj,`:`= 1,2, . . . , j}follows

from the continuity ofy_j⁰.

We are now in a position to fully identify a solution of the QVI.

Theorem 4.13. Suppose thatf` satisfies Hypothesis 3.3for every `∈ J. Then (2.9)has at most one solution satisfying Ansatz 4.1. To be more specific, if uis such a solution, then

u(x) =

(v(x) f or x < sj

y_j(x) f or x≥s_j, (4.16)

wherej is the greatest minimizer ofy`(ζ)with respect to`∈ J for allζ∈R, v(x) = min

v`(x) : 1≤`≤j , (4.17)

v`(x) =yj(Sj,`) +k`+c`(Sj,`−x), (4.18) andS_j,` is the unique number in(s_j,∞)for whichy_j⁰(S_j,`) =−c_`.

Proof. If (2.9) has a solutionusatisfying Ansatz4.1, then Theorem4.2says thats=s_j andu=y_j on [s,∞) for somej ∈ J. Furthermore,u=von (−∞, s] wherevis given by (4.3) withy=y_j. Lemma4.11identifiesjas the greatest minimizer ofy`(ζ) with respect to`∈ J for allζ∈R. By its definition,vis piecewise linear and concave onR. Moreover, (4.4) implies thatv=vj in some proper interval (w, s). Consequently,`∈ {j+ 1, j+ 2, . . . , J} plays no role in the definition of v. Therefore, (4.3) reduces to (4.17) withv`(x) = M`yj

(sj) +c`(sj−x).

By Lemma 4.12, M_`y_j

(s_j) = k_`−c_`s_j+ min{y_j(η) +c_`η:η > s_j}=k_`−c_`s_j+y_j(S_j,`) +c_`S_j,` for every

`∈ {1,2, . . . , j}. Sov` is given by (4.18) for such`.

We note thata priori, it is not possible to decide whichj∈ J may or may not minimizey_`(ζ) with respect to`∈ J for allζ∈R. In this regard the ordering (2.3) and the ordering (2.4) are opposing. This is borne out by the result below.

Proposition 4.14. Lety_±be solutions of equation (3.2)satisfying(4.1)and (4.2)withs,S,cj, andkj replaced by s^±, S^±, c^±, and k^± respectively. Define f^±(x) = f(x) +αc^±x for x ∈ R, and suppose that f^± satisfy Hypothesis 3.3with the same number γ. If c⁺≥c⁻,k⁺≥k⁻, andc⁺+k⁺> c⁻+k⁻, theny₊ > y₋ inR. Proof. Fixx∈R. By the equivalence of the QVI for a single supplier with arbitraryc and withc= 0 outlined at the start of Section3, (3.2), and (3.6),

αy_±(x)−f(x) =−y_±⁰ (x) =c^±− Z x

s^±

e^α(η−x)df^±(η).

So,

α{y+(x)−y₋(x)}=c⁺−c⁻− Z x

s⁺

e^α(η−x)d f⁺−f⁻ (η)−

Z s⁻ s⁺

e^α(η−x)df⁻(η)

= c⁺−c⁻

e^α(^s⁺^−x)− Z s⁻

s⁺

e^α(η−x)df⁻(η).

By Corollaries 3.8and 3.9, s⁺ < s⁻ < γ, which implies that the above expression is positive. Hence y₊(x)>

y−(x).

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4.3. Existence

Having shown that a solution of the QVI satisfying Ansatz 4.1 is necessarily given by (4.16)–(4.18), where j is the greatest minimizer of y_`(ζ) with respect to ` ∈ J for all ζ ∈ R, the task of proving existence is somewhat eased. It suffices to show that the function uso defined is what we seek. To accomplish this, it will be advantageous to first characterize said function in a little more detail.

Let

σ1=sj > σ2>· · ·> σN (4.19) be such thatv is affine on each of the intervals

I1= (σ2, σ1), I2= (σ3, σ2), . . . , IN = (−∞, σN) (4.20) and not on the union of any two of them. Set

M={1,2. . . , N}, (4.21)

and defineκ:M → {1,2, . . . , j}by

v=v_κ(m) on I_m. (4.22)

WhenN ≥2,uwill not be differentiable atσ_m form∈ {2,3, . . . , N}, irrespective of its differentiability at sj in accordance with Lemma 4.7. Thus,Auis not well defined there. We subsequently interpret the condition Au≤f in (−∞, s) asAu≤f in Im for every m∈ M. With this understanding, we formulate our existence result.

Theorem 4.15. Suppose that f` satisfies Hypothesis 3.3 for every ` ∈ J, and j is the greatest minimizer of y_`(ζ) with respect to ` ∈ J for someζ ∈R. Then the function u given by (4.16)–(4.18) satisfies Ansatz 4.1.

Moreover, ifj = 1, then usolves (2.9) and satisfies Ansatz3.1. On the other hand, if j≥2, thenN ≥2, and usolves (2.9)if and only if

αv_κ(m)(σm)≤f(σm) +c_κ(m) f or every m∈ {2,3, . . . , N}. (4.23) The above theorem will be proven with the aid of four further lemmata.

Lemma 4.16. Suppose that j is as stated in Theorem4.15. Then M_`y_j

(x)is well defined for allx∈Rand every `∈ J. Moreover,yj(x)≤ M`yj

(x)for allx≥sj with equality if and only if x=sj and`=j.

Proof. Fix ` ∈ J. Lemma 4.12 states that there is a number S_j,1 such that y⁰_j(η) > −c1 for all η > S_j,1. Consequently,y⁰_j(η)>−c` for all suchη. It follows thatη 7→yj(η) +c`η has an absolute minimum in [x,∞) for all x∈R. Thus M`yj

(x) is well defined for all suchx. By (4.11) with j =`, η 7→y`(η) +c`η is strictly increasing on (−∞, s`], strictly decreasing on [s`, S`], and strictly increasing on [S`,∞). Therefore,

{y`(x) +c_`x} − {y`(z) +c_`z} ≤ {y`(s_`) +c_`s_`} − {y`(S_`) +c_`S_`}=k_` (4.24) for allx≤z, with equality if and only ifx=s`andz=S`. However, by (4.12),

y_j(x)−y_j(z) =y_`(x)−y_`(z)− Z z

x

y⁰_j(η)−y_`⁰(η) dη≤y_`(x)−y_`(z), (4.25) with equality if and only ifyj(ζ) =y`(ζ) orz=x. Combining (4.24) and (4.25), we deduce that

y_j(x)−y_j(z)≤k_`+c_`(z−x) for allx≤z,

with equality if and only ifx=s`,z=S`, andyj(ζ) =y`(ζ). It follows thatyj(x)≤ M`yj

(x) for allx≥sj, with equality if and only if x =s` and yj(ζ) = y`(ζ). The latter criterion implies that yj ≡ y`. The former implies thats`≥sj. Subsequently, recalling that (4.11) withj=`holds, we havec`≥ −y_`⁰(sj) =−y_j⁰(sj) =cj. Thus, ` ≥j. However, j is the greatest minimizer of y_`(ζ) with respect to `∈ J. We are therefore forced to conclude thaty_j(x)≤ M_`y_j

(x) for all x≥s_j with equality as in the statement of the lemma.

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Lemma 4.17. Further to Lemma4.16, the functionudefined by(4.16)to(4.18)is continuous, and continuously differentiable everywhere except atσm form∈ {2,3, . . . , N}when N≥2.

Proof. Lemma4.16implies that u(s_j) = M_ju

(s_j)< M_`u

(s_j) for every`∈ {1,2, . . . , j}. (4.26) By Lemma 4.12, M_`u

(s_j) = v_`(s_j), wherev_` is defined by (4.18), for every such`. By its definition (4.17), v is piecewise linear and concave on R. Thus the numbers (4.19), the intervals (4.20), and the set (4.21) are well defined. Moreover, (4.22) defines a strictly decreasing functionκ:M → J. Noting that (4.26) implies that v=vj in I1, and that (c1−c`) (sj−x)→ −∞asx→ −∞for every`∈ {2,3, . . . , j}, necessarily

κ(1) =j and κ(N) = 1. (4.27)

Becausev=vj and thereforev⁰=−cj onI1, andy_j⁰(sj) =−cj,uis differentiable atsj. Hence,uis continuous

inR, and continuously differentiable as stated.

Lemma 4.18. There holdsv=M uon(−∞, sj].

Proof. We prove thatv=M uonI_mby induction onm. Our induction hypothesis is M`u

(σm) =v`(σm) for 1≤`≤κ(m), (4.28)

and

M u

(σm) =v(σm). (4.29)

Recalling thatσ1=sj, form= 1 (4.28) is a tautology, and (4.29) is true by Lemma4.16. Suppose that (4.28) and (4.29) are true for arbitrarym∈ M. Fixx∈Im. For 1≤`≤κ(m), resurrecting the notation (4.5),u`is non- increasing onIm. Hence, M`u

(x) = M`u

(σm) +c`(σm−x). Substituting (4.28) gives M`u

(x) =v`(σm) + c_`(σ_m−x) =v_`(x)≥v(x), with equality for`=κ(m). On the other hand, forκ(m)≤`≤J,u_`is nondecreasing on I_m. Hence, it has an absolute minimum in [x,∞) at x, or in [σ_m,∞). In the first instance, M_`u

(x) = k`+u(x). In the second instance, M`u

(x) = M`u

(σm) +c`(σm−x)≥ M u

(σm) +cκ(m)(σm−x). So by (4.29), M`u

(x)≥v(σm) +c_κ(m)(σm−x) =v_κ(m)(σm) +c_κ(m)(σm−x) =v_κ(m)(x). Therefore, in both instances, M`u

(x) ≥ v(x). Altogether this implies that M`u

(x) = v`(x) for every ` ∈ {1,2, . . . , κ(m)}, and M u

(x) = v(x). Thus, M u = v in Im, and, (4.28) and (4.29) hold with m replaced by m+ 1 when

m≤N−1.

Lemma 4.19. There holdsAu≤f in(−∞, sj)if and only if (4.23)is true.

Proof. Letm∈ M. Recalling the notation (4.9), forx∈I_m, Au−f

(x) = Av_κ(m)−f

(x) =αv_κ(m)(x)−c_κ(m)−f(x)

=αv_κ(m)(σm)−f(σm)−c_κ(m)+f_κ(m)(x)−f_κ(m)(σm).

Since σ_m ≤ s_j < γ_j ≤γ_` and f_` is strictly decreasing on (−∞, γ_`] for every ` ∈ {1,2. . . , j}, it follows that Au−f ≤0 inI_mif and only ifαv_κ(m)(σ_m)≤f_κ(m)(σ_m) +c_κ(m). However,αv_κ(1)(σ₁) =αv(s_j) =αy_j(s_j) = f(sj)−y_j⁰(sj) =f(sj) +cj=f(σ1) +cκ(1). So, the inequality can be discounted for m= 1.

Given the above four lemmata, verification of Theorem4.15is a sinecure.

Proof of Theorem 4.15. Since y_j solves (3.2), and u=y_j in [s_j,∞),Au=f there. Inasmuchj is the greatest minimizer of {y`(ζ) :`∈ J }, Lemma 4.16 states that u < M u in (sj,∞) and u = M u at sj. Hereupon, Lemma 4.17implies that u has the regularity set out in Ansatz4.1. By Lemma 4.18, u = M u in (−∞, sj].

Thus, u satisfies Ansatz 4.1 in full. Plainly, it satisfies Ansatz 3.1 if and only if N = 1. By (4.27), N = 1 if and only ifj = 1. To conclude whetherusolves (2.9), it remains to ascertain whetherAu≤f in (−∞, s_j).

By Lemma4.19, this will be the case if and only if N = 1, or,N ≥2 and (4.23) holds.

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Together Theorems4.13and4.15reveal a computational procedure for finding a solution of the QVI.

Corollary 4.20. If it exists, the unique solution u of the QVI (2.9) satisfying Ansatz 4.1 is obtainable as follows. Lets`< S` be the unique so-ordered solution pair of the simultaneous equations

Z S_` s_`

e^αηdf`(η) = 0 and f`(s`) =f`(S`) +k`, (4.30) and set

Y`=f(0) +c`− Z 0

s`

e^αηdf`(η) (4.31)

for`∈ J. Let j be the largest number inJ with the property thatY_`≥Y_j for every `∈ J. Define

Bj={fj(sj) +cj}/α. (4.32)

If j= 1then

u(x) = e^−αx (

e^αs^jf(s_j) +c_j

α +

Z x s_j

e^αηf(η) dη )

f or x≥s_j, (4.33)

and

u(x) =B₁−c₁x f or x < s_j. If j≥2, then for every `∈ {1,2, . . . , j−1}, let

B_`=k_`+{f_`(S_j,`) +c_`}/α, (4.34)

whereSj,` is the unique number in (Sj,∞) for which Z Sj,`

Sj

e^αηdf_j(η) = (c_j−c_`) e^αS^j,`. (4.35) Setσ1=sj andκ(1) =j. By induction onm≥2, defineκ(m)as the least maximizer andσmas the maximum value of B_κ(m−1)−B_`

/ c_κ(m−1)−c_`

with respect to`∈ {1,2, . . . , κ(m−1)−1}, and Tm=f_κ(m)(σm) +c_κ(m)−αB_κ(m).

If Tm<0, then (2.9) has no solution satisfying Ansatz 4.1. Otherwise, continue the induction process, ending whenκ(m) = 1andTm≥0. The number mat this stage isN. Hereupon,uis given by (4.33),

u(x) =Bκ(m)−cκ(m)x f or σm+1≤x < σm and 1≤m≤N−1, and

u(x) =B1−c1x f or x < σN.

Proof. By the theory in Section3,s`andS`satisfy (3.7) and (3.9) withfandkreplaced byf`andk`respectively for every`∈ J. In other words, they are given by (4.30). Furthermore, by (3.6),

y_`⁰(x) +c`= Z x

s`

e^α(η−x)df`(η) for all x∈R. (4.36)

So, by (3.2),y_`(0) =Y_`/α. Withj determined, (4.30) and (4.36) with`=j imply that the unique number S_j,`

in (sj,∞) for which y⁰_j(Sj,`) =−c`is given by (4.35) for every`∈ {1,2, . . . , j}. Writing (4.18) as

v_`(x) =B_`−c_`x, (4.37)