OPTIMAL POLICIES FOR A DETERMINISTIC CONTINUOUS-TIME INVENTORY MODEL WITH SEVERAL SUPPLIERS: WHEN A SUPPLIER INCURS NO SET-UP COST

. The subject is a deterministic continuous-time continuous-state inventory control model. Stock is replenished by ordering from one of a number of suppliers incurring a different cost per item and a different set-up cost. Taking the cost of procurement into account, the objective is to minimize the total discounted cost over an infinite planning horizon. The size of the order that is to be placed and the supplier with which it is to be placed are to be decided. Earlier studies of the problem have relied substantially on the assumption that the set-up cost of every supplier is strictly positive. Removing this restriction calls for a significant modification of the adopted approach. This is realized in the present study. It is shown that there is a stable unique optimal policy of a type that encompasses ( 𝑠, 𝑆 ) and generalized ( 𝑠, 𝑆 ) policies. Conditions that are necessary and sufficient for it to reduce to each of these types are established. The case of two suppliers is studied in detail, properties of the solution are investigated, numerical examples illustrating various aspects are included, and the connection with antecedent results is assessed


Introduction
An inventory can be defined as a stock of goods held for use in a production process, the provision of a service, or sale.These goods could be raw materials, components, consumable commodities, or finished products.Carrying such an inventory entails the cost of storage, handling, obsolescence, depreciation, deterioration, insurance, taxation, and miscellaneous other transactions.Not carrying an adequate inventory involves the cost of replenishing stock, lost sales, lost production, loss of good will, overtime, extraordinary administration, and other penalties that might be incurred.The goal of efficient inventory management is to balance these costs.Decisions have to be made on when best to replenish stock and the quantity of goods that should be ordered at these junctures.Mathematical modelling can assist in making these decisions [2,9,13,19,23,30,41].
The subject of the present paper is a mathematical model for an inventory comprising a single item.Shortages are simulated by the admission of negative inventory levels.The model may be categorized as deterministic, continuous-time, and continuous-state.In the absence of intervention, the inventory level decreases according to an evolution process in which changes in inventory level are governed by a differential equation.The level of stock is continuously monitored, and the cost of holding stock or maintaining a backlog is prescribed as a function of the inventory level.Stock can be replenished by ordering from several available suppliers, each of which offers an unlimited supply.Ordering from each supplier incurs a fixed cost per item and a fixed set-up cost.The objective is to minimize the total cost of procurement, holding the inventory, and permitting shortages over an infinite planning horizon.The challenges of managing an inventory when there are multiple potential suppliers have been reviewed in [10,22,33,36].
When there is a single supplier, the decision problem reduces to the search for an optimal policy of a type that is well-documented, namely an (, ) policy.This sets two stock levels,  and  > .If the inventory level is greater than , a manager does not intervene.If it is  or less, the manager orders to bring the inventory level up to .An (, ) policy was introduced by Arrow et al. [1].The further historical development can be traced to [2,20,31,35,40].More contemporary exposés are to be found in [8,28].The complication with the model with several suppliers is that determining an optimal management policy means not only finding the inventory level from which a replenishment should take place, and the quantity of goods that should be ordered, but also identifying the supplier with which the order should be placed.
Given any two suppliers, the first of which incurs both a greater cost per item and a greater set-up cost than the other, an order would naturally be placed with the second.This can also be said if only one of these costs is greater and the other is the same.By the same token, given any two suppliers incurring the same cost per item and set-up cost, it is economically immaterial with which an order would be placed.The essence of the problem is that given any two suppliers, one incurs a lesser cost per item and a greater set-up cost than the other.This enables the suppliers to be ranked in strictly increasing order of cost per item and strictly decreasing order of set-up cost, or vice versa.
The model considered was proposed in [5], where, under the assumption that the set-up cost of every supplier is strictly positive, it was shown that the following alternatives are mutually exclusive.
-There is a unique optimal (, ) policy involving only one predetermined supplier.
-There is a unique optimal generalized (, ) policy involving more than one supplier.-There is no optimal generalized (, ) policy, let alone an optimal (, ) policy.
A generalized (, ) policy involves  suppliers and stock levels for some natural number  ≥ 2. If the inventory level is greater than  (1) , one does not replenish stock.If it is between  (+1) and  () for  from 1 to  − 1, one orders from supplier () up to the level  () .If it is less than  ( ) one orders from supplier ( ) up to the level  ( ) .This policy may deliberately exclude a selection of the available suppliers.For instance, it could involve just three of five available suppliers, with supplier (1) being number 4 in the original ranking, supplier (2) being number 3, supplier (3) being number 1, and, suppliers 2 and 5 dispensed with.A generalized (, ) policy was first proposed as a viable optimal inventory-control policy by Porteus [26,27].Further discourse can be found in [5,6,8,28].
Building upon the investigation of the model in [5] and with retention of the assumption of strict positivity of every set-up cost, it has since been shown [6] that in the event of a generalized (, ) policy not being optimal, nonetheless there is a unique optimal policy.This has a feature not previously documented, and has been termed a hyper-generalized (, ) policy.Like a generalized (, ) policy, it entails  suppliers and stock levels (1.1).In addition, it contains stock levels  () , where for which the following applies.If the inventory level is greater than  (1) , one does not replenish, as with a generalized (, ) policy.If it is between  () and  () for some  indicated, one orders from supplier () up to the level  () .However, if it is between  (+1) and  () for some such , one again does not intervene, just as if the inventory level were greater than  (1) .Regardless, if the inventory level is less than  ( ) , one orders from supplier ( ) up to the level  ( ) , once more as with a generalized (, ) policy.Such a hyper-generalized (, ) policy reduces to a generalized (, ) policy when  () =  (+1) for every  from 1 to  − 1.According to the above definition of a hyper-generalized (, ) policy with  suppliers, when  () =  (−1) for some  ∈ {2, 3, . . .,  − 1}, this policy is indistinguishable from that with  − 1 suppliers in which the inventory levels  () and  () have been removed and supplier ( + 1) has become supplier () for  from  to  − 1.Hence, in addition to (1.1) and (1.2), it can be taken that  ( −1) <  ( −2) < • • • <  (1) . (1.3) In everyday terms, a hyper-generalized (, ) policy arises when there are backlogging levels for which it does not pay to replenish in excess of the conventional cut-off level.If the backlog were less, there would be an optimal course of action in placing an order with a supplier incurring a relatively low set-up cost.However, the high cost per item of such a supplier is prohibitive.On the other hand, if the backlog were greater, there would be an optimal course of action in placing an order with a supplier incurring a relatively low cost per item.However, the high set-up cost of a supplier in this category prohibits this too.The best thing to do is to let the backlog accumulate further until such time that the amount that has to be ordered is so great that is it is indeed worthwhile to replenish from a supplier incurring a low cost per item, whereby the high set-up cost can be set off against the size of the order.
To recapitulate, under the assumption that the set-up cost of every supplier is strictly positive: -When there is no optimal generalized (, ) policy, there is a unique optimal hyper-generalized (, ) policy.
The present paper extends the preceding results to the situation that the set-up cost of one of the available suppliers may be negligible.One may think of a number of suppliers at different locations, whereby a supplier offering a lesser cost per item is situated further away from the customer, entailing a greater delivery cost.In this scenario, an order from a supplier located in close proximity to the customer may incur no transportation cost.Alternatively, when a supplier and the customer are subsidiaries of the same company, transportation of goods from one to the other may be merely a bookkeeping transaction or be covered by sundry overheads.
Extension of the current results to the situation that one of the available suppliers may incur a negligible set-up cost is not achievable by simply considering this to be the limit of the case that the supplier has a positive set-up cost.It requires substantive development of the theory expounded in [5,6], and adaptation of the notion of a generalized and a hyper-generalized (, ) policy.In realizing this, the earlier results will be expanded and consolidated with novel results in a unifying framework.
The problem with a stochastic demand and two suppliers, one of which incurs a negligible set-up cost and the other a significant set-up cost and lesser cost per item, has been investigated heretofore by Fox et al. [11].They concluded that when the sale of unsatisfied demand is lost, an optimal inventory-control policy must be one of three types.The first is an (, ) policy involving only the supplier with a significant set-up cost.The second is a base policy involving only the supplier with a negligible set-up cost.The third is a mixed-ordering policy involving both suppliers.When excess demand is back-ordered, an optimal inventory-control policy is necessarily of the first or third type.In the present setting, the base policy can be viewed as a degenerate (, ) policy in which  = , while the mixed-ordering policy can be seen as a degenerate generalized (, ) policy with  = 2 in which  (2) <  (1) =  (1) <  (2) , supplier (1) is the supplier with no set-up cost, and supplier (2) is the supplier with a significant set-up cost.We shall show that the admissible alternatives are either a conventional (, ) policy involving only the supplier with a significant set-up cost, a degenerate generalized (, ) policy as just described, or a (degenerate) hyper-generalized (, ) policy with  = 2,  (2) <  (1) ≤  (1) =  (1) <  (2) , and a like configuration of suppliers.
More recently, the problem with a stochastic demand, several suppliers, one of which may incur a negligible set-up cost, periodic review, and a finite planning horizon has been studied by Benjaafar et al. [4].They concluded that for each period, except for a bounded interval of inventory levels, a generalized (, ) policy is optimal, and, provided an explicit example demonstrating that this result is best possible as far as the exceptional interval of inventory levels is concerned.They further reported extensive numerical experiments testing the ancillary problem with an infinite planning horizon.The latter discrete-time problem has since been studied by Helal et al. [15].For the case of two available suppliers, they have established conditions under which an (, ) policy involving the supplier with the greater set-up cost is optimal, and, in the event that the demand distribution is exponential, antithetical conditions under which a generalized (, ) involving both suppliers is optimal.
The counterpart to the problem studied in the present paper in which the objective is to minimize the longterm average cost of procurement, holding the inventory, and permitting shortages in continuous time has been examined for a deterministic demand by Perera et al. [24] and for a stochastic demand by a number of authors [3, 14, 16-18, 25, 38, 39].The prevailing conclusion is that the optimal inventory control policy is given by an (, ) policy, which may degenerate into a singular policy with  =  when a supplier incurs no set-up cost.
The organization of the remainder of the paper is as follows.Section 2 reviews the formal statement of the inventory control problem.Section 3 elucidates the difficulty in expanding the current theory in which ordering from every supplier incurs a positive set-up cost to the situation in which ordering from a supplier may incur a negligible set-up cost.Sections 4-7 subsequently develop a theory that encompasses both situations in a comprehensive setting.Incidentally, this simplifies and consolidates aspects of the earlier analysis.This part of the paper culminates in the establishment that the inventory control problem with or without a supplier with a negligible set-up cost has a stable unique solution which depends monotonically on the number of suppliers and the costs of each supplier.Section 8 deals with the determination of the conditions under which, with a slight adjustment of the notions, this solution corresponds to an (, ) policy, a generalized (, ) policy, or a hyper-generalized (, ) policy.Section 9 subsequently delves deeper into the occurrence of these policies for the specific case of two suppliers.Computation of the solution and the attendant optimal policy is the subject of Section 10, which includes numerical examples.The connection between the present results and those of Fox et al. [11], Benjaafar et al. [4] and Helal et al. [15] is discussed in Section 11.The paper closes with a conclusion constituting Section 12.A list of notation is included as Appendix A.

Problem statement
A stock of a single item is considered, whereby the level of stock at time  is ().A level  ≥ 0 corresponds to the number of items held.A level  < 0 indicates a shortage of − items.In the absence of intervention, changes in the stock level are governed by the evolution equation where  is a positive continuous function defined on R accounting for stock-dependent demand and deterioration of items.If  satisfies (2.1) then Conversely, if x is governed by (2.2) and then the reverse transformation leads to (2.1).The condition (2.3) is fulfilled by the commonly used expressions for , which are encapsulated in the generic expression () =  0 +  1 max{, 0} +   max{, 0}  for some  0 > 0,  1 ≥ 0,   ≥ 0 and 0 <  < 1 [12,34].Hence, with nominal loss of generality, it is supposed that  evolves according to (2.2).
To replenish stock there are  available suppliers.Placing an order with supplier  ∈  , where  = {1, 2, . . ., }, entails a fixed cost   per item and set-up cost   .The suppliers are ordered so that and The running cost is given by a continuous nonnegative function  defined on R.This amalgamates the diverse costs of maintaining an inventory when there is stock in hand, and those in the nature of an incurred penalty when there is a shortage.A commonly-used expression is where  > 0 and  > 0 are constants [3,4,8,11,14,15,19,21,23,28,30,32].Variability of money-value in time is considered by the exponential discount of costs at a constant rate This is a well-established method of discounting costs which can be traced to [29] and is a component of the continuous-time inventory models in [7,8,13,15,16,21,28,32,41].
Under the above conditions [5,6], the optimal total future cost () associated with a level of stock  that has evolved according to equation (2.2) will satisfy the equation , where On the other hand, placing an order of size  with supplier  when the level of stock is  will incur a cost of   +   and bring the inventory level up to  + .Hence,  should minimize   +    + ( + ) [3,4,8,11,15,21,26,28,32].

The approach
The key to the successful resolution of many a sophisticated mathematical problem is the notion of a solution of the problem.A restrictive notion makes it difficult to find a solution.With regard to applications, this may result in only being able to establish that there is a solution in limited circumstances.On the other hand, a lax notion makes it difficult to exclude a multiplicity of solutions.This can lead to an inconclusive outcome with regard to applications.There is tradeoff in posing a credible notion.Concerning the problem in hand, one would like a notion of a solution of (2.10) that delivers the kind of optimal inventory control policy that one would intuitively expect and can be applied in practice, and, at the same time, is backed by a conclusive theory.
In the earlier paper devoted to (2.10) with strictness in the rightmost inequality in (2.5) [5], the notion of an admissible solution had too narrow a scope to yield a solution under all circumstances.In the later paper [6], widening the scope led to a successful resolution of the problem.The expanded notion is the following.Ansatz 3.1.The solution of (2.10) is a continuous real function  such that  =   in (−∞, ] ∖ , where  is the union of a finite number of bounded open subintervals of (−∞, ), and,  is differentiable,  =  , and  <   in  ∪ (, ∞), for some number .
The precursor to Ansatz 3.1 limited the set  to the empty set, or, if one so prefers, the number of subintervals comprising  to zero.Given such a limitation, the ansatz partitions R into a stopping region (−∞, ] in which  =  , and a continuation region (, ∞) in which  <   and  =  .This is a primary feature of an (, ) policy and a generalized (, ) policy.Permitting  to comprise one or more nonempty bounded open subintervals opened the way to the realization of a hyper-generalized (, ) policy.
When the rightmost inequality in (2.5) is not strict, the theorem below reveals that Ansatz 3.1 is wanting.The proof of this theorem is given in Appendix B. Theorem 3.2.Suppose that   = 0. Let  be a real function with the property that   is well defined in R.
Then  ≤   in R if and only if  =   =    everywhere in R. Theorem 3.2 implies that when   = 0, asking that a solution  of (2.10) be such that  =   in a subset of R is superfluous.Conversely, asking that a solution  be such that  <   in a subset of R disqualifies it.
To deal with the failure of Ansatz 3.1 when   = 0, we propose the following notion covering   > 0 and   = 0.The theorem below, whose proof is given in Appendix C, confirms that this new ansatz is equivalent to the preceding one when   > 0.
Theorem 3.4.Suppose that   > 0. Then  is a solution of (2.10) satisfying Ansatz 3.1 if and only if it is a solution of (2.10) satisfying Ansatz 3.3.
Armed with Ansatz 3.3, we adopt the strategy previously used to tackle the inventory control problem in [5,6].We start, under the mere assumption that  is continuous, by extracting characteristics of a solution of (2.10) that are concealed in the ansatz.This is the subject of the next section.In the subsequent section we instate the hypothesis that led to the successful resolution of the problem when   > 0, and show that under this hypothesis the QVI has at most one solution with the extracted characteristics (therewith proving the uniqueness of a solution).In the section thereafter, we construct a solution embodying these characteristics (therewith proving existence).A supplementarily section shows that the solution found is stable with respect to perturbations of (2.5), and depends monotonically on  and the components of (2.4) and (2.5).

Preliminary characterization
Inherently a solution of (2.10) satisfying Ansatz 3.3 embodies a number of features.The most significant of these are summarized in the next theorem.The proof of its forerunner in [6] relies heavily on the assumption that   > 0, and cannot be easily modified to contend with   = 0.The search for an alternative has inadvertently uncovered a proof for   ≥ 0 which is simpler than the previous one for   > 0. This is delivered in Appendix D. Theorem 4.1.Suppose that (2.4) and (2.5) hold and  is continuous on R. Let  be a solution of (2.10) satisfying Ansatz 3.3.Then  =  in [, ∞), where  is a solution of the differential equation for some  ∈  and  ≥ .Furthermore, where Theorem 4.1 seems to be as far as one can take the characterization of a solution of (2.10) satisfying Ansatz 3.3 without further hypotheses on the function  or the numbers (2.4), (2.5) and (2.7).To progress, we impose the hypothesis under which it was shown in [5] that when each supplier ℓ is the sole supplier, the QVI has a unique solution corresponding to an (, ) policy no matter how large the set-up cost.This reads as follows.
depend continuously and strictly monotonically on  ℓ ≥ 0,  ℓ =  ℓ =  ℓ when  ℓ = 0, and  ℓ → −∞ when  ℓ → ∞.Furthermore, the concurrent solution  ℓ of equation (4.1) is unique, expressible as depends continuously and strictly decreasingly on  ℓ ≤  ℓ , and is such that (5.4) Proof.Allowing for the omission of  ℓ = 0, expression (5.3) and the dependency of  ℓ on  ℓ , the lemma can be found in Section 3 of [5].Augmentation of the proof accommodates the omissions.With no loss of generality, it can be supposed that  ℓ = 0.By extension of Lemma 3.4 of [5], there exists a unique function Expressing the integrand in terms of  ℓ rather than  , and applying the formula for integration by parts of Riemann-Stieltjes delivers (5.3).The continuous and monotonic dependence of  ℓ on  ℓ is a consequence.
Remark 5.2.When  assumes the classical form (2.6) and (4.7) holds, the right-hand equation in (5.1) can be solved for  ℓ explicitly, yielding Consequently,  ℓ can be eliminated from the left-hand equation in (5.1), making  ℓ the unique solution of the transcendental equation (5.7) Supposing that Hypothesis 4.2 holds for every ℓ ∈  , Lemma 5.1 supplies the existence of a unique solution  ℓ of equation (4.1) satisfying (4.2) and (4.3) with  = ℓ for some  ≥  and the uniqueness of the accompanying numbers,  ℓ and  ℓ , for every ℓ.Theorem 4.1 subsequently tells us that (2.10) has a solution  satisfying Ansatz 3.3 only if  =   ,  =   in [  , ∞), and  ≥   in (−∞,   ) for some  ∈  .The task ahead is to identify .
We begin the quest for  with an observation.
Lemma 5.3.Given any  ∈  and ℓ ∈  , either   <  ℓ and for all  ∈ R, for some constant .The trichotomy follows from whether  is negative, positive or zero.
Proof.When  ℓ > 0, it can be shown that   ≤  ℓ with equality only if   >  ℓ , following the proof of Lemma 4.11 of [5].To deal with the case  ℓ = 0, suppose, contrarily, that   >  ℓ .Then, by Lemma 5.3, Hence, by (5.4) for ℓ = ,  ℓ ≥   .On the other hand, if  ℓ <   , then  ℓ ≥   by (4.6), and   ≥   by (5.2) with ℓ = .So either way,  ℓ ≥   .This means that  is differentiable at  ℓ , and When   > 0, Lemma 5.5 leads to the conclusion that  must be the greatest minimizer of  ℓ with respect to ℓ ∈  .This identifies  precisely.When   = 0 such a definitive conclusion cannot be drawn.Should there be several minimizers, one of which is , the lemma fails to narrow down the selection beyond  and the second greatest minimizer.This presents yet another hurdle in extending the theory from   > 0 to   ≥ 0. Fortunately, it is the last.
Providentially, we can continue the development merely supposing that  is a minimizer.
Remark 5.7.In the specific case that  is given by (2.6) and (4.7) holds, formula (5.9) gives In the light of Theorem 4.1, Lemma 5.6 reveals that (4.5) can be more succinctly expressed where (5.13) Considering (5.12) and (5.13) as definitions for all  ∈ R, the function  is piecewise-linear and concave in R.
Consequently,  has a right derivative  +  and a left derivative  −  everywhere in R. Furthermore, there is a partition such that  is affine in each interval (5.15) Defining N as the smallest number for which such a partition exists, this partition is unique.Moreover, since (2.4), (5.12) and (5.13) imply that ()/ → − 1 as  → −∞, The function  has additional relevant properties stated in the two lemmas below, in which Proof.The leading statement can be verified by the argument used to prove Lemma 4.24 of [6].With regard to the subsidiary statement, suppose that  ℓ <   and ( ℓ ) =   ( ℓ ).Then, by the leading statement, However, by (2.4), (5.12) and (5.13),  −  ≥  +  ≥ −  everywhere.Thus, in this case too,  is differentiable and From Lemma 5.9 it follows that  ≤  in   for  ∈ ℳ if and only if   ≥ 0, where (5.17)By (5.14) and Lemma 5.8, Last but not least, the next lemma concerning  is of importance.For clarification, in the statement of this lemma,  <  is taken in the standard sense in   for  ∈ ℳ.However, at   for  ∈ ℳ ∖ {1}, it is to be interpreted as  +  +  <  −  +  ≤  .By the above, this is equivalent to   ≥ 0.
Lemma 5.10.There exists a uniquely defined nonnegative integer  and uniquely defined Proof.We refer to [6] for the proof of this lemma, which is a synopsis of Lemmas 4.12-4.14and Subsection 4.3 prior to Lemma 4.25 of [6].
The crux of the above is the next lemma.
Lemma 5.11.Continuing on from Lemma 5.10, define ) and   on R by Suppose that  is a solution of (2.Let us recapitulate.We have shown that any solution  of (2.10) satisfying Ansatz 3.3 has the properties stated in Theorem 4.1 for some  ∈  .By Lemma 5.5, we know that  must be a minimizer of  ℓ with respect to ℓ ∈  , and, consequently, by Lemma 5.11, that  =   .It follows that if { ℓ : ℓ ∈  } has a unique minimizer then (2.10) has at most one solution satisfying Ansatz 3.3.If not, rather than attempt to whittle down the number of minimizers by some further means, we adopt a different tactic.We show that whatever the choice of the minimizer, we end up with the same function .The next three lemmas do the trick.Lemma 5.13.Further to Lemma 5.12, let  be the function defined by (5.8), (5.12) and (5.13), N be the smallest natural number for which there is a partition (5.14) with the property that  is affine in each of the intervals (5.15), and   be given by (5.17) for  ∈ ℳ.Then   ∈   and   < 0 for some  ∈ ℳ ∖ {1}.
Proof.By the preamble to Lemma 5.12, a solution  of (2.10) satisfying Ansatz 3.3 is necessarily equal to   where  is a minimizer of { ℓ : ℓ ∈  }.Pursuant to Lemma 5.14, this prescribes  uniquely.

Existence
From the preceding section, we know that if (2.10) has a solution satisfying Ansatz 3.3 then this solution is necessarily the function   , defined by (5.8), (5.12), (5.13), Lemma 5.10 and Lemma 5.11, for a minimizer  of  ℓ with respect to ℓ ∈  .In the light of Lemma 5.14, we may drop the subscript from the notation of  .To prove that (2.10) admits a solution satisfying Ansatz 3.3 it subsequently suffices to verify that  solves (2.10) and possesses the hallmarks of the ansatz.It is convenient to divide this undertaking into five steps, whereby, without further mention, it is supposed that  is the greatest minimizer of { ℓ : ℓ ∈  }.
Proof.Let  be the greatest minimizer of { ℓ : ℓ ∈  }, and  be the function   defined by Lemma 5.11.Then by Lemma 6.3,  ≤   in R. Furthermore, by Lemmas 5.10 and 5.11,  ≤  in R. If now,  <   at some  ∈ R, Lemmas 6.2 and 6.4 imply that either  >   or  ∈   .Hence, by Lemmas 5.10 and 5.11,  is differentiable and  =  at .This confirms that ( −  )( −  ) = 0 in R. Thus we have proven that  solves (2.10).With regard to satisfaction of Ansatz 3.3, let Ω denote the set of  ∈ R for which (3.1) holds.By Lemmas 6.2 and 6.4, This implies that   is the least upper bound of Ω when   > 0. In the light of Lemma 6.5, it likewise implies that   is the least upper bound of Ω when  is the unique minimizer of { ℓ : ℓ ∈  }.However when   = 0 and there is more than one such minimizer, Lemmas 5.8 and 5.14 say that the least upper bound of Ω is   , where  is the second greatest minimizer of { ℓ : ℓ ∈  }.Whatever, Ω has the structure set out in Ansatz 3.3.Lemma 5.10 affirms that  has the required regularity.Theorem 6.6 spawns a number of corollaries.The proof of the first is contained in that of the theorem.That of the second, third and fourth is to be found in Section 5.The fifth is given by Lemmas 5.8, 5.14 and 6.5, and the theorem.Corollary 6.7.If   = 0,  is a minimizer of  ℓ with respect to ℓ ∈  , and there is at least one other such minimizer, then  in Theorem 4.1 is the second greatest of these minimizers.Otherwise,  is the greatest.Corollary 6.8.Except in a finite subset of R, the unique solution  of (2.10) satisfying Ansatz 3.3 is continuously differentiable.Let  be the greatest minimizer of  ℓ with respect to ℓ ∈  .Define  by (5.8), (5.12) and (5.13), the partition (5.14) such that  is affine in each of the intervals (5.15) with N the smallest natural number for which such a partition exists, ℳ by (5.16),   for  ∈ ℳ by (5.17),  and (5.19) by Lemma 5.10, and Ω  by (5.21) and (5.22).Then the finite subset referred to, Ξ say, comprises those   ∈ Ω  for which 2 ≤  ≤ N and those   for which   <  +1 and 1 ≤  ≤ .The derivative of  has a jump discontinuity at every  ∈ Ξ. Corollary 6.9.There holds  +  +  <  −  +  ≤  at every  ∈ Ξ,  <  in the interior of Ω  ∖ Ξ, and  =  elsewhere.

Supplementary properties
The purpose of this section is to ascertain that the unique solution  of (2.10) satisfying Ansatz 3.3 possesses certain properties that can intuitively be expected of an optimal inventory control policy.
The first of the afore-mentioned properties is the stability of  in the sense that small perturbations of (2.5) do not engender large changes in .In other words, under the predominant supposition that every  ℓ satisfies Hypothesis 4.2,  depends continuously on the set-up costs (2.5).Taken together with the existence and uniqueness results, this establishes that the problem of solving (2.10) under Ansatz 3.3 is mathematically well posed in the sense ascribed to Hadamard.The precise stability statement is the following and is proven in Appendix E.
Theorem 7.1.Let  be the unique solution of (2.10) given by Theorem 6.6, and  () the corresponding solution with (2.5) replaced by a like set of numbers furnished with a superscript () for  ∈ N. If  ()  ℓ →  ℓ as  → ∞ for every ℓ ∈  then  () →  uniformly in R.
It might further be expected that, given a selection of available suppliers, should a further supplier appear on the scene, the total future cost of the optimal inventory control policy could fall, at least for some inventory levels if not all.The cost would certainly not be expected to rise.As this property is equivalent to the property that reducing the selection of available suppliers leads to the solution  of (2.10) satisfying Ansatz 3.3 increasing or remaining the same, and the latter property is easier to formulate, we shall view the phenomenon from this perspective.The theorem below summarizes it and is proven in Appendix F. Theorem 7.2.Suppose that  ≥ 2 and ℓ ∈  .Let  be the solution of (2.10) given by Theorem 6.6 and  * the corresponding solution whereby  ℓ and  ℓ are omitted from (2.4) and (2.5) and  is lessened by 1. Then  * ≥ .
In the same vein as the previous property, should the cost per item or the set-up cost of one or more suppliers within a set of available suppliers be increased, then the total future cost of the optimal inventory control policy could be expected to rise for some if not all inventory levels.It is inconceivable that this will provoke a fall in the cost.This monotonicity property is captured by the coming theorem, whose proof is presented in Appendix G.
Theorem 7.3.Let  ± be solutions of (2.10) given by Theorem 6.6 corresponding to (2.4) and (2.5) furnished with a superscript ±.If  + ℓ ≥  − ℓ and  + ℓ ≥  − ℓ for every ℓ ∈  then  + ≥  − .Finally, in the introduction, it was argued that suppliers, that incur both a cost per item and a set-up cost greater than or equal to those of another supplier, do not have to be taken into account.It could equally as well be argued that, as long as no two suppliers have exactly the same costs, it should not be necessary to dismiss such suppliers a priori .If the model were robust, then these suppliers would be excluded from an optimal policy as an outcome.A sift through the theory developed in the preceding sections, verifies that the latter is indeed the case.The theorem below abrogates the previously stated existence, uniqueness and stability results.

Type
In the prototypical circumstance that there is a single supplier, i.e.  = 1, with a set-up cost  1 > 0, the stable unique solution  of (2.10) satisfying Ansatz 3.3 corresponds to a standard (, ) policy.The numbers  and  are  1 and  1 respectively.Confronted with an inventory level  ≤ , the policy is to place an order to bring the inventory level up to .Faced with an inventory level  > , the policy is not to intervene.When the set-up cost  1 = 0, and therefore  1 =  1 , Theorem 6.6 delivers a degenerate (, ) policy with  = .If  <  then the policy is to order up to the level .If  >  then one does not intervene.However, if  =  = , then one maintains the inventory at this level.This is feasible as ordering incurs only the cost per item.
When there are several suppliers, i.e.  ≥ 2, the exact type of the optimal policy propagated by  depends on three facets.Taking it as read that (2.4) and (2.5) apply, the foremost is the greatest minimizer  of  ℓ with respect to ℓ ∈  .In layman's terms, this is the supplier for which the single-supplier (, ) policy involves the least cost when there is a large stock in hand, or, if there is more than one such supplier, that one of these suppliers with the least set-up cost.Of all these suppliers, that is also the supplier for which the single-supplier policy has the least value of  and the greatest value of .When  = 1, i.e. the supplier with the most favourable single-supplier (, ) policy is also the supplier with the least cost per item overall,  corresponds to an (, ) policy.This is indistinguishable from the (, ) policy with supplier 1 as the sole available supplier.Otherwise,  does not represent an (, ) policy.
When the pivotal supplier  is not the supplier with the least cost per item overall, then, in the second instance, the type of policy is determined by the numbers   for  ∈ ℳ ∖ {1}.These numbers are associated with an inventory level   , which has the property that one would ostensibly order from one supplier if one had a slightly smaller backlog and from another if one had a slightly greater backlog.The nonnegativity of   is a way of testing that it is more economical to order from the second supplier than not to place an order at all.The number  1 = 0 irrespective of any other considerations.
The strategy espoused by the generalized (, ) policy is that if the inventory level  is such that  <  ( ) one orders from supplier ( ) to bring the inventory level up to  ( ) .If  (+1) <  <  () for 2 ≤  ≤  or  =  () for  = 1, one orders from supplier () to bring the inventory level up to  () .If  >  (1) one does not intervene.With regard to an inventory level  =  () for 2 ≤  ≤  , one may order from supplier () or supplier ( − 1) to bring the level of inventory up to the appropriate target level.Possibly there are further suppliers, otherwise excluded from the policy, with which one could place an order from this level.
When  ≥ 2,   ≥ 0 for every  ∈ ℳ, and   = 0, then  betokens a degenerate generalized (, ) policy.It is degenerate in the sense that (1.1) is replaced by Except in one detail, the policy has the same characteristics as the generalized (, ) policy just described.The difference is that when  =  (1) =  (1) , the policy is to maintain the inventory at this level.Should there be more than one minimizer of  ℓ with respect to ℓ ∈  , then necessarily   < 0 for some  ∈ ℳ.Thus, without further ado, it can be stated that the solution  of (2.10) satisfying Ansatz 3.3 will not deliver a generalized (, ) policy let alone an (, ) policy.
In a nutshell, the following has emerged.

The case of two suppliers
As much as Theorem 8.1 establishes the necessary and sufficient conditions for the occurrence of a generalized (, ) policy and an (, ) policy, it provides little insight into the way in which this occurrence is regulated by the costs of the suppliers.The present section addresses this lacuna by examining the particular case of two suppliers in more detail.The next three theorems furnish a conspectus.Their proof can be found in Appendices H-J respectively.
Theorem 9.1 establishes that in the case of two available suppliers ordered in terms of increasing cost per item, there are two critical levels for the set-up cost of the first supplier relative to that of the second.When the set-up cost of the first supplier is below the lesser critical level, the second supplier is excluded from the optimal policy.When the set-up cost of the first supplier is at this level or above, both suppliers are involved.When in addition the set-up cost of the first supplier is below the greater critical level, the optimal policy contains shortage levels, between ones where a manager would order from one supplier or the other, for which the policy is to let a backlog grow.When the set-up cost of the first supplier is at or above this greater critical level, the optimal policy entails ordering from one supplier or the other for all inventory levels below the greatest for which this is an option.This is perhaps counterintuitive.An explanation can be sought in looking at what happens to the greatest inventory level,  (1) , from which an order may be placed, as the set-up cost of the first supplier increases.When the set-up cost of the first supplier cost is below the lesser critical level,  (1) has the value of  in the (, ) policy of that supplier.When the cost reaches the critical level, it jumps to the value of  in the single-supplier (, ) policy of the other supplier.Thus, as it were, an interval of inventory levels for which the policy was not to intervene is instantly annihilated.What one observes in the optimal policy is a remnant of this interval in which it still pays not to intervene.Increasing the set-up cost of the first supplier further, traces of the interval disappear.Theorem 9.1 also shows that as the set-up cost of the second supplier increases, the critical levels for the first supplier adjust.The lesser the set-up cost of the first supplier or the greater that of the second, the greater the likelihood that the second supplier is excluded from the policy.Theorems 9.2 and 9.3 confirm that the same applies with regard to the cost per item.These conclusions are as one would expect.Indeed, they have been drawn on the basis of a numerical experiment in [6].When more than two suppliers are available, Example 4.29 of [6] illustrates that the prediction of the interrelation between the costs of several suppliers becomes considerably more problematic.

Computation
The stable unique solution  of problem (2.10) satisfying Ansatz 3.3 is characterized in sufficient detail in Section 5 to inform how to compute it.The algorithm below summarizes the procedure.This is a polished version of Algorithm 4.27 of [6] taking advantage of the new insights.The most noteworthy alteration is that the greatest minimizer of  ℓ with respect to ℓ ∈  is replaced by the least.Since it has been shown that whichever minimizer is used, the end result is the same, this reduces the work when there is more than one.The worst-case complexity of the algorithm is ( 2 ) with regard to the total number of steps that will be executed for a large number  of available suppliers.Step 5.For ℓ = 1, 2, . . .,  − 1, determine  ,ℓ >   from equation (5.9), and set Step 8. Output () =   () for   <  <   and 1 ≤  ≤ .For all other  <   , () and For  ≥   , () is given by the right-hand side of (10.4) with ℓ =  and   =   .End.
Step 6 extracts the partition (5.14) with the property that  defined by (5.12) is affine in each of the intervals (5.15) and N is the smallest number for which such a partition exists.The function  : ℳ = {1, 2, . . ., N} → {1, 2, . . ., } is a device for recording that ℓ for which  =  ℓ in   .Step 7 determines the number , the numbers (5.19), and the functions (5.20) signalled in Lemma 5.10.The function  : {1, 2, . . ., } → ℳ uncovered in this step keeps a tab of that  for which   ∈   .In general, equation (10.5) has a unique solution, a pair of solutions, or no solution in an interval ( n+1 ,  n ].So 'least' in Step 7(g) should be understood as 'unique' or 'lesser'.
Remark 10.2.Regarding Algorithm 10.1 when  is given by (2.6) and (4.7) holds,  ℓ can be computed at Step 1 via (5.6), whereupon   can be found at Step 5 via (5.5).In Step 2, it is convenient to take  = 0, leading to The expression (10.1) simplifies to and (10.2) to Where needed, one can also explicitly solve (10.3) as The thrust of Algorithm 10.1 is the calculation of the solution  of (2.10) satisfying Ansatz 3.3.The corresponding hyper-generalized (, ) policy, and its identification as a generalized (, ) policy or an (, ) policy, is distilled from this.The following sequel, employing the data acquired, outlines the procedure.The worst-case complexity of the algorithm with regard to the total number of computational steps for a large number  of available suppliers is ().Application of Algorithms 10.1 and 10.3 demonstrates that an (, ) policy, a generalized (, ) policy that is not an (, ) policy, and a hyper-generalized (, ) policy that is not a generalized (, ) policy may each occur when there are two suppliers available and one incurs no set-up cost.Each of the following three examples concerns one of these mutually exclusive alternatives.Where transcendental equations have been encountered, they have been solved using readily available propriety software.Example 10.4.Let  be given by (2.6) with  =  = 4,  = 1,  = 2,  1 =  1 = 1,  2 = 3, and  2 = 0.So (4.7) is satisfied.Following Algorithm 10.1, solution of (5.6) yields  1 ≈ −0.814, while it can be verified that  2 = 0. Formula (10.6) subsequently gives Υ 1 ≈ 2.670 and Υ 2 = 3. Therefore Λ = {1},  = 1,  1 =  1 , N = 1 and  = 0. Formula (5.5) leads to  1 ≈ 0.288.Proceeding to Algorithm 10.3,  = 1.Therefore the solution of (2.10) satisfying Ansatz 3.3 corresponds to an (, ) policy involving only supplier 1 with  The outcome is a hyper-generalized (, ) policy that is not a generalized (, ) policy involving both suppliers with 1) is supplier 2, and vice versa.
Figure 1 displays the cost function  given by Examples 10.4-10.6.The lowermost curve is that for Example 10.4 and corresponds to an (, ) policy, the uppermost curve is that for Example 10.5 corresponding to a generalized (, ) policy, while the intermediate curve is that for Example 10.6 corresponding to a hypergeneralized (, ) policy that is not a generalized (, ) policy.Note the discontinuity in the derivative of  appearing in the uppermost curve at  ≈ −2.341, and in the intermediate curve at  ≈ −0.597, reflecting the existence and uniqueness theory of Sections 5 and 6.All three examples have two available suppliers with the same cost per item incurred using supplier 1, the same cost per item incurred using supplier 2, and the same set-up cost incurred using supplier 2. The distinction is in the set-up cost  1 incurred using supplier 1.From top to bottom, the curves in Figure 1 are those for progressively decreasing  1 , as anticipated by Theorem 7.3.Examples where both suppliers have a positive set-up cost, complementary to Examples 10.4-10.6,have been presented in [6].The occurrence of each type of policy is in line with Theorem 9.1.We refer to Example 4.29 of [6] for an illustration of how the occurrence of the different types of policy becomes more opaque when more than a couple of suppliers are available.
The mechanism that leads to the occurrence of  () =  () for  ∈ {1, 2, 3} is the non-uniqueness of a minimizer of  ℓ with respect to ℓ ∈  .In contrast, the occurrence for  ∈ {5, 6} is attributable to functions (5.20), besides have been studied more recently by Benjaafar et al. [4] and Helal et al. [15].This section explores the connection between the conclusions of [4,11,15] and Theorems 6.6 and 9.1.Apart from the demand being deterministic and not stochastic, the most striking difference between the present investigation and its predecessors [4,11,15] is the formulation of the problem as a QVI.The approach in each of the earlier papers is more pragmatic, and can be embedded in the deterministic continuous-time continuous-state setting as follows.
Ansatz 11.1.The function sought is a continuous real function  such that  ≤   in R. The set Ω of  ∈ R for which (3.1) holds is an interval with a finite least upper bound .Furthermore,  is differentiable and  =  in [, ∞).Finally, given any function  with these properties,  ≤  in R.
The merit of Ansatz 11.1 is reflected in the next theorem.
Proof.The proof of Theorem 4.1 is independent of the inequality  ≤  in (2.10).So it carries through.Barring that of Lemma 5.11, so too does the proof of every lemma in Section 5.As a result, it can be concluded that a function  satisfies Ansatz 11.1 only if  =   in [  , ∞) and  =  in (−∞,   ), where  is a minimizer of { ℓ : ℓ ∈  }, and  is defined by (5.8), (5.12) and (5.13).Furthermore, if   = 0,  is a minimizer of { ℓ : ℓ ∈  }, and there is at least one other such minimizer, then  is the second greatest minimizer of { ℓ : ℓ ∈  }.Otherwise, it is the greatest.Appealing to the arguments used to prove Theorem 6.6, it can be verified that  meets the requirements of Ansatz 11.1 with  =   .The considerations leading from Theorem 6.6 to Theorem 8.1, lead to the deductions regarding  representing an (, ) or a generalized (, ) policy.Theorem 11.2 corroborates the conclusions reached for the model with a stochastic demand by Fox et al. [11].When  = 2 and  2 = 0, a function satisfying Ansatz 11.1 corresponds to an (, ) policy involving only supplier 1, or a mixed-ordering policy involving both suppliers with levels  (2) <  (1) =  (1) <  (2) , whereby supplier (1) is the supplier with no set-up cost and supplier (2) is the supplier with a positive set-up cost.This should come as no surprise to those supporting the view that stochastic models are a generalization of deterministic ones, or, alternatively, that deterministic models are special or limiting cases of stochastic ones.
The above naturally raises the question of the equivalence of the function given by Theorem 11.2 and that given by Theorem 6.6.The next theorem provides the answer.
Theorem 11.3.Under the conditions of Theorem 11.2, let  be the solution of (2.10) satisfying Ansatz 3.3, and  be the function satisfying Ansatz 11.1.Furthermore, if   = 0,  is a minimizer of  ℓ with respect to ℓ ∈  , and there is at least one other such minimizer, let  be the second greatest minimizer.Otherwise, let  be the greatest.Denote the number subsequently given by Lemma 5.10 by .Then where  ≺  means that  ≤  everywhere in R and  <  in a nonempty open subset of R.
The phenomenon that the function  found in Theorem 11.2 need not be the solution  of (2.10) provided by Theorem 6.6 has two sides.Notwithstanding that the number  in Ansätze 3.3 and 11.1 fulfils a common role, one side is that  does not satisfy the inequality  ≤  throughout (−∞, ].The reverse is that  is not such that  = , where  is defined by (5.8), (5.12) and (5.13), throughout (−∞, ].What is clear though, whether one prefers the QVI or the more perfunctory approach, is that when  and  do not coincide,  ≺  and  corresponds to a hyper-generalized (, ) policy that is not a generalized (, ) policy.So, under this circumstance, the intuitional generalized (, ) policy is not the optimal inventory control policy, and, a hypergeneralized (, ) policy that is not a generalized (, ) policy is.We speculate that the same is true for the stochastic model.
For the problem with a stochastic demand, several suppliers, one of which may incur a negligible set-up cost, periodic review, and a finite planning horizon, Benjaafar et al. [4] concluded that for each period, except for a bounded interval of inventory levels, a generalized (, ) policy is optimal.A hyper-generalized (, ) policy would account for the exceptional interval of inventory levels.
The results of Helal et al. [15] for the discrete-time problem with a stochastic demand, two available suppliers, one of which may incur a negligible set-up cost, and an infinite planning horizon have a similar character to those of [4].They affirm conditions under which an (, ) policy involving only the supplier with the greater set-up cost is optimal, and, when the demand distribution is exponential, antithetical conditions under which a generalized (, ) policy involving both suppliers is optimal.Beyond technicalities of proof, an explanation of why these two sets of conditions are not complementary is that there are circumstances under which the optimal policy is a hyper-generalized (, ) policy.Noteworthy is that the conditions under which an (, ) policy involving only the supplier with the greater set-up cost is shown to be optimal include the relative closeness of  1 and  2 , which is a defining feature of Theorem 9.1.
A subsidiary result of Benjaafar et al. [4] is that if   is large enough then a generalized (, ) policy is optimal.However, this result relies on a rather strong convexity assumption that does not apply to the model considered in the present paper.Indeed, Theorem 9.1 precludes the analogous result for the problem dealt with in the present paper.
In [4], it is further reported that extensive numerical experiments with normal, log-normal, gamma, and Poisson distributions, and thirty thousand experiments with randomly generated distributions all found that a generalized (, ) policy is optimal.Moreover, even for the examples demonstrating that a generalized (, ) policy is not optimal for a bounded interval of inventory levels, a generalized (, ) policy becomes optimal when the planning horizon is large enough.This has accordingly suggested that a generalized (, ) policy is optimal for the problem with an infinite planning horizon.A rerun with an infinite planning horizon of all the previously reported experiments has found exclusively that a generalized (, ) policy is optimal.We conjecture, as with the analysis of Fox et al., that this comes about because this is what is sought.Theorems 6.6, 11.2 and 11.3 are indicative.Under those circumstances where there is a hyper-generalized (, ) policy that is not a generalized (, ) policy, this policy will turn out to be optimal.

Conclusion
The present paper continues the study of a deterministic continuous-time continuous-state inventory model with several suppliers in [5,6].In [5], the decision problem was formulated as a QVI, and a solution was postulated in the form of an ansatz corresponding to a generalized (, ) policy.Under the premise that every supplier incurs a significant set-up cost, it was shown that there is at most one solution satisfying the ansatz, and the necessary and sufficient conditions for the existence of such a solution were established.In [6] the scope of the ansatz was widened to admit correspondence with a refinement of a generalized (, ) policy, labelled a hyper-generalized (, ) policy.Retaining the premise that every supplier has a positive set-up cost, this led to the successful proof of the unconditional existence and uniqueness of a solution of the QVI.In the present paper, a further adaptation of the ansatz has led to the extension to the situation that a supplier may incur a negligible set-up cost.Moreover, it has been shown that the solution is stable and depends monotonically on the number of potential suppliers and the costs of each suppliers.The case of two suppliers has been scrutinized in some detail, and intuitive ideas, touched upon in the antecedent papers, about the way in which the solution is influenced by the costs of the suppliers have been substantiated.
A stochastic demand is arguably more realistic than a deterministic one.Nevertheless, some similarity between the optimal inventory control policy for the studied problem with a deterministic demand and related problems with a stochastic demand is to be expected.The stochastic problem with two suppliers has been investigated previously by Fox et al. [11], who concluded that the optimal policy is of one of two types.The analogous approach to the deterministic problem leads to the same conclusion.However, approaching the problem through the QVI and admitting the possibility of a hyper-generalized (, ) policy, it transpires that in those situations where the latter occurs, this supersedes the surrogate policy.Comparable models with several suppliers, one of which may incur a negligible set-up cost, have been more recently studied by Benjaafar et al. [4] and Helal et al. [15].The conclusion of the former that, except for a bounded interval of inventory levels, a generalized (, ) policy is optimal, and of the latter that under some conditions an (, ) policy is optimal and under others which are not complementary a generalized (, ) policy is optimal, indicates that a hyper-generalized (, ) policy is appropriate in these situations too.The inference is that a hyper-generalized (, ) policy has a role to play in both of these problems with a stochastic demand and other more elaborate models with a more sophisticated demand.
Avenues for future research include the adaptation of the model to account for lost sales when demand is not met, the extension to suppliers having a limit to the quantity of stock that they can deliver, the extension to set-up costs being an increasing piecewise-constant function of the quantity supplied, and taking supplierdependent lead-times into consideration.Moving to comparable stochastic continuous-time continuous-state inventory models with several suppliers, that with an exponential demand distribution holds promise, as do models with a diffusion demand [16,32,37], and their counterparts with a jump-diffusion demand [7,21].Further possibilities are provided by the corresponding discrete-time models and discrete-demand models with an infinite planning horizon and with a finite planning horizon, complementing the analysis in [4,15].ℓ }︀ for ℓ ∈  , apply Lemma E.6, and note that  ± ℓ =  ℓ when  ± ℓ =  ℓ , while  ± ℓ =  () ℓ when  ± ℓ =  ()  ℓ .This gives  + −  − ≤  () .However, Lemma E.6 further implies that  − ≤  ≤  + and  − ≤  ()

Ansatz 3 . 3 .
The solution of (2.10) is a continuous real function  with the following properties.The set Ω of  ∈ R for which () =  ℓ + ( + ) +  ℓ  for some  > 0 and ℓ ∈ (3.1)isnot empty, Ω has a finite least upper bound , and  = (−∞, ) ∖ Ω is the union of a finite number of bounded open intervals.Furthermore,  is differentiable and  =  at the left endpoint of any subinterval of  ∪ (, ∞).

Figure 1 .
Figure 1.Cost function for Examples 10.4-10.6.That for Example 10.4 is the lowermost curve, for Example 10.5 the uppermost curve, and for Example 10.6 the intermediate curve.

Figure 2 .
Figure 2. Cost function for Examples 10.7 and 10.8.That for Example 10.7 is the lower curve.
Stock level to which inventory should be replenished   Stock level  when supplier  is the sole supplier  ,ℓ Argument at which the derivative of   takes the value − ℓ  () Stock level in optimal policy to which inventory is replenished using supplier()  Set of extraordinary inventory levels from which inventory is not replenished   Set  when supplier  is the key supplier   Test real number for the interval    Necessary form of the unknown function    Function  when supplier  is the key supplier  Upper bound for the difference of two functions  Absolute minimum   Solution of differential equation in subinterval  of the set  Lower case Roman alphabet  Left endpoint of interval   Left endpoint of subinterval of the set  with index   Right endpoint of interval   Right endpoint of subinterval of the set  with index    Cost per item incurred using supplier  d Differential with respect to the variable that follows e Base of the natural logarithms  Function defining running cost with inventory level as input  ℓ Auxiliary function related to  associated with supplier ℓ  0 ,  1 ,   Parameters in expression for the function   Dummy index  Sections 2-4 and Appendix D: index for available suppliers Section 5 and Appendix E: candidate key supplier in optimal policy Sections 6-11 and Appendix F: key supplier in optimal policy   Set-up cost incurred using supplier   * ,  † Least stock level in optimal policy from which inventory is replenished using supplier ()  Stock level from which inventory should be replenished   Stock level  when supplier  is the sole supplier  () Greatest stock level in optimal policy from which inventory is replenished using supplier ()  Time  Section 2: future cost over infinite time horizon Sections 3-5 and Appendices B-D: unknown in QVI Sections 6-11 and Appendices E-G: solution of QVI  ℓ Auxiliary function related to  associated with supplier ℓ  Piecewise-linear function coinciding with  in the set Ω  ℓ Affine function associated with supplier ℓ leading to the construction of   Dummy left endpoint of interval  Inventory level  Solution of ordinary differential equation   Solution  when supplier  is the sole supplier  Dummy right endpoint of interval Real number identifying  ℓ Ω Set of inventory levels from which inventory is replenished Ω  Set of inventory levels related to Ω when supplier  is the key supplier Section 5: alternative candidate key supplier in optimal policy Section 6: least alternative key supplier in optimal policy  Imaginary supplier  Function mapping index  to index ℓ via the criterion  =  ℓ in    Ordering of minimizers  Function mapping index  to index  via the criterion   ∈    Index of open intervals constituting the set   Dummy variable  Maximum of subset of Ω  ℓ Maximum of subset of Ω with index ℓ   Element of partition with index   Alternative to solution  when the optimization problem is not formulated as a QVI  ℓ Function mapping  ℓ to  ℓ   Subset of Ω with index  <   −  −  <   −  −  + Combining (E.9) and (E.10) delivers 0 < − − <   − −  + (︀   − )︀ (  ) in R. Therefore, by Lemmas E.1 and E.2, statement (E.5) holds in (−∞,   ].Lemma E.4.Let  be a minimizer of { ℓ : ℓ ∈  }.Suppose that (2.5) holds with  ℓ replaced by  + ℓ >  ℓ for some ℓ ∈  ∖ {}.Denote the corresponding solution of (2.10) satisfying Ansatz 3.3 by  + .Then 0 ≤  + −  ≤  + ℓ −  ℓ  R. (E.11) Proof.Let  + ℓ be the solution of (4.1) satisfying (4.2) and (4.3) with   =  ℓ and   =  + ℓ for some  ≥ .By Lemma 5.1,  + ℓ >  ℓ .Hence,  is a minimizer of {  :  ∈  } with  ℓ replaced by  + ℓ .Consequently, according to the theory of Sections 4 and 5,  + =  in R when ℓ > .This gives (E.11) immediately.Otherwise,  + =  =   in [  , ∞), and the function  + , fulfilling the role for  + that  does for , differs from  solely in that  ℓ is replaced by ℓ +  + ℓ −  ℓ .Thus,  + =  in [  , ∞) and 0 ≤  + −  ≤  + ℓ −  ℓ in R.Application of Lemmas E.1 and E.2 results in (E.11).Lemma E.5.Suppose that (2.5) holds with  ℓ replaced by  ± ℓ for some ℓ ∈  .Denote the corresponding solutions of (2.10) given by Theorem 6.6 by  ± , and the solution of (4.1) satisfying (4.2) and (4.3) with   =  ℓ and   =  ± ℓ for some  ≥  by ± ℓ .If  + ℓ >  − ℓ then 0 ≤  + −  − ≤  + ℓ −  − ℓ + Proof.Let be a minimizer of   with respect to  ∈  ∖ {ℓ}.By Lemma 5.1,  + ℓ >  − ℓ .Hence, if  + ℓ ≤   , ℓ is a minimizer of {  :  ∈  } with  ℓ replaced by  ± ℓ .Consequently, by Lemma E.3, statement (E.12) holds.Conversely, if  − ℓ ≥   ,  is a minimizer of {  :  ∈  } with  ℓ replaced by  ± ℓ .Hence, by Lemma E.4, statement (E.12) holds in this case too.Finally, if  − <   <  + then, by Lemma 5.1, there is a  ℓ ∈ ( − ℓ ,  + ℓ ) such that  ℓ =   .Let  be the corresponding solution of (2.10) satisfying Ansatz 3.3.By contrivance, ℓ is a minimizer of {  :  ∈  }, and of {  :  ∈  } with  ℓ replaced by  − ℓ , while  is a minimizer of {  :  ∈  }, and of {  :  ∈  } with  ℓ replaced by  + ℓ .Thus, by Lemma E.3, statement (E.5) with  = ℓ holds, and, by Lemma E.4, statement (E.11) holds.Adding these statements yields (E.12) with  ℓ in lieu of  + ℓ .Because  ℓ <  + ℓ , this gives (E.12) as stated.Lemma E.6.Building on Lemma E.5, suppose that (2.5) holds with  ℓ replaced by  ± ℓ for every ℓ ∈  , and  + ℓ ≥  − ℓ for every ℓ ∈  .Then 0 ≤  + −  − ≤ ∑︁ Proof.Starting with ( − 1 ,  − 2 , . . .,  −  ) and replacing  − ℓ with  + ℓ when  + ℓ >  + ℓ in order of increasing ℓ creates a sequence of -tuples preserving the inequalities in (2.5) and ending with ( + 1 ,  + 2 , . . .,  +  ).Synchronous successive application of Lemma E.5 yields the result.