End-of-Life Inventory Management Problem: Results and Insights
EEnd-of-Life Inventory Management Problem:Results and Insights
Emin Ozyoruk ∗ Nesim K. Erkip † C¸ a˘gın Ararat ‡ January 26, 2021
Abstract
We consider a manufacturer who controls the inventory of spare parts in the end-of-life phaseand takes one of three actions at each period: (1) place an order, (2) use existing inventory, (3)stop holding inventory and use an outside/alternative source. Two examples of this source arediscounts for a new generation product and delegating operations. Demand is described by anon-homogeneous Poisson process, and the decision to stop holding inventory is described bya stopping time. After formulating this problem as an optimal stopping problem with addi-tional decisions and presenting its dynamic programming algorithm, we use martingale theoryto facilitate the calculation of the value function. Moreover, we show analytical results to com-pute several metrics of interest including the expected number of orders placed throughout theend-of-life phase. Furthermore, we devise an expandable taxonomy and relate the benchmarkmodels to the literature. Analytical insights from the models as well as an extensive numericalanalysis show the value of our approach. The results indicate that the loss can be high in casethe manufacturer does not exploit flexibility in placing orders or use an outside source. Also, anoutside source can be a significant alternative for the end-of-life inventory management problem.Finally, we show that some counter-intuitive strategies can be valuable for future analysis.
Keywords and phrases:
End-of-life inventory management, optimal stopping with additionaldecisions, non-homogeneous Poisson processes ∗ Booth School of Business, University of Chicago, Chicago, Illinois, 60637, USA, [email protected]. † Department of Industrial Engineering, Bilkent University, Ankara, 06800, Turkey, [email protected]. ‡ Department of Industrial Engineering, Bilkent University, Ankara, 06800, Turkey, [email protected]. a r X i v : . [ m a t h . O C ] J a n Introduction and Literature Review
While rapid technological developments have been shortening the life-cycle of products sold in themarket, competition and customer satisfaction have made the firms increase the warranty periodsof those products. To fix a product in case of failures, a firm holds spare parts inventory for longperiods and even after the product is no longer produced. This leads to a challenging problem ofinventory management of spare parts in the end-of-life phase – a time frame within the product’slife-cycle that begins when the product is no longer produced and that ends at the expiration dateof all customers’ warranties (Fortuin 1980).Original equipment manufacturers strive to properly manage the inventory in the end-of-lifephase since the spare parts are held for long periods although the demand for them can be quitelow. For instance, in electronics industry, a manufacturer may need to keep the spare parts from 4years to 30 years after the product is discontinued from production (Teunter and Haneveld 2002).It might seem tempting to pile up abundant inventory to obey customer warranties; however, thismay result in excessive holding and scrapping costs given that the demand is expected to be low.Indeed, HP suffered from huge obsolescence costs due to end-of-life write-offs (Callioni et al. 2005),and in general, after-sales services can be a significant source of profit for the firms (Shi 2019). As aresult, several strategies have been developed to control inventory and mitigate the risk of over- andunder-stocking of spare parts in the end-of-life phase. Early approaches for inventory control in theend-of-life phase attempt to use classical inventory models while aiming to calibrate the parameterspertinent to this phase. For instance, Silver et al. (2016, p. 363, Subsection 8.5.1) review the studiesthat develop extensions of the economic order quantity (EOQ) model while assuming a deterministicand decreasing demand rate. Those studies find the number of replenishments to make as well asthe timing and sizes of these replenishments. Simultaneously, several studies are motivated by theintermittent demand structure in this phase, devising inventory models with stochastic demand.Extensions of the newsvendor model, for example, are developed where the parameters (e.g., meanand standard deviation of demand) are estimated from available data. Such studies are reviewedby Silver et al. (2016, p. 364, Subsection 8.5.2) as well.The practically oriented approaches often assume that the original equipment manufacturer2an place a single order at the beginning of the end-of-life phase, and they propose complemen-tary business strategies. The motivation behind the single order assumption is that a componentmanufacturer might decide to stop producing certain spare parts, thereby requiring the originalequipment manufacturer to place a final order. This final order is also called last-time buy, finalbuy, end-of-life buy, or all-time requirements. On the other hand, complementary strategies aim tosupport the final buy in case of a discrepancy between the realized demand and the order quantity.The wide literature on business strategies complementing a final order includes, but is notlimited to, repairing defective spare parts collected from customers (Behfard et al. 2015, 2018)while repairing may not be feasible for some of them (van Kooten and Tan 2009), buying backfunctional or dysfunctional used products to take them apart and obtain the recoverable spareparts (Pourakbar et al. 2014, Kleber et al. 2012), considering budget constraints (Hur et al. 2018)or multiple spare parts in the bill-of-materials of a main product (Bradley and Guerrero 2009),extending customer contracts (Pin¸ce et al. 2015, Leifker et al. 2014), designing a new product toreplace the obsolete one (design refresh) (Shen and Willems 2014, Shi and Liu 2020), partiallyscrapping spare parts in case of over-stocking (Pourakbar et al. 2012), differentiating customersbased on demand criticality or service contracts (Pourakbar and Dekker 2012), re-manufacturing(Shi 2019, Bayındır et al. 2007), finding outside/alternative sources (Pourakbar et al. 2012, Frenket al. 2019a, van der Heijden and Iskandar 2013, Jack and Van der Duyn Schouten 2000), andfinally, obviating the need to place a final order at time zero (Cattani and Souza 2003, Teunter andHaneveld 2002, Pin¸ce and Dekker 2011). We focus on the last two strategies in this study.The benefit of complementary strategy that finds an outside/alternative source, instead ofholding spare parts inventory, can be two-fold. On the one hand, in case the demand for spareparts exceeds the available inventory, the manufacturer can start using the outside source as aback-up source and avoid underage costs. On the other hand, it can be used to get rid of excessinventory in case of an insufficient amount of demand, decreasing overage costs. Some examples ofthis outside/alternative source can be to purchase expedited spare parts supply from a third partysupplier, to replace the failed product with a new generation product (Pourakbar and Dekker 2012,Frenk et al. 2019a, van der Heijden and Iskandar 2013, Jack and Van der Duyn Schouten 2000),or to substitute with another spare part having the same functionality (Shi and Liu 2020). If thecost of such a source decreases over time (for instance, due to price erosion of a new generation3roduct), then this strategy can become truly valuable.Within the literature incorporating an outside/alternative source, Pourakbar et al. (2012) con-sider a manufacturer who places a final order at time zero and can decide to use outside/alternativesource at each time period. After providing the dynamic programming formulation, they presentbenchmark models (e.g., one that places a final order but does not use this source) to show thevalue of incorporating this source. Frenk et al. (2019a) assume that the manufacturer makes astatic decision (at time zero) on the final order quantity and on the time to stop holding inventory(called switching time). Under this setting, they show that the objective function is convex in thefinal order quantity for every fixed switching time. Frenk et al. (2019b) extends the model in Frenket al. (2019a) with more general parameters and describe the decision to stop holding inventoryby a stopping time, solving an optimal stopping problem by means of a dynamic programmingalgorithm. Shi and Liu (2020) consider a design refresh program that substitutes an obsolete partwith an alternative part, while modeling this problem as a two-stage stochastic dynamic program.We could find a few recent studies which analyze the benefit of providing flexibility in placingorders in the end-of-life phase, although early inventory control approaches consider such flexibility.It is reasonable to accept the existence of a time point when the manufacturer places a final order.Still, such a time point may need to be found after completing an in-depth analysis since, after all,the component manufacturers might be willing to produce the spare parts as long as it is profitableto do so. Among the studies allowing flexibility in placing orders in the end-of-life phase, Cattaniand Souza (2003) analyze the effects of delaying a final order rather than placing it at time zero,and determine the optimal timing of the final buy from an aggregated supply chain perspective byincluding both the manufacturer and the supplier. They also characterize the benefit of delaying afinal order under different demand scenarios. Inderfurth and Mukherjee (2008) devise a dynamicprogramming model to help manufacturers who can place extra production/procurement orders aswell as remanufacture the recoverable spare parts. Inderfurth and Kleber (2013) further explorethe problem studied by Inderfurth and Mukherjee (2008) and devise an advanced heuristic thatprovides near-optimal solutions and that can quickly solve real-life problem instances. Teunter andHaneveld (2002) devise a continuous-time solution when demand is described by a Poisson processwith constant rate, and find an optimal base-stock policy where order-up-to levels decrease overa finite time horizon. Pin¸ce and Dekker (2011) also provide a continuous-time formulation; their4odel mainly differs from Teunter and Haneveld (2002) in that partial obsolescence is allowed,that is, intensity rate drops to a lower level at a known future time instance. Also see David et al.(1997) for a dynamic programming approach when demand is deterministic.The objective functions considered in the above studies are varying. However, many of themassume that costs are charged at discrete time points (Shi and Liu 2020). Additionally, manyof them assume that there is no fixed cost of ordering (understandable since one ordering in-stance is allowed) (Pourakbar et al. 2012). Hence, having non-zero inventory at time zero wouldnot have a significant effect on the solutions proposed. Many of those the studies, by construc-tion, assume lost-sales for the excess demand or the demand is satisfied from outside sources(Frenk et al. 2019a), though some of them allow for backordering and penalize both the timeand units backordered (Teunter and Haneveld 2002). Finally, some of the costs are assumed tobe constant over time; exceptions are allowing for discounting, a decreasing cost of alternativesource over time (Frenk et al. 2019a), or an increasing unit procurement cost after time zero(Teunter and Haneveld 2002). To the best of our knowledge, none of the above studies combineflexibility in placing orders with an outside/alternative source.
This study analyzes the value of providing flexibility in placing orders while making use of strategiesrelated to the end-of-life phase. In more general terms, we consider a multi-period, lost salesinventory problem where lost sales can be compensated by an outside source, as well as the outsidesource becoming the main source if we decide so (modeled by a stopping time). The novelty of ourstudy is the incorporation of the following main features. • Flexibility in ordering:
Instead of being required to place a single order at time zero, the man-ufacturer has the flexibility to place orders at any time and can limit the number of orders.Additionally, a time point can be found in which the manufacturer does not choose to place anyorders afterward, i.e., the timing of the final order. All these aspects bring in several flexibili-ties: Allowing for multiple orders, chances of delaying the first order and limiting the number oforders. • Strategic switch to an alternative/outside source:
The manufacturer has the opportunity to stop5olding inventory and use an alternative/outside source to satisfy demand (strategically decidenot to use internal sources). This source has a relatively high per-unit cost; however, it can beuseful in avoiding excessive penalty and holding costs in the future. Also, the manufacturer doesnot need to put an effort into the use of this source. Such a source can be another option in caseswhere redesigning products or spare parts may cannibalize design resources that could otherwisebe used for designing new products (Bradley and Guerrero 2008). The manufacturer’s decisionto stop holding inventory is described by a stopping time. • Demand variability:
Demand for spare parts is described by a non-homogeneous Poisson process.We also assume a non-increasing intensity function to fit the problem description, though for mostof our results such an assumption is not necessary. • Cost components:
We consider non-stationary costs and a fixed ordering cost. Additionally,we compute expected costs when these are charged continuously (rather than at discrete timepoints). This allows us to naturally operate with periods which are not necessarily spaced equallyin time. Another motivation for the continuous-time calculation of costs is that the manufacturermay not be able to review the inventory for long periods, so we may miss correct representationof costs. For instance, in our model, we describe the exact time that the inventory on hand hitszero and hence lost sales is observed by using a stopping time.Therefore, the manufacturer’s problem is to make one of the three decisions at each period:(1) place an order for spare parts, (2) do nothing and use existing inventory to satisfy demand, or(3) stop holding inventory permanently and use outside/alternative source until the end. We castthis combined inventory control and optimal stopping problem as an optimal stopping problemwith additional decisions that can be solved by means of a stochastic dynamic programming (DP)algorithm (see Oh and ¨Ozer (2016) for the definition of optimal stopping problems with additionaldecisions).We contribute to the end-of-life inventory management problem in several ways: We propose ageneral framework for the problem by combining several issues raised in the literature and solvingit using dynamic programming. We allow continuous cost computations by using martingale theoryto facilitate the calculation of the value function. We propose a flexible, expandable taxonomy forthe end-of-life inventory management problem which facilitates grouping the existing studies based6n the available decisions. We provide some structural insights on the problem, as well as on somespecial cases. Finally, we present extensive computations to reveal the advantages of such a generalframework and discuss important managerial insights.The rest of this paper is organized as follows. Section 2.1 formally defines the costs consid-ered and their computations, whereas Section 2.2 defines our problem and presents the main DPalgorithm. Section 2.3 presents a taxonomy for related problems. Section 3 provides the analyticalresults to run the DP algorithms as well as several structural results provided for the problemsin the taxonomy. In Section 4.1, we provide possible numerical advantages of using the proposedgeneral framework compared to others presented in the literature. Section 4.2 includes sensitivityanalysis and the resulting managerial insights. We conclude in Section 5.
Let (Ω , H , P ) be a probability space and let T ∈ [0 , ∞ ). We start by assuming that the demand forspare parts is described by a non-homogeneous Poisson process N = ( N t ) t ∈ [0 ,T ] with an intensityfunction λ : [0 , T ] → R + and mean value function Λ( t ) = (cid:82) t λ ( u ) du . We assume that λ is right-continuous with left-limits, piece-wise smooth (that is, differentiable except at finitely many points)and non-increasing. Most of the results in this study can be recovered without the assumption that λ is non-increasing. Still, such assumption can be more appropriate to describe the demand forspare parts in the end-of-life phase. The manufacturer periodically reviews the inventory level andfor brevity of notation, we assume that the lengths of time periods are identical. Our model can beeasily adjusted for non-identical period lengths. At each time point k ∈ T := { , , , . . . , T } , themanufacturer observes the current inventory level x ∈ Z + and decides whether to stop or continueholding inventory. We sometimes refer the time interval [ k, k + 1] as the k th period. After observing inventory x ∈ Z + at time k ∈ T , the manufacturer may decide to continue to holdinventory. If that is the case, an order µ k ( x ) ∈ Z + can be placed, where the function µ k : Z + → Z + c : Z + → R + is given by c ( m ) := K + ¯ c m, if m > , , if m = 0 , (2.1)where ¯ c ∈ R + is the unit purchasing cost and K ∈ R + is the fixed ordering cost. We assume thatthe ordering cost at period k is discounted by e − δk and the lead time is zero, where δ ∈ [0 ,
1] isthe discount rate of continuous compounding. After placing an order at time k , the manufacturercontinues operating during the time interval [ k, k + 1], and the holding cost accrues with rate c ∈ R + . Hence, the expected value of inventory holding cost for the k th period is given by H ( k, x ) := c E (cid:20)(cid:90) k +1 k e − δ ( u − k ) ( x − ( N u − N k )) + du (cid:21) . The following lemma enables us to calculate the holding cost.
Lemma 2.1.
For every k ∈ T , the holding cost H ( k, x ) satisfies H ( k,
0) = 0 when x = 0 , and H ( k, x ) = c x − (cid:88) n =0 n (cid:88) i =0 (cid:90) k +1 k e − δ ( u − k ) e − (Λ( u ) − Λ( k )) (Λ( u ) − Λ( k )) i i ! du, when x ≥ . The proof of Lemma 2.1 is in Appendix B.If the inventory level hits zero during [ k, k +1] and a defective part arrives, then the manufacturerloses the opportunity to replace it from the inventory. Instead, (similar to a lost sales inventorysystem) the part is replaced by paying a time-dependent unit cost c : [0 , T ] → R + at the time ofarrival. Such replacement cost is given by L ( k, x ) := E (cid:34)(cid:90) k +1( k +1) ∧ σ kx e − δ ( u − k ) c ( u ) dN u (cid:35) , where ( k + 1) ∧ σ kx = min (cid:8) k + 1 , σ kx (cid:9) , and σ kx := inf { u > k : N u − N k ≥ x } denotes the arrival time of the x th item after time k . We denote σ x := σ x . The following lemmaenables us to calculate the cost L ( k, x ). Lemma 2.2.
For every k ∈ T and x ∈ Z + , the cost L ( k, x ) incurred after inventory level hits zerocan be expressed by ( k, x ) = (cid:90) k +1 k e − δ ( u − k ) c ( u ) λ ( u ) du − x (cid:88) i =0 (cid:90) k +1 k e − δ ( u − k ) c ( u ) λ ( u ) e − (Λ( u ) − Λ( k )) (Λ( u ) − Λ( k )) i i ! du. The proof of Lemma 2.2 is in Appendix B.We define the one-period operation cost for period k and inventory level x as C ( k, x ) := H ( k, x ) + L ( k, x ) . (2.2)On the other hand, after observing inventory x ∈ Z + at time k ∈ T , the manufacturer maydecide to stop. If that is case, the available inventory is scrapped with unit cost c ∈ R (if c < c : [0 , T ] → R + .Therefore, the cost of stopping holding inventory is given by S ( k, x ) := c x + E (cid:20)(cid:90) Tk e − δ ( u − k ) c ( u ) dN u (cid:21) . (2.3)We assume that ¯ c > − c holds since otherwise, the manufacturer can place an arbitrarily largeorder and scrap inventory at the same time. Moreover, it is natural to assume that c ( u ) ≥ c ( u )for every u ∈ [0 , T ] so that the use of an outside source is meaningful. Now, we are ready to pose our problem and its dynamic programming formulation. To that end,let T denote the set of all stopping times τ : Ω → T of the filtration generated by the demandprocess ( N t ) t ∈ [0 ,T ] . Let π = ( τ, µ , µ , . . . , µ T ) be a policy that specifies an order amount µ k ( x k ) forevery k ∈ T and x k ∈ Z + as well as a stopping time τ ∈ T . Let Π denote the set of all (admissible)policies. Then, under an arbitrary policy π ∈ Π, the inventory level X πk at time k ∈ T can beexpressed by using the recursive relation X πk +1 = (cid:16) X πk + µ k ( X πk ) − ( N k +1 − N k ) (cid:17) + , X π = x ∈ Z + , (2.4)where ( x ) + := max { , x } . We use the notation X k +1 for X πk +1 when π is an arbitrary policy andthere is no risk of confusion, for brevity of notation. The manufacturer’s problem is to determineboth the optimal order amounts and the optimal time to stop in order to minimize the total costs.9e formulate this problem as V ∗ ( x ) = inf π ∈ Π E (cid:34) τ − (cid:88) k =0 e − δk (cid:16) c ( µ k ( X k )) + C ( k, X k + µ k ( X k )) (cid:17) + e − δτ S ( τ, X τ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X = x (cid:35) , (2.5)where x ∈ Z + is called the initial inventory level and the summation over k is set to zero if τ = 0.This formulation yields an optimal stopping problem with additional decisions (Oh and ¨Ozer 2016)and can be solved by means of the following stochastic dynamic programming (DP) algorithm.Define the backward recursion for each k ∈ { T − , T − , . . . , } and x k ∈ Z + by V ( k, x k ) = min { S ( k, x k ) , J ( k, x k ) } , (2.6)where J ( k, x ) := inf m ∈ Z + { c ( m ) + G ( k, x + m ) } , (2.7)and where G ( k, y ) := C ( k, y ) + e − δ E (cid:2) V (cid:0) k + 1 , ( y − ( N k +1 − N k )) + (cid:1)(cid:3) . Also define the terminal condition by V ( T, x T ) = S ( T, x T ) = c x T for each x T ∈ Z + . Then, V (0 , x ) = V ∗ ( x ) for every x ∈ Z + (Oh and ¨Ozer 2016). Moreover, the optimal order amount µ ∗ k ( x k )is equal to m ∗ where m ∗ attains the infimum inf m ∈ Z + { c ( m ) + G ( k, x k + m ) } in (2.7). Furthermore,an optimal stopping time τ ∗ ∈ T is the one that stops the process if S ( k, x k ) ≤ J ( k, x k ) in (2.6),in other words, τ ∗ := min (cid:110) k ∈ T : S ( k, X π ∗ k ) ≤ J ( k, X π ∗ k ) (cid:111) . where π ∗ denotes an optimal policy. The next section provides related problems. In this subsection, we develop benchmark models for the main model in Subsection 2.2. To keeptrack of different formulations, we use the notation V a/b/c to describe the decisions allowed. Inthis way, we introduce levels of flexibility mentioned to be included or excluded in the problemas long as the actual environment allows. The symbol a represents how the time for stopping isdecided: a = D means that the decision is dynamic and it is described by a stopping time; a = S means that the decision is static and made at time zero (we call it switching time); a = T meansthat we decide not to stop until the end of horizon. The symbol b represents the number of orders10llowed: If b = M for some M ∈ Z + , then the manufacturer can place M orders throughout thehorizon; b = ∞ means that the manufacturer can place an order at each period with no restriction.Finally, the symbol c represents alternatives with respect to the timing of the first order: c = Z means that the first order must be placed at time zero and c = F means that the manufactureris free to place the first order at any time. Table 1 summarizes the notation. For instance, ourmain DP model in Section 2.2 can be denoted by D/ ∞ /F . This notation is helpful to keep trackof decisions considered for different formulations while comparing them. In numerical analyses, wecompare these models as well as our main model to show the value of our approach. Note that onecan extend this taxonomy by allowing additional decision features such as disposal of inventory,repairing spare parts, or extending the warranty period. a Time to Stop Holding Inventory S Static decision – made at time 0 D Decision is made dynamically T Do not stop until the end of horizon Tb Maximum Number of Orders M Maximum M ∈ N orders ∞ Unrestricted number of orders c Order Time Z First order must be placed at time zero F First order can be placed at any time
Table 1: Notation for benchmark models.
This benchmark dynamic programming formulation analyzes the effects of delaying a single order,and Cattani and Souza (2003) present this idea in a different setting. Let z ∈ { , } be thenumber of remaining orders that the manufacturer can place. The following dynamic programmingalgorithm describes this formulation. Define the terminal cost for each x T ∈ Z + and z ∈ { , } by V D/ /F ( T, x T , z ) = S ( T, x T ) . If z = 0, then define the backward dynamic programming algorithmfor each k ∈ { T − , T − , . . . , } and x k ∈ Z + by V D/ /F ( k, x k ,
0) = min (cid:110) S ( k, x k ) , C ( k, x k ) + e − δ E (cid:104) V D/ /F ( k + 1 , (cid:0) x k − ( N k +1 − N k ) (cid:1) + , (cid:105)(cid:111) . If z = 1, then define the backward dynamic programming algorithm for each k ∈ { T − , T − , . . . , } x k ∈ Z + by V D/ /F ( k, x k , (cid:26) S ( k, x k ) , C ( k, x k ) + e − δ E (cid:104) V D/ /F ( k + 1 , ( x k − ( N k +1 − N k ) + , (cid:105) , min { inf m ∈ Z + (cid:110) c ( m ) + C ( k, x k + m ) + e − δ E (cid:104) V D/ /F ( k + 1 , ( x k + m − ( N k +1 − N k )) + , (cid:105)(cid:111) (cid:27) . A prevalent assumption in the literature is that a final order has to be placed at time zero. Therefore,we develop a dynamic programming algorithm to reflect the manufacturer’s decision when only oneorder can be placed at time zero and the manufacturer can stop holding inventory at any time.This model resembles the one presented by Pourakbar et al. (2012). Define the backward dynamicprogramming algorithm for each k ∈ { T − , T − , . . . , } and x k ∈ Z + by . V D/ /Z ( k, x k ) = min (cid:110) S ( k, x k ) , C ( k, x k ) + e − δ E (cid:104) . V D/ /Z ( k + 1 , (cid:0) x k − ( N k +1 − N k ) (cid:1) + ) (cid:105)(cid:111) . Also define the terminal condition by . V D/ /Z ( T, x T ) = S ( T, x T ). The optimal order quantity attime zero and the value of this dynamic program is found by calculating V D/ /Z ( x ) = inf m ∈ Z + (cid:110) c ( m ) + . V D/ /Z (0 , x + m ) (cid:111) , x ∈ Z + . (2.8)It is possible to see the following relation between D/ /F and D/ /Z : While solving the model D/ /F , if we decide to place an order at time k , then we solve the model D/ /Z with a differenttime horizon that is equal to T − k . ∞ /F or S/1/F or S/1/Z S/ ∞ /F can be formulated as a special case of D/ ∞ /F . For each switching time k ∈ T , weimplement a restricted version of the dynamic programming algorithm and select the best switchingtime k ∗ .Moreover, S/ /F can be formulated as a special case of D/ /F presented in Subsection 2.3.1.We modify the value function V D/ /F by eliminating the stopping option with cost (cid:101) S ( k, x k ) andsolve the DP algorithm. 12inally, S/ /Z can be formulated as a special case of D/ /Z presented in Subsection 2.3.2. Forevery t ∈ T , we implement D/ /Z without being able to stop. ∞ /F or T/1/F or T/1/Z These benchmark models are further special cases of S/ ∞ /F , S/ /F and T / /F . They resemblethe classical inventory models which can be solved by means of standard DP algorithms. Forinstance, see Beyer et al. (2010, Chapter 4). > These models can be formulated by using a similar idea as for D/ /F in Subsection 2.3.1 byrepresenting the number of setups as a state variable.Finally, Table 2 relates the benchmark models with some of the more relevant literature men-tioned before.Model Explanation Related Studies D/ ∞ /F Multiple orders and stopping time This study D/ /Z Single order at time zero and stopping time (Pourakbar et al. 2012, Frenket al. 2019b,c) S/ /Z Single order at time zero and switching time (Frenk et al. 2019a)
T / ∞ /F Multiple orders without outside source (Teunter and Haneveld 2002,Pin¸ce and Dekker 2011, Inder-furth and Mukherjee 2008, In-derfurth and Kleber 2013)
T / /F A delayed single order without outside source (Cattani and Souza 2003)
Table 2: Benchmark models which resemble the ones presented in previous studies. The Notation for modelsis presented in Table 1.
To reduce the computation of the stopping cost S ( k, x k ) in D/ ∞ /F and other benchmark models,we reformulate V ∗ ( x ) in (2.5). To that end, define for each k ∈ T and x ∈ Z + the new cost ofstopping by (cid:101) S ( k, x ) := c x (cid:101) C ( k, x ) := c E (cid:20)(cid:90) k +1 k e − δ ( u − k ) ( x − ( N u − N k )) + du (cid:21) + E (cid:20)(cid:90) k +1 k e − δ ( u − k ) [ c ( u ) − c ( u )] dN u (cid:21) − E (cid:34)(cid:90) ( k +1) ∧ σ kx k e − δ ( u − k ) c ( u ) dN u (cid:35) . (3.1)Then, the following proposition is used to reformulate V ∗ ( x ) in (2.5). Proposition 3.1.
For every π ∈ Π , E (cid:34) τ − (cid:88) k =0 e − δk (cid:18) c ( µ k ( X k )) + C ( k, X k + µ k ( X k )) (cid:19) + e − δτ S ( τ, X τ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X = x (cid:35) = E (cid:34) τ − (cid:88) k =0 e − δk (cid:18) c ( µ k ( X k )) + (cid:101) C ( k, X k + µ k ( X k )) (cid:19) + e − δτ (cid:101) S ( τ, X τ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X = x (cid:35) + A. where A := E (cid:104)(cid:82) T e − δu c ( u ) dN u (cid:105) ∈ R + . The proof of Proposition 3.1 is in Appendix C.The following corollary defines a new problem (cid:101) V ∗ ( x ) that is equivalent to V ∗ ( x ) in (2.5). Thecorollary directly follows from the Proposition 3.1. Corollary 3.2.
Let x ∈ Z + and define (cid:101) V ∗ ( x ) := inf π ∈ Π E (cid:34) τ − (cid:88) k =0 e − δk (cid:18) c ( µ k ( X k )) + (cid:101) C ( k, X k + µ k ( X k )) (cid:19) + e − δτ (cid:101) S ( τ, X τ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X = x (cid:35) (3.2) Then, V ∗ ( x ) = (cid:101) V ∗ ( x ) + A for every x ∈ Z + . In view of Corollary 3.2, we aim to solve the problem (cid:101) V ∗ ( x ). Again, as stated by Oh and ¨Ozer(2016), the following stochastic dynamic programming algorithm can be used for this purpose.Define the backward recursion for each k ∈ { T − , T − , . . . , } and x k ∈ Z + by (cid:101) V ( k, x k ) = min (cid:110) (cid:101) S ( k, x k ) , (cid:101) J ( k, x k ) (cid:111) (3.3)where (cid:101) J ( k, x ) := inf m ∈ Z + (cid:110) c ( m ) + (cid:101) G ( k, x k + m ) (cid:111) (3.4)and where (cid:101) G ( k, y ) := (cid:101) C ( k, y ) + e − δ E (cid:104) (cid:101) V (cid:0) k + 1 , ( y − ( N k +1 − N k )) + (cid:1)(cid:105) . (cid:101) V ( T, x T ) = (cid:101) S ( T, x T ) = c x T for each x T ∈ Z + . Then, thefollowing corollary states that the DP algorithm described by the recursion in (3.3) can be used tosolve this new re-formulated problem. It follows from the definition of (cid:101) V and Oh and ¨Ozer (2016). Corollary 3.3.
It holds that (cid:101) V (0 , x ) = (cid:101) V ∗ ( x ) for every x ∈ Z + . Moreover, the optimal orderamount (cid:101) µ ∗ k ( x k ) is equal to m ∗ where m ∗ attains the infimum inf m ∈ Z + (cid:110) c ( m ) + (cid:101) G ( k, x k + m ) (cid:111) in (3.4) . Furthermore, an optimal stopping time (cid:101) τ ∗ ∈ T is the one that stops the process if (cid:101) S ( k, x k ) ≤ (cid:101) J ( k, x k ) in (3.3) , in other words, (cid:101) τ ∗ := min (cid:110) k ∈ T : (cid:101) S ( k, X (cid:101) π ∗ k ) ≤ (cid:101) J ( k, X (cid:101) π ∗ k ) (cid:111) , where (cid:101) π ∗ := ( (cid:101) τ ∗ , (cid:101) µ ∗ , . . . , (cid:101) µ ∗ T ) is an optimal policy. Finally, the following proposition enables us to compute (cid:101) C so that we can solve (cid:101) V (0 , x ). Proposition 3.4.
For every k ∈ T , the one-period operation cost (cid:101) C ( k, x ) in (3.1) satisfies (cid:101) C ( k,
0) = (cid:90) k +1 k e − δ ( u − k ) [ c ( u ) − c ( u )] λ ( u ) du when x = 0 , and (cid:101) C ( k, x ) = c x − (cid:88) n =0 n (cid:88) i =0 (cid:90) k +1 k e − δ ( u − k ) e − (Λ( u ) − Λ( k )) (Λ( u ) − Λ( k )) i i ! du + (cid:90) k +1 k e − δ ( u − k ) [ c ( u ) − c ( u )] λ ( u ) du − x (cid:88) i =0 (cid:90) k +1 k e − δ ( u − k ) c ( u ) λ ( u ) e − Λ( u ) − Λ( k ) (Λ( u ) − Λ( k )) i i ! du (3.5) when x ≥ . The proof of Proposition 3.4 is in Appendix C. ∞ /F The main problem D/ ∞ /F falls into the category of an optimal stopping problem with additionaldecisions. More specifically, the inventory situation considered is a multi-period problem withlost sales in a setting with periodic review, finite horizon, non-stationary stochastic demand, non-15tationary costs (continuously charged), and zero ordering lead-time. We present two results fromthe literature which partially gives an understanding of the problem in question. Result 1. (Oh and ¨Ozer 2016, Proposition 11) Let the one-step benefit function be defined as M t ( a t , x t ) := α E [ S t +1 (˜ x t +1 ( a t , x t ))] + C t ( a t , x t ) − S t ( x t ) , where a t denotes the order amount at the beginning of time t , and in our notation, ˜ x t +1 ( a t , x t ) := (cid:0) x t + a t − N tt +1 (cid:1) + . Moreover, S t +1 ( x ) = c x and C t ( a t , x t ) = (cid:101) C ( t, a t + x t ) . Also define B t ( a t , x t ) := α E [ V t +1 (˜ x t +1 ( a t , x t ))] + C t ( a t , x t ) − S t ( x t ) , and ¯ B t ( x t ) := sup a t ∈ Z + B t ( a t , x t ) . When M t ( a t , x t ) is increasing (resp., decreasing) in x t , and ˜ x t +1 ( a t , x t ) is stochastically increasingin x t ∈ Z + for every a t and every t, then the following statements are true for every t: 1. ¯ B t ( x t ) is increasing (resp., decreasing) in x t . 2. A threshold policy that stops the process if x t ≤ ¯ x t (resp., x t ≥ x t ) is optimal. When the one-step benefit function is increasing (or decreasing) in the inventory level, a thresh-old policy that stops the process if inventory level is below a threshold (or above a threshold) isoptimal. In our case, we cannot determine whether the one-step benefit function is increasing ordecreasing as we have fixed ordering cost, and we allow for the timing of lost sales to affect ourcost computations.
Result 2. (Beyer et al. 2010, Theorem 4.2) Consider the inventory situation with multi-periodsand lost sales in a setting with periodic review, finite horizon, non-stationary stochastic demand,non-stationary costs (discretely charged), and non-zero fixed ordering cost with zero ordering leadtime. When there is no stopping time, the optimal order policy is a time-dependent ( s, S ) policy.Note that, other than our problem being a stopping-time problem, the structure mentionedin Beyer et al. (2010, Theorem 4.2) is the same as ours (actually the cost terms in Beyer et al.(2010, Theorem 4.2) are more general, but the costs are not computed continuously). So, given aswitching time (stopping time determined at time zero), Result 2 will hold in our setting.We state following proposition for the structural properties of our problem.16 roposition 3.5.
The optimal solution (cid:101) π ∗ defined in Corollary 3.3 can be summarized by threedisjoint regions for the incoming stock value for every decision epoch as well as order-up-to levels.1. For each time k ∈ T , define the stopping set by R Sk := (cid:110) x ∈ Z + : (cid:101) S ( k, x ) ≤ (cid:101) J ( k, x ) (cid:111) ; define the ordering set by R Ok := (cid:26) x ∈ Z + : inf m ≥ (cid:110) c ( m ) + (cid:101) G ( k, x + m ) (cid:111) < min (cid:110) (cid:101) S ( k, x ) , (cid:101) G ( k, x ) (cid:111)(cid:27) ; define the continuation set by R Ck := (cid:110) x ∈ Z + : (cid:101) V ( k, x ) = (cid:101) G ( k, x ) < (cid:101) S ( k, x ) (cid:111) . Then, the ordering region { ( k, x ) ∈ T × Z + : x ∈ R Ok } , the stopping region { ( k, x ) ∈ T × Z + : x ∈ R Sk } , and the continuation region { ( k, x ) ∈ T × Z + : x ∈ R Ck } partition T × Z + .2. (cid:101) π ∗ is a policy such that if X (cid:101) π ∗ k ∈ R Sk , then we stop; if X (cid:101) π ∗ k ∈ R Ck , then we continue withouttaking an action; if X (cid:101) π ∗ k ∈ R Ok , then we place an order and increase inventory to an order-up-tolevel. Hence, for each inventory level in R Ok , there exists an optimal order-up-to level. The proof of Proposition 3.5 is in Appendix D.
Remark 3.6.
The regions are neither convex nor monotone increasing/decreasing in general, sincethe problem parameters are non-stationary and there is a fixed cost of ordering.To illustrate an optimal solution and substantiate Remark 3.6, we present the solution of dy-namic programming algorithm in (3.3) by using a data set presented by Frenk et al. (2019a). We alsoassume that the setup cost is K = 2000. After calculating the value function for k ∈ { , , . . . , } and x ∈ { , , . . . , } , we show a subset of ordering, stopping, and continuation regions in Table3. A point ( x, k ) is in the continuation, stopping, or ordering region if its value is represented bythe numbers 0, 1, or 2, respectively.By looking at Table 3, we make several observations on the optimal solution as well as theconstructed regions. First, we can see that the regions are not convex or monotone increasing/de-creasing, especially the stopping region. The stopping region is not monotone possibly because ofchanges in the intensity rate and the cost of outside source. On the one hand, when the intensity17ate declines over time, the stopping region (near the points where we face the risk of lost salesand may wish to start using outside source) becomes smaller. On the other hand, when the cost ofoutside source decreases over time, the stopping region gets larger. This is in line with Remark 3.6.Moreover, we can see that the ordering and stopping regions are intertwined. Therefore, there doesnot exist a threshold re-order level such that an order is placed if inventory is below this threshold. x k
20 21 22 23 24 25 39 40 43 44 47846 0 0 0 0 0 0 0 1 1 1 1770 0 0 0 0 0 0 0 0 1 1 1375 0 0 0 0 0 0 0 0 0 1 1345 0 0 0 0 0 0 0 0 0 0 17 2 2 2 0 0 0 0 0 0 0 06 2 2 2 0 0 0 0 0 0 0 05 2 2 2 2 0 0 0 1 1 0 04 2 2 2 2 0 1 1 1 1 0 03 2 2 2 2 1 1 1 1 1 0 02 2 2 2 1 1 1 1 1 1 1 11 2 2 2 1 1 1 1 1 1 1 10 2 2 1 1 1 1 1 1 1 1 1
Table 3: The illustration of an optimal solution and subsets of ordering, stopping and continuation regions,constructed by using data from the literature. Each row and column corresponds to a different inventorylevel x and period k respectively. If a point ( x, k ) has a value 0, 1, or 2, it means that this point belongs tocontinuation, stopping and ordering region respectively. Remark 3.7.
The order-up-to levels specified in the second part of Proposition 3.5 for consecutivetime periods can be increasing or decreasing.The following example substantiates Remark 3.7. We use a data set presented by Frenk et al.(2019a) and assume that setup cost is K = 2000. Figure 1 depicts order-up-to level x t + µ t ( x t ) ∈ Z + at the border of stopping region, that is, at the highest x t value such that ordering is the bestaction. The order-up-to levels are decreasing over time except that at two time points we observepositive jumps. It seems natural to see decreasing order up-to-levels, since expected demand rateis decreasing over time and the time horizon is finite. The positive jumps are most likely due tothe setup cost. At the time point that is right before the jump, you may choose to place a smallorder to utilize the multiple orders opportunity. At the time point that is right after the jump,you may want to place a larger order to avoid future setup costs, considering that future demandis expected to be lower. Indeed, we do not observe these positive jumps when setup cost is K = 0.18 igure 1: Order-up-to level x t + µ ∗ t ( x t ) ∈ Z + at the border of stopping region, that is, at the highest x t suchthat ordering is the best action. This subsection provides an expression for the probability that n ∈ Z + orders are placed throughoutthe end-of-life phase. We show the result when n = 1 and then generalize it to an arbitrary n ∈ Z + .Let us first construct the following setting: Solve D/ ∞ /F and find R Ok , R Ck and R Sk defined in inProposition 3.5. Next, for an ease of notation, assume that initial inventory level x = 0 so that weplace an order at time zero. We describe the case with positive initial inventory level at the end.Let us introduce the notation N tt + s = N t + s − N t for arbitrary t, s ∈ { , , . . . , T − } . Let P t ( t, x t ) ∈ [0 ,
1] denote the probability that no order is placed from period t + 1 to T , given thatwe start period t ∈ T with x t ∈ Z + units of inventory, and given that 1 order is already placed attime t = 0. Then, the function P t : T × Z + → [0 ,
1] is defined by P t ( t, x t ) = P (cid:32) T (cid:91) s = t +1 (cid:18) s − (cid:92) k = t +1 (cid:8) x t − N tk ∈ R Ck (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Continue until time s ∩ (cid:8) x t − N ts ∈ R Ss (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Stop at time s (cid:19)(cid:33) , where, by convention, (cid:84) tk = t +1 { x t − N tk ∈ R Ck } ∩ { x t − N ts ∈ R Ss } = { x t − N ts ∈ R Ss } . The aboveexpression of P t stems from the fact that the inventory process must not hit the ordering region, soit either stays in the continuation region or hits the stopping region (and does not evolve afterward).Next, define P ∈ [0 ,
1] to be the probability that 1 order is placed throughout the horizon, thatis, P := P t (0 , y ), where t = 0 and y ∈ Z + denotes the order-up-to level at time zero. The19ollowing proposition enables us to calculate P t ( t, x t ). Proposition 3.8.
Assume that t = 0 . Then, for each t ∈ T = { , , . . . , T } and x t ∈ Z + , thefunction P t : T × Z + → [0 , satisfies the recursion P t ( t, x t ) = P (cid:8) x t − N tt +1 ∈ R St +1 (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Enter stopping region during [ t, t + 1] + (cid:88) n ∈ x t − R Ct +1 P t ( t + 1 , x t − n ) P (cid:8) N tt +1 = n (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Stay in continue region and move to next period (3.6) with the terminal condition P t ( T, x T ) = 1 for all x T ∈ Z + . Here, x t − R Ct +1 := (cid:8) x t − r : r ∈ R Ct +1 (cid:9) . The proof of Proposition 3.8 is in Appendix D.Next, we proceed with the probability that n orders are placed. Let t , . . . , t n ∈ T with0 = t < · · · < t n denote the candidate times for n orders. Following Proposition 3.5, let y t , . . . , y t n ∈ Z + denote the corresponding order-up-to levels. Let P nt ,...,t n ( t, x t ) denote the probability that n orders are to be given and order times are t , . . . , t n while looking at system from time t ∈ T onwardwith x t ∈ Z + inventory. Then, the function P nt ,...,t n : T × Z + → [0 ,
1] is defined for each t < t and x t ∈ Z + by P nt ,...,t n ( t, x t ) = P (cid:32)(cid:18) t − (cid:92) k = t +1 (cid:8) x t − N tk ∈ R Ck (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Continue until t ∩ (cid:8) x t − N tt ∈ R Ot (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Place an order at t (cid:19) ∩ (cid:18) t − (cid:92) k = t +1 (cid:8) y t − N t k ∈ R Ck (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Continue until t ∩ (cid:8) y t − N t t ∈ R Ot (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Place an order at t (cid:19) ∩ . . .. . . (cid:18) t n − (cid:92) k = t n − +1 (cid:110) y t n − − N t n − k ∈ R Ck (cid:111)(cid:124) (cid:123)(cid:122) (cid:125) Continue until t n ∩ (cid:110) y t n − − N t n − t n ∈ R Ot n (cid:111)(cid:124) (cid:123)(cid:122) (cid:125) Place an order at t n (cid:19) ∩ (cid:18) T (cid:91) s = t n +1 (cid:18) s − (cid:92) k = t n +1 (cid:8) y t n − N t n k ∈ R Ck (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Continue until s ∩ (cid:8) y t n − N t n s ∈ R Ss (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Stop at s (cid:19)(cid:19)(cid:33) . (3.7)Finally, define P n ∈ [0 ,
1] to be the probability that n orders are placed, that is, P n := (cid:88) t ,...,tn ∈{ , ,...,T } t Remark 3.10. In Proposition 3.9, it is possible to remove the assumption that t = 0. In casethe initial inventory is sufficient so that we do not place an order at time zero, we can simply set y ∈ Z + to be the initial inventory level instead of order-up-to level at time zero.We describe the following algorithm to compute the probability that n orders are given.1. Solve D/ ∞ /F and construct the sets R Ok , R Ck , R Sk by using Proposition 3.5.2. Set n = 1 and P t ( T, x T ) = 1 for all x T ∈ Z + . For each t ∈ { T − , T − , . . . , } and x t ∈ Z + ,compute P t ( t, x t ) backwards by using (3.6). Set P = P t (0 , y ).3. For n ∈ { , , . . . , T } ,(a) For t , t , . . . , t n ∈ { , , . . . , T } such that 0 = t < t < · · · < t n ,(i) For each x t − ∈ Z + , set the terminal condition P nt ,...,t n ( t − , x t − ) in (3.10).(ii) For each t ∈ { t − , . . . , } and x t ∈ Z + , compute P nt ,...,t n ( t, x t ) backwards by using (3.9).(b) Compute P n by using (3.8).Next, we state that by using similar conditioning arguments, we can also compute the followinginformation that can be useful for the manufacturer’s operations. Remark 3.11. Expected number of orders given can be computed by using P n . Moreover, we cancompute the distribution of time that next order is placed, in other words, distribution of cyclelength between two consecutive orders. Furthermore, it is possible to compute the probability thatan order is not placed in the next n periods. Similarly, for a fixed initial inventory level, we canfind the expected number of periods before placing the first order.21 .3 Other Useful Structural Results In this subsection, we consider a more specialized environment to analytically characterize therelationship between ordering inventory and using an outside source. To that end, we build ouranalysis on the problem S/ /Z . According to this problem, the manufacturer can place a singleorder at time zero, and stopping decision is static and made at time zero (we call it switchingtime). We also assume that inventory levels take value in I := { , , . . . , I max } where I max < ∞ canbe arbitrarily large. The inventory evolution is described by X k +1 = ( X k − ( N k +1 − N k )) + with X = x ∈ I and the problem S/ /Z is formulated as V S/ /Z ( x ) = inf τ ∈ T ,m ∈ I (cid:40) c ( m ) + E (cid:34) τ − (cid:88) k =0 e − δk C ( k, X k ) + e − δτ S ( τ, X τ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X = x + m (cid:35)(cid:41) , x ∈ I . where the ordering cost function c is defined in (2.1). Since actions are taken only at time 0 andtime τ , we can combine the one-period operation costs as well as the stopping cost. Let us definethe combined cost function C : I × T → by C ( x, τ ) := c E (cid:20)(cid:90) τ e − δu ( x − N u ) + du (cid:21) + E (cid:20)(cid:90) ττ ∧ σ x e − δu c ( u ) dN u (cid:21) + E (cid:20)(cid:90) Tτ e − δu c ( u ) dN u (cid:21) + c e − δτ E [( x − N τ ) + ] . (3.11)Then, we can express the problem S/ /Z as V S/ /Z ( x ) = inf τ ∈ T ,m ∈ I { c ( m ) + C ( m + x , τ ) } , x ∈ I . To analyze changes with respect to switching time τ , we extend the feasible set from T to [0 , T ]and study the following extended problem. V S/ /ZE ( x ) = inf τ ∈ [0 ,T ] (cid:110) V S/ /ZE ( x , τ ) (cid:111) , x ∈ I , (3.12)where V S/ /ZE ( x , τ ) = inf m ∈ I { c ( m ) + C ( m + x , τ ) } , x ∈ I , τ ∈ [0 , T ] . (3.13)22 .3.1 Stopping Time and Order Levels are Alternatives to Manage the InventorySystem with Different Roles and Possible Synergy By proceeding as in Frenk et al. (2019a), we can show that the function x (cid:55)→ C ( x, τ ) is discreteconvex for a fixed τ ∈ [0 , T ] under the following assumptions. NON-INC The functions c and (cid:101) c ( u ) := c ( u ) − c ( u ) are right-continuous with left-limits, piece-wise smooth (that is, differentiable except at finitely many points) and non-increasing. POS The functions c and (cid:101) c are positive (maps from [0 , T ] to R + ) and the quantities c and c − δc are in R + .Let us define the first order difference operator of C by ∆ x C ( x, τ ) := C ( x + 1 , τ ) − C ( x, τ ) andsecond order difference operator of C by ∆ x C ( x, τ ) := ∆ x C ( x + 1 , τ ) − ∆ x C ( x, τ ). By proceedingas in Porteus (2002), we can state in the following lemma that ( s, S )-policy is an optimal policy todescribe ordering decision at time zero, for each fixed switching time τ . Lemma 3.12. Assume NON-INC and POS . Fix x ∈ Z + and τ ∈ [0 , T ] . Let S ( τ ) ∈ I be aglobal minimizer of the discrete convex function x (cid:55)→ ¯ cx + C ( x, τ ) defined by S ( τ ) = min { x ∈ I : ¯ c + ∆ x C ( x, τ ) ≥ } . Let s ( τ ) ∈ I be defined by s ( τ ) = min { x ∈ I : C ( x, τ ) < K + ¯ c ( S ( τ ) − x ) + C ( S ( τ ) , τ ) } . Then, the policy δ ( s ( τ ) ,S ( τ )) : I → I defined by δ ( s ( τ ) ,S ( τ )) ( x ) = (cid:40) S ( τ ) if x < s ( τ ) x otherwiseis optimal for the problem V S/ /ZE ( x , τ ) in (3.13) . The following proposition shows that the order-up-to level S ( τ ) is a non-decreasing function of τ when suitable conditions hold. Proposition 3.13. Assume NON-INC and POS . Also assume that λ is a non-increasing func-tion. Then, for every (cid:15), τ , τ ∈ [0 , T ] such that ( i ) τ < τ (Switching time is delayed) ( ii ) S ( τ ) ≥ S ( τ ) (Order-up-to level has not increased yet) iii ) S ( τ ) ≤ Λ( τ ) (Expected total demand exceeds order-up-to level) ( iv ) c ≤ (cid:20) Λ( u ) − S ( τ )Λ( u ) (cid:21) λ ( u ) c ( u ) , for all u ∈ [ τ , τ + (cid:15) ] (Cost rate of outside source λ ( u ) c ( u ) is still high in the infinitesimal future)we have S ( τ ) ≤ S ( τ + (cid:15) ) . The proof of Proposition 3.13 is in Appendix E. Remark 3.14. If condition ( iv ) of Proposition 3.13 holds, then condition ( iii ) holds as well since c ∈ R + . Still, we write condition ( iii ) to understand the necessary structure behind this insight.Moreover, if condition ( ii ) does not hold, it means that S increases as we delay the switching time,which is the main message of this structural insight. Finally, the expression λ ( u ) c ( u ) denotes thecost rate of using outside source since c ( u ) λ ( u ) = lim (cid:15) ↓ (cid:90) τ + (cid:15)τ c ( u ) λ ( u ) du = lim (cid:15) ↓ E (cid:20)(cid:90) τ + (cid:15)τ c ( u ) dN u (cid:21) Proposition 3.13 enables us to extract the following insights. • If the time before we dispose inventory is delayed, then the order amount that will be used tosatisfy the demand increases. • Since more inventory is held to satisfy the increasing total demand, it may be reasonable to getrid of inventory earlier. • If demand rate is low, then holding inventory may not be reasonable. If, in addition, cost ofoutside source is low, then the usage of it can be a significant alternative. • In case the outside source is not available before some time ( τ ), then most likely we need toincrease the order-up-to level. This shows that stopping time and order levels are alternativesfor managing the inventory system. Proposition 3.13 above shows how switching time τ affects the beginning inventory level S ( τ ).Next, we show xhow the beginning inventory level affects the switching time. Since S ( τ ) is already24 function of τ , we fix the inventory level x and analyze how this fixed x affects τ . Hence, we areinterested in the solution of the following problem.inf τ ∈ [0 ,T ] {C ( x, τ ) } . (3.14)Let τ ∗ ∈ [0 , T ] denote the best switching time that attains the infimum in (3.14). The followingproposition shows an upper bound on τ ∗ . Proposition 3.15. Assume NON-INC and POS . Let τ ub be the smallest τ value satisfying ˜ c ( T ) ≥ P { N τ ≤ x − } [ c ( τ ) + c ] . (3.15) If the inequality does not hold for any τ ∈ [0 , T ] , then let τ ub = T . Then, τ ∗ ≤ τ ub . The proof of Proposition 3.15 is in Appendix F.The following proposition shows a lower bound on the best switching time τ ∗ . Proposition 3.16. Assume NON-INC and POS . Also assume that λ is a non-increasing func-tion. Let τ lb be the largest τ value satisfying λ ( τ ) ≥ and P { N τ ≤ x − } [ c ( τ ) + c ] ≥ x ( c − δc ) + ˜ c (0) . (3.16) If the inequality does not hold for any τ ∈ [0 , T ] , then let τ lb = 0 . Then, τ lb ≤ τ ∗ . The proof of Proposition 3.16 is in Appendix F.Here, we note that P ( N τ ≤ x − 1) [ c ( τ ) + c ] is a decreasing function of τ and an increasingfunction of x . Hence, we are able to extract the following insights for the bounds on the bestswitching time τ ∗ . • The upper bound τ ub can be less than T if the number of arrivals up to τ ub sufficiently exceedsthe beginning inventory level x (so we face the risk of penalty) and if the cost of outside sourcesufficiently declines. In such a case, it may be better to stop holding inventory earlier than theend of horizon. • The lower bound τ lb can be larger than 0 if the number of arrivals before τ lb is considerably lessthan inventory level x and if cost of outside source is still high. In such a case, it may be betterto delay the usage of outside source. 25 Numerical Analysis In Section 3 we demonstrate with some structural results that flexibilities that could be consideredwithin the end-of-life inventory management problem are promising. This section provides numer-ical results regarding the output of dynamic programming algorithms and the analytical resultspresented earlier. For numerical calculations, we further assume that the cost of outside source isgiven as c ( u ) = ¯ c e − γu , where ¯ c ∈ R + is a constant and γ ∈ R + is the decline rate. Moreover, weassume that c ( u ) = ¯ c + c ( u ) for some ¯ c ∈ R + which we interpret as the penalty of lost sales.We allow c to be negative or positive. Finally, to facilitate computations, we assume that theintensity function λ ( t ) is piecewise constant whose value changes at every t ∈ { , , . . . , T − } andit is constant during [ t, t + 1]. may change as a function of t and we assume it is constant during[ t, t + 1] to simplify computations. Table 4 shows the set of parameter values and Table 5 showsthe set of intensity functions used in the numerical analysis. We note that the ranges as stated inTable 4 incorporate in relative terms the case data considered by Frenk et al. (2019a). Additionally,we specify a fixed ordering cost and various forms of the intensity function to represent the rateof decrease, as presented in Table 5. Note that for different cases, the demands are all comparableas the total expected demand over the horizon is kept constant. The parameter settings and theircorresponding numbers are presented in Appendix G.We code our models by using MATLAB and run them on a laptop computer with an Intel(R)Core(TM) i7-7700HQ processor with 2.80GHz CPU. The computation of one-period operation cost C ( k, x ) (for all k and x values) takes approximately 800 and 1750 seconds of CPU time when T = 50 and T = 100 respectively. The computation of DP algorithm takes approximately 25 and60 seconds of CPU time when T = 50 and T = 100 respectively. We verified our code by comparingthe output of benchmark models and the output presented by Frenk et al. (2019a).26arameter Name Set of ValuesUnit Procurement Cost ¯ c = 100Setup Cost K ∈ { , c, c } Holding Cost c = 0 . c Penalty Cost ¯ c = { c, c } c (0) at time zero ¯ c = 2¯ c Discount of c γ ∈ (cid:8) − , . (cid:9) Scrapping Cost c ∈ { ¯ c/ , − ¯ c/ } Planning Horizon T ∈ { , } Time Discount δ ∈ (cid:8) − , . (cid:9) Expected Total Demand (cid:82) T λ ( u ) du = 500Intensity Functions Convex, Concave, Linear, ConstantPresented in Table 5 Table 4: Parameter values used in numerical analysis. The total number of parameter settings is 384. Basecase parameters are ¯ c = 2¯ c, γ = 0 . , c = ¯ c/ , T = 50 , δ = 0 . λ . Function Type Expression T = 50 T = 100Convex λ ( t ) = λ (0 . t λ ( t ) = λ (0 . t Concave λ ( t ) = λ − (0 . t ) λ ( t ) = λ − (0 . t ) Linear λ ( t ) = λ − . t λ ( t ) = λ − . t Constant λ ( t ) = λ λ ( t ) = λ Table 5: Piecewise constant intensity functions λ used in numerical analysis. The value of λ ( t ) changes atevery t ∈ { , , . . . , T − } and it is constant during [ t, t + 1]. Initial point λ ∈ R + is selected such thatexpected total demand (cid:82) T λ ( t ) dt is equal to 500. We present our computational results in two subsections. In Subsection 4.1, we compare thebenefits of our approach with the benchmark models. In Subsection 4.2, we analyze the effects ofproblem parameters. These analyses give us further insights on how our approach can be used tohandle the end-of-life management problem effectively. In this subsection, we compare all models by using the parameter values in Table 4. To be morespecific, we consider the settings (as numbered) which are presented in Appendix G. As comparisonbasis, we consider the percentage increase in the expected discounted total cost over the horizonfor not employing a model which utilizes more flexibility (or flexibilities) over the assumed currentmodel with those flexibilities. We summarize the benefits in four subsections: The first threesubsections assess the contribution of any specific flexibility over the current one in an isolated27anner, whereas the fourth subsection analyzes combination effects.We initially demonstrate the benefits over a base case with the parameters ¯ c = 2¯ c = 200 , γ =0 . , c = ¯ c/ , T = 50 , δ = 0 . λ . The setting number for base case is 1 inAppendix G. The base case is a setting which is one of the closest to the parameters used in Frenket al (2019a). We then show results for all the settings and report minimum, average, and maximumpercentage increase in the expected discounted cost if those flexibilities are not considered. We endeach subsection with a remark summarizing the findings. We first present the comparisons under the base case. Table 6 presents percent loss if the numberof orders is limited to 1 compared to the possibility of multiple ordering under two cases: with nostopping time ( a = T as given in the taxonomy) and with stopping time ( a = D ). The case withstopping time is presented in parentheses for various values of x and K . K x Table 6: Percentage difference 100 × T/ /Z − T/ ∞ /FT/ ∞ /F for different initial inventory x and setup cost K values.The numbers in parentheses show the value of 100 × D/ /Z − D/ ∞ /FD/ ∞ /F under the same parameter setting. Basecase parameters are used. As expected, the use of stopping time is an effective tool as observed with lower percentagesin parentheses. As expected, when x = 0 and K = 5000, we have small percentages indicatingthat the traditional approach, assuming a large fixed ordering cost and hence ordering only once,has a strong logic. However, with some initial inventory, one can observe that the penalty of notemploying a more flexible approach can be significant.We present our results for all our runs in Table 7. Note that we only report the expected percentloss figures under the case where we employ stopping time. For the moderate value of the fixedordering cost ( K = 1000), the average penalty percentages for different initial inventory values areall above 10%. When we analyze the settings where we attain maximum or minimum values, wenotice that most are the settings where we assume a constant intensity rate over time ( K x Table 7: Percentage difference 100 × D/ /Z − D/ ∞ /FD/ ∞ /F for different initial inventory x and setup cost K values. For each x and K , we present maximum, average and minimum values over all parameter settings.Set Remark 4.1. Allowing multiple orders is important for systems with reasonable fixed orderingcost. However, the advantages may be offset by the use of stopping time and/or delaying the firstorder when there is some initial inventory. We present the comparisons under the base case parameters in two tables. Table 8 presents thepercent loss if only one order is given at time zero under two cases: with a switchover time ( a = S as given in the taxonomy) as to no stopping time ( a = T ) and with a stopping time ( a = D ) as toa switchover time ( a = S ). The latter case is presented in parentheses for various values of x and K . Table 9 presents the percent loss if multiple orders are allowed with a stopping time ( a = D )as to no stopping time ( a = T ). 29 x Table 8: Percentage difference 100 × T/ /Z − S/ /ZS/ /Z for different initial inventory x and setup cost K values.The numbers in parentheses show the value of 100 × S/ /Z − D/ /ZD/ /Z under the same parameter setting. Basecase parameters are used. K x Table 9: Percentage difference 100 × T/ ∞ /F − D/ ∞ /FD/ ∞ /F for different initial inventory x and setup cost K values.Base case parameters are used. Note that determining a switching time at the beginning does not constitute much improvementover no stopping time. However, moving to a stopping time improves the results, even undermultiple orders.We present our results for all our runs in Table 10. Note that we only report the expectedpercent loss figures under the case where we employ stopping time as compared to switching time.The results show that the fixed ordering cost does not significantly affect the outcome as we limitourselves to a single order. When we analyze the settings where we attain the least loss values, wenotice that most are the settings where we assume a constant intensity rate over time ( x Table 10: Percentage difference 100 × S/ /Z − D/ /ZD/ /Z for different initial inventory x and setup cost K values.For each x and K , we present maximum, average and minimum values over all parameter settings. Set Remark 4.2. Disposing the available inventory seems to be a critical decision, especially for someremaining time and inventory level combinations (see Section 3.3 for supporting analytical results).Moreover, the dynamic selection of this time to stop (via stopping time) as compared to determiningat the beginning can be valuable in case the manufacturer has such flexibility. However, we noticethat under the case where we allow for multiple orders, the effect of stopping time is reduced thoughnot diminished. We first present the comparisons under the base case. Table 11 presents the percent loss if weallow for only one order but may delay the time to order under two cases: with a switchover time( a = S as given in the taxonomy) and with a stopping time ( a = D ). The case with stopping timeis presented in parentheses for various values of x and K . K x Table 11: Percentage difference 100 × T/ /Z − T/ /FT/ /F for different initial inventory x and setup cost K values.The numbers in parentheses show the value of 100 × D/ /Z − D/ /FD/ /F under the same parameter setting. Basecase parameters are used. 31s we expect, there is no difference when x = 0. Also, the results show that the fixed orderingcost does not significantly affect the outcome as we limit ourselves to a single order. As x grows,we observe an increase in the losses. Note that, stopping time is a powerful tool as it partiallycompensates the mistake in the timing of the first order, and hence the benefits we observe inparentheses are smaller.We present our results for all our runs in Table 12. Note that we only report the expectedpercent loss figures under the case where we also utilize stopping time. The observation made forTable 11 are valid here, as well. However, when looking at the percentages, the maximum valueshere are significant. When we analyze the settings where we attain the maximum loss values,we notice that most are the settings where we assume a constant intensity rate over time ( K x Table 12: Percentage difference 100 × D/ /Z − D/ /FD/ /F for different initial inventory x and setup cost K values. For each x and K , we present maximum, average and minimum values over all parameter settings.Set Remark 4.3. In case the manufacturer is given the opportunity to order at any time, the cutoffinitial inventory level which prevents ordering at time zero can be quite low. Therefore, the prevalentassumption that a final order is to be placed at time zero can be a strong assumption, possiblyleading to significant losses. 32 .1.4 Value of Combining the Features When we combine all the effects, the overall results indicate promising savings. Here we onlypresent the comparisons under the base case.Table 13 presents the cases where we implement the optimal stopping strategy and record thepercent loss if we do not use the flexibility of delaying the first order as well as multiple orderopportunities. As noted before, the optimal stopping time can compensate the advantages whenusing other flexibilities. Nevertheless, we have significant losses if we do not implement otherflexibilities even in the case where we have small K and small x values. Of course, the advantagesreduce (or disappear for x = 0) with larger K values. K x Table 13: Percentage difference 100 × D/ /Z − D/ ∞ /FD/ ∞ /F for different initial inventory x and setup cost K values.Base case parameters are used. Tables 14 and 15, on the other hand, give us another interpretation. If we do not use our fullflexibility scheme compared to the standard final order approach with no stopping, then our lossescan be as much as 32% when x = 250 and K = 0 in Table 14. Almost half of the loss comes fromnot using the flexibility of ordering at any time even if we are going to order at most once; 17.7%when x = 250 and K = 0 in Table 15. Similar deductions can be made when comparing othercases. Of course, as x gets smaller, we see the effect of delaying the order diminishing. K x Table 14: Percentage difference 100 × T/ /Z − D/ ∞ /FD/ ∞ /F for different initial inventory x and setup cost K values.Base case parameters are used. x Table 15: Percentage difference 100 × T/ /Z − D/ /FD/ /F for different initial inventory x and setup cost K values.Base case parameters are used. Remark 4.4. Considering the joint effect of stopping time, order frequency and delaying the firstorder, we conclude that the fixed cost of ordering and the inventory level at time zero play animportant role. If the initial inventory is large (in our numerical experiments, we take the largest x value to be the half of the expected total horizon demand), then the management is advised to searchfor the feasibility of implementing a stopping time, as well as delaying the first order. However, ifinitial inventory level is small, then it is important to consider the possibility of implementing astopping time and the multiple order option concurrently. In this subsection, we analyze the performance of our proposed model in detail. We considerpairwise comparisons of expected discounted total cost given different levels of a parameter utilizedin the model. We summarize these sensitivities in seven subsections: effect of demand structure,effect of outside source, effect of time horizon, effect of penalty cost, effect of time discount andeffect of scrapping cost, and finally effect of incorrectly specifying the demand structure.We only consider limited number settings to demonstrate the sensitivity. We report all thesesensitivities for different K and x values. We end each subsection with a managerial insight sum-marizing and generalizing the findings. The key approach while generating these insights is notnecessarily related to the current decision framework only, but to support further decision makingneeded to handle the complete end-of-life management problem as well. Table 16 shows the effect of demand structure on the model D/ ∞ /F by showing the percentdifference when intensity function λ is concave and convex. If both the initial inventory and thesetup cost are low ( x = 0 , K = 0), then the cost under convex demand is higher. The reason is that34ore demand is satisfied earlier when the intensity is convex. Therefore, the costs are discountedless. On the other hand, if the initial inventory is low yet the setup cost is high ( x = 0 , K = 5000),the cost under concave demand is higher, since more setup might be needed throughout the horizonunder concave demand, as the decline rate of demand is lower. If we start with a very large initialinventory ( x = 400) implying that we may not need much ordering, then holding cost componentdominates and hence we have a much higher cost for the concave case as inventory is depleted muchslower.Table 17 shows the results when T = 100. The trend is similar to what is said for T = 50.However, as the horizon is longer and total expected demand is the same for both time horizons,the expected drop in on-hand inventory for the convex case relative to concave is less, and hencepercent differences for large x values are not as large as the case T = 50. Insight 1: If we have either high initial inventory level x or high setup cost K , it might be wise toencourage (even give incentives) customers to come earlier - hence make the demand rate look likeconvex compared to the original one. On the other hand, with small x and K combinations (northwest part of Table 16), we may look for strategies making customers come later. K x Table 16: Percentage difference 100 × ( V Concave − V Convex ) / V Convex : Comparison of (cid:101) V (0 , x ) + A whendemand is convex and concave (see Table 5 for definitions). The relevant parameters are T = 50 , c = 2¯ c, γ =0 . , c = c/ , δ = 0 . K x Table 17: Percentage difference 100 × ( V Concave − V Convex ) / V Convex : Comparison of (cid:101) V (0 , x ) + A whendemand is convex and concave (see Table 5 for definitions). The relevant parameters are T = 100 , c =2¯ c, γ = 0 . , c = c/ , δ = 0 . .2.2 Effect of Outside Source / Alternative Policy The percentages in Table 18 show the effect of an outside/alternative source by presenting (cid:101) V (0 , x )+ A with a decreasing cost of this source ( γ = 0 . 01) versus and a nearly constant cost ( γ = 10 − ) overtime (time discount is fixed to δ = 0 . x and K .When x = 0 , K = 0, the manufacturer may not use the alternative policy at all since the cost ofprocurement can be sufficiently low. When x = 0 and K = 5000 however, the manufacturer wouldprefer placing a large order at time zero, and then using the alternative policy if needed. Hence,as K increases, the value of having decreasing unit cost in the alternative policy also increases.When x is in the region (350, 450) for any K , it is likely that the manufacturer utilizes initialinventory and then switches to alternative policy, instead of placing an order. Hence, given the coststructure of the alternative policy, cost one can observe the highest percent values in the expectedtotal cost differential in this region of x .When x = 550 or higher, the manufacturer may not use the alternative policy at all until thestopping time, since the initial inventory seems sufficiently high to cover the demand. Note that,for x = 550, it is likely that the optimal stopping time realizes closer to T . Therefore, any changein the unit cost of alternative policy over time has practically no impact on the expected totalcost. Of course, as x goes higher (which might not be very reasonable for the problem structure),we see that the optimal solution may prefer stopping before the end of the horizon (almost at thesame time for all K values) and starting to use the alternative source for the remaining part ofthe horizon. The fact that percentages are higher simply reflect the unit cost difference in thealternative policy considered in the cases compared. Insight 2: It is shown in other parts of the study that the existence of an alternative policy (or anoutside source) can be essential for flexibility needed in the environment. Hence, the cost of thisalternative becomes critical in the effectiveness of the approach. Thus, larger percentages in Table18 demonstrate the fact that it may be better to support the development of an alternative sourceso that it will become more cost-efficient (cheaper) over time. This might be realized by givingincentives to other parties for developing technologies to lower the manufacturing price.36 x Table 18: Percentage difference 100 × ( V γ =10 − − V γ =0 . ) /V γ =0 . : Comparison of (cid:101) V (0 , x )+ A when γ = 10 − and γ = 0 . 01 to show the effect of alternative policy. The relevant parameters are T = 100 , c = 2¯ c, c =¯ c/ , δ = 0 . Table 19 shows the effect of time horizon T by presenting (cid:101) V (0 , x ) + A when T = 50 and T = 100.When x = 0 , K = 0, the expected total cost under T = 100 is lower, since the manufacturer placessmall orders later in time, utilizing time-discount. On the other hand, when x = 0 , K = 5000, asufficiently large order is placed at time zero. Since this purchasing cost occurs at time zero in bothcases, the costs are similar.For a relatively small range for x (300-330) (for instance, when x = 331) and large K , themanufacturer does not place an order and uses the alternative policy. If T = 50, then this policy isused earlier at a time when the unit cost of the alternative policy is relatively higher and discounthas less effect. This results in a 13% difference in relative total expected costs. On the other hand,for x = 331 , K = 0, the manufacturer can place small orders instead of using the alternative policyor facing penalty. This explains a very small percentage difference observed. When x is largerthan the range given above, for smaller K values, we start to observe the negative effects of longerhorizon, as longer horizon brings more carrying cost over time and hence greater expected costs forthe case with T = 100. Insight 3: The selection of time horizon which is equivalent to setting a warranty period is notconsidered in the current work. On the other hand, extending the warranty period will alwaysbe preferable by customers. Hence if one observes benefits of extending, it might be potentiallya beneficial managerial move. If there are moderate to high setup cost K values, and relativelylow initial inventory x values (south middle east part of Table 19) it might be wiser to extend thehorizon to T = 100 under the knowledge that demand will be flatter through 100 periods.37 x Table 19: Percentage difference 100 × ( V T =50 − V T =100 ) /V T =100 : Comparison of (cid:101) V (0 , x )+ A when T = 50 and T = 100 to show the effect of time horizon. The relevant parameters are c = 2¯ c, c = ¯ c/ , γ = 0 . , δ = 0 . Table 20 and Table 21 show the effect of penalty ¯ c by presenting (cid:101) V (0 , x ) + A when ¯ c = 2¯ c and¯ c = 10¯ c . Note that for x values which are in between 0 and the expected total demand (forinstance x = 300), the change in unit penalty cost is expected to have its highest impact, since thefirm takes the risk of penalty for not placing an order. Nevertheless, even if we change penalty costby a factor of 5, the increase in the optimal value of the expected total discounted cost is negligible(less than 2% in all cases). The reason is that the manufacturer can stop holding inventory anduse the alternative policy to avoid penalty cost. Insight 4: As we have an existing alternative (which is much cheaper than the larger penalty costin Table 4), practically there is no significant difference observed after changing the penalty cost.Hence, with the existence of such an alternative, the firm might announce to pay large penaltiesfor not satisfying demand to attract more demand to begin with. This shows the importance ofcreating such an alternative. Additionally, if the cost of alternative decreases over time (periodswhere the risk of paying the penalty is more), then it will be even better for decreasing expectedcosts. K x Table 20: Percentage difference 100 × ( V ¯ c =10¯ c − V ¯ c =2¯ c ) /V ¯ c =2¯ c : Comparison of (cid:101) V (0 , x ) + A when ¯ c = 2¯ c and ¯ c = 10¯ c . The relevant parameters are T = 100 , c = ¯ c/ , γ = 0 . , δ = 0 . 38 100 250 6000 0.4 0.5 0.6 0.01000 0.4 0.5 0.7 0.05000 0.3 0.6 0.7 0.0 Table 21: Percentage difference 100 × ( V ¯ c =10¯ c − V ¯ c =2¯ c ) /V ¯ c =2¯ c : Comparison of (cid:101) V (0 , x ) + A when ¯ c = 2¯ c and ¯ c = 10¯ c . The relevant parameters are T = 100 , c = ¯ c/ , γ = 10 − , δ = 0 . Table 22 shows the effect of time discount δ by presenting (cid:101) V (0 , x )+ A when δ = 0 . 005 and δ = 10 − .The expected total cost is always higher when time discount is close to zero, as can be predicted.The effect of discount decreases in K , for smaller x values, since a large order is placed at timezero rather than later on. However, for intermediate x values (350-450) the effect is reversed ordisappears, since it is likely that an order is needed later in the horizon; hence, the value of K becomes critical. Insight 5: Time discount shows the sensitivity of results on the total discounted expected cost forvarying horizon lengths. In the numerical experiments, the time discount value seems to be effectivefor different x values rather than K ; and hence reiterating the importance of initial inventory. Thislength will be different for industries, and hence the essential insight will be a function of theindustry considered. K x Table 22: Percentage difference 100 × ( V δ =10 − − V δ =0 . ) /V δ =0 . : Comparison of (cid:101) V (0 , x ) + A when δ = 0 . 005 and δ = 10 − . The relevant parameters are T = 100 , c = 2¯ c, c = ¯ c/ , γ = 0 . 01, convex intensity. Table 23 shows the impact of scrapping cost c by presenting (cid:101) V (0 , x ) + A when c = ¯ c/ c = − ¯ c/ 4. If x < x , we stop holding inventory onlyif the inventory level is about to hit zero and the risk of penalty arises. In such a case, only anegligible amount of inventory is scrapped, implying that the scrapping cost has a negligible effect39n the expected total cost. On the other hand, if x > K x Table 23: Percentage difference 100 × ( V c =¯ c/ − V c = − ¯ c/ ) /V c = − ¯ c/ : Comparison of (cid:101) V (0 , x ) + A when c = ¯ c/ c = − ¯ c/ 4. The relevant parameters are T = 100 , c = 2¯ c, γ = 0 . δ = 0 . Insight 6: Scrapping inventory is not a significant burden (or source of income) when the manufac-turer wishes to start using outside source rather than holding inventory, unless the initial inventorylevel is excessively large. So, this parameter seems to be less effective for decision making purposes. We analyze the impact of an error in selecting the intensity function of the non-homogeneousPoisson process ( N t ) t ∈ [0 ,T ] . Suppose that the manufacturer chooses a linear intensity function, butthe true intensity function is convex (recall Table 5 for definitions). To calculate the cost of makingsuch assumption, we first solve (cid:101) V (0 , x ) + A when the intensity function is convex and linear to findthe best decision variables (ordering, stopping, continuation regions and order-up-to levels). Next,with those fixed decision variables, we compute the objective function (cid:101) V (0 , x ) + A when intensityis convex.Table 24 shows the percent difference in expected total cost. If x = 0 , K = 0 and intensity islinearly decreasing, then the manufacturer places small orders more frequently. On the other hand,if the intensity is decreasing in a convex manner, then the manufacturer tends to place larger ordersat the beginning and smaller orders towards the end. Hence, by presuming a linearly decreasingintensity function and taking actions based on this assumption, the manufacturer can observe excesspenalty cost at the beginning and excess holding cost towards the end, resulting in a significantloss (above 30%). We believe that this is a motivation to study a problem where intensity rate israndom itself. Also note that for other x and K combinations, the loss can be as high as 116%.The largest percent loss can be observed when there is an initial inventory level which is close to40he expected demand throughout the horizon. For instance when x = 450, it is likely that we waitfor an amount of time and then place an order. If intensity is linearly decreasing, then this futureorder can be large. Hence, if the manufacturer places a large order in the future, then it is mostlikely that excess holding and procurement costs incur as the arrival rate under decreasing-convexcase is much lower towards the end of horizon.When x is large (above expected demand), one may be less willing to stop towards the endunder the presumption that the intensity rate is decreasing linearly. Hence, this difference in thestopping region create small, but meaningful percent difference in expected total costs indicatingthe importance of the stopping time. Insight 7: One of the critical inputs to the end-of-life management problem is the estimation ofthe demand rate over the time horizon. The results presented in Table 24 indicate that the penaltyof this misspecification can be drastic, especially if the fixed ordering cost is large. Hence an initialstudy to analyze underlying demand structure seems to be a reasonable way for management touse her resources. Similarly, an agreement with the consumers on the possible timing of demandarrivals may further help in quantifying the demand intensity over time, decreasing the risk ofmisspecification. 0 100 250 450 550 6500 36 22 6 10 0 31000 11 14 16 85 1 35000 29 5 36 116 0 3 Table 24: Percentage difference 100 × ( V linear − decvar − V convex − decvar ) / V convex − decvar : The cost V linear − decvar denotes the expected total cost of (cid:101) V (0 , x )+ A calculated with convex intensity, but the best de-cision variables are found with linear intensity. V Convex − decvar denotes the expected total cost of (cid:101) V (0 , x ) + A with convex intensity. The relevant parameters are T = 50 , c = 2¯ c, γ = 0 . , c = c/ , δ = 0 . This study analyzes the value of providing flexibility in the end-of-life management problem.Namely, we allow for multiple orders as well as a change in the timing of the first order, andwe utilize stopping times to decide on when to dispose all the available inventory. To that end, weconsider a manufacturer whose problem is to make one of the three decisions at each period: (1)place an order for spare parts, (2) do nothing and use existing inventory to satisfy demand, or (3)41top holding inventory permanently and use an outside/alternative source. We cast this problemas an optimal stopping problem with additional decisions so that it can be solved by means ofa stochastic dynamic programming. After providing the dynamic programming formulation, weuse martingale theory to facilitate the calculation of the value function. We devise a taxonomyfor benchmark models to show the value of our approach as well as compare our results with thecurrent literature. Several analytical results for our models and benchmark models are presented tofurther enhance our understanding of the problem. Finally, we present our computational results,generating several managerial insights.The originality of the study comes from the fact that several dimensions of the end-of-lifeproblem are considered. The first dimension is related to the decision-making environment. Weconsider possible decisions which give us the benefit of using all flexibilities. Remarks 4.1 through 4.4summarize when and how to exploit these flexibilities. Accordingly, approaches with the premisethat a final order must be placed at time zero can be a strong assumption leading to losses.Moreover, the dynamic selection of the time to stop (via a stopping time) and additionally allowingfor multiple orders can be valuable.Another dimension is related to the practicalities during the implementation phase of thesepolicies. Operationally, additional information we obtain can be used more effectively to managethe decisions within the horizon. The related information is summarized in Remark 3.11. Moreprecisely, one can compute the distribution of number of orders and the expected number of ordersgiven for the selected solution, as demonstrated in Subsection 3.2.2. This knowledge would allowproactive agreements with the possible supplier. Similarly, one can compute the distribution ofthe time that the next order is placed, in other words, the distribution of cycle length betweentwo consecutive orders. The manufacturer can share this distribution (or some summary statistics)with the supplier so that the supplier can plan its operations in advance. One can also computethe probability that an order is not placed in the next n periods. For instance, if the manufacturercalculates that the probability of having no orders in the next n periods is greater than 0.99, thenit might be beneficial to review the inventory for some number of periods, avoiding review costs.This information can be particularly useful if the cost of reviewing inventory is high. Finally, fora fixed initial inventory level, the expected number of periods before placing the first order can becalculated to further inform the supplier. 42 final dimension we consider is regarding the setting of the parameters which effect the problemoutcome. In Subsection 4.2, seven managerial insights are proposed. The insights are mainly to-wards controlling the environment of the end-of-life management problem. The insights are relatedto affecting the customer arrival rates, making monetary arrangements to support development ofthe alternative source, extending the warranty period, and announcing favorable parameter val-ues to attract more customers so that demand rates increase, making the expected profits of theend-of-life period even more attractive for the manufacturing firm.There are a few straightforward extensions that can follow: Use of time-varying unit procure-ment cost, use of time-varying fixed ordering cost, use of unequal review periods, and use of coststo review inventory. Except the last one, the current dynamic programming formulation can beadjusted. For the case with review costs, state-space reducing properties can be studied. Finally,observing Insight 7, which is related to the significant cost of misspecification of demand intensityrate function, the case with random intensity might be a reasonable future direction for research.One such example is Ararat et al. (2021), which considers a problem environment with a singleorder, and demand following a conditionally Poisson process whose intensity rate is random. A Auxiliary Results This subsection provides auxiliary results for the other proofs. Recall that (Ω , H , P ) denotes the un-derlying probability space which hosts the non-homogeneous Poisson process N . Let F = ( F t ) t ∈ [0 ,T ] denote the filtration generated by N , that is, F t := σ ( N s , s ∈ [0 , t ]) for every t ∈ [0 , T ]. LemmaA.1 below introduces the martingale property for N and it helps us convert Poisson integrals intoLebesgue integrals. Lemma A.1. (C¸ ınlar 2011, p. 299, VI.6.4) Let ( H t ) t ∈ [0 ,T ] be a non-negative F -predictable processsuch that E (cid:104)(cid:82) t H u λ ( u ) du (cid:105) < ∞ for every t ∈ [0 , T ] . Then, the process ( L t ) t ∈ [0 ,T ] defined by L t := (cid:90) t H u dN u − (cid:90) t H u λ ( u ) du, t ∈ [0 , T ] , is a martingale with respect to F . Moreover, for each F -stopping time τ ∈ T , E (cid:20)(cid:90) τ H u dN u (cid:21) = E (cid:20)(cid:90) τ H u λ ( u ) du (cid:21) . Lemma A.2. For every x ∈ Z + and k ∈ T , E [( x − N k ) + ] = x − (cid:88) n =0 P { N k ≤ n } . where the sum is defined to be 0 when x = 0 . Proof. Note that E [( x − N k ) + ] − E [( x − − N k ) + ] = P { N k ≤ x − } . Iterating this equality yieldsthat E [( x − N k ) + ] = E [( x − − N k ) + ] + P { N k ≤ x − } = E [( x − − N k ) + ] + P { N k ≤ x − } + P { N k ≤ x − } = · · · = (cid:80) x − n =0 P { N k ≤ n } . Recall that σ kx = inf { u > k : N u − N k ≥ x } and σ x = σ x . The following lemma shows that thestopping time σ x is conditionally independent from the past given the present. This enables us towrite dynamic programming algorithms when the objective function includes σ x . For an F -stoppingtime τ , we define the stopped σ -algebra F τ := { E ∈ H : E ∩ { τ ≤ t } ∈ F t for each t ∈ [0 , T ] } . Lemma A.3. For every F -stopping time τ ∈ T and x ∈ Z + , on { τ < σ x } , σ x = σ τx − N τ + τ, where σ τx − N τ = inf { t > N t + τ − N τ ≥ x − N τ } . Moreover, σ τx − N τ is conditionally independent of F τ given N τ . Proof. On the set { τ < σ x } , we see that N τ < x . Then, on { τ < σ x } , we have σ x = inf { t > N t ≥ x } = inf { t > τ : N t ≥ x } = inf { u > N u + τ − N τ + N τ ≥ x } + τ = σ τx − N τ + τ. For the second claim, we know from the strong Markov property of non-homogeneous Poissonprocesses (C¸ ınlar 2011, p. 296, VI.5.18) that N u + τ is conditionally independent of F τ given N τ forany u > 0. Hence, by definition, σ τx − N τ is conditionally independent of F τ given N τ . B Proofs of the Results in Section 2 Proof of Lemma 2.1. Clearly, H ( k, 0) = 0. Let x ≥ 1. Then,44 ( k, x ) = c E (cid:20)(cid:90) k +1 k e − δ ( u − k ) (cid:0) x − ( N u − N k ) (cid:1) + du (cid:21) = c (cid:90) k +1 k e − δ ( u − k ) E (cid:104)(cid:0) x − ( N u − N k ) (cid:1) + (cid:105) du (Fubini’s Theorem)= c (cid:90) k +1 k e − δ ( u − k ) x − (cid:88) n =0 P { N u − N k ≤ n } du (Lemma A.2)= c x − (cid:88) n =0 (cid:90) k +1 k e − δ ( u − k ) P { N u − N k ≤ n } du = c x − (cid:88) n =0 n (cid:88) i =0 (cid:90) k +1 k e − δ ( u − k ) P { N u − N k = i } du = c x − (cid:88) n =0 n (cid:88) i =0 (cid:90) k +1 k e − δ ( u − k ) e − (Λ( u ) − Λ( k )) (Λ( u ) − Λ( k )) i i ! du, where Λ( u ) − Λ( k ) = (cid:82) uk λ ( s ) ds . Proof of Lemma 2.2. We first note that L ( k, x ) can be expressed as L ( k, x ) = E (cid:20)(cid:90) k +1 k e − δ ( u − k ) c ( u ) dN u (cid:21) − E (cid:34)(cid:90) ( k +1) ∧ σ x k e − δ ( u − k ) c ( u ) dN u (cid:35) It follows from Lemma A.1 that E (cid:20)(cid:90) k +1 k e − δ ( u − k ) c ( u ) dN u (cid:21) = (cid:90) k +1 k e − δ ( u − k ) c ( u ) λ ( u ) du Moreover, we have E (cid:34)(cid:90) ( k +1) ∧ σ kx k e − δ ( u − k ) c ( u ) dN u (cid:35) = E (cid:34)(cid:90) ( k +1) ∧ σ kx k e − δ ( u − k ) c ( u ) λ ( u ) du (cid:35) (Lemma A.1)= (cid:90) k +1 k E (cid:104) { u<σ kx } (cid:105) e − δ ( u − k ) c ( u ) λ ( u ) du (Fubini’s Theorem)= (cid:90) k +1 k P { N u − N k ≤ x } e − δ ( u − k ) c ( u ) λ ( u ) du (Definition of σ kx )= x (cid:88) i =0 (cid:90) k +1 k e − δ ( u − k ) c ( u ) λ ( u ) e − (Λ( u ) − Λ( k )) (Λ( u ) − Λ( k )) i i ! du. Hence, the result of the lemma follows. 45 Proof of the Results in Subsection 3.1 Proof of Proposition 3.1. Note that for every t ∈ T and every x t ∈ Z + , we can write e − δt S ( t, x t ) = e − δt c x t + e − δt E (cid:20)(cid:90) Tt e − δ ( u − t ) c ( u ) dN u (cid:21) = e − δt c x t + E (cid:20)(cid:90) Tt e − δu c ( u ) dN u (cid:21) = e − δt c x t + E (cid:20)(cid:90) T e − δu c ( u ) dN u (cid:21) − E (cid:20)(cid:90) t e − δu c ( u ) dN u (cid:21) = e − δt c x t + E (cid:20)(cid:90) T e − δu c ( u ) dN u (cid:21) − t − (cid:88) k =0 e − δk E (cid:20)(cid:90) k +1 k e − δ ( u − k ) c ( u ) dN u (cid:21) = e − δt c x t + A − t − (cid:88) k =0 e − δk E (cid:20)(cid:90) k +1 k e − δ ( u − k ) c ( u ) dN u (cid:21) . Moreover, for every t ∈ T and every x , x , . . . , x t − ∈ Z + , we can write t − (cid:88) k =0 e − δk L ( k, x k ) = t − (cid:88) k =0 e − δk E (cid:34)(cid:90) k +1( k +1) ∧ σ kxk e − δ ( u − k ) c ( u ) dN u (cid:35) = t − (cid:88) k =0 e − δk (cid:32) E (cid:20)(cid:90) k +1 k e − δ ( u − k ) c ( u ) dN u (cid:21) − E (cid:34)(cid:90) ( k +1) ∧ σ kxk k e − δ ( u − k ) c ( u ) dN u (cid:35)(cid:33) . Then, for every t ∈ T and every x , x , . . . , x t ∈ Z + , we have t − (cid:88) k =0 e − δk C ( k, x k ) + e − δt S ( t, x t )= t − (cid:88) k =0 e − δk ( H ( k, x k ) + L ( k, x k )) + e − δt S ( t, x t )= t − (cid:88) k =0 e − δk c E (cid:20)(cid:90) k +1 k e − δ ( u − k ) ( x k − ( N u − N k )) + du (cid:21) + t − (cid:88) k =0 e − δk (cid:32) E (cid:20)(cid:90) k +1 k e − δ ( u − k ) c ( u ) dN u (cid:21) − E (cid:34)(cid:90) ( k +1) ∧ σ kxk k e − δ ( u − k ) c ( u ) dN u (cid:35)(cid:33) + e − δt c x t + A − t − (cid:88) k =0 e − δk E (cid:20)(cid:90) k +1 k e − δ ( u − k ) c ( u ) dN u (cid:21) = t − (cid:88) k =0 e − δk c E (cid:20)(cid:90) k +1 k e − δ ( u − k ) ( x k − ( N u − N k )) + du (cid:21) + t − (cid:88) k =0 e − δk (cid:32) E (cid:20)(cid:90) k +1 k e − δ ( u − k ) [ c ( u ) − c ( u )] dN u (cid:21) − E (cid:34)(cid:90) ( k +1) ∧ σ kxk k e − δ ( u − k ) c ( u ) dN u (cid:35)(cid:33) e − δt c x t + A = t − (cid:88) k =0 e − δk (cid:101) C ( k, x k ) + e − δt (cid:101) S ( t, x t ) + A. Therefore, for every π ∈ Π, we have E (cid:34) τ − (cid:88) k =0 e − δk (cid:18) c ( µ k ( X k )) + C ( k, X k + µ k ( X k )) (cid:19) + e − δτ S ( τ, X τ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X = x (cid:35) = E (cid:34) τ − (cid:88) k =0 e − δk (cid:18) c ( µ k ( X k )) + (cid:101) C ( k, X k + µ k ( X k )) (cid:19) + e − δτ (cid:101) S ( τ, X τ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X = x (cid:35) + A. Proof of Proposition 3.4. Let k ∈ T and x ∈ Z + . If x = 0, then σ kx = k and ( x − ( N u − N k )) + =0 for every u > k . Therefore, we apply Lemma A.1 to the definition of (cid:101) C ( k, x ) in (3.1) to get (cid:101) C ( k, 0) = E (cid:20)(cid:90) k +1 k e − δ ( u − k ) [ c ( u ) − c ( u )] dN u (cid:21) = (cid:90) k +1 k e − δ ( u − k ) [ c ( u ) − c ( u )] λ ( u ) du. Next, if x ≥ 1, then from Lemma 2.1, we have E (cid:20)(cid:90) k +1 k e − δ ( u − k ) ( x − ( N u − N k )) + du (cid:21) = x − (cid:88) n =0 n (cid:88) i =0 (cid:90) k +1 k e − δ ( u − k ) e − (Λ( u ) − Λ( k )) (Λ( u ) − Λ( k )) i i ! du. Moreover, from Lemma A.1, we have E (cid:20)(cid:90) k +1 k e − δ ( u − k ) [ c ( u ) − c ( u )] dN u (cid:21) = (cid:90) k +1 k e − δ ( u − k ) [ c ( u ) − c ( u )] λ ( u ) du. Finally, by proceeding as in the proof of Lemma 2.2, it is possible to show that E (cid:34)(cid:90) ( k +1) ∧ σ kx k e − δ ( u − k ) c ( u ) dN u (cid:35) = x (cid:88) i =0 (cid:90) k +1 k e − δ ( u − k ) c ( u ) λ ( u ) e − Λ( u ) − Λ( k ) (Λ( u ) − Λ( k )) i i ! du. Combining the above terms concludes the proof. D Proof of the Results in Subsection 3.2 Proof of Proposition 3.5. asd To show the partitioning of T × Z + into three regions, let an arbitrary time k ∈ T and aninventory level x ∈ Z + be given (recall that T = { , , . . . , T } ).47ase 1: Assume that (cid:101) S ( k, x ) ≤ (cid:101) J ( k, x ) = inf m ∈ Z + { c ( m ) + (cid:101) G ( k, x + m ) } . Then x ∈ R Sk by definition.It also holds that (cid:101) S ( k, x ) ≤ (cid:101) G ( k, x ) because c (0) = 0. Then, x / ∈ R Ck . Moreover, it holds that (cid:101) S ( k, x ) ≤ inf m ≥ { c ( k ) + (cid:101) G ( k, x + m ) } , so x / ∈ R Ok .Case 2: Assume that (cid:101) S ( k, x ) > (cid:101) J ( k, x ) = inf m ∈ Z + { c ( m ) + (cid:101) G ( k, x + m ) } and that (cid:101) G ( k, x ) ≤ inf m ≥ { c ( m ) + (cid:101) G ( k, x + m ) } . Then, from (3.3), it is possible to see that (cid:101) V ( k, x ) = (cid:101) G ( k, x ). More-over, (cid:101) G ( k, x ) < (cid:101) S ( k, x ) due to the assumption of this case. Hence, x ∈ R Ck . Next, x / ∈ R Sk since (cid:101) S ( k, x ) > (cid:101) G ( k, x ). Finally, (cid:101) G ( k, x ) ≤ inf m ≥ { c ( m ) + (cid:101) G ( k, x + m ) } and therefore x / ∈ R Ok .Case 3: Assume that (cid:101) S ( k, x ) > (cid:101) J ( k, x ) = inf m ∈ Z + { c ( m ) + (cid:101) G ( k, x + m ) } and that (cid:101) G ( k, x ) > inf m ≥ { c ( m ) + (cid:101) G ( k, x + m ) } . Then, inf m ≥ { c ( m ) + (cid:101) G ( k, x + m ) } < min { (cid:101) S ( k, x ) , (cid:101) G ( k, x ) } andso x ∈ R Ok . Moreover, from (3.3), it is possible to see that (cid:101) V ( k, x ) < (cid:101) G ( k, x ) so x / ∈ R Ck . Finally,we have x / ∈ R Sk since (cid:101) S ( k, x ) > inf m ∈ Z + { c ( m ) + (cid:101) G ( k, x + m ) } from the assumption of the case. Let (cid:101) π ∗ be given as in the proposition. For ease of notation, define for each k ∈ T and x ∈ Z + the function (cid:101) J ( k, x ) := inf m ≥ { c ( m ) + (cid:101) G ( k, x + m ) } . If X (cid:101) π ∗ k ∈ R Sk , then (cid:101) S ( k, X (cid:101) π ∗ k ) ≤ (cid:101) J ( k, X (cid:101) π ∗ k ).Then, we stop due to the definition of (cid:101) τ ∗ in Corollary 3.3. If X (cid:101) π ∗ k ∈ R Ok , then (cid:101) J ( k, X (cid:101) π ∗ k ) < min { (cid:101) S ( k, X (cid:101) π ∗ k ) , (cid:101) G ( k, X (cid:101) π ∗ k ) } . Then, from (3.3), we have (cid:101) V ( k, X (cid:101) π ∗ k ) = (cid:101) J ( k, X (cid:101) π ∗ k ) , so it is optimal toplace an order of (cid:101) µ ∗ k ◦ X (cid:101) π ∗ k ≥ (cid:101) µ ∗ k is defined as in Corollary 3.3. Hence, it is op-timal to increase inventory to the level S ∗ k ◦ X (cid:101) π ∗ k where S ∗ k ( x ) := x + (cid:101) µ ∗ k ( x ) for each x ∈ R Ok . Finally,if X (cid:101) π ∗ k ∈ R Ck , then (cid:101) V ( k, X (cid:101) π ∗ k ) = (cid:101) G ( k, X (cid:101) π ∗ k ), so the decision of not placing an order ( (cid:101) µ ∗ k ◦ X (cid:101) πk = 0 )can attain the infimum in the definition of (cid:101) J in (3.4). Moreover, X (cid:101) π ∗ k / ∈ R Ok and X (cid:101) π ∗ k / ∈ R Sk since R Ck , R Ok and R Sk are disjoint from the first claim of the proposition. Then, it is optimal to continuewithout taking an action. Proof of Proposition 3.8. For each s ∈ { t + 1 , t + 2 , . . . , T } , define the set A s ∈ H by A s := (cid:8) x t − N ts ∈ R Ss (cid:9) , if s = t + 1 , s − (cid:92) k = t +1 (cid:8) x t − N tk ∈ R Ck (cid:9) ∩ (cid:8) x t − N ts ∈ R Ss (cid:9) , if s ≥ t + 2 . Then, for any s and s in { t + 1 , t + 2 , . . . , T } such that s < s , we have A s ⊂ (cid:8) x t − N ts ∈ R Ss (cid:9) and A s ⊂ (cid:8) x t − N ts ∈ R Cs (cid:9) . Therefore, A s ∩ A s ⊂ (cid:8) x t − N ts ∈ R Ss (cid:9) ∩ (cid:8) x t − N ts ∈ R Cs (cid:9) = (cid:8) x t − N ts ∈ R Ss ∩ R Cs (cid:9) = ∅ R Ss ∩ R Cs = ∅ by the first claim of Proposition 3.5. This implies that A t +1 , A t +2 , . . . , A T aredisjoint sets. Hence, P t ( t, x t ) = P (cid:32) T (cid:91) s = t +1 (cid:32) s − (cid:92) k = t +1 (cid:8) x t − N tk ∈ R Ck (cid:9) ∩ (cid:8) x t − N ts ∈ R Ss (cid:9)(cid:33)(cid:33) = P (cid:32) T (cid:91) s = t +1 A s (cid:33) = T (cid:88) s = t +1 P ( A s )= T (cid:88) s = t +1 P (cid:32) s − (cid:92) k = t +1 (cid:8) x t − N tk ∈ R Ck (cid:9) ∩ (cid:8) x t − N ts ∈ R Ss (cid:9)(cid:33) = P (cid:8) x t − N tt +1 ∈ R St +1 (cid:9) + T (cid:88) s = t +2 P (cid:32) s − (cid:92) k = t +1 (cid:8) x t − N tk ∈ R Ck (cid:9) ∩ (cid:8) x t − N ts ∈ R Ss (cid:9)(cid:33) . Moreover, T (cid:88) s = t +2 P (cid:32) s − (cid:92) k = t +1 (cid:8) x t − N tk ∈ R Ck (cid:9) ∩ (cid:8) x t − N ts ∈ R Ss (cid:9)(cid:33) = T (cid:88) s = t +2 P (cid:32)(cid:8) x t − N tt +1 ∈ R Ct +1 (cid:9) ∩ s − (cid:92) k = t +2 (cid:8) x t − N tk ∈ R Ck (cid:9) ∩ (cid:8) x t − N ts ∈ R Ss (cid:9)(cid:33) = T (cid:88) s = t +2 (cid:88) n ∈ x t − R Ct +1 P (cid:32) s − (cid:92) k = t +2 (cid:8) x t − N tk ∈ R Ck (cid:9) ∩ (cid:8) x t − N ts ∈ R Ss (cid:9) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:8) N tt +1 = n (cid:9)(cid:33) P (cid:8) N tt +1 = n (cid:9) = T (cid:88) s = t +2 (cid:88) n ∈ x t − R Ct +1 P (cid:32) s − (cid:92) k = t +2 (cid:8) x t − n − N t +1 k ∈ R Ck (cid:9) ∩ (cid:8) x t − n − N t +1 s ∈ R Ss (cid:9)(cid:33) P (cid:8) N tt +1 = n (cid:9) (independent increments property of non-homogeneous Poisson process)= (cid:88) n ∈ x t − R Ct +1 P (cid:32) T (cid:91) s = t +2 (cid:32) s − (cid:92) k = t +2 (cid:8) x t − n − N t +1 k ∈ R Ck (cid:9) ∩ (cid:8) x t − n − N t +1 s ∈ R Ss (cid:9)(cid:33)(cid:33) P (cid:8) N tt +1 = n (cid:9) = (cid:88) n ∈ x t − R Ct +1 P t ( t + 1 , x t − n ) P (cid:8) N tt +1 = n (cid:9) . As a result, P t ( t, x t ) = P (cid:8) x t − N tt +1 ∈ R St +1 (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Enter stopping region during [ t, t + 1] + (cid:88) n ∈ x t − R Ct +1 P t ( t + 1 , x t − n ) P (cid:8) N tt +1 = n (cid:9) . (cid:124) (cid:123)(cid:122) (cid:125) Stay in continue region and move to next period roof of Proposition 3.9. For brevity of notation, let E n − t ,...,t n ( t , y t ) ∈ H denote the event thatstarting at t with the inventory level y t , we place n − t , t , . . . , t n and no ordersbetween t n + 1 and T , that is, E n − t ,...,t n ( t , y t ) := (cid:18) t − (cid:92) k = t +1 (cid:8) y t − N t k ∈ R Ck (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Continue until t ∩ (cid:8) y t − N t t ∈ R Ot (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Place an order at t (cid:19) ∩ . . . (cid:18) t n − (cid:92) k = t n − +1 (cid:110) y t n − − N t n − k ∈ R Ck (cid:111)(cid:124) (cid:123)(cid:122) (cid:125) Continue until t n ∩ (cid:110) y t n − − N t n − t n ∈ R Ot n (cid:111)(cid:124) (cid:123)(cid:122) (cid:125) Place an order at t n (cid:19) ∩ (cid:18) T (cid:91) s = t n +1 s − (cid:92) k = t n +1 (cid:8) y t n − N t n k ∈ R Ck (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Continue until s ∩ (cid:8) y t n − N t n s ∈ R Ss (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Stop at s (cid:19) . Moreover, P ( E n − t ,...,t n ( t , y t )) = P n − t ,t ,...,t n ( t , y t )and this can be seen from the definition of P n − t ,t ,...,t n : When we start looking at the system at time t onward, the events before t are excluded. Therefore, the probability that n − t , t , . . . , t n is same as the probability that n − t , t , t , . . . , t n as long as we start looking at the system at time t > t (recall t = 0). Furthermore, for each t ∈ { , , , . . . , t − } , we can write P nt ,...,t n ( t, x t ) = P (cid:18) t − (cid:92) k = t +1 (cid:8) x t − N tk ∈ R Ck (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Continue until t ∩ (cid:8) x t − N tt ∈ R Ot (cid:9)(cid:124) (cid:123)(cid:122) (cid:125) Place an order at t ∩ E n − t ,...,t n ( t , y t ) (cid:19) . Hence, we can write for each t ∈ { , , , . . . , t − } that P nt ,...,t n ( t, x t )= P (cid:32) t − (cid:92) k = t +1 (cid:8) x t − N tk ∈ R Ck (cid:9) ∩ (cid:8) x t − N tt ∈ R Ot (cid:9) ∩ E n − t ,...,t n ( t , y t ) (cid:33) = P (cid:32)(cid:8) x t − N tt +1 ∈ R Ct +1 (cid:9) ∩ t − (cid:92) k = t +2 (cid:8) x t − N tk ∈ R Ck (cid:9) ∩ (cid:8) x t − N tt ∈ R Ot (cid:9) ∩ E n − t ,...,t n ( t , y t ) (cid:33) = (cid:88) n ∈ x t − R Ct +1 P (cid:32) t − (cid:92) k = t +2 (cid:8) x t − N tk ∈ R Ck (cid:9) ∩ (cid:8) x t − N tt ∈ R Ot (cid:9) ∩ E n − t ,...,t n ( t , y t ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:8) N tt +1 = n (cid:9)(cid:33) P (cid:8) N tt +1 = n (cid:9) (cid:88) n ∈ x t − R Ct +1 P (cid:32) t − (cid:92) k = t +2 (cid:8) x t − n − N t +1 k ∈ R Ck (cid:9) ∩ (cid:8) x t − n − N t +1 t ∈ R Ot (cid:9) ∩ E n − t ,...,t n ( t , y t ) (cid:33) P (cid:8) N tt +1 = n (cid:9) = (cid:88) n ∈ x t − R Ct +1 P nt ,...,t n ( t + 1 , x t − n ) P (cid:8) N tt +1 = n (cid:9) , which shows the recursion. Now, let us show the terminal condition. P nt ,...,t n ( t − , x t − )= P (cid:16)(cid:110) x t − − N t − t ∈ R Ot (cid:111) ∩ E n − t ,...,t n ( t , y t ) (cid:17) = P (cid:16) E n − t ,...,t n ( t , y t ) (cid:12)(cid:12) (cid:110) x t − − N t − t ∈ R Ot (cid:111)(cid:17) P (cid:110) x t − − N t − t ∈ R Ot (cid:111) = (cid:88) n ∈ x t − − R Ot P (cid:16) E n − t ,...,t n ( t , y t ) (cid:12)(cid:12) (cid:110) N t − t = n (cid:111)(cid:17) P (cid:110) N t − t = n (cid:111) = (cid:88) n ∈ x t − − R Ot P n − t ,t ,...,t n ( t , y t ) P (cid:110) N t − t = n (cid:111) , where y t is the order-up-to level of x t − − n and it is also the new inventory level after orderingat t . E Proof of Proposition 3.13 In this section, we prove two lemmata and Proposition 3.13. In the sequel, we use the followingforms of C ( x, τ ) , ∆ x C ( x, τ ) and ∆ x C ( x, τ ) shown by Frenk et al. (2019a) in relations (4)-(6) and(14): C ( x, τ ) = c x + E (cid:18) (cid:90) τ e − δu λ ( u )[ − c − c ( u )] P { N u ≤ x − } du (cid:19) + (cid:90) τ e − δu λ ( u )˜ c ( u ) du + ( c − δc ) (cid:90) τ e − δu E [( x − N u ) + ] du + (cid:90) T e − δu c ( u ) λ ( u ) du, (E.1) C (0 , τ ) = (cid:90) τ e − δu λ ( u )˜ c ( u ) du + (cid:90) T e − δu c ( u ) λ ( u ) du, (E.2)∆ x C ( x, τ ) = c + (cid:90) τ e − δu λ ( u )[ − c − c ( u )] P { N u = x } du + ( c − δc ) (cid:90) τ e − δu P { N u ≤ x } du, (E.3)51 x C ( x − , τ ) = e − δτ (cid:0) c ( τ ) + c (cid:1) P { N τ = x } + (cid:90) τ e − δu (cid:2) c − c (cid:48) ( u ) + δc ( u ) (cid:3) P { N u = x } du − (cid:88) i ≤ m, l i ≤ τ e − δl i ∆ c ( l i ) P { N l i = x } . (E.4)Lemma E.1, Lemma E.2 and Proposition 3.13 essentially utilize the idea that ∆ x C ( x, τ ) is adecreasing function of τ under the conditions of Proposition 3.13. Therefore, if τ increases, then S ( τ ) increases as well since it is the minimum x value satisfying the first order condition, namely S ( τ ) = min { x ∈ Z + : ¯ c + ∆ x C ( x, τ ) ≥ } . (E.5) Lemma E.1. For every (cid:15) ∈ [0 , T ] and every τ ∈ [0 , T ] such that c ≤ λ ( τ + (cid:15) ) c ( τ + (cid:15) ) , we have ∆ x C (0 , τ + (cid:15) ) < ∆ x C (0 , τ ) . Proof. By using the expression for ∆ x C ( x, τ ) in (E.3), we obtain∆ x C (0 , τ + (cid:15) ) − ∆ x C (0 , τ )= (cid:90) τ + (cid:15)τ e − δu λ ( u ) [ − c − c ( u )] P { N u = 0 } du ( c − δc ) (cid:90) τ + (cid:15)τ e − δu P { N u = 0 } du = (cid:90) τ + (cid:15)τ e − δu P { N u = 0 } (cid:18) c − δc + λ ( u )[ − c − c ( u ) − ˜ c ( u )] (cid:19) du = (cid:90) τ + (cid:15)τ e − δu P { N u = 0 } (cid:16) − δc + λ ( u )[ − c − ˜ c ( u ) (cid:124) (cid:123)(cid:122) (cid:125) < ] (cid:17) du + (cid:90) τ + (cid:15)τ e − δu P { N u = 0 } (cid:16) c − λ ( u ) c ( u ) (cid:124) (cid:123)(cid:122) (cid:125) ≤ (cid:17) du< , where the inequality − c − ˜ c ( u ) < u ∈ [ τ, τ + (cid:15) ] since c and ˜ c ( u ) are positive.Moreover, the inequality c − λ ( u ) c ( u ) ≤ u ∈ [ τ, τ + (cid:15) ] since c ≤ λ ( τ + (cid:15) ) c ( τ + (cid:15) ) ≤ λ ( u ) c ( u ) , where first inequality is due to the condition of the lemma and the second inequality is because λ and c are non-increasing.The next lemma is helpful while stating in Proposition 3.13 that if expected total demandexceeds the order amount and cost rate of outside source does not decline sufficiently, then theorder amount should increase. 52 emma E.2. For every x ∈ Z + , every (cid:15) ∈ [0 , T ] and every τ ∈ [0 , T ] such that ( i ) x < Λ( τ ) , ( ii ) c ≤ (cid:20) Λ( u ) − x Λ( u ) (cid:21) λ ( u ) c ( u ) for all u ∈ [ τ , τ + (cid:15) ] , we have ∆ x C ( x − , τ + (cid:15) ) < ∆ x C ( x − , τ ) . (E.6) Proof. For the non-homogeneous Poisson process N with right-continuous intensity function λ ,the right-directional derivative of the function u (cid:55)→ ψ ( u ) = P { N u = x } exists and it is given by ψ (cid:48) ( u +) := lim (cid:15) ↓ (cid:15) [ ψ ( u + (cid:15) ) − ψ ( u )]= − λ ( u ) e − Λ( u ) Λ( u ) x x ! + e − Λ( u ) Λ( u ) x − ( x − λ ( u ) = − λ ( u ) P { N u = x } (cid:20) − x Λ( u ) (cid:21) . Moreover, we observe that the function Λ is strictly increasing and ψ (cid:48) ( u +) < u ∈ [0 , T ] such that Λ( u ) > x. (E.7)After applying chain rule to the function τ → e − δτ (cid:124)(cid:123)(cid:122)(cid:125) c ( τ ) + c ) (cid:124) (cid:123)(cid:122) (cid:125) P { N τ = x } (cid:124) (cid:123)(cid:122) (cid:125) x C ( x − , τ + (cid:15) ) − ∆ x C ( x − , τ )= − (cid:90) τ + (cid:15)τ δe − δu (cid:124) (cid:123)(cid:122) (cid:125) (cid:0) c ( u ) + c (cid:1) P { N u = x } du + (cid:90) τ + (cid:15)τ e − δu c (cid:48) ( u ) (cid:124) (cid:123)(cid:122) (cid:125) P { N u = x } du + (cid:88) i ≤ m, l i ≤ τ e − δl i ∆ c ( l i ) P { N l i = x }− (cid:90) τ + (cid:15)τ e − δu (cid:0) c ( u ) + c (cid:1) λ ( u ) P { N u = x } (cid:20) − x Λ( u ) (cid:21)(cid:124) (cid:123)(cid:122) (cid:125) du + (cid:90) τ + (cid:15)τ e − δu (cid:2) c − c (cid:48) ( u ) + δc ( u ) (cid:3) P { N u = x } du − (cid:88) i ≤ m, τ ≤ l i ≤ τ + (cid:15) e − δl i ∆ c ( l i ) P { N l i = x } . Notice that all the integrals include the expression e − δu P { N u = x } . Grouping them gives∆ x C ( x − , τ + (cid:15) ) − ∆ x C ( x − , τ )53 (cid:90) τ + (cid:15)τ e − δu P { N u = x }× (cid:20) − δ (cid:0) c ( u ) (cid:124) (cid:123)(cid:122) (cid:125) + c (cid:1) + c (cid:48) ( u ) (cid:124) (cid:123)(cid:122) (cid:125) − λ ( u ) (cid:20) − x Λ( τ ) (cid:21) ( c ( u ) + c )+ c − c (cid:48) ( u ) (cid:124) (cid:123)(cid:122) (cid:125) + δc ( u ) (cid:124) (cid:123)(cid:122) (cid:125) ) (cid:21) du (underbraced terms cancel each other)= (cid:90) τ + (cid:15)τ e − δu P { N u = x } (cid:20) ( c − δc ) − λ ( u ) (cid:20) Λ( u ) − x Λ( u ) (cid:21) ( c ( u ) + c ) (cid:21) du Next, after separating the remaining terms, we can see that∆ x C ( x − , τ + (cid:15) ) − ∆ x C ( x − , τ )= (cid:90) τ + (cid:15)τ e − δu P { N u = x } ( − δc ) du + (cid:90) τ + (cid:15)τ e − δu P { N u = x } (cid:20) − λ ( u ) (cid:124)(cid:123)(cid:122)(cid:125) ≥ λ ( τ + (cid:15) ) (cid:20) Λ( u ) − x Λ( u ) (cid:21) c (cid:21) du ( λ is non-increasing)+ (cid:90) τ + (cid:15)τ e − δu P { N u = x } − λ ( u ) (cid:124)(cid:123)(cid:122)(cid:125) ≥ λ ( τ + (cid:15) ) (cid:20) Λ( u ) − x Λ( u ) (cid:21) ( ˜ c ( u ) (cid:124) (cid:123)(cid:122) (cid:125) ≥ ˜ c ( T ) ) du (˜ c is non-increasing)+ (cid:90) τ + (cid:15)τ e − δu P { N u = x } (cid:124) (cid:123)(cid:122) (cid:125) = ψ ( u ) ≥ ψ ( τ + (cid:15) ) (cid:20) c − λ ( u ) (cid:20) Λ( u ) − x Λ( u ) (cid:21) c ( u ) (cid:21) du (Relation (E.7) and condition ( i )) ≤ − (cid:90) τ + (cid:15)τ e − δu P { N u = x } ( δc ) du − λ ( τ + (cid:15) ) c (cid:90) τ + (cid:15)τ e − δu P { N u = x } (cid:20) Λ( u ) − x Λ( u ) (cid:21) du − λ ( τ + (cid:15) )(˜ c ( T )) (cid:90) τ + (cid:15)τ e − δu P { N u = x } (cid:20) Λ( u ) − x Λ( u ) (cid:21) du + e − δ ( τ + (cid:15) ) P { N τ + (cid:15) = x } (cid:90) τ + (cid:15)τ (cid:20) c − λ ( u ) (cid:20) Λ( u ) − x Λ( u ) (cid:21) c ( u ) (cid:21) du (E.8) < . In (E.8), the first and second terms are negative due to the assumption c ∈ R + and condition ( i )of the lemma. The third term is negative since ˜ c ( T ) ≥ i ) of the lemma. Lastterm is negative due to condition ( ii ) of the lemma. This concludes the proof. Proof of Proposition 3.13. The function x (cid:55)→ C ( x, τ ) being discrete-convex implies that thefunction x (cid:55)→ ∆ x C ( x, τ ) is non-decreasing. Moreover, by the definition of S ( τ ), the first order54ondition in equation (E.5) has to be satisfied by S ( τ ) and τ as well as S ( τ + (cid:15) ) and τ + (cid:15) ,meaning that ¯ c + ∆ x C ( S ( τ ) , τ ) ≥ c + ∆ x C ( S ( τ + (cid:15) ) , τ + (cid:15) ) ≥ . If we can show that ∆ x C ( S ( τ ) , τ + (cid:15) ) < ∆ x C ( S ( τ ) , τ ) , (E.9)then S ( τ ) ≤ S ( τ + (cid:15) ) must hold. To show (E.9), we proceed with three steps. First, condition( iv ) implies that c ≤ (cid:124)(cid:123)(cid:122)(cid:125) ( iv ) (cid:20) Λ( τ + (cid:15) ) − S ( τ )Λ( τ + (cid:15) ) λ ( τ + (cid:15) ) c ( τ + (cid:15) ) (cid:21) ≤ λ ( τ + (cid:15) ) c ( τ + (cid:15) ) . By using Lemma E.1, we obtain ∆ x C (0 , τ + (cid:15) ) < ∆ x C (0 , τ ) . Next, observe from conditions ( iv ) and ( ii ) that for every u ∈ [ τ , τ + (cid:15) ], c ≤ (cid:124)(cid:123)(cid:122)(cid:125) ( iv ) (cid:20) Λ( u ) − S ( τ )Λ( u ) (cid:21) λ ( u ) c ( u ) ≤ (cid:124)(cid:123)(cid:122)(cid:125) ( ii ) (cid:20) Λ( u ) − S ( τ )Λ( u ) (cid:21) λ ( u ) c ( u ) . By using Lemma E.2, we obtain∆ x C ( S ( τ ) − , τ + (cid:15) ) < ∆ x C ( S ( τ ) − , τ )and similarly, for all x ∈ { , , . . . , S ( τ ) − } , we have∆ x C ( x − , τ + (cid:15) ) < ∆ x C ( x − , τ ) . Finally, we obtain∆ x C ( S ( τ ) , τ + (cid:15) ) =∆ x C (0 , τ + (cid:15) ) + S ( τ ) − (cid:88) x =0 ∆ x C ( x, τ + (cid:15) ) < ∆ x C (0 , τ ) + S ( τ ) − (cid:88) x =0 ∆ x C ( x, τ ) = ∆ x C ( S ( τ ) , τ ) . F Proof of Results in Subsubsection 3.3.2 In this section, we prove Proposition 3.15 and Proposition 3.16. For the brevity of notation, wetake derivative of integrals when the functions are right-continuous. To ensure the existence, it is55ossible to take right-derivative and obtain the same expressions. We characterize an upper bound τ ub and a lower bound τ lb by stating, respectively, that ∂ C ( x,τ ) ∂τ ≥ τ ≥ τ ub and ∂ C ( x,τ ) ∂τ ≤ τ ≤ τ lb . Lemma F.1 shows a condition which makes the first derivative of C ( x, τ ) positive. Lemma F.1. Let x ∈ Z + and τ ∈ [0 , T ] . If P ( N τ ≥ x )˜ c ( τ ) ≥ P ( N τ ≤ x − c ( τ ) + c ] , then ∂ C ( x,τ ) ∂τ ≥ . Proof. By using equation (E.1), we see that ∂ C ( x, τ ) ∂τ = e − δτ λ ( τ )[ − c − c ( τ )] P { N τ ≤ x − } + e − δτ λ ( τ )˜ c ( τ ) + e − δτ ( c − δc ) E [( x − N τ ) + ]= e − δτ λ ( τ )[ − c − c ( τ ) − ˜ c ( τ )] P { N τ ≤ x − } + e − δτ λ ( τ )˜ c ( τ ) + e − δτ ( c − δc ) x − (cid:88) k =0 P { N τ ≤ k } = e − δτ λ ( τ )[ − c − c ( τ )] P { N τ ≤ x − } + e − δτ λ ( τ )˜ c ( τ ) P { N τ ≥ x } + e − δτ ( c − δc ) x − (cid:88) k =0 P { N τ ≤ k } = e − δτ λ ( τ ) (cid:16) P { N τ ≥ x } ˜ c ( τ ) − P { N τ ≤ x − } [ c ( τ ) + c ] (cid:17) + e − δτ ( c − δc ) x − (cid:88) k =0 P { N τ ≤ k } , where the second equality uses Lemma A.2. Proof of Proposition 3.15. We first note that P { N τ ≥ x } = 1 − P { N τ ≤ x − } , so P { N τ ≥ x } ˜ c ( τ ) ≥ P { N τ ≤ x − } [ c ( τ ) + c ] ⇐⇒ ˜ c ( τ ) ≥ P { N τ ≤ x − } [ c ( τ ) + ˜ c ( τ ) + c ] ⇐ = ˜ c ( T ) ≥ P { N τ ≤ x − } [ c ( τ ) + ˜ c ( τ ) + c ] , where last implication is due to ˜ c ( . ) being non-increasing. Since τ ub satisfies inequality (3.15), itfollows from Lemma F.1 that ∂ C ( x, τ ) ∂τ (cid:12)(cid:12)(cid:12)(cid:12) τ = τ ub ≥ . Therefore, it suffices to show that for any τ ≥ τ ub , ∂ C ( x,τ ) ∂τ ≥ 0. To see this, we show that thefunction τ (cid:55)→ f ( τ ) := P { N τ ≤ x − } [ c ( τ ) + c ] is non-increasing. Taking the derivative of f yields that ∂f ( τ ) ∂τ = − λ ( τ ) P { N τ = x − } [ c ( τ ) + c ] + P { N τ ≤ x − } c (cid:48) ( τ ) . c ( τ ) + c ≥ λ ( τ ) ≥ 0. The secondterm is negative since ˜ c ( τ ) is non-increasing. Therefore, f is non-increasing. Lemma F.2. If λ ( τ ) ≥ and P { N τ ≤ x − } [ c ( τ ) + c ] ≥ x ( c − δc ) + ˜ c (0) , then ∂ C ( x,τ ) ∂τ ≤ . Proof. Using the same steps in Lemma F.1 yields ∂ C ( x, τ ) ∂τ = e − δτ λ ( τ )˜ c ( τ ) + e − δτ ( c − δc ) x − (cid:88) k =0 P { N τ ≤ k } − e − δτ λ ( τ )[ c ( τ ) + c ] P { N τ ≤ x − } . Moreover, it is possible to see that x − (cid:88) k =0 P { N τ ≤ k } ≤ x and ˜ c ( τ ) ≤ ˜ c (0) (F.1)since the function ˜ c ( . ) is non-increasing. Therefore, ∂ C ( x, τ ) ∂τ ≤ ⇐⇒ e − δτ λ ( τ )˜ c ( τ ) + e − δτ ( c − δc ) x − (cid:88) k =0 P { N τ ≤ k } ≤ e − δτ λ ( τ )[ c ( τ ) + c ] P { N τ ≤ x − }⇐ = λ ( τ )˜ c (0) + ( c − δc ) x ≤ λ ( τ )[ c ( τ ) + c ] P { N τ ≤ x − } (by (F.1)) ⇐ = λ ( τ )˜ c (0) + λ ( τ )( c − δc ) x ≤ λ ( τ )[ c ( τ ) + c ] P { N τ ≤ x − } (since λ ≥ ⇐⇒ ˜ c (0) + ( c − δc ) x ≤ [ c ( τ ) + c ] P { N τ ≤ x − } . Proof of Proposition 3.16. It suffices to show that any τ ∈ [0 , τ lb ] satisfy the inequality in(3.16), so that ∂ C ( x,τ ) ∂τ ≤ λ is non-increasing,therefore, any τ ≤ τ lb satisfies λ ( τ ) ≥ 1. Next, we show that the function τ (cid:55)→ g ( τ ) := P { N τ ≤ x − } [ c ( τ ) + c ]is a non-increasing function. Taking derivative of g yields that ∂g ( τ ) ∂τ = − λ ( τ ) P { N τ ≤ x − } [ c ( τ ) + c ] + P { N τ ≤ x − } c (cid:48) ( τ )The first term is negative due to the assumptions that c ( τ ) + c ≥ λ ( τ ) ≥ 0. Thesecond term is negative since both c and ˜ c is non-increasing.57 Numbering for the Parameter Settings This section assigns a number for each parameter setting where the parameters take values in thesets presented in Table 4. The Table 25 below show the assigned numbers. Also recall that ¯ c = 100, c = 0 . c and ¯ c = 2¯ c . We present the setup cost K and initial inventory x values alongside therelated result. λ T c γ δ ¯ c λ T c γ δ ¯ c λ T c γ δ ¯ c − 200 85 Lin 100 25 10 − − − − 200 45 Conc 50 -25 10 − − − − − − − − − − 200 89 Lin 100 -25 0.01 0.005 2006 Conv 50 25 10 − − − − − 200 49 Conc 100 25 0.01 0.005 200 91 Lin 100 -25 0.01 10 − − − − − 200 93 Lin 100 -25 10 − − − − 200 53 Conc 100 25 10 − − − − − − − − − − 200 97 Cons 50 25 0.01 0.005 20014 Conv 50 -25 10 − − − − − 200 57 Conc 100 -25 0.01 0.005 200 99 Cons 50 25 0.01 10 − − − − − 200 101 Cons 50 25 10 − − − − 200 61 Conc 100 -25 10 − − − − − − − − − − 200 105 Cons 50 -25 0.01 0.005 20022 Conv 100 25 10 − − − − − 200 65 Lin 50 25 0.01 0.005 200 107 Cons 50 -25 0.01 10 − − − − − 200 109 Cons 50 -25 10 − − − − 200 69 Lin 50 25 10 − − − − − − − − − − 200 113 Cons 100 25 0.01 0.005 20030 Conv 100 -25 10 − − − − − 200 73 Lin 50 -25 0.01 0.005 200 115 Cons 100 25 0.01 10 − − − − − 200 117 Cons 100 25 10 − − − − 200 77 Lin 50 -25 10 − − − − − − − − − − 200 121 Cons 100 -25 0.01 0.005 20038 Conc 50 25 10 − − − − − 200 81 Lin 100 25 0.01 0.005 200 123 Cons 100 -25 0.01 10 − − − − − 200 125 Cons 100 -25 10 − − − − − − − Table 25: Numbers for each parameter setting. The columns with the symbol λ is convex (Conv), concave (Conv), linear (Lin), or constant (Cons). eferences Ararat, C¸ ., N. Erkip, E. Ozyoruk. 2021. End-of-life inventory control under delay and stochasticintensity. Working Paper .Bayındır, Z. P., N. Erkip, R. G¨ull¨u. 2007. Assessing the benefits of remanufacturing option underone-way substitution and capacity constraint. Computers & Operations Research , 34 (2), 487-514.Behfard, S., M. C. van der Heijden, A. Al Hanbali, W. H. Zijm. 2015. Last time buy and repairdecisions for spare parts. European Journal of Operational Research , 244 (2), 498-510.Behfard, S., A. Al Hanbali, M. C. van der Heijden, W. H. Zijm. 2018. Last time buy and repairdecisions for fast moving parts. International Journal of Production Economics , 197, 158-173.Beyer, D., F. Cheng, S. P. Sethi, M. Taksar. 2010. Markovian Demand Inventory Models. NewYork: Springer .Bradley, J. R., H. H. Guerrero. 2008. Product design for life-cycle mismatch. Production and Op-erations Management , 17 (5), 497-512.Bradley, J. R., H. H. Guerrero. 2009. Lifetime buy decisions with multiple obsolete parts. Productionand Operations Management , 18 (1), 114-126.Callioni, G., X. de Montgros, R. Slagmulder, L. N. Van Wassenhove, L. Wright. 2005. Inventory-driven costs. Harvard Business Review , 83 (3), 135-141.Cattani, K. D., G. C. Souza. 2003. Good buy? Delaying end-of-life purchases. European Journal ofOperational Research , 146 (1), 216-228.C¸ ınlar, E. 2011. Probability and Stochastics. Springer Science & Business Media .David, I., E. Greenshtein, A. Mehrez. 1997. A dynamic-programming approach to continuous-reviewobsolescent inventory problems. Naval Research Logistics (NRL) , 44 (8), 757-774.Fortuin, L. 1980. The all-time requirement of spare parts for service after sales—theoretical analysisand practical results. International Journal of Operations & Production Management .59ortuin, L. 1984. Initial supply and re-order level of new service parts. European Journal of Oper-ational Research , 15 (3), 310-319.Frenk, J. B. G., S. Javadi, M. Pourakbar, S. O. Sezer. 2019. An exact static solution approach forthe service parts end-of-life inventory problem. European Journal of Operational Research , 272(2), 496-504.Frenk, J. B. G., S. Javadi, S. O. Sezer. 2019. An optimal stopping approach for the end-of-lifeinventory problem. Mathematical Methods of Operations Research , 90 (3), 329-363.Frenk, J. B. G., C. Pehlivan, S. O. Sezer. 2019. Order and exit decisions under non-increasing pricecurves for products with short life cycles. Mathematical Methods of Operations Research , 90 (3),365-397.Hur, M., B. B. Keskin, C. P. Schmidt. 2018. End-of-life inventory control of aircraft spare partsunder performance based logistics. International Journal of Production Economics , 204, 186-203.Inderfurth, K., R. Kleber. 2013. An advanced heuristic for multiple-option spare parts procurementafter end-of-production. Production and Operations Management , 22 (1), 54-70.Inderfurth, K., K. Mukherjee. 2008. Decision support for spare parts acquisition in post productlife cycle. Central European Journal of Operations Research , 16 (1), 17-42.Jack, N., F. Van der Duyn Schouten. 2000. Optimal repair–replace strategies for a warrantedproduct. International Journal of Production Economics , 67 (1), 95-100.Kleber, R., T. Schulz, G. Voigt. 2012. Dynamic buy-back for product recovery in end-of-life spareparts procurement. International Journal of Production Research , 50 (6), 1476-1488.Leifker, N. W., P. C. Jones, T. J. Lowe. 2014. Determining optimal order amount for end-of-life partsacquisition with possibility of contract extension. The Engineering Economist , 59 (4), 259-281.Leifker, N. W., P. C. Jones, T. J. Lowe. 2012. A continuous-time examination of end-of-life partsacquisition with limited customer information. The Engineering Economist , 57 (4), 284-301.Oh, S., ¨O. ¨Ozer. 2016. Characterizing the structure of optimal stopping policies. Production andOperations Management , 25 (11), 1820-1838. 60in¸ce, C¸ ., J. B. G. Frenk, R. Dekker. 2015. The role of contract expirations in service parts man-agement. Production and Operations Management , 24 (10), 1580-1597.Pin¸ce, C¸ ., R. Dekker. 2011. An inventory model for slow moving items subject to obsolescence. European Journal of Operational Research , 213 (1), 83-95.Porteus, E. L. 2002. Foundations of Stochastic Inventory Theory. Stanford University Press .Pourakbar, M., E. van der Laan, R. Dekker. 2014. End-of-life inventory problem with phaseoutreturns. Production and Operations Management , 23 (9), 1561-1576.Pourakbar, M., J. B. G. Frenk, R. Dekker. 2012. End-of-life inventory decisions for consumerelectronics service parts. Production and Operations Management , 21 (5), 889-906.Pourakbar, M., R. Dekker. 2012. Customer differentiated end-of-life inventory problem. EuropeanJournal of Operational Research , 222 (1), 44-53.Shi, Z., S. Liu. 2020. Optimal inventory control and design refresh selection in managing partobsolescence. European Journal of Operational Research . 287 (1), 133-144.Shi, Z., 2019. Optimal remanufacturing and acquisition decisions in warranty service consideringpart obsolescence. Computers & Industrial Engineering , 135, 766-779.Shen, Y., S. P. Willems. 2014. Modeling sourcing strategies to mitigate part obsolescence. EuropeanJournal of Operational Research , 236 (2), 522-533.Silver, E. A., D. F. Pyke, D. J. Thomas. 2016. Inventory and production management in supplychains. CRC Press .Teunter, R. H., W. K. K. Haneveld. 2002. Inventory control of service parts in the final phase. European Journal of Operational Research , 137 (3), 497-511.van der Heijden, M., B. P. Iskandar. 2013. Last time buy decisions for products sold under warranty. European Journal of Operational Research , 224 (2), 302-312.van Kooten, J. P., T. Tan. 2009. The final order problem for repairable spare parts under condem-nation.