Robust Multi-product Newsvendor Model with ... - Optimization Online

Robust Multi-product Newsvendor Model with Substitution under Cardinality-constrained Uncertainty Set Jie Zhang Department of Industrial and Systems Engineering, Virginia Tech, Blacksburg, VA 24061, [email protected]

Weijun Xie Department of Industrial and Systems Engineering, Virginia Tech, Blacksburg, VA 24061, [email protected]

This paper studies Robust Multi-product Newsvendor Model with Substitution (R-MNMS), where the demand is stochastic and is subject to cardinality-constrained uncertainty set. The goal of this work is to determine the optimal order quantities of multiple products to maximize the worst-case total profit. To achieve this, we first show that for given order quantities, computing the worst-case total profit in general is NP-hard. Therefore, we derive the closed-form optimal solutions for the following three special cases: (1) if there are only two products, (2) if there is no substitution among different products, and (3) if the budget of uncertainty is equal to the number of products. For a general R-MNMS, we formulate it as a mixed integer linear program with an exponential number of constraints, and develop a branch and cut algorithm to solve it. For large-scale problem instances, we further propose a conservative approximation of R-MNMS and prove that under some certain conditions, this conservative approximation yields an exact optimal solution to R-MNMS. The numerical study demonstrates the effectiveness of the proposed approaches and the robustness of our model. Key words : Newsvendor Model, Robust, Cardinality-constrained Uncertainty Set, Mixed Integer Program, Branch and Cut Algorithm

1. Introduction This paper studies Multi-product Newsvendor Model with Substitution (MNMS) under demand uncertainty, in which a retailer determines the optimal order quantity for each product to maximize its total profit. Due to similarity among different products and their occasional unavailability, the phenomenon of substitution among different products is quite common and has been observed in many studies (cf., Bassok et al. 1999, Rajaram and Tang 2001, Chopra and Meindl 2007, Shumsky and Zhang 2009, Stavrulaki 2011, Choi 2012, Yu et al. 2015). For instance, when shopping at Amazon.com, a customer might turn to the blue hat if his or her first-choice green hat were currently unavailable. The existence of substitution somehow increases the profit of the retailer, however, on the other hand, significantly complicates the problem and makes the problem very challenging to handle. In addition, due to the stochasticity of customers’ demand, it might be hard to forecast 1

Jie Zhang and Weijun Xie: Robust Multi-product Newsvendor Model with Substitution

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the demand accurately. Therefore, many works (cf., Erlebacher 2000, Schweitzer and Cachon 2000, Rajaram and Tang 2001, Rao et al. 2004, Zhang et al. 2018) proposed stochastic programming models to tackle the demand uncertainty by assuming that the probability distribution of the demand is known. However, in many cases, a good estimation of probability distribution might be very challenging and a misleading probability distribution can result in a much worse solution than the true optimal one (cf., Xie and Ahmed 2018a). Therefore, to foster a more reliable decision, instead we study the “Robust” Multi-product Newsvendor Model with Substitution (R-MNMS) subject to cardinality-constrained uncertainty set. A R-MNMS encounters the following technical features. First of all, due to the substitution effect, it has been shown in Zhang et al. (2018), even when the demand is deterministic, MNMS can be NPhard. Second, most of the existing researches assumed that the customers’ demand follows a given probability distribution, which, however, might result in a loss of sales due to inaccurate demand forecasting. Third, although many existing works illustrated interesting properties of MNMS, the closed-form optimal solutions are rarely known, therefore very limited managerial insights have been discovered so far. In this paper, we will show that under some conditions, all of these features can be appropriately addressed. 1.1. Relevant Literature There are two types of relevant literature on how to handle the uncertainty of MNMS: (1) using stochastic programming, and (2) using robust optimization. In this section, we will review both approaches. Many works on MNMS assumed that the probability distribution of the demand is known, for example, Huang et al. 2011, Netessine and Rudi 2003, Zhang et al. 2018. Huang et al. (2011) analyzed the decentralized MNMS, gave the conditions in which the Nash equilibrium exists and provided an iterative algorithm to solve the model. However, its centralized counterpart becomes highly non-convex, which will be studied in this paper. Netessine and Rudi (2003) demonstrated that the profit function could be quasi-concave or bi-modal when the demand is deterministic. Recently, Zhang et al. (2018) formulated stochastic MNMS as a mixed integer linear program and developed polynomial-time approximation algorithms with performance guarantee to solve it. Different from these works, this paper studies centralized R-MNMS under cardinality-constrained uncertainty set. However, in practice, it might not be easy to learn the distribution of the random demand completely, in particular, when the random demand is not stationary, i.e., the probability distribution of the random demand is subject to change from time to time. In addition, the inaccurate probability distribution might result in unreliable or misleading decisions. Under these circumstances,


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alternatively, one can choose the robust approach to formulate the model with partial information of the demand, which can be easily characterized or will stay the same at a relatively long period (i.e., mean, variance, or support). Therefore, some works applied the robust optimization to the ¨ newsvendor problems (Scarf 1957, Vairaktarakis 2000, Ozler et al. 2009, Lin and Ng 2011, Raza 2014, Hanasusanto et al. 2015, Carrizosa et al. 2016, Chen and Zhang 2009, Ardestani-Jaafari and Delage 2016). Especially, Scarf (1957) was the pioneer to introduce the robust idea to analyze single-product newsvendor problem with known mean and variance of the demand. Vairaktarakis (2000) studied several minimax regret formulations for robust multi-item newsvendor models with a budget constraint when the support of demand is known. They developed efficient algorithms to solve the proposed robust models. Similarly, when the demand is subject to a given interval, Lin and Ng (2011) determined the optimal order quantity as well as the market selection for a minimax regret multi-market newsvendor model. They further developed an approximation algorithm for solving the large-sized problem instances. With known first and second moments and the shape of the demand distribution, Perakis and Roels (2008) derived the optimal order policy by minimizing the maximum regret of the newsvendor problem. Ardestani-Jaafari and Delage (2016) studied the robust optimization with sum of piecewise linear functions and polyhedral uncertainty set, which can be applied to solve the robust multi-product newsvendor problem under budget uncertainty set. However, all of these works either studied robust single-product newsvendor problem or multiproduct newsvendor problem without substitution, while different from these existing works, this paper will study robust multi-product newsvendor problem with substitution, i.e., R-MNMS. There are very limited works on R-MNMS. For decentralized R-MNMS, Jiang et al. (2011) used the absolute regret criterion to obtain the unique Nash equilibrium. In their work, only the support of the demand is known, and they also showed that the robust model tended to be more tractable than the stochastic counterpart. Recent work in Li and Fu (2017) studied a robust two-product newsvendor model with substitution, when the first two moments of demand are known. However, the authors were only able to provide the optimal solution for the following two extreme cases: (1) if there exists no substitution, or (2) if there is a perfect substitution between products. Different from these works, this paper studies centralized R-MNMS, and is not only restricted to the two-product cases. Many robust optimization problems become NP-hard although their stochastic counterparts can be solved relatively easily (cf., Bertsimas et al. 2011). Therefore, as described in Chen et al. 2007, Xie and Ahmed 2018b, constructing a suitable uncertainty set is an effective way to address the issue of tractability and over-conservatism. In this paper, we study R-MNMS by using the cardinalityconstrained uncertainty set to characterize the random demand. The cardinality-constrained uncertainty set was first introduced by Bertsimas and Sim (2004) into robust optimization to reduce


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over-conservatism while at the same time, still achieve the robustness. This framework has been successfully applied to many different areas, for example, healthcare (cf., Lanzarone and Matta 2012, Carello and Lanzarone 2014, Addis et al. 2015), manufacturing (cf., Lugaresi 2016, Lugaresi et al. 2017), inventory management (cf. Bertsimas et al. 2004, Bertsimas and Thiele 2006, Solyalı et al. 2012), portfolio optimization (cf. Moon and Yao 2011), scheduling (cf., HazıR and Dolgui 2013, Lu et al. 2014, Moreira et al. 2015), etc. Since no much work has been done on R-MNMS, this paper will fill this gap and apply cardinality uncertainty set into it. We show that under certain conditions (for example, for the two-product case), we can derive closed-form optimal solutions, which allow us to draw interesting managerial insights. 1.2. Summary of Main Contributions The objective of this paper is to help a retailer determine optimal order quantities of a singleperiod multi-product newsvendor model with substitution, which optimizes the worst-case total profit under the cardinality-constrained uncertainty set. The main contributions of this paper are summarized as below: (i) We develop an equivalent reformulation of R-MNMS and prove that computing the worst-case total profit in general is NP-hard for given order quantities. (ii) We derive closed-form solutions for the following three special cases of R-MNMS: (1) if there are only two products; (2) if there is no substitution among different products; or (3) if the budget of uncertainty is equal to the number of products. (iii) We further reformulate R-MNMS as a mixed-integer linear program (MILP) with an exponential number of constraints, and develop branch and cut algorithm to solve it. (iv) We provide a conservative approximation of R-MNMS, which can be solved more efficiently, and also prove that under certain conditions, the proposed conservative approximation is equivalent to R-MNMS. The remainder of the paper is organized as follows. Section 2 introduces the problem setting and the model. Section 3 presents the properties of the model and proves the complexity of computing the worst-case total profit. In Section 4, we derive the optimal order quantities for three special cases of the model. Section 5 reformulates the R-MNMS as an MILP, and develops a branch and cut algorithm and a conservative approximation to solve it. Section 6 presents the results of our numerical investigation on the proposed algorithms.

Notation: The following notation is used throughout the paper. We use bold-letters (e.g., x, A) to denote vectors and matrices, and use corresponding non-bold letters to denote their components. Given a vector x, its zero norm kxk0 denotes the number of its nonzero elements. We let e be


5

the vector of all ones, and let ei be the ith standard basis vector. Given an integer n, we let [n] := {1, 2, . . . , n}, and use Rn+ := {x ∈ Rn : xi ≥ 0, ∀i ∈ [n]}. Given a real number t, we let (t)+ := max{t, 0}. Given a finite set I, we let |I | denote its cardinality. We let ξ˜ denote a random vector and denote its realizations by ξ. Additional notation will be introduced as needed.

2. Model Formulation In this section, we present the model formulation for R-MNMS. To begin with, suppose that there is a retailer selling n similar products in the market indexed by [n] := {1, · · · , n} at a given time period. For each product i ∈ [n], its cost is ci , price is pi , and salvage value is si , where by convention, we assume that pi ≥ ci ≥ si . Each product also bears with ˜ i for each i ∈ [n]. Ideally, the retailer would like to determine the optimal a random demand D order quantity for each product i ∈ [n], denoted as Qi . Due to the substitution effect, the effective demand of each product will be affected by its realized demand, its order quantity as well as other products’ conditions (i.e., whether out-of-stock or not). To formulate this effect, we suppose that the demand of product j ∈ [n] can be proportionally substituted by another product i ∈ [n] and i 6= j, once the part of the demand of product j cannot be satisfied by its order quantity Qi . In particular, we let αji be the substitution rate, which is the proportion of the unmet demand of product j substituted by product i. In this paper, we assume that all the products have the same unit, therefore, substitution rate satisfies αji ∈ [0, 1] for each pair of products i, j ∈ [n]. Also, by ˜ is (Q) denote the effective demand function default, we let αii = 0 for each product i ∈ [n]. We let D of product i ∈ [n] as below: ˜ is (Q) = D ˜i + D

X

˜ j − Qj )+ , ∀i ∈ [n], αji (D

(1)

j∈[n]

where the second term in the sum is due to its substitution to the unavailable products. As shown in Zhang et al. (2018), the retailer’s total profit for given order quantities Q and ˜ can be formulated as: demand D X s s ˜ ˜ ˜ Π(Q, D) := pi min Qi , Di (Q) − ci Qi + si Qi − Di (Q) . (2) i∈[n]

+

Oftentimes, the demand of products is stochastic and the distribution is unknown. To well address the demand uncertainty, we will use robust optimization. In particular, we will study R-MNMS under cardinality-constrained uncertainty set. In this uncertainty set, suppose that the demand of ˜ is within a box, e.g., D ˜ ∈ [D − l, D + u]), where D denotes the nominal the n products (i.e., D) demand, l, u denote the lower and upper deviations of the demand respectively satisfying l ∈ [0, D] and u ≥ 0. We also assume that at most k ∈ [n] ∪ {0} products are allowed to deviate from their


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nominal demand D, where k is also known as the budget of uncertainty in Bertsimas et al. (2011), which is used to leverage the price of robustness in Bertsimas and Sim (2004). Therefore, the uncertainty set of the demand can be written as o n ˜ :D ˜ i = Di + ∆i , −li ≤ ∆i ≤ ui , ∀i ∈ [n], k∆k ≤ k , U0 = D 0

(3)

where k · k0 denotes the zero-norm. With the notation introduced above, R-MNMS can be formulated as:     X ˜ := ˜ is (Q) − ci Qi + si Qi − D ˜ is (Q) v ∗ = maxn min Π(Q, D) pi min Qi , D . ˜ Q∈R+ D∈U  +  0

(4)

i∈[n]

In Model (4), the objective is to find optimal order quantities to maximize the worst-case total profit over the uncertainty set U0 . For each product i ∈ [n], we let P i = pi − ci ≥ 0 and S i = pi − si ≥ 0. Note that P i can be interpreted as the marginal profit or underage cost of product i ∈ [n], while S i is the sum of the underage cost (pi − ci ) and overage cost (ci − si ) of product i ∈ [n], where their Pi Si

is known as the critical ratio of newsvendor model (c.f., Nahmias and Olsen (2015)). Since ˜ is (Q))+ for each i ∈ [n], the above Model (4) is equivalent to ˜ is (Q)) = Qi − (Qi − D min(Qi , D     X X ˜ := ˜ is (Q) v ∗ = maxn min Π(Q, D) . (5) P i Qi − S i Qi − D ˜ Q∈R+ D∈U  + 0

ratio

i∈[n]

i∈[n]

For notational convenience, throughout this paper, we will let Q∗ denote an optimal solution to R-MNMS (5).

3. Equivalent Reformulation and Model Properties In this section, we study R-MNMS under cardinality-constrained uncertainty set and derive its equivalent reformulation. We also provide upper bounds of optimal order quantities and show that computing the worst-case total profit for given order quantities in general is NP-hard. 3.1. Equivalent Reformulation In this subsection, we provide an alternative formulation for Model (5). First, we make the following observation. ˜ is monotone nondecreasing in D. ˜ Lemma 1 For any Q ∈ Rn+ , the profit function Π Q, D ˜ is nondecreasing in D ˜ is (Q) Proof: According to Model (5), we see that the profit function Π Q, D ˜ is (Q) is also nondecreasing in D ˜ i for each product i ∈ [n]. and from (1), the effective demand D ˜ is nondecreasing in the demand D. ˜ Therefore, the profit function Π Q, D


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Now we are ready to show our equivalent reformulation. The main idea of the derivation is to show that in the worst-case, the uncertainty set U0 can be restricted to the following mixed integer set: U=

 

˜: D



X

˜ i = Di − li zi , zi ∈ {0, 1}, ∀i ∈ [n] zi ≤ k, D

 

.

(6)



i∈[n]

˜ z) satisfying constraints in (6), let us define Clearly, set U ⊆ U0 , since for any feasible point (D, ˜ ∆) satisfies the constraints in (3). Indeed, we can show that ∆i = −li zi for each i ∈ [n], then (D, Proposition 1 R-MNMS is equivalent to ˜ , v ∗ = maxn min Π Q, D

(7)

˜ Q∈R+ D∈U

where U is defined in (6). Proof: Let v1 denote the optimal value of Model (7), then we only need to show v1 = v ∗ . ˜ ∈ U0 , there exists ∆ such that k∆k ≤ k, D ˜ i = Di + ∆i , −li ≤ ∆i ≤ ui . (i) v1 ≥ v ∗ . For any D 0 0, if ∆i = 0 Let us define binary variable zi = for each i ∈ [n]. Since k∆k0 ≤ k, thus we 1, if ∆i 6= 0 P ˜ i∗ = Di − li zi for each i ∈ [n]. Clearly, we have D ˜∗ ∈ must have i∈[n] zi ≤ k. Let us define D ˜ ∗ ≤ D. ˜ For any fixed Q ∈ Rn+ , by Lemma 1, we know that the profit function U and D ˜ ∗ , which implies ˜ is nondecreasing in the demand D. ˜ ≥ Π Q, D ˜ Thus, Π Q, D Π Q, D ˜ ≥ min ˜ Π Q, D ˜ for any Q ∈ Rn . This proves v1 ≥ v ∗ . minD∈U Π Q, D ˜ + D∈U 0 ∗ n ˜ ≤ min ˜ Π Q, D ˜ , thus, Π Q, D (ii) v1 ≤ v . Since U0 ⊇ U , thus for any Q ∈ R+ , minD∈U ˜ D∈U 0 v1 ≤ v ∗ . ˜ i = Di − li zi in (5) and defining the following cardinality From Proposition 1, by substituting D set

   X  X= z: zi ≤ k, zi ∈ {0, 1} ,  

(8)

i∈[n]

then we can have the following equivalent formulation of R-MNMS:     X P i Qi − R(Q) , v ∗ = maxn f (Q) := Q∈R+  

(9a)

i∈[n]

where

 R(Q) := max

X

z∈X

S i Qi − Di + li zi −

i∈[n]

 X j∈[n]

αji (Dj − lj zj − Qj )+  .

(9b)

+

This new equivalent formulation (9) allows us to compute the worst-case profit function via an integer program rather than a nonconvex program, which can be further reduced to a mixed integer linear program (MILP) in Section 5.


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One direct benefit of formulation (9) is that we can easily derive upper bounds of optimal order quantities. The result can be proved by contradiction. Proposition 2 There exists an optimal solution Q∗ to R-MNMS such that for each product i ∈ [n], P Q∗i ≤ Mi , where Mi = Di + j∈[n] αji Dj . Proof: See Appendix A.1.

This result is very useful to derive an equivalent MILP formulation of R-MNMS in Section 5. 3.2. Complexity of the Inner Maximization Problem (9b) In this subsection, we will show that the inner maximization problem (9b) of R-MNMS is NP-hard. First, observe that ( (Dj − lj − Qj )+ , (Dj − lj zj − Qj )+ = (Dj − Qj )+ ,

if zj = 1 if zj = 0

for each j ∈ [n], thus this observation allows us to linearize nonlinear expressions {(Dj − lj zj − Qj )+ }j∈[n] and to rewrite (9b) as

R (Q) = max z∈X

 X 

 S i Qi − Di + li zi −

i∈[n]

X j∈[n]

αji

   . (Dj − lj − Qj )+ zj + (Dj − Qj )+ (1 − zj )   +

(10) Next, we show that the inner maximization problem (10) is NP-hard via a reduction to the well known clique problem. Theorem 1 The inner maximization problem (10) in general is NP-hard. Proof: See Appendix A.2.

Theorem 1 shows that unlike many robust optimization problems, it might be difficult to derive a tractable form for the general inner maximization problem (10). Thus, instead, in Section 4, we propose three special cases such that both inner maximization (10) and R-MNMS are tractable. For general R-MNMS, we propose an equivalent MILP reformulation and develop exact and approximate algorithms to solve it, which will be presented in Section 5.

4. Three Special Cases: Closed-form Optimal Solutions In this section, we will study three different special cases of R-MNMS (9) and derive their closedform optimal solutions.


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4.1. Special Case I: n = 2, k = 1 In this section, we study R-MNMS with only two products (i.e., n = 2) and the budget of uncertainty is equal to 1 (i.e., k = 1 in set X defined in (8)). Note that if k = 0 or 2, it reduces to Special Case III, which will be discussed in Section 4.3. Under this setting, R-MNMS (9) becomes:     X  X X ∗   v = max , P i Qi − max S i Qi − Di + li zi − αji (Dj − lj zj − Qj )+ z∈X   Q∈R2 + i∈[2]

i∈[2]

j∈[2]

(11)

+

and X = {z : z1 + z2 ≤ 1, zi ∈ {0, 1}, ∀i ∈ [2]}. To simplify our closed-form solutions, we further make the following assumption. Assumption 1 Suppose that D2 α21 ≥ l1 ≥ α21 l2 , D1 α12 ≥ l2 ≥ α12 l1 . Assumption 1 postulates that the demand deviation of one product cannot be smaller than the substitution part of the other product’s demand deviation and cannot be larger than the substitution part of the other product’s nominal demand. Please note that our analysis is general and can be also applied to the other parametric settings without satisfying Assumption 1. However, for the brevity of this paper, we will stick to this assumption. The next theorem presents our main findings of the optimal order quantities for this special case under Assumption 1. The key ideas to these results are: (1) to divide the feasible regions into 9 subregions by comparing Qi with Di − li and Di for each i ∈ [2]; (2) for each subregion, R-MNMS (11) becomes a concave maximization problem with a piecewise linear objective function, thus one of its optimal solutions can be achieved by an extreme point; and (3) for each subregion, there are not too many potential optimal solutions, thus, we enumerate all the candidate solutions and find the one which achieves the highest total profit across all the 9 subregions. Theorem 2 Suppose n = 2, k = 1, and Assumption 1 holds, then the optimal order quantities Q∗ = (Q∗1 , Q∗2 ) are characterized by the following three cases: Case 1: If P 1 ≤ P 2 α12 P 2 ≥ P 1 α21 , then (Q∗1 , Q∗2 ) = (0, D2 − l2 + α12 D1 ). Case 2: If P 2 ≤ P 1 α21 P 1 ≥ P 2 α12 , then (Q∗1 , Q∗2 ) = (D1 − l1 + α21 D2 , 0). Case 3: If P 1 ≥ P 2 α12 , P 2 ≥ P 1 α21 , then we have the following two sub-cases: l1 −α21 l2 l2 −α12 l1 Sub-case 3.1: If S 1 l1 ≥ S 2 l2 , then (Q∗1 , Q∗2 ) = D1 − 1−α , D − or (Q∗1 , Q∗2 ) = 2 1−α12 α21 12 α21 2 −S 1 l1 D1 , D2 − SS2 l−S . 2 1 α21 l1 −α21 l2 l2 −α12 l1 Sub-case 3.2: If S 1 l1 ≤ S 2 l2 , then (Q∗1 , Q∗2 ) = D1 − 1−α , D − or (Q∗1 , Q∗2 ) = 2 1−α12 α21 12 α21 1 −S 2 l2 D1 − SS1 l−S , D2 . α 1

2 12


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Proof: See Appendix A.3.

Theorem 2 provides a complete characterization of optimal order quantities of the two-product case, which highly depend on the comparison between the marginal profit of product i and the profit generated by using product j to substitute product i. In particular, we make the following remarks. Remark 1

(i) Suppose that the marginal profit of product 1 is lower than the profit generated

by using product 2 to substitute product 1, but the marginal profit of product 2 is higher than the profit generated by using product 1 to substitute product 2, i.e., product 2 is much more profitable than product 1. Thus, in this case, the decision makers should only order product 2 to satisfy their customers’ demand as well as to satisfy part of the customers’ demand of product 1 by substitution. In this case, the worst-case demand of product 2 is D2 − l2 while the worst-case demand of product 1 is equal to the nominal demand D1 . (ii) Similarly, suppose that the marginal profit of product 2 is lower than the profit generated by using product 1 to substitute product 2, but the marginal profit of product 1 is higher than the profit generated by using product 2 to substitute product 1, i.e., product 1 is much more profitable than product 2. Thus, in this case, the decision makers should only order product 1 to satisfy their customers’ demand as well as part of the customers’ demand of product 2 by substitution. In this case, the worst-case demand of product 1 is D1 − l1 while the worst-case demand of product 2 is equal to the nominal demand D2 . (iii) If the marginal profit of one product is higher than the profit generated by using the other product to substitute this product (i.e., both products are similarly profitable), then the optimal order quantities depend on the relationship between S 1 l1 and S 2 l2 . One special case is that when si = ci for each product i ∈ [2], i.e., the salvage value of each product is equal to l1 −α21 l2 its unit production cost, the optimal order quantity of product 1 is Q∗1 = D1 − 1−α and 12 α21 l2 −α12 l1 , while the worst-case demand the optimal order quantity of product 2 is Q∗2 = D2 − 1−α 12 α21

of products 1 and 2 can be (D1 , D2 − l2 ) or (D1 − l1 , D2 ), respectively. (iv) It is impossible to have the case that P 1 < P 2 α12 , P 2 < P 1 α21 , which implies 1 < α12 α21 , contradicting the assumption that all the substitution rates are between 0 and 1. 4.2. Special Case II: α = 0 In this subsection, we analyze robust multi-product newsvendor problem without substitution, i.e., ˜ is (Q) = D ˜ i = Di − li zi . Thus, R-MNMS (9) α = 0. In this setting, the effective demand becomes D reduces to: v ∗ = maxn

 

Q∈R+ 

f (Q) :=

X

P i Qi − max

X

z∈X

i∈[n]

i∈[n]

S i (Qi − Di + li zi )+

  

,

(12)


11

where set X is defined in (8). We first make the following observation. Lemma 2 There exists an optimal solution Q∗ of Model (12) such that Di − li ≤ Q∗i ≤ Di for all i ∈ [n]. Proof: For notational convenience, let us define Q−i = [Q1 , · · · , Qi−1 , Qi+1 , · · · , Qn ]> to be the vector n−1 of the remaining elements of Q. It is sufficient to show that for any fixed Q−i ∈ R+ , the objective

function of Model (12), f (Qi , Q−i ), is monotone nondecreasing in Qi when Qi ∈ [0, Di − li ] and monotone nonincreasing in Qi when Qi ∈ [Di , +∞). Indeed, we note that f (Qi , Q−i )  X

=

P τ Qτ + P i Qi − max 

S τ (Qτ − Dτ + lτ zτ )+ + S i (Qi − Di + li zi )+ 

z∈X

τ ∈[n]\{i}

τ ∈[n]\{i}

   

P

  

P

P

P τ Qτ − max

τ ∈[n]\{i}

=

 X

z∈X

S τ (Qτ − Dτ + lτ zτ )+ + P i Qi ,

if Qi ∈ [0, Di − li ],

τ ∈[n]\{i}

! P

P τ Qτ − max

τ ∈[n]\{i}

z∈X

S τ (Qτ − Dτ + lτ zτ )+ − Di + li zi

+ P i − S i Qi ,

if Qi ∈ [Di , +∞).

τ ∈[n]\{i}

Clearly, from the above equation, we know that if Qi ∈ [0, Di − li ], the coefficient of Qi is P i , which is nonnegative, while if Qi ∈ [Di , +∞), the coefficient of Qi is P i − S i , which is nonpositive. Thus, f (Qi , Q−i ) is nondecreasing on Qi when Qi ∈ [0, Di − li ] and nonincreasing on Qi when Qi ∈ [Di , +∞). This completes the proof.

According to Lemma 2, without loss of generality, we can assume in Model (12), Q ∈ [D − l, D]. ( 0, if zi = 0 Thus, for each i ∈ [n], (Qi − Di + li zi )+ = = (Qi − Di + li ) zi . Therefore, Qi − Di + li , if zi = 1 Model (12) is equivalent to   X X v ∗ = max  P i Qi − max S i (Qi − Di + li ) zi  , (13) Q∈[D−l,D]

z∈X

i∈[n]

i∈[n]

where X is defined in (8). Suppose that {(1), (2), · · · , (n)} is a permutation of [n] such that S (1) l(1) ≥ S (2) l(2) ≥ · · · ≥ S (n) l(n) . We can obtain a closed-form optimal solution to Model (13) as follows. Theorem 3 When α = 0, the optimal solutions Q∗ of Model (13) are characterized as follows: P P (i) If i∈[n] PS i ≤ k, then Q∗i = Di − li , and v ∗ = i∈[n] P i (Di − li ). i P (ii) If i∈[n] PS i > k, i ( S l Di − li + (t+1)S (t+1) , if i ∈ T ∗ i Qi = , Di , if i ∈ [n] \ T and v∗ =

X i∈[n]\T

P i li − S (t+1) l(t+1) k +

X Pi i∈T

Si

S (t+1) l(t+1) +

X i∈[n]

P i (Di − li ) ,


12

where set T := {(1), (2), · · · , (t)} satisfying

Pi i∈T S i

P

≤ k,

Pi i∈T ∪{(t+1)} S i

P

> k.

Proof: See Appendix A.4. Theorem 3 reveals the impact of the budget of uncertainty on the optimal order quantities. P Indeed, if i∈[n] PS i ≤ k, i.e., the budget of uncertainty k is no smaller than the sum of the critical i

ratios of all the products, then in this case, the optimal order quantity for each product is equal to the lower bound of the demand, i.e., Q∗i = Di − li for each i ∈ [n]. Hence, this implies that when the products are not very profitable or the accuracy of demand forecasting is relatively low, then the decision of the retailer should be conservative to hedge against unnecessary loss from P demand forecasting. Suppose that i∈[n] PS i > k, i.e, the budget of uncertainty is smaller than the i

sum of critical ratios of all the products, or equivalently, relatively a small number of demand can be allowed to deviate from the nominal demand D. Also, note that for each product i ∈ [n], the value of S i li can be interpreted as the risk of lost sales for product i when its order quantity is Di with the worst-case demand Di − li (i.e, the sum of underage cost and overage cost multiplies the demand deviation). In this case, for each product i ∈ T whose risk of lost sales is larger than a threshold S (t+1) l(t+1) , its order quantities should be equal to Di − li +

S (t+1) l(t+1) Si

; otherwise, it

should be Di . The threshold S (t+1) l(t+1) can be determined by searching for the product such that sum of the critical ratios of the products whose risk is higher than product (t + 1) is no larger than the budget of uncertainty k, but including the critical ratio of this product into the sum will be above k. This result tells that the products with lower risk of lost sales should be ordered up to the nominal demand, while those with higher risk should be ordered less than the nominal demand. 4.3. Special Case III: k = n When the budget of uncertainty is equal to n, i.e., k = n, the uncertainty set U becomes     X ˜ :D ˜ i = Di − li zi , U= D zi ≤ n, zi ∈ {0, 1}, ∀i ∈ [n] .   i∈[n]

˜ is nonincreasing in D, ˜ thus at the From Lemma 1, we know that the profit function Π(Q, D) optimality, we must have zi = 1 for all i ∈ [n] in the inner maximization problem (9b), i.e., the worst-case demand in this special case will always be equal to D − l. Thus, Model (9) becomes   X X X P i Qi − S i Qi − Di + li − αji (Dj − lj − Qj )+  . (14) v ∗ = maxn Q∈R+

i∈[n]

i∈[n]

j∈[n]

+

Note that Model (14) is a multi-product newsvendor model with substitution when the demand is deterministic and is equal to D − l. According to the recent work in Zhang et al. (2018), the optimal


13

order quantities of Model (14) can be completely characterized as follows (For more details, please refer to Zhang et al. (2018)). Theorem 4 (Theorem 1, Zhang et al. 2018) When k = n, the optimal order quantities Q∗ and the optimal total profit v ∗ are characterized as follows: (i) Q∗j =

 P Djs (Q∗ ) = Dj − lj + αij (Di − li ), i∈Γ∗

0,

P

if P j −

αji P i ≥ 0

i∈[n]\Γ∗

(15)

otherwise

for each j ∈ [n], where [n] \ supp(Q∗ ) = Γ∗ , i.e., Γ∗ = {i ∈ [n] : Q∗i = 0}; and (ii)     X X X v ∗ = max f (Γ) := P i (Di − li ) := f (Γ∗ ), αji P i (Dj − lj ) + Γ⊆[n]   j∈Γ i∈[n]\Γ

(16)

i∈[n]\Γ

In Theorem 4, if the budget of uncertainty is equal to the number of products, then for each product j ∈ [n], its optimal order quantity Q∗j is equal to its effective demand if its marginal profit P j is larger than or equal to the sum of the profits generated by using other products to substitute it, and 0, otherwise. This suggests that the retailer does not need to order a product if its marginal profit is relatively low and should order up to its effective demand, otherwise. Also, in (16), the first term is the sum of the total profit for selling product i ∈ [n] \ Γ to meet the demand of its substitutable products j ∈ Γ and the second term is the profit of selling product i ∈ [n] \ Γ to meet its own demand. Finally, please note that although we completely characterize the optimal order quantities for all the products, obtaining these value is in general NP-hard (cf., Zhang et al. 2018). Another interesting observation from Theorem 4 is that the optimal order quantity for each product can be equal to their worst-case demand, i.e., Q∗j = Dj − lj for each product j ∈ [n], under the following assumptions. Corollary 1 Suppose (1) P i = P j , ∀i, j ∈ [n] and (2) for each product j ∈ [n],

P

αji < 1. Then

i∈[n]

Q∗j = Dj − lj for all j ∈ [n]. Proof: Note that from Theorem 4, the optimal subset Γ∗ = ∅. Therefore, Q∗j = Djs (Q∗ ) = Dj − lj for all j ∈ [n].

Corollary 1 shows that if all the products share the same underage cost and cannot be completely substituted by the others, then the optimal order quantities are equal to the worst-case demand, i.e., Q∗j = Dj − lj for each product j ∈ [n]. Finally, we remark that if k = 0, then the results in Theorem 4 will also hold simply by replacing li = 0 for each i ∈ [n].


14

5. Solution Approaches Note that the inner maximization Model (9b) is a nonconvex and nonsmooth optimization problem. In this section, we will introduce equivalent MILP formulations for R-MNMS (9) and its inner maximization Model (9b) by linearizing the nonconvex terms in the profit function. These equivalent formulations allow us to develop an effective branch and cut algorithm and an alternative conservative approximation to solve R-MNMS. 5.1. An Equivalent MILP Formulation of the Inner Maximization Problem In this subsection, we will present an MILP formulation, which is equivalent to the inner maximization problem (9b). To begin with, in (9b), let us define two new variables uj = (Dj − lj − Qj )+ ,

ψj = (Dj − Qj )+

for each j ∈ [n]. Clearly, we have ψj ≥ uj for each j ∈ [n]. For simplicity, we still use the function R(Q, u, ψ) to denote the optimal value of inner maximization problem (9b) for any given Q, u, ψ, i.e., the inner maximization problem becomes   X X S i Qi − Di + li zi − αji (uj zj + ψj (1 − zj )) , R(Q, u, ψ) = max z

i∈[n]

s.t.

X

j∈[n]

(17a)

+

zi ≤ k,

(17b)

i∈[n]

zi ∈ {0, 1}, ∀i ∈ [n].

(17c)

Note that Model (17) is a convex integer maximization problem. Thus, we will further linearize the objective function into a linear form. To do so, for each i ∈ [n], let us define a binary variable P xi = 1, if Qi − Di + li zi − j αji (uj zj + ψj (1 − zj )) ≥ 0, and 0, otherwise. Thus, Model (17) is equivalent to  X

R(Q, u, ψ) = max x,z

S i Qi − Di + li zi −

i∈[n]

s.t.

X

 X

αji (uj zj + ψj (1 − zj )) xi ,

(18a)

j∈[n]

zi ≤ k,

(18b)

i∈[n]

xi , zi ∈ {0, 1}, ∀i ∈ [n].

(18c)

The above Model (18) now becomes a binary bilinear program, which can be further linearized by introducing new variables representing the bilinear terms. The final reformulation result is shown below.


15

Proposition 3 The inner maximization problem (17) is equivalent to   X X R(Q, u, ψ) = max αji (uj yji + ψj (xi − yji )) S i (Qi − Di )xi + li yii −

(19a)

x,y,z

i∈[n]

s.t.

X

j∈[n]

zi ≤ k.

(19b)

i∈[n]

yji ≤ xi , ∀i, j ∈ [n],

(19c)

yji ≤ zj , ∀i, j ∈ [n],

(19d)

zi , xi ∈ {0, 1}, yji ≥ 0, ∀i, j ∈ [n].

(19e)

Proof: See Appendix A.5.

5.2. Reformulation of R-MNMS and branch and cut algorithm Next we are going to investigate an MILP reformulation for R-MNMS (9), which is amenable for a branch and cut algorithm. First, from Proposition 2, without loss of generality, we can assume that the order quantities Q can be upper bounded by M . Thus, for each product i ∈ [n], its order quantity Qi must belong to one of the following three intervals: [0, Di − li ], [Di − li , Di ], [Di , Mi ] (0)

(we break the boundary points arbitrarily). For notational convenience, let us denote Di (1)

(2)

= 0,

(3)

Di = Di − li , Di = Di , and Di = Mi . Next, we introduce one binary variable for each interval (e)

(e−1)

to indicate whether Qi is in this interval or not, i.e., we let χi = 1 if Qi ∈ [Di

(e)

, Di ] for each

e ∈ [3]; and 0, otherwise. And we let X

(e)

χi = 1,

(20a)

e∈[3]

to enforce that Qi indeed belongs to only one interval. Correspondingly, for each product i ∈ [n] (e)

and e ∈ [3], we further introduce another variable wi

(e−1)

to be equal to Qi if Qi ∈ [Di

(e)

, Di ], and

0, otherwise. That is, (e−1)

Di X

(e)

(e)

(e)

(e)

χi ≤ wi ≤ Di χi , ∀i ∈ [n], e ∈ [3],

(20b)

(e)

(20c)

wi = Qi , ∀i ∈ [n].

e∈[3]

Next, we can express ui and ψi (recall that ui = (Di − li − Qi )+ and ψi = (Di − Qi )+ ) as linear (e)

(e)

functions of variables {χi }e∈[2] and {wi }e∈[2] for each product i ∈ [n], i.e., (1)

(1)

ui = (Di − li )χi − wi , ∀i ∈ [n], X (e) X (e) ψ i = Di χi − wi , ∀i ∈ [n], e∈[2]

e∈[2]

(20d) (20e)


16

(1)

(1)

Clearly, in (20d), if Qi > Di − li , then ui is equal to 0 since both χi = 0, wi = 0 and otherwise, it (1)

(2)

(1)

is equal to Di − li − Qi . And in (20e), if Qi > Di , then ψi is equal to 0 since χi = χi = 0, wi = (2)

wi = 0, and otherwise, it is equal to Di − Qi . For the inner maximization problem (19), let us also define function g (Q, u, ψ, x, y, z) to be its objective function, i.e.,   X X αji (uj yji + ψj (xi − yji )) , g (Q, u, ψ, x, y, z) = S i (Qi − Di ) xi + li yii − j∈[n]

i∈[n]

and set Ξ to be its feasible region, i.e., Ξ = {(x, y, z) : (19b) − (19e)} . In view of the above development, we have the following equivalent MILP formulation of RMNMS (9): v∗ =

max

X

Q,u,ψ,χ,η

P i Qi − η,

(21a)

i∈[n]

s.t.

η ≥ g (Q, u, ψ, x, y, z) , ∀(x, y, z) ∈ Ξ, (e)

(e)

wi , ui , ψi ≥ 0, χi ∈ {0, 1}, ∀i ∈ [n], e ∈ [3].

(21b) (21c)

(20a) − (20e). Note that in (21b), there can be exponentially many constraints. Therefore, we propose a branch b ⊆ Ξ, which and cut algorithm to solve Model (21). To begin with, suppose we are given a subset Ξ can be empty, then the master problem is formulated as below:    X b (20e), (21b) − (21c) . P i Qi − η : η ≥ g (Q, u, ψ, x, y, z) , ∀(x, y, z) ∈ Ξ, max Q,u,v,χ,η  

(22)

i∈[n]

b ⊆ Ξ. Given an optimal solution Clearly, Model (22) is a relaxation of Model (21), since Ξ b χ, b u b , ψ, b ηb to the master problem (22) to check whether this solution is optimal to original Q,

Model (21) or not, it is sufficient to check whether it satisfies constraints (21b), i.e., solve the inner b as below: b u b , ψ) maximization problem (19) by letting (Q, u, ψ) = (Q, b = max b u b , ψ) R(Q,

(x,y,z)∈Ξ

n o b x, y, z , b u b , ψ, g Q,

b or not. If ηb ≥ R(Q, b then b u b u b , ψ) b , ψ), and check if ηb ≥ R(Q,

(23)

b χ, b u b , ψ, b ηb is optimal to Model Q,

(21). Otherwise, let (b x, yb, zb) be an optimal solution to Model (23). Then add a new constraint b, yb, zb) η ≥ g(Q, u, ψ, x


17

into the master problem (22) and continue. Note that this solution procedure can be integrated with branch and bound, which is known as “branch and cut” (cf. Padberg and Rinaldi 1991, Sen and Sherali 2006). Below, we summarize the proposed branch and cut algorithm to solve Model (21), i.e., at each branch and bound node, we proceed the following solution procedure. b = ∅. Step 0: Initialize set Ξ b χ, b u b , ψ, b ηb . Step 1: Solve the proposed master problem (22) with an optimal solution Q, b b u b , ψ). Step 2: Solve Model (23), denote its optimal solution by (b x, yb, zb) and optimal value R(Q, Step 3: There are two cases: b set Q∗ ← Q, b χ∗ ← χ, b u b u∗ ← u b , ψ), b , ψ ∗ ← ψ, b η ∗ ← ηb, stop and output the Csse 1: If ηb ≥ R(Q, optimal solution (Q∗ , u∗ , ψ ∗ , χ∗ , η ∗ ). b then augment set Ξ b u b=Ξ b ∪ (b b , ψ), Csse 2: If ηb < R(Q, x, yb, zb), and go to Step 1. Note that although this branch and cut algorithm will terminate in a finite number of steps since there are only a finite number of points in set Ξ, as well as finite number of binary variables in the master problem. However, to generate a new constraint at Step 2 might be very time-consuming since it involves solving an MILP (23), i.e., the inner maximization problem (19). In the remaining part of this section, we will replace this MILP (23) by its continuous relaxation and derive a conservative approximation for R-MNMS. 5.3. Conservative Approximation In practice, branch and cut algorithm might not be efficiently to solve very large-scale problem instances. In this section, we propose a simple but very effective conservative approximation to solve R-MNMS (21), i.e., the optimal solution from conservative approximation is a feasible solution to R-MNMS (21). We also provide some sufficient conditions under which this conservative approximation yields an exact optimal solution to R-MNMS (21). To derive the conservative approximation, we simply relax variables (x, y, z) in set Ξ to be continuous in R-MNMS (21), then we can obtain the following lower bound, i.e., a conservative approximation to Model (21):   X  v CA = max P i Qi − η : η ≥ g (Q, u, ψ, x, y, z) , ∀(x, y, z) ∈ ΞC , (20a) − (20e), (21c) , Q,u,v,χ,η   i∈[n]

(24) where ΞC denotes the continuous relaxation of set X. Note that the constraints η ≥ g(Q, u, ψ, x, y, z), ∀(x, y, z) ∈ ΞC is equivalent to η≥

max (x,y,z)∈ΞC

{g (Q, u, ψ, x, y, z)} ,


18

where the right-hand side is a linear program with nonempty and bounded feasible region for any given (Q, u, ψ). Therefore, according to the strong duality of linear program, we can replace the max operator by its dual, i.e., an equivalent min operator, and further change the min operator with the existence one. Let $, σ, ρ, ζ, ξ be the dual variables associated with constraints (19b),(19c), (19d), z ≤ e and x ≤ e, respectively. Then the conservative approximation (24) is equivalent to the following MILP: v CA =

X

max

Q,u,v,η,$,σ,ρ,ζ,ξ

P i Qi − η,

(25a)

X

(25b)

i∈[n]

s.t. η ≥ k$ +

(ζi + ξi ),

i∈[n]

$ + ζj −

X

ρji ≥ 0, ∀j ∈ [n],

(25c)

σji + ρji ≥ −S i αji (uj − ψj ), ∀i, j ∈ [n], i 6= j,

(25d)

σjj + ρjj ≥ S j lj , ∀j ∈ [n], X X αji ψj , ∀i ∈ [n], ξi − σji ≥ S i (Qi − Di ) − S i

(25e)

i∈[n]

j∈[n]

(25f)

j∈[n]

ζi , ξi , σij , ρij ≥ 0, ∀i, j ∈ [n],

(25g)

(20a) − (20e), (21b) − (21c). The following result summarizes the above development of the conservative approximation and also shows that under some sufficient conditions, this approximation can be exact, i.e., v CA = v ∗ . Theorem 5 Let v CA denote the optimal value of Model (25). Then (i) v CA ≤ v ∗ ; and (ii) v CA = v ∗ , if one of the following conditions holds: (1) α = 0, or (2) n = k. Proof: See Appendix A.6.

From Theorem 5, we see that the conservative approximation (25) provides a feasible solution to R-MNMS (21). In addition, Theorem 5 tells that the conservative approximation can find a very good-quality solution, which can even be optimal to R-MNMS (21). We will illustrate these facts in Section 6.

6. Computational Study In this section, we test the performances of branch and cut algorithm and conservative approximation to solve R-MNMS (21). We considered instances with n = 10 and n = 20 products. For each n ∈ {10, 20}, we generated 10 random instances, where for each product i ∈ [n], the nominal demand Di is between 5 and 100, the


19

unit price pi ranged from 85 to 95, unit cost ci varied from 40 to 50, and the salvage value si was between 22 and 30. All the products were assumed to be similar, and thus, the substitution rates P were generated uniformly between 0 and 1, satisfying i∈[n] αji = 0.8 and αjj = 0 for each j ∈ [n]. The lower bound of the demand was set to be proportional to the nominal demand, i.e., l = θD, where θ ∈ (0, 1) is called “deviation ratio”. We tested these instances with the deviation ratio θ ∈ {0.2, 0.4} and the budget of uncertainty k ∈ {5, 10}. Both approaches were coded in Python 2.7 with calls to Gurobi 7.5 on a personal computer with 2.3 GHz Intel Core i5 processor and 8G of memory. The CPU time limit of Gurobi was set to be 3600 seconds. Table 1 and Table 2 display the computational results of branch and cut algorithm and conservative approximation method with n = 10 and n = 20, respectively. For the branch and cut algorithm, Opt.val denotes the optimal value if available, LB and UB denote best lower and upper bounds, Gap denotes the optimality gap, computed as (UB-LB)/UB; for conservative approximation, C.val denotes its output objective value and A-Gap represents its optimality gap, computed as (UB-C.val)/UB (note that UB is equal to the Opt.val if available). From Table 1, we see that when n = 10, both approaches finds good-quality feasible solutions. As explained in Section 5.3, the solutions obtained from conservative approximation method are equal to the solutions calculated from branch and cut algorithm when k = n = 10, i.e., A-Gap=0. In Table 2, we note that when the number of products increases to 20, the conservative approximation method can find good-quality feasible solutions within the time limit. Oftentimes, the conservative approximation solution can be even better than that obtained by the branch and cut algorithm. Also from Table 2, we see that if the budget of uncertainty k increases and k ≤ n2 , i.e., the number of possible realizations of products’ demand grows, then the computational time tends to be longer. Also, since a larger k implies that a larger number of products whose worst-case demand can be equal to their lower bounds, therefore we can anticipate that the total profit becomes smaller. Since θ denotes how much the worst-case demand can deviate from the nominal demand, the increase of θ implies that the variance of random demand grows, which means a chance of being understock or overstock becomes larger, and leads to a smaller total profit. We also tested the reliability of the solution from robust Model (21) versus the risk neutral one studied in Zhang et al. (2018), which has the following form:    X  X ˜ is (Q)  , v rn = maxn P i Qi − EP  S i Qi − D Q∈R+  +  i∈[n]

(26)

i∈[n]

where P denotes a particular probability distribution. For the comparison purpose, we considered n = 2 products, where the nominal demand of each product is 50 and the maximum possible deviation of the random demand is 30, i.e., the random


20 Table 1

Computational results of branch and cut algorithm and conservative approximation method with n = 10

k

5

5

10

10

Branch and Cut θ Instances Time Opt.val 7.41 29366.4 1 2 1.45 30903.2 3 7.39 27824.7 1.97 27509.3 4 5 8.62 33482.8 0.2 1.10 30826.7 6 7 7.69 30652.2 1.23 29820.5 8 9 7.81 31935.9 10 1.20 33228.9 Average 4.59 30555.1 7.57 24225.8 1 1.24 25434.4 2 3 10.16 22961.9 1.16 22841.6 4 5 5.85 27626.2 0.4 6 1.32 25459.5 7 7.96 25480.5 8 1.15 24566.1 9 13.32 26447.3 10 2.06 27463.7 Average 5.18 25250.7 1 10.29 27605.6 1.44 29097.6 2 3 6.67 26152.8 1.39 25741.6 4 5 9.76 31471.2 0.2 1.28 28955.2 6 7 11.58 28659.2 8 1.14 28060.0 9 9.13 29938.4 10 1.13 31195.2 Average 5.38 28687.7 1 6.62 20704.2 2 1.23 21823.2 3 7.75 19614.6 4 1.31 19306.2 5 7.84 23603.4 0.4 6 1.26 21716.4 7 10.12 21494.4 8 1.66 21045.0 9 7.91 22453.8 10 1.49 23396.4 Average 4.72 21515.8

Conservative Approximation Time C.val A-Gap (%) 0.54 29246.3 0.41 0.48 30738.8 0.53 0.58 27742.2 0.30 0.45 27431.1 0.28 0.52 33362.9 0.36 0.36 30730.1 0.31 0.42 30614.1 0.12 0.35 29651.1 0.57 0.46 31876.4 0.19 0.33 33163.8 0.20 0.45 30455.7 0.33 0.45 23985.6 0.99 0.44 25105.6 1.29 0.45 22793.5 0.74 0.39 22685.1 0.69 0.35 27386.8 0.87 0.43 25266.2 0.76 0.33 25404.2 0.30 0.31 24227.1 1.38 0.33 26329.8 0.44 0.36 27333.6 0.47 0.38 25051.8 0.79 0.45 27605.6 0 0.37 29097.6 0 0.32 26152.8 0 0.34 25741.6 0 0.46 31471.2 0 0.30 28955.2 0 0.34 28659.2 0 0.35 28060.0 0 0.40 29938.4 0 0.33 31195.2 0 0.37 28687.7 0 0.33 20704.2 0 0.41 21823.2 0 0.34 19614.6 0 0.39 19306.2 0 0.49 23603.4 0 0.34 21716.4 0 0.46 21494.4 0 0.38 21045.0 0 0.29 22453.8 0 0.35 23396.4 0 0.38 21515.8 0


Computational results of branch and cut algorithm and conservative approximation method

Table 2

with n = 20

k

5

5

10

10

21

θ

Instances

1 2 3 4 5 0.2 6 7 8 9 10 Average 1 2 3 4 5 0.4 6 7 8 9 10 Average 1 2 3 4 5 0.2 6 7 8 9 10 Average 1 2 3 4 5 0.4 6 7 8 9 10 Average

Time 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600 3600

Branch and Cut Conservative Approximation LB UB Gap (%) Time C.val A-Gap (%) 62696.6 64708.6 3.21 426 63598.0 1.72 57808.9 59814.1 3.47 376 58676.9 1.90 67405.0 69220.8 2.69 386 68271.3 1.37 63595.0 65665.0 3.26 474 64289.4 2.10 68212.6 70193.6 2.90 483 68970.1 1.74 63162.1 65226.0 3.27 389 64158.1 1.64 57161.3 58977.5 3.18 443 57873.8 1.87 59865.9 61533.3 2.79 399 60482.0 1.71 63039.5 64734.2 2.69 297 64015.2 1.11 63736.1 65823.4 3.28 445 64440.0 2.10 62668.3 64589.6 2.98 412 63477.5 1.73 56885.9 59198.5 4.07 209 57605.3 2.69 52545.4 55277.5 5.20 164 53302.7 3.57 61266.0 63481.5 3.62 210 62016.9 2.31 57894.0 60005.2 3.65 171 58491.9 2.52 62189.9 64349.5 3.47 214 62787.7 2.43 57600.8 59577.6 3.43 162 58411.6 1.96 51958.4 53876.0 3.69 188 52501.2 2.55 54290.0 56740.3 4.51 166 54681.1 3.63 57439.3 60157.6 4.73 162 58052.8 3.50 57968.5 60021.7 3.54 193 58443.9 2.63 57003.8 59268.5 4.54 184 57629.5 3.96 58550.7 61391.8 4.85 677 58760.5 4.29 53961.4 55228.0 2.35 676 54308.9 1.66 62883.8 66951.2 6.47 674 62988.8 5.92 59129.6 62307.3 5.37 744 59423.0 4.63 63215.9 65995.8 5.37 775 63774.9 3.37 58661.4 61816.1 5.38 588 59310.5 4.05 53268.3 56034.5 4.40 679 53631.6 4.29 55993.3 58219.7 3.98 584 56143.3 3.57 58682.7 61631.3 5.03 519 59298.9 3.79 59474.8 62217.1 4.61 667 59677.8 4.08 58382.2 61179.3 3.84 658 58731.8 2.78 48742.5 50596.0 3.80 192 48090.8 4.95 44928.1 46423.0 3.33 132 44626.2 3.87 52462.0 54713.1 4.29 162 51612.7 5.67 49276.9 51014.4 3.53 149 48758.0 4.42 52576.1 55947.0 6.41 137 52511.7 6.14 48995.8 50832.1 3.75 113 48792.0 4.01 44435.2 45413.2 2.20 133 44066.6 2.97 46552.2 47417.1 1.86 141 45977.6 3.04 49148.4 50407.4 2.56 117 48861.3 3.07 49663.0 50784.3 2.26 183 48956.5 3.60 48678.0 50354.8 3.27 146 48225.3 4.20


22

˜ i ∈ [20, 80] for each i ∈ [2]. α12 = 0.48 and α21 = 0.42. In the risk neutral model, we assume demand D that the random demand for each product is independently uniformly distributed between 20 and 80. Under this setting, we can solve both robust model and sample average approximation (SAA) with 1000 samples of risk neutral model to the optimality, where we let QR and QN denote their corresponding optimal solutions, respectively. To evaluate the quality of both solutions, we plug in them into the risk neutral model (26) under different “true” probability distributions, i.e., Uniform, Laplace, Gaussian, and Logistic, where we generated 1000 i.i.d. samples under each probability distribution. Note that for the Laplace, Gaussian, and Logistic distributions, we assumed that their means are equal to 25 and standard deviations are 4, and for each sample D, we truncated it into the range of 20 and 80 using the function D = max{min{D, 80}, 20}, i.e., we do not allow the demand to be outside of the interval [20, 80]. For the uniform distribution, we assume the true demand is uniformly distributed in the range between 20 and 30. The computational results are presented in Table 3. From Table 3, we see that if the underlying probability distribution is not the same as the one we predict, then the total profit from the solution of robust Model (21) can be much higher (i.e., about 20%) than that of the risk neutral model. Therefore, this demonstrates that Model (9a) can be indeed more reliable and robust than the risk neutral one. Table 3

The objective value of Model (9a) and risk neutral model under different distributions

Distribution Value of robust solution Value of risk neutral solution Percent (%) of differences Gaussian(25,16) 1649.50 1285.32 22.1 Laplace(25,16) 1733.84 1322.40 23.7 1791.80 1366.28 23.7 Logistic(25, 16) Uniform[20,30] 2020.49 1639.30 18.9

7. Conclusion This paper studies the robust multi-product newsvendor problem with substitution (R-MNMS) under cardinality-constrained uncertainty set. We first prove that evaluating the worst-case total profit for given order quantities in general is NP-hard. Next, we identify three solvable special cases of R-MNMS and derive their closed-form optimal solutions. For a general R-MNMS, we propose a mixed integer linear program formulation which can be solved by a branch and cut algorithm. We also develop a conservative approximation method to solve R-MNMS, and show that under certain conditions, its optimal solution can also be optimal to R-MNMS. Finally, we conduct numerical studies to illustrate the effectiveness and solution quality of the proposed algorithms. One possible future direction is to incorporate pricing decision into R-MNMS, i.e., to study joint inventory and pricing optimization in R-MNMS.


23

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26

Appendix A. Proofs A.1

Proof of Proposition 2

Proposition 2 There exists an optimal solution Q∗ to R-MNMS such that for each product i ∈ [n], P Q∗i ≤ Mi , where Mi = Di + j∈[n] αji Dj . Proof: We prove the result by contradiction. Suppose for any optimal solution Q∗ , there exists a product i ∈ [n] such that Q∗i > ( Mi . Let set B := {i ∈ [n] : Q∗i > Mi }. Hence, B 6= ∅. Let us define a b such that Q b i = Mi , if i ∈ B for each product i ∈ [n]. Clearly, Q b i ≤ Mi for each new solution Q Q∗i , otherwise b is equal to i ∈ [n]. Then the objective value f (Q) b = f (Q)

X

X

P i Mi +

i∈B

P i Q∗i

i∈[n]\B

− max z∈X

 X 





S i Mi − Di + li zi −

X

i∈B

X

αji (Dj − lj zj − Mj )+ −

j∈B

X

S i Q∗i − Di + li zi −

=

X

P i Mi +

i∈B

P i Q∗i

z∈X

 X 

+

X

+

 X


i∈B

=


z∈X

 X

+

αji

 Dj − lj zj − Q∗j + 

Si − P i




(Q∗i

αji

   Dj − lj zj − Q∗j +   +



i∈[n]

i∈[n]

+

X j∈[n]\B

P i Q∗i − max

X

αji

+

   Dj − lj zj − Q∗j +  

j∈[n]\B

i∈[n]\B

X



− Mi )

 X

X j∈[n]\B

S i (Q∗i

+

i∈B

i∈[n]\B

− max

αji (Dj − lj zj − Mj )+ −

j∈B

i∈[n]\B

X

X

j∈[n]\B

 +

αji Dj − lj zj − Q∗j

X

αji

j∈[n]

   Dj − lj zj − Q∗j +   +

− Mi )

i∈B

=v ∗ +

X

S i − P i (Q∗i − Mi )

i∈B ∗

≥v ,

where the second equality is because of Mj = Dj +

P

i∈[n] αji Di

for all j ∈ B, the third equality

is because for each j ∈ B, we have Q∗j > Mj = Dj +

P

i∈[n] αij Di ,

the forth equality is due to the

optimality of Q∗ , and the last inequality holds because S i ≥ P i and Q∗i > Mi for each i ∈ [n]. This b is also an optimal solution, a contradiction. implies Q


A.2

27

Proof of Theorem 1

Theorem 1 The inner maximization problem (10) in general is NP-hard. Proof: We prove this result from a reduction of clique problem to be a special case of Model (10). (Clique Problem) Given an undirected graph G(V, E), does it have a size-τ clique? Let us consider a special instance of the inner maximization problem (10): suppose that there are n = |V | + |E | products, and for each product i ∈ E, we let S i = 1, Qi = Di , li = 1, while for each product j ∈ V , we let S j = 1, Qj = Dj − lj , lj = 1. Additionally, the substitution rate matrix α is defined as

( 1, αji = 0,

if edge i ∈ E contains node j ∈ V otherwise τ (τ +1) . 2

for all i, j ∈ V ∪ E. Let the budget of uncertainty k =

Under this setting, the inner maxi-

mization problem (10) reduces to R (Q) = max

X

z

(E) zi

−

i∈E

s.t.

X

X

αji

(V ) 1 − zj

!

j∈V (V ) zj

+

j∈V

X

(27a)

+

(E) zi

i∈E

(V ) (E) zj , z i

,

τ (τ + 1) , ≤ 2

(27b)

∈ {0, 1}.

(27c)

It is sufficient to show that the Clique Problem is equivalent to Model (27), i.e., we only need to show the following claim. Claim 1 There is a clique with τ nodes in the undirected graph G(V, E) if and only if R (Q) = τ (τ −1) . 2

Proof: Before we prove the result, let us denote z ∗ as an optimal solution of Model (27), and also (V ) (E) b ∗, E∗) define the following two sets: V ∗ = {j ∈ V : (zj )∗ = 1}, E ∗ = {i ∈ E : (zi )∗ = 1}, i.e., G(V b ∗ , E ∗ ) might not be a graph since we might not choose is a substructure of G(V, E). Note that G(V enough nodes to cover all the edges, i.e., there might exist an edge in set E ∗ but not both of its two nodes are selected in set V ∗ . Thus, R (Q) is equal to R (Q) =

X i∈E ∗

1−

X

(V )

αji 1 − (zj )∗

j∈V



! = +

X

1 −

i∈E ∗

 X j∈V \V ∗

αji  . +

From (28), we have the following inequality:     ∗ X X X X |V |     R (Q) = 1− αji ≤ 1− αji ≤ , 2 ∗ ∗ i∈E i∈E ∗ j∈V ∈V /

+

j∈V ∈V /

+

(28)

(29)


28

where the first inequality is due to E ∗ ⊆ E, and the second inequality is because of P = 0 if at least one of the two nodes from edge i is not covered by set V ∗ . α 1 − j∈V ∈V ∗ ji / +

Now we are ready to prove the main results. “only if ”. Suppose that there exists a size-τ clique (Vτ , Eτ ) in the graph G(V, E). Let us denote a binary vector zb as ( 1, (V ) zbj = 0,

j ∈ Vτ , ∀i ∈ V, otherwise

( 1, (E) zbi = 0,

i ∈ Eτ , ∀j ∈ E. otherwise

Clearly, zb is a feasible solution to Model (27), with an objective value equal to R (Q) ≥

τ (τ −1) . 2

Thus,

τ (τ −1) . 2

Now suppose that R (Q) >

τ (τ −1) . 2

|E ∗ | =

According to the objective function (27a), we must have

X

(E)

(zi )∗ ≥ R (Q) >

i∈E

τ (τ − 1) . 2

Also, the constraint (27b) implies that |V ∗ | + |E ∗ | ≤

τ (τ + 1) , 2

hence, we must have |V ∗ | < τ , thus |V ∗ | ≤ τ − 1. Thus, in the graph G(V, E), the subset of τ (τ −1) ∗ nodes V ∗ can cover at most |V2 | ≤ τ −1 < 2 − 1 < R(Q) edges. Hence, according to (29), 2 we have R (Q) ≤

|V ∗ | < R(Q), 2

a contradiction. τ (τ −1) . 2

According to (28), we must have   X X τ (τ − 1) 1 − αji  ≤ |E ∗ | = R (Q) = 2 ∗ i∈E ∗

“if ”. Suppose that R (Q) =

j∈V ∈V /

and by (29), we also have

i.e., |V ∗ | ≥ τ, |E ∗ | ≥

+

∗ τ (τ − 1) |V | = R (Q) ≤ , 2 2

τ (τ −1) . 2

On the other hand, the constraint (27b) implies that V ∗ | + |E ∗ | ≤ τ (τ +1) b ∗, E∗) . Thus, we must have |V ∗ | = τ, |E ∗ | = τ (τ2−1) . Suppose that the substructure G(V 2

is not a clique, then there exists i0 = (u0 , v0 ) ∈ E ∗ such that at least one of its nodes is not P = 0. Thus, by (28), we have chosen, i.e., 1 − j∈V ∈V / ∗ αji0 +

τ (τ − 1) = R (Q) = 2



 X i∈E ∗ \{i0 }

1 −

X j∈V ∈V / ∗

αji  ≤ |E ∗ | − 1
max min(c ) x : 0 ≤ x ≤ U . x

i∈[τ ]

Then an optimal solution of the above optimization problem can be one of the following points: (i) the extreme points of the box [0, U ]; and (ii) the intersection point of any affine system (ci )> x = (cj )> x for all i, j ∈ B ⊆ [τ ] with 2 ≤ |B| ≤ n and the boundary of the box [0, U ]. (iii) the unique solution of the affine system (ci )> x = (cj )> x for all i, j ∈ B with |B| = n + 1, which is in the box [0, U ]. Proof: Note that the piecewise concave optimization program can be written as the following linear program: max w : w ≤ (ci )> x, ∀i ∈ [τ ], 0 ≤ x ≤ U . x

The conclusion follows by the fact that one optimal solution of the above linear program must be an extreme point, and condition (i), (ii), and (iii) exactly characterize all the extreme points.

Figure 1

Decomposition

of

feasible

solution

regions

into

9

sub-regions,

where

Ω1

=

{(Q1 , Q2 ) : 0 ≤ Q1 ≤ D1 − l1 , 0 ≤ Q2 ≤ D2 − l2 }, Ω2 = {(Q1 , Q2 ) : 0 ≤ Q1 ≤ D1 − l1 , D2 − l2 ≤ Q2 ≤ D2 }, Ω3 = {(Q1 , Q2 ) : 0 ≤ Q1 ≤ D1 − l1 , D2 ≤ Q2 }, Ω4 = {(Q1 , Q2 ) : D1 − l1 ≤ Q1 ≤ D1 , D2 − l2 ≤ Q2 ≤ D2 }, Ω5

= {(Q1 , Q2 ) : D1 − l1 ≤ Q1 ≤ D1 , D2 ≤ Q2 },

{(Q1 , Q2 ) : D1 − l1 ≤ Q1 ≤ D1 , 0 ≤ Q2 ≤ D2 − l2 },

Ω6 Ω30

= {(Q1 , Q2 ) : D1 ≤ Q1 , D2 ≤ Q2 }, =

=

{(Q1 , Q2 ) : D1 ≤ Q1 , 0 ≤ Q2 ≤ D2 − l2 },

Ω50 = {(Q1 , Q2 ) : D1 ≤ Q1 , D2 − l2 ≤ Q2 ≤ D2 }.

Now we are ready to discuss the following 9 cases. Case 1. Suppose (Q1 , Q2 ) ∈ Ω1 , i.e., 0 ≤ Q1 ≤ D1 − l1 , 0 ≤ Q2 ≤ D2 − l2 . In this case, R(Q) = 0 and Model (30) becomes: max f (Q) = P 1 Q1 + P 2 Q2 : 0 ≤ Q1 ≤ D1 − l1 , 0 ≤ Q2 ≤ D2 − l2 . Q

Ω20


31

Clearly, the optimal solution of the above linear program is t11 = (D1 − l1 , D2 − l2 ), and its optimal total profit is f (t11 ) = P 1 (D1 − l1 ) + P 2 (D2 − l2 ). Case 2. Suppose (Q1 , Q2 ) ∈ Ω2 , i.e., 0 ≤ Q1 ≤ D1 − l1 , D2 − l2 ≤ Q2 ≤ D2 .

Figure 2

Illustration of possible solutions of Case 2.

In this case, we have R(Q) = S 2 (Q2 − D2 + l2 − α12 (D1 − Q1 ))+ and Model (30) becomes: max f (Q) = P 1 Q1 + P 2 Q2 − S 2 (Q2 − D2 + l2 − α12 (D1 − Q1 ))+ : 0 ≤ Q1 ≤ D1 − l1 , D2 − l2 ≤ Q2 ≤ D2 . Q

According to Observation 1, the optimal solution of above optimization problem can be one of the following points: (1) extreme points of Ω2 i.e., t12 , t22 , t32 , t42 ; and (2) the intersection points of linear equation Q2 − D2 + l2 − α12 (D1 − Q1 ) = 0 with feasible regions, i.e., t52 , t62 (see Figure 2 for an illustration). These potential optimal solutions and their corresponding total profits are listed in Table 4. Table 4

The possible solutions and their total profits in Case 2

Possible solutions Total profit 1 1 t2 = (0, D2 − l2 ) f (t2 ) = P 2 (D2 − l2 ) t22 = (0, D2 ) f (t22 ) = P 2 D2 t32 = (D1 − l1 , D2 ) f (t32 ) = P 1 (D1 − l1 ) + P 2 D2 − S 2 (l2 − α12 l1 ) 1 4 t1 = t2 = (D1 − l1 , D2 − l2 ) f (t42 ) = f (t11 ) = P 1 (D1 − l1 ) + P 2 (D2 − l2 ) t52 = (D1 − l2 /α12 , D2 ) f (t52 ) = P 1 (D1 − l2 /α12 ) + P 2 D2 6 6 t2 = (D1 − l1 , D2 − l2 + α12 l1 ) f (t2 ) = P 1 (D1 − l1 ) + P 2 (D2 − l2 + α12 l1 )

It remains to compare these solutions. Clearly, we have • f (t22 ) ≥ f (t12 ), since P 2 , l2 ≥ 0, • f (t52 ) − f (t22 ) = P 1 (D1 − l2 /α12 ) ≥ 0 due to Assumption 1 that α12 D1 − l2 ≥ 0,


32

• f (t62 ) − f (t32 ) = P 2 (−l2 + α12 l1 ) + S 2 (l2 − α12 l1 ) = (−P 2 + S 2 )(l2 − α12 l1 ) ≥ 0 due to S 2 ≥ P 2

and Assumption 1 that l2 − α12 l1 ≥ 0, • f (t62 ) − f (t42 ) = P 2 α12 l1 ≥ 0, and 2 • f (t62 ) − f (t52 ) = −P 1 l1 + P 2 (−l2 + α12 l1 ) + P 1 αl12 = α112 (P 1 − P 2 α12 )(l2 − ( ≥ 0, if P 1 ≥ P 2 α12 α12 l1 ) , due to Assumption 1 that l2 − α12 l1 ≥ 0. ≤ 0, if P 1 ≤ P 2 α12 From the above comparison, we can draw the following conclusion on the best solution in the

subregions Ω1 and Ω2 : (i) If P 1 − P 2 α12 ≤ 0, then the point t52 dominates the other points in the subregions Ω1 and Ω2 , since f (t52 ) ≥ f (t22 ) ≥ f (t12 ), f (t52 ) ≥ f (t62 ) ≥ f (t42 ), and f (t52 ) ≥ f (t62 ) ≥ f (t32 ). (ii) If P 1 − P 2 α12 ≥ 0, then the point t62 dominates the other points in the subregions Ω1 and Ω2 , since f (t62 ) ≥ f (t52 ) ≥ f (t22 ) ≥ f (t12 ), f (t62 ) ≥ f (t42 ), and f (t62 ) ≥ f (t32 ). Case 3. Suppose (Q1 , Q2 ) ∈ Ω3 , i.e., 0 ≤ Q1 ≤ D1 − l1 , D2 ≤ Q2 . In this case, R(Q) = S 2 (Q2 − D2 + l2 − α12 (D1 − Q1 ))+ and Model (30) becomes:

max f (Q) = P 1 Q1 + P 2 Q2 − S 2 (Q2 − D2 + l2 − α12 (D1 − Q1 ))+ : 0 ≤ Q1 ≤ D1 − l1 , D2 ≤ Q2 . Q

According to Observation 1, the optimal solution can be one of the following points: (1) the extreme points in Ω3 , i.e., t13 , t23 ; and (2) the intersection points of linear Q2 − D2 + l2 − α12 (D1 − Q1 ) = 0 and the boundary of Ω3 , i.e., t33 , t43 (see Figure 3 for an illustration). These solutions and their corresponding total profits are listed in Table 5.

Figure 3

Possible solutions in Case 3


Table 5

33

Possible solutions and their total profits in Case 3

Possible solutions Total profit t13 = t22 = (0, D2 ) f (t13 ) = f (t22 ) = P 2 D2 t23 = t32 = (D1 − l1 , D2 ) f (t23 ) = f (t32 ) = P 1 (D1 − l1 ) + P 2 D2 − S 2 (l2 − α12 l1 ) 3 t3 = (0, D2 − l2 + α12 D1 ) f (t33 ) = P 2 (D2 − l2 + α12 D1 ) t43 = t52 = (D1 − l2 /α12 , D2 ) f (t43 ) = f (t52 ) = P 1 (D1 − l2 /α12 ) + P 2 D2

In view of the results in Case 1 and Case 2, it remains to compare f (t52 ), f (t33 ) and also f (t62 ), f (t33 ). Clearly, we have • f (t52 ) − f (t33 ) = P 1 (D1 − l2 /α12 ) − P 2 (−l2 + α12 D1 ) = ( ≥ 0, if P 1 ≥ P 2 α12 l2 ) , due to Assumption 1 that α12 D1 ≥ l2 , ≤ 0, if P 1 ≤ P 2 α12

1 (P 1 α12

− P 2 α12 )(α12 D1 −

( ≥ 0, if P 1 ≥ P 2 α12 • f (t62 ) − f (t33 ) = P 1 (D1 − l1 ) + P 2 α12 (l1 − D1 ) = (P 1 − P 2 α12 )(D1 − l1 ) ≤ 0, if P 1 ≤ P 2 α12 since Dl ≥ l1 , From the above comparison as well as the results of Case 1 and Case 2, we can draw the following conclusion on the best solution in subregions Ω1 , Ω2 and Ω3 : (i) If P 1 ≥ P 2 α12 , then t62 dominates all the other points in Ω1 , Ω2 , and Ω3 since f (t62 ) ≥ f (t52 ) ≥ f (t33 ). (ii) If P 1 ≤ P 2 α12 , then t33 dominates all the other points in Ω1 , Ω2 , and Ω3 since f (t62 ) ≤ f (t52 ) ≤ f (t33 ). Case 4. Suppose (Q1 , Q2 ) ∈ Ω4 , i.e., D1 − l1 ≤ Q1 ≤ D1 , D2 − l2 ≤ Q2 ≤ D2 . In this case, R(Q) = max S 1 (Q1 − D1 + l1 − α21 (D2 − Q2 ))+ , S 2 (Q2 − D2 + l2 − α12 (D1 − Q1 ))+ and Model (30) becomes:

max f (Q) = P 1 Q1 + P 2 Q2 − R(Q) : D1 − l1 ≤ Q1 ≤ D1 , D2 − l2 ≤ Q2 ≤ D2 . Q

According to Observation 1, the optimal solution can be one of the following points: (1) the extreme points of Ω4 , i.e., t14 , t24 , t34 , t44 ; (2) the intersection points of the line Q1 − D1 + l1 − α11 (D2 − Q2 ) = 0 with the boundary of Ω4 , i.e., t24 , t54 ; (3) the intersection points of the line Q2 − D2 + l2 − α12 (D1 − Q1 ) = 0 with the boundary of Ω4 , i.e., t44 , t64 ; and (4) the intersection point of two lines S 1 (Q1 − D1 + l1 − α21 (D2 − Q2 ))+ = 0 and S 2 (Q2 − D2 + l2 − α12 (D1 − Q1 ))+ = 0, i.e., t74 (see Figure 4 for an illustration). These solutions and their total profits are listed in Table 6.


34

Figure 4


Table 6


Possible solutions Total profit 1 4 1 1 4 t4 = t2 = t1 = (D1 − l1 , D2 − l2 ) f (t4 ) = f (t2 ) = f (t11 ) = P 1 (D1 − l1 ) + P 2 (D2 − l2 ) 2 2 3 2 t4 = t3 = t2 = (D1 − l1 , D2 ) f (t4 ) = f (t23 ) = f (t32 ) = P 1 (D1 − l1 ) + P2 D2 − S 2 (l 2 − α12 l1 ) 3 3 t4 = (D1 , D2 ) f (t4 ) = P 1 D1 + P 2 D2 − max S 1 l1 , S 2 l2 t44 = (D1 , D2 − l2 ) f (t44 ) = P 1 D1 + P 2 (D2 − l2 ) − S 1 (l1 − α21 l2 ) t54 = (D1 − l1 + α21 l2 , D2 − l2 ) f (t54 ) = P 1 (D1 − l1 + α21 l2 ) + P 2 (D2 − l2 ) 6 6 6 6 t4 =t2 = (D1 − l1 , D2 − l2 + α12 l1 ) f (t4 ) = f (t ) + P 2 (D 2 ) = P 1 (D1 − l1 2 − l2 + α12 l1)

l1 −α21 l2 l2 −α12 l1 t74 = D1 − 1−α , D2 − 1−α 12 α21 12 α21

l1 −α21 l2 l2 −α12 l1 f (t74 ) = P 1 D1 − 1−α + P 2 D2 − 1−α 12 α21 12 α21

In view of the results in Case 1- Case 3, we know that point t62 or t33 dominates all the other points of Ω1 , Ω2 , and Ω3 , so we only need to compare f (t62 ), f (t33 ), f (t34 ), f (t44 ), f (t54 ), f (t74 ). • f (t44 ) − f (t54 ) = (P 1 − S 1 )(l1 − α21 l2 ) ≤ 0 due to Assumption 1 that l1 − α21 l2 ≥ 0 and the fact

that P 1 ≤ S 1 , • If P 1 ≤ P 2 α12 , then

f (t34 ) − f (t33 ) = P 1 D1 − max{S 1 l1 , S 2 l2 } + P 2 l2 − P 2 α12 D1 = (P 1 − P 2 α12 )D1 + (P 2 l2 − max{S 1 l1 , S 2 l2 }) ≤ 0, where the inequality is because of D1 ≥ 0 and P 2 l2 ≤ S 2 l2 ≤ max{S 1 l1 , S 2 l2 }. Otherwise, f (t34 ) and f (t33 ) are incomparable. • Compare f (t62 ) with f (t74 )

l1 − α21 l2 l2 − α12 l1 f (t62 ) − f (t74 ) = P 2 (−l2 + α12 l1 ) + P 1 (−l1 ) + P 1 +P2 1 − α α 1 − α12 α21 12 21 ( (P 1 − P 2 α12 )(l2 − l1 α12 ) ≤ 0, if P 1 ≥ P 2 α12 = −α21 , 1 − α21 α21 ≥ 0, if P 1 ≤ P 2 α12 where the inequalities are due to Assumption 1 that l2 ≥ α12 l1 and the fact that 0 ≤ α21 ≤ 1, 0 ≤ α12 ≤ 1.


35

• If P 1 ≤ P 2 α12 , then

f (t54 ) − f (t33 ) = P 1 (D1 − l1 + α21 l2 ) − P 2 α12 D1 = (P 1 − P 2 α12 )D1 − P 1 (l1 − α21 l2 ) ≤ 0, where the inequality is due to Assumption 1 that l1 ≥ α21 l2 and the fact that P 1 ≥ 0, D1 ≥ 0. • Compare f (t62 ) with f (t33 )

f (t62 ) − f (t33 ) = P 1 (D1 − l1 ) + P 2 α12 l1 − P 2 α12 D1 = (P 1 − P 2 α12 )(D1 − l1 ) ( ≥ 0, if P 1 ≥ P 2 α12 , ≤ 0, if P 1 ≤ P 2 α12 where the inequalities are due to the fact that D1 ≥ l1 . • Compare f (t54 ) with f (t74 )

l1 − α21 l2 l2 − α12 l1 f (t54 ) − f (t74 ) = P 1 (−l1 + α21 l2 ) + P 2 (−l2 ) + P 1 +P2 1 − α12 α21 (1 − α12 α21 (P 2 − P 1 α21 )(l1 − l2 α21 ) ≤ 0, if P 2 ≥ P 1 α21 = −α12 , 1 − α12 α21 ≥ 0, if P 2 ≤ P 1 α21 where the inequalities are due to Assumption 1 that l1 ≥ α21 l2 and the fact that 0 ≤ α21 ≤ 1, 0 ≤ α12 ≤ 1. • Compare f (t52 ) with f (t74 )

(P 1 − P 2 α12 )(l2 − l1 α12 ) l2 l1 − α21 l2 l2 − α12 l1 +P1 +P2 =− f (t52 ) − f (t74 ) = −P 1 α12 1 − α12 α21 1 − α12 α21 α12 (1 − α21 α21 ) ( ≤ 0, if P 1 ≥ P 2 α12 , ≥ 0, if P 1 ≤ P 2 α12 where the inequalities are due to Assumption 1 that l2 ≥ α12 l1 and the fact that 0 ≤ α21 ≤ 1, 0 ≤ α12 ≤ 1. From the above comparison as well as the results of Case 1-Case 3, we can draw the following conclusion on the best solution in subregions Ω1 , Ω2 , Ω3 , and Ω4 : (i) If P 1 ≤ P 2 α12 , then we must have P 2 ≥ P 1 α21 since α12 , α21 ∈ [0, 1], and t33 dominates all the other points in subregions Ω1 , Ω2 , Ω3 , and Ω4 , since f (t33 ) ≥ f (t62 ) ≥ f (t74 ), f (t54 ) ≤ f (t33 ), and f (t34 ) ≤ f (t33 ). (ii) If P 1 ≥ P 2 α12 and P 2 ≤ P 1 α21 , then t34 or t54 dominates all the other points in subregions Ω1 , Ω2 , Ω3 , and Ω4 , since f (t54 ) ≥ f (t74 ) ≥ f (t52 ), and f (t74 ) ≥ f (t62 ) ≥ f (t33 ). (iii) If P 1 ≥ P 2 α12 and P 2 ≥ P 1 α21 , then t34 or t74 dominates all the other points in subregions Ω1 , Ω2 , Ω3 , and Ω4 , since f (t74 ) ≥ f (t33 ), f (t74 ) ≥ f (t62 ) ≥ f (t33 ), and f (t74 ) ≥ f (t54 ).


36

Case 5. Suppose (Q1 , Q2 ) ∈ Ω5 , i.e., D1 − l1 ≤ Q1 ≤ D1 , D2 ≤ Q2 . In this case, R(Q) = max S 1 (Q1 − D1 + l1 ) + S 2 (Q2 − D2 ), S 2 (Q2 − D2 + l2 − α12 (D1 − Q1 )) and Model (30) becomes: max f (Q) = P 1 Q1 + P 2 Q2 − R(Q) : D1 − l1 ≤ Q1 ≤ D1 , D2 ≤ Q2 . Q

According to Observation 1, the optimal solution can be one of the following points: (1) the extreme points of Ω5 , i.e, t15 , t25 ; (2) the intersection point of line S 1 (Q1 − D1 + l1 ) + S 2 (Q2 − D2 ) = S 2 (Q2 − D2 + l2 − α12 (D1 − Q1 )) and the boundary of Ω5 , i.e., t35 (See Figure 5 for an illustration). Note that t35 ∈ Ω5 if S 1 l1 ≥ S 2 l2 , otherwise, t35 6∈ Ω5 . These possible solutions are listed in Table 7.

Figure 5


Table 7

t15


Possible solutions Total profit 2 3 1 2 2 3 = = t3 = t2 = (D1 − l1 , D2 ) f (t5 ) = f (t4 ) = f (t3 ) = f (t2 ) = P 1 (D1 − l1 ) + P2 D2 − S 2 (l 2 − α12 l1 ) t25 = t34 = (D1 , D2 ) f (t25 ) = f (t34 ) = P 1 D1 + P 2 D2 − max S 1 l1 , S 2 l2 1 α12 1 −S 2 l2 1 −S 2 l2 t35 = (D1 − SS1 l−S , D2 ) f (t35 ) = P 1 D1 + P 2 D2 − P 1 SS1 l−S − S 1 S 2 Sl2 −l α α −S α t24

1

2 12

1

2 12

1

2 12

In view of the results in Case 1- Case 4, the only new point is t35 , which is in subregion 5 if S 1 l1 ≥ S 2 l2 . Thus, suppose that S 1 l1 ≥ S 2 l2 , we will compare f (t35 ) with f (t33 ), f (t34 ). • Compare f (t34 ) with f (t35 ):

f (t34 ) − f (t35 ) = P 1 D1 + P 2 D2 − S 1 l1 − P 1 D1 − P 2 D2 + P 1 =−

(S 1 − P 1 )(S 1 l1 − S 2 l2 ) ≤ 0, S 1 − S 2 α12

where the inequality is due to S 1 ≥ P 1 and S 1 l1 ≥ S 2 l2 .

S 1 l1 − S 2 l2 l2 − l1 α12 + S1S2 S 1 − S 2 α12 S 1 − S 2 α12


37

• Compare f (t33 ) with f (t35 )

l2 − l1 α12 S 1 l1 − S 2 l2 + S1S2 S 1 − S 2 α12 S 1 − S 2 α12 l2 − l1 α12 S 1 l1 − S 2 l2 ≥ P 2 (−l2 + α12 D1 ) − P 1 D1 + P 1 + S1P 2 S 1 − S 2 α12 S 1 − S 2 α12 S 1 l1 − S 2 l2 = P 2 α12 − P 1 D1 − . S 1 − S 2 α12

f (t33 ) − f (t35 ) = P 2 (D2 − l2 + α12 D1 ) − P 1 D1 − P 2 D2 + P 1

1 −S 2 l2 ≥ D1 − l1 ≥ 0 and S 1 l1 ≥ S 2 l2 , thus, f (t33 ) ≥ f (t35 ), if P 1 ≤ P 2 α12 . Since D1 − SS1 l−S α 1

2 12

From the above comparison results as well as the results of Case 1-Case 4, we can draw the following conclusion on the best solution in subregions Ω1 , Ω2 , Ω3 , Ω4 , and Ω5 : (i) If P 1 ≤ P 2 α12 and P 2 ≥ P 1 α21 , then t33 dominates all the other points in subregions Ω1 , Ω2 , Ω3 , Ω4 , and Ω5 , since f (t33 ) ≥ f (t35 ). (ii) If P 1 ≥ P 2 α12 and P 2 ≤ P 1 α21 , then (a) if S 1 l1 ≤ S 2 l2 , then t34 or t54 dominates all the other points in subregions Ω1 , Ω2 , Ω3 , Ω4 , and Ω5 , and (b) if S 1 l1 ≥ S 2 l2 , then t35 or t54 dominates all the other points in subregions Ω1 , Ω2 , Ω3 , Ω4 , and Ω5 . (iii) If P 1 ≥ P 2 α12 and P 2 ≥ P 1 α21 , then (a) if S 1 l1 ≤ S 2 l2 , then t34 or t74 dominates all the other points in subregions Ω1 , Ω2 , Ω3 , Ω4 , and Ω5 , and (b) if S 1 l1 ≥ S 2 l2 , then t35 or t74 dominates all the other points in subregions Ω1 , Ω2 , Ω3 , Ω4 , and Ω5 . Case 6. Suppose (Q1 , Q2 ) ∈ Ω6 , i.e., D1 ≤ Q1 , D2 ≤ Q2 . In this case, R(Q) = max{S 1 (Q1 − D1 + l1 ) + S 2 (Q2 − D2 ), S 1 (Q1 − D1 ) + S 2 (Q2 − D2 + l2 )} and Model (30) becomes: max f (Q) = P 1 Q1 + P 2 Q2 − R(Q) : D1 ≤ Q1 , D2 ≤ Q2 . Q

According to Observation 1, the optimal solution can only be t16 , which is listed in Table 8. Table 8


Possible solutions Total profit t16 = t25 = t34 = (D1 , D2 ) f (t16 ) = f (t25 ) = f (t34 ) = P 1 D1 + P 2 D2 − max S 1 l1 , S 2 l2

Note that there is no new optimal solution generated in the case, thus the conclusion in Case 5 still follows.


38

Figure 6


Next, for the Cases 20 , 30 , 50 , since they are symmetric to Cases 2, 3, 5, thus we will directly write down the possible solutions. Case 20 . Suppose (Q1 , Q2 ) ∈ Ω20 , i.e., D1 ≤ Q1 , D2 ≤ Q2 . Case 20 is symmetric to Case 2, and its possible solutions are listed in Table 9. Table 9

Possible solutions and their total profits in Case 2’

Possible solutions Total profit t120 = (D1 − l1 , 0) f (t120 ) = P 1 (D1 − l1 ) t220 = (D1 , 0) f (t220 ) = P 1 D1 3 3 t20 = (D1 , D2 − l2 ) f (t20 ) = P 2 (D2 − l2 ) + P 1 D1 − S 1 (l1 − α21 l2 ) 4 1 4 1 4 t20 = t4 = t2 = t1 = (D1 − l1 , D2 − l2 ) f (t20 ) = f (t14 ) = f (t42 ) = f (t11 ) = P 2 (D2 − l2 ) + P 1 (D1 − l1 ) t520 = (D1 , D2 − l1 /α21 ) f (t520 ) = P 2 (D2 − l1 /α21 ) + P 1 D1 t620 = (D1 − l1 + α21 l2 , D2 − l2 ) f (t620 ) = P 2 (D2 − l2 ) + P 1 (D1 − l1 + α21 l2 )

Case 30 . Suppose (Q1 , Q2 ) ∈ Ω30 , i.e., D1 ≤ Q1 , 0 ≤ Q2 ≤ D2 − l2 . Case 30 is symmetric to Case 3 and its possible solutions are listed in Table 10. Table 10

Possible solutions and their total profits in Case 3’

Possible solutions Total profit t130 = t220 = (D1 , 0) f (t130 ) = f (t220 ) = P 1 D1 t230 = t320 = (D1 , D2 − l2 ) f (t230 ) = f (t320 ) = P 2 (D2 − l2 ) + P 1 D1 − S 1 (l1 − α21 l2 ) 3 t30 = (D1 − l1 + α21 D2 , 0) f (t330 ) = P 1 (D1 − l1 + α21 D2 ) 4 5 4 t30 = t20 = (D1 , D2 − l1 /α21 ) f (t30 ) = f (t520 ) = P 2 (D2 − l1 /α21 ) + P 1 D1

Case 50 . Suppose (Q1 , Q2 ) ∈ Ω50 , i.e., D1 ≤ Q1 , D2 − l2 ≤ Q2 ≤ D2 . Case 50 is symmetric to Case 5 and its possible solutions are listed in Table 11.


Table 11

Possible solutions and their total profits for Case 5’

Possible solutions t150

39

= t24 = t230 = t320 = (D1 , D2 − l2 ) t250 = t25 = t14 = (D1 , D2 ) 2 −S 1 l1 ) t350 = (D1 , D2 − SS2 l−S 2 1 α21

Total profit f (t150 )

= f (t24 ) = f (t230 ) = f (t320 ) = P 2 (D2 − l2 ) + P 1 D1 − S 1 (l1 − α21 l2 ) f (t250 ) = f (t25 ) = f (t14 ) = P 2 D2 + P 1 D1 − max S 2 l2 , S 1 l1 2 −S 1 l1 2 α21 − S 1 S 2 Sl1 −l f (t350 ) = P 1 D1 + P 2 D2 − P 2 SS2 l−S 2 1 α21 2 −S 1 α21

Based on the results in Case 1-Case 6, thus symmetricly, we can also draw the following conclusions in the subregions Ω1 , Ω20 , Ω30 , Ω4 , Ω50 and Ω6 : (i) If P 1 ≤ P 2 α12 and P 2 ≥ P 1 α21 , then (a) if S 1 l1 ≥ S 2 l2 , then t34 or t54 dominates all the other points in subregions Ω1 , Ω20 , Ω30 , Ω4 , and Ω50 . (b) if S 1 l1 ≤ S 2 l2 , then t350 or t54 dominates all the other points in subregions Ω1 , Ω20 , Ω30 , Ω4 , and Ω50 , since f (t34 ) ≤ f (t350 ). (ii) If P 1 ≥ P 2 α12 and P 2 ≤ P 1 α21 , then t330 dominates all the other points in subregions Ω1 , Ω20 , Ω30 , Ω4 , and Ω50 . (iii) If P 1 ≥ P 2 α12 and P 2 ≥ P 1 α21 , then (a) if S 1 l1 ≥ S 2 l2 , then t34 or t74 dominates all the other points in subregions Ω1 , Ω20 , Ω30 , Ω4 , and Ω50 , and (b) if S 1 l1 ≤ S 2 l2 , then t350 or t74 dominates all the other points in subregions Ω1 , Ω20 , Ω30 , Ω4 , and Ω50 , since f (t34 ) ≤ f (t350 ). Thus, combining all the comparison results, we can conclude that Case 1: If P 1 ≤ P 2 α12 and P 2 ≥ P 1 α21 , then t33 dominates all the other points of the 9 subregions, since f (t54 ) ≤ f (t33 ), f (t34 ) ≤ f (t33 ) from the results of Case 4, and when S 1 l1 ≤ S 2 l2 , we have S 2 l2 − S 1 l1 l1 − l2 α21 − S1S2 − P 2 (D2 − l2 + α12 D1 ) S 2 − S 1 α21 S 2 − S 1 α21 l2 α21 − l1 = P 1 − P 2 α12 D1 + S 1 S 2 − P 2 ≤ 0, S 2 − S 1 α21

f (t350 ) − f (t33 ) = P 1 D1 + P 2 D2 − P 2

where the inequality is due to P 1 ≤ P 2 α12 , S 1 ≥ 0, S 2 − P 2 ≥ 0, l2 α21 ≤ l1 and S 2 ≥ S 1 α21 (due to S 1 l1 ≤ S 2 l2 and l2 α21 ≤ l1 ). Case 2: If P 2 ≤ P 1 α21 and P 1 ≥ P 2 α12 , then t330 dominates all the other points of the 9 subregions, since f (t54 ) ≤ f (t330 ), f (t34 ) ≤ f (t330 ) which is due to the symmetric results from Case 4 that f (t54 ) ≤ f (t33 ), f (t34 ) ≤ f (t33 ), and when S 1 l1 ≥ S 2 l2 , by the results that f (t350 ) ≤ f (t33 ), we must have f (t35 ) ≤ f (t330 ).


40

Case 3: If P 1 ≥ P 2 α12 , P 2 ≥ P 1 α21 , there we can separate the results into two sub-cases: Sub-case 3.1: If S 1 l1 ≥ S 2 l2 , t35 or t74 dominates all the other points, since f (t35 ) ≥ f (t34 ) from Case 5; Sub-case 3.2: If S 1 l1 ≤ S 2 l2 , t350 or t74 dominates all the other points, since f (t350 ) ≥ f (t34 ) by symmetry. This completes the proof. A.4

Proof of Theorem 3

Theorem 3 When α = 0, the optimal solutions Q∗ of Model (13) are characterized as follows: P P (i) If i∈[n] PS i ≤ k, then Q∗i = Di − li , and v ∗ = i∈[n] P i (Di − li ). i P (ii) If i∈[n] PS i > k, i ( S l Di − li + (t+1)S (t+1) , if i ∈ T ∗ i , Qi = Di , if i ∈ [n] \ T and v∗ =

X

P i li − S (t+1) l(t+1) k +

X Pi i∈T

i∈[n]\T

where set T := {(1), (2), · · · , (t)} satisfying

Pi i∈T S i

P

Si

S (t+1) l(t+1) +

≤ k,

X

P i (Di − li ) ,

i∈[n] Pi i∈T ∪{(t+1)} S i

P

> k.

b = {z : P Proof: Let X i∈[n] zi ≤ k, zi ∈ [0, 1], ∀i ∈ [n]}, which is a well-known integral polytope. Thus, b and the inner maximization problem of (12) is equivalent to maximize a linear conv(X) = X b Thus, we have function of set X.  v∗ =

max Q∈[D−l,D]

 X



P i Qi − max b z∈X

i∈[n]

X

S i (Qi − Di + li ) zi  .

(31)

i∈[n]

Let qi = Qi − Di + li for each i ∈ [n], then Model (31) is equivalent to  v ∗ = max min  b q∈[0,l] z∈X

 X

X

P i − S i zi qi +

P i (Di − li ) .

(32)

i∈[n]

i∈[n]

Let λ be the dual variable associated with constraint

P

i∈[n] zi

≤ k and βi be the dual variable

associated with constraint zi ≤ 1 for each i ∈ [n]. Then by reformulating the inner maximization into its dual form, Model (32) is equivalent to

v ∗ = max kλ +

X

q,λ,β

i∈[n]

βi +

X

P i qi +

i∈[n]

s.t. λ + βi ≤ −S i qi , i ∈ [n],

X

P i (Di − li )

(33a)

i∈[n]

(33b)


41

0 ≤ qi ≤ li , i ∈ [n],

(33c)

λ, βi ≤ 0, i ∈ [n].

(33d)

b is In Model (32), since the objective function is concave in q and convex in z, and set X convex compact set, thus according to Sion’s minimax theorem (cf., Sion 1958), we can equivalently reformulate Model (32) by switching the min with max operators as follows:   X  X ∗ v = min max P i − S i zi qi + P i (Di − li ) . b q∈[0,l]   z∈X i∈[n]

Note that maxqi ∈[0,li ]

P i∈[n]

i∈[n]

P P i − S i zi qi = i∈[n] P i − S i zi + li for each i ∈ [n]. Thus, Model

(32) is equivalent to v ∗ = min b z∈X

 X

P i − S i zi

l + i

+



i∈[n]

X

P i (Di − li )

 

.

(34)



i∈[n]

we also observe that Claim 2 Model (34) is equivalent to   X  X v ∗ = min P i − S i zi li + P i (Di − li ) , b1   z∈X i∈[n]

n b1 = z : P where X i∈[n] zi ≤ k, 0 ≤ zi ≤

Pi Si

(35)

i∈[n]

o

.

Proof: Let v1∗ denote the optimal value of Model (35). To prove Model (34) is equivalent to Model (35), we only need to show v ∗ = v1∗ . v ∗ ≥ v1∗ . Given an optimal solution z ∗ of Model (34), let us define set J1 = {i ∈ [n] : 0 ≤ zi∗ ≤ and J2 = {i ∈ [n] :

P¯i ¯i S

P¯i ¯i } S

< zi∗ ≤ 1}. Clearly, J1 ∪ J2 = [n] and J1 ∩ J2 = ∅. Next, we define a new

solution zb such that

( z∗, zbi = P¯ii ¯i , S

if i ∈ J1 , otherwise

b1 . We also have for each i ∈ [n]. Clearly, zb ∈ X v∗ =

X

P i − S i zi∗

l + i

+

X

X P i − S i zi∗ li + P i (Di − li )

i∈J1

=

X i∈[n]

P i (Di − li )

i∈[n]

i∈[n]

=

X

i∈[n]

P i − S i zbi li +

X

P i (Di − li ) ,

i∈[n]

where the third equality is due to the definition of zb. Therefore, zb is feasible to Model (35) with the same objective value v ∗ . Thus, we have v ∗ ≥ v1∗ .


42

b1 ⊆ X, b thus, v ∗ ≤ v1∗ . v ∗ ≤ v1∗ . Since X

Note that Model (35) is a continuous knapsack minimization problem and can be solved by greedy procedure (c.f., Dantzig 1957, Levi et al. 2014). Let z ∗ denote an optimal solution to Model (35). To obtain z ∗ , we first sort {S i li }i∈[n] in the descending order S (1) l(1) ≥ S (2) l(2) ≥ · · · ≥ S (n) l(n) . Next, we discuss two cases: P Case 1. If i∈[n] PS i ≤ k, then we have i

Pi , ∀i ∈ [n], Si P and v ∗ = i∈[n] P i (Di − li ). On the other hand, in (33), let us consider the following feasible P solution λ∗ = 0, βi∗ = 0, qi∗ = 0 for all i ∈ [n] with objective value equal to i∈[n] P i (Di − li ). zi∗ =

Therefore, (Q∗ , λ∗ , β ∗ ) is optimal to (33). Hence, the optimal order quantity for each product i ∈ [n] is Q∗i = qi∗ + Di − li = Di − li . Case 2. If P

Pi i Si

P

> k, then let us define set T := {(1), (2), . . . , (t)} such that

Pi i∈T ∪(t+1) S i

> k. Then we have P i   Si P ∗ zi = k − τ ∈T   0

Pi i∈T S i

P

≤ k,

if i ∈ T Pτ Sτ

if i = (t + 1) , otherwise

for each i ∈ [n], and X X v∗ = P i − S i zi∗ li + P i (Di − li ) i∈[n]\T

i∈[n]

= P (t+1) l(t+1) − S (t+1) k −

X Pi i∈T

=

X

P i li − S (t+1) l(t+1) k +

Si

X Pi i∈T

i∈[n]\T

!

Si

X

l(t+1) +

P i li +

i∈{[n]\T ∪(t+1)}

X

S (t+1) l(t+1) +

X

P i (Di − li )

i∈[n]

P i (Di − li )

i∈[n]

Next λ∗ = −S t+1 lt+1 , βi∗ = 0 and ( qi∗

=

S t+1 lt+1 , Si

if i ∈ T

li ,

if i ∈ [n] \ T

for each product i ∈ [n]. Clearly, (Q∗ , λ∗ , β ∗ ) is feasible to (33) with objective value equal to v ∗ . Therefore, (Q∗ , λ∗ , β ∗ ) is optimal to (33). Hence, the optimal order quantity for each product i ∈ [n] is Q∗i

=

qi∗

+ Di − li =

( S l Di − li + t+1S t+1 , i

Di ,

if i ∈ T if i ∈ [n] \ T

.


A.5

43

Proof of Proposition 3

Proposition 3 The inner maximization problem (17) is equivalent to   X X αji (uj yji + ψj (xi − yji )) R(Q, u, ψ) = max S i (Qi − Di )xi + li yii − x,y,z

s.t.

X

(19a)

j∈[n]

i∈[n]

zi ≤ k.

(19b)

i∈[n]

yji ≤ xi , ∀i, j ∈ [n],

(19c)

yji ≤ zj , ∀i, j ∈ [n],

(19d)

zi , xi ∈ {0, 1}, yji ≥ 0, ∀i, j ∈ [n].

(19e)

b Proof: Let R(Q, u, ψ) denote the optimal value of Model (19). It is sufficient to show that b R(Q, u, ψ) = R(Q, u, ψ) for any feasible (Q, u, ψ) ∈ R3n + . b R(Q, u, ψ) ≤ R(Q, u, ψ). Suppose (x∗ , y ∗ ) ∈ {0, 1}2n is an optimal solution of Model (18). Define ∗ yji = zj∗ x∗i for each i, j ∈ [n]. Clearly, (x∗ , y ∗ , z ∗ ) is feasible to Model (19) and " # X X ∗ ∗ ∗ ∗ R(Q, u, ψ) = S i Qi − Di + li zi − αji uj zj + vj (1 − zj ) xi j

i∈[n]

 =

X

S i Qi − Di )x∗i + li yii∗ −

X

αji

 ∗ ∗ uj yji + ψj (x∗i − yji ) ,

j∈[n]

i∈[n]

b i.e., it yields the same objective value as R(Q, u, ψ). Thus, R(Q, u, ψ) ≤ R(Q, u, ψ). b R(Q, u, ψ) ≥ R(Q, u, ψ). Suppose (x∗ , y ∗ , z ∗ ) is an optimal solution to Model (19). Since both x∗ ∗ and z ∗ are binary, thus according to constraints (19c) and (19d), we must have yji ≤ zj∗ x∗i =

min{zj∗ , x∗i } for each i, j ∈ [n]. Therefore,   X X ∗ ∗ b R(Q, u, ψ) = S i (Qi − Di )x∗i + li yii∗ − αji uj yji + ψj (x∗i − yji )  i∈[n]

j∈[n]

" ≤

X i∈[n]

Si

# Qi − Di + li zi∗

−

X

αji

uj zj∗

+ vj (1 − zj∗ )

x∗i

j

∗ where the first inequality is due to the coefficients of {yji }j,i∈[n] are all nonnegative, i.e.,

S i li ≥ 0 and ψj ≥ uj for all i, j ∈ [n]. Hence, (x∗ , z ∗ ) is feasible to Model (18) and yields an b b objective value at least as large as R(Q, u, ψ), which implies that R(Q, u, ψ) ≥ R(Q, u, ψ). A.6

Proof of Theorem 5

Theorem 5 Let v CA denote the optimal value of Model (25). Then


44

(i) v CA ≤ v ∗ ; and (ii) v CA = v ∗ , if one of the following conditions holds: (1) α = 0, or (2) n = k. Proof: v ∗ ≥ v CA holds since in (25), we replace set Ξ to be its continuous relaxation ΞC . It remains to show that v ∗ = v CA if α = 0 or n = k. To prove this result, it is sufficient to show that for any given (Q, u, ψ) satisfying constraints (20a) − (20e), (21b) − (21c), the following linear program max

g (Q, u, ψ, x, y, z)

(x,y,z)∈ΞC

has an integral optimal solution, i.e., the continuous relaxation of Model (19) has an integral optimal solution. α = 0. In this case, Model (19) is equivalent to R(Q, u, ψ) = max

X

S i [(Qi − Di )xi + li yii ]

(38a)

zi ≤ k.

(38b)

yii ≤ xi , ∀i ∈ [n],

(38c)

yii ≤ zi , ∀i ∈ [n],

(38d)

zi , xi ∈ {0, 1}, yii ≥ 0, ∀i ∈ [n].

(38e)

x,y,z

i∈[n]

s.t.

X i∈[n]

b C denote the continuous relaxation of the feasible region of Model (38), where we We let Ξ b C is an integral polytope. relax x, z to be continuous. Then, it is sufficient to show that Ξ First of all, let us write the constraints (38b) − (38e) in the matrix form as below:      z k e> 0 0  0 I −I  y  ≤ 0 , (39) x 0 −I I 0       z1 y11 x1  z2   y22   x2        where z =  . , y =  . , x =  . . By Theorem 19.3 on Page 269 of Schrijver (1998), to . . .  .   ..  zn ynn xn   e> 0 0 prove P is a integral polytope, it is sufficient to prove A =  0 I −I  is totally unimodular −I I 0   k (TU). Indeed, if A is TU, and since 0 is integral, thus, P is an integral polytope. Hence, 0 the continuous relaxation of Model (38), which is a linear program, has an integral optimal solution. According to Theorem 19.3 on Page 269 of Schrijver (1998), to prove A is a totally unimodular matrix, it is sufficient to prove that for any S ⊆ [3n], there exist S1 and S2 such P P 2n+1 that S1 ∩ S2 = ∅, S1 ∪ S2 = S, i∈S1 A.i − i∈S2 A.i ∈ {−1, 0, 1} .


45

We let Sb1 = S ∩ {1, 2, · · · , n} = {i1 , · · · , i|S 1 | }, Tz1 = {iτ }τ ≤|S 1 |,τ is odd and Tz2 = {iτ }τ ≤|S 1 |,τ is even . Also, we let Sb2 = S ∩ {n + 1, · · · , 2n}, Sb3 = S ∩ {2n + 1, · · · , 3n}, Ty1 = n o n o j ∈ Sb2 : j − n ∈ Tz1 , Ty2 = Sb2 \ Ty1 , Tx1 = j ∈ Sb3 : j − n ∈ Ty1 , Tx2 = S 3 \ Tx1 . Clearly, we P P have S1 = Tz1 ∪ Ty1 ∪ Tx1 , S2 = S \ S1 . For such S1 and S2 , we have i∈S1 A.i − i∈S2 A.i ∈ {−1, 0, 1}

2n+1

.

k = n. From the discussion in Section 4.3, we already know at the optimality, we must have zi∗ = 1 P for all i ∈ [n] when k = n. In (19a), the coefficient of yji is j∈[n] αji (ψj − uj ) ≥ 0, since ψj ≥ uj for each j, i ∈ [n] and j 6= i. Also, the coefficient of yii is li , which is nonnegative, for each i ∈ [n]. Thus, at the optimality of the continuous relaxation of Model (19), we must have yji = min(xi , zj ) = min(xi , 1) = xi for all i, j ∈ [n]. Then, the continuous relaxation of Model (19) is equivalent to  max

x∈[0,1]n

X



S i (Qi − Di )xi + li yii −

i∈[n]

X

αji (uj yji + ψj (xi − yji ))

j∈[n]

 = maxn x∈[0,1]

X i∈[n]

S i (Qi − Di + li ) +

 X

αji (Qj − Dj + lj )+  xi ,

j∈[n]

which is a linear program over a unit box. Thus, there exists an optimal solution x∗ of the above linear program, which corresponds to an extreme point of the box [0, 1]n , i.e., x∗i ∈ {0, 1} ∗ for all i ∈ [n]. Thus, yji = x∗i ∈ {0, 1} for all i, j ∈ [n]. Therefore, the continuous relaxation of

Model (19) has an integral optimal solution (x∗ , y ∗ , z ∗ ).