Constrained Inventory Allocation

Transcription

1 Constrained Inventory Allocation Phillip G. Bradford Michael N. Katehakis January 25, 2006 SYNOPTIC ABSTRACT Supply constraints may include minimum supply commitments and maximum delivery limits. Given such constraints, this paper shows how to allocate a limited supply of resources to retailers in one time period by building on a classical result of Derman. This analysis yields solutions to inventory allocation for a centralized distribution system with multiple suppliers and retailers. The focus is to ensure suppliers maintain minimal supply commitments and maximum delivery limits. The objective is to maximize the number of units sold. The expected supply and expected consumption are described by random probability density functions. Department of Computer Science, The University of Alabama, Box , Tuscaloosa, AL pgb@cs.ua.edu Management and Information Sciences Dept., Rutgers Business School, Newark and New Brunswick, New Brunswick, New Jersey, mnk@andromeda.rutgers.edu 1

2 1 Introduction Suppliers may have incentives (penalties) to keep supplies above some minimum level. Likewise, retailers may have maximum delivery limits to restrain capital and other inventory costs. This paper considers a single period for distributing and consuming goods or services with upper and lower bound constraints. This extends Derman s model for optimal resource allocation (Derman, 1959). This model assumes one time period and random consumption by each retailer s consumers. This random consumption may be modeled using known distributions. Also, optimal resource supply may be modeled using known distributions. Different from the news-vendor problem (Zipkin, 2000), (Khouja, 1999), this problem assumes unused resources lose all value if they are not allocated. This assumption models issues such as the allocation of services. This paper s central motivation follows (Zipkin, 1982), in that there is a clear desire to have simple and effective models to enhance management reasoning. How may we balance expected supply and expected demand of a retailer given contract constraints? What if the supplier has a different expected supply function than the retailer s expected demand function? The paper (Zipkin, 1980) gives a generalized framework for interpreting and solving nonlinear-additive functions constrained by the sum of their inputs. The focus is on the algorithmic solution to such problems. Variations on these problems are very close to the one considered by the current paper. Furthermore, (Dentcheva, et al., 2001) characterize and extend probabilistically constrained convex programming on discrete distributions. Single source supplies to several retailers or locations are considered in (Federgruen and Zipkin, 1984a) with finite time periods. The focus is on minimizing cost for the system. An infinite horizon case is considered in (Federgruen and Zipkin, 1984b). In (Federgruen and Zipkin, 1984b) it is shown it may be expressed as single location problems. The paper (Zipkin, 1984) considers the balancing of inventory with a horizon constraint. 2

3 The remainder of Section 1 gives the background on the problems this paper focuses on. Section 2 discusses how to balance supply and demand. Finally, Section 3 gives an extended example and future directions. 1.1 Background Retailer i has the demand function X i 0 that is random and unconstrained with probability density function h i (x). Similarly, supplier j has the supply function X j 0 that is random and unconstrained with density function l j (x). There may be constraints on the amount of assets given to each retailer in each time period. Assume the supply and demand functions are differentiable at least twice. Under these constraints our results show equalizing sell-out (or stock-out) for as many retailers as possible while equalizing the surplus for as many suppliers as possible gives an optimal solution. This would maximize consumption or sales. Assume there is no salvage value for unused resources and consider the upper and lower bounds for each retailer. The goal is to maximize the total sales. Each retailer wants a lower bound on the supplies they are given. They desire to satisfy their demand. The supplier wants an upper bound on each delivery. They also desire to satisfy the expected demand of the retailers they supply. Supply is the negative of demand. So, if s is supply and d is the same amount of demand, then s + d = 0. This paper shows that one should equalize the sell-out of as many retailers as possible while satisfying all constraints. Analogously, the supplier should equalize the surplus for as many retailers as possible while satisfying all constraints. Finally, this paper indicates one way to maximize sales and keep expected supply equal to expected demand is to equalize the probability of surplus and the probability of sell-out as much as possible. Consider the total expected sale function G(y 1,, y k1, z k1 +1,, z k1 +k 2 ) = H(y 1,, y k1 ) + L(z k1 +1,, z k1 +k 2 ). 3

4 The function H = H(y 1,, y k1 ) is the expected demand of k 1 retailers in one time period so that retailer i is initially allotted y i l i 0, for i : k 1 i 1. Where l i 0 is a lower-bound on the delivery of y i to retailer i. Following (Derman, 1959), let H(y 1,, y k1 ) be a non-decreasing function of y 1,, y k1. Additionally, there is a total supplier constraint of y y k1 T y. Given H, and these constraints, the objective is to maximize the total sold to all of these k 1 retailers. The function L = L(z k1 +1,, z k1 +k 2 ) is the expected supply for k 2 retailers where supplier i initially allots z i 0, for retailer i : k 1 + k 2 i k and each retailer may be supplied with at most u i z i. Moreover, z k z k1 +k 2 T z is the total supply constraint. Given L, and these constraints, the objective is to maximize the total consumed by these k 2 retailers. In many cases, both the supplier as well as the retailers may agree on the same expected demand functions. However, there may be times when there is disagreement to what the demand to expect. Generally, we expect consumer and supplier to agree on upper and lower bounds for each location. Both the supplier and the retailer may be aware of the total resources available T. This type of situation is not unusual in consulting: a known budget for a project with minimum and maximum time limits for different activities. In some cases, H(y i ) and L(z i+k1 ) may refer to the same product or service. That is, H(y i ) is the expected demand for retailer i and L(z i+k1 ) is the expected supply given to retailer i. This allows retailer i to have a different expected demand function from the supplier s expected consumption function for retailer i. This may happen when, for example, the retailer and supplier have different information about consumption. This paper shows, even in the case of different expected consumption and expected supply functions, it is optimal to equal sell-out and surplus. Finally, the structure of the function L is the negative of the function H, though H has lower bounds and L has upper bounds. This means if expected demand is larger than expected supply, then there is an expected shortage 4

5 represented as a positive G. If expected supply is larger than expected demand, then there is an overall surplus represented as a negative G. Perhaps these negative and positive values are best interpreted as answers to the question: How much more is needed? Given these relationships of H and L, the objective is to finding inputs y1,, yk 1, zk 1 +1,, zk 1 +k 2 so that G(y1,, yk 1, zk 1 +1,, zk 1 +k 2 ) = 0. Additionally, it is interesting to understand how to maximize and minimize the function G(y 1,, y k1, z k1 +1,, z k1 +k 2 ). The non-negative values y1,, yk 1, zk 1 +1,, zk 1 +k 2 are the initial allocations of the resource to consumers 1,, k 1 + k 2, respectively. This is done so each y i is lower bounded and each z j is upper bounded. The next derivations are based directly on Derman (1959). If X i y i, for all i : k 1 i 1, then the expected demand is H(y 1,, y k1 ) = IE[X 1 ] + + IE[X k1 ]. On the other hand, suppose X i > y i, for all i : k 1 i 1, then we must add (y i x)f i (x) dx, y i to H(y 1,, y k1 ). Since if y i X i < 0, then X i is in the range (y i,, + ) and consumer i has a shortage. Thus, the objective is to maximize the concave function (Derman, 1959), H(y 1,, y k1 ) = = = k 1 (y i x)f i (x) dx + xf i (x)dx i=1 y i 0 k 1 yi ] [y i f i (x) dx + xf i (x) dx y i i=1 k 1 i=1 y i IP[X i > y i ] + IE[X i y i ], while assuming IP[X i = 0] = 0, for all i : k 1 i 1. Note that H(y 1,, y k1 ) is concave by the second-derivative test by applying Leibniz s rule to compute 5 0

6 d dy i H(y 1,, y k1 ), for each y i while noting IP[X i > y i ] = y i f i (x) dx = 1 yi 0 f i(x) dx. Thus, y i IP[X i > y i ] = y i y i IP[X i y i ]. Initially allocating y i to retailer i, then the probability of sell-out is IP[X i > y i ] = y i f i (x) dx. This paper assumes there is no salvage value, so there is only opportunity cost from sell-out. In this case, Derman (1959) shows optimally allocating y 1,, y k1 is done by equalizing the probabilities of sell-out for these k 1 retailers. Theorem 1 (Derman) Assume H is differentiable in all y i, where i : k 1 i 1 while y y k1 T y. If the bounds hold, for j : k 1 j 1, H H y i yi =l i y j then a maximum of H is at an interior point. li <y j T y Theorem 1 is a tiny bit different from Derman s equivalent theorem. The only difference is here the boundary is not 0 but l i. The proof does not change. Theorem 1 is used in the proof of the next theorem to show there is an optimal allocation in which every consumer gets some initial allotment. Theorem 2 (Derman) Assume H(y 1,, y k1 ) is differentiable in all y i, where i : k 1 i 1 while y y k1 T y. If IP[X i l i ] = 0, for all i : k 1 i 1, then the maximum of H is at an interior point (y1,, yk 1 ) satisfying the constraint y1 + + yk 1 T y and: IP[X 1 > y1] = IP[X 2 > y2] = = IP[X k1 > yk 1 ]. Each initial allotment y i has a lower bound l i so that y i l i, for all i : k 1 i 1. Suppose, sell-out cannot be equalized as suggested in Theorem 2 6

7 while yi = l i, for all i : k 1 i 1. However, sell-outs may be equalized by adding a value y i to each lower bound l i : IP[X 1 > l 1 + y 1 = y 1] = IP[X 2 > l 2 + y 2 = y 2] = = IP[X k1 > l k1 + y k1 = y k 1 ], while y y k1 T y l 1 l k1 or y y k 1 T y. Let L(z k1 +1,, z k1 +k 2 ) be a non-increasing function of z k1 +1,, z k1 +k 2. The function L(z k1 +1,, z k1 +k 2 ) is the expected supply provided to all k 2 retailers k 1 + 1,, k 1 + k 2 in one time period. Each z i is bounded above by a contract constraint. The objective is to minimize the overall surplus given these constraints. The initial allocations to retailers k 1 + 1,, k 1 + k 2 are z k1 +1,, z k1 +k 2 where z i 0 for all i : k 1 + k 2 i k Further, say each z i is bounded above by a contract constraint u i z i, for all i : k 1 + k 2 i k Let X i be the expected demand by retailer i. If X j z j, then the expected supply is L(z k1 +1,, z k1 +k 2 ) = IE[X k1 +1] IE[X k1 +k 2 ]. If z i < X i, then X i is in the range (z i,, + ). Thus, for each retailer i, subtract IE[X i ] from the function z i (x z i )f i (x) dx. Now, the objective is to minimize the next function, L(z k1 +1,, z k1 +k 2 ) = = = k 1 +k 2 i=k 1 +1 z i k 1 +k 2 [ i=k 1 +1 k 1 +k 2 i=k 1 +1 z i (x z i )f i (x) dx xf i (x) dx z i z i 0 xf i (x) dx f i (x) dx z i IP[X i > z i ] IE[X i z i ]. 0 ] xf i (x) dx 7

8 Note that L is a convex function of z k1 +1,, z k1 +k 2 by its relationship to H or by the second-derivative test from applying Leibniz s rule to d dz i L(z k1 +1,, z k1 +k 2 ), for each z i. As with H, for L there is no salvage value, but there is opportunity cost: if retailer i receives excess while j has unsatisfied demand, then there is lost opportunity. The next result is symmetric to Derman s Theorems (Theorems 1 and 2). Theorem 3 Assume L is differentiable in all z i, where i : k 1 + k 2 i k and z k z k1 +k 2 T z, for j : k 1 j 1, where L L. z i zi =u i z j li <z j <u i If IP[X i l i ] = 0, for all i : k i 1, then a minimum of L is at an interior point (zk 1 +1,, zk 1 +k 2 ) satisfying: IP[X k1 +1 zk 1 +1] = IP[X k1 +2 zk 1 +2] = = IP[X k1 +k 2 zk 1 +k 2 ]. The proof is skipped since the statement of this result is the compliment of Theorems 1 and 2. Theorem 3 indicates to minimize L then the objective is to equalize the probability of having surpluses. Since, IP[X k1 +1 u k1 +1 z k1 +1 = zk 1 +1] = = IP[X k1 +k 2 u k1 +k 2 z k1 +k 2 = zk 1 +k 2 ], then by the properties of probability functions it must also be that IP[X k1 +1 > u k1 +1 z k1 +1 = zk 1 +1] = = IP[X k1 +k 2 > u k1 +k 2 z k1 +k 2 = zk 1 +k 2 ]. The next calculation is straight-forward using Leibniz s rule. 8

9 Fact 1 Let G = G(x 1,, x k1 +k 2 ), then G f i (x) dx if i : k 1 i 1 and x i = a i a = i x i f i (x) dx if i : k 1 + k 2 i k and x i = b i. b i Additionally, y i y i f i (x) dx = f i (y i ). An example of applying Theorem s 1, 2 and 3 is given next. Take the next case of one retailer with an initial allotment of y and their supplier starts with an initial allotment of z, giving G(y, z) = y IP[X 1 > y] + IE[X 1 y] z IP[X 2 > z] IE[X 2 z]. Figure 1 gives an example of the trade-off of the expected demand and expected supply. It uses the exponential distribution with probability density function f i (x) = e λix, for i : 2 i 1 where λ 1 = 0.25 and λ 2 = 0.5. This assumes the constraint x = 20 z, for z goes from 0 to 19. Figure 1 shows x = 3 units should be allocated to the retailer giving 20 x = 17 units to the supplier thus balancing expected supply and expected demand. 2 Balancing Expected Supply and Demand This section provides key relationships to balance expected supply and expected demand. Given G(y 1,, y k1, z k1 +1,, z k1 +k 2 ) = H(y 1,, y k1 ) + L(z k1 +1,, z k1 +k 2 ) it is desirable to understand which relationships must hold among the sell-out and surpluses to find the maximum of G. In other words, what initial values 9

10 G(x,20-x) x=0 to x=19 G(x,20-x) Figure 1: An Example of G(x, 20 x) = y IP[X 1 > x] + IE[X 1 x] (20 x) IP[X 2 > (20 x)] IE[X 2 (20 x)]. 10

11 y 1,, y k 1, z k 1 +1,, z k 1 +k 2 can be chosen so G(y 1,, y k 1, z k 1 +1,, z k 1 +k 2 ) is maximal or minimal while y y k 1 T y and z k z k 1 +k 2 T z. Generally, this paper assumes a single T so that y y k 1 + z k z k 1 +k 2 T. These maximal (minimal) values indicate maximal and minimal surpluses. Thus, it is also important to understand what initial values y 1,, y k 1, z k 1 +1,, z k 1 +k 2 can be chosen so G(y 1,, y k 1, z k 1 +1,, z k 1 +k 2 ) = 0. This will indicate an optimal expected balance between supply and demand. Both H(y i ) and L(z i+k1 ) may refer to the same product or service. That is, H(y i ) is the expected demand for retailer i and L(z i+k1 ) is the expected supply for retailer i. This allows retailer i to have a different expected demand function from the supplier s expected consumption function for retailer i. For example, the retailer and supplier may have different information on consumption. Clearly, this paper assumes yi zi yi for all i. If T y u u k1 l i and u i z i. Assume, u i l i and and since H is non-decreasing, then H maximizes when y i = u i, for all i : k 1 i 1, giving H(u 1,, u k1 ). Likewise, if T y = l 1 + +l k1, then this constraint means that H maximizes when y i = l i, for all i : k 1 i 1, giving H(l 1,, l k1 ) since each y i l i. That is, parameter i of H cannot be less than l i, thus here the constraints require H(l 1,, l k1 ) to be the maximum possible. Given any integer n, this paper uses the notation: [n] = { 1, 2,, n }. Lemma 1 Consider a maximal H(y1,, yk 1 ) where y1 + + yk 1 T y and u u k1 > T y > l l k1. If there is a non-empty I 1 [k 1 ] so that holds for all i I 1, where dh dy i y i =yi { h min i I 1 y i = l i + y i : u i > y i > l i, equals h, then max { IP[X i > l i + y i = y u i >yi >l i ] } i 11 }.

12 Proof: The proof is in several cases. Since I 1 [k 1 ], this result is not much more than Derman s Theorem (Theorem 2), except for the min max result. In this case, apply Lagrange multipliers. Let g i (x) = y f i i (x) dx 1 0. This gives, by Fact 1. L(x, λ) = H + λ i g i (x) i=1 = k 1 k 1 f i (x) dx λ i f i (yi ) i=1 yi i=1 = k 1 (IP[X i > yi ] λ i f i (yi )), i=1 k 1 To find the maximum value of H(x, λ) we find the values of λ i 0 so that L(x, λ) = 0. This can be done since H is concave and has one maximum. So, one way to maximize H given the boundaries u i and l i, for all i : k 1 i 1, is first to equalizing the sell-outs, IP[X 1 > l 1 + y 1 = y1] = = IP[X k1 > l k1 + y k1 = yk 1 ]. Then for each f i (y i ) choose appropriate λ i values equalizing IP[X i > y i ] and λ i f i (yi ), then these values of yi T y y1 + + yk 1. maximize H with the given constraint Our objective is to first find the smallest y 1,, y k 1, where y i > l i, so that the sell-outs are equalized. To minimize each of the values y i > l i, then we must maximize IP[X i > y i ]. That is, for each i : k 1 i 1, compute, max { IP[X i > l i + y i = y u i >yi >l i ] }. i 12

13 The value IP[X i > yi ] can only be decreased by increasing yi above l i, thus the smallest IP[X i > l i ] is the largest value all sell-outs can be equalized to. Therefore, in this case, minimizing each yi > l i, for all i : k 1 i 1, ensures that we may set all IP[X i > yi ] equal to at most: min k 1 i 1 { max { IP[X i > l i + y i = y u i >yi >l i ] } i }. This completes the proof. By Fact 1, it must be that H(x 1,,x k1 ) x i = IP[X i > x i ] and we are assuming that H is non-decreasing on all its parameters. Say H(x i ) H(x j ) and let IP[X i > x i ] > IP[X j > x j ], for all x i : u i x i l i and x j : u j x j l j where x i + x j = T x. Now, H(x i ) grows faster than H(x j ). This means, to maximize H(x i ) + H(x j ) we must set x i = T x l j and x j = l j. Lemma 2 Consider a maximal H(y1,, yk 1 ) where y1 + + yk 1 T y and u u k1 > T y > l l k1. If there is a non-empty I 1 [k 1 ] and I 1 = k 1 1, so that yi = l i + y i : u i > yi > l i, holds for all i I 1, where dh dy i y i =y i equals h, then If H(yj ) H(yi ) and H(y 1,,y k1 ) yj y j > h, for all i I 1 and j [k 1 ] I 1, =u j then yj = T y (k 1 1) max { l t }. t [k 1 ] {j} If H(y j ) < H(y i ) and h > H(y 1,,y k1 ) y j yj =l j, for all i I 1 and j [k 1 ] I 1, then y j = l j. 13

14 Proof: must be that, Since, H is non-decreasing in each parameter, then for all i [k 1 ], it where y i = l i + y i and u i > y i > l i. H(u i ) H(y i ) H(l i ), Recall that H(x 1,,x k1 ) x i = IP[X i > x i ]. Moreover, because u i l i, then IP[X i > l i ] IP[X i > u i ] always holds. By Derman s Theorem (Theorem 2), if it is possible to equalize all sellouts, then this maximizes all H(y 1,, y k 1 ). Thus, in the next cases assume I 1 [k 1 ] is a maximal set that equalizes all sell-outs where I 1 = k 1 1. Case 1: IP[X j > l j ] IP[X j > u j ] > h. Here we are assuming H(yj ) H(yi ) for all i I 1 and j [k 1 ] I 1. Additionally, the function H(y j ) grows faster than any H(yi ), where i I 1 and j [k 1 ] I 1. Moreover, H could have been made larger by reducing all yi to their minimal values as allowed by the lower-bounds l i while maintaining equality for all IP[X i > yi ]. By assumption, it could never be that IP[X j > yj ] = h, even though all yi are reduced to their minimal values, thus potentially increasing h. Case 2: h > IP[X j > l j ] IP[X j > u j ]. Here we are assuming H(yi ) > H(yj ) for all i I 1 and j [k 1 ] I 1. Additionally, for any yj : u j > yj > l j, H(yj ), grows slower than H(yi ). Therefore, all H(yi ) for i I 1, must have grown faster by dispersing the value yj > l i to each of yi while maintaining equality for all IP[X i > yi ]. By assumption, it could never be that IP[X j > yj ] = h, even though all yi may be increased, thus potentially decreasing h. Thus, under these conditions, H(l j ) must have maximized H. These cases complete the proof. 14

15 The next definition allows us to apply the results from the last lemma to numerous classes of supply and demand functions. Definition 1 Let Q {, }. Two lists x 1,, x n and y 1,, y n are comonatonic iff x i Q y i holds for any i [n], implies x j Q y j holds for all j [n]. For example, consider two lists of exponentially distributed variables. The first list has parameters λ 1,i and random variables X 1,i and the second list has parameters λ 2,i and random variables X 2,i. It is well-known that, for s { 1, 2 }, and also IE[X s,i x] = 1 ( x + 1 ) e λ s,i x λ s,i λ s,i IP[X s,i > x] = 1 e λ s,i x. Now, suppose the two lists λ 1,1,, λ 1,n and λ 2,1,, λ 2,n are comonotonic and λ 1,1 λ 2,1. This implies, that IE[X 1,i x i ] IE[X 2,i x i ] and IP[X 1,i > x i ] IP[X 2,i > x i ]. Therefore, applying Lemma 2 indicates x t = l t for all t [k 1 ] I 1 to maximize H. The next Lemma is symmetric to Lemma 1. Since d dy i H is non-increasing in its parameters, see Theorem 1, and L is non-decreasing in its parameters, see Theorem 3, the next lemma is expressed in terms of IP[X i > u i z i = z i ] since u i z i decreases when z i increases. Lemma 3 Consider a minimal L(zk 1 +1,, zk 1 +k 2 ) while zk zk 1 +k 2 T z and u u k1 > T z > l l k1. If there is a non-empty I 2 [k 1 + k 2 ] [k 1 ] so that zi = u i z i : u i > zi > l i, 15

16 holds for all i I 2, and dl dz i z i =zi { l max i I 2 equals l, then min { IP[X i > u i z i = z u i >zi >l i ] } i }. A proof is symmetric to that of Lemma 1, thus it is skipped. Lemma 4 Consider a minimal L(y1,, yk 1 ) where zk zk 1 +k 2 T z and u k u k1 +k 2 > T z > l k l k1 +k 2. If there is a non-empty I 2 [k 1 + k 2 ] [k 1 ] and I 2 = k 2 1, so that zi = u i z i : u i > zi > l i, holds for all i I 2, and dh dz i z i =z i equals l. Then, If L(z j ) L(z i ) and L(z k 1 +1,,z k1 +k 2 ) z j zj =u j > l, for all i I 2 and j [k 1 + k 2 ] [k 1 ] I 2, then z j = T z max t [k1 +k 2 ] [k 1 ] {j}{ l t }. If L(zj ) > L(zi ) and l > L(z k 1 +1,,z k1 +k 2 ) zj z j, for all i I 2 and j =l j [k 1 + k 2 ] [k 1 ] I 2, then zj = l j. The supply and demand may be modeled using different functions H(y i ) and L(z i+k1 ) each with different probability densities and inputs, for all i : k 1 i 1. The maximum of G(y1,, yk 1, zk 1 +1,, zk 1 +k 2 ) is the maximum expected surplus that may be achieved given the constraints. The minimum of G(y1,, yk 1, zk 1 +1,, zk 1 +k 2 ) is the minimum expected shortage that may be achieved given the constraints. 16

17 Theorem 4 Consider maximizing or minimizing the total expected sale function G(y1,, yk 1, zk 1 +1,, zk 1 +k 2 ) = H(y1,, yk 1 ) + L(zk 1 +1,, zk 1 +k 2 ) where y1 + + yk 1 + zk zk 1 +k 2 T, and u u k1 +k 2 > T > l l k1 +k 2. If G is maximal, then there is a h and a set I 1 [k 1 ] so that, h = IP[X i > l i + y i = yi ] = IP[X j > l j + y j = yj ], i, j I 1, and all zi = l i for i [k 1 + k 2 ] [k 1 ]. If G is minimal, then there is a l and a set I 2 [k 1 + k 2 ] [k 1 ] so that, l = IP[X i u i z i = zi ] = IP[X j u j z j = z j ], i, j I 2, and all yi = l i for i [k 1 ]. Proof: There are two cases. This proof only considers the case of maximizing G. The case of minimizing G is symmetric. Given G(y1,, yk 1, zk 1 +1) = H(y1,, yk 1 )+L(zk 1 +1). Then let H (yk 1 +1) = L(zk 1 +1) giving G(y1,, yk 1, zk 1 +1) = H(y1,, yk 1 ) + H (yk 1 +1) = H(y1,, yk 1, yk 1 +1). This means, H(yk 1 +1) < H(yi ) for all i [k 1 ]. In addition, H(y k 1 +1) y k1 +1 IP[X k1 +1 > x k1 +1] and IP[X k1 +1 > x k1 +1] IP[X i > x i ] for all i [k 1 ]. = Therefore, applying Lemma 2 indicates that setting y k1 +1 = l k1 +1 maximizes the function H(y1,, yk 1, yk 1 +1). Moreover, Lemma 1 indicates there is value h and a set I 1 [k 1 ] so that h = IP[X i > l i + y i = yi ] = IP[X j > l j + y j = yj ], i, j I 1. This same argument can be applied to H (y k 1 +i) = L(z k 1 +i) for each i [k 1 + k 2 ] [K 1 ], completing the proof. 17

18 Expression Lower Bound Maximizing Value Upper Bound z 1 IP[X 1 > z 1 ] IE[X 1 z 1 ] 5 z 1 = 5 50 z 2 IP[X 2 > z 2 ] IE[X 2 z 2 ] 5 z 2 = 5 50 z 3 IP[X 3 > z 3 ] IE[X 3 z 3 ] 5 z 3 = 5 50 z 4 IP[X 4 > z 4 ] IE[X 4 z 4 ] 5 z 4 = 5 50 z 5 IP[X 5 > z 5 ] IE[X 5 z 5 ] 5 z 5 = 5 50 y 6 IP[X 6 > y 6 ] + IE[X 6 y 6 ] 10 y 6 = y 7 IP[X 7 > y 7 ] + IE[X 7 y 7 ] 10 y 7 = y 8 IP[X 8 > y 8 ] + IE[X 8 y 8 ] 10 y 8 = y 9 IP[X 9 > y 9 ] + IE[X 9 y 9 ] 10 y 9 = y 10 IP[X 10 > y 10 ] + IE[X 10 y 10 ] 10 y 10 = Figure 2: Maximizing G Giving a Shortage by Minimizing the Expected Supply and Maximizing Expected Demand Figure 2 has 10 instances of the exponential distribution where λ 1 = 1/4, λ 2 = 1/24, λ 3 = 1/34, λ 4 = 1/35, λ 5 = 1/36, λ 6 = 1/46, λ 7 = 1/56, λ 8 = 1/66, λ 9 = 1/67, λ 10 = 1/68. Figure 2 shows the maximum shortage when z z 5 + y y 10 T = 105. Building on Derman s Theorem 2, Theorem 4 shows, y 6 IP[X i > y 6 ] + IE[X i y 6 ] = = y 10 IP[X 10 > y 10 ] + IE[X 10 y 10 ]. Since this is a maximum shortage the sum 5 10 z i IP[X i > z i ] IE[X i z i ] + y i IP[X i > y i ] + IE[X i y i ] i=1 i=6 is maximized given the constraints. 18

19 The next result looks at characteristics of the inputs y 1,, y k 1, z k 1 +1,, z k 1 +k 2 that make G(y 1,, y k 1, z k 1 +1,, z k 1 +k 2 ) = 0. Given the bounds u i and l i and y y k 1 + z k z k 1 +k 2 T, consider the case where there are no inputs y 1,, y k 1, z k 1 +1,, z k 1 +k 2 where G(y 1,, y k 1, z k 1 +1,, z k 1 +k 2 ) = 0. Suppose for all valid bounded inputs, This means G(y 1,, y k 1, z k 1 +1,, z k 1 +k 2 ) < 0. L(z k 1 +1,, z k 1 +k 2 ) > H(y 1,, y k 1 ), for all valid inputs. In this case, since G(y 1,, y k 1, z k 1 +1,, z k 1 +k 2 ) < 0 for any valid inputs, then by Theorem 4, to get as close to 0 as possible, then we want to maximize G on these inputs. Alternatively, if for all valid bounded inputs, This means G(y 1,, y k 1, z k 1 +1,, z k 1 +k 2 ) > 0. L(z k 1 +1,, z k 1 +k 2 ) < H(y 1,, y k 1 ), for all valid inputs. In this case, since G(y 1,, y k 1, z k 1 +1,, z k 1 +k 2 ) > 0 for any valid inputs, then by Theorem 4, to get as close to 0 as possible, then we want to minimize G on these inputs. 3 Discussion This section gives a computational example. Figure 3 has five instances of the exponential distribution where the instances have the following parameters: λ 1 = 1/4, λ 2 = 1/12, λ 3 = 1/24, λ 4 = 1/36, λ 5 = 1/48. 19

20 Figure 3 maximizes the total expected supply by considering the H functions only. Moreover, the total available supply is limited to at most 10 units. H Maximizing Value Upper Bound x 1 IP[X 1 > x 1 ] + IE[X 1 x 1 ] x 1 = x 2 IP[X 2 > x 2 ] + IE[X 2 x 2 ] x 2 = x 3 IP[X 3 > x 3 ] + IE[X 3 x 3 ] x 3 = x 4 IP[X 4 > x 4 ] + IE[X 4 x 4 ] x 4 = x 5 IP[X 5 > x 5 ] + IE[X 5 x 5 ] x 5 = 1 1 Figure 3: Maximizing H where x x Generalizing this situation, consider the seven cases of different upper bounds on the total units available. Figure 4 shows the affects of increasing the total inventory T. T = 10 T = 20 T = 40 T = 60 T = 90 T = 140 T = 200 x x x x x Figure 4: Maximizing H where x x 5 T. Given the values for x 1,, x 5, Figure 5 shows the actual values of the functions x i IP[X i > x i ] + IE[X i x i ] for each i : 5 i 1. The upper bounds for each x i are in the last column of the table in Figure 3. For all values of T 90, the upper bounds in Figure 4 are all satisfied, thus maximizing the objective functions in Figure 5. 20

21 Objective Objective Figure 5: The Objective Functions. 21

22 Along the same lines, Derman s Theorem 2 indicates to maximize the objective function, then we should equalize the sell-outs. However, for all cases in this example, x 5 1, thus in Figure 6 then IP[X 5 x 5 ] forms the top horizontal line. This is because, in all of the seven cases, the objective function H is maximized while x 5 = 1. On the other hand, the sell-outs of x 1,, x 4 are all the same, as required by Derman s Theorem, until the last three cases where T 90, see Figure 6. This is because, in the last three cases each x i is at its upper limit. This allows the sell-outs to take on different values P[X1 >= x1] P[X2 >= x2] P[X3 >= x3] P[X4 >= x4] P[X5 >= x5] Figure 6: An Example of the Sell Outs. 22

23 3.1 Future Directions Efficient and effective supply chain management is the result of a great deal of in-depth knowledge. Supply chain management includes areas such as inventory management, finance, information systems all converging to make a firm more globally efficient. For overviews, see (Khouja, 1999), (Lariviere, 1999) and (Tsay, et al., 1999). The issue of a supplier and retailer having asymmetric information seems interesting. Additionally, it may be useful to understand the value of the information of how different suppliers and retailers may evaluate demand. References F. Bernstein, F. Chen, and A. Federgruen (2003): Vendor Managed Inventories and Supply Chain Coordination: The case with one supplier and competing retailers, Working Paper, Columbia Business School. D. Dentcheva, A. Prékopa, A. Ruszczyński (2001): On Convex Probabilistic Programming with Discrete Distributions, Nonlinear Analysis, 47, C. Derman (1959): A Simple Allocation Problem, Management Science, 27, A. Federgruen and P. H. Zipkin (1984a). Approximation of dynamic, multilocation production and inventory problems, Management Science, 30, A. Federgruen and P. H. Zipkin (1984b). Computational issues in an infinitehorizon, multi-echelon inventory model, Oper. Res., 32, M. Khouja (1999): The single-period (news-vendor) problem: literature review and suggestions for further research, Omega, International Journal of Management Science, 27,

24 M. A. Lariviere (1999): Supply Chain Contracting and Coordination with Stochastic Demand, in Quantative Models for Supply Chain Management, S. Tayur and R. Ganeshan, and M. Magazine, Kluwer Academic Publishers. A. L. Peressini, F. E. Sullivan, J. J. Uhl, Jr. (1988): The Mathematics of Nonlinear Programming, Springer Verlag. A. A. Tsay, S. Nahmias and N. Aggarwal (1999): Modeling Supply Chain Contracts: A Review in Quantative Models for Supply Chain Management, S. Tayur and R. Ganeshan, and M. Magazine, Kluwer Academic Publishers. P. H. Zipkin (1980). Simple ranking methods for allocation of one-resource, Management Science, 26, P. H. Zipkin (1982). Exact and Approximate Cost Function for Product Aggregates, Management Science, 28, P. H. Zipkin (1984). On the imbalance of inventories in multi-echelon systems, Math. Oper. Res., 9, 402. P. H. Zipkin (2000). Foundations of Inventory Management, McGraw-Hill,