SCN : 1. Inventory. 1 1-Period Stochastic Demand: Newsvendor Model

Transcription

1 SCN : 1. Inventory 1 1-Period Stochastic Demand: Newsvendor Model Although we will work with a different cost notation, the initial results should be familiar to the reader. Let D denote the demand with pdf and cdf f and F ; also let µ := E(D). c unit ordering cost c H unit holding cost c P unit shortage cost Let x be the initial inventory level and y be the level after the order is received but before the demand is materialized. For convenience, it is better to write the inventory costs in terms of y L(y) := c H E(y D) + + c P E(D y) + Using the identity (y D) + = (D y) + + (y D), the cost can be written in alternative forms L(y) = c H (y µ) + (c P + c H )E(D y) + We can also add the ordering costs to obtain the total costs g(y) := cy + L(y) = cµ + (c + c H )(y µ) + (c P + c H )E(D y) + Since (D y) + = max(d y, 0), it is an upper envelope of convex functions so it is convex. E(D y) + is the expected value of a convex function so it is convex. Consequently, g(y) is a convex function and is minimized at y which solves F (y) = c P c c P + c H = c P c (c P c) + (c H + c) = underage cost (underage cost)+(overage cost) In the optimality condition ζ := (c P c)(c P + c H ) is known as the critical fractile. Also note that the critical fractile is the probability of not stocking out (sometimes known as Cycle Service Level). A similar but different concept is the fill rate, defined as the ratio of expected unsatisfied demand to the expected demand, i.e. E(D y) + /µ. Since cµ is not a relevant cost, it can be dropped from the total cost expression. Note that cµ is the cost of operating the inventory system if it were possible to choose the order quantity after the demand observation. Because of this,g(y) cµ is termed as the cost of not having perfect information on the demand. Computational note: When the demand is N(µ, σ 2 ), the computation of expected shortages can be streamlined as E(D y) + = (y µ) (1 Φ((y µ)/σ)) + σφ((y µ)/σ) (1) 1

2 where Φ and φ are cdf and pdf for standard normal variate. After setting z = (y µ)/σ, it follows that E(D y) + /σ = z (1 Φ(z)) + φ(z) =: I N (z) where I N (z) is known as the unit normal loss function and is tabulated at the end of the books. Manipulations with normal density is not easy. For example, Φ cannot be put into a closed form expression. However, it is possible to obtain it with numerical integration and all modern software -including Excel- has built-in functions to compute these probabilities. Computing probabilities for X N(µ, σ 2 ) Functions Matlab Excel pdf, φ(x) normpdf(x, µ, σ) normdist(x, µ, σ, 0) cdf, Φ(x) normcdf(x, µ, σ) normdist(x, µ, σ, 1) inverse cdf, Φ 1 (p) norminv(p, µ, σ) norminv(p, µ, σ) For a given p (0 p 1), inverse cdf finds the outcome value x such that the probability of observing smaller outcomes than x is exactly p. Note that if p = Φ(x) then x = Φ 1 (p). Note that Excel uses a single function to compute both pdf and cdf, use the last argument of normdist to specify whether you want pdf or cdf. Constraints: In general, there can be constraints on the inventory level y and the order size y x 1. Inventory cannot be disposed off: y x 2. Storage capacity: y M 3. Limitations on order sizes, perhaps because of transportation economies: a y x b where 0 a b No matter how many restrictions are posed on y, WLOG we assume that there are only two: a lower lb and an upper ub bound. Then we have a linearly constrained convex optimization problem min g(y) lb y ub where we can apply the KKT conditions. There are three possibilities: i) arg min y g(y) lb, then set y = lb. ii) arg min y g(y) ub, then set y = ub. iii) lb arg min y g(y) ub, then set y =arg min y g(y). Multiple products with a single capacity constraint: When there are I products whose demands and costs are indexed by i, we first compute the fractile solutions y i = F 1 ((c P i c i )/(c P i + C Hi )). If these inventory levels satisfy the capacity constraint I y i M i=1 we are done. Otherwise we know from KKT conditions that the constraint will be binding. Let λ 0 be the dual variable for the constraint, KKT conditions require g i(y) + λ = 0 for all i = 1... I, ( I y i M)λ = 0 i=1 2

3 Then it follows that λ should be increased starting from 0 until I i=1 F 1 i ( ) cp i (c i + λ) = M c P i + c Hi Note that now the underage cost is c P i (c i + λ) and is smaller by λ units from the original underage cost. The underage becomes chepaer once the dual price of the limited capacity is factored in. 1.1 Pricing in the Newsvendor Problem The basic principle of inventory management is matching the demand and the stocks. Traditionally the stock levels were the only decision variables. Recently, with ideas borrowed from economics theory, prices also become decision variables. By adjusting the price p, the demand can be regulated to better match the stocks. Naturally the demand mean becomes a function of theprice and we write µ(p). In almost all practical contexts, the demand mean decreases in p. Some examples are 1. µ(p) = a bp with a > 0 and b > µ(p) = ap b with a > 0 and b > 1. There generally are two ways to introduce randomness into the demand 1. Additive Randomness: D(p) = µ(p) + ξ where µ(p) = a bp. 2. Multiplicative Randomness: D(p) = µ(p)ξ where µ(p) = ap b. In order to keep D(p = 0) 0, ξ is lower bounded by a in the additive case and by 0 in the multiplicative case. In general, we assume that ξ [A, B] 1. Whether the demand randomness is additive or multiplicative, the revenue is pe(y D(p)). The costs are still g(y) so the expected profit becomes Π(y, p) = pe(y D(p)) g(y, p) where g(y, p) is the obvious extension of g(y), obtained by replacing D with D(p) in the expressions. Additive Randomness: We use the additive form to simplify Π(y, p). First let define the safety stock z := y µ(p). With this transformation and using E((µ(p) + ξ) (µ(p) + z)) = µ(p) + E(ξ) E(ξ z) +, the profit can be manipulated Π(z, p) = pe((µ(p) + ξ) (µ(p) + z)) c H E(z ξ) + c P E(ξ z) + c(µ(p) + z) = p(µ(p) + E(ξ)) c H E(z ξ) + (c P + p)e(ξ z) + c(µ(p) + z + E(ξ) E(ξ)) = (p c)(µ(p) + E(ξ)) (c H + c)e(z ξ) + (c P + p c)e(ξ z) + = Ψ(p) L(z, p) 1 When we compare a random variable against a real number, we think of inequalities in the almost sure sense, e.g. P (A D(p = 0) B) = 1 3

4 where Ψ(p) := (p c)(µ(p) + E(ξ)) is the profit function with perfect information. L(z, p) is almost the same function as we defined before except that the overage and underage costs are increased by c and p c in the current profit maximization framework. Finally, our problem is max Π(z, p). z,p As a first cut, consider the problem under perfect information i.e. max p Ψ(p). This problem has a concave objective function ( 2 Ψ/ p 2 = 2b < 0) and the optimal price is p 0 = a + bc + E(ξ) 2b The objective function Π(z, p) is concave in z and in p separately but not jointly. Thus, when p is fixed we obtain the newsvendor problem. When z is fixed, we set Π(z, p)/ p = 0 to find p(z) = p 0 E(ξ z)+ 2b Substituting this price back into the profit, we obtain a one dimensional optimization problem max z Π(z, p(z)). The objective function is not necessarily concave or not even unimodal. But it can be made unimodal with the following restrictions (Petruzzi and Dada 1999) 1. Condition 1: 2r(z) 2 + dr(z)/dz > 0 for A z B where r(z) is the failure rate function for ξ. 2. Condition 2: a b(c 2c P ) + A > 0. If the second condition fails, there could be multiple critical z values but the largest maximizes Π(z, p(z)). Multiplicative Randomness: We follow the same steps as in the additive randomness but with transformation z := y/µ(p) Π M (z, p) = pe((µ(p)ξ) (µ(p)z)) c H µ(p)e(z ξ) + c P µ(p)e(ξ z) + cµ(p)z = (p c)µ(p)e(ξ) (c H + c)µ(p)e(z ξ) + (c P + p c)µ(p)e(ξ z) + = Ψ M (p) L M (z, p) where superscript M is used to differentiate current functions from their additive counterparts which take superscript A later on. Our problem is to maximize the profit. The profit is concave in z and unimodal in p. We first find the optimal price for a fixed z by setting Π M / p = 0. p M (z) = p 0,M + b b 1 ( (c + ch )E(z ξ) + + c P E(ξ z) + ) µ E(ξ z) + where p 0,M is the optimal price under perfect information and p 0,M = bc/(b 1). It is easy to establish that p M (z) p 0,M. Similar to the additive randomness case, optimization over p, z is reduced to optimization over z by inserting p(z) instead of p: max z Π M (z, p(z)). Once more Π M (z, p(z)) can be made unimodal with extra conditions. Indeed Condition 1 of the additive case is still necessary while Condition 2 simplifies to b 2. 4

5 Thus, under these conditions, the optimal z can be found via dπ M (z, p(z))/dz = 0. Managerially, the curious issue is p A (z) p 0,A while p M (z) p 0,M. Additive or multiplicative randomness modelling yields seemingly conflicting results: in the former (latter) case the optimal price for a given z is always smaller (larger) than or equal to the price under perfect information. A possible explanation requires first observing that the coefficient of variation of the demand is increasing (nonincreasing) in p in the additive (multiplicative) case. When switching from perfect information and p 0 to a stocahstic model, the price is used to control the variability in the demand. If the variability is measured by the coefficient of variation, then price should decrease (increase) in the additive (multiplicative) case with respect to their full information price. 2 Rɛαδ Chapter 1 of Graves et al 3 Infinte Horizon Constant Demand: EOQ Model 3.1 Shelf-Inventory Dependent Demand A store experiences shelf-inventory I(t) dependent but predictable demand rate D(t) for a product as follows, { } D1 if I(t) J D(t) = if I(t) J D 2 where J is a given number of units and t denotes time. The store incurs usual ordering cost S (as oppposed to usual notation in K) and holding costs H. a) Do you expect D 1 > D 2 or D 1 < D 2 if the product is bread, why? Fresh food items have larger demand. High inventory indicates freshness so D 2 > D 1. b) Do you expect D 1 > D 2 or D 1 < D 2 if the product is a stylish necklace, why? Stylish products are more valuable when they are rare so D 2 < D 1. c) If the store orders Q units and Q J, what is the optimal Q that minimizes the sum of ordering and holding costs? Since Q J, D = D 1 always. This is the setting for the EOQ model so Q = min{ 2D 1 S/H, J}. d) Suppose that Q > J for the remaining parts except for part i). Draw the inventory level I (y-axis) as a function of time (x-axis). Clearly indicate Q, J, D 1 and D 2 on your graph. You could start the inventory level at Q at time zero, it drops with a slope of D 2 until the level reaches J. Then the inventory drops with a slope of D 1 until it reaches 0. As soon as the inventory reaches 0, it is brought back to Q. And the cycles continue in this manner. e) What is the length of an inventory cycle for a given Q? 5

6 From the graph, it takes (Q J)/D 2 to reach inventory level J and an additional J/D 1 to reach 0. Then the cycle time is (Q J)/D 2 + J/D 1. f) Write down the per time ordering cost? Check that your expression reduces to the traditional EOQ cost when D 1 = D 2 or J = 0. Per cycle ordering cost is S, per time ordering cost is S/((Q J)/D 2 ) + J/D 1 ). This expression reduces to the traditionl cost SD 2 /Q when D 1 = D 2 or J = 0. g) A friend of yours claims that the inventory holding cost is ( (Q + J)(Q J) H + J 2 ). 2D 2 2D 1 Determine if this per cycle or per time cost. If you say per cycle cost, then convert it to per time cost. This is per cycle cost, the two terms inside the parantheses are total inventory held during [0, (Q J)/D 2 ] and [(Q J)/D 2, J/D 1 ]. To convert it to per time cost, divide by the cycle length: ( ) (Q+J)(Q J) 2D 2 + J2 2D 1 H Q J D 2 + J. D 1 h) Write down the total cost T C in terms of Q. Take the derivative to establish that optimal Q solves 2SD 1 H = Q2 α + 2QJ(1 α) J 2 (1 α) (2) where α := D 1 /D 2. Specialize this to the case of α = 1/2 to establish that Q = 2J 2 + 4SD 1 /H J. T C(Q) = Taking the derivative: S + H (Q + J)(Q J)/(2D 2) + J 2 /(2D 1 ) (Q J)/D 2 + J/D 1 (Q J)/D 2 + J/D 1 = SD 1 Qα + J(1 α) + H 2 Q 2 α + J 2 (1 α) Qα + J(1 α) dt C(Q) dq = α SD 1 (Qα + J(1 α)) 2 + H 2 2αQ(Qα + J(1 α)) α(q 2 α + J 2 (1 α)) (Qα + J(1 α)) 2 Rearranging we obtain (2): 2SD 1 H When α = 1/2 = 2Q(Qα + J(1 α)) (Q2 α + J 2 (1 α)) = Q 2 α + 2QJ(1 α) J 2 (1 α). 4SD 1 H = Q2 + 2QJ J 2 = (Q + J) 2 2J 2. This yields Q = 2J 2 + 4SD 1 /H J. Note that when J = 0, Q = 4SD 1 /H = 2SD 2 /H as expected. i) Find the optimal lot size when J = 4, D 1 = 1, D 2 = 2, H = 1, S = 16. 6

7 For Q J = 4, Q = /1 > 4. We use Q = J = 4 in this region, total cost is S D 1 Q + H Q 2 = 16/4 + 2 = 6 For Q J = 4, (2) yields Q = /1 = 4( 6 1) = 5.8. Then the cycle time is (Q J)/D 2 + J/D 1 = (5.8 4)/2 + 4/1 = 4.9 and the total cost is T C(Q = 5.8) = ( )(5.8 4)/(2 2) + 42 /2 = = Thus, the overall optimal lot size is Q = Renewal Reward Theorem Let {X i } i 1 be times between events i 1 and i. Suppose that {X i } i 1 are iid. Let {R i } i 1 reward earned at the end (begining or during) event i. Suppose that {R i } i 1 are iid. We allow for R i to depend on X i but it should be independent of {X j : j 1, j i}. Define the cumulative number N(t) of events by time t as N(t) := i 0 1[ i X j t] where 1[.] is the indicator function taking events as arguments. Associated with N(t), we can define cumulative reward earned by t as C(t) := C(t) does not have to be monotone; it is nondecreasing (nonincreasing) if R i denotes revenue (cost) or zigzagging if R i is profit. C(t) can have discontinuities when rewards happen only at discrete time periods. A quantity of interest is the expected cumulative reward by time t, i.e. c(t) := E(C(t)). Theorem 1. i) Suppose E(X 1 ) <, then j=1 N(t) i=1 R i N(t) lim = 1 t t E(X 1 ) ii) Renewal reward theorem: In addition, if E R i <, then the limit below exists almost surely and 3.3 Supply Uncertainty Example C(t) lim t t = E(R 1) 2 E(X 1 ) Consider the EOQ model under two types of supply uncertainty: Random yield and random supply availability. Suppose that K is the ordering cost, h is the holding cost and demand rate is constant at D. 2 Warning: E(R 1)/E(X 1) E(R 1/X 1) 7

8 a) In the random yield case, Q R units are received when Q units are ordered where R is the random yield and E(R) = 1. Write down the expected total cost T C R (Q) incurred per time if order quantity is Q. Find the optimal order quantity Q R. We use the renewal reward theorem by dividing E(cost per cycle) to E(cycle length): T C R (Q) = K + he(q2 R 2 )/(2D) E(QR)/D = KD Q + hq 2 E(R2 ) and Q R = 2KD h 1 E(R 2 ) If you make a mistke here and set costs equal to E(cost per cycle/cycle length), you get ( K + h(q 2 R 2 ) )/(2D) E = KD QR/D Q E( 1 R ) + hq 2. Cleraly you get an incorrect expression that differs from the correct costs T C R (Q). b) Is Q R EOQ or Q R EOQ, prove your answer. Does your answer change when E(R) < 1? Clearly E(R 2 ) E(R) 2 = 1, Q R/EOQ = 1 E(R 2 ) 1 = 1 E(R) where the inequality follows from Jensen s inequality. On the other hand, when E(R) < 1 a counterexample can be constructed with R = 1/2 with probability 1 to show that Q R = 2EOQ. c) In the random supply availability case, min{q, C} units are received when Q units are ordered where C 0 is the random supply. Establish that the total cost T CC(Q) is given by T CC(Q) = KD E(min{Q, C}) + h E[(min{Q, C}) 2 ] 2 E(min{Q, C}) Now obtain the equation for optimal lot size, which should solve an equation of the form KD +? E[(min{Q, C}) 2 ] +? E(min{Q, C})+? = 0 where? are to be determined. Also compare the optimal solution to this equation with EOQ lot size. If you need, let the pdf and cdf of C be f and F. The total cost expression follows from computing total costs per cycle and cycle length and dividing them while using Renewal Reward Theorem. Now write where T CC(Q) = 1 E(min{Q, C}) {KD + h 2 E[(min{Q, C})2 ]} E(min{Q, C}) = E[(min{Q, C}) 2 ] = Q 0 Q 0 cf(c)dc + Q F (Q) c 2 f(c)dc + Q 2 F (Q) 8

9 Now take the derivative of the cost by noting that d 1 dq E(min{Q, C}) = F (Q) [E(min{Q, C})] 2 d dq E[(min{Q, C})]2 = 2Q F (Q) ( d dq T CC(Q) = F (Q) [E(min{Q, C})] 2 KD h ) 2 {2QE(min{Q, C}) + E[(min{Q, C})2 ]} Therefore the critical Q solves Now note that KD h 2 {2QE(min{Q, C}) E[(min{Q, C})2 ]} = 0 2QE(min{Q, C}) E[(min{Q, C}) 2 ] = 2Q after some algebra Thus we obtain that Q 2QE(min{Q, C}) E[(min{Q, C}) 2 ] = Q 2 0 Q cf(c)dc + 2Q 2 F (Q) c 2 f(c)dc Q 2 F (Q) Q 0 0 (Q c) 2 f(c)dc Q 2 KD h 2 {2EOQ E(min{EOQ, C}) E[(min{EOQ, C})2 ]} 0 (+) because EOQ solves KD h 2 {Q2 } = 0 Equality in Inequality (+) can be obtained by increasing Q if we show that Q 2 Q 0 (Q c)2 f(c)dc is increasing in Q. This can be shown by taking the derivative of Q 2 Q 0 (Q c)2 f(c)dc with respect to Q: 2Q Q 0 2(Q c)f(c)dc = 2Q 2QF (Q) + As a result the critical Q value with random supply is larger than the EOQ. Q 0 2cf(c)dc 0 4 Finite Horizon Dynamic and Deterministic Demand: Wagner-Whitin Model We will actually study a more generalized model than Wagner-Whitin. But we first need an important result from optimization. Given a concave objective function f : R N R, a matrix A of size M N and column vector b of size m 1, consider min{f(x) : Ax = b, x 0}. 9

10 This is a problem of minimizing a concave function over a convex set. It is well known that the optimal solution can be found at one of the extreme points of the feasible set {Ax = b, x 0}. This result can be applied directly to network flow formulations with concave costs. Given a network with node and arc sets N and A. Suppose we have concave arc costs c i,j (x) : R R and consider min c i,j (x i,j ) : x i,j x j,k = b j, x i,j 0. (i,j) A (i,j) A (j,k) A We call the extreme points of the network flow polytope as extreme flows. It is easy to see that concave cost network flow problem will be optimized by an extreme flow solution. A cycle (not necessarily directional) in the residual network (with respect to flows x i,j ) can be written as a sum of two flows. Thus, extreme flows cannot contain cycles. For example, eliminating cycles in a network flow problem with linear costs drives the solution towards optimality and such algorithms are known as cycle cancelling algorithms. Many production and inventory problems can be formulated as concave cost network flows. In these formulations, it may be possible to generate a super supply node with the only negative b j value. Then all the flows initially come out of the super supply node. If there is a node with incoming flows on two arcs, then the associated solution is not an extreme flow; because the two paths cominig from the super supply node to this node constitute a cycle. This fact is apparent in Wagner-Whitin network but we will illustrate it on a problem with N serial facilities which reduces to the Wagner-Whitin model when N = 1. Consider a production operation that requires N serial steps each of which is performed at a single facility. Also let {d t } T t=1 be given deterministic demands for the final product. We use i and t for production step and time indices. Concave production cost function c i t is for facility i and time t. It is possible to hold work in process inventories after finishing step i in period t, let the associated concave cost function be h i t. These data is sufficient to set up the problem. Figure 1 contains an example with N = 4 and T = 5 and d 1 = 5, d 2 = 1,d 3 = 3, d 4 = 2 and d 5 = 4. It also illustrates an extreme flow solution. Now let x i t be the production at step i in period t, similarly define inventory y i t held at that step. It is possible to express inventory y i t in terms of production. For that collapse all the nodes {(τ, i) : 1 τ t} into a single node and write conservation of flow for that node: inflow = outflow t x i τ = τ=1 t τ=1 x i+1 τ + y i t We assume that ending inventories should be zero y i T = 0. Inventory shortages are not allowed, i.e. yi t 0. Production amount must be nonnegative,i.e. x i t 0. Putting all these four types of constraints along with the objective objective function below, we formalize the model: min T t=1 i=1 N c i t(x i t) + h i t(yt) i Now the extreme flow idea can be reiterated by requiring y i t 1 xi t = 0: If we bring in inventory (product that has gone through step i) to step i from period t 1, we should not perform any step i operation in period t. Or if we perform this operation, the incoming inventory must be zero. Once the superscript i is dropped, we 10

11 Super Supply Node Facilities: Demands: Time periods: Figure 1: A numerical example. Vertical (horizontal) arcs are for production steps (inventory). Only flows on the thick arcs are positive. actually obtain the familiar Wagner-Whitin valid (for optimal solution) complementarity equality y t 1 x t = 0. Let us suppose that costs are stationary in time and further suppose that h i t is nondecreasing in i. Under this conditions extreme flows have the following nestedness property: Nestedness property: A feasible flow is nested if x i t = 0 implies x j t = 0 for all j < i in all time periods t. It is possible to prove under the assumptions of stationary costs and nondecreasing h i t in i that a nested flow can give the optimal solution to the serial production system with concave cost formulated as a network flow problem. The flow in Figure 1 is also nested in addition to being extreme. Note that with the valid complementarity constraint we have restricted the set of solutions to search for optimality. With nestedness property, we are imposing yet another restriction. However, we luckily do not have to search for the optimal solution, instead we use an efficient dynamic program. Let C f q,s be the optimal cost of producing at stage f in period q and using that production to meet demands of periods from q to s 1. Observe that how this cost definition uses extreme flow and nestedness properties. Because of the extreme flow property, we know that if the production at step f in time q is used to meet demand d s 1, it will also be used to meet all the demands between q and s 1. Without this property we need to define a cost for each subset of time periods, clearly a nonpolynomial affair. Because of nestedness property, we know that the production at step f in time q must be used to meet d q ; otherwise production can be delayed until the next period without increasing the costs. The cost C f q,s is actually for a subsytem with f... N steps and q... s 1 time periods; see Figure 2. For simplicity let us suppose that production costs have two components a fixed cost and a proportinal (to production quantity) cost. It is possible to argue that production costs are irrelevant for our objective function and set them equal to zero without loss of generality. Let K f denote the remaining fixed costs for 11

12 Facilities f C f qs K f h f h f h f C f rs C f+1 qr f+1 N Time q-1 q q+1 r-1 r r+1 s-1 s periods Figure 2: Justification of DP recursion under extreme flow and nestedness properties. step f. Also let h f be the linear (still concave) holding costs for work in process inventory at step i. In computing C f q,s, we know that flow x f q will be split into two. The first part will go into step f + 1 for production in period q. The second part will be kept in the inventory until period r, p < r s. Note that r = p violates the nestedness property. On the other hand, r = s means no inventory is kept as the second part of the flow x f q. Looking at Figure 2, we can decompose C f q,s as follows Cq,s f = min K s 1 f + (r q)h f d t + Cq,r f+1 + Cr,s f for 1 f N, 1 q < s T + 1 q<r s t=r where minimization over r ensures that Cq,s f is the optimal cost as long as Cq,r f+1 finish the DP formulation after specifying the boundary conditions and C f r,s are so. We can Now we are in a position to provide an algorithm: C N+1 q,s = 0 for all q, s; C f s,s = 0 for all s, f. Set q = T, s = T + 1, f = N and C f q,s = K F (start at the bottom right). Repeat If s > q + 1, compute C f q,s 1 and set s := s 1 (move to right). else if s = q + 1 and q > 1, compute C f q 1,T +1 and set q := q 1 (consider periods q... T + 1). else if s = q + 1, q = 1 and f > 1, compute C f,f 1 T,T +1 and set f := f 1 (move up). until s = T + 1, q = 1 and f = 1 (arrival at the top left). 12

13 Stop C1,T 1 +1 is the optimal solution. Finally observe that there are O(N T 2 ) cost computations for {C f q,s : 1 f N, 1 q < s T + 1}. Each cost computation involves a minimization (actually comparison of T numbers, one for each possible r) so it takes O(T ). In total the algorithm takes O(N T 3 ). This DP algorithm isvery efficient and can be used in practical situations. Its efficiency is inherited from the Wagner-Whitin algorithm which owes its efficiency to extreme flow property. However, the current algorithm exploits another property, nestedness, which does not arise in Wagner-Whitin context. The discussion here is mainly based on Love 1972, more details can be found in there. For a moree recent and general treatment of minimum concave-cost network flow problems see Guisewite and Pardalos Infinite Horizon Stationary and Stochastic Demand Consider a continuous time inventory model. Let t be the continuous time variable. Let NI(t) be the (net) inventory level at time t. Inventory level is the on-hand (physically availabale at the warehouse) inventory minus all the backorders at any time. Inventory related costs are based on inventory level. It can be negative with backorders. Also suppose that NI(0) = 0 and Q 1 is placed at t = 0. Let L be the lead time for deliveries; time from when an order is placed until it arrives. L is an uncontrollable deterministic parameter for most of our discussion. If you are at time t, the orders that will be placed after t cannot affect the costs in [t, t + L). This is because such orders will all arrive after t + L. Suppose that cost incurred at time t is computed by function g : R R +. It is very commnon to take g(y) = hy + + by where h and b are holding and backorder costs, both measured per unit and per time. Then g(ni(t)) is the rate at which the inventory costs are incurred at time t. In addition, every time an order is placed a fixed cost of K is incurred. Let {D(t), t 0} denote the demand stochastic process. Demand over an interval (s, t] is denoted by D(s, t]. We also define demand rate as ED(s, t] λ := t s Identically distributed demand means for any s, t, u 0, D(t, t + u] and D(s, s + u] have the same distribution. Note that we allow for (t, t + u] (s, s + u]. For stochastic processes, identically distributed process and stationary process are usually used synonmously. To define independence, we pick s, t, u 0 such that (t, t + u] (s, s + u] = and require D(t, t + u] and D(s, s + u] be independent. Let IIT (t) be the inventory in-transit (pipeline) that has been shipped before t but has not arrived yet at t. However, IIT (t) will definetely arrive by t + L. Some of IIT (t) may actually arrive before t + L. Inventory position IP (t) is defined as IP (t) := NI(t) + IIT (t). Although IP is not used for cost accounting, it has nice properties that will allow us to handle NI. Observe that IP (t) includes all the orders placed by t. Until after t+l, no orders will arrive except for those already 13

14 accounted for in IP (t). However, demand of D(t, t + L] will be materialized until then. These observations let us to obtain the distribution of NI(t + L) given that IP (t) is known NI(t + L) = IP (t) D(t, t + L] (3) Since the demand can be a general process, we must be careful in writing demands over intervals to avoid double counting. Our convention is to exclude (include) the left (right) end point of an interval so we write D(s, t]. By (3), the costs at time t are then incurred after observing the demand at time t. We can now define two closely related policies for continuous inventory review. (R, Q) policy: Whenever IP (t) R, order Q units. (s, S) policy: Whenever IP (t) s, order S IP (t) units. Although (s, S) policies are generally discussed in the context of periodic review and proven to be optimal, they can be used in continuous review. Note that the diffrenece between (R, Q) and (s, S) policies are very small indeed. For example, when the demand happens in units of 1 (as opposed to batch demand arrivals) and s = R, S = Q + s, the sample paths of NI(t) will exactly be the same under both policies. 5.1 (R, Q) Cost formulas The challenging part in computing the expected cost per time is finding the long run average holding and backorder costs 1 T Eg(NI) := lim sup g(ni(t))dt T T Note that we are defining Eg(NI) from the above equation. It is not always necessary that Eg(NI(t)) will converge to Eg(NI) as t. For example, g should not jump around much, e.g. some fancy requirements, such as uniformly (in t) upper boundable Eg(NI(t)) and Eg(NI(t)) 2, are used. Under general conditions, it can be shown that lim Eg(NI(t)) = Eg(NI) t Thus, the next step should be obtaining the limiting distribution of NI(t) as t. This can be achieved with (3), if the limiting distribution of IP (t) is known. At this point we must appreciate why IP is useful, its limiting distribution is easy to find and that can be used as a stepping stone towards NI(t) limiting distribution. To compute IP (t) as t, first note its uniform (in t) range: IP (t) (R, R + Q]. For simplicity, suppose that demand is in units of one so that R and R + Q must be integers. Consider the continuous time Markov Chain {IP (t), t 0} which decreases from R+Q to R with demands and jumps back to R+Q. The important observation is that state transition times and probabilities are totally governed by the demand process {D(t) : t 0}. This is because the customers cannot know the inventory position. Because of the transition probability symmetry, in the long run all the states have the same limiting probability: t=0 lim P (IP (t) = i) = 1/Q for all states i t Then we can compute Eg(NI) by conditioning on the inventory position IP as Eg(NI) = E IP [Eg(NI) IP ] = E IP [E D g(ni = IP D(0, L])] 14

15 For convenience let G(y) := E D (g(y D)). Then Eg(NI) = E IP [G(IP )] = R+Q R G(y)dy/Q. The last expression is the expected holding and backorder cost accimulation rate (per time) in the long run. If we add to this the ordering cost per time K/(Q/λ), we obtain the total cost per time: C(R, Q) = Kλ + R+Q R G(y)dy Q where the B(R,Q) is average backorders per time B(R, Q) = = Kλ/Q + h(q/2 + R Lλ) + (h + b)b(r, Q) (4) R+Q R E(D y) + dy. (5) Q When the cumulative demand is a nondecreasing stochastic process with stationary increments and continuous sample paths (see Serfozo and Stidham 1978), the expected per time cost is given by (4). 5.2 (R, Q) optimization From the discussion in Zipkin 1986, C(R, Q) is jointly convex in their (R, Q). conditions C(R, Q) = G(R + Q) G(R) = 0 R C(R, Q) Q = Kλ/Q 2 R+Q In other words, we need R and Q to respectively solve Kλ + R G(R + Q) = G(R) R+Q R G(y)dy/Q 2 + G(R + Q)/Q = 0 G(y)dy = QG(R + Q). Consider the first order These equations have geometric interpretations, see Figure 3. The optimality condition on R says that the line piece between (R, G(R)) and (R + Q, G(R + Q)) is parallel to 0 y axis. The condition on Q says that Kλ is exactly the area between the curve G(y) and the line piece between (R, G(R)) and (R +Q, G(R +Q)). 6 Exercises 1. Establish the identity in (1). 2. If D N(25, 5 2 ) and y = 25, use (1) to compute the fill rate. 3. Either prove or find a counterexample for the following inequality: µ E(D x) + for a random variable D, µ = E(D) and x 0. If you can construct a counterexample, think of an additional condition on D to make the inequality valid. 4. Establish that when the randomness is multiplicative the optimal price under perfect information is bc/(b 1). 15

16 G(y) G(y) R+Q R Kλ G ( y) dy G ( R + Q) R Q R + Q y Figure 3: A geometric interpretation of optimal R and Q. 5. Sensitivity of EOQ costs to a finite planning horizon. Average inventory costs over an infinite planning horizon (T = ) in EOQ setting is known to be 2KDh. Consider a finite horizon problem T < during which n orders are made. The begining (t = 0) and ending (t = T ) inventory levels are constrained to be zero. a) Argue that optimal order quantites must be equal to each other. b) With n Z + orders, argue that the per time cost over [0, T ] is given by T C(n, T ) = nk/t + hdt/(2n). c) Show that optimal n(t ) for given T is found by searching for the minimum value of n that satisfies n(n + 1) hdt 2 /(2K). Note that n(t ) is a step-wise increasing function of T. d) Let C(T ) be the per time minimum cost if the planning horizon is T. Observe that C(T ) = min n Z + T C(n, T ) = n(t ) K T + hdt 2n(T ). 2K Dh ) = 2KDh, Draw C(T ) and point out how it relates to T C(n, T ) for n 1. Prove that C(T = k i.e. planning horizons that are integer multiples of an optimal EOQ cycle yield optimal EOQ costs. (Hint: Choose n = k to keep T/n constant and equal to EOQ cycle length.) e) As we observed in c), n(t ) switches from a k to k + 1 as T grows. Find the largest value of T (k) where k cycles are still optimal, i.e. at T (k) + ɛ, k + 1 cycles are optimal as discussed in c). Show that at the value of T (k) where the switch from k to k + 1 cycles happens ) k + 1 2KDh C(T = T (k)) = 1 2 ( k k f) Note that switching T values are local maxima of C(T ) and argue that if C(T = T (k)) C 0 for some C 0 then C(T ) C 0 for T T (k). Find the smallest value of k(2.5%) such that C(T = T (k)) KDh k 16

17 Compute the planning horizon T (k = k(2.5%)) value (in terms of EOQ cycle length 2K/(Dh)) corresponding to the k(2.5%) value. We have just proved that by using a planning horizon longer than T (k = k(2.5%)), the per time optimal cost of finite planning horizon problem can at most be 2.5% worse than the optimal cost of the infinite horizon problem (EOQ costs). 6. Prove that in order to minimize a concave function over a polytope, it sufficient to consider only the extreme points of the polytope. Hint: Use proof by contradiction, suppose x is the optimal solution but can be written as λx 1 + (1 λ)x 2 where x 1 and x 2 are in the polytope, show that f(x ) min{f(x 1 ), f(x 2 )}. Generalize this argument. 7. Prove that in order to minimize a concave function over a closed convex set, it sufficient to consider only the boundary of the convex set. 8. Refer to serial production system with concave costs. Suppose that a given (x, y) is an extreme flow and y i t 1 + xi t > 0, then argue that there exist integers τ 1 and τ 2, t τ 1 τ 2 T such that τ 2 yt 1 i + x i t = d τ. τ=τ 1 Express this property in a sentence. How (if) will you modify this equality if it is also known that (x, y) is a nested flow? 9. Refer to serial production system with concave costs. Motivate the fact that h i t is likely to be nondecreasing in i. Suppose that N = 2 and i = 1 is a production activity in the factory while i = 2 is a packaging activity at a distribution center. You may choose to give an argument for this special case. 10. Refer to serial production system with concave costs. Under the assumptions of stationary costs and nondecreasing h i t in i prove that a nested flow can give the optimal solution. Do you really need stationary cost assumption? For example if c i t is nonincreasing in time (no inflation), can you repeat the proof? 11. For the example in Figure 1, draw a feasible flow that is nested but not extreme. 12. Refer to serial production system with concave costs. When production costs have a fixed and proportional cost component, prove that proportinal costs can be set to zero without loss of generality. Does your proof depend on extreme flow or nestedness properties. If not, find a feature of the model that is necessary for your proof in the sense that your proof fails without this feature. References [1] G.M. Guisewite and P.M. Pardalos (1990). Minimum concave-cost network flow problems: Applications, complexity, and algorithms. Annals of Operations Research 25: [2] S.F. Love (1972). A facilities in series inventory model with nested schedules. ManSci Vol.18 No.5:

18 [3] N.C. Petruzzi and M. Dada (1999). Pricing and the newsvendor problem: a review with extensions. Operations Research Vol.47 No.2: [4] R. Serfozo and S. Stidham (1978). Semi-Stationary Clearing Processes. Stochastic Process Applications, Vol.6: [5] P. Zipkin (1986). Inventory Service-Level Measures: Convexity and Approximation. Management Science, Vol.32, No.8: HW Solutions Establish the identity in (1). E(D y) + = = y ( ξ µ (ξ y)φ σ (y µ)/σ ) dξ (σx + µ y)φ(x)dx = (µ y)(1 Φ((y µ)/σ)) + = (µ y)(1 Φ((y µ)/σ)) + σ (y µ)/σ ((y µ)/σ) 2 /2 σx 1 2π exp( x 2 /2)dx 1 2π exp( z)dz = (µ y)(1 Φ((y µ)/σ)) σ 1 exp( z) 2π ((y µ)/σ) 2 /2 ( ( )) ( ) y µ y µ = (µ y) 1 Φ + σφ. σ σ Either prove or find a counterexample for the following inequality: µ E(D x) + for a random variable D, µ = E(D) and x 0. If you can construct a counterexample, think of an additional condition on D to make the inequality valid. Clearly with D = 1 wp 1 fails µ E(D x) +. Then we require D 0 wp 1 so that µ = 0 Df(D)dD x (D x)f(d)dd = E(D x) + Thus, D 0 is a sufficient condition. However, D 0 is not necessary. To see this let D = 1 and D = 11 with equal probabilities and set x = 11, then µ = 5 > 0 = E(D 11) +. Establish that when the randomness is multiplicative the optimal price under perfect information is bc/(b 1). First recall So that Ψ M (p) p Ψ M (p) = (p c)ap b E(ξ) = ap b 1 (p (p c)b) = 0 p 0,M = bc b 1. 18

19 Sensitivity of EOQ costs to a finite planning horizon. Average inventory costs over an infinite planning horizon (T = ) in EOQ setting is known to be 2KDh. Consider a finite horizon problem T < during which n orders are made. The begining (t = 0) and ending (t = T ) inventory levels are constrained to be zero. a) Argue that optimal order quantites must be equal to each other. Suppose that there are two consecutive orders Q 1 and Q 2 with Q 1 Q 2. Then the corresponding cycle times are Q 1 /D and Q 2 /D respectively, and the total cost in the two cycles are C(Q 1, Q 2 ) = 2K + hq2 1 2D + hq2 2 2D. Now consider an alternative: order Q = (Q 1 + Q 2 )/2 twice. Then we have to identical cycles with cycle time (Q 1 + Q 2 )/(2D) (note the total time of two cycle does not change), and the resulting total cost in the two cycles are C(Q, Q) = 2K + 2 h[(q 1 + Q 2 )/2] 2 2D = 2K + h(q 1 + Q 2 ) 2. 4D Since Q Q2 2 (Q 1 + Q 2 ) 2 /2 = (Q 1 Q 2 ) 2 /2 > 0 for Q 1 Q 2, we deduce that C(Q 1, Q 2 ) > C(Q, Q). Thus the order sizes must be equal. b) With n Z + orders, argue that the per time cost over [0, T ] is given by T C(n, T ) = nk/t + hdt/(2n). Since beginning (t = 0) and ending (t = T ) inventories are restricted to be zero, and the order quantities are the same, T is divided into n identical cycles. That is, the cycle time is T/n. It follows that the order quantity is T D/n. Then the cost per cycle is K + hdt 2 /(2n 2 ). Hence we can derive the cost per time is T C(n, T ) = K + hdt 2 /(2n 2 ) T/n = nk/t + hdt/2n. (6) c) Show that optimal n(t ) for given T is found by searching for the minimum value of n that satisfies n(n + 1) hdt 2 /(2K). Note that n(t ) is a step-wise increasing function of T. We take the first order difference with respect to n in (6) and get n T C(n, T ) = T C(n + 1, T ) T C(n, T ) = K/T + hdt/(2(n + 1)) hdt/(2n) = K T n(n + 1) [n(n + 1) hdt 2 /(2K)]. (7) It is easily checked that (6) is convex in n. Thus the optimal n(t ) is an integer such that n T C(n, T ) < 0 for n < n(t ) and n T C(n, T ) > 0 for n > n(t ). Then, from (7), n(t ) can be chosen as the smallest number which satisfy n(n + 1) hdt 2 /(2K). 19

20 C(T ) 2KDh q 2K Dh 2 q 2K Dh T Figure 4: The per time cost C(T ) of finite horizon EOQ model as a function of the horizon length T. d) Let C(T ) be the per time minimum cost if the planning horizon is T. Observe that C(T ) = min T C(n, T ) = n(t ) K n Z + T + hdt 2n(T ). 2K Draw C(T ) and point out how it relates to T C(n, T ) for n 1. Prove that C(T = k Dh ) = 2KDh, i.e. planning horizons that are integer multiples of an optimal EOQ cycle yield optimal EOQ costs. (Hint: Choose n = k to keep T/n constant and equal to EOQ cycle length.) C(T ) is drawn in Figure 4. As you may expect C(T ) converges to 2KDh as T. Now put T = k (2K)/(Dh) in the condition derived in part c), then n(n + 1) hdt 2 /(2K) = k 2. So n(t ) = k. Then ( C T = k ) K (2K)/(Dh) = k k (2K)/(Dh) + hdk (2K)/(Dh) 2k = 2KDh. Note that horizon lengths T that are integer multiples of an optimal EOQ cycle length yield optimal EOQ costs in Figure 4. e) As we observed in c), n(t ) switches from a k to k + 1 as T grows. Find the largest value of T (k) where k cycles are still optimal, i.e. at T (k) + ɛ, k + 1 cycles are optimal as discussed in c). Show that at the value of T (k) where the switch from k to k + 1 cycles happens ( C(T = T (k)) = 1 ) k k + 1 2KDh 2 k k 20

21 Since T (k) is the largest where k cycles are still optimal, Solving this yields T (k) = k(k + 1) 2K/(hD). Thus, k(k + 1) = hd[t (k) 2 ]/(2K). (8) K C(T = T (k)) = k + hd k(k + 1) 2K/(hD) k(k + 1) 2K/(hD) 2k ( = 1 ) k k + 1 2KDh. 2 k (9) k f) Note that switching T values are local maxima of C(T ) and argue that C(T = T (k)) is nonincreasing in k. Find the smallest value of k(2.5%) such that C(T = T (k)) KDh Compute the planning horizon T (k = k(2.5%)) value (in terms of EOQ cycle length 2K/(Dh)) corresponding to the k(2.5%) value. We have just proved that by using a planning horizon longer than T (k = k(2.5%)), the per time optimal cost of finite planning horizon problem can at most be 2.5% worse than the optimal cost of the infinite horizon problem (EOQ costs). We first show that local minima of C(T ) is between the switching T values. Fix k 1 arbitrarily. We only examine {T : T (k 1) T T (k)} where n(t ) = k We argue that T (k 1) arg min T C(k, T ) T (k) T Note that arg min T T C(k, T ) satisfies k 2 = hdt 2 /2K while T (k) satisfies k(k+1) = hdt 2 /2K. Thus, min T T C(k, T ) T (k). Similiarly argue for T (k 1) arg min T T C(k, T ). Since C(T ) is convex and its local minima are between switching T values, C(T ) is locally maximized at the switching T values. It is instructive to go over these arguments in view of Figure 4. From (9), C(T = T (k + 1)) C(T = T (k)) = 1 2 2KDh ( k + 1 k k + 2 k + 1 k k + 1 ) k k The last inequality follows from the fact that. is a concave function and k/(k +1) (k +1)(k +2) (k + 2)(k + 1) (k + 1)(k). Consequently, C(T = T (k)) is nonincreasing in k. Combining this with the fact that local maxima happens only at the switching points, we establish that for T 1 T 2 C(T 2 ) C(T (k(t 2 ) 1)) C(T (k(t 1 ) 1)) In English, cost at any T is smaller than or equal to costs at all switching T values that are smaller than T. This upper bound on the cost helps us to obtain worst case performance. ( ) To get k(2.5%), find the minimum k such that 1 k 2 k+1 + k+1 k That gives k = 2. From (8), T (2) = 2 2K/(Dh). As a result, if you have a finite horizon longer than 2 EOQ cycles, per time costs will be only 2.5% worse than EOQ costs: Per time costs are not sensitive to horizon length T. 21

22 Prove that in order to minimize a concave function over a polytope, it sufficient to consider only the extreme points of the polytope. Suppose x is the minimizer of a concave function f( ) over a polytope P. If x is not an extreme point, then x is a convex combination of extreme points of P, i.e., x = i I λ ix i where λ i 0 and i I λ i = 1, and {x i P i I} is the set of extreme points of P. Then by concavity f(x ) i I a i f(x i ) i I a i min i I f(x i) = min i I f(x i) If f is strictly concave, we have a contradiction. If f is not strictly concave, the minimality of x implies f(x ) = min i I f(x i ). That is, there is an extreme point in P which also minimizes f. Hence, it is sufficient to check the minimizer only at the extreme points. Refer to serial production system with concave costs. Suppose that a given (x, y) is an extreme flow and y i t 1 + xi t > 0, then argue that there exist integers τ 1 and τ 2, t τ 1 τ 2 T such that τ 2 yt 1 i + x i t = d τ. τ=τ 1 Express this property in a sentence. How (if) will you modify this equality if it is also known that (x, y) is a nested flow? First, observe that there cannot exist any demand d t which is only partially satisfied by the inflow yt 1 i + xi t into node (i, t) due to extreme flow property. Then let T be a subset of {1,..., T } and we can always write yt 1 i + x i t = d τ. τ T But in that case attempt to pick a τ / T such that τ 1 < τ < τ 2 and τ 1, τ 2 T. If no such τ can be found, the proof is complete. If there is such a τ then the flow from the source to node (N, τ) and the flow from (i, t) to (N, τ 2 ) collide at a node because the graph is planar. But this collison also violates the extreme flow property, establishing a contradiction no such τ can exist. In English: Inflow to any node is the sum of demands of some adjacent periods starting at or after the period the node belongs to. In case of nestedness, τ 1 = t. In English: Inflow to any node is the sum of demands of some adjacent periods starting at the period the node belongs to. 22