IEOR 4000: Professor Guillermo Gallego Production Management December 2, 2003 1 Stochastic Demand In this section we discuss the problem of controlling the inventory of a single item with stochastic demands. We start by studying the single period problem, also known as the Newsvendor Problem and then extend it to multi-period and in¯nite-horizon problems with and without setup costs. 1.1 The Newsvendor Problem Let D denote the one period random demand, with mean ¹ = E[D] and variance σ2 = Var[D]. Let c be the unit cost, c(1+m) the selling price and c(1¡d) the salvage value. You can interpret m as the retail markup and d as the salvage discount. If Q units are ordered, then min(Q; D) units are sold 1 and (Q¡D)+ units are salvaged. The pro¯t is given by c(1+m) min(Q; D)+c(1¡d)(Q¡D)+ ¡cQ. Taking expectations we ¯nd the expected pro¯t: ¼(Q) = c(1 + m)E min(Q; D) + c(1 ¡ d)E(Q ¡ D)+ ¡ cQ Using the fact that min(Q; D) = D ¡ (D ¡ Q)+ we can write the expected pro¯t as ¼(Q) = cm¹ ¡ G(Q) where G(Q) = cdE(Q ¡ D)+ + cmE(D ¡ Q)+ ¸ 0: This allow us to view the problem of maximizing ¼(Q) as the problem of minimizing the cost G(Q). For convenience let h = cd and p = cm. It is convenient to think of h as the per unit overage cost and of p as the per unit underage cost. Sometimes the underage cost is inflated to take into account the ill-will cost associated with unsatis¯ed demand. Later, in the study of multi-period problems, we will call h the holding cost rate and p the shortage cost rate per. Let g(x) = hx+ + px¡, then G(Q) can be written as G(Q) = E[g(Q ¡ D)]. Since g is convex and convexity is preserved by linear transformations and by the expectation operator it follows that G is also convex. det Let G (Q) = h(¹¡Q)+ +p(Q¡¹)+. This represents the cost when D is deterministic. Clearly Q = ¹ minimizes Gdet and Gdet(¹) = 0, so ¼det(¹) = cm¹. By Jensen's inequality G(Q) ¸ Gdet(Q). As a result, ¼(Q) · ¼det(Q) · ¼det(¹) = cm¹. If the distribution of D is continuous, we can ¯nd an optimal solution by taking the derivative of G and setting it to zero. Since we can interchange the derivative and the expectation operators, it follows that G0(Q) = hE±(Q ¡ D) ¡ pE±(D ¡ Q) where ±(x) = 1 if x ¸ 0 and zero otherwise. Consequently, G0(Q) = hP r(Q ¡ D ¸ 0) ¡ pP r(D ¡ Q ¸ 0): Setting the derivative to zero reveals that P r(D · Q) = ¯; (1) p where ¯ = h+p . If F is continuous then there is at least one Q satisfying Equation (1). We can select the smallest such solution by letting Q¤ = inffQ ¸ 0 : P r(D · Q) ¸ ¯g: (2) 1 + ¡ We will use x+ = x = max(x; 0) and x¡ = x = max(¡x; 0) to denote the positive and the negative part of a number. IEOR 4000: Production Management page 2 Professor Guillermo Gallego If F is strictly increasing then F has an inverse and there is a unique optimal solution given by Q¤ = F ¡1(¯): (3) In practice, D often takes values in the set of natural numbers N = f0; 1;:::g. In this case it is useful to work with the forward di®erence ¢G(Q) = G(Q + 1) ¡ G(Q), Q 2 N . By writing P1 E(D ¡ Q)+ = j=Q P r(D > j), it is easy to see that ¢G(Q) = h ¡ (h + p)P r(D > Q) is non-decreasing in Q, and that limQ!1 ¢G(Q) = h > 0, so an optimal solution is given by Q = minfQ 2 N : ¢G(Q) ¸ 0g, or equivalently, Q¤ = minfQ 2 N : P r(D · Q) ¸ ¯g; (4) The origin of the Newsvendor model appears to date back to the 1888 paper by Edgeworth [2] who used the Central Limit Theorem to determine the amount of cash to keep at a bank to satisfy random cash withdrawals from depositors. The fractile solution (1) appeared in 1951 in the classical paper by Arrow, Harris and Marchak [1]. The newsvendor solution can be interpreted as providing the smallest supply quantity that guar- antees that all demand will be satis¯ed with probability at least 100¯%. Thus, the pro¯t maximizing solution results in a service level 100¯%. In practice, managers often specify ¯ and then ¯nd Q ac- cordingly. The service level measure implied by the Newsvendor problem should not be confused with the fraction of demand served, or ¯ll-rate, which is de¯ned as ® = E min(D; Q)=ED. 1.2 Normal Demand Distribution An important special case arises when the distribution D is normal. The normal assumption is justi¯ed by the Central Limit Theorem when the demand comes from many di®erent independent or weakly dependent customers. If D is normal, then we can write D = ¹ + σZ where Z is a standard normal random variable. Let ©(z) = P r(Z · z) be the cumulative distribution function of the standard normal random variable. Although the function © is not available in closed form, it is available in Tables and also in electronic spreadsheets. Let z¯ be such that ©(z¯) = ¯. In Microsoft Excel, for example, the command NORMSINV(0.75) returns 0.6745 so z:75 = 0:6745. Since P r(D · ¹ + z¯σ) = ©(z¯) = ¯, it follows that ¤ Q = ¹ + z¯σ (5) satis¯es Equation (3), so Equation (5) gives the optimal solution for the case of normal demand. ¤ The quantity z¯ is known as the safety factor and Q ¡ ¹ = z¯σ is known as the safety stock. ¤ It can be shown that E(D ¡ Q )+ = σE(Z ¡ z¯)+ = σ[Á(z¯) ¡ (1 ¡ ¯)z¯] where Á is the density of the standard normal random variable. As a consequence, ¤ ¤ ¤ G(Q ) = hE(Q ¡ D)+ + pE(D ¡ Q )+ ¤ ¤ = h(Q ¡ ¹) + (h + p)E(D ¡ Q )+ = hz¯σ + (h + p)σE(Z ¡ z¯)+ = hz¯σ + (h + p)σ[Á(z¯) ¡ (1 ¡ ¯)z¯] = (h + p)σÁ(z¯); so ¤ ¼(Q ) = cm¹ ¡ (h + p)σÁ(z¯) = cm¹ ¡ c(d + m)σÁ(z¯): IEOR 4000: Production Management page 3 Professor Guillermo Gallego ¤ ¤ In addition, since E min(D; Q ) = ED¡E(D¡Q )+, we can divide by ED and write the ¯ll-rate as ® = 1 ¡ cv[Á(z¯ ¡ (1 ¡ ¯)z¯] where cv = σ=¹ is the coe±cient of variation of demand. Since Á(z¯ ¡ (1 ¡ ¯)z¯ ¸ 0 is decreasing in ¯, it follows that the ® is increasing in ¯ and decreasing in cv. Numerical results show that ® ¸ ¯ for all reasonable values of cv, including cv · 1=3, which is about the highest cv value for which the normal model is appropriate. Example Normal Demand: Suppose that D is normal with mean ¹ = 100 and standard deviation σ = 20. If c = 5, h = 1 and p = 3, then ¯ = 0:75 and Q¤ = 100 + 0:6745 ¤ 20 = 113:49. Notice that the order is for 13.49 units (safety stock) more than the mean. Typing NORMDIST(.6574,0,1,0) in Microsoft Excel, returns Á(:6745) = 0:3178 so G(113:49) = 4 ¤ 20 ¤ :3178 = 25:42, and ¼(113:49) = 274:58, with ® = 97%. Table 1 gives z¯, Á(z¯) and ® (at cv = :2) for di®erent values of ¯. 100¯% z¯ Á(z¯) 100®% 50% 0 0.3989 92.0% 75% 0.6745 0.3178 97.0% 90% 1.2816 0.1755 99.1% 95% 1.6499 0.1031 99.6% 97.5% 1.9600 0.0584 99.8% 99% 2.3263 0.0267 99.9% Table 1: Normal solution for several values of ¯ 1.3 Poisson Distribution Another distribution that arises often in practice is the Poisson distribution. D is said to be Poisson with parameter ¸ > 0 if ¸k P r(D = k) = exp(¡¸) k = 0; 1; 2;::: k! The Poisson distribution arises as a limit of the binomial distribution with large n and small p via the relationship ¸ = np. For example, the number of customers that enter a store and makep a purchase can often be modeled as a Poisson distribution. It is well known that ¹ = ¸ and σ = ¸ so the coe±cient of variation σ=¹ becomes small for large ¸. When ¸ is large, the Poisson distributionp can be approximated by the Normal distribution with mean ¹ = ¸ and standard deviation σ = ¸. The following recursions, starting from P r(D = 0) = e¡¸ and E[D] = ¸, are useful in tabulating and solving problems involving the Poisson distribution: P r(D = k) = P r(D = k ¡ 1)¸=k; k = 1; 2;::: P r(D · k) = P r(D · k ¡ 1) + P r(D = k); k = 1; 2;::: E[(D ¡ k)+] = E[(D ¡ k + 1)+] ¡ P r(D ¸ k) k = 1; 2;::: An optimal value of Q is given by the smallest integer such that P (D · Q) ¸ ¯. Example Poisson: If D is Poisson with parameter ¸ = 25, and c = 5, h = 1 and p = 3, then ¤ ¤ ¯ = 0:75 and Q = 28 is optimal. To compute G(Q ) notice that G(Q) = h(Q¡¸)+(h+p)E(D¡Q)+, so G(28) = 6:48. Table 2 provides some of the values associated with the Poisson distribution. At Q = 28, E(D ¡ 28)+ = 0:87 so ® = 1 ¡ 0:87=25 = :97. IEOR 4000: Production Management page 4 Professor Guillermo Gallego k P r(D = k) P r(D · k) E(D ¡ k)+ G(k) ® 22 0.07 0.32 3.80 12.21 .85 23 0.08 0.39 3.12 10.48 .88 24 0.08 0.47 2.51 9.06 .90 25 0.08 0.55 1.99 7.95 .