Inventory Control Subject to Uncertain Demand Solutions To Problems From Chapter 5

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1 Inventory Control Subject to Uncertain Demand Solutions To Problems From Chapter A newsboy keeps careful records of the number of papers he sells each day and the various costs that are relevant to his decision regarding the optimal number of newspapers to purchase. For what reason might his results be inaccurate? What would he need to do in order to accurately measure the daily demand for newspapers? 5.7 If he only keeps track of the number of sales, he has no way to accurately estimate the demand since demand = sales + lost sales. He would need some way to gauge the lost sales. One method would be to increase his supply for a period of time so that he would be able to meet all demand. 5.8 Billy s Bakery bakes fresh bagels each morning. The daily demand for bagels is a random variable with a distribution estimated from prior experience given by Number of bagels Probability sold in one day The bagels cost Billy s 8 cents to make, and they are sold for 35 cents each. Bagels unsold at the end of the day are purchase by a nearby charity soup kitchen for 3 cents each. a. Based on the given discrete distribution, how many bagels should Billy s bake at the start of each day? (Your answer should be a multiple of 5) b. If you were to approximate the discrete distribution with a normal distribution, would you expect the resulting solution to be close to the answer that you obtained in part (a)? Why or why not? c. Determine the optimal number of bagels to bake each day using a normal approximation. (Hint: you must compute the mean µ and the variance σ 2 of the demand from the given discrete distribution) 5.8 a) c 0 = =.05 c u = = Critical ratio = = From the given distribution, we have: 5-1

2 Q f(q) F(Q) < Since the critical ratio falls between 20 and 25 the optimal is Q = 25 bagels. b) The answers should be close since the given distribution appears to be close to the normal. c) µ = xf(x) = (0)(.05) + (5)(.10) +...+(35)(.05) = 18 σ 2 = x 2 f(x) - µ 2 = (18) 2 = 78.5 σ = (2)(32)(1032).36 = 8.86 The z value corresponding to a critical ratio of is Hence, Q* = σz + µ = (8.86)(1.01) + 18 = ~ Happy Henry s car dealership sells Spyker cars. Once every three months, a shipment of the cars is made. Emergency shipments can be made between these three-month intervals to resupply the cars when inventory falls short of demand. The emergency shipments require two weeks and buyers are willing to wait this long for the cars, but will generally go elsewhere before the next three-month shipment is due. From experience, it appears that the demand for the cares over a three-month interval is normally distributed with a mean of 60 and a variance of 36. The cost of holding a car for one year is $500. Emergency shipments cost $250 per car over and above normal shipping costs. a. How many cars should Happy Henry s be purchasing every three months? b. Repeat the calculations, assuming that excess demands are back-ordered from one threemonth period to the next. Assume a loss-of-goodwill cost of $100 for customers having to wait until the next three-month period and a cost of $50 per customer for bookkeeping expenses. c. Repeat calculations, assuming that when Happy Henry s is out of stock, the customer will purchase the car elsewhere. In this case, assume that the cars cost Henry an average of $10,000 and sell for an average of $13,500. Ignore loss-of-goodwill costs for this 5-2

3 calculation a) A period is three months. Holding cost per year is $500, which means that the cost for a 3-month period is: 500/4 = $125 = c 0 c u = 250 (emergency shipment cost) Critical ratio = = 2 3 =.667 z =.44. Hence Q * = σz + µ = (6)(.44) + 60 = cars b) c u = 150 Critical ratio = c 0 = 125 =.5454 F(Q * ) =.5454 z =.11 Q * = σz + µ = (6)(.11) + 60 = ~ 61 cars. c) This corresponds to an infinite horizon problem with lost sales. From part 3 of Appendix B at the end of the chapter. c u = lost profit = $3,500 c 0 = holding cost = 125 Critical ratio = =.9655 z = 1.82 which gives Q * = (6)(1.82) + 60 = ~ 71 cars Weiss s paint store uses a (Q, R) inventory system to control its stock levels. For a particularly popular white latex paint, historical data show that the distribution of monthly demand is approximately normal, with mean 28 and standard deviation of 8. Replenishment lead time for this paint is about 14 weeks. Each can of paint cost the store $6. Although excess demands are back-ordered, the store owner estimates that unfulfilled demands costs about $10 each in bookkeeping and loss-of-goodwill costs. Fixed costs of replenishment are $15 per order, and holding costs are based on a 30 percent annual rate of interest. a. what are the optimal lot sizes and reorder point for this brand of paint? b. b. What is the optimal safety stock for this paint? 5-3

4 5.14 a) Note: We assume 4 weeks/month and 48 weeks/year. Monthly demand is normal (µ = 28, σ = 8) γ = 14 weeks = 3.5 months LTD ~ normal with µ = (28)(3.5) = 98 σ = (8) 3.5 = 15 h = Ic = (.3)(6) = 1.8 λ = (28)(12) = 336 year p = 10 K = 15 µ = 98 Q 0 = EOQ = (2)(336)(15) 1.8 = 75 F(R 1 ) = (75)(1.8) (10)(336) =.04 z = 1.75 giving R 1 = σz + µ = 124 and n(r 1 ) = σl(z) =.2426 Q 1 = (2)(336) 1.8 (15 + (10)(.2426)) = 81 F(R 2 ) + (81)(1.8) (10)(336) =.0434 z = 1.71 R 2 = σz + µ = 124. Since R 2 = R 1, we stop. Conclude that (Q,R) = (81,124) b) S = R - µ = = 26 units After taking a production seminar, Al Weiss, the owner of Weiss s paint store mentioned in Problem 5.14, decides that his stock-out costs of $10 may not be very accurate and switches to a service model. He decides to set his lot size by the EOQ formula and determines his order point so that there is no stock-out in 90 percent of the order cycles. a. a. find the resulting (Q,R) values. b. Suppose that, unfortunately, he really wanted to satisfy 90 percent of his demands (i.e., achieve a 90 percent fill rate). What fill rate did he actually achieve from the policy determined in part (a)? 5-4

5 5.15 a) Type 1 service of 90% EOQ = 75 (from 13 (a)) F(R) =.10, z = 1.28, R = σz + µ = (15)(1.28) + 98 = 117 Q,R) = (75,117) b) Find Type II service level achieved in part (a). n(r) Q = 1 β = σl(z) Q = (15)(.0475) 75 =.0095 β =.9905 (99.05% service level) 5-5