数据模型与决策(运筹学)课后习题和案例答案(8).docx
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1、CHAPTER 19 INVENTORY MANAGEMENT WITH UNCERTAIN DEMANDReview Questions19.1-1 Freddie should consider the trade-off between too much inventory and too little inventory.19.1- 2 Based on Freddies data, 9, 10, and 11 are the only numbers of copies that he has sold in the past.19.1-3 The state of nature f
2、br each day is the number of requests to purchase a copy that will occur. The number of requests is the state of nature because it is the random variable.19.2-1 Only a single time period is needed because perishable products cannot be sold later.19.2- 2 The only decision to be made is how many units
3、 to order so they can be placed into inventory at the beginning of the period.19.2- 3 It is assumed that the demand during the period is uncertain but that the probability distribution of demand is known.19.2- 4 The unit cost of underordering is the decrease in profit that results from failing to or
4、der a unit that could have been sold during the period. The unit cost of overordering is the decrease in profit that results from ordering a unit that could not be sold during the period.19.2- 5 Bayes decision rule will make the same decision since both approaches are applying Bayes decision rule, b
5、ut with different payoffs, where one is to be maximized and the other minimized.19.2- 6 The service level is the probability that no shortage will occur.19.2- 7 Optimal Service level = Cunder / (Cunder + Cover).19.2- 8 The point at which the optimal service level hits the cumulative distribution fun
6、ction gives the optimal order quantity.19.2- 9 Yes.19.3- 1 There have been four stockouts of panoramic disposable cameras during the year with durations ranging from a few days to a couple weeks.19.4- 2 Recent complaints from distributors about delays in shipping the disposable panoramic camera have
7、 concerned the Vice President fbr Marketing and she is suggesting having more frequent production runs to keep the inventory better stocked.b)Purchase 4 additional $100 checks, for a total of $1,600.c) Service level fbr buying 0 = 0.3Service level fbr buying 1 = 0.55Service level fbr buying 2 = 0.75
8、Service level for buying 3 = 0.85Service level fbr buying 4 = 0.95Service level fbr buying 5 = 1Optimal service level = Cunder / (Cunder + Cover) = $49 / ($49 + $3) = 0.94.Buy 4 additional checks.19.7 a) Optimal service level = Cunder / (Cunder + Cover) = $3,000 / ($3,000 + $1,000) = 0.75.b) When th
9、e demand is less-than-or-equal-to Q. no shortage will occur. Therefore the probability that the demand is less-than-or-equal-to Q gives us the service level that corresponds to choosing the order quantity Q. Increasing Q increases this probability. Therefore the smallest order quantity that provides
10、 the service level L is the value of Q such that (demand Q) = L.c) 0=|i + Klg = 50+ (0.675)(15) = 60.19.8 a) When interpreting this problem as an inventory problem, overbooked reservations are the perishable products that are being placed into inventory.b) Cunder = lost fare = $250.Cover = cost of c
11、ertificate = $150.c) Service level for accepting 0 = 0.05Service level for accepting 1 = 0.15Service level fbr accepting 2 = 0.3Service level fbr accepting 3 = 0.45Service level fbr accepting 4 = 0.6Service level fbr accepting 5 = 0.75Service level for accepting 6 = 0.85Service level for accepting 7
12、 = 0.95Service level fbr accepting 8 = 1Optimal service level = Cunder / (Cunder + Cover) = $250 / ($250 + $150) = 0.625.Accept 5 overbooked reservation.19.9 a)。* is cut in half. R is unchanged. This will reduce the average monthly holding cost. However, it will increase the average monthly shortage
13、 cost (since the number of order cycles per month increases). The smaller setup cost reduces the average monthly setup cost.b)IIIIII2is unchanged. R is reduced from 12,000 to 10,000. This will reduce the average monthly holding cost (less safety stock to hold). The average monthly shortage costs and
14、 setup costs will be unchanged.c)0* is unchanged. R is reduced from 12,000 to 2,300. This will reduce the average monthly holding cost. The average monthly shortage costs and setup costs will be unchanged.IIIIII20* is cut in half. R is reduced from 12,000 to 2,300. This will reduce the average month
15、ly holding cost. However, it will increase the average monthly shortage cost (since the number of order cycles per month increases). The smaller setup cost reduces the average monthly setup cost.19.10 a)hUP2KD$3,000+ $L000$1,0002($L500)(900)$3,000d) Average inventory just before an order is received
16、 = 2,300 - 1,600 = 700. Average inventory just after an order is received = 700 + 12,837 = 13,537. Average inventory = (700 + 13,537) / 2 = 7,119.Average monthly holding cost = 7,119 ($0.30) = $2,136.Average number of orders per year = (8,000)/(12,837) = 0.62.Probability of stockout before order rec
17、eived = 0.25.Expected number of stockouts per month = (0.25)(0.62) = 0.155.Average stockout size = 700 / 2 = 350.Estimate of average delay per camera delayed = 1/15 week = 0.015 months. Average monthly shortage cost = (0.155)(350)(0.015)($ 10) = $8.Average number of setups per month = 8000 / 12,837
18、= 0.62.Average monthly setup cost = (0.62)($35000) = $1,860.Average monthly cost = $2/36 + $8 + $1,860 = $4,004.This is less than half the cost before the changes ($8,896).19.11 a)hUP2KD $8 + $l /2($40)(40)$8=60./? = a + L(b-a) = 5 + (0.8)(15 - 5)=13.b) R = pi+ Klq = 50+ (0.675)(15) = 60.d) Safety S
19、tock = R- mean = 60 - 50 = 10.e) If demand during the delivery time exceeds 60 (the order quantity), then the reorder point will be hit again before the order arrives, triggering another order.b)c) Average number of orders per year = (40)(12) / 60 = 8. Probability of a stockout before order received
20、 = 0.2.Average number of stockouts per year = (8)(0.2) = 1.6.19.12 a)LC = KloCase 1 =$1, 0=1Case 2 /z=$1009 o=lCase 3 =$1, 5=100Case 4 =$100, o=l 000.5$0$0$0$00.750.67567.567.56,7500.91.282128.2128.212,8200.951.645164.5164.516,4500.992.327232.7232.723,2700.9993.098309.8309.830,980b)ALACCase 1 h=$l,C
21、ase 2 h=$100,Case 3 h=$l,o=100Case 4 h=$100, o=l 000.25$0,675$67.5$67.5$6,7500.150.60760.760.76,0700.050.36336.336.33,6300.040.68268.268.26,8200.0090.77177.177.17,710c) As the service level gets higher, increasing the service level further costs more for smaller increases. Thus, there will be dimini
22、shing returns when raising the service level further and further. You should balance the cost of the safety stock with the cost of stockouts to determine the best service level.19.13 a) C = hKLo = ($ 100)( 1.282)( 100) = $ 12,820.b) o =( 100 = (4)cti =5i = 50.If the lead time were 1 day, then C = hK
23、hS = ($100)(1.282)(50) = $6,410.This is a 50% reduction in the cost of the safety stock.c) o = (8)(50) = 141.4.C = hKLs = ($ 100)(1.282)(141.4) = $18,127. This isa41% increase.d) The lead time would need to quadruple to 16 days.19.14 a) The safety stock would drop to zero.b) The safety stock would d
24、ecrease.c) The safety stock is unchanged for a given service level. However, with higher shortage costs, there will be an incentive to increase the service level (with a correspondingly higher level of safety stock).d) The safety stock increases.e) The safety stock doubles.f) The safety stock double
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