# Pdfballou09

Páginas: 46 (11288 palabras) Publicado: 28 de noviembre de 2011
CHAPTER 9
INVENTORY POLICY DECISIONS 1 The probability of finding all items in stock is the product of the individual probabilities. That is, (0.95)×(0.93) ×(0.87) ×(0.85) ×(0.94) ×(0.90) = 0.55 2 (a) The order fill rate is the weighted average of filling the item mix on an order. We can setup the following table. (2) Frequency Order Item mix probabilities of order 1 0.20 .95×.95×.95×.90×.90 =.69 2 0.15 .95×.95×.95 = .86 3 0.05 .95×.95×.90×.90 = .73 4 0.15 .95×.95×.95×.95×.95×.90×.90 = .62 5 0.30 .95×.95×.90×.90×.90×.90 = .59 6 0.15 .95×.95×.95×.95×.95 = .77 Order fill rate (1) (3)=(1)×(2) Marginal probability 0.139 0.129 0.037 0.094 0.178 0.116 0.693

Since 69.3 percent is less than 92 percent, the target order fill rate is not met. (b) The item service levels that will give an orderfill rate of 92 percent must be found by trial and error. Although there are many combinations of item service levels that can achieve the desired service level, a service level of 99 percent for items A, B, C, D, E, and F, and 97 percent to 98 percent for the remaining items would be about right. The order fill rates can be found as follows. (1) Order 1 2 3 4 5 6 Item mix probabilities(.99)3×(.975)2 = .922 (.99)3 = .970 (.99)2×(.975)2 = .932 (.99)5×(.975)2 = .904 (.99)2×(.975)4 = .886 (.99)5 = .951 (2) Frequency of order 0.20 0.15 0.05 0.15 0.30 0.15 Order fill rate (3)=(1) ×(2) Marginal probability 0.184 0.146 0.047 0.136 0.266 0.143 0.922

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3 This is a problem of push inventory control. The question is one of finding how many of 120,000 sets to allocate to each warehouse. Webegin by estimating the total requirements for each warehouse. That is, Total requirements = Forecast + z×Forecast error From Appendix A, we can find the values for z corresponding to the service level at each warehouse. Therefore, we have: (1) Demand forecast, sets 10,000 15,000 35,000 25,000 85,000 (2) Forecast error, sets 1,000 1,200 2,000 3,000 (3) Values for z 1.28 1.04 1.18 1.41(4)=(1)+(2)×(3) Total requirements, sets 11,280 16,248 37,360 29,230 94,118

Warehouse 1 2 3 4 Total

We can find the net requirements for each warehouse as the difference between the total requirements and the quantity on hand. The following table can be constructed:

Warehouse 1 2 3 4

(1) Total requirements 11,280 16,248 37,360 29,230 94,118

(2) On hand quantity 700 0 2,500 1,800

(3)=(1)−(2) Netrequirements 10,580 16,248 34,860 27,430 89,118

(4) Proration of excess 3,633 5,450 12,716 9,083 30,882

(5)=(3)+(4) Allocation 14,213 21,698 47,576 36,513 120,000

There is 120,000 − 89,118 = 30,882 sets to be prorated. This is done by assuming that the demand rate is best expressed by the forecast and proportioning the excess in relation to each warehouse's forecast to the total forecastquantity. That is, for warehouse 1, the proration is (10,000/85,000)×30,882 = 3,633 sets. Prorations to the other warehouses are carried out in a similar manner. The allocation to each warehouse is the sum of its net requirements plus a proration of the excess, as shown in the above table. 4 (a) The reorder point system is defined by the order quantity and the reorder point quantity. Since thedemand is known for sure, the optimum order quantity is:

Q * = 2 DS / IC = 2(3,200)(35) / ( 015)(55) = 164.78, or 165 cases .

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The reorder point quantity is:
ROP = d × LT = (3,200 / 52) × 15 = 92 units .

(b) The total annual relevant cost of this design is: TC = D × S / Q + I × C × Q * / 2 . = (3,200)(35) / 164.78 + ( 015)(55)(164.78) / 2 = 679.69 + 679.97 = \$1,359.66 (c) The revisedreorder point quantity would be:
ROP = (3,200 / 52 ) × 3 = 185 units .

The ROP is greater than Q*. It is possible under these circumstances the reorder quantity may not bring the stock level above the ROP quantity. In deciding whether the ROP has been reached, we add any quantities on order or in transit to the quantity on hand as the effective quantity in inventory. Of course, we start...

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