# Inventory Management IE 314: Operations Management KAMAL Lecture 7.

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Inventory Management IE 314: Operations Management KAMAL Lecture 7

EXERCISE 12.27

Solution: a)  = 60;  LT = 7 Safety stock for 90% service level = Z  LT = 7  1.28 = 8.96  9 b) ROP =  + Z  LT = 60 + 9 = 69 BX-5

EXERCISE 12.30

The safety stock which minimizes total incremental cost is 100 kilos. The reorder point then is 200 kilos + 100 kilos = 300 kilos. Annual Stockout Cost = Sum of units short*probability*stockout cost/unit*number of orders per year Safety StockCarrying CostStockout CostTotal Cost 0070(100*0.4+200*0.2)= 5,600\$5,600 100100*15= 1,50070*100*0.2= 1,400\$2,900 200200*15= 3,0000\$3,000

EXERCISE 12.32

(a)ROP = (Average daily demand  Lead time in days)  Z  dLT = 1,000 * 2 + (2.055)* 100 √ 2 = 2,291 towels (b) Safety Stock = 291 towels

EXERCISE 12.33

ROP  (Daily demand  Average lead time in days)    Daily demand   LT = 12,500 * 4 + (1.88)* 12,500*1 = 73,500 pages

EXERCISE 12.34

ROP  (200  6)  1.28  dLT  dLT = √(6*(25^2)+((200^2)*(2^2)  405 (a)ROP  1,200  (1.28)(405)  1,200  518  1,718 cigars (b) For 95% service level, Z = 1.65 ROP = (200 × 6) + 1.65(405)  1,200 + 668  1,868 cigars

EXERCISE 12.34 (c) A higher service level means a lower probability of stocking out. Hence, the ROP increases from 1,718 to 1,868 when the service level increases 5%.

EXERCISE 12.38

μ = (2/3)*90,000 = 60,000 programs,  = 5,000 (a)C s = 4 – 1 = \$3 (b)C 0 = 1 – 0.1 = \$0.9 (c)Service Level = (C s /C s +C 0 ) = 3/3.9 = 0.7692 Z  0.735 Q* = μ + Z  = 60,000 + (0.735)(5,000) = 63,675 programs (d) Stockout risk = 1 – Service level = 1 – 0.7692 = 0.2308 or23.1%

HW 12.31 12.37 12.40