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IEEM 341

Global Supply Chain Management

Fall 03

Assignment 8 Due date: Nov. 6, 2003 (Thursday) The full mark of assignment 8 is 15 points. All the meanings of the following symbols follow the definitions in the textbook. Exercise 1 Weekly demand for Motorola cellular phones at a Best Buy store is normally distributed with a mean of 300 and a standard deviation 200. Motorola takes two weeks to supply a Best Buy order. Best Buy is targeting a cycle service level of 95 percent and monitors its inventory continuously. What level of safety inventory of cellular phones should Best Buy carry? What should its reorder point be? Solution: R L = 2 × 300 = 600, σ L = 200 2 CSL = 0.95 ss = Fs−1 (CSL) × σ L == 1.644853 × 200 2 = 465.2348 ROP = RL + ss = 1065.2348

Exercise 2 Reconsider the Best Buy store in problem 1. The store manager has decided to follow a periodic review policy to manage inventory of cellular phones. She plans to order every three weeks. Given a desired cycle service level of 95 percent, how much safety inventory should the store carry? What should its order up to level be? Solution: Note that we have to also consider the average demand in the leadtime besides the average demand in the order cycle.

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IEEM 341

Global Supply Chain Management

Fall 03

R = 300, σ R = 200 2 , RT + L = (T + L) × 300 = (3 + 2) × 300 = 1500,

σ L = 200 T + L = 200 5 , CSL = 0.95, ss = Fs−1 (CSL) × σ L = 1.644853 × 200 5 = 735.6008, OUL = RL + ss = 2235.6008.

Exercise 3 Assume that the Best Buy store has a continuous review policy of ordering cellular phones from Motorola in lots of 500. Weekly demand for Motorola cellular phones at the store is normally distributed with a mean of 300 and a standard deviation of 200. Motorola takes two weeks to supply an order. If the store manager is targeting a fill rate of 99 percent, what safety inventory should Best Buy carry? What should its reorder point be? Solution: RL = 2 × 300 = 600, σ L = 200 × 2 fr = 0.99, Q = 500, ESC = (1 − fr ) × Q = (1 − 0.99) × 500 = 5 ESC = − ss[1 − Fs (

ss

σL

)] + σ L f s (

ss

σL

)=5

ss = 484.76; ROP = RL + ss = 1084.76

By IEEM 310’s knowledge,

we have

n( R) = (1 − β )Q = 5

L( z ) = n( R ) / σ L = 0.017678 z = 1.71 ss = z × σ L = 1.71 × 200 × 2 = 483.661, ROP = RL + ss = 1083.661

Exercise 4 Weekly demand for HP printers at a Sam’s Club store is normally distributed with a mean of 250 and a standard deviation of 150. The store manager continuously monitors inventory and currently orders 1,000 printers each time the inventory drops to 600 printers. HP currently takes two weeks to fill an order. How much safety inventory -2-

IEEM 341

Global Supply Chain Management

Fall 03

does the store carry? What cycle service level does Sam’s Club achieve as a result of this policy? What fill rate does the store achieve? Solution:

ROP = 600 Q = 1000 RL = 2 × 250 = 500

σ L = 150 2 ss = ROP − RL = 100 CSL = Fs (

ROP − RL

600 − 500

) = 0.681324 150 2 ss ss ESC = − ss[1 − Fs ( )] + σ L f s ( ) = 43.86126

σL

) = Fs (

σL

fr = 1 −

σL

ESC 43.86126 = 1− = 0.956139 Q 1000

Note that many students obtain the wrong value of ESC.

Exercise 5 Return to the Sam’s Club store in problem 4. Assume that the supply lead time from HP is normally distributed with a mean of 2 weeks and a standard deviation of 1.5 weeks. How much safety inventory should Sam’s Club carry if it wants to provide a cycle service level of 95 percent? How does the required safety inventory change as the standard deviation of lead time is reduced from 1.5 weeks to 0 in intervals of 0.5 weeks? Solution:

R = 250, σ R = 150, L = 2, s L = 1.5, CSL = 0.95, RL = 2 × 250 = 500

σ L = Lσ R2 + R 2 s L2 = 2 × 150 2 + (250 × 1.5) 2 = 430.8422 ss = Fs−1 (CSL) × σ L = 708.672

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IEEM 341

Global Supply Chain Management

Fall 03

By following the above procedure, we can calculate the required safety inventory change as the standard deviation of lead time is reduced from 1.5 weeks to 0 in intervals of 0.5 weeks in the table below. SL

σL

ss

1 0.5 0

327.8719 246.2214 212.132

539.3013 404.9982 348.9261

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