Based on available information, lead time demand for PC jump drives averages 53 units (normally distributed), with a standard deviation of 5 drives. Management wants a 95%
service level. Refer to the standard normal table for z-values.
a) What value of Z should be applied? (1.04/1.28/2.06/1.65)
b) How many drives should be carried as safety stock? __________ units (round your response to the nearest whole number).
c) What is the appropriate reorder point? __________ units (round your response to the nearest whole number).
Homework Answers
1. Z value for 95% service level = 1.65
2. SS = Z * STANDARD DEVIATION * SQRT(LEAD TIME)
Standard deviation during lead time = 5
Z value for 95% service level = 1.64
Safety Stock = 1.65 * 5 = 8
3. ROP = Demand during lead time + SS = 53 + 8 = 61
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