Question

A manager is reordering lubricant when the amount on-hand reaches 422 pounds. Average daily usage is 63 pounds and the standard deviation of demand during lead-time is σL = 32.84 pounds. Lead-time is six days.

12. What is the z value?

A. 0.55

B. 1.34

C. 2.90

D. 3.26

E. 11.16

13. What is the safety stock?

A. 422

B. 378

C. 108

D. 80

E. 44

14. What is the probability of stock-out during lead-time?

A. 0.9099

B. 0.9032

C. 0.7088

D. 0.2912

E. 0.0901

15. What is the optimal annual total cost?

A. $16,143

B. $16,236

C. $17,673

D. $16,758

E. $17,722

Answer 1

As given,

Reorder point, R = 422 pounds

Average daily usage, d-bar= 63 pounds

Standard deviation of demand during lead time, σL = 32.84 pounds

Lead time, LT = 6 days.

Reorder point, R = d-bar * LT + z*σ_d* √LT

where d-bar= average daily demand

LT= lead time

σ_d = standard deviation of daily demand

Z= number of standard deviation corresponding to service level probability

a)

R = d-bar * LT + z*σ_d* √LT

422 = 63*6 + z *32.84* √6

z *80.441 = 44

z = 0.55

Correct answer is "A. 0.55"

b)

Safety Stock = z*σ_d* √LT = 0.55*32.84* √6 =44.24 ~ 44 (round off)

Correct answer is "E. 44"

c)

For z value of 0.55, to find the probability of stock out we will look at z table as given below. Look for positive z value graph on right- row 0.5 and column 0.05, we get probability value 0.7088 which is the target service level

Hence, Probability of stock out = 1- 0.7088 = 0.2912

Correct answer is "D. 0.2912"

(Need costing details related to holding cost and ordering cost etc. to calculate annual optimal total cost to answer part 15.)

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