Answer:
a) 749
b) 4.073
Explanation:
Given:
Mean = demand = 80 pounds
Standard deviation of demand = 10 pounds
Lead time = 8 days
Standard deviation of lead time = 1 day
a) What ROP would provide a stock out risk of 10 percent during lead time.
To find this re-order point (ROP) quantity, take the formula:
[tex] ROP = d(LT) + z \sqrt{ LT \sigma_d ^2 + LT^2 \sigma_L_T ^2} [/tex]
Here, service level = 100%-10% = 90%,
Thus z at 90% = ±1.28
[tex] ROP = 80(8) + 1.28 \sqrt{8* 10^2 + (8)^2*(1)^2} [/tex]
[tex] ROP = 640 + 1.28\sqrt{800 + 64} [/tex]
= 640 + 1.28* 84.85
= 748.61
≈ 749 units
b) What is the expected number of units (pounds) short per cycle.
Find the number of units shorts per cycle. Take the formula:
[tex] E(n) = E(z) * \sigma d_L_T [/tex]
[
Where E(z) = standardized number of shorts = 0.048
[tex] \sigma d_L_T [/tex] = standard deviation of lead time demand = 84.85
Therefore,
E(n) = 0.048 * 84.85
= 4.073
