The question is incomplete.
The complete question is:
Marlo Stanfield's operation uses large quantities of prepaid cell phones, on average 1500 per week with a standard deviation of 145. The lead time for their own brand of prepaid cell phones is three weeks and they have a lot size of 350 phones. To ensure they never run out, they keep a safety stock of 500 phones with Proposition Joe.
a. What is the standard deviation of demand during lead time?
b. What is the expected shortage per cycle under this policy?
c. What is the fill rate under this policy?
d. What safety stock should be held to put the expected shortage per cycle at 5 phones?
e. What safety stock should be held to put the fill rate at 0.99?
Answer:
a. 251.15
b. 2.185
c. 99.38%
d. 419 units
e. 452 units
Explanation:
The average weekly demand, (d) is equal to 1,500
Standard Deviation of weekly demand, (σ) is equal to 145.
Lead time, (L) is equal to 3 weeks.
Lot size, (Q) is equal to 350.
Safety stock, (SS) is equal to 500.
A. Standard Deviation of demand during lead time, σLTD = σ√L = 145×√3 = 251.15
B. Expected shortage per cycle, ESC = E(z) ××σLTD
Safey stock = 500 = Z × σLTD
So, z = 500 / σLTD = 500 / 251.15 = 1.991
Corresponding loss function E(z) = NORMDIST(1.991,0,1,0) - 1.991 × (1-NORMSDIST(1.991)) = 0.0087
ESC = 0.0087 × 251.15 = 2.185
C. Fill rate = 1 - ESC / Q = 1 - 2.185/350 = 99.38%
D. ESC = 5
Therefore,
E(z) × σLTD = 5
or
E(z) = 5 / σLTD = 5 / 251.15 = 0.02
The nearest value of z from the standard normal table is z = 1.67
So, Safety stock = Z × σLTD = 1.67*251.15 = 419 units
E.
Fill rate = 99%
Therefore,
1 - ESC / Q = 0.99
or
ESC = 0.01 × Q = 0.01 × 350 = 3.5
So,
E(z) × σLTD =3.5
or
E(z) = 3.5 / σLTD = 3.5 / 251.15 = 0.0139
The nearest value of z from the standard normal table is z equal to 1.8
The Safety stock = z*σLTD = 1.8 × 251.15 = 452 units
