In: Statistics and Probability
Daily demand for cat litter at the Cat Café in Jones is 1500
ounces with a standard deviation of
300 ounces. The average lead time is 5 days and the standard
deviation of lead time is 2 days.
Use this information to answer questions 44 to 46.
Suppose the café wants to peg their service level at 99.5%. What
is the level of safety
inventory they should carry?
A) between 7500 and 7510 ounces
B) between 3070 and 3080 ounces
C) between 1950 and 1960 ounces
D) between 7910 and 7920 ounces
Suppose the café wishes to hold 7162 ounces of safety inventory.
This is equal to how many
days of inventory?
A) 4.77 days
B) 5.12 days
C) 5.33 days
D) 5.67 days
Suppose the café wishes to carry 10,728 ounces as their safety
inventory. What service level
would they achieve if standard deviation of demand over lead time
is 3,074 ounces?
A) 93.9%
B) 99.98%
C) 95.2%
D) 95.7%
Concepts
Safety stock is given by
Q1
At Service level at 99.5%, Z = 2.576. We can derive this from normal table, or we can use excel function to find
NORMSINV(0.995) = 2.576
Substituting the above values in the safety stock formula
Safety Stock is 7918.9, So the answer is Option D
Q2.
Safety inventory = 7162
Mean demand = 1500
Days of inventory = 7162/1500 = 4.77
Inventory days is 4.77. So the answer is Option A
Q3
Z = Safety stock/SD of demand over lead time
Z = 10728/3074 = 3.49
If we see the Z table or Excel function we see that at 99.98% , we get Z close to 3.49
So, the answer is 99.98%, So the answer is Option B