Question

In: Statistics and Probability

Daily demand for cat litter at the Cat Café in Jones is 1500 ounces with a...

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%

Solutions

Expert Solution

Concepts

Safety stock is given by

  • where alpha is the service level
  • E(L) is the mean of lead time, SigmaL is the standard deviation of lead time
  • E(D) is the mean of demand, SigmaD is the standard deviation of demand

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


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