A bakery prepares all its cakes between 4am and 6am so that they are fresh when

Question

A bakery prepares all its cakes between 4am and 6am so that they are fresh when customers arrive. The cost of baking a cake is $2 and selling price is $10. Demand for the day is estimated to be normally distributed with a mean of 20 and a standard deviation of 4.

a. If cakes are worthless if they are not sold on that day, what is the optimal number of cakes to bake?

b. If all day old cakes can be sold at a deeply discounted price of $1, how does that change the overage cost? What is the optimal number to bake?

Explanation / Answer

Answer to question a :

Given are following data :

Selling price of cake = P = $10

Cost of cake = C = $2

Salvage price ( when cakes are worthless ) = S = 0

Underage cost , Cu = P – C = $8 / unit

Overage cost , Co = C – S = $2/ unit

Critical ratio = Cu/ Cu + Co = 8 / ( 8 + 2) = 8/10 = 0.8

Critical ratio is the in stock probability.

Corresponding Z value for in stock probability 0.8 = NORMSINV ( 0.80) = 0.8416

Optimal number of cakes to bake

= Mean demand + z value x standard deviation of demand

= 20 + 0.8416 x 4

= 20 + 3.36

= 23.36

OPTIMAL NUMBER OF CAKES TO BAKE = 23.36

Answer to question b :

If the cakes are sold at discounted price of $1,

Then salvage price = S = $1

The revised overage cost =Co = C – S = $2 - $1 = $1

The revised , Critical ratio = Cu / Cu + Co = 8/ ( 8 + 1 ) = 9/9 = 0.8888

Corresponding Z value = NORMSINV ( 0.8888) = 1.2201

Optimal number of cakes to bake

= Mean demand + z value x Standard deviation of demand

= 20 + 1.2201 x 4

= 20 + 4.8804

= 24.8804

OPTIMAL NUMBER OF CAKES TO BAKE = 24.8804

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