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Decision Making

Megley Cheese Company is a small manufacturer of several different cheese products. One of the products is a cheese spread that is sold to retail outlets. Jason Megley must decide how many cases of cheese spread to manufacture each month. The probability that the demand will be six cases is 0.1, for 7 cases, 0.3, for 8 cases is 0.5, and for 9 cases is 0.1. The cost of every case is $45, and the price that Jason gets for each case is $95. Unfortunately, any cases not sold by the end of the month are of no value, due to spoilage. How many cases of cheese should Jason manufacture each month?

Solution Preview

Please refer attached file for better clarity of tables

Let us first make the Conditional Profit table.

In case demand is for 6 cases, and Jason keeps a stocks of 6 cases.
Profit =6*95-6*45=300
In case demand is for 6 cases, and Jason keeps a stocks of 7 cases.
6 cases will be sold and 1 case will be spoiled.
Profit =6*95-7*45=255
In case demand is for 6 cases, and Jason keeps a stocks of 8 cases.
6 cases will be sold and 2 case will be spoiled.
Profit =6*95-8*45=210
In case demand is for 6 cases, and Jason keeps a stocks of 9 cases.
6 cases will be sold and 3 case will be spoiled.
Profit =6*95-9*45=165

In case demand is for 7 cases, and Jason keeps a stocks of 6 cases.
He will buy 6 cases and 6 cases will be sold
Profit =6*95-6*45=300
In case demand is for 7 cases, and Jason keeps a stocks of 7 cases.
He will buy 7 cases and 7 cases will be sold
Profit =7*95-7*45=350
In case demand is for 7 cases, and Jason keeps a stocks of 8 cases.
7 cases will be sold and 1 case will be spoiled.
Profit =7*95-8*45=305
In case demand is for 7 cases, and Jason keeps a stocks of 9 cases.
7 cases will be sold and 2 case will be spoiled.
Profit =7*95-9*45=260

In case demand is for 8 cases, and Jason keeps a stocks of 6 cases.
He will buy 6 cases and 6 cases will be sold
Profit ...

Solution Summary

Solution determines the number of cases of cheese to be ordered each month in the given case.

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