Question

In: Economics

A machine shop owner is attempting to decide whether to purchase a new drill press, a...

A machine shop owner is attempting to decide whether to purchase a new drill press, a lathe, or a grinder. The profit or loss from each purchase are shown in the following table where one of two states of nature could occur (the company succeeds in getting a military contract, or it does not get a contract).

Contract status

Purchase             Get Contract      No contract

Drill press            30,000                       -10,000

Lathe                     25,000                        4,000

Grinder                 12,000                     10,000

  1. What is the expected monetary value of each purchase option when the probability of getting contract is .4 and which is the best option under this probability?

  1. Since .4 may be incorrect as the probability of getting a military contract, over what range of values for the probability of getting a contract would each purchase option be the most favored? Show your work!

Solutions

Expert Solution

a. Expected monetary value of each purchase option is found by: since either the owner gets or doesn't get the contract, that is total probability of these two events is 1.

Hence, the expected monetary value of each purchase is found to be in the sheet below:

This shows that the 'lathe' is the best purchase option under the given probability of getting a military contract.

b. We now compare the expected monetary value of each contract for a range of x from 0 to 1.

We do this by comparing these three expressions, where x is the probability of getting the military contract.

If we increase x in steps of 0.1 from 0 to 1, we get the following table, where the highlighted cell shows the purchase with maximum expected profit.

To find the exact ranges of x, we first equate the expressions for lathe and grinder: to get x as 0.316

On equating the expressions for the lathe and drill press, we get x as 0.737.

Hence, the ranges for maximum expected profit are:

Hope this helps!


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