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# Applications of Cost Functions

4. A hospital has seven large identical heat pumps. If one of the heat pumps malfunctions, it could seriously impact hospital operations. If preventive maintenance is performed weekly, it costs \$1200 to perform preventive maintenance (PM) on all seven of them. If one of the heat pumps breaks down between PM inspections, the average cost to the company of a breakdown is \$2200. The historical data for the heat pumps is as follows:

Weeks between Preventive Maintenance
Expected Number of Breakdowns
between Preventive Maintenance

1
1.10

2
2.52

3
4.63

4
8.27

a) Suppose that the cost of preventive maintenance increases as the weeks between the preventive maintenance increases. What is the cost for performing preventive maintenance with two weeks between PM that makes the cost of the two week strategy equal to the one week strategy?

b) Now suppose that for each week that preventive maintenance is delayed, the cost to the company of a breakdown increases by \$200 (for example, if preventive maintenance is every two weeks, the cost of a breakdown is \$2400. Based on the above information, what Preventive Maintenance policy do you recommend? Why?

#### Solution Preview

We have the cost of a breakdown (\$2200), which remains constant, and the cost of maintenance, which increases.

Given one week between maintenance, the expected weekly cost is:

EWC = \$1200 + 1.1(\$2200) = \$3620

(That is, the cost of doing the maintenance, plus the cost of a breakdown times the expected number of breakdowns)

Switching to two ...

#### Solution Summary

Applications of Cost Functions are investigated. The solution is detailed and well presented.

\$2.19