For a particular piece of equipment, the probability of failure during a given w

Question

For a particular piece of equipment, the probability of failure during a given week is as follows:

Week of Operation Probaility of failure

For a particular piece of equipment, the probability of failure during a given week is as follows: Management is considering a preventive maintenance program that would be implemented at the end of a given week of production. The production loss and downtime costs associated with an equipment failure are estimated to be $2,500 per failure. Assume 52 weeks of operation per year. If it costs $500 to perform the preventive maintenance, how much time between inspections should there be for the least expensive program?

1 .25 2 .08 3 .07 4 .10 5 .20 6 .30

Explanation / Answer

Solution :

Total probability of failure is after weeks . So failure cost + down time cost = $ 2500 at the sixth week.

We have total 52 weeks

Annual cost of preventive maintenance cost = 500 * 52 weeks

= $ 26000

Failure cost = (52/6) *2500

= $ 21667

In above case it seems that maintenance cost is more than failure cost.

So we have to increase the preventive maintenance period from per week to 2 weeks .

So Maintenance cost will be = 26 weeks * 500

= $ 13000

So 2 weeks time can be enough for preventive maintenance.

If our equipment is reliable , preventive system is strong in that case we can increase the period from 2 weeks to 3 weeks.

Week of Operation    Probability of failure Cost of Failure in $ Actual cost 1 0.25 2500 625 2 0.08 2500 200 3 0.07 2500 175 4 0.1 2500 250 5 0.2 2500 500 6 0.3 2500 750

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