Suppose the cost of a unit failure is $200 and we need to maintain production for 3,000 hours. How much (in $) can we save if we improve the mean time between failures (mtbf) from 6 hours to 8 hours? Round the answer to two decimal places. G

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Lozano

Mathematics

18 August, 02:52

Suppose the cost of a unit failure is $200 and we need to maintain production for 3,000 hours. How much (in $) can we save if we improve the mean time between failures (mtbf) from 6 hours to 8 hours? Round the answer to two decimal places. G

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Alexzander Cortez · Answer 1

The number of failures in 3000 hours with an MTBF of 6 hours is 500. When the MTBF is 8 hours, the number of failures drops to 375, a savings of the cost of 125 failures, or $25,000.