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A designer estimates that she can increase the average time between failures of a part by 5 percent of a cost of $450
A designer estimates that she can increase the average time between failures of a part by 5 percent of a cost of $450. Currently, the average time between failures is 100 hours and the average repair time is 4 hours.
B. A designer estimates that she can reduce the average repair time by 10 percent at a cost of $200. Currently, the average time between failures is 100 hours and the average repair time is 4 hours.
C. What option would be more effective?
Expert Solution
MeanTimeFailure(MTF)=100HoursMeanTimeRepair(MTR)=4HoursMeanTimeFailure(MTF)=100HoursMeanTimeRepair(MTR)=4Hours
Therefore,
Currentavailability=MTFMTF+MTR=100100+4=100104=0.96Currentavailability=MTFMTF+MTR=100100+4=100104=0.96
a) When a 5% increase in time between failures
Given -
Cost= $450MTF=100+(5%of100)=100+5=105hoursCost= $450MTF=100+(5%of100)=100+5=105hours
Therefore,
Availabilty=MTFMTF+MTR=105105+4=105109=0.9633Availabilty=MTFMTF+MTR=105105+4=105109=0.9633
b) Fall in 10/units in repair time
Given-
Cost= $200MTR=4−(10%of4)=4−0.4=3.6hoursCost= $200MTR=4−(10%of4)=4−0.4=3.6hours
Therefore,
Availability=MTFMTF+MTR=100100+3.6=100103.6=0.9652Availability=MTFMTF+MTR=100100+3.6=100103.6=0.9652
c) Thus, after comparing both the options the second option is a better option because it is more efficient and cost effective.
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