Consider a humdinger with 4 components in series (Components A, B, C, & D) with
ID: 386329 • Letter: C
Question
Consider a humdinger with 4 components in series (Components A, B, C, & D) with reliabilities 0.9, 0.92, 0.87, 0.99 respectively. If the humdinger fails it will cost the company $1300. (Keep all reliabilities to 4 decimal places)
(4 points) What is the system reliability and the expected failure cost per part?
(9 points) Consider 3 improvement options:
Improvement 1: Add a 90% reliable backup for component A. The backup costs $75 per part.
Improvement 2: Add a 75% reliable backup for component C. The backup costs $65 per part.
Improvement 3: Implement both Improvement 1 and Improvement 2.
For each of the three improvements:
Find the system reliability
Find the total cost (failure + improvement) per part.
Finally, which of the three improvements (if any) would you recommend?
Explanation / Answer
When there are n components in series, where the system reliability of the i-th component is denoted by ri , the
System reliability is R(s) = (r1)(r2)......(ri)
System Reliability = (0.9)(0.92)(0.87)(0.99) = 0.713
Expected failure cost = failure cost * probability of failure
= $1300 * (1 - reliability)
= $1300 * (1 - 0.713) = $373.10
Improvement 1: Add a 90% reliable backup for component A
New system reliability of A = 1 - {(1 - reliability original)*(1 - reliability backup)}
= 1 - {(1 - 0.9)*(1 - 0.9)}
= 1 - 0.01
= 0.99
New System Reliability = (0.99)(0.92)(0.87)(0.99) = 0.784
New Expected failure cost = failure cost * probability of failure
= $1300 * (1 - reliability)
= $1300 * (1 - 0.784) = $280.8
Reduction in failure cost = $373.1 - $280.8
= $92.3
Improvement 2: Add a 75% reliable backup for component C
New system reliability of C = 1 - {(1 - reliability original)*(1 - reliability backup)}
= 1 - {(1 - 0.9)*(1 - 0.75)}
= 1 - 0.025
= 0.975
New System Reliability = (0.9)(0.92)(0.975)(0.99) = 0.79
New Expected failure cost = failure cost * probability of failure
= $1300 * (1 - reliability)
= $1300 * (1 - 0.79) = $261
Reduction in failure cost = $373.1 - $261
= $112.1
Improvement 3: Implement both Improvement 1 and Improvement 2
New System Reliability = (0.99)(0.92)(0.975)(0.99) = 0.88
New Expected failure cost = failure cost * probability of failure
= $1300 * (1 - reliability)
= $1300 * (1 - 0.88)
= $157.1
Reduction in failure cost = $373.1 - $157.1
= $216
Since the reduction in failure cost is highest in Improvement 3, I would recommend it.
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