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Chapter 5 – Part 3 Solutions to Text book HW Problems 2, 7, 6

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Review: Components in Series R 1 =.90R 2 =.87 Part 1Part 2 Both parts needed for system to work. R S = R 1 x R 2 = (.90) x (.87) =.783

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Review: (Some) Components in Parallel R 1 =.93 R BU =.85 R 2 =.90

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Review: (Some) Components in Parallel System has 2 main components plus a BU Component. First component has a BU which is parallel to it. For system to work, –Both of the main components must work, or –BU must work if first main component fails and the second main component must work.

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A = Probability that 1st component or its BU works when needed B = Probability that 2nd component works or its BU works when needed = R 2 R S = A x B Review: (Some) Components in Parallel

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A = R 1 + [(R BU ) x (1 - R 1 )] =.93 + [(.85) x (1 -.93)] =.9895 B = R 2 =.90 R s = A x B =.9895 x.90 =.8906 Review: (Some) Components in Parallel

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Problem 2 A jet engine has ten components in series. The average reliability of each component is.998. What is the reliability of the engine?

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R S = reliability of the product or system R 1 = reliability of the first component R 2 = reliability of the second component and so on R S = (R 1 ) (R 2 ) (R 3 )... (R n ) Solution to Problem 2

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R 1 = R 2 = … =R 10 =.998 R S = R 1 x R 2 x … x R 10 = (.998) x (.998) x x (.998) = (.998) 10 =.9802

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Problem 7 An LCD projector in an office has a main light bulb with a reliability of.90 and a backup bulb, the reliability of which is.80. R 1 =.90 R BU =.80

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Solution to Problem 7 R S = R 1 + [(R BU ) x (1 - R 1 )] 1 - R 1 = Probability of needing BU component = Probability that 1 st component fails

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Solution to Problem 7 R BU =.80 R 1 =.90 R S = R 1 + [(R BU ) x (1 - R 1 )] R S =.90 + [(.80) x (1 -.90)] =.90 + [(.80) x (.10)] =

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Problem 6 What would the reliability of the bank system above if each of the three components had a backup with a reliability of.80? How would the total reliability be different?

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Problem 6 R BU =.80 R 1 =.90 R 3 =.95 R 2 =.89

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First BU is in parallel to first component and so on. Convert to a system in series by finding the probability that each component or its backup works. Then find the reliability of the system. Solution to Problem 6 – With BU

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A = Probability that 1st component or its BU works when needed B = Probability that 2nd component or its BU works when needed C = Probability that 3rd component or its BU works when needed R S = A x B x C

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Solution to Problem 6 – With BU A = R 1 + [(R BU ) x (1 - R 1 )] =.90 + [(.80) x (1 -.90)] =.98

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Solution to Problem 6 – With BU B = R 2 + [(R BU ) x (1 - R 2 )] =.89 + [(.80) x (1 -.89)] =.978

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Solution to Problem 6 – With BU C = R 3 + [(R BU ) x (1 - R 3 )] =.95 + [(.80) x (1 -.95)] =.99

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R S = A x B x C =.98 x.978 x.99 =.9489 Solution to Problem 6 – With BU

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Solution to Problem 6– No BUs R 1 =.90 R 3 =.95 R 2 =.89 R S = R 1 x R 2 x R 3 = (.90) x (.89) x (.95) =.7610

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Solution to 6 - BU vs. No BU Reliability of system with backups =.9489 Reliability of system with backups =.7610 Reliability of system with backups is 25% greater than reliability of system with no backups: ( )/.7610 =.25

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