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

Review: Components in Series Part 1 Part 2 Both parts needed for system to work. RS = R1 x R2 = (.90) x (.87) =.783

Review: (Some) Components in Parallel RBU =.85 R2 =.90

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.

Review: (Some) Components in Parallel A = Probability that 1st component or its BU works when needed B = Probability that 2nd component works or its BU works when needed = R2 RS = A x B

Review: (Some) Components in Parallel A = R1 + [(RBU) x (1 - R1)] = .93 + [(.85) x (1 - .93)] = .9895 B = R2 = .90 Rs = A x B = .9895 x .90 = .8906

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?

Solution to Problem 2 RS = reliability of the product or system R1 = reliability of the first component R2 = reliability of the second component and so on RS = (R1) (R2) (R3) . . . (Rn)

Solution to Problem 2 R1 = R2 = … =R10 = .998 RS = R1 x R2 x … x R10 = (.998) x (.998) x    x (.998) = (.998)10 =.9802

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. R1 =.90 RBU =.80

Solution to Problem 7 RS = R1 + [(RBU) x (1 - R1)] 1 - R1 = Probability of needing BU component = Probability that 1st component fails

Solution to Problem 7 RS = R1 + [(RBU) x (1 - R1)] RS = .90 + [(.80) x (1 - .90)] = .90 + [(.80) x (.10)] = .9802 .98 RBU = .80 R1 = .90

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?

Problem 6 RBU = .80 R1 = .90 R3 = .95 R2 = .89

Solution to Problem 6 – With BU 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 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 RS = A x B x C

Solution to Problem 6 – With BU A = R1 + [(RBU) x (1 - R1)] = .90 + [(.80) x (1 - .90)] = .98

Solution to Problem 6 – With BU B = R2 + [(RBU) x (1 - R2)] = .89 + [(.80) x (1 - .89)] = .978

Solution to Problem 6 – With BU C = R3 + [(RBU) x (1 - R3)] = .95 + [(.80) x (1 - .95)] = .99

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

Solution to Problem 6– No BUs RS = R1 x R2 x R3 = (.90) x (.89) x (.95) = .7610

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