Chapter 5 – Part 3

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Chapter 5 – Part 3 . Solutions to Text book HW Problems 2, 7, 6. Part 1. Part 2. R 1 =.90. R 2 =.87. Review: Components in Series. Both parts needed for system to work. R S = R 1 x R 2 = (.90) x (.87) =.783. R 1 =.93. R 2 =.90. R BU =.85. Review: (Some) Components in Parallel.

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Chapter 5 – Part 3

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Chapter 5 – Part 3

Solutions to Text book HW

Problems 2, 7, 6

Part 1

Part 2

R1 =.90

R2 =.87

Review: Components in Series

Both parts needed for system to work.

RS = R1 x R2 = (.90) x (.87) =.783

R1 =.93

R2 =.90

RBU =.85

Review: (Some) Components in Parallel

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

RBU =.80

R1 =.90

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.
Solution to Problem 7

RS = R1 + [(RBU) x (1 - R1)]

1 - R1 = Probability of needing BU component

= Probability that 1st component fails

R1 = .90

RS = R1 +[(RBU) x (1 - R1)]

RS = .90 +[(.80) x (1 - .90)]

= .90 +[(.80) x (.10)]

RBU = .80

= .9802 .98

Solution to Problem 7

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?

R1 = .90

R2 = .89

R3 = .95

RBU = .80

RBU = .80

RBU = .80

Problem 6
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

.98

.978

.99

Solution to Problem 6 – With BU

RS = A x B x C

= .98 x .978 x .99

=.9489

R1 = .90

R2 = .89

R3 = .95

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 =.7610
• Reliability of system with backups is 25% greater than reliability of system with no backups:

(.9489 - .7610)/.7610 = .25