Chapter 5 Part 3

1
Chapter 5 Part 3
Solutions to Text book HW Problems 2, 7, 6
2
Review Components in Series

• Both parts needed for system to work.
• RS R1 x R2 (.90) x (.87) .783

3
Review (Some) Components in Parallel
4
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.

5
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
6
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

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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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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)
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Solution to Problem 2
R1 R2 R10 .998 RS R1 x R2 x x R10
(.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.

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Solution to Problem 7
RS R1 (RBU) x (1 - R1) 1 - R1
Probability of needing BU component
Probability that 1st component
fails
12
Solution to Problem 7
13
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?

14
Problem 6
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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.

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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
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Solution to Problem 6 With BU

• A R1 (RBU) x (1 - R1)
• .90 (.80) x (1 - .90)
• .98

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Solution to Problem 6 With BU

• B R2 (RBU) x (1 - R2)
• .89 (.80) x (1 - .89)
• .978

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Solution to Problem 6 With BU

• C R3 (RBU) x (1 - R3)
• .95 (.80) x (1 - .95)
• .99

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Solution to Problem 6 With BU
RS A x B x C .98 x .978 x .99 .9489
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Solution to Problem 6 No BUs
RS R1 x R2 x R3 (.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

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