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Review: Components in Series
R1 =.90 R2 =.87
Part 1 Part 2
Both parts needed for system to work.
RS = R1 x R2 = (.90) x (.87) =.783
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.
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
Review: (Some) Components in Parallel
Problem 2
A jet engine has ten components in series.The average reliability of each componentis.998. What is the reliability of the engine?
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
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
RBU = .80
R1 = .90RS = R1 + [(RBU) x (1 - R1)]
RS = .90 + [(.80) x (1 - .90)]= .90 + [(.80) x (.10)]
= .9802 .98
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?
• 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
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– No BUs
R1 = .90 R3 = .95R2 = .89
RS = R1 x R2 x R3
= (.90) x (.89) x (.95)
= .7610