Transcript – Part 3 Chapter 5 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
R1 =.90
R2 =.87
Both parts needed for system to work.
RS = R1 x R2 = (.90) x (.87) =.783
Review: (Some) Components in Parallel
R1 =.93
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.
RBU =.80
R1 =.90
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)]
R1 = .90
RS = .90 + [(.80) x (1 - .90)]
= .90 + [(.80) x (.10)]
= .9802
.98
RBU = .80
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
R1 = .90
R2 = .89
R3 = .95
RBU = .80
RBU = .80
RBU = .80
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
RS = A x B x C
= .98 x .978 x .99
=.9489
.99
Solution to Problem 6– No BUs
R1 = .90
R2 = .89
R3 = .95
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