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

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

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

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)]
= [(.85) x ( )]
= .9895
B = R2 = .90
Rs = A x B
= x .90
= .8906

7. Problem 2 A jet engine has ten components in series.
The average reliability of each component
is What is the reliability of the engine?

8. 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)

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

10. 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

11. 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 RS = R1 + [(RBU) x (1 - R1)]
RS = [(.80) x ( )]
= [(.80) x (.10)] =
RBU = .80
R1 = .90

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

15. 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.

16. 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

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

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

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

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

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

22. 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:
( )/ = .25