Chapter 5 – Part 3

Solutions to Text book HW

Problems 2, 7, 6

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

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.

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?

Problem 6

RBU = .80 RBU = .80 RBU = .80

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

• 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 – 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

RS = A x B x C

= .98 x .978 x .99

=.9489

Solution to Problem 6 – With BU

Solution to Problem 6– No BUs

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

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