Reliability of Networks

1 2

B

E

D

A C

Simple 2 Terminal Networks

Reliability of a 2 terminal network is the probability there is a connection between the 2 terminals.. It is common to assume that components of a network behave independently in their reliabilities. Sometimes this assumption is unjustified.

Components in Series

1 2BA

Use the notation A for the event that there is a connection through A.

Then P(A) is the probability that there is a connection through A ie A is working.

There is a connection between the two terminals when both A and B are working.

Rel = P(AB) = P(A).P(B) = (P(A),P(B))

Components in Parallel

There is a connection between the two terminals when either A or B is working.

Rel = 1 - (1-P(A ))(1- P(B ))

(OR is inclusive = and/or)

There is no connection if both A and B are not working

P(A B ) = P(A ) P(B ) =

(1-P(A ))(1- P(B ))

The probability of either A or B is working is

= P(A),P(B))

1 2

B

A

1 - P(A B )

Want to find the reliability: P(AB)

1 - (1-P(A ))(1- P(B ))

P(A),P(B)) =

1 2

B

A

1 2BA

Series Parallel

Rel = (P(A),P(B)) = P(A),P(B))

Rel

(P(A),P(B)) = P(A)P(B)where where

IMPORTANT: We have assumed independant events.

Note: This generalises eg

Rel = (P(A),P(B),P(C)) = P(A)P(B)P(C)

1 2CA B

= P(A),P(B),P(C)) = 1 - (1- P(A))(1- P(B))(1- P(C))

Rel

C

A

1 2B

The operator is a symbol for the calculation of the probability of the union of independent events.

The operator is a symbol for the calculation of the probability of the intersection of independent events.

Example

1 2A B

C

D E

Components A and B have reliability 0.9 and components C, D and E have reliability 0.8. All components perform independently. What is the reliability of the connection between terminals 1 and 2?

1 20.9 0.9

0.8

0.8 0.8

A B

C

D E

1 2

0.8

0.81

0.64

C

20.81 0.928

0.9×0.9 = 0.81

0.8×0.8 = 0.64

1 - (1-0.8)(1-0.64)

= 0.928

0.81×0.928 = 0.75168

20.75168

Bridge Networks

A bridge network is the simplest network that can’t be broken down into a series-parallel system. To calculate the through reliability of this network we will need to use conditional probability.

1 2

pC

pDpB

pA

pE E

B D

C

A

Component E is the problem.

Break the system up according two the two outcomes of E working or not.

Under each of the outcomes the system becomes a series/parallel system.

Rel(network) = Rel(network working|E working) pE +

Rel(network working |E not working) (1pE)

E

E

PEworking).|workingP(network

Eworking)workingP(network

PEworking)workingP(network

Eworking)|workingP(network

Similarly

)P-g).(1not workin E|workingP(network

g)not workin EworksP(network

E

Case 2: E working

1 2

pC

pDpB

pA

B D

C

A

1 2

pC

pDpB

pA

B D

C

A

1 2

pC

pDpB

pA

B D

C

A

1 2

pC

pDpB

pA

B D

C

A

Case 2: E not working

What is the reliability of the following network given all reliabilities are 0.9?

0.9 1 2 E

B

DC

A

0.9 0.9

0.9 0.9

Example

E works: 1 2

B

DC

A

0.9 0.9

0.9 0.9

1-(1-0.9)(1-0.9) = 0.99 1-(1-0.9)(1-0.9) = 0.99

20.99 0.99

20.9801

0.99*0.99

1 2

B

DC

A

0.9 0.9

0.9 0.9

E does not work:

20.9639

0.9*0.9

0.9*0.9

1 2

0.81

0.81

1-(1-0.81)(1-0.81)

Rel(network) = Rel(network working|E working) pE +

Rel(network working |E not working) (1pE)

= 0.9801 0.9 + 0.9639 0.1

= 0.97848

What is the reliability of the following network given all reliabilities are 0.9?

Example

FE

0.9 E

B

DC

A

0.9 0.9

0.9 0.9

0.9 0.9

FE

0.9 0.9

0.97848

0.81

0.97848

0.99591

0.75

Example : All components have reliability 0.5

Strategy: Reduce to a simple bridge circuit

0.375

All components have reliability 0.5 unless otherwise shown

0.375

0.75 0.75

0.5625 0.4375

0.375 0.5625 + 0.6255 0.4375

0.25

0.25

Rel = = 0.4846

0.375Bridge working 0.625

Bridge not working

Reduce the following to a workable circuit.Example