Reliability of Networks

1 2 B E D AC 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 2 B A 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(A  B) = 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(A  B)

1 - (1-P(A ))(1- P(B )) P(A),P(B)) =  12 B A 1 2 B A SeriesParallel Rel =  (P(A),P(B)) = P(A),P(B))  Rel  (P(A),P(B)) = P(A)P(B) 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 2 C A B = P(A),P(B),P(C)) = 1 - (1- P(A))(1- P(B))(1- P(C))  Rel C A 1 2 B

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 12 AB C DE 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?

AB C DE C  ×0.9 = ×0.8 = (1-0.8)(1-0.64) = ×0.928 =   

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. 12 pCpC pDpD pBpB pApA p E 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)  p E + Rel(network working |E not working)  (1  p E ) Similarly

Case 2: E working 12 pCpC pDpD pBpB pApA B D C A 12 pCpC pDpD pBpB pApA B D C A 12 pCpC pDpD pBpB pApA B D C A

12 pCpC pDpD pBpB pApA B D C A Case 2: E not working

What is the reliability of the following network given all reliabilities are 0.9? E B DC A Example

E works: 12 B DC A 0.9  1-(1-0.9)(1-0.9) = 0.99   *0.99

12 B DC A 0.9 E does not work:  0.9*0.9   1-(1-0.81)(1-0.81)

Rel(network) = Rel(network working|E working)  p E + Rel(network working |E not working)  (1  p E ) =   0.1 =

What is the reliability of the following network given all reliabilities are 0.9? Example FE 0.9 E B DC A FE

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   Rel = = Bridge working Bridge not working

Reduce the following to a workable circuit. Example