1. Reliability of Networks

2. 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.

3. 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))

4. 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)

5. 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.

6. 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

7. 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.

8. 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?

9. 12 0.9 0.8 AB C DE 12 0.81 0.64 C  2 0.81 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 2 0.75168   

10. 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.

11. Rel(network) = Rel(network working|E working)  p E + Rel(network working |E not working)  (1  p E ) Similarly

12. 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

13. 12 pCpC pDpD pBpB pApA B D C A Case 2: E not working

14. What is the reliability of the following network given all reliabilities are 0.9? 0.9 12 E B DC A Example

15. E works: 12 B DC A 0.9  1-(1-0.9)(1-0.9) = 0.99  2 0.99  2 0.9801 0.99*0.99

16. 12 B DC A 0.9 E does not work: 2 0.9639  0.9*0.9  12 0.81  1-(1-0.81)(1-0.81)

17. Rel(network) = Rel(network working|E working)  p E + Rel(network working |E not working)  (1  p E ) = 0.9801  0.9 + 0.9639  0.1 = 0.97848

18. What is the reliability of the following network given all reliabilities are 0.9? Example FE 0.9 E B DC A FE 0.97848 0.81 0.97848 0.99591

19. 0.75 Example : All components have reliability 0.5 Strategy: Reduce to a simple bridge circuit 0.375

20. All components have reliability 0.5 unless otherwise shown 0.375 0.75 0.56250.4375 0.375  0.5625 + 0.6255  0.4375 0.25 Rel = = 0.4846 0.375 Bridge working 0.625 Bridge not working

21. Reduce the following to a workable circuit. Example