‘equivalent’ reliability

Example of ‘Equivalence Analysis’ for a Risk Network [Updated 10-2007]
Given the Risk Network below, find its equivalent overall RDF:
Assuming that the numbers given are reliabilities, one can develop an ‘equivalent’ reliability then find the RDF using
the complement rule. Begin by combining the various series and parallel combinations using the addition and
multiplication rules for probabilities.
1. D and I are simple ‘series’ or serial combinations so we can use the multiplication rule to combine D and I into D’
D’ = D X I = 0.95 X 0.99 = 0.9405
2. F and H are a parallel combination (‘OR’) so we can use the addition rule here where there is joint probability that
both F and H will function simultaneously.
F’ = F + H – (F X H) = 1.5 – (0.56) = 0.94
0.9 0.7
0.7
0.8
0.7
0.8
0.95 0.99
OUT
IN A B C
D
E F
H
I
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Risk Network Example0.9 0.7
0.7
0.8
0.94
0.9405 OUT
IN A B C
D’
E F’
Redrawing the network to show D’ and F’ equivalents:
3. Now combine B and C as simple serial equivalents as:
B’ = B X C = 0.7 X 0.8 = 0.56
4. And, E and F’ likewise as:
E’ = E X F’ = 0.7 X 0.94 = 0.658
Redrawing showing B’ and E’:
5. Now, B’ and E’ can be combined as a parallel combination as:
B’’ = B’ + E’ – (B’ X E’) = 0.56 + 0.658 – (0.56 X 0.658) = 1.232 – 0.3685 = 0.84952
0.9 0.56
0.658
IN A B’
0.9405 D’ OUT
E’
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Risk Network Example0.9 A 0.8495 B” 0.9405 D’ OUT
Redrawing now showing B’’ :
Finally, combining A with B’’ and D’ as a simple serial combination as:
A’ = A X B’’ X D’ = 0.9 X 0.84952 X 0.9405 = 0.7191
This is the equivalent Reliability of the network. Now, to find the RDF we subtract the Reliability from 1 as:
RDF = 1 – R = 1 – 0.7191 = 0.2809
Therefore, the Final network looks like this where the RDF equivalent is shown:
IN 0.2809 A’ OUT
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Risk Network Exampl