Reliability of Networks
Dec 20, 2015
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
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