The above circuit operates if and only if there is a path of functional devices
ID: 445107 • Letter: T
Question
The above circuit operates if and only if there is a path of functional devices from left to right. The reliability level (probability that device perform its intended function) of each device (A, B, C, D and E) is shown in the graph. Assume that devices (A, B,C, D and E) fail independently. The reliability level of the circuit, i.e. what is the probability that the circuit operates? A device called F can perform the functions of devices D and E at the same time and hence can replace them is now available on the market; At what conditions on device F's reliability level will you consider it to change D and E? What should be the reliability level of device F that let the circuit reach 75% of reliability? In order to increase the circuit's reliability, you are allowed to duplicate (*) only one of the device used: which device would you duplicate? And why? (*)Duplication is equivalent to redundancy, which simply means to insert an identical device in parallel to the used one. This is also called, active redundancy (see Figure below for an example)Explanation / Answer
1)
Given P(A)=0.5, P(B)=0.4, P(C)=0.8, P(D)=0.3, P(E)=0.4
P(circuit operates) = P(A)*P(B)*P(D) + P(A)*P(B)*P(E) + P(C)*P(D) + P(C)*P(E)
= (0.5*0.4*0.3) + (0.5*0.4*0.4) + (0.8*0.3) + (0.8*0.4)
= 0.06+0.08+0.24+0.32
= 0.7
2) On replacing D and E with F, P(circuit operates) > 0.7
P(circuit operates) = P(A)*P(B)*P(F) + P(C)*P(F)
=> 0.2*P(F) + 0.8*P(F) > 0.7
=> P(F) >0.7
On condition that F has reliability which is more than 0.7, it can replace D and E
If circuit reaches 75% of reliability, then P(circuit operates) = 0.75
=> P(A)*P(B)*P(F) + P(C)*P(F) = 0.75
=> 0.2P(F) + 0.8P(F) =0.75
=> P(F) = 0.75
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