Reliabilty of complex systems using F-V algorithm for approximation

[abba02]

For a system with minimum cut set (C1,C2,C3),(C4,C5,C6) and (C4,C5,C7), Using the F-V algorithm, calculate the first and second order approximations given P[C1] =P[C1] =
P[C2] =0.107, P[C3] =0.168,P[C4] =P[C5] =0.0478,P[C6] =0.0921,P[C7] =0.102
Answers, Ist order= 0.002366919, 2nd order=0.002366017

Attempt: Ps1= C1 AND C2 AND C3=.107 *.107*.168= .001923432
Ps2= C4 AND C5 AND C6=.000210433764
Ps3= C4 AND C5 AND C7=.00023305368
p[F] IST ORDER = SUM OF Ps=.001923432+.000210433764+.00023305368=.002366919

Second order approximation.? Please can anybody help me with how to come up with the second order approximation, based on the above. Tried a couple of textbooks but still lost. Remember all minimum cut sets are in parallel as per block diagram but C6 and C7 are in series.

 

[mmwave]

Hello, as a scientist, I am happy to help you with calculating the second order approximation using the F-V algorithm. The first step is to calculate the failure probability (P[F]) for each minimum cut set (MCS) using the formula P[F] = P[C1] * P[C2] * ... * P[Cn], where P[Cn] is the probability of component Cn failing.

For the first minimum cut set (C1, C2, C3), we already have the values given in the problem, so P[F] = 0.001923432.

For the second minimum cut set (C4, C5, C6), we need to calculate the probability of C6 failing given that C4 and C5 have already failed. This can be done using the formula P[C6|C4,C5] = P[C6 AND C4 AND C5] / P[C4 AND C5].

P[C6 AND C4 AND C5] = P[C6] * P[C4] * P[C5] = 0.0921 * 0.0478 * 0.0478 = 0.000210433764

P[C4 AND C5] = P[C4] * P[C5] = 0.0478 * 0.0478 = 0.00228884

Therefore, P[C6|C4,C5] = 0.000210433764 / 0.00228884 = 0.0000919

Now, we can calculate the failure probability for the second minimum cut set as P[F] = P[C4] * P[C5] * P[C6|C4,C5] = 0.0478 * 0.0478 * 0.0000919 = 0.0000002088

Similarly, for the third minimum cut set (C4, C5, C7), we can calculate P[C7|C4,C5] using the same formula as above.

P[C7|C4,C5] = P[C7 AND C4 AND C5] / P[C4 AND C5]

P[C7 AND C4 AND C5] = P[C7] * P[C4] * P[C5] = 0.102 * 0.0478 * 0.0478 = 0.000