Fault Tree Analysis
i. The following is the fault tree diagram for the given problem
ii. Minimal Cut Sets
TE = X1 + X2
TE = (X11 + X12) + (X21 • X22)
TE = ((A • B • C • E) + (A • B)) + ((A • C) • (A • C • D))
Te following is the result with the application of absorption and idempotent law:
TE = A • B + A • C • D (minimal cut set)
The complete derivation is as follows:
X1 = (A • B • C • E) + (A • B)
Assume A • B = Y, and C • E = Z; then,
X1 = Y • Z + Y
By absorption law,
X1 = Y
So X1 = A • B
Now, X2 = ((A • C) • (A • C • D)
Assume A • C = Y, then
X2 = Y • (Y•D)
By idempotent law,
X2 = (Y•D)
Thus X2 = (A • C • D)
Thus, TE = A • B + A • C • D
Thus TE occurs when events A and B occur or events A, C and D occur.
IV. Probability of occurrence of TE at t=1,000
P (A) = 1 – e-λt = 1 – e-0.0003 x 1,000 = 0.259
P (B) = 1 – e-λt = 1 – e-0.0004 x 1,000 = 0.33
P (C) = 1 – e-λt = 1 – e-0.0001 x 1,000 = 0.095
P (D) = 1 – e-λt = 1 – e-0.0006 x 1,000 = 0.451
P (TE) = P (AB + ACD)
= P (AB) + P (ACD) – P (ABACD)
= P (A) P (B) + P (A) P (C) P (D) – P (A) P (B) P (C) P (D)
= 0.259 x 0.33 + 0.259 x 0.095 x 0.451 – 0.259 x 0.33 x 0.095 x 0.451
= 0.08547 + 0.0111 – 0.00366
= 0.06591
Thus the probability of TE occurring at t=1,000 is 0.06591
