Fault tree analysis & minimum cut set question

by FizziCystWanna
Tags: analysis, fault, minimum, tree
FizziCystWanna is offline
Jan23-13, 06:40 PM
P: 1
I'm picking up FTA theory from "An Introduction to Reliability and Maintainability Enginenering", Charles Ebeling. In his final FTA example, he has an event consisting of the intersection of 3 events:

event = {wheel subsystem failure, b3, b4}

"Wheel subsystem failure" (wsf) is further expressible as:

wsf = (w1+b1)(w2+b2)(w3+b3)(w4+b4)

where I use algebraic multiplication and addition to represent set intersection and union, respectively. The b's represent 4 brake-pad assemblies (1 per wheel) and the w's represent 4 wheel cylinders. I re-expressed this as:

event = wsf b3 b4

= (w1+b1) (w2+b2) x
[ w3 w4 + b3 w4 + w3 b4 + b3 b4 ] b3 b4

The 2nd line can be simplified using the set theory identify (A+B+C)C = C, with C being b3.b4:

event = (w1+b1) (w2+b2) b3 b4

Since it decomposes into 4 terms in sum-of-products form, this should yield 4 minimum cut sets. But this disagrees with the statements in the textbook:

"The wheel subsystem failure can be decomposed into 16 combinations of wheel cylinder and brake-pad assembly failures. In the latter case, four of these decompositions include failure of both b3 and b4. Therefore, only 12 unique cut sets are formed."

I actually expanded out the wsf expression, and indeed I get 16 terms, 4 of which contain b3.b4. The only way that I know of to drop only those 4 terms is if b3.b4 was a cut set that was identified from another part of the tree. This is definitely not the case. So from my newbie perspective, my reduction of cut sets using (A+B+C)C=C is the only way, yielding far fewer minimum cut sets than implied by the author.

What am I missing?
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