Thursday, September 8, 2011

Minimal cut set and Fault tree analysis

Fault tree: Fault tree diagrams represent the logical relationship between sub-system and component failures and how they combine to cause system failures. The TOP event of a fault tree represents a system event of interest and is connected by logical gates to component failures known as basic events.


Figure is a fault tree: a tree where A,B,C,D,E,F are sets of fault events that can cause software to fail.
Here, minimal events for software to fail is {A,C},{B,C}.... These {A,C},{B,C} are the minimal cut set [a minimal set of events to cause failure, whose further sub set do not cause failure]

Conclusion: We should always strive to make minimal cut set as big as possible. In above figure  the desirable minimal cut set would be {A,B,C,D,E,F}
Why? :
Say was have minimal cut sets : {A,C},{B,C},{A,C},{A,E},{A,F}
So, Probability to fail = P{A,C}+P{B,C}+P{A,C}+P{A,F}
                                =P(A)*P(B)+P(B)*P(C)+P(A)*P(C)+P(A)*P(F)
When mininal cut set is{A,B,C,D,E,F}
Probability to fail = P{A,B,C,D,E,F}
                          = P(A)*P(B)*P(C)*P(D)*P(E)*P(F)

Therefor, P(A)*P(B)*P(C)*P(D)*P(E)*P(F) < P(A)*P(B)+P(B)*P(C)+P(A)*P(C)+P(A)*P(F)



2 comments:

  1. I have a question here. How {A,B,C,D,E,F} is a minimal cut set here? because minimal cut set means if I remove any one of the set member then remaining event should not be a cut set. If I remove C and D then still remaining are cut set..!! can you explain this..!!

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  2. {A,B,C,D,E,F} is the supposition cut set it self,, to derive a minimal cut set , A,B,C,D... notations are not sufficient we need a real life scenartio.

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