Fault Tree and Event Tree Analysis

Solution Preview

---

Hello please find attached. Note I haven't had time to check my work as yet

For the fault tree given below

Find the minimum cut sets

The system will fail if any of the following scenarios occur

A and C fail (A∩C)=A.C
B and C fail (B∩C)=B.C

A and X fail (A∩X)=A.X
B and X fail(B∩X)=B.X

Further
X will fail if either just B fails and D fails at the same time or E (on its own) fails whilst simultaneously A or B fails

Thus MCS are therefore from inspection

MCS∈{A.C,B.D,A.E,B.C,B.E}

 

If the probability of each basic event is 0.10, what is the approximate top event probability? Using the minimum cut sets, what is the exact top event probability?

Approximate top event probability
To approximate the system probability we will consider the probability of an OR
function follows mutually exclusive axiom (it is only an estimation after all) that
〖P(X+Y)=P(X)+P(Y)〗_
Probability of an AND function follows
P(X.Y)=P(X).P(Y)
Probability of Switch event is
P(Switch)=P(B)+P(E)=0.1+0.1=0.2
Probability of Emergency valve failure event is
P(EV)=P(D)+P(E)=0.1+0.1=0.2
Probability of secondary event X is thus
P(X)= P(Switch). P(EV)=0.2×0.2=0.04
Probability of Backup event is
P(BU)=P(X)+P(C)=0.04+0.1=0.14
Probability of Main event is
P(Main)= P(A)+P(B)=0.1+0.1=0.2
Finally probability of the top even failure is
P(Tank Rupture)= P(Main). P(BU)=0.2×0.14≃0.028

Exact top event probability using MCS (considering non mutual exclusivity)
Let P_1=P(A.C)=P(A).P(C)=0.1×0.1=0.01
Let P_2=P(B.D)=P(B).P(D)=0.1×0.1=0.01
Let P_3=P(A.E)=P(A).P(E)=0.1×0.1=0.01
Let P_4=P(B.C)=P(B).P(C)=0.1×0.1=0.01
Let P_5=P(B.E)=P(B).P(E)=0.1×0.1=0.01

Since considering the events are non-mutually exclusive we apply the probability axiom
for added probabilities

P(A.C+B.D+A.E+B.C+B.E)=1-∏_(i=1)^5▒〖(1-P_i)〗

P(A.C+B.D+A.E+B.C+B.E)=1-(1-0.01)^5=1-0.〖99〗^5=0.049

P(Tank Rupture)=0.049
 

Find a set of mutually exclusive cut sets (using a ...