Post date: Feb 23, 2014 8:30:51 PM
Hello Scott,
I think the structure function given in slide 23 is wrong because it must be simplified (using boolean simplification : (x_i)^2=x_i) before to be used to compute the system’s reliability.
The exact formula (obtained from the system’s structure function) to compute the reliability of the bridge system (using Risk Calc) in the independent case:
r1=0.7
r2=0.75
r3=0.9
r4=0.95
r5=0.82
R=r1*r3+r2*r4+r1*r4*r5+r2*r3*r5-r1*r2*r3*r5-r1*r3*r4*r5-r1*r2*r4*r5-r2*r3*r4*r5-r1*r2*r3*r4+2*r1*r2*r3*r4*r5
R
0.91556
Which corresponds to the lower value obtained in the paper that I give you.
But when I use minimal paths sets, I obtain for the independent case :
r1*r3 ||| r2*r4 ||| r1*r4*r5 ||| r2*r3*r5
0.9784034
This value corresponds to the formulas that you have used :
R=1-((1-r1*r3)*(1-r2*r4)*(1-r1*r4*r5)*(1-r2*r3*r5))
R
0.9784034
And
r1*r3 | r2*r4 | r1*r4*r5 | r2*r3*r5
[ 0.7124999, 1]
for the total ignorance of dependency
Note that using interval values of r_i, I obtain :
r1*r3+r2*r4+r1*r4*r5+r2*r3*r5-r1*r2*r3*r5-r1*r3*r4*r5-r1*r2*r4*r5-r2*r3*r4*r5-r1*r2*r3*r4+2*r1*r2*r3*r4*r5
[ 0.1478099, 1.743078]
And using minimal paths :
r1*r3 ||| r2*r4 ||| r1*r4*r5 ||| r2*r3*r5
[ 0.9784033, 0.9960807]
The reuslt I obtained in the pape is the exact result because I compute only upper and lower bounds of system’s reliability R from upper and lower bounds r_i of components and thus I obtain : R=[0.9156, 0.9753]
All the best
Mohamed