System contains 3 branches in parallel. Branch 1 holds unit 1&2 in series. Branch 2 holds unit 3&4 in series. Branch 3 holds unit 5 alone. The hazard rates are as follows: 1: 0.3/year 2: 0.2/year 3: 0.25/year 4: 0.22/year 5: 0.35/year Find the reliability Solution relaiabilty of each branch is r1=e^(-0.3t) similarly others now since 1&2 in series then r11=r1*r2=e^(-0.5t) now since 3&4 in series then r12=r3*r4=e^(-0.47t) since the three brqaches are in parallel then R=1-(1-e^(-0.5t))(1-e^(-0.47t))(1-e^(-0.35t)).

1. System contains 3 branches in parallel. Branch 1 holds unit 1&2 in series. Branch 2 holds unit
3&4 in series. Branch 3 holds unit 5 alone.
The hazard rates are as follows:
1: 0.3/year
2: 0.2/year
3: 0.25/year
4: 0.22/year
5: 0.35/year
Find the reliability
Solution
relaiabilty of each branch is r1=e^(-0.3t)
similarly others
now since 1&2 in series then r11=r1*r2=e^(-0.5t)
now since 3&4 in series then r12=r3*r4=e^(-0.47t)
since the three brqaches are in parallel
then R=1-(1-e^(-0.5t))(1-e^(-0.47t))(1-e^(-0.35t))