Question

there has to be at least one operating path. Once a component fails on a path, that path is no longer operating. The reliability of component 1 is 0.95, the reliability of component 2 is 0.95, the reliability of component 3 is 0.5, the reliability of component 4 is 0.79 and the reliability of component 5 is 0.6. What is the reliability of the entire system

Answer 1

Answer:

$R = P(R_1) P(R_2|R_1) P(R_3|R_2)*....*P(R_n |R_1,..., R_{n-1}$

For this case the reliability of the sytem would be given by:

$R= \prod_{i=1}^n R_i$

R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6

And replacing we got:

$R = 0.95*0.95*0.5*0.79*0.6= 0.21389$

Step-by-step explanation:

We can assume that the system work in series

If we have in general n units the reliability of the system is the probability that unit 1 succeeds and unit 2 succeeds and all of the other units in the system succeed. fror n units must succeed for the system to succeed. The reliability of the system is then given by:

$R = P(R_1) P(R_2|R_1) P(R_3|R_2)*....*P(R_n |R_1,..., R_{n-1}$

For this case the reliability of the sytem would be given by:

$R= \prod_{i=1}^n R_i$

R1= 0.95, R2 =0.95, R3= 0.5, R4 = 0.79 , R5 = 0.6

And replacing we got:

$R = 0.95*0.95*0.5*0.79*0.6= 0.21389$