Question

A system, which is composed of two components, will func- tion as long as at least one of the two components func- tions. When both components are operating, the lifetime distribution of each is exponential with mean 1. However, the distribution of the remaining lifetime of the good com- ponent, after one fails, is exponential mean 1 . (The idea 2 is that after one component fails the other component car- ries twice the load and hence has only half the expected life time.) Find the lifetime distribution of the system.

Answer 1

The lifetime distribution of each is exponential with mean 1. and the distribution of the remaining lifetime of the good component, after one fails, is exponential mean 1/2 .

Let X1 and X2 be the time till components 1 and 2 are working.

Let X be the time till both components are operating. Let Y be the time when only one component is working.

Then, X1 ~ Exp( = 1) and X2 ~ Exp( = 1)

X ~ min(X1, X2)

and Y ~ Exp( = 2) { = 1/mean = 1/(1/2) = 2}

The distribution of X is,

n(1 - F(x))^n-1 f(x) where n = 2, F(x) = 1- exp(-x) and f(x) = exp(-x) as X1, X2 ~ Exp(1)

= 2 exp(-x) * exp(-x)

= 2 exp(-2x)

Thus, X ~ exp( = 2)

and Y ~ exp( = 2)

Now, lifetime distribution of the system is X + Y.

We know that sum of 'n' exponential distribution with same parameter follows Gamma distribution with parameters n and .

Thus, X + Y ~ Gamma(2, 2)