Stochastic process (renewal process)

[ewup]

A component in a manufacturing process breaks down regulary and needs to be replaced by a new component. Assume that the lifetimes of components are i.i.d. random variables. The company adopts this policy: a component is replaced when it breaks down or after it has operated for time "a", whichever comes first. "a" is a fixed positive parameter.

Question> Assume the lifetime of the component is exponentially distributed with rate "alpha". Compute the mean time between replacements. Let B(t) be the forward recurrence time so that P{B(t)>x} is the probability that there will be no replacement for another x time units. Find the lim P{B(t)>x} when t goes to infinity.

I thought that this would be solved by inspection paradox. Right?

 

[lippyka]

Answer: Yes, the inspection paradox can be used to solve this problem. The mean time between replacements is given by 1/α (the inverse of the rate parameter). The limit of P{B(t)>x} as t goes to infinity is given by the probability that a component has a lifetime greater than x+a. This can be calculated using the cumulative distribution function for an exponential random variable (F(x) = 1 - e^(-α(x+a))).