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Stochastic Processes

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Spare Parts for an Alarm System.

Consider an alarm system that operates continuously. The alarm system contains a minicomputer that is critical for the alarm system to function properly. Accordingly, a large number of spare minicomputers are maintained, so that the minicomputer can be instantaneously replaced whenever it fails.

Each minicomputer contains four components. The minicomputer remains functioning as long as at least two of the four components are functioning, but these components cannot be repaired or replaced when they fail. The alarm system is kept operational by instantaneously replacing the minicomputer with a spare minicomputer whenever a minicomputer fails.
Assume that the lifetimes of the components of each minicomputer are mutually independent random variables with exponential distributions. Let the mean lifetime of component j be 1/ j years, 1 <= j <= 4.
(a) What is the probability that no component of a minicomputer fails during the &#64257;rst 0.2 years of operation of that minicomputer?
(b) What is the probability that component 1 of a minicomputer is the &#64257;rst component of that minicomputer to fail?
(c) What is the expected lifetime of a minicomputer?
(d) What is the variance of the lifetime of a minicomputer?
(e) Let N(t) be the number of replaced minicomputers in the alarm system during the time interval [0,t], assuming that the alarm system starts with a new minicomputer at time 0. Is the stochastic process {N(t): t >= 0} a Poisson process? Explain.
(f) What is the approximate probability that at least 50 minicomputers have to be replaced during the &#64257;rst 10 years of operation of the alarm system?

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The solution contains the determination of reliability, mean and variance of a k out of n system. The spare parts for an alarm systems are determined.

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Suppose and denote the lifetimes of the four components of the minicomputer. Here it is given that the lifetimes of the components are mutually independent exponentially distributed random variables with j-th component has mean lifetime 1/j, j = 1,2,3,4.
Thus the survival functions are , , and
a) The probability that no component of a minicomputer fails during the first 0.2 years of operation is given by

= , since the lifetimes of the components are mutually independent.
=
= 0.135335
b) At any time 't' the probability that component 1 of a minicomputer is the first component of that minicomputer to fail is ...

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