Abstract
This paper solves the machine interference problem in which N machines grouped by r groups and each group are looked after by an operative. It is assumed the machines of each group are identical and the run times of the machines in i th group have a general distribution, and the repair times i th group are assumed to have a negative exponential distribution with mean 1/ μi and furthermore we suppose the (i − 1) th group has priority over i th group and whenever an operative becomes free, then he will pay to next group. An explicit expression for the probability that a particular of each is found running in the steady state is derived. For this other useful results for the system can be obtained by using the steady-state probability distribution. It is shown that, these depend on the run time distribution only through the means of this distribution.
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