Discussion
A queueing system has one single server workstation that admits an infinitely long queue. The rate of arrival of jobs to the queueing system follows the Poisson distribution with a mean of 5 jobs/hour. The service time of the server is exponentially distributed with a mean of 6 minutes. In steady state operation of the queueing system, the probability that the server is not busy at any point in time is
Answer: C
The probability that the server is idle is 0.50.
This is an M/M/1 queue. The arrival rate is \(\lambda = 5\) jobs/hour; a mean service time of 6 minutes gives a service rate \(\mu = 60/6 = 10\) jobs/hour.
\[ \begin{aligned} \rho &= \frac{\lambda}{\mu} = \frac{5}{10} = 0.5 \\ P_{0} &= 1 - \rho = 0.50 \end{aligned} \]
\(\rho\) is the fraction of time the server is busy, so the idle fraction is \(1 - \rho\). Here both happen to be 0.5. 0.20, 0.17 and 0.83 would each need a different arrival or service rate from those given; with \(\rho = 0.5\) the idle probability can only be 0.5.
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