Queuing Theory Questions and Answers

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.

Answer: D

The relationship that is not true is \(L_{s} = L_{q} + 1/\lambda\).

The correct relation is \(L_{s} = L_{q} + \lambda/\mu\): the mean number in the system exceeds the mean number in the queue by the mean number in service, which for a single server is the utilisation \(\rho = \lambda/\mu\). It follows from multiplying \(W_{s} = W_{q} + 1/\mu\) by \(\lambda\) and applying Little's law.

The other three are true. \(W_{s} = W_{q} + 1/\mu\) says time in system is waiting time plus mean service time. \(L_{s} = \lambda W_{s}\) and \(L_{q} = \lambda W_{q}\) are Little's law for the system and for the queue.

Answer: D

The average waiting time in the queue is \(L_{q}/\lambda\).

Little's law applied to the queue alone states \(L_{q} = \lambda W_{q}\): the average number waiting equals the arrival rate times the average wait. It holds for any infinite-source queue in steady state, whatever the arrival and service distributions. So \(W_{q} = L_{q}/\lambda\).

\(L_{q}/\mu\) divides by the wrong rate. \(\lambda/\mu\) is the traffic intensity \(\rho\), a pure number, and \(\mu/\lambda\) is its reciprocal; neither is a time.