For an $M/M/c$ queue where arrivals follow a Poisson distribution with rate $\lambda$ and are iid, service follows a Poisson process with rate $\mu$ and there are $c$ parallel servers, we can estimate the average queue time by $$ E[QT]\approx\frac{\rho^{\sqrt{2\left(c + 1\right)} - 1}}{c\left(1 - \rho\right)}\cdot\frac{1}{\mu}, $$ where $\rho= \lambda/c\mu$.

We also know that since $\rho$ is the utilization rate, it must be less than equal to 1 and for system stability, we need to have $\rho\le1-\epsilon$.

From the structure of $E[QT]$, we can see that it's nonlinear in $\rho$ and both high and low utilization rate mean we are going to have a long average queue time.

But let us assume that customers arrive at a very high rate and the value of ratio $\lambda/c\mu$ is more than 1.

How would that impact the average queue time? In the estimation formula above, it doesn't make any difference if the ratio $\lambda/c\mu$ is 2 or 1000. But in reality, wouldn't a large $\lambda$ mean a very long waiting time?


1 Answer 1


I do not agree with the assertion that "both high and low utilization rate mean we are going to have a long average queue time". If $\rho$ is low (close to 0), $E[QT]$ is close to 0. Low utilization makes for short waits.

As far as $\rho > 1$ is concerned, that will cause the queue to "explode". In theory, wait times will diverge to infinity. In practice, either you will run out of arrivals because all possible customers are stuck in the system or, more likely, the system will become clogged and either turn off arrivals or shut down entirely.

  • $\begingroup$ Thank you. Where in the formulation do we see the queue exploding or having a long waiting time? $\endgroup$
    – user9659
    May 13, 2022 at 22:27
  • $\begingroup$ Operating continuously (meaning the queue never empties), the expected rate at which services are completed (customers leave) is $c \mu.$ If $\rho > 1,$ $\lambda > c \mu,$ meaning the average arrival rate exceeds the average departure rate. So customers are entering on average faster than customers are leaving. $\endgroup$
    – prubin
    May 14, 2022 at 16:26
  • $\begingroup$ What confuses me is the average queue time which remains the same for any $\rho>1$. But if $\lambda\gg c\mu$, wouldn't that mean some customers wait for a very very long time? If so, why don't we see that in the $E[QT]$ formulation? $\endgroup$
    – user9659
    May 14, 2022 at 16:54
  • $\begingroup$ @Sigma, what mentioned by Prof. Rubin in the last sentence of the first paragraph should answer your question. Just for a long waiting time, the rule is vice versa. $\endgroup$
    – A.Omidi
    May 14, 2022 at 18:35
  • 1
    $\begingroup$ The formula you have for $E[QT]$ does not apply when $\rho > 1.$ $\endgroup$
    – prubin
    May 14, 2022 at 21:57

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