Consider a $M/M/1$ queue with a constant arrival rate $\lambda$ and service rate $\mu$ with $\lambda < \mu$. We know that in this case the limiting distribution exists and it is a geometric distribution given by $$ P_n = P_0 \, \frac{\lambda^n}{\mu^n},$$ with $P_0 = \frac{1}{\sum\limits_{n=0}^{\infty} \frac{\lambda^n}{\mu^n }}.$

Now consider a time-varying $M_t/M_t/1$ queue, with arrival rate $\lambda(t)=4 +2\,sin(t)$ and service rate $\mu=5,$ so that in some time-interval we have $\lambda(t) \ge \mu(t)$ and hence the utilization becomes greater than or equal to $1$ priodically: $$\frac{\lambda(t)}{\mu }\ge1,$$ For all $t$ such that $ \frac{1}{2} \le sin(t) \le 1$.

My question is, for a time-varying queue in above, is the periodic limiting distribution calculated by the same formula?:

$$ P_n = P_0 \, \frac{\lambda(t)^n}{\mu^n},$$ with $P_0 = \frac{1}{\sum\limits_{n=0}^{\infty} \frac{\lambda(t)^n}{\mu^n }}.$

If so, how can we deal with time $t$ for which $\frac{\lambda(t)}{\mu }\ge1?$

I would like to know how you find the limiting distribution of the above time-varying queue.

  • $\begingroup$ Limiting distribution from what? $\endgroup$ Jun 14, 2021 at 8:58

1 Answer 1


No, that equation is not the correct limiting probability. With a periodic arrival, the limiting distribution will also be time-dependent, that is you should think of $P_n(t)$ for $t\in[0,p]$ where $p$ is the period of the arrival rate.

You can characterize the distribution using differential equations and solve them numerically, but there is no closed form solution. When the system transitions between underloaded and overloaded regimes fluid models provide good approximations. See https://wmassey.princeton.edu/20th%20Century/MtMt1%20sample%20path.pdf for details.

  • $\begingroup$ Thank you, this is what I expected. $\endgroup$
    – moonlight
    Jun 15, 2021 at 8:42
  • $\begingroup$ Fell free to accept the answer! $\endgroup$ Jul 21, 2021 at 2:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.