# Limiting distribution in $M_t/M_t/1$ queue

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.

• Limiting distribution from what? Jun 14 at 8:58

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.