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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.

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 $\frac{\lambda(t)}{\mu }\ge1?$

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.

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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 $\frac{\lambda(t)}{\mu }\ge1?$