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Suppose for a $M/M/\infty$ queue with arrival rate $\lambda$ and service rate $\mu$, with probability $p$ the service is successful and the person will leave the queue, and otherwise she will go back to the end of the queue. I am wondering if it is equivalent to another $M/M/\infty$ queue? If this is the case, what is this queue's arrival and service rates and how to prove it?

Thanks in advance!

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    $\begingroup$ since each server is memoryless, then sending a customer back into the queue is just like delaying the service a bit. $\endgroup$
    – user619894
    Aug 9, 2023 at 16:07
  • $\begingroup$ Got it, thank you! $\endgroup$
    – Zona
    Aug 10, 2023 at 8:37

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Let $S$ be the (total) service time of a customer and $N$ the number of times a customer is served. Then $S\mid N=n\sim\mathsf{Erlang}(n,\mu)$ and $N\sim\mathsf{Geo}(p)$, so by the law of total probability \begin{align} f_S(t) &= \sum_{n=1}^\infty f_{S\mid N=n} (t)\mathbb P(N=n)\\ &= \sum_{n=1}^\infty \frac{\mu(\mu t)^{n-1}}{(n-1)!} e^{-\mu t}(1-p)^{n-1}p\\ &= pe^{-\mu t}\sum_{n=0}^\infty \frac{((1-p)(\mu t))^n}{n!}\\ &= p\mu e^{-p\mu t}, \end{align} and hence $S\sim\mathsf{Expo}(p\mu)$.

The resulting detailed balance equations $\pi_n = \frac{\lambda}{np\mu}\pi_{n-1}$, $n\geqslant 1$ differ from those of a regular $M/M/\infty$ queue only by the factor of $\frac 1p$ on the right-hand side and accordingly the stationary distribution has Poisson distribution with mean $\frac\lambda{p\mu}$.

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