Consider an $G/G/1$ First In First Out (FIFO) queue in which $m$ people are waiting to be processed at time $t=0$. Then, people arrive according to i.i.d inter-arrival times with pdf $p(x)$ and rate $\lambda=(\int_{x=0}^{\infty} x p(x) dx)^{-1} $ and they are processed according to i.i.d processing times with pdf $q(x)$ and rate $\mu=(\int_{x=0}^{\infty} x q(x) dx)^{-1}$. Note that $\lambda=\mu$. What is the probability that from $t=0$ until the moment that the $n^{th}$ person leaves the queue, the queue never goes empty, as $n ,m \to \infty$, we know that $m<n$.

I can state this problem differently:

A Gambler receives dollars according to a renewal process (\$1 per arrival) with inter-arrival time pdf $p(x)$ and spends single dollars according to another renewal process (\$1 per arrival) with inter-arrival time pdf $q(x)$. Assume the rate of earning $\lambda$ is equal to the rate of spending $\mu$. If he starts (at time $t=0$) with $m$ dollars, what is the probability that his money never reaches zero between time $t=0$ and the moment at which he spends $n^{th}$ dollar bill (we know that $m<n$)?

Is there anywhere I can find the answer? That should be a problem already discussed. Any useful keywords for search?

  • $\begingroup$ What is meant by "he rate of arrival is the same as the rate of the process"? Does this mean $p(x)=q(x)$? $\endgroup$ – alphacapture Mar 21 '17 at 22:09
  • $\begingroup$ I just updated the problem statement to clarify this. $\endgroup$ – Susan_Math123 Mar 21 '17 at 22:12
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    $\begingroup$ Is there any reason to believe that the answer to this is known? I am not an expert in queueing theory, but my impression has been that we can say practically nothing about $G/G/1$ queues. $\endgroup$ – Misha Lavrov Mar 23 '17 at 2:18
  • $\begingroup$ @michael any idea how to solve this? $\endgroup$ – Susan_Math123 May 7 '17 at 0:16
  • $\begingroup$ @mikespivey any idea? $\endgroup$ – Susan_Math123 May 7 '17 at 0:20

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