Suppose there are 2 independent queues A and B having m and n customers respectively. The service time of each of the queues are Exponentially distributed with service rate as $\mu_{A}$ and $\mu_{B}$. There are external Poisson ($\lambda$) arrivals to only queue A, i.e., queue A behaves like an M/M/1 queue while queue B has no external arrivals. What is the probability that queue A becomes empty before queue B? What is the probability that queue B empties before queue A?

My response: The mean time to empty a queue with $m$ customers and external arrivals of customers is $m\left(\mu_{A}-\lambda\right)^{-1}$. Thus, to determine the required probability, I adjust the value of $\mu_{A}$ to $\mu_{A} - \lambda$, which is the net rate at which the queue length of type $i$ customers decreases at station 1. Now, the M/M/1 queue becomes static queue with no arrivals but net service rate as $\mu_{A} - \lambda$. Now, I compare the probability that sum of m exponential random variable is less than the sum of n random variable. Is this approach correct? If so, why? If not, why (and how should I do it otherwise)?

  • $\begingroup$ What have you done so far? $\endgroup$ – saulspatz Jun 12 '18 at 20:56
  • $\begingroup$ @saulspatz, I've edited the question to show my approach. $\endgroup$ – RSA Jun 12 '18 at 21:04

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