3
$\begingroup$

I am working on two queues in tandem; the first queue is M/G/1 with hyper-exponential service times and the second queue has exponential service times. I want to know if the second queue can be modeled as an M/M/1 queue, or equivalently, the departure process of the first queue has Poisson distribution.

I have run simulations for two tandem queues and noticed that the simulation results for total delay match the sum of the analytical values for delays in M/G/1 and M/M/1 queues. I want to know if there is a good reference to show that the departure process of M/G/1 queue with hyper-exponential service times has Poisson distribution.

$\endgroup$

1 Answer 1

2
$\begingroup$

It is not true. You can refer to the following two papers, which provide a general method to analyze the departure process.

  1. The Departure Process of the GI/G/1 Queue and Its MacLaurin Series.
  2. Correlated queues with service times depending on inter-arrival times.

Using the methods in the two papers, it is not difficult to obtain that $$E[D^k]=\rho E[S^k]+(1-\rho) E[(A+S)^k],$$ where $D$ is the generic inter-departure time, $S$ is the generic service time, $A$ is inter-arrival times and $\rho=E[S]/E[A]$ is the traffic intensity. From this formula, we can conclude that the departure process of M/G/1 queue is not a Poisson process.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.