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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.

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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.

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  • $\begingroup$ Thanks. I know in general, the departure process of M/G/1 is not Poisson, but this is a special case of M/G/1 where the service times are hyper-exponential meaning that different customers experience an exponential service time with different mean value. That's why I am not sure. $\endgroup$
    – Cror2014
    Commented May 19 at 6:56
  • $\begingroup$ I found a lecture notes file which says the departure process is Poisson. refer to page 35 at cse.wustl.edu/~jain/cse567-13/ftp/k_30iqt.pdf $\endgroup$
    – Cror2014
    Commented May 19 at 7:27
  • $\begingroup$ I found no statement about the departure process is a Poisson process in the provided notes. Can you explicitly point out that? $\endgroup$
    – Weimin Dai
    Commented May 21 at 1:55
  • $\begingroup$ Check the 35th page of the pdf file not the slide numbers. Actually, it seems to be about the case with multiple servers, each having exponential service time, which means M/M/m. But I think we may be able to approximate the departure process with Poisson when at each time, only one of the servers is active. $\endgroup$
    – Cror2014
    Commented May 21 at 8:36

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