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.


1 Answer 1


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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