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so: I have a M/G/1-queue with Poisson arrivals with rate lambda=1 and the service time being the sum of two exp-distributed variables vith rates u1=1 and u2=2.

If we let Wq be the time an average customer spends in the queue, what is the probability that Wq equals 0, i.e. P(Wq=0)=?.

Formulas for M/M/1-queues can't be used as we have a M/G/1-queue, hence something else must be tried.

I tried to look at it as a regenerative process, with renewal moment being when the queue empties, but I couldn't get that approach to work.

Another idea was to model it as a markov-chain, but the equilibrium equations aren't very nice to solve, at leat not from what I can see.

Is either of these approaches the right way to go, or are there any other ideas how to beat this problem?

Thanks, Niklas

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  • $\begingroup$ Why the close votes? Maybe not entirely research level, but this looks to me like a reasonable question. $\endgroup$
    – Daniel Moskovich
    Commented Dec 16, 2013 at 1:42

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Did you look at the Pollaczek–Khinchine transform? The distribution is hypo-exponential, which is simple in the bi-variate case, $$2(\exp(-x)-\exp(-2x)).$$ The Laplace transform of this is simple to find which should enable the Pollaczek–Khinchine transform.

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  • $\begingroup$ It is better to write the symbol of the exponential function as $\exp$ \exp than as $exp$ exp. $\endgroup$
    – user43961
    Commented Dec 15, 2013 at 21:11
  • $\begingroup$ Thank you for the answer. If I should believe the simulations I've been running, I got it right in the end! $\endgroup$
    – Niklas Andersson
    Commented Dec 16, 2013 at 12:28

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