# M/G/1 queue - probability that waiting time is zero

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

• Why the close votes? Maybe not entirely research level, but this looks to me like a reasonable question. – Daniel Moskovich Dec 16 '13 at 1:42

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.
• It is better to write the symbol of the exponential function as $\exp$ \exp than as $exp$ exp. – user43961 Dec 15 '13 at 21:11