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



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


