I want to compute the distribution of the waiting time and the number of jobs for M/GI/1 where the service time is Heavy-Tailed or more specifically Pareto. I found this paper http://dl.acm.org/citation.cfm?id=1340307. However, I cannot understand at the end of the day how I can compute these distributions. Can someone explain it to me simple?
M/GI/1 service time distribution
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$\begingroup$
probability-distributions
queueing-theory
2 Answers
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You can use the Pollaczek–Khinchine formula to compute the transform of the response time W as $W^\ast(s) = \frac{(1-\rho)s g(s)}{s-\lambda(1-g(s))}$ (where g(s) is the transform of service time pdf). See Diagle's book for more details
- Daigle, N. The Basic M/G/1 Queueing System. Queueing Theory with Applications to Packet Telecommunication (2005): 159-223 (pages 165-66 available on Google Books).
These may also be of interest, in particular the explicit results for your case (heavy tailed distributions).
- Takacs, L. Delay distributions for one line with Poisson input, general holding times, and various orders of service. Bell System Technical Journal 42 (1963): 487-504.
- Abate, Joseph, and Ward Whitt. Explicit M/G/1 waiting-time distributions for a class of long-tail service-time distributions. Operations Research Letters 25.1 (1999): 25-31.
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Is it M/G/1 or M/GI/1? Moreover, are you asking about just the time spent receiving service or total delay (service+wait time).
If it is just service time, then isn't it given by G?!
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1Hello Mahdi, welcome to math.stackexchange! As what you have posted is not really answer, but rather a request of clarification, it is better to post it as a comment and not as an answer. – 2012-11-06