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I have an inter arrival rate of a queue described by a continuous uniform distribution [0,freeParam]. I am trying to model such a queue using Kendall notation and confused whether continuous uniform distribution comes under Poisson process? Basically what to consider M/M/1 or G/M/1 if my service time is exponential. Thank you in advance for your help.

I did lot of study but couldn't find any justification or a clear understanding of what to be considered. Might be i am not understanding the basics right but any help on this would be very helpful. Thank you

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  • $\begingroup$ Can an inter arrival time assume any value from zero to infinity? If not, I doubt that it is a Poisson arrival process, but I'm not sure. $\endgroup$ – Bobson Dugnutt Aug 20 '16 at 23:21
  • $\begingroup$ Definitely G/M/1. Each M in M/M/1 stands for "Markov" and indicates an exponential distribution. Then the rate of arrivals is constant. $\endgroup$ – Did Aug 21 '16 at 9:31
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What do you mean by "inter arrival rate"?

If you mean that inter-arrival times are uniformly distributed, then the answer is no. The arrival process is not Poisson and the queue is a G/M/1.

The arrival process is only Poisson if the inter-arrival times follow an exponential distribution.

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  • $\begingroup$ Yes, the inter-arrival times are uniformly distributed. Thank you $\endgroup$ – Vijay Krishna Chalamalasetty Aug 21 '16 at 15:08

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