Why is the Poisson distribution chosen to model arrival processes in Queueing theory problems?

When we consider Queueing theory scenarios where individuals arrive to a serving node and queue up, usually a Poisson process is used to model the arrival times. These scenarios come up in network routing problems. I’d appreciate an intuitive explanation as to why a Poisson process is best suited to model the arrivals.

Answer

The Poisson process involves a “memoryless” waiting time until the arrival of the next customer. Suppose the average time from one customer to the next is $\theta$. A memoryless continuous probability distribution until the next arrival is one in which the probability of waiting an additional minute, or second, or hour, etc., until the next arrival, does not depend on how long you’ve been waiting since the last one. That you’ve already waited five minutes since the last arrival does not make it more likely that a customer will arrive in the next minute, than it would be if you’d only waited 10 seconds since the last arrival.

This automatically implies that the waiting time $T$ until the next arrival satisfies $\Pr(T>t) = e^{-t/\theta}$, i.e., it’s an exponential distribution.

And that in turn can be shown to imply that the number $X$ of customers arriving during any time interval of length $t$ satisfies $\Pr(X=x) = \dfrac{e^{-t/\theta} (t/\theta)^x}{x!}$, i.e. it has a Poisson distribution with expected value $t/\theta$. Moreover, it implies that the numbers of customers arriving in non-overlapping time intervals are probabilistically independent.

So memorylessness of waiting times leads to the Poisson process.

Attribution
Source : Link , Question Author : Vighnesh , Answer Author : Michael Hardy

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