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. 
 A: Pretty much any intro to queuing theory or stochastic processes book will cover this, e.g., Ross, Stochastic Processes, or the Kleinrock, Queuing Theory.  
For an outline of a proof that memoryless arrivals lead to an exponential dist'n:
Let G(x) = P(X > x) = 1 - F(x).  Now, if the distribution is memoryless,
G(s+t) = G(s)G(t)  
i.e., the probability that x > s+t = the probability that it is greater than s, and that, now that it is greater than s, it's greater than (s+t).  The memoryless property means that the second (conditional) probability is equal to the probability that a different r.v. with the same distribution > t.
To quote Ross:
"The only solutions of the above equation that satisfy any sort of reasonable conditions, (such as monotonicity, right or left continuity, or even measurability), are of the form:"
G(x) = exp(-ax)  for some suitable value of a.
and we are at the Exponential distribution.
A: 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.
