I have some general uncertainties regarding the rejection process from an Erlang-B loss queue ($M/M/c/c$), where the total capacity of the queue is equal to the number $c$ of servers.

Consider the simplest case: $M/M/1/1$ with arrival rate $\lambda$ and service rate $\mu$. This is a birth-death process with two states $0$ and $1$, where the steady state probabilities are:

$P_{0} = \frac{1}{1 + \frac{\lambda}{\mu}}$

$P_{1} = \frac{\lambda}{\lambda + \mu}$

Call the Poisson input process $N(t)$. By the arrival theorem (PASTA) I claim that the output processes $N_1(t)$ (those who are served) and $N_2(t)$ (those who were rejected) is a subdivision of $N(t)$ where the rate of $N_1(t)$ is $P_{0} \lambda$ and $N_2(t)$ is $(1 - P_0)\lambda$.

Further, both $N_1(t)$ and $N_2(t)$ are both Poisson processes (because I can subdivide Poisson processes according to $Bern(P_0)$ and the resulting processes are Poisson).

My questions:

1) Are $N_1(t)$ and $N_2(t)$ indeed Poisson?

2) If question 1 is true, does Burke's Theorem apply to finite capacity queues? I'm pretty sure it doesn't. Clearly:

$N(t) \neq N_1(t) + N_2(t)$

because $N_1(t)$ is subject to service times, while $N_2(t)$ is not. I'm definitely misunderstanding something here, but I'm sort of ending up with the same conclusion as Burke's Theorem: the output processes are Poisson.

3) Because I'm only relying on the arrival theorem, if I let the service rate be some fixed value, I'll again end up with some $P_0$ and $P_1$ by the Erlang-B loss formula (turns out it'll be a higher probabilty of rejecting). If I use the same logic as above for the exponential service case, I very counterintuitively end up with the same conclusion, that $N_1(t)$ and $N_2(t)$ are both Poisson. What did I miss?

  • $\begingroup$ My current thinking is that this relies on the validity of the arrival theorem in this case, and that I just need to show that it holds pubsonline.informs.org/doi/abs/10.1287/opre.38.1.156 $\endgroup$ – bobbyzhivago Oct 12 '16 at 23:20
  • $\begingroup$ This may be relevant: cambridge.org/core/services/aop-cambridge-core/content/view/… $\endgroup$ – Math1000 Oct 14 '16 at 17:45
  • $\begingroup$ Thank you, this is definitely interesting. So far I think I've convinced myself that the arrival theorem holds, but although arrivals see time averages, successive rejections/blockings are not independent of one another. It looks like if you calculate the marginal distributions of the served output process and the rejected output process the dependence pops out. $\endgroup$ – bobbyzhivago Oct 15 '16 at 22:08

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