Loss networks and Poisson processes

Question

Consider a shop with capacity C, where customers spend an i.i.d exponential(1) amount of time before leaving through the back door. Customers arrive through the front door as a Poisson($\lambda$) process. If a customer arrives when the shop is already full, then they immediately leave through the side door.

Say the equilibrium probability that the shop is full is $p$, and that the system is in equilibrium.

I have seen some notes on Erlang alternative routing that refer to essentially the setup above, and claim that the process of exits through the side door is just a Poisson($p \lambda$) process.

Is this even true? To me it seems very far from obvious, as the side-door process is not a simple "thinning" of the front-door process. I suspect the PASTA property is useful here, but even then it seems more work is needed since arrivals do not see the state of the shop independently of each other.

I understand that some renewal-reward argument can be made to say the long-term rate of flow through the side door is $p \lambda$, but that is all I can manage to say. This setup seems a very general one, so I would really like to get things straight in my head.

If indeed the statement is true, please could someone offer a proof?

Answer 1

The statement seems to be false, indeed. Let us consider the case $C=1$, so the shop is either full or empty. Then $0<p=\frac{\lambda}{1+\lambda}<1$. Consider the time intervals from $0$ to $T$ and from $T$ to $2T$ with small $T>0$.

Consider the provability that there was at least one exit through the side door during each of these intervals. If the claim were true, it would be $(p\lambda T)^2 + O(T^3)$. On the other hand, it is at least the probability $p$ that the shop was full at time $0$ (or any other time, since we are at the stationary state) times the probability $e^{-2T}$ that the shop customer that was present at time $0$ hasn't left by the time $2T$ times the probability $\lambda T+O(T^2)$ of an arrival between $0$ and $T$ times the same probability of an arrival between $T$ and $2T$, which gives $p\lambda^2 T^2+O(T^3)$. For small $T>0$ this quantity is clearly above the conjectured one.

Answer 2

In my opinion, @fedja is right. The fact that the flow of lost arrivals is not Poisson can also be seen from the tansform domain.

Let $C=1$.

Let $\tau_0(s)$ - Laplace-Stieltjes transform of the time until next loss if right now the shop is empty.

Let $\tau_1(s)$ - Laplace-Stieltjes transform of the time until next loss if right now the shop is full.

Since arrivals and service are independent, then we can use the properties of Laplace-Stieltjes transforms to get: $$ \tau_0(s)={1 \over \lambda + s} \tau_1(s), $$ $$ \tau_1(s)={\lambda \over \lambda+\mu} {1 \over \lambda+\mu + s} + {\lambda \over \lambda+\mu} {1 \over \lambda+\mu + s}\tau_0(s). $$ One can solve this system for $\tau_0(s)$ and $\tau_1(s)$.

Let $\alpha_0 (\alpha_1)$ be the fraction of time that the shop is empty (full). Then the time between two consequtive losses is $$ \alpha_0 \tau_0(s) + \alpha_1 \tau_1(s). $$

If this sum would be the Laplace--Stietjes tarnsform of the exponential distribution, it would be equal to ${1\over \alpha_1 \lambda + s}$. But it is not.