Splitting a Poisson Process: M/G/Infinity Queue

1. Notation

• M: "Markovian" or "Memoryless" arrival process (i.e., Poisson Process)
• G: General service time (not necessarily exponential)
• $\infty$: Infinite number of servers

Let 

• $X(t)$ be the number of customers who have completed service by time t
• $Y(t)$ be the number of customers who are being served at time t
• $N(t)$ be the total number of customers who have arrived by time t

2. Splitting the arrival process

• Fix a reference time T.
• Consider the process of customers arriving prior to time T. 
• A customer arriving at time $t \leq T$ is
• Type I: if service is completed before T
• occur with probability $P_i(t) = G(T-t)$
• Type-II: if customer still is service at time T
• occur with probability $P_{II}(t) = G^c(T-t)$

Since arrival times and services times are all independent, the type assignments are independent. Therefore, 

• $X(T)$ is a Poisson random variable with mean $\lambda \int^T_0 P_I(t) dt = \lambda \int^T_0 G(T-t) dt = \lambda \int^T_0 G(t)dt$.
• $Y(T)$ is a Poisson random variable with mean $\lambda \int^T_0 P_{II}(t)dt = \lambda \int^T_0 G^c(T-t)dt = \lambda int^T_0 G^c(t)dt$
• $X(T)$ and $Y(T)$ are independent

What happens when $T \to \infty$

• $G(t) \approx 1$ for large t. Therefore, $X(T)$ is a Poisson random variable with mean $\lambda t$
• $Y(T)$ is a Poisson random variable with mean $\lambda \int^T_0 G^ct)dt = \lambda E[G]$

Summary: Number of customers in service in an $M/G/\infty$ queue, in steady state, is a Poisson random variable with mean $\lambda E[G]$.

3. Example