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

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