Customers come to receive a service at a Poisson process of intensity $\lambda$. They are served one at a time and the service time is exponentially distributed parameter $\mu$. In addition, customers waiting in line to be served become impatient and they leave to file of rate $\sigma$ independently of each other. Determining the stationary distribution in the case where $\sigma = \mu$.

$X_t$ : The number of customers in the queue (the number of states is infinite)

$A$ : The time before the arrival of the next customer - $A \sim Exp(\lambda)$

$B$ : The service time with a customer - $B \sim Exp(\mu)$

$C$ : The time before a customer in the queue become impatient - $C \sim Exp(\lambda)$

$S_i$ The residence time in the state i.

Here we have a birth and death process with $S_0 = A \sim Exp(\lambda) $ and $S_i = \min \{A,B,C\} \sim Exp (\mu + \sigma + \lambda)$. Hence, we got $\mu_i=\mu + \sigma = 2 \sigma$ and $\nu_i=\lambda$.

Question : In the answer key, the value of $\mu_i=\mu+(i-1)\sigma$? Is anyone could explain to me in details why this answer is true instead of $\mu_i=2 \sigma$? I think I don't understand something in the theory of Markov process, particularly the birth and death process.

  • $\begingroup$ If there are $i$ customers in the system, then one is in service and $i-1$ are waiting and subject to leaving. Hence, $\mu + (i-1)\sigma$. $\endgroup$ – Math1000 Mar 22 '16 at 21:05
  • $\begingroup$ I already know that, but could you explain that in using the definition of $S_i$ and the exponential distribution? $\endgroup$ – user320554 Mar 22 '16 at 21:08

In state $i$ there are $(i-1)$ customers in the queue. Each of them will leave the queue with a rate of $\sigma$. So it is more likely to transition to a lower state if there are many customers in the queue, hence $\mu_i$ has to take the number of customers in the queue into account.


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