Customers arrive at a single-server station with Poisson rate $\lambda$. A customer enters the bank if the server is available; otherwise, the customer leaves. The service times of successive customers are independent and have a common distribution $G$ and mean $\mu$. What is the rate at which the customer enters the system?

I am unable to figure out the answer. I assume that renewal reward process is to be applied here with regeneration happening at every service completion. Could anyone please help. Thanks in advance!

  • $\begingroup$ I've never done one of these problems, but it would seem that the sum of two poisson random variables is another poisson variable. E.g $X= X_{1} + X_{2}$ with mean $ \lambda = \lambda_{1} + \lambda_{2} $. It seems idealized but they enter the computer when with rate $ \lambda + \mu$ $\endgroup$ – Shogun Jul 6 '18 at 23:42
  • $\begingroup$ answer was give as λ/(1+λμ) . this is logic I could assume to arrive at the answer given . since arrival happens after each service completion (as customer who enters when server is busy leaves the system), inter-arrival for an accepted customer would be service time μ + 1/λ (exponential arrival has memoryless property), therefore arrival rate is λ/(1+λμ). $\endgroup$ – ranya Jul 7 '18 at 17:22

In your question it is implicitly assumed that the system operates in the steady state. For the steady state to exist the system's load (denote it by $\rho$) must be less than one i.e. $\rho=\lambda \mu<1$.

The system you are asking about, when operating in the steady state, follows the cycle $empty \rightarrow busy \rightarrow empty \rightarrow busy \rightarrow ...$. Thus the proportion of time that the system is empty is $$ {(mean \,\, empty \,\, time) \over (mean \,\, empty \,\, time) + (mean \,\, busy \,\, time)}= {{1 \over \lambda} \over {1 \over \lambda} + \mu}. $$

Thus the arrival rate of customers which find the system empty is $\lambda \times {{1 \over \lambda} \over {1 \over \lambda} + \mu}$.

p.s. The latter conclusion uses the PASTA property of the Poisson process. So in general it is not valid.

p.p.s. in Kendall's notation, you are asking about the $M/G/1/0$ queueing system. In any classic queueing theory book, there are many results on this queue.


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