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Write a computer programme that by means of stochastic simulation finds an approximation of the variance of a typical waiting time W(q) (in the queue) before service for a typical customer arriving to a steady-state M(1)/M(2)/1/2 queuing system. (In other words, the queuing system has exp(1)-distributed times between arrivals of new customers and exp(2)-distributed service times. Further, the system has one server and one queuing place.)

attempt at solution:

N = 100000

wait = vector(length=N)

for (i in 1:N) {

   t1 = rexp(1,1)           # arrival times
   s1 = rexp(1,2)           # service times

   if (s1<t1) {

      wait[i] = 0

    } else {
      wait[i] = s1
    }
}

VarWait = var(wait)

cat("Variance of a typical waiting time W(q) = ", VarWait, "\n")

I get 0.2704 but the answer should be 0.1413

This should be super simple but im stuck... Can anyone spot my mistake?

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I read the logic to be that at the start of each loop, we are in the state where the queue is empty and a customer is being serviced. Then $s1$ is his service time and $t1$ the time till the next customer arrival.

If $s1\lt t1$ then the time to wait for the next customer will be $0$ - so you have that right.

If $t1\lt s1$ then we have a new customer who has to wait. You have him waiting for time $s1$ but I think that's wrong. He begins waiting from the time he arrives, so we have to draw a new random service time, say $s2,$ and that will be his wait time; that is, the remaining service time of the previous customer.

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  • $\begingroup$ Yes, you are of course right. Thank you! $\endgroup$ – JKnecht Jan 6 '16 at 0:02

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