Customers arrive at a system according to a Poisson process with rate $\lambda$. If server 1 is free when a customer arrives then they enter service with server 1. Otherwise they form a queue and wait. Service times at server 1 are i.i.d random variables with cdf $G_1$. Upon completion of service at server 1 the customer enters service at server 2, provided that server 2 is free. If server 2 is busy the customer will wait at server 1 and block it until server 2 is free. Service times at server 2 are i.i.d random variables with cdf $G_2$. The system is said to be busy whenever server 1 is busy or blocked, and is idle otherwise. A busy period is defined as the time period that starts when server 1 first becomes busy and ends when server 1 becomes free (at which point there will be exactly one customer in the system, and they will be starting service at server 2). Assume that at time t = 0 a single customer starts service at server 2 and no customers are at server 1 (so the system is initially idle). Find the expected length of the first busy period.

I just define $B_{(2)}$ as the length of busy period that starts with a customer in server 1 and a customer in server 2. Let $S_1$ denote the random variable of service time of server 1 and $S_2$ denote the random variable of service time of server 2. Then we can get $B_{(2)} \sim q(S_1 + \sum\limits_{i=1}^{N(S_1)}) + p(S_2 + \sum\limits_{i=1}^{N(S_2)})$, Where $q=P\{S_1 < S_2\}$ and $q = 1-p$. Then, we can get the expression of $E[B_{(2)}]$ by taking expectation on both side.

Then, Let $B$ denote the random variable of length of busy period starting with only one customer in server 2. By condition on the time of first arrival, we can get $B \sim \int_{0}^{\infty} (q^{\prime}(S_1 + \sum\limits_{i=1}^{N(S_1)}B_{(2)}) + p^{\prime}(S_2 - t + \sum\limits_{i=1}^{N(S_2 - t)}B_{(2)}) ) \lambda e^{-\lambda t}dt$, where $q^{\prime} = P\{S_2 < t +S_1\}$ and $p=1-q$.

I think I can get a very complex expression of $E[B]$ by using this method. But I cannot think of any other method.

I want to know if there are any other solutions or simple expression for the expectation.

  • $\begingroup$ I highly doubt there will be a simple expression for this considering the general service times. $\endgroup$ – Math1000 Oct 6 '19 at 4:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.