# Sequential-server busy period with poisson arrival and general service time distribution (M/G/sequential 2 server)

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.

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