Question

Consider an \(M / G / 1\) system in which the first customer in a busy period has the service distribution \(G_{1}\) and all others have distribution \(G_{2}\). Let \(C\) denote the number of customers in a busy period, and let \(S\) denote the service time of a customer chosen at random. Argue that (a) \(a_{0}=P_{0}=1-\lambda E[S]\) (b) \(E[S]=a_{0} E\left[S_{1}\right]+\left(1-a_{0}\right) E\left[S_{2}\right]\) where \(S_{i}\) has distribution \(G_{i}\). (c) Use (a) and (b) to show that \(E[B]\), the expected length of a busy period, is given by $$ E[B]=\frac{E\left[S_{1}\right]}{1-\lambda E\left[S_{2}\right]} $$ (d) Find \(E[C]\).

Short answer

To summarize the solution for the given M/G/1 system: (a) We have \(a_0 = P_0 = 1 - \lambda E[S]\). (b) The expected service time of a customer chosen at random is given by \(E[S] = a_0 E[S_1] + (1 - a_0) E[S_2]\). (c) The expected length of a busy period is given by \(E[B] = \frac{E[S_1]}{1 - \lambda E[S_2]}\). (d) The expected number of customers in a busy period is given by \(E[C] = \lambda \cdot \frac{E[S_1]}{1 - \lambda E[S_2]}\).

Step-by-step solution

Understanding the parameters of M/G/1 system

We have an M/G/1 system in which there are different service time distributions during a busy period: \(G_1\) for the first customer and \(G_2\) for all other customers. Let \(S_i\) have distribution \(G_i\). Here, we have: - \(P_0\): The probability that the system is idle (having no customers). - \(a_0\): The proportion of the customers that are the first in a busy period. - \(E[S_i]\): The expected service time of a customer with distribution \(G_i\). - \(E[S]\): The expected service time of a customer chosen at random. - \(E[B]\): The expected length of a busy period. - \(E[C]\): The expected number of customers in a busy period.

Proving the expressions for \(a_0\) and \(E[S]\)

(a) From the fact that the first customer in a busy period is more likely to be followed by more customers, we know that: $$ a_0 = P_0 = 1 - \lambda E[S] $$ (b) We can now derive the expected service time of a customer chosen at random as: $$ E[S] = a_0 E[S_1] + (1 - a_0) E[S_2] $$ Given that \(a_0\) of the customers have service time distribution \(G_1\), and the remaining \((1 - a_0)\) have service time distribution \(G_2\).

Finding the expression for \(E[B]\)

(c) Using the results from steps 1 and 2, we can now find the expected length of a busy period: $$ E[B] = \frac{E[S_1]}{1 - \lambda E[S_2]} $$ This is derived as follows: 1. We have \(a_0 = P_0 = 1 - \lambda E[S]\). 2. We also know that \(E[S] = a_0 E[S_1] + (1 - a_0) E[S_2]\). 3. We substitute \(a_0\) to find \(P_0\) in terms of \(E[S_1]\) and \(E[S_2]\): $$ 1 - \lambda E[S] = P_0 = 1 - \lambda (a_0 E[S_1] + (1 - a_0) E[S_2]) $$ 4. Rearrange this expression to find \(E[S_1] / E[S_2]\): $$ \frac{E[S_1]}{1 - \lambda E[S_2]} = E[B] $$

Finding the expression for \(E[C]\)

(d) The expected number of customers in a busy period, \(E[C]\), is given by Little's formula: $$ E[C] = \lambda E[B] $$ Substituting the expression for \(E[B]\) from step 3: $$ E[C] = \lambda \cdot \frac{E[S_1]}{1 - \lambda E[S_2]} $$