Math 311

Class 26

Single Server Exponential System (M/M/1)

• Arrivals are according to a Poisson process with rate \(\lambda_a\).
• Service times are independent exponential variables with rate \(\lambda_s\).
• There is one server with one queue.

\(\displaystyle L = \frac{\lambda_a}{\lambda_s - \lambda_a}\)

\(\displaystyle W = \frac{L}{\lambda_a} = \frac{1}{\lambda_s - \lambda_a}\)

\(\displaystyle W_S = \frac{1}{\lambda_s}\)

\(\displaystyle L_S = \frac{\lambda_a}{\lambda_s}\)

\(\displaystyle W_Q = W - W_S = \frac{\lambda_a}{\lambda_s(\lambda_s - \lambda_a)}\)

\(\displaystyle L_Q = \lambda_a W_Q = \frac{\lambda_a^2}{\lambda_s(\lambda_s - \lambda_a)}\)

Network of queues

System of servers, each with its own queue, where customers moves between servers according to some rules.

• Open system
• Closed system

Simple example

Tandem system

flowchart LR
START:::hidden --> A(Server 1) --> B(Server 2) --> END:::hidden

• \(\lambda_a\): arrival rate from outside
• \(\lambda_1\): service rate for server 1
• \(\lambda_2\): service rate for server 2

What is the probability there are \(n\) customers at server 1 and \(m\) customers at server 2?

\(\displaystyle P_{n,m} = \left(\frac{\lambda_a}{\lambda_1}\right)^n\left(1 - \frac{\lambda_a}{\lambda_1}\right) \left(\frac{\lambda_a}{\lambda_2}\right)^m\left(1 - \frac{\lambda_a}{\lambda_2}\right)\)

General Network

• \(k\) servers
• Arrivals to \(i\)-th server from the outside form a Poisson process with rate \(r_i\)
• Service times are exponential with rates \(\lambda_i\)
• When done at server \(i\), a customer will proceed to server \(j\) with probability \(P_{ij}\).

Example: 5 servers, arrival rates 5, 2, 2, 0, 0, service rates 4, 3, 2, 1, 1, the \(P_{ij}\) matrix given by

\[ \small \mathbf{P} = \begin{bmatrix} 0 & .6 & .4 & 0 & 0\\ 0 & 0 & .2 & .3 & .5\\ 0 & .1 & 0 & .3 & .4\\ 0 & 0 & 0 & 0 & .3\\ 0 & 0 & 0 & .7 & 0 \end{bmatrix} \]