EEL6507 Spring 2009, Lecture 15, Wednesday 2009/02/09 (Notes created by Ajay Jain)

Finish M/M/m
M/M/I/k Queue

State Space Representation

Recall from previous result

$P_n=P_0 \prod_{k=0}^{n-1}\frac{\lambda_k}{\mu_{k+1}},$

and since we should have

$\sum_{n=0}^{m} F_n = \sum_{n=0}^{k} F_n + \sum_{n=k+1}^{m} F_n,$

hence,

$\prod_{n=0}^{k-1}\frac{\lambda_n}{\mu_{n+1}}=\prod_{n=0}^{r-1}\frac{\lambda_n}{\mu_{n+1}} \prod_{n=r}^{k-1}\frac{\lambda_n}{\mu_{n+1}},$

$P_k =P_0 \prod_{n=0}^{r-1}\frac{\lambda_n}{\mu_{n+1}} \prod_{n=r}^{k-1}\frac{\lambda_n}{\mu_{n+1}},$

$P_k =P_r \prod_{n=r}^{k-1}\frac{\lambda_n}{\mu_{n+1}},$

CASE 1: K < m-1

So, $λ k = λ,μ k + 1 = (k + 1)μ$

$P_k=P_0 \prod_{n=0}^{k-1}\frac{\lambda_k}{\mu_{k+1}} \frac{1}{n+1},$

$P_k=P_0 (\frac{\lambda_k}{\mu_{k+1}})^k \frac{1}{k!},$

CASE 2: K > m

So, $λ k = λ,μ k + 1 = m μ$

$P_k=P_m \prod_{n=m}^{k-1}\frac{\lambda}{m\mu},$

$P_k=P_m (\frac{\lambda}{m\mu})^{k-m} ,$

Where,

$P_k=P_0 (\frac{\lambda}{\mu})^m \frac{1}{m!},$

Solving Further with Pm value we get,

$P_k=P_0 (\frac{(m)^m}{m!})(\frac{\lambda}{m\mu})^{k} ,$

$\sum_{n=0}^{\infin} P_n = 1,$

$\sum_{n=0}^{m-1} P_n +\sum_{n=m}^{\infin} P_n= 1,$

$1=P_0 \sum_{n=0}^{m-1}(\frac{\lambda_k}{\mu_{k+1}})^k \frac{1}{k!}+P_0 \sum_{n=0}^{m-1} (\frac{(m)^m}{m!})(\frac{\lambda}{m\mu})^{k} ,$

$1=P_0 \sum_{n=0}^{m-1}(\frac{\lambda_k}{\mu_{k+1}})^k \frac{1}{k!}+P_0(\frac{(m)^m}{m!})\frac {(\frac{\lambda}{m\mu})^{m}}{1-\frac{\lambda}{m\mu}} ,$

Let,

$\rho=\frac{\lambda}{m\mu},$

$1=P_0 \sum_{n=0}^{m-1}(m\rho)^k \frac{1}{k!}+P_0\frac{(m\rho)^m}{m!}\frac{1}{1-\rho} ,$

$P_0=\frac{1}{ \sum_{n=0}^{m-1}(m\rho)^k \frac{1}{k!}+P_0\frac{(m\rho)^m}{m!}\frac{1}{1-\rho} },$

Then

$Prob\left[All\ Busy \right]=\frac{\frac{(m\rho)^m}{m!}\frac{1}{1-\rho}}{ \sum_{n=0}^{m-1}(m\rho)^k \frac{1}{k!}+P_0\frac{(m\rho)^m}{m!}\frac{1}{1-\rho} }$

This is ERLANG-C Formula