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EEL6507 Spring 2009, Lecture 28, Friday 2009/03/23 (Notes created by Xiaoyuan Li)

Closed Jackson Networks
Cyclic Queues

Closed Jackson Networks

An example

Example of Closed Jackson Networks

$K i$ = number of customers in system i.

To solve the problem, recall that

$\vec \lambda = \vec \gamma+\vec \lambda\vec R$

If $\vec \gamma=0$ ,then $\vec \lambda =\vec \lambda\vec R$

Therefore,

$\vec\lambda$ is +1 eigenvector of $\vec R$ .

Let us define

$X_i = \frac {\lambda_i}{\mu_i}$

$\beta_i(k_i)= \begin{cases} K_i !,&k_i<m_i\\ m_i ! m_{i}^{k_i-m_i},&k_i \ge m_i \end{cases}$

$m i$ = number of servers at node i

Closed Network has almost product solutions.

Product Solution

$P \left( K_1,K_2,...,K_n\right)= P (K_1)P (K_1)...P (K_n)$

It doesn't work for Closed Networks because $\sum_{i=1}^N K_i = K$

For M/M/m (Review in new notation)

$\begin{align} P(k_i) &= \frac{\frac{X_{i}^{k_i}}{\beta_i(k_i)}}{\sum_{j=0}^{\infty}\frac{X_{i}^{j}}{\beta_i(j)}}\\ &= \begin{cases} \left(\frac{\lambda_i}{\mu_i}\right)^k_i\frac{1}{k_i !},&k_i<m_i\\ \frac{m_i^{m_i}}{m_i !}\left(\frac{\lambda_i}{m_i \mu_i}\right)^{k_i},&k_i \ge m_i \end{cases} \end{align}$

Back to Closed Jackson Networks

Product solution holds except when

$\sum_{i}k_i \ne k, P(k_1,...,k_n)=0$

$\vec k = (k_1,k_2,...,k_n)$

$P(\vec k) = \frac{1}{G(\vec k)} \left(\frac{X_1^{k_1}}{\beta_1(k_1)}\right) \left(\frac{X_2^{k_2}}{\beta_2(k_2)}\right) ...\left(\frac{X_n^{k_n}}{\beta_n(k_n)}\right)$

Due to normalization

$\sum_{k,|\vec k|=K} P(\vec k)=1$

We can derive that

$G(\vec k) = \sum_{|\vec k|=K}\prod_{i=1}^{N} \left(\frac{X_i^{k_i}}{\beta_i(k_i)}\right)$

If $|\vec k| \ne K$ , then $G(|\vec k|) = \infty$

Bottleneck:

The nodes with greatest intensity

Node i is bottleneck if

$\frac{\lambda_i}{m_i \mu_i} > \frac{\lambda_j}{m_j\mu_j}$ for all $j \ne i$

then as $k \to \infty$

$P(\vec k) \to 0$ for all $\vec k$ with $k_i \to \infty$

or

customers all bunch up at the bottleneck

Cyclic Queues

The following is an example of K=1

Its Markov Chain Model is

$P 01 + P 10 = 1$

$μ 2 P 01 = μ 1 P 10$

According to the two equations, we get

$\begin{cases} &P_{10} = \frac{\mu_2}{\mu_1+\mu_2}\\ &P_{01} = \frac{\mu_1}{\mu_1+\mu_2} \end{cases}$

If K=2,N=2, the Markov Chain Model is

$\begin{cases} P_{02}+P_{11}+P_{20}=1\\ \mu_2 P_{02} = \mu_1 P_{11}\\ \mu_1 P_{20} = \mu_2 P_{11} \end{cases}$

$\begin{cases} P_{11} = \frac {\mu_2 \mu_1}{\mu_2 \mu_1 + \mu_2^2 + \mu_1^2}\\ P_{02} = \frac {\mu_1^2}{\mu_2 \mu_1 + \mu_2^2 + \mu_1^2}\\ P_{20} = \frac {\mu_2^2}{\mu_2 \mu_1 + \mu_2^2 + \mu_1^2} \end{cases}$