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==Theorem==
In an open [[Jackson network]] of ''m'' queues where the [[utilization]] <math>\rho_i</math> is less than 1 at every queue, the equilibrium state probability distribution exists and for state <math>\scriptstyle{(k_1,k_2,\ldots,k_m)}</math> is given by the product of the individual queue equilibrium distributions
Assuming an open queueing network of single-server queues with the following notation and assumptions:
:<math>\pi (k_1,k_2,\ldots,k_m) = \prod_{i=1}^{m} \pi_i(k_i) = \prod_{i=1}^{m} [\rho_i^{k_i} (1-\rho_i)].</math>
* M = # of queues in the system, not counting queue 0 which represents the outside world
* <math>\mu_i</math> = service rate at queue ''i''
* <math>\lambda_i</math> = total rate at which jobs arrive at queue ''i''
* <math>\forall i,1\leq i\leq M:\rho_i = </math> utilization of the service at queue <math>i = \frac {\lambda_i}{\mu_i} < 1</math>
* <math>n_i(t)</math> =# of jobs in queue ''i'' at time ''t''
* <math>n(t)=(n_1(t), n_2(t), \dots, n_M(t))^T</math>= the system state at time ''t''
* <math>P(k_1, k_2, \dots, k_M, t) = \Pr(n(t)=(k_1, k_2, \dots, k_M)^T)</math>
* <math>P(k_1, k_2, \dots, k_M)=\lim_{t\to\infty}P(k_1,k_2,\dots,k_M,t)</math>
* Arrivals from the outside world are Poisson. All queues have exponential service time distributions.
then a unique equilibrium state probability distribution exists and is given, for some state vector <math>\scriptstyle{(k_1,k_2,\ldots,k_M)}</math>
:<math>P(k_1,k_2,\dots,k_M)=\prod_{i=1}^{M}
\left[\left(\frac{\lambda_i}{\mu_i}\right)^{k_i}\left(1-\frac{\lambda_i}{\mu_i}\right)\right]
= \prod_{i=1}^{M}[(1-\rho_i)\rho_i^{k_i}]</math><br>
(where <math>\scriptstyle{\rho_i=\frac{\lambda_i}{\mu_i}}</math>).
==See also==
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