Markov Models and Its Evaluation

In a queueing system, if the service time and inter-arrival time are both expoentially distributed, denoted by $\mu$ and $\lambda$ respectively plus

• the system is in steady state
• the population is infinite

then we can evaluate the queueing system using the so called Markov Model.

Let $P_i$ represent the probability the system has i customers (including the ones in queue and ones in server).

Using the principle that the flow into a state is balanced by the flow out of the state, we have

$P_0 \lambda$	=	$P_1 \mu$
$P_1 \lambda$	$=$	$P_2 \mu$
...

solve this system of equations we get

$$P_i = P_0 (\frac{\lambda}{\mu})^{i} ~~~ {\rm for} ~ i > 0$$

Using the relation

$$\sum_{i=0}^{\infty} P_i = 1$$

and let

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

we have

$$\sum_{i=0}^{\infty} P_0 \rho^i = 1$$

thus

$$P_0 = \frac{1}{\sum_{i=0}^{\infty} \rho^i}$$

where

$$\sum_{i=0}^{\infty} \rho^i = \frac{1}{1-\rho}$$

so

$$P_0 = 1 - \rho$$

The meaning of the probability with no customer in the system is the same as the server is idle. So the measure $\rho$ can be considered as the probability that the server is busy, which is the utilization of the server. With $P_0$ solved, all other items can be solved.

$$P_i = (1 - \rho) \rho^i$$

Use $P_i$s, we can obtain all other measures of interest.

Average number of customers in system :

$$L = \sum_{i=0}^{\infty} i * P_i = \sum_{i=0}^{\infty} i * (1-\rho) * \rho^i = \frac{\rho}{1-\rho}$$

waiting time in system:

use Little's Law

$$W = \frac{L}{\lambda} = \frac{1}{\mu (1-\rho)}$$

waiting time in the queue:

this can be calculated using the fact that waiting time in the queue is the total waiting time less the time spent in the server. The time spent in the server is the average service time $\frac{1}{\mu}$ (note that $\mu$ is the service rate here) so

$$W_Q = W - \frac{1}{\mu} = \frac{\rho}{\mu (1-\rho)}$$

Average queue length:

use the Little's Law again

$$L_Q = \lambda W_Q = \frac{\rho^2}{1-\rho}$$