Queuing models

There are various kinds of queuing models. These queuing models have a set of defined characteristics like some arrival and service distribution, queue discipline, etc. The queuing models are represented by using a notation which is discussed in the following section of queue notation.

M/M/1 model

In this model the arrival times and service rates follow Markovian distribution or exponential distribution which are probabilistic distributions, so this is an example of stochastic process. In this model there is only one server. The important results of this model are:

Average number of customers in the system = $L= \frac{\rho}{1-\rho}$
Average number of customers in the system = $L_q = \frac{\rho^2}{1-\rho}$
Expected waiting time in the system $W = \frac{L}{\lambda} = (1/\lambda)\frac{\lambda}{\mu - \lambda} = \frac{1}{\mu-\lambda}$
Expected waiting time in the queue $W_q = \frac{L_q}{\lambda} = \frac{1}{\lambda}\times\frac{\lambda^2}{\mu(\mu-\lambda)} = \frac{\lambda}{\mu(\mu-\lambda)}$

Numerical example

Vehicles arrive at a toll booth at an average rate of 300 per hour. Average waiting time at the toll booth is 10s per vehicle. If both arrivals and departures are exponentially distributed, what is the average number of vehicles in the system, average queue length, the average delay per vehicle, the average time a vehicle is in the system?

Solution

Mean arrival rate $\lambda$ = 300 vehicles/hr. Mean service rate $\mu=\frac{3600}{10}$ vehicles/hr. Utilization factor = traffic intensity = $\rho=\frac{\lambda}{\mu}=\frac{300}{360}$ = 0.833. Percent of time the toll booth will be idle = P(0) = P(X=0) = $\rho^0(1-\rho) = (0.833)^0(1-0.833)=0.139(60 min)$ =8.34 min. Average number of vehicles in the system = $E[X]=\frac{\rho}{1-\rho}$ =4.98. Average number of vehicles in the queue = $E[L_q] = \frac{\rho^2}{1-\rho}$ = 4.01. Average a vehicle spend in the system = $E[T] = \frac{1}{\mu-\lambda}$ = 0.016 hr = 0.96 min = 57.6 sec. Average time a vehicle spends in the queue = $E[T_q] = \frac{\lambda}{\mu(\mu-\lambda)}$ = 0.013hr = 0.83 min = 50 sec.