Module 8 : Specialized Traffic Studies

Lecture 45 : Queuing Analysis

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	M/M/N model The difference between the earlier model and this model is the number of servers. This is a multi -server model with N number of servers whereas the earlier one was single server model. The assumptions stated in M/M/1 model are also assumed here. Figure 1: Multi-server model Here $\mu$ is the average service rate for N identical service counters in parallel. For x=0 $$P(0)=\left[\sum_{x=0}^{N-1}\left(\frac{\rho^x}{x!}+\frac{\rho^N}{(N-1)!(N-\rho)}\right)\right]^{-1}$$ (1) The probability of x number of customers in the system is given by P(x). For $1 \leq x \leq N$ $$P(x)=\frac{\rho^x}{x!}*P(0)$$ (2) For $x>N$ $$P(x)=\frac{\rho^x}{N!N^{x-N}}*P(0)$$ (3) The average number of customers in the system is $$E[X] = \rho+[\frac{\rho^{N+1}}{(N-1)!(N-\rho)^2}]P(0)$$ (4) The average queue length $$E[L_q] = [\frac{\rho^{N+1}}{(N-1)!(N-\rho)^2}]P(0)$$ (5) The expected time in the system $$E[T] = \frac{E[X]}{\lambda}$$ (6) The expected time in the queue $$E[T_q] = \frac{E[L_q]}{\lambda}$$ (7) Numerical example Consider the earlier problem as a multi-server problem with two servers in parallel. Solution Average arrival rate = $\lambda$ = 300 vehicles/hr. Average service rate = $\mu=\frac{3600}{10}$ vehicles/hr. Utilization factor = traffic intensity = $\rho=\frac{\lambda}{\mu}=\frac{300}{360}$ = 0.833. $$P(0) = \left[\sum_{x=0}^{N-1}\left(\frac{\rho^x}{x!}+\frac{\rho^N}{(N-1)!(N-\rho)}\right)\right]^{-1}$$ $$= 0.92(60)=55.2 min$$ Average number of vehicles in the system is = L = $E[X]= \rho+[\frac{\rho^{N+1}}{(N-1)!(N-\rho)^2}]P(0)$ = 1.22. The average number of customers in the queue = $L_q = E[L_q] = [\frac{\rho^{N+1}}{(N-1)!(N-\rho)^2}]P(0)$= 0.387. Expected time in the system = $W = \frac{E[X]}{\lambda}$= 0.004 hr = 14 sec. The expected time in the queue = $W_q = \frac{L_q}{\lambda}$= 0.00129 hr = 4.64 sec.