Chapter 5, Problem 3.3Q

Summary Introduction

Interpretation: Probability of highest and lowest of a vehicle arrival.

Concept introduction:The M/M/1 queue assumes poison arrivals, exponential service times, and a single server serving customers in a FCFS fashion. Poisson arrivals are a reasonably good assumption for unscheduled systems. Further if there is a mix of many different types of jobs the exponential distribution can be realistic for service times. Otherwise it tends to be too variable of a distribution.

Answer to Problem 3.3Q

Highest in overnight hours and lowest in commuters’ hours.

Explanation of Solution

Given information:

  $\begin{array}{l}\text{average cars per hour}(\lambda )=100\text{per hour}\\ \text{service rate(}\mu )\text{ = 240 per hour}\\ \text{no}{\text{.of vehicles in the queue = L}}_q\le 4\\ \text{ traffic rate(}\rho )=\left(\frac{\lambda }{\mu }\right)\\ =\frac{100}{240}\\ =0.42\\ \text{no: of vehicles in queue(}{L}_q)=\frac{{\rho }^2}{1-\rho }\\ =0.30\\ {\rho }_0=1-\rho \\ =1-0.42\\ =\mathrm{0.58..........................2}\\ \end{array}$

(references from 97136-5-3.1Q)

(references from 97136-5-3.1Q)

  $\begin{array}{c}\lambda =75\text{ per hour}\\ \mu \text{=240 per hour}\\ \text{no}{\text{.of vehicles in the queue = L}}_q\le 4\\ \rho =\left(\frac{75}{240}\right)\\ =0.3125\\ L_q=\frac{{\rho }^2}{\begin{array}{l}1-\rho \\ =0.142\\ P_0=1-\rho \\ =1-0.3125\\ =0.6875\end{array}}\end{array}$

0.103 [References from 97136-5-3.1Q]

0.58 [References from 97136-5-3.2Q]

0.6875 [References from 97136-5-3.3Q]

Hence, the probability that a vehicle finds a lane idle and gets instant service is highest for the overnight hours and lowest for the commuter hours.