Chapter 5, Problem 37P

a)

Summary Introduction

Interpretation:customer officer during time period.

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 37P

The customer officer is busy 75% of the time.

Explanation of Solution

Given information:the scenario meets the assumptions of M/M/1 model

  $\begin{array}{c}\text{arrival rate}\left(\lambda \right)\text{ = 9 cars per hour}\\ \text{as the officer takes 5 minutes to inspect and release the car,}\\ \text{the number of cars relaesed per minute is }\frac{1}{5}\\ \text{service rate}\left(\mu \right)=60\times \frac{1}{5}\\ =12\text{ cars per hour}\end{array}$

Explanation:

  $\begin{array}{c}\text{we need to find the utilization,}\rho \text{.}\\ \text{this is given as the ratio of arrival rate to the service rate}\text{.}\\ \text{the utilization}\text{.}\\ \rho =\frac{\lambda }{\mu }\\ =\frac{9}{12}\\ =0.75\\ =75\%.\end{array}$

b)

Summary Introduction

Interpretation: average wait to speak to the officer andthe probability that someone will be able to speak immediately upon 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 37P

average wait to speak to the officer 15 minutes.

the probability that someone will be able to speak immediately upon arrival is 0.25.

Explanation of Solution

Given information:

  $\begin{array}{c}\text{arrival rate}\left(\lambda \right)\text{ = 9 cars per hour}\\ \text{as the officer takes 5 minutes to inspect and release the car,}\\ \text{the number of cars relaesed per minute is }\frac{1}{5}\\ \text{service rate}\left(\mu \right)=60\times \frac{1}{5}\\ =12\text{ cars per hour}\end{array}$

Explanation:

Average wait time to speak to the office, ${\text{W}}_q$  is given using below formula

  $\begin{array}{c}{W}_q=\frac{\lambda }{\mu \left(\mu -\lambda \right)}\\ \text{average wait time is as given below}\\ W_q=\frac{9}{12\left(12-9\right)}\\ =\frac{9}{36}\\ =\frac{1}{4}\\ =0.25\text{ hours}\\ \text{=0}\text{.25}\times \text{60 mins}\\ \text{=15 mins}\end{array}$

The probability that someone will be able to speak immediately upon arrival is the same as the probability of there being no cars in the system. This probability $P_0$ is given as:

  $\begin{array}{c}{P}_0=1-\frac{\lambda }{\mu }\\ =1-\frac{9}{12}\\ =1-\frac{3}{4}\\ =1-0.75\\ =0.25\end{array}$

c)

Summary Introduction

Interpretation:the average cars present at the border crossing plaza during that time period.

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 37P

the average cars present at the border crossing plaza during that time period is 3 cars.

Explanation of Solution

Given information:

  $\begin{array}{c}\text{arrival rate}\left(\lambda \right)\text{ = 9 cars per hour}\\ \text{as the officer takes 5 minutes to inspect and release the car,}\\ \text{the number of cars relaesed per minute is }\frac{1}{5}\\ \text{service rate}\left(\mu \right)=60\times \frac{1}{5}\\ =12\text{ cars per hour}\end{array}$

Explanation:

The average cars present at the plaza during that period, L is given using the below formula

  $\begin{array}{c}L=\frac{{\lambda }^2}{\mu \left(\mu -\lambda \right)}+\frac{\lambda }{\mu }\\ \text{hence the average cars present at the plaza is as below}\\ L=\frac{9^2}{12\left(12-9\right)}+\frac{9}{12}\\ =\frac{81}{12\times 3}+0.75\\ =2.25+0.75\\ =3\end{array}$