Chapter 5, Problem 35P

a)

Summary Introduction

Interpretation: Average total time truck spends waiting.

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 35P

average total time truck spends waiting20 mins.

Explanation of Solution

Given information:

  $\begin{array}{l}\left(\lambda \right)=15\text{ per hour}\\ \left(\mu \right)=18\text{ trucks per hour}\end{array}$

Explanation:

Average number of waiting

  $\begin{array}{c}\text{LQ = }\frac{{\lambda }^2}{\mu \left(\mu -\lambda \right)}\\ =\frac{{15}^2}{18\left(18-15\right)}\\ =\frac{225}{18\left(3\right)}\\ =4.16\\ \\ \frac{\text{average wait time}}{\text{delay}}\\ \text{WQ =}\frac{\text{LQ}}{\lambda }\\ =\frac{4.16}{15}\\ =0.27\\ \text{average system time}\\ \text{WS = }\frac{\text{1}}{\left(\mu -\lambda \right)}\\ =\frac{1}{\left(18-15\right)}\\ =0.33\times 60\\ =20\text{ mins}\text{.}\end{array}$

b)

Summary Introduction

Interpretation: Probability of arriving truck must wait.

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 35P

probability of arriving truck must wait 0.1416 or 14.12%.

Explanation of Solution

Given information:

  $\begin{array}{l}\left(\lambda \right)=15\text{ per hour}\\ \left(\mu \right)=18\text{ trucks per hour}\end{array}$

Explanation:

Probability of 0 in the system

  $\begin{array}{c}P0\text{ = 1-}\left(\frac{\lambda }{\mu }\right)\\ P0=1-\left(\frac{15}{18}\right)\\ =0.17\text{ or 17\%}\end{array}$

Probability of N number in a queue at least 1 in the queue

  $\begin{array}{c}PN=P0\times {\left(\frac{\lambda }{\mu }\right)}^N\\ N=\text{ number of customer}\\ N=1\\ P1=0.17\times {\left(\frac{15}{18}\right)}^1\\ =0.1416\text{ or 14}\text{.12\%}\end{array}$

c)

Summary Introduction

Interpretation: Trucks waiting in line.

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 35P

4.17 trucks waiting in line

Explanation of Solution

Given information:

  $\begin{array}{l}\left(\lambda \right)=15\text{ per hour}\\ \left(\mu \right)=18\text{ trucks per hour}\end{array}$

Explanation:

Average number of customers waiting

  $\begin{array}{c}\text{LQ = }\frac{{\lambda }^2}{\mu \left(\mu -\lambda \right)}\\ =\frac{{15}^2}{18\left(18-15\right)}\\ =4.166\text{ or 4}\text{.17}\end{array}$

d)

Summary Introduction

Interpretation: Trucks at the weigh station

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 35P

3 trucks at the weigh station.

Explanation of Solution

Given information:

  $\begin{array}{l}\left(\lambda \right)=15\text{ per hour}\\ \left(\mu \right)=18\text{ trucks per hour}\end{array}$

Explanation:

Average number of trucks in the system

  $\begin{array}{c}\text{LS =}\frac{\lambda }{\left(\mu -\lambda \right)}\\ =\frac{15}{\left(18-15\right)}\\ =3\end{array}$

e)

Summary Introduction

Interpretation: Probability of no more than 3 trucks at weigh station.

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 35P

0.096 probability of no more than 3 trucks at weigh station.

Explanation of Solution

Given information:

  $\begin{array}{l}\left(\lambda \right)=15\text{ per hour}\\ \left(\mu \right)=18\text{ trucks per hour}\end{array}$

Explanation:

N = not more than 3 trucks

  $\begin{array}{c}{P}_N={\left(\frac{\lambda }{\mu }\right)}^N\left(1-\frac{\lambda }{\mu }\right)\\ ={\left(\frac{15}{18}\right)}^3\left(1-\frac{15}{18}\right)\\ =0.096\end{array}$