Chapter 5, Problem 34P

a)

Summary Introduction

Interpretation: Average length of the 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 34P

Average length of the line is 2.33

Explanation of Solution

Given information:

  $\begin{array}{l}\text{customer arrival }\left(\lambda \right)=14\text{ per hr}\\ \text{served customer }\left(\mu \right)=\frac{60}{3}\end{array}$

  $\begin{array}{c}\lambda =14\text{ per hour}\\ \mu \text{= }\frac{60}{3}\\ =20\text{ per hour}\\ \rho \text{ =}\frac{\lambda }{\begin{array}{l}\mu \\ =\frac{14}{\begin{array}{c}20\\ =0.7\end{array}}\end{array}}\end{array}$

Explanation:

  $\begin{array}{c}\text{average no}\text{.of customers waiting =}\frac{\rho }{1-\rho }\\ =\frac{0.7}{1-0.7}\\ =\frac{0.7}{0.3}\\ =2.33\end{array}$

B)

Summary Introduction

Interpretation:customer wait to buy tickets.

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 34P

Customer wait to buy tickets is 7 minutes.

Explanation of Solution

Given information:

Average waiting time in queue = 10 min -service time

Explanation:

  $\begin{array}{c}10-\frac{1}{\mu }\\ =10-\frac{1}{20}\\ =10-\frac{60}{20}\\ =10-3\\ =7\min \end{array}$

C)

Summary Introduction

Interpretation:time to buy tickets.

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 34P

10 minutes to buy tickets.

Explanation of Solution

Given information:

  $\begin{array}{l}\lambda =14\\ \mu =20\end{array}$

Explanation:

average waiting time to buy tickets $\begin{array}{c}=\frac{1}{\mu -\lambda }\\ =\frac{1}{20-14}\\ =\frac{1}{6}\\ =\frac{60}{60}\\ =10\min \end{array}$

D)

Summary Introduction

Interpretation: probability of arriving customers 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 34P

0.7 probability of arriving customers wait

Explanation of Solution

Given information:Probability that there are zero customers in the system

  $\rho =0.7$

Explanation:

Probability that there are zero customers in the system

  $\begin{array}{c}=1-\rho \\ =1-0.7\\ =0.3\\ \text{probability of waiting = 1-P}\left(0\right)\\ =1-0.3\\ =0.7\end{array}$

E)

Summary Introduction

Interpretation: probability of having four customers.

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 34P

probability of having four customers is 0.07203

Explanation of Solution

Given information:

P = 4 customers

  $\begin{array}{l}\rho =0.7\\ \rho =0.3\end{array}$

Explanation:

  $\begin{array}{l}P\left(4\right)={\left(\rho \right)}^4\left(1-\rho \right)\\ =0.7\times 0.7\times 0.7\times 0.7\left(1-0.7\right)\\ =0.2401\times 0.3\\ =0.07203\end{array}$