Practical Operations Management
Practical Operations Management
2nd Edition
ISBN: 9781939297136
Author: Simpson
Publisher: HERCHER PUBLISHING,INCORPORATED
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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.

a)

Expert Solution
Check Mark

Answer to Problem 34P

Average length of the line is 2.33

Explanation of Solution

Given information:

  customer arrival (λ)=14 per hrserved customer (μ)=603

  λ=14 per hourμ603=20 per hourρ =λ μ = 14 20 =0.7

Explanation:

  average no.of customers waiting =ρ1ρ=0.710.7=0.70.3=2.33

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.

B)

Expert Solution
Check Mark

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:

  101μ=10120=106020=103=7min

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.

C)

Expert Solution
Check Mark

Answer to Problem 34P

10 minutes to buy tickets.

Explanation of Solution

Given information:

  λ=14μ=20

Explanation:

average waiting time to buy tickets =1μλ=12014=16=6060=10min

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.

D)

Expert Solution
Check Mark

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

  ρ=0.7

Explanation:

Probability that there are zero customers in the system

  =1ρ=10.7=0.3probability of waiting = 1-P(0)=10.3=0.7

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.

E)

Expert Solution
Check Mark

Answer to Problem 34P

probability of having four customers is 0.07203

Explanation of Solution

Given information:

P = 4 customers

  ρ=0.7ρ=0.3

Explanation:

  P(4)=(ρ)4(1ρ)=0.7×0.7×0.7×0.7(10.7)=0.2401×0.3=0.07203

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