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.

Question

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Answer 1

Question

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

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 hourμ= 603=20 per hourρ =λ μ = 14 20 =0.7

Explanation:

average no.of customers waiting =ρ1−ρ=0.71−0.7=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

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:

10−1μ=10−120=10−6020=10−3=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

Answer to Problem 34P

10 minutes to buy tickets.

Explanation of Solution

Given information:

λ=14μ=20

Explanation:

average waiting time to buy tickets =1μ−λ=120−14=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

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

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

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(1−ρ)=0.7×0.7×0.7×0.7(1−0.7)=0.2401×0.3=0.07203

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Students have asked these similar questions

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Answer 2

Question

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

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 hourμ= 603=20 per hourρ =λ μ = 14 20 =0.7

Explanation:

average no.of customers waiting =ρ1−ρ=0.71−0.7=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

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:

10−1μ=10−120=10−6020=10−3=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

Answer to Problem 34P

10 minutes to buy tickets.

Explanation of Solution

Given information:

λ=14μ=20

Explanation:

average waiting time to buy tickets =1μ−λ=120−14=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

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

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

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(1−ρ)=0.7×0.7×0.7×0.7(1−0.7)=0.2401×0.3=0.07203

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Answer 3

Question

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

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 hourμ= 603=20 per hourρ =λ μ = 14 20 =0.7

Explanation:

average no.of customers waiting =ρ1−ρ=0.71−0.7=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

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:

10−1μ=10−120=10−6020=10−3=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

Answer to Problem 34P

10 minutes to buy tickets.

Explanation of Solution

Given information:

λ=14μ=20

Explanation:

average waiting time to buy tickets =1μ−λ=120−14=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

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

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

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(1−ρ)=0.7×0.7×0.7×0.7(1−0.7)=0.2401×0.3=0.07203

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Answer 4

Question

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

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 hourμ= 603=20 per hourρ =λ μ = 14 20 =0.7

Explanation:

average no.of customers waiting =ρ1−ρ=0.71−0.7=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

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:

10−1μ=10−120=10−6020=10−3=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

Answer to Problem 34P

10 minutes to buy tickets.

Explanation of Solution

Given information:

λ=14μ=20

Explanation:

average waiting time to buy tickets =1μ−λ=120−14=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

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

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

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(1−ρ)=0.7×0.7×0.7×0.7(1−0.7)=0.2401×0.3=0.07203

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Answer 5

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

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 hourμ= 603=20 per hourρ =λ μ = 14 20 =0.7

Explanation:

average no.of customers waiting =ρ1−ρ=0.71−0.7=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

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:

10−1μ=10−120=10−6020=10−3=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

Answer to Problem 34P

10 minutes to buy tickets.

Explanation of Solution

Given information:

λ=14μ=20

Explanation:

average waiting time to buy tickets =1μ−λ=120−14=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

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

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

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(1−ρ)=0.7×0.7×0.7×0.7(1−0.7)=0.2401×0.3=0.07203

Answer 6

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

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 hourμ= 603=20 per hourρ =λ μ = 14 20 =0.7

Explanation:

average no.of customers waiting =ρ1−ρ=0.71−0.7=0.70.3=2.33

Answer 7

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 8

a)

Expert Solution

Answer to Problem 34P

Average length of the line is 2.33

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

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

λ=14 per hourμ= 603=20 per hourρ =λ μ = 14 20 =0.7

Explanation:

average no.of customers waiting =ρ1−ρ=0.71−0.7=0.70.3=2.33

Answer 13

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

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:

10−1μ=10−120=10−6020=10−3=7min

Answer 14

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 15

B)

Expert Solution

Answer to Problem 34P

Customer wait to buy tickets is 7 minutes.

Answer 16

B)

Expert Solution

Answer 17

B)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:

Average waiting time in queue = 10 min -service time

Explanation:

10−1μ=10−120=10−6020=10−3=7min

Answer 20

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

Answer to Problem 34P

10 minutes to buy tickets.

Explanation of Solution

Given information:

λ=14μ=20

Explanation:

average waiting time to buy tickets =1μ−λ=120−14=16=6060=10min

Answer 21

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 22

C)

Expert Solution

Answer to Problem 34P

10 minutes to buy tickets.

Answer 23

C)

Expert Solution

Answer 24

C)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given information:

λ=14μ=20

Explanation:

average waiting time to buy tickets =1μ−λ=120−14=16=6060=10min

Answer 27

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

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

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

Answer 28

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 29

D)

Expert Solution

Answer to Problem 34P

0.7 probability of arriving customers wait

Answer 30

D)

Expert Solution

Answer 31

D)

Expert Solution

Answer 32

Expert Solution

Answer 33

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

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

Answer 34

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

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(1−ρ)=0.7×0.7×0.7×0.7(1−0.7)=0.2401×0.3=0.07203

Answer 35

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 36

E)

Expert Solution

Answer to Problem 34P

probability of having four customers is 0.07203

Answer 37

E)

Expert Solution

Answer 38

E)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given information:

P = 4 customers

ρ=0.7ρ=0.3

Explanation:

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

Answer 41

Ch. 5 - Prob. 1DQ Ch. 5 - Prob. 2DQ Ch. 5 - Prob. 3DQ Ch. 5 - Prob. 4DQ Ch. 5 - Prob. 5DQ Ch. 5 - Prob. 6DQ Ch. 5 - Prob. 7DQ Ch. 5 - Prob. 1P Ch. 5 - Prob. 2P Ch. 5 - Prob. 3P

Concept explainers

Answer to Problem 34P

Explanation of Solution

Answer to Problem 34P

Explanation of Solution

Answer to Problem 34P

Explanation of Solution

Answer to Problem 34P

Explanation of Solution

Answer to Problem 34P

Explanation of Solution

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Chapter 5 Solutions