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

Want to see more full solutions like this?

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

A branch office of a large engineering firm has one on-line terminal connected to a central computer system for 16 hours each day. Engineers drive to the branch office to use the terminal to make routine calculations with an average exponential distribution of 30 minutes per use. The daily arrival pattern of engineers is random (Poisson) with an average of 20 persons. The branch manager is starting to receive complaints from the engineers about the length of time many of them have to wait to use the terminal. Question: 1. What is and u measured in per hour intervals? 2. On the average, how many minutes does each engineer have to wait? 3. Using increments of 0.1 hours, what would be the least value of u so that the waiting time will be not exceed 30 minutes?

The vehicles are entering the parking area of the a mall.There is only one ticket line to get and pay a parkingticket. Each ticket purchase takes an average of 18seconds. The average arrival rate is 3 vehicles/minute.Find the average length of queue, average waiting timein queue and average time spent by the vehicles in thesystem assuming M/M/1 queuing.

The new accounts loan officer of the Millennium Commercial Bank interviews all customers for new accounts. The customers desiring to open new accounts arrive at the rate of 4 per hour, according to a Poisson distribution, and the accounts officer spends an average of 12 minutes with each customer, setting up a new account. Determine the operating characteristics (P0, L, Lq, W, Wq and Pw) for this system.

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Want to see more full solutions like this?

Chapter 5 Solutions