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.

Question

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Answer 1

Question

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.

a)

Expert Solution

Answer to Problem 35P

average total time truck spends waiting20 mins.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of waiting

LQ = λ2μ( μ−λ)= 15218( 18−15)=22518(3)=4.16average wait timedelayWQ =LQλ=4.1615=0.27average system timeWS = 1( μ−λ)=1( 18−15)=0.33×60=20 mins.

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.

b)

Expert Solution

Answer to Problem 35P

probability of arriving truck must wait 0.1416 or 14.12%.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Probability of 0 in the system

P0 = 1-(λμ)P0=1−( 15 18)=0.17 or 17%

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

PN=P0×( λ μ)NN= number of customerN=1P1=0.17×( 15 18)1=0.1416 or 14.12%

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

c)

Expert Solution

Answer to Problem 35P

4.17 trucks waiting in line

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of customers waiting

LQ = λ2μ( μ−λ)= 15218( 18−15)=4.166 or 4.17

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

d)

Expert Solution

Answer to Problem 35P

3 trucks at the weigh station.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of trucks in the system

LS =λ( μ−λ)=15( 18−15)=3

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.

e)

Expert Solution

Answer to Problem 35P

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

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

N = not more than 3 trucks

PN=( λ μ)N(1−λμ)=( 15 18)3(1− 15 18)=0.096

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Answer 2

Question

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.

a)

Expert Solution

Answer to Problem 35P

average total time truck spends waiting20 mins.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of waiting

LQ = λ2μ( μ−λ)= 15218( 18−15)=22518(3)=4.16average wait timedelayWQ =LQλ=4.1615=0.27average system timeWS = 1( μ−λ)=1( 18−15)=0.33×60=20 mins.

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.

b)

Expert Solution

Answer to Problem 35P

probability of arriving truck must wait 0.1416 or 14.12%.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Probability of 0 in the system

P0 = 1-(λμ)P0=1−( 15 18)=0.17 or 17%

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

PN=P0×( λ μ)NN= number of customerN=1P1=0.17×( 15 18)1=0.1416 or 14.12%

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

c)

Expert Solution

Answer to Problem 35P

4.17 trucks waiting in line

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of customers waiting

LQ = λ2μ( μ−λ)= 15218( 18−15)=4.166 or 4.17

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

d)

Expert Solution

Answer to Problem 35P

3 trucks at the weigh station.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of trucks in the system

LS =λ( μ−λ)=15( 18−15)=3

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.

e)

Expert Solution

Answer to Problem 35P

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

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

N = not more than 3 trucks

PN=( λ μ)N(1−λμ)=( 15 18)3(1− 15 18)=0.096

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Answer 3

Question

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.

a)

Expert Solution

Answer to Problem 35P

average total time truck spends waiting20 mins.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of waiting

LQ = λ2μ( μ−λ)= 15218( 18−15)=22518(3)=4.16average wait timedelayWQ =LQλ=4.1615=0.27average system timeWS = 1( μ−λ)=1( 18−15)=0.33×60=20 mins.

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.

b)

Expert Solution

Answer to Problem 35P

probability of arriving truck must wait 0.1416 or 14.12%.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Probability of 0 in the system

P0 = 1-(λμ)P0=1−( 15 18)=0.17 or 17%

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

PN=P0×( λ μ)NN= number of customerN=1P1=0.17×( 15 18)1=0.1416 or 14.12%

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

c)

Expert Solution

Answer to Problem 35P

4.17 trucks waiting in line

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of customers waiting

LQ = λ2μ( μ−λ)= 15218( 18−15)=4.166 or 4.17

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

d)

Expert Solution

Answer to Problem 35P

3 trucks at the weigh station.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of trucks in the system

LS =λ( μ−λ)=15( 18−15)=3

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.

e)

Expert Solution

Answer to Problem 35P

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

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

N = not more than 3 trucks

PN=( λ μ)N(1−λμ)=( 15 18)3(1− 15 18)=0.096

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Answer 4

Question

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.

a)

Expert Solution

Answer to Problem 35P

average total time truck spends waiting20 mins.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of waiting

LQ = λ2μ( μ−λ)= 15218( 18−15)=22518(3)=4.16average wait timedelayWQ =LQλ=4.1615=0.27average system timeWS = 1( μ−λ)=1( 18−15)=0.33×60=20 mins.

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.

b)

Expert Solution

Answer to Problem 35P

probability of arriving truck must wait 0.1416 or 14.12%.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Probability of 0 in the system

P0 = 1-(λμ)P0=1−( 15 18)=0.17 or 17%

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

PN=P0×( λ μ)NN= number of customerN=1P1=0.17×( 15 18)1=0.1416 or 14.12%

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

c)

Expert Solution

Answer to Problem 35P

4.17 trucks waiting in line

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of customers waiting

LQ = λ2μ( μ−λ)= 15218( 18−15)=4.166 or 4.17

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

d)

Expert Solution

Answer to Problem 35P

3 trucks at the weigh station.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of trucks in the system

LS =λ( μ−λ)=15( 18−15)=3

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.

e)

Expert Solution

Answer to Problem 35P

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

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

N = not more than 3 trucks

PN=( λ μ)N(1−λμ)=( 15 18)3(1− 15 18)=0.096

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Answer 5

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.

a)

Expert Solution

Answer to Problem 35P

average total time truck spends waiting20 mins.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of waiting

LQ = λ2μ( μ−λ)= 15218( 18−15)=22518(3)=4.16average wait timedelayWQ =LQλ=4.1615=0.27average system timeWS = 1( μ−λ)=1( 18−15)=0.33×60=20 mins.

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.

b)

Expert Solution

Answer to Problem 35P

probability of arriving truck must wait 0.1416 or 14.12%.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Probability of 0 in the system

P0 = 1-(λμ)P0=1−( 15 18)=0.17 or 17%

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

PN=P0×( λ μ)NN= number of customerN=1P1=0.17×( 15 18)1=0.1416 or 14.12%

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

c)

Expert Solution

Answer to Problem 35P

4.17 trucks waiting in line

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of customers waiting

LQ = λ2μ( μ−λ)= 15218( 18−15)=4.166 or 4.17

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

d)

Expert Solution

Answer to Problem 35P

3 trucks at the weigh station.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of trucks in the system

LS =λ( μ−λ)=15( 18−15)=3

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.

e)

Expert Solution

Answer to Problem 35P

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

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

N = not more than 3 trucks

PN=( λ μ)N(1−λμ)=( 15 18)3(1− 15 18)=0.096

Answer 6

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.

a)

Expert Solution

Answer to Problem 35P

average total time truck spends waiting20 mins.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of waiting

LQ = λ2μ( μ−λ)= 15218( 18−15)=22518(3)=4.16average wait timedelayWQ =LQλ=4.1615=0.27average system timeWS = 1( μ−λ)=1( 18−15)=0.33×60=20 mins.

Answer 7

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 8

a)

Expert Solution

Answer to Problem 35P

average total time truck spends waiting20 mins.

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of waiting

LQ = λ2μ( μ−λ)= 15218( 18−15)=22518(3)=4.16average wait timedelayWQ =LQλ=4.1615=0.27average system timeWS = 1( μ−λ)=1( 18−15)=0.33×60=20 mins.

Answer 13

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.

b)

Expert Solution

Answer to Problem 35P

probability of arriving truck must wait 0.1416 or 14.12%.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Probability of 0 in the system

P0 = 1-(λμ)P0=1−( 15 18)=0.17 or 17%

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

PN=P0×( λ μ)NN= number of customerN=1P1=0.17×( 15 18)1=0.1416 or 14.12%

Answer 14

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 15

b)

Expert Solution

Answer to Problem 35P

probability of arriving truck must wait 0.1416 or 14.12%.

Answer 16

b)

Expert Solution

Answer 17

b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Probability of 0 in the system

P0 = 1-(λμ)P0=1−( 15 18)=0.17 or 17%

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

PN=P0×( λ μ)NN= number of customerN=1P1=0.17×( 15 18)1=0.1416 or 14.12%

Answer 20

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

c)

Expert Solution

Answer to Problem 35P

4.17 trucks waiting in line

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of customers waiting

LQ = λ2μ( μ−λ)= 15218( 18−15)=4.166 or 4.17

Answer 21

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 22

c)

Expert Solution

Answer to Problem 35P

4.17 trucks waiting in line

Answer 23

c)

Expert Solution

Answer 24

c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of customers waiting

LQ = λ2μ( μ−λ)= 15218( 18−15)=4.166 or 4.17

Answer 27

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

d)

Expert Solution

Answer to Problem 35P

3 trucks at the weigh station.

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of trucks in the system

LS =λ( μ−λ)=15( 18−15)=3

Answer 28

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 29

d)

Expert Solution

Answer to Problem 35P

3 trucks at the weigh station.

Answer 30

d)

Expert Solution

Answer 31

d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

Average number of trucks in the system

LS =λ( μ−λ)=15( 18−15)=3

Answer 34

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.

e)

Expert Solution

Answer to Problem 35P

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

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

N = not more than 3 trucks

PN=( λ μ)N(1−λμ)=( 15 18)3(1− 15 18)=0.096

Answer 35

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 36

e)

Expert Solution

Answer to Problem 35P

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

Answer 37

e)

Expert Solution

Answer 38

e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

Given information:

(λ)=15 per hour(μ)=18 trucks per hour

Explanation:

N = not more than 3 trucks

PN=( λ μ)N(1−λμ)=( 15 18)3(1− 15 18)=0.096

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 35P

Explanation of Solution

Answer to Problem 35P

Explanation of Solution

Answer to Problem 35P

Explanation of Solution

Answer to Problem 35P

Explanation of Solution

Answer to Problem 35P

Explanation of Solution

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