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Description

I want a solution to the same problem with the Excel program, but changed the data with the QM tool, and a picture of the problem was attached, and I attached Chapter 13 .

Chapter 13

Waiting Lines and Queuing

Theory Models

To accompany

Quantitative Analysis for Management, Eleventh Edition,

by Render, Stair, and Hanna

Power Point slides created by Brian Peterson

Learning Objectives

After completing this chapter, students will be able to:

1.

2.

3.

4.

5.

Describe the trade-off curves for cost-ofwaiting time and cost of service.

Understand the three parts of a queuing

system: the calling population, the queue

itself, and the service facility.

Describe the basic queuing system

configurations.

Understand the assumptions of the common

models dealt with in this chapter.

Analyze a variety of operating characteristics

of waiting lines.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-2

13.1

13.2

13.3

13.4

13.5

Chapter Outline

Introduction

Waiting Line Costs

Characteristics of a Queuing System

Single-Channel Queuing Model with

Poisson Arrivals and Exponential Service

Times (M/M/1)

Multichannel Queuing Model with Poisson

Arrivals and Exponential Service Times

(M/M/m)

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-3

Chapter Outline

13.6

13.7

13.8

13.9

Constant Service Time Model (M/D/1)

Finite Population Model (M/M/1 with Finite

Source)

Some General Operating Characteristic

Relationships

More Complex Queuing Models and the

Use of Simulation

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-4

Introduction

Queuing theory is the study of waiting lines.

It is one of the oldest and most widely used

quantitative analysis techniques.

The three basic components of a queuing

process are arrivals, service facilities, and the

actual waiting line.

Analytical models of waiting lines can help

managers evaluate the cost and effectiveness

of service systems.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-5

Waiting Line Costs

Most waiting line problems are focused on

finding the ideal level of service a firm should

provide.

In most cases, this service level is something

management can control.

When an organization does have control, they

often try to find the balance between two

extremes.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-6

Waiting Line Costs

There is generally a trade-off between cost of

providing service and cost of waiting time.

A large staff and many service facilities generally results

in high levels of service but have high costs.

Having the minimum number of service facilities keeps

service cost down but may result in dissatisfied

customers.

Service facilities are evaluated on their total

expected cost which is the sum of service costs

and waiting costs.

Organizations typically want to find the service

level that minimizes the total expected cost.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-7

Queuing Costs and Service Levels

Figure 13.1

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13-8

Three Rivers Shipping Company

Three Rivers Shipping operates a docking facility

on the Ohio River.

An average of 5 ships arrive to unload their

cargos each shift.

Idle ships are expensive.

More staff can be hired to unload the ships, but

that is expensive as well.

Three Rivers Shipping Company wants to

determine the optimal number of teams of

stevedores to employ each shift to obtain the

minimum total expected cost.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-9

Three Rivers Shipping Company

Waiting Line Cost Analysis

NUMBER OF TEAMS OF STEVEDORES

WORKING

1

7

(b) Average time each ship waits to

be unloaded (hours)

5

(a) Average number of ships arriving

per shift

(c) Total ship hours lost per shift

(a x b)

(d) Estimated cost per hour of idle

ship time

(e)

(f)

cost (c x d)

Stevedore team salary or service

cost

(g) Total expected cost (e + f)

2

3

5

5

3

20

$1,000

$20,000

$12,000

$32,000

4

5

4

35

$1,000

$35,000

$6,000

$41,000

2

15

$1,000

$15,000

$18,000

$33,000

10

$1,000

$10,000

$24,000

$34,000

Optimal cost

Table 13.1

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13-10

Characteristics of a Queuing System

There are three parts to a queuing system:

1. The arrivals or inputs to the system

(sometimes referred to as the calling

population).

2. The queue or waiting line itself.

3. The service facility.

These components have their own

characteristics that must be examined

before mathematical models can be

developed.

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13-11

Characteristics of a Queuing System

Arrival Characteristics have three major

characteristics: size, pattern, and behavior.

The size of the calling population can be either

unlimited (essentially infinite) or limited (finite).

The pattern of arrivals can arrive according to

a known pattern or can arrive randomly.

Random arrivals generally follow a Poisson

distribution.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-12

Characteristics of a Queuing System

Behavior of arrivals

Most queuing models assume customers are

patient and will wait in the queue until they are

served and do not switch lines.

Balking refers to customers who refuse to join

the queue.

Reneging customers enter the queue but

become impatient and leave without receiving

their service.

That these behaviors exist is a strong

argument for the use of queuing theory to

managing waiting lines.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-13

Characteristics of a Queuing System

Waiting Line Characteristics

Waiting lines can be either limited or unlimited.

Queue discipline refers to the rule by which

customers in the line receive service.

The most common rule is first-in, first-out (FIFO).

Other rules are possible and may be based on other

important characteristics.

Other rules can be applied to select which

customers enter which queue, but may apply

FIFO once they are in the queue.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-14

Characteristics of a Queuing System

Service Facility Characteristics

Basic queuing system configurations:

Service systems are classified in terms of

the number of channels, or servers, and the

number of phases, or service stops.

A single-channel system with one server is

quite common.

Multichannel systems exist when multiple

servers are fed by one common waiting line.

In a single-phase system, the customer

receives service form just one server.

In a multiphase system, the customer has to

go through more than one server.

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13-15

Four basic

queuing

system

configurations

Figure 13.2

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13-16

Characteristics of a Queuing System

Service time distribution

Service patterns can be either constant or

random.

Constant service times are often machine

controlled.

More often, service times are randomly

distributed according to a negative

exponential probability distribution.

Analysts should observe, collect, and plot

service time data to ensure that the

observations fit the assumed distributions

when applying these models.

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13-17

Identifying Models Using

Kendall Notation

D. G. Kendall developed a notation for queuing

models that specifies the pattern of arrival, the

service time distribution, and the number of

channels.

Notation takes the form:

Arrival

distribution

Service time

distribution

Number of service

channels open

Specific letters are used to represent probability

distributions.

M = Poisson distribution for number of occurrences

D = constant (deterministic) rate

G = general distribution with known mean and variance

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-18

Identifying Models Using

Kendall Notation

A single-channel model with Poisson arrivals and

exponential service times would be represented

by:

M/M/1

If a second channel is added the notation would

read:

M/M/2

A three-channel system with Poisson arrivals and

constant service time would be

M/D/3

A four-channel system with Poisson arrivals and

normally distributed service times would be

M/G/4

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-19

Single-Channel Model, Poisson Arrivals,

Exponential Service Times (M/M/1)

Assumptions of the model:

Arrivals are served on a FIFO basis.

There is no balking or reneging.

Arrivals are independent of each other but the

arrival rate is constant over time.

Arrivals follow a Poisson distribution.

Service times are variable and independent but

the average is known.

Service times follow a negative exponential

distribution.

Average service rate is greater than the

average arrival rate.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-20

Single-Channel Model, Poisson Arrivals,

Exponential Service Times (M/M/1)

When these assumptions are met, we

can develop a series of equations that

operating

characteristics.

Queuing Equations:

Let

= mean number of arrivals per time period

= mean number of customers or units

served per time period

The arrival rate and the service rate must be

defined for the same time period.

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13-21

Single-Channel Model, Poisson Arrivals,

Exponential Service Times (M/M/1)

1. The average number of customers or units in the

system, L:

2. The average time a customer spends in the

system, W:

3. The average number of customers in the queue, Lq:

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13-22

Single-Channel Model, Poisson Arrivals,

Exponential Service Times (M/M/1)

4. The average time a customer spends waiting in

the queue, Wq:

5. The utilization factor for the system, , the

probability the service facility is being used:

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13-23

Single-Channel Model, Poisson Arrivals,

Exponential Service Times (M/M/1)

6. The percent idle time, P0, or the probability no one

is in the system:

7. The probability that the number of customers in

the system is greater than k, Pn>k:

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-24

3 per hour.

Customers arrive at a rate of 2 per hour.

So:

= 2 cars arriving per hour

= 3 cars serviced per hour

2 cars in the system

on average

1 hour that an average car

spends in the system

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13-25

1.33 cars waiting in line

on average

40 minutes average

waiting time per car

percentage of time

mechanic is busy

probability that there

are 0 cars in the system

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13-26

Probability of more than k cars in the system

0.039

7

0.058

6

0.088

5

0.132

4

0.198

3

0.296

2

0.444

1

0.667

0

Pn>k = (2/3)k+1

k

Note that this is equal to 1

P0 = 1

0.33 = 0.667

Implies that there is a 19.8% chance that more

than 3 cars are in the system

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-27

Muffler Example

Program 13.1

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13-28

Introducing costs into the model:

Arnold wants to do an economic analysis of

the queuing system and determine the waiting

cost and service cost.

The total service cost is:

Total

(Number of channels)

=

service cost

x (Cost per channel)

Total

= mCs

service cost

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13-29

Waiting cost when the cost is based on time in the

system:

Total

(Total time spent waiting by all

=

waiting cost

arrivals) x (Cost of waiting)

(Number of arrivals) x

=

(Average wait per arrival)Cw

Total

= ( W)Cw

waiting cost

If waiting time cost is based on time in the queue:

Total

= ( Wq)Cw

waiting cost

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13-30

So the total cost of the queuing system when based

on time in the system is:

Total cost = Total service cost + Total waiting cost

Total cost = mCs + WCw

And when based on time in the queue:

Total cost = mCs + WqCw

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13-31

Arnold estimates the cost of customer waiting

time in line is $50 per hour.

Total daily

waiting cost = (8 hours per day) WqCw

= (8)(2)(2/3)($50) = $533.33

Arnold has identified the mechanics wage $7 per

hour as the service cost.

Total daily

service cost = (8 hours per day)mCs

= (8)(1)($15) = $120

So the total cost of the system is:

Total daily cost of

the queuing system = $533.33 + $120 = $653.33

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-32

Arnold is thinking about hiring a different

mechanic who can install mufflers at a faster rate.

The new operating characteristics would be:

= 2 cars arriving per hour

= 4 cars serviced per hour

1 car in the system

on the average

1/2 hour that an average car

spends in the system

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13-33

1/2 car waiting in line

on the average

15 minutes average

waiting time per car

percentage of time

mechanic is busy

probability that there

are 0 cars in the system

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13-34

Probability of more than k cars in the system

0.004

7

0.008

6

0.016

5

0.031

4

0.062

3

0.125

2

0.250

1

0.500

0

Pn>k = (2/4)k+1

k

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13-35

The customer waiting cost is the same $50 per

hour:

Total daily

waiting cost = (8 hours per day) WqCw

= (8)(2)(1/4)($50) = $200.00

The new mechanic is more expensive at $20 per

hour:

Total daily

service cost = (8 hours per day)mCs

= (8)(1)($20) = $160

So the total cost of the system is:

Total daily cost of

the queuing system = $200 + $160 = $360

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13-36

The total time spent waiting for the 16 customers

per day was formerly:

(16 cars per day) x (2/3 hour per car) = 10.67 hours

It is now is now:

(16 cars per day) x (1/4 hour per car) = 4 hours

The total daily system costs are less with the new

mechanic resulting in significant savings:

$653.33

$360 = $293.33

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13-37

Enhancing the Queuing Environment

Reducing waiting time is not the only way

to reduce waiting cost.

Reducing the unit waiting cost (Cw) will

also reduce total waiting cost.

This might be less expensive to achieve

than reducing either W or Wq.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-38

Multichannel Queuing Model with Poisson

Arrivals and Exponential Service Times

(M/M/m)

Assumptions of the model:

Arrivals are served on a FIFO basis.

There is no balking or reneging.

Arrivals are independent of each other but the

arrival rate is constant over time.

Arrivals follow a Poisson distribution.

Service times are variable and independent but

the average is known.

Service times follow a negative exponential

distribution.

The average service rate is greater than the

average arrival rate.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-39

Multichannel Queuing Model with Poisson

Arrivals and Exponential Service Times (M/M/m)

Equations for the multichannel queuing model:

Let

m = number of channels open

= average arrival rate

= average service rate at each channel

1. The probability that there are zero customers in

the system is:

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-40

Multichannel Model, Poisson Arrivals,

Exponential Service Times (M/M/m)

2. The average number of customers or units in the

system

3. The average time a unit spends in the waiting line or

being served, in the system

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13-41

Multichannel Model, Poisson Arrivals,

Exponential Service Times (M/M/m)

4. The average number of customers or units in line

waiting for service

5. The average number of customers or units in line

waiting for service

6. The average number of customers or units in line

waiting for service

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-42

Arnold wants to investigate opening a second

garage bay.

He would hire a second worker who works at the

same rate as his first worker.

The customer arrival rate remains the same.

probability of 0 cars in the system

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13-43

Average number of cars in the system

Average time a car spends in the system

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13-44

Average number of cars in the queue

Average time a car spends in the queue

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-45

Adding the second service bay reduces the

waiting time in line but will increase the service

cost as a second mechanic needs to be hired.

Total daily waiting cost = (8 hours per day) WqCw

= (8)(2)(0.0415)($50) = $33.20

Total daily service cost = (8 hours per day)mCs

= (8)(2)($15) = $240

So the total cost of the system is

Total system cost = $33.20 + $240 = $273.20

This is the cheapest option: open the second bay

and hire a second worker at the same $15 rate.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-46

Operating Characteristics

LEVEL OF SERVICE

OPERATING

CHARACTERISTIC

ONE

MECHANIC

=3

40 minutes

Average time spent in the

queue (Wq)

1.33 cars

Average number of cars in

the queue (Lq)

60 minutes

Average time spent in the

system (W)

2 cars

Average number of cars in

the system (L)

0.33

Probability that the system

is empty (P0)

TWO

MECHANICS

= 3 FOR BOTH

ONE FAST

MECHANIC

=4

0.50

0.50

0.75 cars

22.5 minutes

1 car

30 minutes

0.083 car

2.5 minutes

0.50 car

15 minutes

Table 13.2

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13-47

Muffler Multichannel Example

Program 13.2

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13-48

Constant Service Time Model (M/D/1)

Constant service times are used when

customers or units are processed

according to a fixed cycle.

The values for Lq, Wq, L, and W are always

less than they would be for models with

variable service time.

In fact both average queue length and

average waiting time are halved in

constant service rate models.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-49

Constant Service Time Model (M/D/1)

1. Average length of the queue

2. Average waiting time in the queue

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13-50

Constant Service Time Model (M/D/1)

3. Average number of customers in the system

4. Average time in the system

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13-51

Garcia-Golding Recycling, Inc.

The company collects and compacts

aluminum cans and glass bottles.

Trucks arrive at an average rate of 8 per hour

(Poisson distribution).

Truck drivers wait about 15 minutes before

they empty their load.

Drivers and trucks cost $60 per hour.

A new automated machine can process

truckloads at a constant rate of 12 per hour.

A new compactor would be amortized at $3

per truck unloaded.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-52

Constant Service Time Model (M/D/1)

Analysis of cost versus benefit of the purchase

Current waiting cost/trip = (1/4 hour waiting time)($60/hour cost)

= $15/trip

New system: = 8 trucks/hour arriving

= 12 trucks/hour served

Average waiting

time in queue = Wq = 1/12 hour

Waiting cost/trip

with new compactor = (1/12 hour wait)($60/hour cost) = $5/trip

Savings with

new equipment = $15 (current system) $5 (new system)

= $10 per trip

Cost of new equipment

amortized = $3/trip

Net savings = $7/trip

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-53

Excel QM Solution for Constant Service

Time Model with Garcia-Golding Recycling

Example

Program 13.3

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13-54

Finite Population Model

(M/M/1 with Finite Source)

When the population of potential customers is

limited, the models are different.

There is now a dependent relationship between

the length of the queue and the arrival rate.

The model has the following assumptions:

1. There is only one server.

2. The population of units seeking service is

finite.

3. Arrivals follow a Poisson distribution and

service times are exponentially distributed.

4. Customers are served on a first-come, firstserved basis.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-55

Finite Population Model

(M/M/1 with Finite Source)

Equations for the finite population model:

Using = mean arrival rate, = mean service

rate, and N = size of the population, the

operating characteristics are:

1. Probability that the system is empty:

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13-56

Finite Population Model

(M/M/1 with Finite Source)

2. Average length of the queue:

3. Average number of customers (units) in the system:

4. Average waiting time in the queue:

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13-57

Finite Population Model

(M/M/1 with Finite Source)

5. Average time in the system:

6. Probability of n units in the system:

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13-58

Department of Commerce

The Department of Commerce has five printers

that each need repair after about 20 hours of work.

Breakdowns will follow a Poisson distribution.

The technician can service a printer in an average

of about 2 hours, following an exponential

distribution.

Therefore:

= 1/20 = 0.05 printer/hour

= 1/2 = 0.50 printer/hour

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-59

Department of Commerce Example

1.

2.

3.

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13-60

Department of Commerce Example

4.

5.

If printer downtime costs $120 per hour and the

technician is paid $25 per hour, the total cost is:

Total

hourly

cost

(Average number of printers down)

= (Cost per downtime hour)

+ Cost per technician hour

= (0.64)($120) + $25 = $101.80

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-61

Excel QM For Finite Population Model with

Department of Commerce Example

Program 13.4

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13-62

Some General Operating

Characteristic Relationships

Certain relationships exist among specific

operating characteristics for any queuing system

in a steady state.

A steady state condition exists when a system is

in its normal stabilized condition, usually after an

initial transient state.

Equations:

L= W

Lq = Wq

(or W = L/ )

(or Wq = Lq/ )

And

W = Wq + 1/

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13-63

More Complex Queuing Models and

the Use of Simulation

In the real world there are often variations from

basic queuing models.

Computer simulation can be used to solve these

more complex problems.

Simulation allows the analysis of controllable

factors.

Simulation should be used when standard

queuing models provide only a poor

approximation of the actual service system.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-64

Copyright

All rights reserved. No part of this publication may be

reproduced, stored in a retrieval system, or transmitted, in

any form or by any means, electronic, mechanical,

photocopying, recording, or otherwise, without the prior

written permission of the publisher. Printed in the United

States of America.

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

13-65

Chapter 7

Linear Programming Models:

Graphical and Computer

Methods

To accompany

Quantitative Analysis for Management, Eleventh Edition,

by Render, Stair, and Hanna

Power Point slides created by Brian Peterson

A

B C

1 Garcia-Golding Recycling

12345

Excel QM Solution for Constant Service

Time Model with Garcia-Golding Recycling

3 Waiting Lines

9

10

11

12

6 Data

7 Arrival rate (2)

8

Service rate (μ)

M/D/1 (Constant Service Times)

The arrival RATE and service RATE both must be rates and use the same time unit. Given a time

Program 13.3

Example

▬▬▬▬▬▬▬▬▬▬▬ ٧٩٤ / ۳۲۲-۳۲۱

Results

8 Average server utilization (p)

12 Average number of customers in the queue(La)

Average number of customers in the system(L)

Average waiting time in the queue(W₁)

Average time in the system(W)

Probability (% of time) system is empty (Po)

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

E

0.66667

0.66667

1.33333

0.08333

0.16667

0.33333

●●●

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