STAT 201 SEU Deterministic and Probabilistic Models Discussion

PURPOSE

To assess your ability to:

apply the terminology of decision making to describe business problems

compare and contrast deterministic and probabilistic models

ACTION ITEMS

A probabilistic model; and

A situation in which you could use post optimality analysis (also known as sensitivity analysis).

Chapter 1
Introduction to
Quantitative Analysis
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. Describe the quantitative analysis approach
2. Understand the application of quantitative
analysis in a real situation
3. Describe the use of modeling in quantitative
analysis
4. Use computers and spreadsheet models to
perform quantitative analysis
5. Discuss possible problems in using
quantitative analysis
6. Perform a break-even analysis
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1-2
Chapter Outline
1.1 Introduction
1.2 What Is Quantitative Analysis?
1.3 The Quantitative Analysis Approach
1.4 How to Develop a Quantitative Analysis
Model
1.5 The Role of Computers and Spreadsheet
Models in the Quantitative Analysis
Approach
1.6 Possible Problems in the Quantitative
Analysis Approach
1.7 Implementation — Not Just the Final Step
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1-3
Introduction
◼ Mathematical tools have been used for
thousands of years.
◼ Quantitative analysis can be applied to
a wide variety of problems.
◼ It’s not enough to just know the
mathematics of a technique.
◼ One must understand the specific
applicability of the technique, its
limitations, and its assumptions.
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1-4
Examples of Quantitative Analyses
◼ In the mid 1990s, Taco Bell saved over $150
million using forecasting and scheduling
quantitative analysis models.
◼ NBC television increased revenues by over
$200 million between 1996 and 2000 by using
quantitative analysis to develop better sales
plans.
◼ Continental Airlines saved over $40 million in
2001 using quantitative analysis models to
quickly recover from weather delays and other
disruptions.
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1-5
What is Quantitative Analysis?
Quantitative analysis is a scientific approach
to managerial decision making in which raw
data are processed and manipulated to
produce meaningful information.
Raw Data
Quantitative
Analysis
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Meaningful
Information
1-6
What is Quantitative Analysis?
◼ Quantitative factors are data that can be
accurately calculated. Examples include:
◼ Different investment alternatives
◼ Interest rates
◼ Inventory levels
◼ Demand
◼ Labor cost
◼ Qualitative factors are more difficult to
quantify but affect the decision process.
Examples include:
◼ The weather
◼ State and federal legislation
◼ Technological breakthroughs.
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The Quantitative Analysis Approach
Defining the Problem
Developing a Model
Acquiring Input Data
Developing a Solution
Testing the Solution
Analyzing the Results
Implementing the Results
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Figure 1.1
1-8
Defining the Problem
Develop a clear and concise statement that
gives direction and meaning to subsequent
steps.
◼ This may be the most important and difficult
step.
◼ It is essential to go beyond symptoms and
identify true causes.
◼ It may be necessary to concentrate on only a
few of the problems – selecting the right
problems is very important
◼ Specific and measurable objectives may have
to be developed.
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Developing a Model
$ Sales
Quantitative analysis models are realistic,
solvable, and understandable mathematical
representations of a situation.
$ Advertising
There are different types of models:
Scale
models
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Schematic
models
1-10
Developing a Model
Models generally contain variables
(controllable and uncontrollable) and
parameters.
◼ Controllable variables are the decision
variables and are generally unknown.
◼
How many items should be ordered for inventory?
◼ Parameters are known quantities that are a
part of the model.
◼
What is the holding cost of the inventory?
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Acquiring Input Data
Input data must be accurate – GIGO rule:
Garbage
In
Process
Garbage
Out
Data may come from a variety of sources such as
company reports, company documents, interviews,
on-site direct measurement, or statistical sampling.
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1-12
Developing a Solution
The best (optimal) solution to a problem is
found by manipulating the model variables
until a solution is found that is practical
and can be implemented.
Common techniques are
◼ Solving equations.
◼ Trial and error – trying various approaches
and picking the best result.
◼ Complete enumeration – trying all possible
values.
◼ Using an algorithm – a series of repeating
steps to reach a solution.
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Testing the Solution
Both input data and the model should be
tested for accuracy before analysis and
implementation.
◼ New data can be collected to test the model.
◼ Results should be logical, consistent, and
represent the real situation.
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Analyzing the Results
Determine the implications of the solution:
◼ Implementing results often requires change in
an organization.
◼ The impact of actions or changes needs to be
studied and understood before
implementation.
Sensitivity analysis determines how much
the results will change if the model or
input data changes.
◼ Sensitive models should be very thoroughly
tested.
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Implementing the Results
Implementation incorporates the solution
into the company.
◼ Implementation can be very difficult.
◼ People may be resistant to changes.
◼ Many quantitative analysis efforts have failed
because a good, workable solution was not
properly implemented.
Changes occur over time, so even
successful implementations must be
monitored to determine if modifications are
necessary.
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1-16
Modeling in the Real World
Quantitative analysis models are used
extensively by real organizations to solve
real problems.
◼ In the real world, quantitative analysis
models can be complex, expensive, and
difficult to sell.
◼ Following the steps in the process is an
important component of success.
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1-17
How To Develop a Quantitative
Analysis Model
A mathematical model of profit:
Profit = Revenue – Expenses
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How To Develop a Quantitative
Analysis Model
Expenses can be represented as the sum of fixed and
variable costs. Variable costs are the product of unit
costs times the number of units.
Profit = Revenue – (Fixed cost + Variable cost)
Profit = (Selling price per unit)(number of units
sold) – [Fixed cost + (Variable costs per
unit)(Number of units sold)]
Profit = sX – [f + vX]
Profit = sX – f – vX
where
s = selling price per unit
f = fixed cost
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v = variable cost per unit
X = number of units sold
1-19
How To Develop a Quantitative
Analysis Model
Expenses can be represented as the sum of fixed and
variable costs and variable
costs are the
product
of
The parameters
of this
model
unit costs times the number
units
are f, v,of
and
s as these are the
inputscost
inherent
in the cost)
model
Profit = Revenue – (Fixed
+ Variable
The
decision
variable
Profit = (Selling price
per
unit)(number
of of
units
interest
X
sold) – [Fixed
cost +is(Variable
costs per
unit)(Number of units sold)]
Profit = sX – [f + vX]
Profit = sX – f – vX
where
s = selling price per unit
f = fixed cost
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v = variable cost per unit
X = number of units sold
1-20
Pritchett’s Precious Time Pieces
The company buys, sells, and repairs old clocks.
Rebuilt springs sell for $10 per unit. Fixed cost of
equipment to build springs is $1,000. Variable cost
for spring material is $5 per unit.
s = 10
f = 1,000
v=5
Number of spring sets sold = X
Profits = sX – f – vX
If sales = 0, profits = -f = –$1,000.
If sales = 1,000, profits = [(10)(1,000) – 1,000 – (5)(1,000)]
= $4,000
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1-21
Pritchett’s Precious Time Pieces
Companies are often interested in the break-even
point (BEP). The BEP is the number of units sold
that will result in $0 profit.
0 = sX – f – vX,
or
0 = (s – v)X – f
Solving for X, we have
f = (s – v)X
f
X= s–v
Fixed cost
BEP = (Selling price per unit) – (Variable cost per unit)
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1-22
Pritchett’s Precious Time Pieces
Companies are often interested in their break-even
point (BEP). The BEP is the number of units sold
BEP for Pritchett’s Precious Time Pieces
that will result in $0 profit.
= –200
0 BEP
= sX –= f$1,000/($10
– vX, or – 0$5)
= (s
v)Xunits
–f
Salesfor
of less
200 units of rebuilt springs
Solving
X, wethan
have
will result in a loss.
f = (s – v)X
Sales of over 200 unitsfof rebuilt springs will
result in a profit. X =
s–v
Fixed cost
BEP = (Selling price per unit) – (Variable cost per unit)
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1-23
Advantages of Mathematical Modeling
1. Models can accurately represent reality.
2. Models can help a decision maker
formulate problems.
3. Models can give us insight and information.
4. Models can save time and money in
decision making and problem solving.
5. A model may be the only way to solve large
or complex problems in a timely fashion.
6. A model can be used to communicate
problems and solutions to others.
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Models Categorized by Risk
◼ Mathematical models that do not involve
risk are called deterministic models.
◼ All of the values used in the model are
known with complete certainty.
◼ Mathematical models that involve risk,
chance, or uncertainty are called
probabilistic models.
◼ Values used in the model are estimates
based on probabilities.
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1-25
Computers and Spreadsheet Models
QM for Windows
◼ An easy to use
decision support
system for use in
POM and QM
courses
◼ This is the main
menu of
quantitative
models
Program 1.1
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Computers and Spreadsheet Models
Excel QM’s Main Menu (2010)
◼ Works automatically within Excel spreadsheets
Program 1.2
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Computers and Spreadsheet Models
Selecting
Break-Even
Analysis in
Excel QM
Program 1.3A
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Computers and Spreadsheet Models
BreakEven
Analysis
in Excel
QM
Program 1.3B
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Computers and Spreadsheet Models
Using Goal
Seek in the
BreakEven
Problem
Program 1.4
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Possible Problems in the
Quantitative Analysis Approach
Defining the problem
◼ Problems may not be easily identified.
◼ There may be conflicting viewpoints
◼ There may be an impact on other
departments.
◼ Beginning assumptions may lead to a
particular conclusion.
◼ The solution may be outdated.
Developing a model
◼ Manager’s perception may not fit a textbook
model.
◼ There is a trade-off between complexity and
ease of understanding.
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Possible Problems in the
Quantitative Analysis Approach
Acquiring accurate input data
◼ Accounting data may not be collected for
quantitative problems.
◼ The validity of the data may be suspect.
Developing an appropriate solution
◼ The mathematics may be hard to understand.
◼ Having only one answer may be limiting.
Testing the solution for validity
Analyzing the results in terms of the whole
organization
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1-32
Implementation –
Not Just the Final Step
There may be an institutional lack of
commitment and resistance to change.
◼ Management may fear the use of formal
analysis processes will reduce their
decision-making power.
◼ Action-oriented managers may want
“quick and dirty” techniques.
◼ Management support and user
involvement are important.
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Implementation –
Not Just the Final Step
There may be a lack of commitment
by quantitative analysts.
◼ Analysts should be involved with the
problem and care about the solution.
◼ Analysts should work with users and
take their feelings into account.
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1-34
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.
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Chapter 3
Decision Analysis
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. List the steps of the decision-making
process.
2. Describe the types of decision-making
environments.
3. Make decisions under uncertainty.
4. Use probability values to make decisions
under risk.
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3-2
Learning Objectives
After completing this chapter, students will be able to:
5. Develop accurate and useful decision
trees.
6. Revise probabilities using Bayesian
analysis.
7. Use computers to solve basic decisionmaking problems.
8. Understand the importance and use of
utility theory in decision making.
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3-3
Chapter Outline
3.1 Introduction
3.2 The Six Steps in Decision Making
3.3 Types of Decision-Making
Environments
3.4 Decision Making under Uncertainty
3.5 Decision Making under Risk
3.6 Decision Trees
3.7 How Probability Values Are
Estimated by Bayesian Analysis
3.8 Utility Theory
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3-4
Introduction
◼ What is involved in making a good
decision?
◼ Decision theory is an analytic and
systematic approach to the study of
decision making.
◼ A good decision is one that is based
on logic, considers all available data
and possible alternatives, and the
quantitative approach described here.
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3-5
The Six Steps in Decision Making
1. Clearly define the problem at hand.
2. List the possible alternatives.
3. Identify the possible outcomes or states
of nature.
4. List the payoff (typically profit) of each
combination of alternatives and
outcomes.
5. Select one of the mathematical decision
theory models.
6. Apply the model and make your decision.
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3-6
Thompson Lumber Company
Step 1 – Define the problem.
◼ The company is considering
expanding by manufacturing and
marketing a new product – backyard
storage sheds.
Step 2 – List alternatives.
◼ Construct a large new plant.
◼ Construct a small new plant.
◼ Do not develop the new product line
at all.
Step 3 – Identify possible outcomes.
◼ The market could be favorable or
unfavorable.
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3-7
Thompson Lumber Company
Step 4 – List the payoffs.
◼ Identify conditional values for the
profits for large plant, small plant, and
no development for the two possible
market conditions.
Step 5 – Select the decision model.
◼ This depends on the environment and
amount of risk and uncertainty.
Step 6 – Apply the model to the data.
◼ Solution and analysis are then used to
aid in decision-making.
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3-8
Thompson Lumber Company
Decision Table with Conditional Values for
Thompson Lumber
STATE OF NATURE
FAVORABLE
MARKET ($)
UNFAVORABLE
MARKET ($)
Construct a large plant
200,000
–180,000
Construct a small plant
100,000
–20,000
0
0
ALTERNATIVE
Do nothing
Table 3.1
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3-9
Types of Decision-Making
Environments
Type 1: Decision making under certainty
◼ The decision maker knows with
certainty the consequences of every
alternative or decision choice.
Type 2: Decision making under uncertainty
◼ The decision maker does not know the
probabilities of the various outcomes.
Type 3: Decision making under risk
◼ The decision maker knows the
probabilities of the various outcomes.
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3-10
Decision Making Under
Uncertainty
There are several criteria for making decisions
under uncertainty:
1. Maximax (optimistic)
2. Maximin (pessimistic)
3. Criterion of realism (Hurwicz)
4. Equally likely (Laplace)
5. Minimax regret
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3-11
Maximax
Used to find the alternative that maximizes the
maximum payoff.
◼ Locate the maximum payoff for each alternative.
◼ Select the alternative with the maximum number.
STATE OF NATURE
FAVORABLE
MARKET ($)
UNFAVORABLE
MARKET ($)
MAXIMUM IN
A ROW ($)
Construct a large
plant
200,000
–180,000
200,000
Construct a small
plant
100,000
–20,000
100,000
0
0
0
ALTERNATIVE
Maximax
Do nothing
Table 3.2
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3-12
Maximin
Used to find the alternative that maximizes
the minimum payoff.
◼ Locate the minimum payoff for each alternative.
◼ Select the alternative with the maximum
number.
STATE OF NATURE
FAVORABLE
MARKET ($)
UNFAVORABLE
MARKET ($)
MINIMUM IN
A ROW ($)
Construct a large
plant
200,000
–180,000
–180,000
Construct a small
plant
100,000
–20,000
–20,000
0
0
0
ALTERNATIVE
Do nothing
Table 3.3
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Maximin
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Criterion of Realism (Hurwicz)
This is a weighted average compromise
between optimism and pessimism.
◼ Select a coefficient of realism , with 0≤α≤1.
◼ A value of 1 is perfectly optimistic, while a
value of 0 is perfectly pessimistic.
◼ Compute the weighted averages for each
alternative.
◼ Select the alternative with the highest value.
Weighted average = (maximum in row)
+ (1 – )(minimum in row)
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3-14
Criterion of Realism (Hurwicz)
◼ For the large plant alternative using  = 0.8:
(0.8)(200,000) + (1 – 0.8)(–180,000) = 124,000
◼ For the small plant alternative using  = 0.8:
(0.8)(100,000) + (1 – 0.8)(–20,000) = 76,000
STATE OF NATURE
FAVORABLE
MARKET ($)
UNFAVORABLE
MARKET ($)
CRITERION
OF REALISM
( = 0.8) $
Construct a large
plant
200,000
–180,000
124,000
Construct a small
plant
100,000
–20,000
76,000
0
0
0
ALTERNATIVE
Realism
Do nothing
Table 3.4
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3-15
Equally Likely (Laplace)
Considers all the payoffs for each alternative
◼ Find the average payoff for each alternative.
◼ Select the alternative with the highest average.
STATE OF NATURE
FAVORABLE
MARKET ($)
UNFAVORABLE
MARKET ($)
ROW
AVERAGE ($)
Construct a large
plant
200,000
–180,000
10,000
Construct a small
plant
100,000
–20,000
40,000
0
0
ALTERNATIVE
Do nothing
Equally likely
0
Table 3.5
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3-16
Minimax Regret
Based on opportunity loss or regret, this is
the difference between the optimal profit and
actual payoff for a decision.
◼ Create an opportunity loss table by determining
the opportunity loss from not choosing the best
alternative.
◼ Opportunity loss is calculated by subtracting
each payoff in the column from the best payoff
in the column.
◼ Find the maximum opportunity loss for each
alternative and pick the alternative with the
minimum number.
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3-17
Minimax Regret
Determining Opportunity Losses for Thompson Lumber
STATE OF NATURE
FAVORABLE MARKET ($)
UNFAVORABLE MARKET ($)
200,000 – 200,000
0 – (–180,000)
200,000 – 100,000
0 – (–20,000)
200,000 – 0
0–0
Table 3.6
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3-18
Minimax Regret
Opportunity Loss Table for Thompson Lumber
STATE OF NATURE
ALTERNATIVE
FAVORABLE
MARKET ($)
UNFAVORABLE
MARKET ($)
Construct a large plant
0
180,000
Construct a small plant
100,000
20,000
Do nothing
200,000
0
Table 3.7
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3-19
Minimax Regret
Thompson’s Minimax Decision Using Opportunity Loss
STATE OF NATURE
ALTERNATIVE
FAVORABLE
MARKET ($)
UNFAVORABLE
MARKET ($)
MAXIMUM IN
A ROW ($)
Construct a large
plant
0
180,000
180,000
Construct a small
plant
100,000
20,000
100,000
Do nothing
200,000
0
Minimax
200,000
Table 3.8
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3-20
Decision Making Under Risk
◼ This is decision making when there are several
possible states of nature, and the probabilities
associated with each possible state are known.
◼ The most popular method is to choose the
alternative with the highest expected monetary
value (EMV).
◼ This is very similar to the expected value calculated in
the last chapter.
EMV (alternative i) = (payoff of first state of nature)
x (probability of first state of nature)
+ (payoff of second state of nature)
x (probability of second state of nature)
+ … + (payoff of last state of nature)
x (probability of last state of nature)
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3-21
EMV for Thompson Lumber
◼ Suppose each market outcome has a probability of
occurrence of 0.50.
◼ Which alternative would give the highest EMV?
◼ The calculations are:
EMV (large plant) = ($200,000)(0.5) + (–$180,000)(0.5)
= $10,000
EMV (small plant) = ($100,000)(0.5) + (–$20,000)(0.5)
= $40,000
EMV (do nothing) = ($0)(0.5) + ($0)(0.5)
= $0
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3-22
EMV for Thompson Lumber
STATE OF NATURE
FAVORABLE
MARKET ($)
UNFAVORABLE
MARKET ($)
EMV ($)
Construct a large
plant
200,000
–180,000
10,000
Construct a small
plant
100,000
–20,000
40,000
Do nothing
0
0
0
Probabilities
0.50
0.50
ALTERNATIVE
Table 3.9
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Largest EMV
3-23
Expected Value of Perfect
Information (EVPI)
◼ EVPI places an upper bound on what you should
pay for additional information.
EVPI = EVwPI – Maximum EMV
◼ EVwPI is the long run average return if we have
perfect information before a decision is made.
EVwPI = (best payoff for first state of nature)
x (probability of first state of nature)
+ (best payoff for second state of nature)
x (probability of second state of nature)
+ … + (best payoff for last state of nature)
x (probability of last state of nature)
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3-24
Expected Value of Perfect
Information (EVPI)
◼ Suppose Scientific Marketing, Inc. offers
analysis that will provide certainty about
market conditions (favorable).
◼ Additional information will cost $65,000.
◼ Should Thompson Lumber purchase the
information?
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3-25
Expected Value of Perfect
Information (EVPI)
Decision Table with Perfect Information
STATE OF NATURE
FAVORABLE
MARKET ($)
UNFAVORABLE
MARKET ($)
EMV ($)
Construct a large
plant
200,000
-180,000
10,000
Construct a small
plant
100,000
-20,000
40,000
Do nothing
0
0
0
With perfect
information
200,000
0
100,000
Probabilities
0.5
ALTERNATIVE
0.5
EVwPI
Table 3.10
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3-26
Expected Value of Perfect
Information (EVPI)
The maximum EMV without additional information is
$40,000.
EVPI = EVwPI – Maximum EMV
= $100,000 – $40,000
= $60,000
So the maximum Thompson
should pay for the additional
information is $60,000.
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3-27
Expected Value of Perfect
Information (EVPI)
The maximum EMV without additional information is
$40,000.
EVPI = EVwPI – Maximum EMV
= $100,000 – $40,000
= $60,000
So the maximum Thompson
should pay for the additional
information is $60,000.
Therefore, Thompson should not
pay $65,000 for this information.
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3-28
Expected Opportunity Loss
◼ Expected opportunity loss (EOL) is the
cost of not picking the best solution.
◼ First construct an opportunity loss table.
◼ For each alternative, multiply the
opportunity loss by the probability of that
loss for each possible outcome and add
these together.
◼ Minimum EOL will always result in the
same decision as maximum EMV.
◼ Minimum EOL will always equal EVPI.
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3-29
Expected Opportunity Loss
STATE OF NATURE
ALTERNATIVE
Construct a large plant
Construct a small
plant
Do nothing
Probabilities
FAVORABLE
MARKET ($)
0
UNFAVORABLE
MARKET ($)
180,000
EOL
90,000
100,000
20,000
60,000
200,000
0.50
0
0.50
100,000
Table 3.11
Minimum EOL
EOL (large plant) = (0.50)($0) + (0.50)($180,000)
= $90,000
EOL (small plant) = (0.50)($100,000) + (0.50)($20,000)
= $60,000
EOL (do nothing) = (0.50)($200,000) + (0.50)($0)
= $100,000
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3-30
Sensitivity Analysis
◼ Sensitivity analysis examines how the decision
might change with different input data.
◼ For the Thompson Lumber example:
P = probability of a favorable market
(1 – P) = probability of an unfavorable market
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3-31
Sensitivity Analysis
EMV(Large Plant) = $200,000P – $180,000)(1 – P)
= $200,000P – $180,000 + $180,000P
= $380,000P – $180,000
EMV(Small Plant) = $100,000P – $20,000)(1 – P)
= $100,000P – $20,000 + $20,000P
= $120,000P – $20,000
EMV(Do Nothing) = $0P + 0(1 – P)
= $0
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3-32
Sensitivity Analysis
EMV Values
$300,000
$200,000
$100,000
EMV (large plant)
Point 2
EMV (small plant)
Point 1
0
EMV (do nothing)
.167
–$100,000
.615
1
Values of P
–$200,000
Figure 3.1
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3-33
Sensitivity Analysis
Point 1:
EMV(do nothing) = EMV(small plant)
0 = $120,000 P − $20,000
P=
20,000
= 0.167
120,000
Point 2:
EMV(small plant) = EMV(large plant)
$120,000 P − $20,000 = $380,000 P − $180,000
160,000
P=
= 0.615
260,000
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3-34
Sensitivity Analysis
RANGE OF P
VALUES
BEST
ALTERNATIVE
Do nothing
Less than 0.167
EMV Values
Construct a small plant
0.167 – 0.615
$300,000
Construct a large plant
Greater than 0.615
$200,000
$100,000
EMV (large plant)
Point 2
EMV (small plant)
Point 1
0
EMV (do nothing)
.167
–$100,000
.615
1
Values of P
–$200,000
Figure 3.1
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3-35
Using Excel
Input Data for the Thompson Lumber Problem
Using Excel QM
Program 3.1A
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3-36
Using Excel
Output Results for the Thompson Lumber Problem
Using Excel QM
Program 3.1B
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3-37
Decision Trees
◼ Any problem that can be presented in a
decision table can also be graphically
represented in a decision tree.
◼ Decision trees are most beneficial when a
sequence of decisions must be made.
◼ All decision trees contain decision points
or nodes, from which one of several alternatives
may be chosen.
◼ All decision trees contain state-of-nature
points or nodes, out of which one state of
nature will occur.
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3-38
Five Steps of
Decision Tree Analysis
1. Define the problem.
2. Structure or draw the decision tree.
3. Assign probabilities to the states of
nature.
4. Estimate payoffs for each possible
combination of alternatives and states of
nature.
5. Solve the problem by computing
expected monetary values (EMVs) for
each state of nature node.
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3-39
Structure of Decision Trees
◼ Trees start from left to right.
◼ Trees represent decisions and outcomes
in sequential order.
◼ Squares represent decision nodes.
◼ Circles represent states of nature nodes.
◼ Lines or branches connect the decisions
nodes and the states of nature.
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3-40
Thompson’s Decision Tree
A State-of-Nature Node
Favorable Market
A Decision Node
1
Unfavorable Market
Favorable Market
Construct
Small Plant
2
Unfavorable Market
Figure 3.2
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3-41
Thompson’s Decision Tree
EMV for Node
1 = $10,000
= (0.5)($200,000) + (0.5)(–$180,000)
Payoffs
Favorable Market (0.5)
Alternative with best
EMV is selected
1
Unfavorable Market (0.5)
Favorable Market (0.5)
Construct
Small Plant
2
Unfavorable Market (0.5)
EMV for Node
2 = $40,000
Figure 3.3
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
$200,000
–$180,000
$100,000
–$20,000
= (0.5)($100,000)
+ (0.5)(–$20,000)
$0
3-42
Thompson’s Complex Decision Tree
First Decision
Point
Second Decision
Point
Payoffs
Favorable Market (0.78)
2
Small
Plant
3
1
–$190,000
$90,000
–$30,000
No Plant
–$10,000
Favorable Market (0.27)
4
Small
Plant
5
Figure 3.4
Small
Plant
7
–$190,000
$90,000
–$30,000
No Plant
–$10,000
Unfavorable Market (0.50)
Favorable Market (0.50)
Unfavorable Market (0.50)
No Plant
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
$190,000
Unfavorable Market (0.73)
Favorable Market (0.27)
Unfavorable Market (0.73)
Favorable Market (0.50)
6
$190,000
Unfavorable Market (0.22)
Favorable Market (0.78)
Unfavorable Market (0.22)
$200,000
–$180,000
$100,000
–$20,000
$0
3-43
Thompson’s Complex Decision Tree
1. Given favorable survey results,
EMV(node 2) = EMV(large plant | positive survey)
= (0.78)($190,000) + (0.22)(–$190,000) = $106,400
EMV(node 3) = EMV(small plant | positive survey)
= (0.78)($90,000) + (0.22)(–$30,000) = $63,600
EMV for no plant = –$10,000
2. Given negative survey results,
EMV(node 4) = EMV(large plant | negative survey)
= (0.27)($190,000) + (0.73)(–$190,000) = –$87,400
EMV(node 5) = EMV(small plant | negative survey)
= (0.27)($90,000) + (0.73)(–$30,000) = $2,400
EMV for no plant = –$10,000
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3-44
Thompson’s Complex Decision Tree
3. Compute the expected value of the market survey,
EMV(node 1) = EMV(conduct survey)
= (0.45)($106,400) + (0.55)($2,400)
= $47,880 + $1,320 = $49,200
4. If the market survey is not conducted,
EMV(node 6) = EMV(large plant)
= (0.50)($200,000) + (0.50)(–$180,000) = $10,000
EMV(node 7) = EMV(small plant)
= (0.50)($100,000) + (0.50)(–$20,000) = $40,000
EMV for no plant = $0
5. The best choice is to seek marketing information.
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Thompson’s Complex Decision Tree
First Decision
Point
Second Decision
Point
Payoffs
$106,400
$106,400 Favorable Market (0.78)
Small
Plant
$63,600
–$190,000
$90,000
–$30,000
No Plant
–$10,000
Small
Plant
Figure 3.5
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
$40,000
$49,200
$2,400
–$87,400 Favorable Market (0.27)
Small
Plant
$2,400
$190,000
Unfavorable Market (0.22)
Favorable Market (0.78)
Unfavorable Market (0.22)
$190,000
Unfavorable Market (0.73)
Favorable Market (0.27)
Unfavorable Market (0.73)
–$190,000
$90,000
–$30,000
No Plant
–$10,000
$10,000
Favorable Market (0.50)
$40,000
Unfavorable Market (0.50)
Favorable Market (0.50)
Unfavorable Market (0.50)
No Plant
$200,000
–$180,000
$100,000
–$20,000
$0
3-46
Expected Value of Sample Information
◼ Suppose Thompson wants to know the
actual value of doing the survey.
Expected value
with sample
EVSI = information, assuming –
no cost to gather it
Expected value
of best decision
without sample
information
= (EV with sample information + cost)
– (EV without sample information)
EVSI = ($49,200 + $10,000) – $40,000 = $19,200
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3-47
Sensitivity Analysis
◼ How sensitive are the decisions to
changes in the probabilities?
◼ How sensitive is our decision to the
probability of a favorable survey result?
◼ That is, if the probability of a favorable
result (p = .45) where to change, would we
make the same decision?
◼ How much could it change before we would
make a different decision?
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3-48
Sensitivity Analysis
p = probability of a favorable survey result
(1 – p) = probability of a negative survey result
EMV(node 1) = ($106,400)p +($2,400)(1 – p)
= $104,000p + $2,400
We are indifferent when the EMV of node 1 is the
same as the EMV of not conducting the survey,
$40,000
$104,000p + $2,400 = $40,000
$104,000p = $37,600
p = $37,600/$104,000 = 0.36
If p0.36,
conduct the survey.
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3-49
Bayesian Analysis
◼ There are many ways of getting
probability data. It can be based on:
Management’s experience and intuition.
◼ Historical data.
◼ Computed from other data using Bayes’
theorem.
◼
◼ Bayes’ theorem incorporates initial
estimates and information about the
accuracy of the sources.
◼ It also allows the revision of initial
estimates based on new information.
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3-50
Calculating Revised Probabilities
◼ In the Thompson Lumber case we used these four
conditional probabilities:
P (favorable market(FM) | survey results positive) = 0.78
P (unfavorable market(UM) | survey results positive) = 0.22
P (favorable market(FM) | survey results negative) = 0.27
P (unfavorable market(UM) | survey results negative) = 0.73
◼ But how were these calculated?
◼ The prior probabilities of these markets are:
P (FM) = 0.50
P (UM) = 0.50
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Calculating Revised Probabilities
◼ Through discussions with experts Thompson has
learned the information in the table below.
◼ He can use this information and Bayes’ theorem
to calculate posterior probabilities.
STATE OF NATURE
RESULT OF
SURVEY
FAVORABLE MARKET
(FM)
UNFAVORABLE MARKET
(UM)
Positive (predicts
favorable market
for product)
P (survey positive | FM)
= 0.70
P (survey positive | UM)
= 0.20
Negative (predicts
unfavorable
market for
product)
P (survey negative | FM)
= 0.30
P (survey negative | UM)
= 0.80
Table 3.12
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Calculating Revised Probabilities
◼ Recall Bayes’ theorem:
P ( B | A)  P ( A)
P( A | B) =
P ( B | A)  P ( A) + P ( B | A )  P ( A)
where
A, B = any two events
A = complement of A
For this example, A will represent a favorable
market and B will represent a positive survey.
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3-53
Calculating Revised Probabilities
◼ P (FM | survey positive)
P ( survey positive | FM )  P ( FM )
=
P(survey positive |FM)  P(FM) + P(survey positive |UM)  P(UM)
=
(0.70)(0.50)
0.35
=
= 0.78
(0.70)(0.50) + (0.20)(0.50) 0.45
◼ P (UM | survey positive)
P ( survey positive | UM )  P (UM )
=
P(survey positive |UM)  P(UM) + P(survey positive |FM)  P(FM)
=
(0.20)(0.50)
0.10
=
= 0.22
(0.20)(0.50) + (0.70)(0.50) 0.45
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Calculating Revised Probabilities
Probability Revisions Given a Positive Survey
POSTERIOR PROBABILITY
CONDITIONAL
PROBABILITY
P(SURVEY
POSITIVE | STATE
OF NATURE)
PRIOR
PROBABILITY
FM
0.70
X 0.50
=
0.35
0.35/0.45 = 0.78
UM
0.20
X 0.50
=
0.10
0.10/0.45 = 0.22
P(survey results positive) =
0.45
1.00
STATE OF
NATURE
JOINT
PROBABILITY
P(STATE OF
NATURE |
SURVEY
POSITIVE)
Table 3.13
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3-55
Calculating Revised Probabilities
◼ P (FM | survey negative)
P ( survey negative | FM )  P ( FM )
=
P(survey negative |FM)  P(FM) + P(survey negative |UM)  P(UM)
=
(0.30)(0.50)
0.15
=
= 0.27
(0.30)(0.50) + (0.80)(0.50) 0.55
◼ P (UM | survey negative)
P ( survey negative | UM )  P (UM )
=
P(survey negative |UM)  P(UM) + P(survey negative |FM)  P(FM)
=
(0.80)(0.50)
0.40
=
= 0.73
(0.80)(0.50) + (0.30)(0.50) 0.55
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3-56
Calculating Revised Probabilities
Probability Revisions Given a Negative Survey
POSTERIOR PROBABILITY
CONDITIONAL
PROBABILITY
P(SURVEY
NEGATIVE | STATE
OF NATURE)
PRIOR
PROBABILITY
FM
0.30
X 0.50
=
0.15
0.15/0.55 =
0.27
UM
0.80
X 0.50
=
0.40
0.40/0.55 =
0.73
P(survey results positive) =
0.55
STATE OF
NATURE
JOINT
PROBABILITY
P(STATE OF
NATURE |
SURVEY
NEGATIVE)
1.00
Table 3.14
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3-57
Using Excel
Formulas Used for Bayes’ Calculations in Excel
Program 3.2A
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3-58
Using Excel
Results of Bayes’ Calculations in Excel
Program 3.2B
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3-59
Potential Problems Using
Survey Results
◼ We can not always get the necessary
data for analysis.
◼ Survey results may be based on cases
where an action was taken.
◼ Conditional probability information
may not be as accurate as we would
like.
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3-60
Utility Theory
◼ Monetary value is not always a true
indicator of the overall value of the
result of a decision.
◼ The overall value of a decision is called
utility.
◼ Economists assume that rational
people make decisions to maximize
their utility.
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3-61
Utility Theory
Your Decision Tree for the Lottery Ticket
$2,000,000
Accept
Offer
$0
Reject
Offer
Heads
(0.5)
Tails
(0.5)
Figure 3.6
EMV = $2,500,000
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$5,000,000
3-62
Utility Theory
◼ Utility assessment assigns the worst outcome a
utility of 0, and the best outcome, a utility of 1.
◼ A standard gamble is used to determine utility
values.
◼ When you are indifferent, your utility values are
equal.
Expected utility of alternative 2 = Expected utility of alternative 1
Utility of other outcome = (p)(utility of best outcome, which is 1)
+ (1 – p)(utility of the worst outcome,
which is 0)
Utility of other outcome = (p)(1) + (1 – p)(0) = p
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3-63
Standard Gamble for Utility
Assessment
(p)
(1 – p)
Best Outcome
Utility = 1
Worst Outcome
Utility = 0
Other Outcome
Utility = ?
Figure 3.7
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3-64
Investment Example
◼ Jane Dickson wants to construct a utility curve
revealing her preference for money between $0
and $10,000.
◼ A utility curve plots the utility value versus the
monetary value.
◼ An investment in a bank will result in $5,000.
◼ An investment in real estate will result in $0 or
$10,000.
◼ Unless there is an 80% chance of getting $10,000
from the real estate deal, Jane would prefer to
have her money in the bank.
◼ So if p = 0.80, Jane is indifferent between the bank
or the real estate investment.
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3-65
Investment Example
p = 0.80
$10,000
U($10,000) = 1.0
(1 – p) = 0.20
$0
U($0.00) = 0.0
$5,000
U($5,000) = p = 0.80
Utility for $5,000 = U($5,000) = pU($10,000) + (1 – p)U($0)
= (0.8)(1) + (0.2)(0) = 0.8
Figure 3.8
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3-66
Investment Example
◼ We can assess other utility values in the same way.
◼ For Jane these are:
Utility for $7,000 = 0.90
Utility for $3,000 = 0.50
◼ Using the three utilities for different dollar amounts,
she can construct a utility curve.
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3-67
Utility Curve
1.0 –
0.9 –
0.8 –
U ($10,000) = 1.0
U ($7,000) = 0.90
U ($5,000) = 0.80
0.7 –
Utility
0.6 –
0.5 –
U ($3,000) = 0.50
0.4 –
0.3 –
0.2 –
0.1 –
U ($0) = 0
|
|
$0
$1,000
|
|
$3,000
Figure 3.9
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
|
|
|
$5,000
|
$7,000
|
|
|
$10,000
Monetary Value
3-68
Utility Curve
◼ Jane’s utility curve is typical of a risk avoider.
◼ She gets less utility from greater risk.
◼ She avoids situations where high losses might occur.
◼ As monetary value increases, her utility curve increases
at a slower rate.
◼ A risk seeker gets more utility from greater risk
◼ As monetary value increases, the utility curve increases
at a faster rate.
◼ Someone with risk indifference will have a linear
utility curve.
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3-69
Preferences for Risk
Utility
Risk
Avoider
Risk
Seeker
Figure 3.10
Monetary Outcome
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3-70
Utility as a
Decision-Making Criteria
◼ Once a utility curve has been developed
it can be used in making decisions.
◼ This replaces monetary outcomes with
utility values.
◼ The expected utility is computed instead
of the EMV.
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3-71
Utility as a
Decision-Making Criteria
◼ Mark Simkin loves to gamble.
◼ He plays a game tossing thumbtacks in
the air.
◼ If the thumbtack lands point up, Mark wins
$10,000.
◼ If the thumbtack lands point down, Mark
loses $10,000.
◼ Mark believes that there is a 45% chance
the thumbtack will land point up.
◼ Should Mark play the game (alternative 1)?
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3-72
Utility as a
Decision-Making Criteria
Decision Facing Mark Simkin
Tack Lands
Point Up (0.45)
Tack Lands
Point Down (0.55)
Figure 3.11
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
Mark Does Not Play the Game
$10,000
–$10,000
$0
3-73
Utility as a
Decision-Making Criteria
◼ Step 1– Define Mark’s utilities.
U (–$10,000) = 0.05
U ($0) = 0.15
U ($10,000) = 0.30
◼ Step 2 – Replace monetary values with
utility values.
E(alternative 1: play the game) = (0.45)(0.30) + (0.55)(0.05)
= 0.135 + 0.027 = 0.162
E(alternative 2: don’t play the game) = 0.15
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3-74
Utility Curve for Mark Simkin
1.00 –
Utility
0.75 –
0.50 –
0.30 –
0.25 –
0.15 –
Figure 3.12
0.05 –
0 |–
–$20,000
|
–$10,000
|
$0
|
$10,000
|
$20,000
Monetary Outcome
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3-75
Utility as a
Decision-Making Criteria
Using Expected Utilities in Decision Making
E = 0.162
Tack Lands
Point Up (0.45)
Tack Lands
Point Down (0.55)
Figure 3.13
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
Don’t Play
Utility
0.30
0.05
0.15
3-76
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.
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3-77
Chapter 5
Forecasting
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. Understand and know when to use various
families of forecasting models.
2. Compare moving averages, exponential
smoothing, and other time-series models.
3. Seasonally adjust data.
4. Understand Delphi and other qualitative
decision-making approaches.
5. Compute a variety of error measures.
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5-2
Chapter Outline
5.1 Introduction
5.2 Types of Forecasts
5.3 Scatter Diagrams and Time Series
5.4 Measures of Forecast Accuracy
5.5 Time-Series Forecasting Models
5.6 Monitoring and Controlling Forecasts
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5-3
Introduction
 Managers are always trying to reduce
uncertainty and make better estimates of what
will happen in the future.
 This is the main purpose of forecasting.
 Some firms use subjective methods: seat-of-the
pants methods, intuition, experience.
 There are also several quantitative techniques,
including:
 Moving averages
 Exponential smoothing
 Trend projections
 Least squares regression analysis
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5-4
Introduction
 Eight steps to forecasting:
1. Determine the use of the forecast—what
objective are we trying to obtain?
2. Select the items or quantities that are to be
forecasted.
3. Determine the time horizon of the forecast.
4. Select the forecasting model or models.
5. Gather the data needed to make the
forecast.
6. Validate the forecasting model.
7. Make the forecast.
8. Implement the results.
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5-5
Introduction
 These steps are a systematic way of initiating,
designing, and implementing a forecasting
system.
 When used regularly over time, data is
collected routinely and calculations performed
automatically.
 There is seldom one superior forecasting
system.
 Different organizations may use different techniques.
 Whatever tool works best for a firm is the one that
should be used.
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5-6
Forecasting Models
Forecasting
Techniques
Qualitative
Models
Time-Series
Methods
Causal
Methods
Delphi
Methods
Moving
Average
Regression
Analysis
Jury of Executive
Opinion
Exponential
Smoothing
Multiple
Regression
Sales Force
Composite
Trend
Projections
Figure 5.1
Consumer
Market Survey
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Decomposition
5-7
Qualitative Models
 Qualitative models incorporate judgmental
or subjective factors.
 These are useful when subjective factors
are thought to be important or when
accurate quantitative data is difficult to
obtain.
 Common qualitative techniques are:
 Delphi method.
 Jury of executive opinion.
 Sales force composite.
 Consumer market surveys.
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5-8
Qualitative Models
 Delphi Method – This is an iterative group
process where (possibly geographically
dispersed) respondents provide input to decision
makers.
 Jury of Executive Opinion – This method collects
opinions of a small group of high-level managers,
possibly using statistical models for analysis.
 Sales Force Composite – This allows individual
salespersons estimate the sales in their region
and the data is compiled at a district or national
level.
 Consumer Market Survey – Input is solicited from
customers or potential customers regarding their
purchasing plans.
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5-9
Time-Series Models
 Time-series models attempt to predict
the future based on the past.
 Common time-series models are:
 Moving average.
 Exponential smoothing.
 Trend projections.
 Decomposition.
 Regression analysis is used in trend
projections and one type of
decomposition model.
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5-10
Causal Models
 Causal models use variables or factors
that might influence the quantity being
forecasted.
 The objective is to build a model with
the best statistical relationship between
the variable being forecast and the
independent variables.
 Regression analysis is the most
common technique used in causal
modeling.
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
5-11
Scatter Diagrams
Wacker Distributors wants to forecast sales for
three different products (annual sales in the table,
in units):
YEAR
Table 5.1
1
2
3
4
5
6
7
8
9
10
TELEVISION
SETS
250
250
250
250
250
250
250
250
250
250
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
RADIOS
300
310
320
330
340
350
360
370
380
390
COMPACT DISC
PLAYERS
110
100
120
140
170
150
160
190
200
190
5-12
Scatter Diagram for TVs
Annual Sales of Televisions
(a)
 Sales appear to be
330 –
250 –          
200 –
150 –
100 –
constant over time
Sales = 250
 A good estimate of
sales in year 11 is
250 televisions
50 –
|
|
|
|
|
|
|
|
|
|
0 1 2 3 4 5 6 7 8 9 10
Time (Years)
Figure 5.2a
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5-13
Scatter Diagram for Radios
(b)
Annual Sales of Radios
420 –
 Sales appear to be
400 –
380 –
360 –
340 –
320 –
300 –  








280 –
|
|
|
|
|
|
|
|
|
|
0 1 2 3 4 5 6 7 8 9 10
increasing at a
constant rate of 10
radios per year
Sales = 290 + 10(Year)
 A reasonable
estimate of sales in
year 11 is 400
radios.
Time (Years)
Figure 5.2b
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5-14
Scatter Diagram for CD
Players
(c)
Annual Sales of CD Players
 This trend line may
200 –

180 –

160 –
140 –
120 –
100 –



|
 



|
|
|
|
|
|
|
|
|
0 1 2 3 4 5 6 7 8 9 10
not be perfectly
accurate because
of variation from
year to year
 Sales appear to be
increasing
 A forecast would
probably be a
larger figure each
year
Time (Years)
Figure 5.2c
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5-15
Measures of Forecast Accuracy
 We compare forecasted values with actual values
to see how well one model works or to compare
models.
Forecast error = Actual value – Forecast value
 One measure of accuracy is the mean absolute
deviation (MAD):
forecast error

MAD 
n
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5-16
Measures of Forecast Accuracy
Using a naïve forecasting model we can compute the MAD:
Table 5.2
YEAR
ACTUAL
SALES OF
CD
PLAYERS
FORECAST
SALES
ABSOLUTE VALUE OF
ERRORS (DEVIATION),
(ACTUAL – FORECAST)
1
110
—
—
2
100
110
|100 – 110| = 10
3
120
100
|120 – 110| = 20
4
140
120
|140 – 120| = 20
5
170
140
|170 – 140| = 30
6
150
170
|150 – 170| = 20
7
160
150
|160 – 150| = 10
8
190
160
|190 – 160| = 30
9
200
190
|200 – 190| = 10
10
190
200
|190 – 200| = 10
11
—
190
—
Sum of |errors| = 160
MAD = 160/9 = 17.8
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5-17
Measures of Forecast Accuracy
Using a naïve forecasting model we can compute the MAD:
YEAR
ACTUAL
SALES OF CD
PLAYERS
FORECAST
SALES
ABSOLUTE VALUE OF
ERRORS (DEVIATION),
(ACTUAL – FORECAST)
1
110
—
—
2
100
110
|100 – 110| = 10
3
120
100
|120 – 110| = 20
4
140
120
|140 – 120| = 20
5
170
140
|170 – 140| = 30
6
150
7
160
150
|160 – 150| = 10
8
190
160
|190 – 160| = 30
9
200
190
|200 – 190| = 10
10
190
200
|190 – 200| = 10
11
—
190
—
forecast error 160

MAD 

 17.8
n
170
9
|150 – 170| = 20
Sum of |errors| = 160
MAD = 160/9 = 17.8
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Table 5.2
5-18
Measures of Forecast Accuracy
 There are other popular measures of forecast
accuracy.
 The mean squared error:
2
(
error)

MSE 
n
 The mean absolute percent error:
error
 actual
MAPE 
100%
n
 And bias is the average error.
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5-19
Time-Series Forecasting Models
 A time series is a sequence of evenly
spaced events.
 Time-series forecasts predict the future
based solely on the past values of the
variable, and other variables are
ignored.
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5-20
Components of a Time-Series
A time series typically has four components:
1. Trend (T) is the gradual upward or downward
movement of the data over time.
2. Seasonality (S) is a pattern of demand
fluctuations above or below the trend line that
repeats at regular intervals.
3. Cycles (C) are patterns in annual data that
occur every several years.
4. Random variations (R) are “blips” in the data
caused by chance or unusual situations, and
follow no discernible pattern.
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5-21
Decomposition of a Time-Series
Demand for Product or Service
Product Demand Charted over 4 Years, with Trend
and Seasonality Indicated
Figure 5.3
Trend
Component
Seasonal Peaks
Actual
Demand
Line
Average Demand
over 4 Years
|
|
|
|
Year
1
Year
2
Year
3
Year
4
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Time
5-22
Decomposition of a Time-Series
 There are two general forms of time-series
models:
 The multiplicative model:
Demand = T x S x C x R
 The additive model:
Demand = T + S + C + R
 Models may be combinations of these two
forms.
 Forecasters often assume errors are
normally distributed with a mean of zero.
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5-23
Moving Averages
 Moving averages can be used when
demand is relatively steady over time.
 The next forecast is the average of the
most recent n data values from the time
series.
 This methods tends to smooth out shortterm irregularities in the data series.
Moving average forecast 
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Sum of demands in previous n periods
n
5-24
Moving Averages
 Mathematically:
Ft 1 
Yt  Yt 1  …  Yt  n1
n
Where:
Ft 1 = forecast for time period t + 1
Yt = actual value in time period t
n = number of periods to average
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5-25
Wallace Garden Supply
 Wallace Garden Supply wants to
forecast demand for its Storage Shed.
 They have collected data for the past
year.
 They are using a three-month moving
average to forecast demand (n = 3).
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5-26
Wallace Garden Supply
MONTH
ACTUAL SHED SALES
THREE-MONTH MOVING AVERAGE
January
10
February
12
March
13
April
16
(10 + 12 + 13)/3 = 11.67
May
19
(12 + 13 + 16)/3 = 13.67
June
23
(13 + 16 + 19)/3 = 16.00
July
26
(16 + 19 + 23)/3 = 19.33
August
30
(19 + 23 + 26)/3 = 22.67
September
28
(23 + 26 + 30)/3 = 26.33
October
18
(26 + 30 + 28)/3 = 28.00
November
16
(30 + 28 + 18)/3 = 25.33
December
14
(28 + 18 + 16)/3 = 20.67
January
—
(18 + 16 + 14)/3 = 16.00
Table 5.3
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5-27
Weighted Moving Averages
 Weighted moving averages use weights to put
more emphasis on previous periods.
 This is often used when a trend or other pattern is
emerging.
Ft 1
( Weight in period i )( Actual value in period)


 ( Weights)
 Mathematically:
w1Yt  w2Yt 1  …  wnYt  n1
Ft 1 
w1  w2  …  wn
where
wi = weight for the ith observation
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5-28
Wallace Garden Supply
 Wallace Garden Supply decides to try a
weighted moving average model to forecast
demand for its Storage Shed.
 They decide on the following weighting
scheme:
WEIGHTS APPLIED
PERIOD
3
2
1
Last month
Two months ago
Three months ago
3 x Sales last month + 2 x Sales two months ago + 1 X Sales three months ago
6
Sum of the weights
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5-29
Wallace Garden Supply
THREE-MONTH WEIGHTED
MOVING AVERAGE
MONTH
ACTUAL SHED SALES
January
10
February
12
March
13
April
16
[(3 X 13) + (2 X 12) + (10)]/6 = 12.17
May
19
[(3 X 16) + (2 X 13) + (12)]/6 = 14.33
June
23
[(3 X 19) + (2 X 16) + (13)]/6 = 17.00
July
26
[(3 X 23) + (2 X 19) + (16)]/6 = 20.50
August
30
[(3 X 26) + (2 X 23) + (19)]/6 = 23.83
September
28
[(3 X 30) + (2 X 26) + (23)]/6 = 27.50
October
18
[(3 X 28) + (2 X 30) + (26)]/6 = 28.33
November
16
[(3 X 18) + (2 X 28) + (30)]/6 = 23.33
December
14
[(3 X 16) + (2 X 18) + (28)]/6 = 18.67
January
—
[(3 X 14) + (2 X 16) + (18)]/6 = 15.33
Table 5.4
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5-30
Wallace Garden Supply
Selecting the Forecasting Module in Excel QM
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Program 5.1A
5-31
Wallace Garden Supply
Initialization Screen for Weighted Moving Average
Program 5.1B
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5-32
Wallace Garden Supply
Weighted Moving Average in Excel QM for Wallace
Garden Supply
Program 5.1C
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5-33
Exponential Smoothing
 Exponential smoothing is a type of moving
average that is easy to use and requires little
record keeping of data.
New forecast = Last period’s forecast
+ (Last period’s actual demand
– Last period’s forecast)
Here  is a weight (or smoothing constant) in
which 0≤≤1.
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5-34
Exponential Smoothing
Mathematically:
Ft 1  Ft   (Yt  Ft )
Where:
Ft+1 = new forecast (for time period t + 1)
Ft = pervious forecast (for time period t)
 = smoothing constant (0 ≤  ≤ 1)
Yt = pervious period’s actual demand
The idea is simple – the new estimate is the old
estimate plus some fraction of the error in the
last period.
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5-35
Exponential Smoothing Example
 In January, February’s demand for a certain
car model was predicted to be 142.
 Actual February demand was 153 autos
 Using a smoothing constant of  = 0.20, what
is the forecast for March?
New forecast (for March demand) = 142 + 0.2(153 – 142)
= 144.2 or 144 autos
 If actual demand in March was 136 autos, the
April forecast would be:
New forecast (for April demand) = 144.2 + 0.2(136 – 144.2)
= 142.6 or 143 autos
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5-36
Selecting the Smoothing Constant
 Selecting the appropriate value for  is key
to obtaining a good forecast.
 The objective is always to generate an
accurate forecast.
 The general approach is to develop trial
forecasts with different values of  and
select the  that results in the lowest MAD.
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5-37
Exponential Smoothing
Port of Baltimore Exponential Smoothing Forecast
for =0.1 and =0.5.
QUARTER
ACTUAL
TONNAGE
UNLOADED
1
180
175
175
2
168
175.5 = 175.00 + 0.10(180 – 175)
177.5
3
159
174.75 = 175.50 + 0.10(168 – 175.50)
172.75
4
175
173.18 = 174.75 + 0.10(159 – 174.75)
165.88
5
190
173.36 = 173.18 + 0.10(175 – 173.18)
170.44
6
205
175.02 = 173.36 + 0.10(190 – 173.36)
180.22
7
180
178.02 = 175.02 + 0.10(205 – 175.02)
192.61
8
182
178.22 = 178.02 + 0.10(180 – 178.02)
186.30
9
?
178.60 = 178.22 + 0.10(182 – 178.22)
184.15
FORECAST
USING  =0.10
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Table 5.5
FORECAST
USING  =0.50
5-38
Exponential Smoothing
Absolute Deviations and MADs for the Port of
Baltimore Example
QUARTER
ACTUAL
TONNAGE
UNLOADED
FORECAST
WITH  = 0.10
ABSOLUTE
DEVIATIONS
FOR  = 0.10
1
180
175
5…..
175
5….
2
168
175.5
7.5..
177.5
9.5..
3
159
174.75
15.75
172.75
13.75
4
175
173.18
1.82
165.88
9.12
5
190
173.36
16.64
170.44
19.56
6
205
175.02
29.98
180.22
24.78
7
180
178.02
1.98
192.61
12.61
8
182
178.22
3.78
186.30
4.3..
Sum of absolute deviations
MAD =
Table 5.6
FORECAST
WITH  = 0.50
ABSOLUTE
DEVIATIONS
FOR  = 0.50
82.45
Σ|deviations|
n
=
Best choice
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10.31
98.63
MAD =
12.33
5-39
Port of Baltimore Exponential
Smoothing Example in Excel QM
Program 5.2
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5-40
Exponential Smoothing with
Trend Adjustment
 Like all averaging techniques, exponential
smoothing does not respond to trends.
 A more complex model can be used that
adjusts for trends.
 The basic approach is to develop an
exponential smoothing forecast, and then
adjust it for the trend.
Forecast including trend (FITt+1) = Smoothed forecast (Ft+1)
+ Smoothed Trend (Tt+1)
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5-41
Exponential Smoothing with
Trend Adjustment
 The equation for the trend correction uses a new
smoothing constant  .
 Tt must be given or estimated. Tt+1 is computed by:
Tt 1  (1   )Tt   ( Ft 1  FITt )
where
Tt = smoothed trend for time period t
Ft = smoothed forecast for time period t
FITt = forecast including trend for time period t
α =smoothing constant for forecasts
 = smoothing constant for trend
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5-42
Selecting a Smoothing Constant
 As with exponential smoothing, a high value of 
makes the forecast more responsive to changes
in trend.
 A low value of  gives less weight to the recent
trend and tends to smooth out the trend.
 Values are generally selected using a trial-anderror approach based on the value of the MAD for
different values of .
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5-43
Midwestern Manufacturing
 Midwest Manufacturing has a demand for
electrical generators from 2004 – 2010 as given
in the table below.
 To forecast demand, Midwest assumes:
 F1 is perfect.
 T1 = 0.
YEAR
ELECTRICAL GENERATORS SOLD
 α = 0.3
2004
74
 β = 0.4.
2005
79
Table 5.7
2006
2007
2008
2009
2010
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
80
90
105
142
122
5-44
Midwestern Manufacturing
 According to the assumptions,
FIT1 = F1 + T1 = 74 + 0 = 74.
 Step 1: Compute Ft+1 by:
= Ft + α(Yt – FITt)
= 74 + 0.3(74-74) = 74
 Step 2: Update the trend using:
Tt+1 = Tt + β(Ft+1 – FITt)
T2
= T1 + .4(F2 – FIT1)
= 0 + .4(74 – 74) = 0
FITt+1
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5-45
Midwestern Manufacturing
 Step 3: Calculate the trend-adjusted
exponential smoothing forecast (Ft+1)
using the following:
FIT2
= F2 + T2
= 74 + 0 = 74
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5-46
Midwestern Manufacturing
 For 2006 (period 3) we have:
 Step 1: F3
 Step 2: T3
 Step 3: FIT3
= FIT2 + 0.3(Y2 – FIT2)
= 74 + .3(79 – 74)
= 75.5
= T2 + 0.4(F3 – FIT2)
= 0 + 0.4(75.5 – 74)
= 0.6
= F3 + T3
= 75.5 + 0.6
= 76.1
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5-47
Midwestern Manufacturing Exponential
Smoothing with Trend Forecasts
Time
(t)
Demand
(Yt)
FITt+1 = Ft + 0.3(Yt– FITt)
Tt+1 = Tt + 0.4(Ft+1 – FITt)
FITt+1 = Ft+1 + Tt+1
1
74
74
0
74
2
79
74=74+0.3(74-74)
0 = 0+0.4(74-74)
74 = 74+0
3
80
75.5=74+0.3(79-74)
0.6 = 0+0.4(75.5-74)
76.1 = 75.5+0.6
4
90
77.270=76.1+0.3(80-76.1)
1.068 = 0.6+0.4(77.27-76.1)
78.338 =
77.270+1.068
5
105
81.837=78.338+0.3(9078.338)
2.468 = 1.068+0.4(81.83778.338)
84.305 =
81.837+2.468
6
142
90.514=84.305+0.3(10584.305)
4.952 = 2.468+0.4(90.51484.305)
95.466 =
90.514+4.952
7
122
109.426=95.466+0.3(14295.466)
10.536 =
4.952+0.4(109.426-95.466)
119.962 =
109.426+10.536
120.573=119.962+0.3(122119.962)
10.780 =
10.536+0.4(120.573119.962)
131.353 =
120.573+10.780
8
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
Table 5.8
5-48
Midwestern Manufacturing
Midwestern Manufacturing Trend-Adjusted
Exponential Smoothing in Excel QM
Program 5.3
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5-49
Trend Projections
 Trend projection fits a trend line to a
series of historical data points.
 The line is projected into the future for
medium- to long-range forecasts.
 Several trend equations can be
developed based on exponential or
quadratic models.
 The simplest is a linear model developed
using regression analysis.
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5-50
Trend Projection
The mathematical form is
Yˆ  b0  b1 X
Where
Ŷ = predicted value
b0 = intercept
b1 = slope of the line
X = time period (i.e., X = 1, 2, 3, …, n)
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5-51
Midwestern Manufacturing
Excel Input Screen for Midwestern Manufacturing
Trend Line
Program 5.4A
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5-52
Midwestern Manufacturing
Excel Output for Midwestern Manufacturing Trend Line
Program 5.4B
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5-53
Midwestern Manufacturing
Company Example
 The forecast equation is
Yˆ  56.71  10.54 X
 To project demand for 2011, we use the coding
system to define X = 8
(sales in 2011) = 56.71 + 10.54(8)
= 141.03, or 141 generators
 Likewise for X = 9
(sales in 2012) = 56.71 + 10.54(9)
= 151.57, or 152 generators
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5-54
Midwestern Manufacturing
Electrical Generators and the Computed Trend Line
Figure 5.4
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5-55
Midwestern Manufacturing
Excel QM Trend Projection Model
Program 5.5
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5-56
Seasonal Variations
 Recurring variations over time may
indicate the need for seasonal
adjustments in the trend line.
 A seasonal index indicates how a
particular season compares with an
average season.
 When no trend is present, the seasonal
index can be found by dividing the
average value for a particular season by
the average of all the data.
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5-57
Eichler Supplies
 Eichler Supplies sells telephone
answering machines.
 Sales data for the past two years has
been collected for one particular model.
 The firm wants to create a forecast that
includes seasonality.
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5-58
Eichler Supplies Answering Machine
Sales and Seasonal Indices
SALES DEMAND
AVERAGE TWOYEAR DEMAND
MONTHLY
DEMAND
AVERAGE
SEASONAL
INDEX
MONTH
YEAR 1
YEAR 2
January
80
100
90
94
0.957
February
85
75
80
94
0.851
March
80
90
85
94
0.904
April
110
90
100
94
1.064
May
115
131
123
94
1.309
June
120
110
115
94
1.223
July
100
110
105
94
1.117
August
110
90
100
94
1.064
September
85
95
90
94
0.957
October
75
85
80
94
0.851
November
85
75
80
94
0.851
December
80
80
80
94
0.851
Total average demand = 1,128
Average monthly demand =
Table 5.9
1,128
= 94
12 months
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Average two-year demand
Seasonal index = Average monthly demand
5-59
Seasonal Variations
 The calculations for the seasonal indices are
July
1,200
 1.117  112
12
Feb.
1,200
 0.851  85
12
Aug.
1,200
 1.064  106
12
Mar.
1,200
 0.904  90
12
Sept.
1,200
 0.957  96
12
Apr.
1,200
 1.064  106
12
Oct.
1,200
 0.851  85
12
May
1,200
 1.309  131
12
Nov.
1,200
 0.851  85
12
June
1,200
 1.223  122
12
Dec.
1,200
 0.851  85
12
Jan.
1,200
 0.957  96
12
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5-60
Seasonal Variations with Trend
 When both trend and seasonal components are
present, the forecasting task is more complex.
 Seasonal indices should be computed using a
centered moving average (CMA) approach.
 There are four steps in computing CMAs:
1. Compute the CMA for each observation
(where possible).
2. Compute the seasonal ratio =
Observation/CMA for that observation.
3. Average seasonal ratios to get seasonal
indices.
4. If seasonal indices do not add to the number
of seasons, multiply each index by (Number
of seasons)/(Sum of indices).
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5-61
Turner Industries
 The following table shows Turner Industries’
quarterly sales figures for the past three years, in
millions of dollars:
QUARTER
YEAR 1
YEAR 2
YEAR 3
AVERAGE
1
108
116
123
115.67
2
125
134
142
133.67
3
150
159
168
159.00
4
141
152
165
152.67
Average
131.00
140.25
149.50
140.25
Table 5.10
Definite trend
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Seasonal
pattern
5-62
Turner Industries
 To calculate the CMA for quarter 3 of year 1 we
compare the actual sales with an average quarter
centered on that time period.
 We will use 1.5 quarters before quarter 3 and 1.5
quarters after quarter 3 – that is we take quarters
2, 3, and 4 and one half of quarters 1, year 1 and
quarter 1, year 2.
0.5(108) + 125 + 150 + 141 + 0.5(116)
CMA(q3, y1) =
= 132.00
4
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5-63
Turner Industries
Compare the actual sales in quarter 3 to the CMA to
find the seasonal ratio:
Seasonal ratio 
Sales in quarter 3 150

 1.136
CMA
132
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5-64
Turner Industries
YEAR
1
2
3
QUARTER
1
2
3
4
1
2
3
4
1
2
3
4
SALES
108
125
150
141
116
134
159
152
123
142
168
165
CMA
SEASONAL RATIO
132.000
134.125
136.375
138.875
141.125
143.000
145.125
147.875
1.136
1.051
0.851
0.965
1.127
1.063
0.848
0.960
Table 5.11
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5-65
Turner Industries
There are two seasonal ratios for each quarter so
these are averaged to get the seasonal index:
Index for quarter 1 = I1 = (0.851 + 0.848)/2 = 0.85
Index for quarter 2 = I2 = (0.965 + 0.960)/2 = 0.96
Index for quarter 3 = I3 = (1.136 + 1.127)/2 = 1.13
Index for quarter 4 = I4 = (1.051 + 1.063)/2 = 1.06
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5-66
Turner Industries
Scatterplot of Turner Industries Sales Data and
Centered Moving Average
CMA
200 –
Sales
150 –
 

100 – 
50 –
0–




 


Original Sales Figures
|
|
|
|
1
2
3
4
|
|
|
|
|
|
|
|
5
6
7
Time Period
8
9
10
11
12
Figure 5.5
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5-67
The Decomposition Method of Forecasting
with Trend and Seasonal Components
 Decomposition is the process of isolating linear
trend and seasonal factors to develop more
accurate forecasts.
 There are five steps to decomposition:
1. Compute seasonal indices using CMAs.
2. Deseasonalize the data by dividing each
number by its seasonal index.
3. Find the equation of a trend line using the
deseasonalized data.
4. Forecast for future periods using the trend
line.
5. Multiply the trend line forecast by the
appropriate seasonal index.
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5-68
Deseasonalized Data for Turner
Industries
 Find a trend line using the deseasonalized data:
b1 = 2.34
b0 = 124.78
 Develop a forecast using this trend and multiply
the forecast by the appropriate seasonal index.
Ŷ = 124.78 + 2.34X
= 124.78 + 2.34(13)
= 155.2 (forecast before adjustment for
seasonality)
Ŷ x I1 = 155.2 x 0.85 = 131.92
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5-69
Deseasonalized Data for Turner
Industries
SALES
($1,000,000s)
108
125
150
141
116
134
159
152
123
142
168
165
SEASONAL
INDEX
0.85
0.96
1.13
1.06
0.85
0.96
1.13
1.06
0.85
0.96
1.13
1.06
DESEASONALIZED
SALES ($1,000,000s)
127.059
130.208
132.743
133.019
136.471
139.583
140.708
143.396
144.706
147.917
148.673
155.660
Table 5.12
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5-70
San Diego Hospital
A San Diego hospital used 66 months of adult
inpatient days to develop the following seasonal
indices.
MONTH
SEASONALITY INDEX
MONTH
SEASONALITY INDEX
January
1.0436
July
1.0302
February
0.9669
August
1.0405
March
1.0203
September
0.9653
April
1.0087
October
1.0048
May
0.9935
November
0.9598
June
0.9906
December
0.9805
Table 5.13
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5-71
San Diego Hospital
Using this data they developed the following
equation:
Ŷ = 8,091 + 21.5X
where
Ŷ = forecast patient days
X = time in months
Based on this model, the forecast for patient days
for the next period (67) is:
Patient days = 8,091 + (21.5)(67) = 9,532 (trend only)
Patient days = (9,532)(1.0436)
= 9,948 (trend and seasonal)
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5-72
San Diego Hospital
Initialization Screen for the Decomposition method
in Excel QM
Program 5.6A
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5-73
San Diego Hospital
Turner Industries Forecast Using the Decomposition
Method in Excel QM
Program 5.6B
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5-74
Using Regression with Trend
and Seasonal Components
 Multiple regression can be used to forecast both
trend and seasonal components in a time series.
 One independent variable is time.
 Dummy independent variables are used to represent the
seasons.
 The model is an additive decomposition model:
Yˆ  a  b1 X 1  b2 X 2  b3 X 3  b4 X 4
where
X1 = time period
X2 = 1 if quarter 2, 0 otherwise
X3 = 1 if quarter 3, 0 otherwise
X4 = 1 if quarter 4, 0 otherwise
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5-75
Regression with Trend and
Seasonal Components
Excel Input for the Turner Industries Example Using
Multiple Regression
Program 5.7A
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5-76
Using Regression with Trend
and Seasonal Components
Excel Output for
the Turner
Industries
Example Using
Multiple
Regression
Program 5.7B
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5-77
Using Regression with Trend
and Seasonal Components
 The resulting regression equation is:
Yˆ  104.1  2.3 X 1  15.7 X 2  38.7 X 3  30.1X 4
 Using the model to forecast sales for the first two
quarters of next year:
Ŷ  104.1 2.3(13)  15.7(0)  38.7(0)  30.1(0)  134
Ŷ  104.1 2.3(14)  15.7(1)  38.7(0)  30.1(0)  152
 These are different from the results obtained
using the multiplicative decomposition method.
 Use MAD or MSE to determine the best model.
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5-78
Monitoring and Controlling Forecasts
 Tracking signals can be used to monitor
the performance of a forecast.
 A tracking signal is computed as the
running sum of the forecast errors (RSFE),
and is computed using the following
equation:
Tracking signal 
RSFE
MAD
where
forecast error

MAD 
n
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5-79
Monitoring and Controlling Forecasts
Plot of Tracking Signals
Signal Tripped
Upper Control Limit
+
Tracking Signal
Acceptable
Range
0 MADs
–
Lower Control Limit
Figure 5.6
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
Time
5-80
Monitoring and Controlling Forecasts
 Positive tracking signals indicate demand is
greater than forecast.
 Negative tracking signals indicate demand is less
than forecast.
 Some variation is expected, but a good forecast
will have about as much positive error as
negative error.
 Problems are indicated when the signal trips
either the upper or lower predetermined limits.
 This indicates there has been an unacceptable
amount of variation.
 Limits should be reasonable and may vary from
item to item.
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5-81
Kimball’s Bakery
Quarterly sales of croissants (in thousands):
TIME
PERIOD
FORECAST
DEMAND
ACTUAL
DEMAND
ERROR
RSFE
|FORECAST |
| ERROR |
CUMULATIVE
ERROR
MAD
TRACKING
SIGNAL
1
100
90
–10
–10
10
10
10.0
–1
2
100
95
–5
–15
5
15
7.5
–2
3
100
115
+15
0
15
30
10.0
0
4
110
100
–10
–10
10
40
10.0
–1
5
110
125
+15
+5
15
55
11.0
+0.5
6
110
140
+30
+35
35
85
14.2
+2.5
forecast error 85

MAD 

 14.2
n
6
RSFE
35
Tracking signal 

 2.5MADs
MAD 14.2
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5-82
Adaptive Smoothing
 Adaptive smoothing is the computer
monitoring of tracking signals and selfadjustment if a limit is tripped.
 In exponential smoothing, the values of 
and  are adjusted when the computer
detects an excessive amount of variation.
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5-83
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.
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5-84
Chapter 4
Regression 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. Identify variables and use them in a regression
model.
2. Develop simple linear regression equations.
from sample data and interpret the slope and
intercept.
3. Compute the coefficient of determination and
the coefficient of correlation and interpret their
meanings.
4. Interpret the F-test in a linear regression model.
5. List the assumptions used in regression and
use residual plots to identify problems.
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4-2
Learning Objectives
After completing this chapter, students will be able to:
6. Develop a multiple regression model and use it
for prediction purposes.
7. Use dummy variables to model categorical
data.
8. Determine which variables should be included
in a multiple regression model.
9. Transform a nonlinear function into a linear
one for use in regression.
10. Understand and avoid common mistakes made
in the use of regression analysis.
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4-3
Chapter Outline
4.1 Introduction
4.2 Scatter Diagrams
4.3 Simple Linear Regression
4.4 Measuring the Fit of the Regression
Model
4.5 Using Computer Software for Regression
4.6 Assumptions of the Regression Model
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4-4
Chapter Outline
4.7
4.8
4.9
4.10
4.11
4.12
Testing the Model for Significance
Multiple Regression Analysis
Binary or Dummy Variables
Model Building
Nonlinear Regression
Cautions and Pitfalls in Regression
Analysis
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4-5
Introduction
 Regression analysis is a very valuable
tool for a manager.
 Regression can be used to:
 Understand the relationship between
variables.
 Predict the value of one variable based on
another variable.
 Simple linear regression models have
only two variables.
 Multiple regression models have more
variables.
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4-6
Introduction
 The variable to be predicted is called
the dependent variable.
 This is sometimes called the response
variable.
 The value of this variable depends on
the value of the independent variable.
 This is sometimes called the explanatory
or predictor variable.
Dependent
variable
=
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Independent
variable
+
Independent
variable
4-7
Scatter Diagram



A scatter diagram or scatter plot is
often used to investigate the
relationship between variables.
The independent variable is normally
plotted on the X axis.
The dependent variable is normally
plotted on the Y axis.
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4-8
Triple A Construction
 Triple A Construction renovates old homes.
 Managers have found that the dollar volume of
renovation work is dependent on the area
payroll.
TRIPLE A’S SALES
($100,000s)
6
8
9
5
4.5
9.5
LOCAL PAYROLL
($100,000,000s)
3
4
6
4
2
5
Table 4.1
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4-9
Triple A Construction
Scatter Diagram of Triple A Construction Company Data
Figure 4.1
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4-10
Simple Linear Regression
 Regression models are used to test if there is a
relationship between variables.
 There is some random error that cannot be
predicted.
Y   0  1X  e
where
Y = dependent variable (response)
X = independent variable (predictor or explanatory)
 0 = intercept (value of Y when X = 0)
 1 = slope of the regression line
e = random error
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4-11
Simple Linear Regression
 True values for the slope and intercept are not
known so they are estimated using sample data.
Yˆ  b0  b1 X
where
Y^ = predicted value of Y
b0 = estimate of β0, based on sample results
b1 = estimate of β1, based on sample results
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4-12
Triple A Construction
Triple A Construction is trying to predict sales
based on area payroll.
Y = Sales
X = Area payroll
The line chosen in Figure 4.1 is the one that
minimizes the errors.
Error = (Actual value) – (Predicted value)
e  Y  Yˆ
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4-13
Triple A Construction
For the simple linear regression model, the values
of the intercept and slope can be calculated using
the formulas below.
Yˆ  b0  b1 X
X

X
 average (mean) of X values
n
Y

Y
 average (mean) of Y values
n
( X  X )(Y  Y )

b 
(X  X )
1
2
b0  Y  b1 X
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4-14
Triple A Construction
Regression calculations for Triple A Construction
Y
X
(X – X)2
(X – X)(Y – Y)
6
8
9
5
4.5
3
4
6
4
2
(3 – 4)2 = 1
(4 – 4)2 = 0
(6 – 4)2 = 4
(4 – 4)2 = 0
(2 – 4)2 = 4
(3 – 4)(6 – 7) = 1
(4 – 4)(8 – 7) = 0
(6 – 4)(9 – 7) = 4
(4 – 4)(5 – 7) = 0
(2 – 4)(4.5 – 7) = 5
9.5
5
(5 – 4)2 = 1
(5 – 4)(9.5 – 7) = 2.5
Σ(X – X)2 = 10
Σ(X – X)(Y – Y) = 12.5
ΣY = 42
Y = 42/6 = 7
ΣX = 24
X = 24/6 = 4
Table 4.2
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4-15
Triple A Construction
Regression calculations
X 24

X

4
6
6
Y 42

Y

7
6
6
( X  X )(Y  Y ) 12.5

b 

 1.25
10
(X  X )
1
2
b0  Y  b1 X  7  (1.25)(4)  2
Therefore Yˆ  2  1.25 X
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
4-16
Triple A Construction
Regression calculations
X 24

X

4
6
6
sales = 2 + 1.25(payroll)
If the payroll next
Y 42

Y

 7 year is $600 million
6
6
ˆ
 1..525(6)  9.5 or $ 950,000
( X  X )(Y YY
) 2 12

b 

 1.25
10
(X  X )
1
2
b0  Y  b1 X  7  (1.25)(4)  2
Therefore Yˆ  2  1.25 X
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4-17
Measuring the Fit
of the Regression Model
 Regression models can be developed
for any variables X and Y.
 How do we know the model is actually
helpful in predicting Y based on X?
 We could just take the average error, but
the positive and negative errors would
cancel each other out.
 Three measures of variability are:
 SST – Total variability about the mean.
 SSE – Variability about the regression line.
 SSR – Total variability that is explained by
the model.
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4-18
Measuring the Fit
of the Regression Model
 Sum of the squares total:
SST   (Y  Y )2
 Sum of the squared error:
SSE   e 2   (Y  Yˆ )2
 Sum of squares due to regression:
SSR   (Yˆ  Y )2
 An important relationship:
SST  SSR  SSE
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
4-19
Measuring the Fit
of the Regression Model
Sum of Squares for Triple A Construction
Y
X
(Y – Y)2
Y
^
(Y – Y)2
(Y – Y)2
6
3
(6 – 7)2 = 1
2 + 1.25(3) = 5.75
0.0625
1.563
8
4
(8 – 7)2 = 1
2 + 1.25(4) = 7.00
1
0
9
6
(9 – 7)2 = 4
2 + 1.25(6) = 9.50
0.25
6.25
5
4
(5 – 7)2 = 4
2 + 1.25(4) = 7.00
4
0
4.5
2
(4.5 – 7)2 = 6.25
2 + 1.25(2) = 4.50
0
6.25
9.5
5
(9.5 – 7)2 = 6.25
2 + 1.25(5) = 8.25
1.5625
1.563
∑(Y – Y)2 = 22.5
Y=7
^
∑(Y – Y)
^2
= 6.875
∑(Y
^ – Y)2 = 15.625
SSE
= 6.875
SSR = 15.625
SST = 22.5
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^
Table 4.3
4-20
Measuring the Fit
of the Regression Model
 Sum of the squares total
For Triple
A Construction
2
SST   (Y  Y )
SST = 22.5
 Sum of the squared error SSE = 6.875
SSRˆ=215.625
2
SSE   e   (Y  Y )
 Sum of squares due to regression
SSR   (Yˆ  Y )2
 An important relationship
SST  SSR  SSE
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4-21
Measuring the Fit
of the Regression Model
Deviations from the Regression Line and from the Mean
Figure 4.2
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4-22
Coefficient of Determination
 The proportion of the variability in Y explained by
the regression equation is called the coefficient
of determination.
 The coefficient of determination is r2.
SSR
SSE
r 
 1
SST
SST
2
 For Triple A Construction:
15.625
r 
 0.6944
22.5
2
 About 69% of the variability in Y is explained by
the equation based on payroll (X).
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4-23
Correlation Coefficient
 The correlation coefficient is an expression of the
strength of the linear relationship.
 It will always be between +1 and –1.
 The correlation coefficient is r.
r  r2
 For Triple A Construction:
r  0.6944  0.8333
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4-24
Four Values of the Correlation
Coefficient
Y
Y
*
*
*
* **
** *
* *
* *
*
*
*
(a) Perfect Positive X
Correlation:
r = +1
Y
X
Y
*
*
* *
* * * *
*
* *** *
Figure 4.3
(b) Positive
Correlation:
0
MSR/MSE)
4-44
ANOVA for Triple A Construction
Program 4.1C
(partial)
P(F > 9.0909) = 0.0394
Because this probability is less than 0.05, we reject
the null hypothesis of no linear relationship and
conclude there is a linear relationship between X
and Y.
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4-45
Multiple Regression Analysis
 Multiple regression models are
extensions to the simple linear model
and allow the creation of models with
more than one independent variable.
Y =  0 + 1X1 + 2X2 + … +  kXk + e
where
Y = dependent variable (response variable)
Xi = ith independent variable (predictor or explanatory
variable)
 0 = intercept (value of Y when all Xi = 0)
 i = coefficient of the ith independent variable
k = number of independent variables
e = random error
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4-46
Multiple Regression Analysis
To estimate these values, a sample is taken the
following equation developed
Yˆ  b0  b1 X 1  b2 X 2  …  bk X k
where
Ŷ = predicted value of Y
b0 = sample intercept (and is an estimate of  0)
bi = sample coefficient of the ith variable (and is
an estimate of  i)
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4-47
Jenny Wilson Realty
Jenny Wilson wants to develop a model to determine
the suggested listing price for houses based on the
size and age of the house.
Yˆ  b0  b1 X1  b2 X 2
where
Ŷ = predicted value of dependent variable (selling
price)
b0 = Y intercept
X1 and X2 = value of the two independent variables (square
footage and age) respectively
b1 and b2 = slopes for X1 and X2 respectively
She selects a sample of houses that have sold
recently and records the data shown in Table 4.5
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4-48
Jenny Wilson Real Estate Data
Table 4.5
SELLING
PRICE ($)
SQUARE
FOOTAGE
AGE
CONDITION
95,000
1,926
30
Good
119,000
2,069
40
Excellent
124,800
1,720
30
Excellent
135,000
1,396
15
Good
142,000
1,706
32
Mint
145,000
1,847
38
Mint
159,000
1,950
27
Mint
165,000
2,323
30
Excellent
182,000
2,285
26
Mint
183,000
3,752
35
Good
200,000
2,300
18
Good
211,000
2,525
17
Good
215,000
3,800
40
Excellent
219,000
1,740
12
Mint
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4-49
Jenny Wilson Realty
Input Screen for the Jenny Wilson Realty Multiple
Regression Example
Program 4.2A
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4-50
Jenny Wilson Realty
Output for the Jenny Wilson Realty Multiple
Regression Example
Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall
Program 4.2B
4-51
Evaluating Multiple Regression Models
 Evaluation is similar to simple linear
regression models.
 The p-value for the F-test and r2 are
interpreted the same.
 The hypothesis is different because there
is more than one independent variable.
 The F-test is investigating whether all
the coefficients are equal to 0 at…