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Chapter 1

To accompany Quantitative Analysis for Management, Eleventh Edition, by Render, Stair, and Hanna Power Point slides created by Brian Peterson

Introduction to Quantitative Analysis

Learning Objectives

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

After completing this chapter, students will be able to:

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

Introduction

n Mathematical tools have been used for thousands of years.

n Quantitative analysis can be applied to a wide variety of problems. n It’s not enough to just know the

mathematics of a technique.

n One must understand the specific applicability of the technique, its limitations, and its assumptions.

Examples of Quantitative Analyses

n In the mid 1990s, Taco Bell saved over $150 million using forecasting and scheduling quantitative analysis models.

n NBC television increased revenues by over $200 million between 1996 and 2000 by using quantitative analysis to develop better sales plans.

n Continental Airlines saved over $40 million in 2001 using quantitative analysis models to quickly recover from weather delays and other disruptions.

Meaningful Information

Quantitative Analysis

Quantitative analysis is a scientific approach to managerial decision making in which raw data are processed and manipulated to produce meaningful information.

What is Quantitative Analysis?

Raw Data

n Quantitative factors are data that can be accurately calculated. Examples include: n Different investment alternatives

n Interest rates

n Inventory levels

n Demand

n Labor cost

n Qualitative factors are more difficult to quantify but affect the decision process. Examples include: n The weather

n State and federal legislation

n Technological breakthroughs.

What is Quantitative Analysis?

Implementing the Results

Analyzing the Results

Testing the Solution

Developing a Solution

Acquiring Input Data

Developing a Model

The Quantitative Analysis Approach

Defining the Problem

Figure 1.1

Defining the Problem

Develop a clear and concise statement that gives direction and meaning to subsequent steps.

n This may be the most important and difficult step.

n It is essential to go beyond symptoms and identify true causes.

n It may be necessary to concentrate on only a few of the problems – selecting the right problems is very important

n Specific and measurable objectives may have to be developed.

Developing a Model

Quantitative analysis models are realistic, solvable, and understandable mathematical representations of a situation.

There are different types of models:

$ Advertising

$ S

a le

Schematic models

Scale models

Developing a Model

Models generally contain variables (controllable and uncontrollable) and parameters.

n Controllable variables are the decision variables and are generally unknown. n How many items should be ordered for inventory?

n Parameters are known quantities that are a part of the model. n What is the holding cost of the inventory?

Acquiring Input Data

Input data must be accurate – GIGO rule:

Data may come from a variety of sources such as company reports, company documents, interviews, on-site direct measurement, or statistical sampling.

Garbage In

Process

Garbage Out

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 n Solving equations. n Trial and error – trying various approaches

and picking the best result. n Complete enumeration – trying all possible

values. n Using an algorithm – a series of repeating

steps to reach a solution.

Testing the Solution

Both input data and the model should be tested for accuracy before analysis and implementation.

n New data can be collected to test the model.

n Results should be logical, consistent, and represent the real situation.

Analyzing the Results

Determine the implications of the solution: n Implementing results often requires change in

an organization.

n 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.

n Sensitive models should be very thoroughly tested.

Implementing the Results

Implementation incorporates the solution into the company.

n Implementation can be very difficult.

n People may be resistant to changes.

n 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.

Modeling in the Real World

Quantitative analysis models are used extensively by real organizations to solve real problems.

n In the real world, quantitative analysis models can be complex, expensive, and difficult to sell.

n Following the steps in the process is an important component of success.

How To Develop a Quantitative Analysis Model

A mathematical model of profit:

Profit = Revenue – Expenses

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 v = variable cost per unit f = fixed cost X = number of units sold

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 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 v = variable cost per unit f = fixed cost X = number of units sold

The parameters of this model are f, v, and s as these are the inputs inherent in the model

The decision variable of interest is X

Pritchett’s Precious Time Pieces

Profits = sX – f – vX

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

If sales = 0, profits = -f = –$1,000.

If sales = 1,000, profits = [(10)(1,000) – 1,000 – (5)(1,000)]

= $4,000

Pritchett’s Precious Time Pieces

0 = sX – f – vX, or 0 = (s – v)X – f

Companies are often interested in the break-even point (BEP). The BEP is the number of units sold that will result in $0 profit.

Solving for X, we have

f = (s – v)X

X = f

s – v

BEP = Fixed cost

(Selling price per unit) – (Variable cost per unit)

Pritchett’s Precious Time Pieces

0 = sX – f – vX, or 0 = (s – v)X – f

Companies are often interested in their break-even point (BEP). The BEP is the number of units sold that will result in $0 profit.

Solving for X, we have

f = (s – v)X

X = f

s – v

BEP = Fixed cost

(Selling price per unit) – (Variable cost per unit)

BEP for Pritchett’s Precious Time Pieces

BEP = $1,000/($10 – $5) = 200 units

Sales of less than 200 units of rebuilt springs will result in a loss.

Sales of over 200 units of rebuilt springs will result in a profit.

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.

Models Categorized by Risk

n Mathematical models that do not involve risk are called deterministic models. n All of the values used in the model are

known with complete certainty.

n Mathematical models that involve risk, chance, or uncertainty are called probabilistic models. n Values used in the model are estimates

based on probabilities.

Computers and Spreadsheet Models

QM for Windows

n An easy to use decision support system for use in POM and QM courses

n This is the main menu of quantitative models

Program 1.1

Computers and Spreadsheet Models

Excel QM’s Main Menu (2010)

n Works automatically within Excel spreadsheets

Program 1.2

Computers and Spreadsheet Models

Selecting Break-Even Analysis in Excel QM

Program 1.3A

Computers and Spreadsheet Models

Break- Even Analysis in Excel QM

Program 1.3B

Computers and Spreadsheet Models

Using Goal Seek in the Break- Even Problem

Program 1.4

Possible Problems in the Quantitative Analysis Approach

Defining the problem n Problems may not be easily identified.

n There may be conflicting viewpoints

n There may be an impact on other departments.

n Beginning assumptions may lead to a particular conclusion.

n The solution may be outdated.

Developing a model n Manager’s perception may not fit a textbook

model.

n There is a trade-off between complexity and ease of understanding.

Possible Problems in the Quantitative Analysis Approach

Acquiring accurate input data n Accounting data may not be collected for

quantitative problems.

n The validity of the data may be suspect.

Developing an appropriate solution n The mathematics may be hard to understand.

n Having only one answer may be limiting.

Testing the solution for validity

Analyzing the results in terms of the whole organization

Implementation – Not Just the Final Step

There may be an institutional lack of commitment and resistance to change.

n Management may fear the use of formal analysis processes will reduce their decision-making power.

n Action-oriented managers may want “quick and dirty” techniques.

n Management support and user involvement are important.

Implementation – Not Just the Final Step

There may be a lack of commitment by quantitative analysts.

n Analysts should be involved with the problem and care about the solution.

n Analysts should work with users and take their feelings into account.

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Chapter 3

To accompany Quantitative Analysis for Management, Eleventh Edition, by Render, Stair, and Hanna Power Point slides created by Brian Peterson

Decision Analysis

Learning Objectives

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.

After completing this chapter, students will be able to:

Learning Objectives

5. Develop accurate and useful decision trees.

6. Revise probabilities using Bayesian analysis.

7. Use computers to solve basic decision- making problems.

8. Understand the importance and use of utility theory in decision making.

After completing this chapter, students will be able to:

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

Introduction

n What is involved in making a good decision?

n Decision theory is an analytic and systematic approach to the study of decision making.

n A good decision is one that is based on logic, considers all available data and possible alternatives, and the quantitative approach described here.

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.

Thompson Lumber Company

Step 1 – Define the problem.

n The company is considering expanding by manufacturing and marketing a new product – backyard storage sheds.

Step 2 – List alternatives.

n Construct a large new plant.

n Construct a small new plant.

n Do not develop the new product line at all.

Step 3 – Identify possible outcomes.

n The market could be favorable or unfavorable.

Thompson Lumber Company

Step 4 – List the payoffs.

n 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.

n This depends on the environment and amount of risk and uncertainty.

Step 6 – Apply the model to the data.

n Solution and analysis are then used to aid in decision-making.

Thompson Lumber Company

STATE OF NATURE

ALTERNATIVE FAVORABLE MARKET ($)

UNFAVORABLE MARKET ($)

Construct a large plant 200,000 –180,000

Construct a small plant 100,000 –20,000

Do nothing 0 0

Table 3.1

Decision Table with Conditional Values for Thompson Lumber

Types of Decision-Making Environments

Type 1: Decision making under certainty

n The decision maker knows with certainty the consequences of every alternative or decision choice.

Type 2: Decision making under uncertainty

n The decision maker does not know the probabilities of the various outcomes.

Type 3: Decision making under risk

n The decision maker knows the probabilities of the various outcomes.

Decision Making Under Uncertainty

1. Maximax (optimistic)

2. Maximin (pessimistic)

3. Criterion of realism (Hurwicz)

4. Equally likely (Laplace)

5. Minimax regret

There are several criteria for making decisions under uncertainty:

Maximax

Used to find the alternative that maximizes the maximum payoff.

n Locate the maximum payoff for each alternative.

n Select the alternative with the maximum number.

STATE OF NATURE

ALTERNATIVE 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

Do nothing 0 0 0

Table 3.2

Maximax

Maximin

Used to find the alternative that maximizes the minimum payoff.

n Locate the minimum payoff for each alternative.

n Select the alternative with the maximum number.

STATE OF NATURE

ALTERNATIVE 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

Do nothing 0 0 0

Table 3.3 Maximin

Criterion of Realism (Hurwicz)

This is a weighted average compromise between optimism and pessimism.

n Select a coefficient of realism a, with 0≤α≤1.

n A value of 1 is perfectly optimistic, while a value of 0 is perfectly pessimistic.

n Compute the weighted averages for each alternative.

n Select the alternative with the highest value.

Weighted average = a(maximum in row) + (1 – a)(minimum in row)

Criterion of Realism (Hurwicz)

n For the large plant alternative using a = 0.8:

(0.8)(200,000) + (1 – 0.8)(–180,000) = 124,000

n For the small plant alternative using a = 0.8:

(0.8)(100,000) + (1 – 0.8)(–20,000) = 76,000

STATE OF NATURE

ALTERNATIVE FAVORABLE MARKET ($)

UNFAVORABLE MARKET ($)

CRITERION OF REALISM

(a = 0.8) $

Construct a large plant

200,000 –180,000 124,000

Construct a small plant

100,000 –20,000 76,000

Do nothing 0 0 0

Table 3.4

Realism

Equally Likely (Laplace)

Considers all the payoffs for each alternative n Find the average payoff for each alternative.

n Select the alternative with the highest average.

STATE OF NATURE

ALTERNATIVE 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

Do nothing 0 0 0

Table 3.5

Equally likely

Minimax Regret

Based on opportunity loss or regret, this is the difference between the optimal profit and actual payoff for a decision.

n Create an opportunity loss table by determining the opportunity loss from not choosing the best alternative.

n Opportunity loss is calculated by subtracting each payoff in the column from the best payoff in the column.

n Find the maximum opportunity loss for each alternative and pick the alternative with the minimum number.

Minimax Regret

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

Determining Opportunity Losses for Thompson Lumber

Minimax Regret

Table 3.7

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

Opportunity Loss Table for Thompson Lumber

Minimax Regret

Table 3.8

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 0 200,000 Minimax

Thompson’s Minimax Decision Using Opportunity Loss

Decision Making Under Risk

n This is decision making when there are several possible states of nature, and the probabilities associated with each possible state are known.

n The most popular method is to choose the alternative with the highest expected monetary value (EMV). n 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)

EMV for Thompson Lumber

n Suppose each market outcome has a probability of

occurrence of 0.50.

n Which alternative would give the highest EMV?

n 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

EMV for Thompson Lumber

STATE OF NATURE

ALTERNATIVE 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

Table 3.9 Largest EMV

Expected Value of Perfect Information (EVPI)

n EVPI places an upper bound on what you should pay for additional information.

EVPI = EVwPI – Maximum EMV

n 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)

Expected Value of Perfect Information (EVPI)

n Suppose Scientific Marketing, Inc. offers analysis that will provide certainty about market conditions (favorable).

n Additional information will cost $65,000.

n Should Thompson Lumber purchase the information?

Expected Value of Perfect Information (EVPI)

STATE OF NATURE

ALTERNATIVE 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 0.5

Table 3.10

EVwPI

Decision Table with Perfect Information

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.

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.

Expected Opportunity Loss

n Expected opportunity loss (EOL) is the cost of not picking the best solution.

n First construct an opportunity loss table.

n For each alternative, multiply the opportunity loss by the probability of that loss for each possible outcome and add these together.

n Minimum EOL will always result in the same decision as maximum EMV.

n Minimum EOL will always equal EVPI.

Expected Opportunity Loss

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

Table 3.11

STATE OF NATURE

ALTERNATIVE FAVORABLE MARKET ($)

UNFAVORABLE MARKET ($) EOL

Construct a large plant 0 180,000 90,000

Construct a small plant

100,000 20,000 60,000

Do nothing 200,000 0 100,000

Probabilities 0.50 0.50

Minimum EOL

Sensitivity Analysis

n Sensitivity analysis examines how the decision might change with different input data.

n For the Thompson Lumber example:

P = probability of a favorable market

(1 – P) = probability of an unfavorable market

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

Sensitivity Analysis

$300,000

$200,000

$100,000

–$100,000

–$200,000

EMV Values

EMV (large plant)

EMV (small plant)

EMV (do nothing)

Point 1

Point 2

.167 .615 1

Values of P

Figure 3.1

Sensitivity Analysis

Point 1:

EMV(do nothing) = EMV(small plant)

000200001200 ,$,$ -= P 1670 000120

00020 .

, ==P

00018000038000020000120 ,$,$,$,$ -=- PP

6150 000260

000160 .

, ==P

Point 2:

EMV(small plant) = EMV(large plant)

Sensitivity Analysis

$300,000

$200,000

$100,000

–$100,000

–$200,000

EMV Values

EMV (large plant)

EMV (small plant)

EMV (do nothing)

Point 1

Point 2

.167 .615 1

Values of P

BEST ALTERNATIVE

RANGE OF P VALUES

Do nothing Less than 0.167

Construct a small plant 0.167 – 0.615

Construct a large plant Greater than 0.615

Figure 3.1

Using Excel

Program 3.1A

Input Data for the Thompson Lumber Problem Using Excel QM

Using Excel

Program 3.1B

Output Results for the Thompson Lumber Problem Using Excel QM

Decision Trees

n Any problem that can be presented in a decision table can also be graphically represented in a decision tree.

n Decision trees are most beneficial when a sequence of decisions must be made.

n All decision trees contain decision points or nodes, from which one of several alternatives may be chosen.

n All decision trees contain state-of-nature points or nodes, out of which one state of nature will occur.

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.

Structure of Decision Trees

n Trees start from left to right.

n Trees represent decisions and outcomes in sequential order. n Squares represent decision nodes.

n Circles represent states of nature nodes.

n Lines or branches connect the decisions nodes and the states of nature.

Thompson’s Decision Tree

Favorable Market

Unfavorable Market

Favorable Market

Unfavorable Market

Construct

Small Plant 2

Figure 3.2

A Decision Node

A State-of-Nature Node

Thompson’s Decision Tree

Favorable Market

Unfavorable Market

Favorable Market

Unfavorable Market

Construct

Small Plant 2

Alternative with best EMV is selected

Figure 3.3

EMV for Node 1 = $10,000

= (0.5)($200,000) + (0.5)(–$180,000)

EMV for Node 2 = $40,000

= (0.5)($100,000) + (0.5)(–$20,000)

Payoffs

$200,000

–$180,000

$100,000

–$20,000

(0.5)

Thompson’s Complex Decision Tree

First Decision Point

Second Decision Point

Favorable Market (0.78)

Unfavorable Market (0.22)

Favorable Market (0.78)

Unfavorable Market (0.22)

Favorable Market (0.27)

Unfavorable Market (0.73)

Favorable Market (0.27)

Unfavorable Market (0.73)

Favorable Market (0.50)

Unfavorable Market (0.50)

Favorable Market (0.50)

Unfavorable Market (0.50) Small Plant

No Plant

Small Plant

No Plant

Small Plant

No Plant

Payoffs

–$190,000

$190,000

$90,000

–$30,000

–$10,000

–$180,000

$200,000

$100,000

–$20,000

–$190,000

$190,000

$90,000

–$30,000

–$10,000

Figure 3.4

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

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.

Thompson’s Complex Decision Tree

Figure 3.5

First Decision Point

Second Decision Point

Favorable Market (0.78)

Unfavorable Market (0.22)

Favorable Market (0.78)

Unfavorable Market (0.22)

Favorable Market (0.27)

Unfavorable Market (0.73)

Favorable Market (0.27)

Unfavorable Market (0.73)

Favorable Market (0.50)

Unfavorable Market (0.50)

Favorable Market (0.50)

Unfavorable Market (0.50) Small Plant

No Plant

Small Plant

No Plant

Small Plant

No Plant

Payoffs

–$190,000

$190,000

$90,000

–$30,000

–$10,000

–$180,000

$200,000

$100,000

–$20,000

–$190,000

$190,000

$90,000

–$30,000

–$10,000

$ 4 0 ,0

0 0

$ 2 ,4

0 0

$ 1 0 6 ,4

0 0

$ 4 9 ,2

0 0

$106,400

$63,600

–$87,400

$2,400

$10,000

$40,000

Expected Value of Sample Information

n Suppose Thompson wants to know the actual value of doing the survey.

EVSI = –

Expected value with sample

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

Sensitivity Analysis

n How sensitive are the decisions to changes in the probabilities? n How sensitive is our decision to the

probability of a favorable survey result?

n That is, if the probability of a favorable result (p = .45) where to change, would we make the same decision?

n How much could it change before we would make a different decision?

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 p<0.36, do not conduct the survey. If p>0.36, conduct the survey.

Bayesian Analysis

n There are many ways of getting probability data. It can be based on: n Management’s experience and intuition.

n Historical data.

n Computed from other data using Bayes’ theorem.

n Bayes’ theorem incorporates initial estimates and information about the accuracy of the sources.

n It also allows the revision of initial estimates based on new information.

Calculating Revised Probabilities

n 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

n But how were these calculated?

n The prior probabilities of these markets are:

P (FM) = 0.50

P (UM) = 0.50

Calculating Revised Probabilities

n Through discussions with experts Thompson has learned the information in the table below.

n 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

Calculating Revised Probabilities

n Recall Bayes’ theorem:

)()|()()|(

)()|( )|(

APABPAPABP

APABP BAP

¢´¢+´

´ =

where

events two any=BA,

AA of complement=¢

For this example, A will represent a favorable market and B will represent a positive survey.

Calculating Revised Probabilities

n P (FM | survey positive)

P(UM)|UM)P(P(FM) |FM)P(

FMPFMP

´+´

´ =

positive surveypositive survey

positive survey )()|(

780 450

350

500200500700

500700 .

).)(.().)(.(

).)(.( ==

+ =

P(FM)|FM)P(P(UM) |UM)P(

UMPUMP

´+´

´ =

positive surveypositive survey

positive survey )()|(

220 450

100

500700500200

500200 .

).)(.().)(.(

).)(.( ==

+ =

n P (UM | survey positive)

Calculating Revised Probabilities

POSTERIOR PROBABILITY

STATE OF NATURE

CONDITIONAL PROBABILITY

P(SURVEY POSITIVE | STATE

OF NATURE) PRIOR

PROBABILITY JOINT

PROBABILITY

P(STATE OF NATURE | SURVEY

POSITIVE)

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

Table 3.13

Probability Revisions Given a Positive Survey

Calculating Revised Probabilities

n P (FM | survey negative)

P(UM)|UM)P(P(FM) |FM)P(

FMPFMP

´+´

´ =

negative surveynegative survey

negative survey )()|(

270 550

150

500800500300

500300 .

).)(.().)(.(

).)(.( ==

+ =

P(FM)|FM)P(P(UM) |UM)P(

UMPUMP

´+´

´ =

negative surveynegative survey

negative survey )()|(

730 550

400

500300500800

500800 .

).)(.().)(.(

).)(.( ==

+ =

n P (UM | survey negative)

Calculating Revised Probabilities

POSTERIOR PROBABILITY

STATE OF NATURE

CONDITIONAL PROBABILITY

P(SURVEY NEGATIVE | STATE

OF NATURE) PRIOR

PROBABILITY JOINT

PROBABILITY

P(STATE OF NATURE | SURVEY

NEGATIVE)

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 1.00

Table 3.14

Probability Revisions Given a Negative Survey

Using Excel

Program 3.2A

Formulas Used for Bayes’ Calculations in Excel

Using Excel

Program 3.2B

Results of Bayes’ Calculations in Excel

Potential Problems Using Survey Results

n We can not always get the necessary data for analysis.

n Survey results may be based on cases where an action was taken.

n Conditional probability information may not be as accurate as we would like.

Utility Theory

n Monetary value is not always a true indicator of the overall value of the result of a decision.

n The overall value of a decision is called utility.

n Economists assume that rational people make decisions to maximize their utility.

Heads (0.5)

Tails (0.5)

$5,000,000

Utility Theory

Accept Offer

Reject Offer

$2,000,000

EMV = $2,500,000Figure 3.6

Your Decision Tree for the Lottery Ticket

Utility Theory

n Utility assessment assigns the worst outcome a utility of 0, and the best outcome, a utility of 1.

n A standard gamble is used to determine utility values.

n 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

Standard Gamble for Utility

Assessment

Best Outcome Utility = 1

Worst Outcome Utility = 0

Other Outcome Utility = ?

(p)

(1 – p)

Figure 3.7

Investment Example

n Jane Dickson wants to construct a utility curve revealing her preference for money between $0 and $10,000.

n A utility curve plots the utility value versus the monetary value.

n An investment in a bank will result in $5,000.

n An investment in real estate will result in $0 or $10,000.

n 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.

n So if p = 0.80, Jane is indifferent between the bank or the real estate investment.

Investment Example

Figure 3.8

p = 0.80

(1 – p) = 0.20

$10,000 U($10,000) = 1.0

$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

Investment Example

Utility for $7,000 = 0.90

Utility for $3,000 = 0.50

n We can assess other utility values in the same way.

n For Jane these are:

n Using the three utilities for different dollar amounts, she can construct a utility curve.

Utility Curve

U ($7,000) = 0.90

U ($5,000) = 0.80

U ($3,000) = 0.50

U ($0) = 0

Figure 3.9

1.0 –

0.9 –

0.8 –

0.7 –

0.6 –

0.5 –

0.4 –

0.3 –

0.2 –

0.1 –

| | | | | | | | | | |

$0 $1,000 $3,000 $5,000 $7,000 $10,000

Monetary Value

U ti

li ty

U ($10,000) = 1.0

Utility Curve

n Jane’s utility curve is typical of a risk avoider. n She gets less utility from greater risk.

n She avoids situations where high losses might occur.

n As monetary value increases, her utility curve increases at a slower rate.

n A risk seeker gets more utility from greater risk n As monetary value increases, the utility curve increases

at a faster rate.

n Someone with risk indifference will have a linear utility curve.

Preferences for Risk

Figure 3.10

Monetary Outcome

U ti

li ty

Risk Avoider

Risk Seeker

Utility as a Decision-Making Criteria

n Once a utility curve has been developed it can be used in making decisions.

n This replaces monetary outcomes with utility values.

n The expected utility is computed instead of the EMV.

Utility as a Decision-Making Criteria

n Mark Simkin loves to gamble.

n He plays a game tossing thumbtacks in the air.

n If the thumbtack lands point up, Mark wins $10,000.

n If the thumbtack lands point down, Mark loses $10,000.

n Mark believes that there is a 45% chance the thumbtack will land point up.

n Should Mark play the game (alternative 1)?

Utility as a Decision-Making Criteria

Figure 3.11

Tack Lands Point Up (0.45)

$10,000

–$10,000

Tack Lands Point Down (0.55)

Mark Does Not Play the Game

Decision Facing Mark Simkin

Utility as a Decision-Making Criteria

n Step 1– Define Mark’s utilities.

U (–$10,000) = 0.05

U ($0) = 0.15

U ($10,000) = 0.30

n 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

Utility Curve for Mark Simkin

Figure 3.12

1.00 –

0.75 –

0.50 –

0.30 –

0.25 –

0.15 –

0.05 –

0 – | | | | |

–$20,000 –$10,000 $0 $10,000 $20,000

Monetary Outcome

U ti

li ty

Utility as a Decision-Making Criteria

Figure 3.13

Tack Lands Point Up (0.45)

0.30

0.05

0.15

Tack Lands Point Down (0.55)

Don’t Play

UtilityE = 0.162

Using Expected Utilities in Decision Making

Chapter 5

To accompany Quantitative Analysis for Management, Eleventh Edition, by Render, Stair, and Hanna Power Point slides created by Brian Peterson

Forecasting

Learning Objectives

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.

After completing this chapter, students will be able to:

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

Introduction

n Managers are always trying to reduce uncertainty and make better estimates of what will happen in the future. n This is the main purpose of forecasting.

n Some firms use subjective methods: seat-of-the pants methods, intuition, experience.

n There are also several quantitative techniques, including:

n Moving averages

n Exponential smoothing

n Trend projections

n Least squares regression analysis

Introduction

n 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.

Introduction

n These steps are a systematic way of initiating, designing, and implementing a forecasting system.

n When used regularly over time, data is collected routinely and calculations performed automatically.

n There is seldom one superior forecasting system. n Different organizations may use different techniques.

n Whatever tool works best for a firm is the one that should be used.

Regression Analysis

Multiple Regression

Moving Average

Exponential Smoothing

Trend Projections

Decomposition

Delphi Methods

Jury of Executive Opinion

Sales Force Composite

Consumer Market Survey

Time-Series Methods

Qualitative Models

Causal Methods

Forecasting Models

Forecasting Techniques

Figure 5.1

Qualitative Models

n Qualitative models incorporate judgmental or subjective factors.

n These are useful when subjective factors are thought to be important or when accurate quantitative data is difficult to obtain.

n Common qualitative techniques are: n Delphi method.

n Jury of executive opinion.

n Sales force composite.

n Consumer market surveys.

Qualitative Models

n Delphi Method – This is an iterative group process where (possibly geographically dispersed) respondents provide input to decision makers.

n Jury of Executive Opinion – This method collects opinions of a small group of high-level managers, possibly using statistical models for analysis.

n Sales Force Composite – This allows individual salespersons estimate the sales in their region and the data is compiled at a district or national level.

n Consumer Market Survey – Input is solicited from customers or potential customers regarding their purchasing plans.

Time-Series Models

n Time-series models attempt to predict the future based on the past.

n Common time-series models are: n Moving average.

n Exponential smoothing.

n Trend projections.

n Decomposition.

n Regression analysis is used in trend projections and one type of decomposition model.

Causal Models

n Causal models use variables or factors that might influence the quantity being forecasted.

n The objective is to build a model with the best statistical relationship between the variable being forecast and the independent variables.

n Regression analysis is the most common technique used in causal modeling.

Scatter Diagrams

Wacker Distributors wants to forecast sales for three different products (annual sales in the table, in units):

YEAR TELEVISION

SETS RADIOS

COMPACT DISC PLAYERS

1 250 300 110

2 250 310 100

3 250 320 120

4 250 330 140

5 250 340 170

6 250 350 150

7 250 360 160

8 250 370 190

9 250 380 200

10 250 390 190

Table 5.1

Scatter Diagram for TVs

Figure 5.2a

l l l l l l l l ll

330 –

250 –

200 –

150 –

100 –

50 –

| | | | | | | | | |

0 1 2 3 4 5 6 7 8 9 10

Time (Years)

A n

n u

a l

S a le

s o

f T e le

v is

io n

(a) n Sales appear to be

constant over time

Sales = 250

n A good estimate of sales in year 11 is 250 televisions

Scatter Diagram for Radios

l l

n Sales appear to be increasing at a constant rate of 10 radios per year

Sales = 290 + 10(Year)

n A reasonable estimate of sales in year 11 is 400 radios.

420 –

400 –

380 –

360 –

340 –

320 –

300 –

280 –

| | | | | | | | | |

0 1 2 3 4 5 6 7 8 9 10

Time (Years)

A n

n u

a l

S a le

s o

f R

a d

io s

(b)

Figure 5.2b

Scatter Diagram for CD

Players

l l

n This trend line may not be perfectly accurate because of variation from year to year

n Sales appear to be increasing

n A forecast would probably be a larger figure each year

200 –

180 –

160 –

140 –

120 –

100 –

| | | | | | | | | |

0 1 2 3 4 5 6 7 8 9 10

Time (Years)

A n

n u

a l

S a le

s o

f C

D P

la y e rs

(c)

Figure 5.2c

Measures of Forecast Accuracy

n We compare forecasted values with actual values to see how well one model works or to compare models.

Forecast error = Actual value – Forecast value

n One measure of accuracy is the mean absolute deviation (MAD):

å =

errorforecast MAD

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

6 150 170 |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

Table 5.2

Measures of Forecast Accuracy

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

Table 5.2

817 9

160errorforecast .MAD ===

å n

Using a naïve forecasting model we can compute the MAD:

Measures of Forecast Accuracy

n There are other popular measures of forecast accuracy.

n The mean squared error:

å =

2 error)(

MSE

n The mean absolute percent error:

%MAPE 100 actual

error

å =

n And bias is the average error.

Time-Series Forecasting Models

n A time series is a sequence of evenly spaced events.

n Time-series forecasts predict the future based solely on the past values of the variable, and other variables are ignored.

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.

Decomposition of a Time-Series

Average Demand over 4 Years

Trend Component

Actual Demand

Line

Time

D e m

a n

d f

o r

P ro

d u

c t

o r

S e rv

ic e

| | | |

Year Year Year Year 1 2 3 4

Seasonal Peaks

Figure 5.3

Product Demand Charted over 4 Years, with Trend and Seasonality Indicated

Decomposition of a Time-Series

n There are two general forms of time-series models: n The multiplicative model:

Demand = T x S x C x R

n The additive model:

Demand = T + S + C + R

n Models may be combinations of these two forms.

n Forecasters often assume errors are normally distributed with a mean of zero.

Moving Averages

n Moving averages can be used when demand is relatively steady over time.

n The next forecast is the average of the most recent n data values from the time series.

n This methods tends to smooth out short- term irregularities in the data series.

nperiods previous in demands of Sum forecast average Moving =

Moving Averages

n Mathematically:

YYY F

nttt

+--

+++ =

...

Where:

= forecast for time period t + 1

= actual value in time period t

n = number of periods to average t Y

1+t F

Wallace Garden Supply

n Wallace Garden Supply wants to forecast demand for its Storage Shed.

n They have collected data for the past year.

n They are using a three-month moving average to forecast demand (n = 3).

Wallace Garden Supply

Table 5.3

MONTH ACTUAL SHED SALES THREE-MONTH MOVING AVERAGE

January 10

February 12

March 13

April 16

May 19

June 23

July 26

August 30

September 28

October 18

November 16

December 14

January —

(12 + 13 + 16)/3 = 13.67

(13 + 16 + 19)/3 = 16.00

(16 + 19 + 23)/3 = 19.33

(19 + 23 + 26)/3 = 22.67

(23 + 26 + 30)/3 = 26.33

(26 + 30 + 28)/3 = 28.00

(30 + 28 + 18)/3 = 25.33

(28 + 18 + 16)/3 = 20.67

(18 + 16 + 14)/3 = 16.00

(10 + 12 + 13)/3 = 11.67

Weighted Moving Averages

n Weighted moving averages use weights to put more emphasis on previous periods.

n This is often used when a trend or other pattern is emerging.

å å

= +

)(

))((

Weights

period in value Actual period inWeight 1

i F t

n Mathematically:

ntntt

www

YwYwYw F

+++

+++ =

+--

...

1121

where

wi = weight for the i th observation

Wallace Garden Supply

n Wallace Garden Supply decides to try a weighted moving average model to forecast demand for its Storage Shed.

n They decide on the following weighting scheme:

WEIGHTS APPLIED PERIOD

3 Last month

2 Two months ago

1 Three months ago

3 x Sales last month + 2 x Sales two months ago + 1 X Sales three months ago

Sum of the weights

Wallace Garden Supply

Table 5.4

MONTH ACTUAL SHED SALES THREE-MONTH WEIGHTED

MOVING AVERAGE

January 10

February 12

March 13

April 16

May 19

June 23

July 26

August 30

September 28

October 18

November 16

December 14

January —

[(3 X 13) + (2 X 12) + (10)]/6 = 12.17

[(3 X 16) + (2 X 13) + (12)]/6 = 14.33

[(3 X 19) + (2 X 16) + (13)]/6 = 17.00

[(3 X 23) + (2 X 19) + (16)]/6 = 20.50

[(3 X 26) + (2 X 23) + (19)]/6 = 23.83

[(3 X 30) + (2 X 26) + (23)]/6 = 27.50

[(3 X 28) + (2 X 30) + (26)]/6 = 28.33

[(3 X 18) + (2 X 28) + (30)]/6 = 23.33

[(3 X 16) + (2 X 18) + (28)]/6 = 18.67

[(3 X 14) + (2 X 16) + (18)]/6 = 15.33

Wallace Garden Supply

Program 5.1A

Selecting the Forecasting Module in Excel QM

Wallace Garden Supply

Program 5.1B

Initialization Screen for Weighted Moving Average

Wallace Garden Supply

Program 5.1C

Weighted Moving Average in Excel QM for Wallace Garden Supply

Exponential Smoothing

n 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 + a(Last period’s actual demand – Last period’s forecast)

Here a is a weight (or smoothing constant) in which 0≤a≤1.

Exponential Smoothing

Mathematically:

)( tttt FYFF -+=

+ a

Where:

Ft+1 = new forecast (for time period t + 1)

Ft = pervious forecast (for time period t)

a = smoothing constant (0 ≤ a ≤ 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.

Exponential Smoothing Example

n In January, February’s demand for a certain car model was predicted to be 142.

n Actual February demand was 153 autos

n Using a smoothing constant of a = 0.20, what is the forecast for March?

New forecast (for March demand) = 142 + 0.2(153 – 142) = 144.2 or 144 autos

n 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

Selecting the Smoothing Constant

n Selecting the appropriate value for a is key to obtaining a good forecast.

n The objective is always to generate an accurate forecast.

n The general approach is to develop trial forecasts with different values of a and select the a that results in the lowest MAD.

Exponential Smoothing

QUARTER

ACTUAL TONNAGE

UNLOADED FORECAST

USING a =0.10 FORECAST

USING a =0.50

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

Table 5.5

Port of Baltimore Exponential Smoothing Forecast for a=0.1 and a=0.5.

Exponential Smoothing

QUARTER

ACTUAL TONNAGE

UNLOADED FORECAST

WITH a = 0.10

ABSOLUTE DEVIATIONS FOR a = 0.10

FORECAST WITH a = 0.50

ABSOLUTE DEVIATIONS FOR a = 0.50

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 82.45 98.63

MAD = Σ|deviations|

= 10.31 MAD = 12.33 n

Table 5.6 Best choice

Absolute Deviations and MADs for the Port of Baltimore Example

Port of Baltimore Exponential Smoothing Example in Excel QM

Program 5.2

Exponential Smoothing with Trend Adjustment

n Like all averaging techniques, exponential smoothing does not respond to trends.

n A more complex model can be used that adjusts for trends.

n The basic approach is to develop an exponential smoothing forecast, and then adjust it for the trend.

Forecast including trend (FIT t+1

) = Smoothed forecast (F t+1

) + Smoothed Trend (T

t+1 )

Exponential Smoothing with Trend Adjustment

n The equation for the trend correction uses a new smoothing constant b .

n T t must be given or estimated. Tt+1 is computed by:

)()1( 11 tttt FITFTT -+-=

++ bb

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

b = smoothing constant for trend

Selecting a Smoothing Constant

n As with exponential smoothing, a high value of b makes the forecast more responsive to changes in trend.

n A low value of b gives less weight to the recent trend and tends to smooth out the trend.

n Values are generally selected using a trial-and- error approach based on the value of the MAD for different values of b.

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. § α = 0.3 § β = 0.4.

YEAR ELECTRICAL GENERATORS SOLD

2004 74

2005 79

2006 80

2007 90

2008 105

2009 142

2010 122 Table 5.7

Midwestern Manufacturing

n According to the assumptions,

FIT1 = F1 + T1 = 74 + 0 = 74.

n Step 1: Compute Ft+1 by:

FITt+1 = Ft + α(Yt – FITt)

= 74 + 0.3(74-74) = 74

n Step 2: Update the trend using:

Tt+1 = Tt + β(Ft+1 – FITt)

T2 = T1 + .4(F2 – FIT1)

= 0 + .4(74 – 74) = 0

Midwestern Manufacturing

n Step 3: Calculate the trend-adjusted exponential smoothing forecast (Ft+1) using the following:

FIT2 = F2 + T2 = 74 + 0 = 74

Midwestern Manufacturing

n For 2006 (period 3) we have:

n Step 1: F3 = FIT2 + 0.3(Y2 – FIT2)

= 74 + .3(79 – 74)

= 75.5

n Step 2: T3 = T2 + 0.4(F3 – FIT2)

= 0 + 0.4(75.5 – 74)

= 0.6

n Step 3: FIT3 = F3 + T3 = 75.5 + 0.6

= 76.1

Midwestern Manufacturing Exponential Smoothing with Trend Forecasts

Table 5.8

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(90- 78.338)

2.468 = 1.068+0.4(81.837- 78.338)

84.305 = 81.837+2.468

6 142 90.514=84.305+0.3(105- 84.305)

4.952 = 2.468+0.4(90.514- 84.305)

95.466 = 90.514+4.952

7 122 109.426=95.466+0.3(142- 95.466)

10.536 = 4.952+0.4(109.426-95.466)

119.962 = 109.426+10.536

8 120.573=119.962+0.3(122- 119.962)

10.780 = 10.536+0.4(120.573-

119.962)

131.353 = 120.573+10.780

Midwestern Manufacturing

Program 5.3

Midwestern Manufacturing Trend-Adjusted Exponential Smoothing in Excel QM

Trend Projections

n Trend projection fits a trend line to a series of historical data points.

n The line is projected into the future for medium- to long-range forecasts.

n Several trend equations can be developed based on exponential or quadratic models.

n The simplest is a linear model developed using regression analysis.

Trend Projection

The mathematical form is

XbbY 10

+=ˆ

Where = predicted value

b0 = intercept b1 = slope of the line X = time period (i.e., X = 1, 2, 3, …, n)

Ŷ

Midwestern Manufacturing

Program 5.4A

Excel Input Screen for Midwestern Manufacturing Trend Line

Midwestern Manufacturing

Program 5.4B

Excel Output for Midwestern Manufacturing Trend Line

Midwestern Manufacturing Company Example

n The forecast equation is

XY 54107156 ..ˆ +=

n 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

n Likewise for X = 9

(sales in 2012) = 56.71 + 10.54(9) = 151.57, or 152 generators

Midwestern Manufacturing

Figure 5.4

Electrical Generators and the Computed Trend Line

Midwestern Manufacturing

Program 5.5

Excel QM Trend Projection Model

Seasonal Variations

n Recurring variations over time may indicate the need for seasonal adjustments in the trend line.

n A seasonal index indicates how a particular season compares with an average season.

n 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.

Eichler Supplies

n Eichler Supplies sells telephone answering machines.

n Sales data for the past two years has been collected for one particular model.

n The firm wants to create a forecast that includes seasonality.

Eichler Supplies Answering Machine Sales and Seasonal Indices

MONTH

SALES DEMAND AVERAGE TWO- YEAR DEMAND

MONTHLY DEMAND

AVERAGE SEASONAL

INDEXYEAR 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

Seasonal index = Average two-year demand

Average monthly demand Average monthly demand = = 94

1,128

12 months Table 5.9

Seasonal Variations

n The calculations for the seasonal indices are

Jan. July969570 12

2001 =´ .

, 1121171

2001 =´ .

Feb. Aug.858510 12

2001 =´ .

, 1060641

2001 =´ .

Mar. Sept.909040 12

2001 =´ .

, 969570

2001 =´ .

Apr. Oct.1060641 12

2001 =´ .

, 858510

2001 =´ .

May Nov.1313091 12

2001 =´ .

, 858510

2001 =´ .

June Dec.1222231 12

2001 =´ .

, 858510

2001 =´ .

Seasonal Variations with Trend

n When both trend and seasonal components are present, the forecasting task is more complex.

n Seasonal indices should be computed using a centered moving average (CMA) approach.

n 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).

Turner Industries

n 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 Seasonal pattern

Turner Industries

n To calculate the CMA for quarter 3 of year 1 we compare the actual sales with an average quarter centered on that time period.

n 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.

CMA(q3, y1) = = 132.00 0.5(108) + 125 + 150 + 141 + 0.5(116)

Turner Industries

Compare the actual sales in quarter 3 to the CMA to find the seasonal ratio:

1361 132

1503 quarter in Sales ratio Seasonal .

CMA ===

Turner Industries

YEAR QUARTER SALES CMA SEASONAL RATIO

1 1 108

2 125

3 150 132.000 1.136

4 141 134.125 1.051

2 1 116 136.375 0.851

2 134 138.875 0.965

3 159 141.125 1.127

4 152 143.000 1.063

3 1 123 145.125 0.848

2 142 147.875 0.960

3 168

4 165

Table 5.11

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

Turner Industries

Scatterplot of Turner Industries Sales Data and Centered Moving Average

l l

CMA

Original Sales Figures

200 –

150 –

100 –

50 –

0 –

S a le

| | | | | | | | | | | |

1 2 3 4 5 6 7 8 9 10 11 12

Time Period

Figure 5.5

The Decomposition Method of Forecasting with Trend and Seasonal Components

n Decomposition is the process of isolating linear trend and seasonal factors to develop more accurate forecasts.

n 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.

Deseasonalized Data for Turner Industries

n Find a trend line using the deseasonalized data:

b1 = 2.34 b0 = 124.78

n 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

Deseasonalized Data for Turner Industries

SALES ($1,000,000s)

SEASONAL INDEX

DESEASONALIZED SALES ($1,000,000s)

108 0.85 127.059

125 0.96 130.208

150 1.13 132.743

141 1.06 133.019

116 0.85 136.471

134 0.96 139.583

159 1.13 140.708

152 1.06 143.396

123 0.85 144.706

142 0.96 147.917

168 1.13 148.673

165 1.06 155.660

Table 5.12

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

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)

San Diego Hospital

Program 5.6A

Initialization Screen for the Decomposition method in Excel QM

San Diego Hospital

Program 5.6B

Turner Industries Forecast Using the Decomposition Method in Excel QM

Using Regression with Trend and Seasonal Components

n Multiple regression can be used to forecast both trend and seasonal components in a time series. n One independent variable is time.

n Dummy independent variables are used to represent the seasons.

n The model is an additive decomposition model:

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

44332211 XbXbXbXbaY ++++=ˆ

Regression with Trend and Seasonal Components

Program 5.7A

Excel Input for the Turner Industries Example Using Multiple Regression

Using Regression with Trend and Seasonal Components

Program 5.7B

Excel Output for the Turner Industries Example Using Multiple Regression

Using Regression with Trend and Seasonal Components

n The resulting regression equation is:

4321 130738715321104 XXXXY .....ˆ ++++=

n Using the model to forecast sales for the first two quarters of next year:

n These are different from the results obtained using the multiplicative decomposition method.

n Use MAD or MSE to determine the best model.

13401300738071513321104 =++++= )(.)(.)(.)(..Ŷ

15201300738171514321104 =++++= )(.)(.)(.)(..Ŷ

Monitoring and Controlling Forecasts

n Tracking signals can be used to monitor the performance of a forecast.

n A tracking signal is computed as the running sum of the forecast errors (RSFE), and is computed using the following equation:

MAD

RSFE =signal Tracking

å =

errorforecast MAD

where

Monitoring and Controlling Forecasts

Acceptable Range

Signal Tripped

Upper Control Limit

Lower Control Limit

0 MADs

–

Time Figure 5.6

Tracking Signal

Plot of Tracking Signals

Monitoring and Controlling Forecasts

n Positive tracking signals indicate demand is greater than forecast.

n Negative tracking signals indicate demand is less than forecast.

n Some variation is expected, but a good forecast will have about as much positive error as negative error.

n Problems are indicated when the signal trips either the upper or lower predetermined limits.

n This indicates there has been an unacceptable amount of variation.

n Limits should be reasonable and may vary from item to item.

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

214 6

85errorforecast .MAD ===

å n

MADs5.2 2.14

35 signal Tracking ===

MAD

RSFE

Adaptive Smoothing

n Adaptive smoothing is the computer monitoring of tracking signals and self- adjustment if a limit is tripped.

n In exponential smoothing, the values of a and b are adjusted when the computer detects an excessive amount of variation.

Chapter 4

To accompany Quantitative Analysis for Management, Eleventh Edition, by Render, Stair, and Hanna Power Point slides created by Brian Peterson

Regression Models

Learning Objectives

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.

After completing this chapter, students will be able to:

Learning Objectives

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.

After completing this chapter, students will be able to:

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

Chapter Outline

4.7 Testing the Model for Significance

4.8 Multiple Regression Analysis

4.9 Binary or Dummy Variables

4.10 Model Building

4.11 Nonlinear Regression

4.12 Cautions and Pitfalls in Regression Analysis

Introduction

n Regression analysis is a very valuable tool for a manager.

n Regression can be used to: n Understand the relationship between

variables.

n Predict the value of one variable based on another variable.

n Simple linear regression models have only two variables.

n Multiple regression models have more variables.

Introduction

n The variable to be predicted is called the dependent variable. n This is sometimes called the response

variable.

n The value of this variable depends on the value of the independent variable. n This is sometimes called the explanatory

or predictor variable.

Independent variable

Dependent variable

Independent variable

= +

Scatter Diagram

n A scatter diagram or scatter plot is often used to investigate the relationship between variables.

n The independent variable is normally plotted on the X axis.

n The dependent variable is normally plotted on the Y axis.

Triple A Construction

n Triple A Construction renovates old homes.

n Managers have found that the dollar volume of renovation work is dependent on the area payroll.

TRIPLE A’S SALES ($100,000s)

LOCAL PAYROLL ($100,000,000s)

6 3

8 4

9 6

5 4

4.5 2

9.5 5

Table 4.1

Triple A Construction

Figure 4.1

Scatter Diagram of Triple A Construction Company Data

Simple Linear Regression

where

Y = dependent variable (response)

X = independent variable (predictor or explanatory)

b0 = intercept (value of Y when X = 0)

b1 = slope of the regression line

e = random error

n Regression models are used to test if there is a relationship between variables.

n There is some random error that cannot be predicted.

ebb ++= XY 10

Simple Linear Regression

n True values for the slope and intercept are not known so they are estimated using sample data.

XbbY 10

+=ˆ

where

Y = predicted value of Y

b0 = estimate of β0, based on sample results

b1 = estimate of β1, based on sample results

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)

YYe ˆ-=

Triple A Construction

For the simple linear regression model, the values of the intercept and slope can be calculated using the formulas below.

XbbY 10

+=ˆ

values of (mean) average X n

X X ==

values of (mean) average Y n

Y Y == å

å å

-- =

21 )(

))((

YYXX b

XbYb 10

Triple A Construction

Y X (X – X)2 (X – X)(Y – Y)

6 3 (3 – 4)2 = 1 (3 – 4)(6 – 7) = 1

8 4 (4 – 4)2 = 0 (4 – 4)(8 – 7) = 0

9 6 (6 – 4)2 = 4 (6 – 4)(9 – 7) = 4

5 4 (4 – 4)2 = 0 (4 – 4)(5 – 7) = 0

4.5 2 (2 – 4)2 = 4 (2 – 4)(4.5 – 7) = 5

9.5 5 (5 – 4)2 = 1 (5 – 4)(9.5 – 7) = 2.5

ΣY = 42

Y = 42/6 = 7

ΣX = 24

X = 24/6 = 4

Σ(X – X)2 = 10 Σ(X – X)(Y – Y) = 12.5

Table 4.2

Regression calculations for Triple A Construction

Triple A Construction

4 6

6 ===

åX X

7 6

6 ===

åY Y

251 10

512

21 .

)(

))(( ==

-- =

å å

YYXX b

242517 10

=-=-= ))(.(XbYb

Regression calculations

XY 2512 .ˆ +=Therefore

Triple A Construction

4 6

6 ===

åX X

7 6

6 ===

åY Y

251 10

512

21 .

)(

))(( ==

-- =

å å

YYXX b

242517 10

=-=-= ))(.(XbYb

Regression calculations

XY 2512 .ˆ +=Therefore

sales = 2 + 1.25(payroll)

If the payroll next year is $600 million

000950 $ or 5962512 ,.)(.ˆ =+=Y

Measuring the Fit of the Regression Model

n Regression models can be developed for any variables X and Y.

n How do we know the model is actually helpful in predicting Y based on X? n We could just take the average error, but

the positive and negative errors would cancel each other out.

n Three measures of variability are: n SST – Total variability about the mean.

n SSE – Variability about the regression line.

n SSR – Total variability that is explained by the model.

Measuring the Fit of the Regression Model

n Sum of the squares total: 2 )(å -= YYSST

n Sum of the squared error:

å å -== 22 )ˆ( YYeSSE

n Sum of squares due to regression:

å -= 2 )ˆ( YYSSR

n An important relationship:

SSESSRSST +=

Measuring the Fit of the Regression Model

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 – Y)2 = 6.875 ∑(Y – Y)2 = 15.625

Y = 7 SST = 22.5 SSE = 6.875 SSR = 15.625

Table 4.3

Sum of Squares for Triple A Construction

n Sum of the squares total 2 )(å -= YYSST

n Sum of the squared error

å å -== 22 )ˆ( YYeSSE

n Sum of squares due to regression

å -= 2 )ˆ( YYSSR

n An important relationship

SSESSRSST +=

Measuring the Fit of the Regression Model

For Triple A Construction

SST = 22.5 SSE = 6.875 SSR = 15.625

Measuring the Fit of the Regression Model

Figure 4.2

Deviations from the Regression Line and from the Mean

Coefficient of Determination

n The proportion of the variability in Y explained by the regression equation is called the coefficient of determination.

n The coefficient of determination is r2.

SST

SSE

SST

SSR r -== 1 2

n For Triple A Construction:

69440 522

625152 .

. ==r

n About 69% of the variability in Y is explained by the equation based on payroll (X).

Correlation Coefficient

n The correlation coefficient is an expression of the strength of the linear relationship.

n It will always be between +1 and –1.

n The correlation coefficient is r.

2 rr =

n For Triple A Construction:

8333069440 .. ==r

Four Values of the Correlation Coefficient

* *

* (a) Perfect Positive

Correlation: r = +1

* *

* **

(d) Perfect Negative Correlation: r = –1

* *

* * * *

* (b) Positive

Correlation: 0 < r < 1

* *

Figure 4.3

Using Computer Software for Regression

Program 4.1A

Accessing the Regression Option in Excel 2010

Using Computer Software for Regression

Program 4.1B

Data Input for Regression in Excel

Using Computer Software for Regression

Program 4.1C

Excel Output for the Triple A Construction Example

Assumptions of the Regression Model

1. Errors are independent.

2. Errors are normally distributed.

3. Errors have a mean of zero.

4. Errors have a constant variance.

n If we make certain assumptions about the errors in a regression model, we can perform statistical tests to determine if the model is useful.

n A plot of the residuals (errors) will often highlight any glaring violations of the assumption.

Residual Plots

Pattern of Errors Indicating Randomness

Figure 4.4A

E rr

o r

Residual Plots

Nonconstant error variance

Figure 4.4B

E rr

o r

Residual Plots

Errors Indicate Relationship is not Linear

Figure 4.4C

E rr

o r

Estimating the Variance

n Errors are assumed to have a constant variance (s 2), but we usually don’t know this.

n It can be estimated using the mean squared error (MSE), s2.

-- ==

SSE MSEs

where n = number of observations in the sample k = number of independent variables

Estimating the Variance

n For Triple A Construction:

71881 4

87506

116

87506

2 .

.. ==

-- =

-- ==

SSE MSEs

n We can estimate the standard deviation, s.

n This is also called the standard error of the estimate or the standard deviation of the regression.

31171881 .. === MSEs

Testing the Model for Significance

n When the sample size is too small, you can get good values for MSE and r2 even if there is no relationship between the variables.

n Testing the model for significance helps determine if the values are meaningful.

n We do this by performing a statistical hypothesis test.

Testing the Model for Significance

n We start with the general linear model

ebb ++= XY 10

n If b1 = 0, the null hypothesis is that there is no relationship between X and Y.

n The alternate hypothesis is that there is a linear relationship (b1 ≠ 0).

n If the null hypothesis can be rejected, we have proven there is a relationship.

n We use the F statistic for this test.

Testing the Model for Significance

n The F statistic is based on the MSE and MSR:

SSR MSR =

where k = number of independent variables in the model

n The F statistic is:

MSE

MSR F =

n This describes an F distribution with: degrees of freedom for the numerator = df1 = k

degrees of freedom for the denominator = df2 = n – k – 1

Testing the Model for Significance

n If there is very little error, the MSE would be small and the F-statistic would be large indicating the model is useful.

n If the F-statistic is large, the significance level (p-value) will be low, indicating it is unlikely this would have occurred by chance.

n So when the F-value is large, we can reject the null hypothesis and accept that there is a linear relationship between X and Y and the values of the MSE and r2 are meaningful.

Steps in a Hypothesis Test

1. Specify null and alternative hypotheses:

0 10 =b:H

0 11 ¹b:H

2. Select the level of significance (a). Common values are 0.01 and 0.05.

3. Calculate the value of the test statistic using the formula:

MSE

MSR F =

Steps in a Hypothesis Test

4. Make a decision using one of the following methods:

a) Reject the null hypothesis if the test statistic is greater than the F-value from the table in Appendix D. Otherwise, do not reject the null hypothesis:

21 ifReject

dfdfcalculated FF

,,a >

kdf = 1

1 2

--= kndf

b) Reject the null hypothesis if the observed significance level, or p-value, is less than the level of significance (a). Otherwise, do not reject the null hypothesis:

)( statistictest calculatedvalue- >= FPp

a<value- ifReject p

Triple A Construction

Step 1.

H0: b1 = 0 (no linear relationship between X and Y)

H1: b1 ≠ 0 (linear relationship exists between X and Y)

Step 2.

Select a = 0.05

625015 1

625015 .

. ===

SSR MSR

099 71881

625015 .

. ===

MSE

MSR F

Step 3.

Calculate the value of the test statistic.

Triple A Construction

Step 4.

Reject the null hypothesis if the test statistic is greater than the F-value in Appendix D.

df1 = k = 1

df2 = n – k – 1 = 6 – 1 – 1 = 4

The value of F associated with a 5% level of significance and with degrees of freedom 1 and 4 is found in Appendix D.

F0.05,1,4 = 7.71

Fcalculated = 9.09

Reject H0 because 9.09 > 7.71

F = 7.71

0.05

9.09

Triple A Construction

Figure 4.5

n We can conclude there is a statistically significant relationship between X and Y.

n The r2 value of 0.69 means about 69% of the variability in sales (Y) is explained by local payroll (X).

Analysis of Variance (ANOVA) Table

n When software is used to develop a regression model, an ANOVA table is typically created that shows the observed significance level (p-value) for the calculated F value.

n This can be compared to the level of significance (a) to make a decision.

DF SS MS F SIGNIFICANCE

Regression k SSR MSR = SSR/k MSR/MSE P(F > MSR/MSE)

Residual n - k - 1 SSE MSE = SSE/(n - k - 1)

Total n - 1 SST

Table 4.4

ANOVA for Triple A Construction

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.

Program 4.1C (partial)

P(F > 9.0909) = 0.0394

Multiple Regression Analysis

n Multiple regression models are extensions to the simple linear model and allow the creation of models with more than one independent variable.

Y = b0 + b1X1 + b2X2 + … + bkXk + e

where

Y = dependent variable (response variable)

Xi = i th independent variable (predictor or explanatory variable)

b0 = intercept (value of Y when all Xi = 0)

bi = coefficient of the i th independent variable

k = number of independent variables

e = random error

Multiple Regression Analysis

To estimate these values, a sample is taken the following equation developed

kk XbXbXbbY ++++= ...ˆ

22110

where

= predicted value of Y

b0 = sample intercept (and is an estimate of b0)

bi = sample coefficient of the ith variable (and is an estimate of bi)

Ŷ

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.

22110

ˆ XbXbbY ++= where

= 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

Ŷ

She selects a sample of houses that have sold recently and records the data shown in Table 4.5

Jenny Wilson Real Estate Data

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 MintTable 4.5

Jenny Wilson Realty

Program 4.2A

Input Screen for the Jenny Wilson Realty Multiple Regression Example

Jenny Wilson Realty

Program 4.2B

Output for the Jenny Wilson Realty Multiple Regression Example

Evaluating Multiple Regression Models

n Evaluation is similar to simple linear regression models.

n The p-value for the F-test and r2 are interpreted the same.

n The hypothesis is different because there is more than one independent variable.

n The F-test is investigating whether all the coefficients are equal to 0 at the same time.

Evaluating Multiple Regression Models

n To determine which independent variables are significant, tests are performed for each variable.

0 10 =b:H

0 11 ¹b:H

n The test statistic is calculated and if the p-value is lower than the level of significance (a), the null hypothesis is rejected.

Jenny Wilson Realty

n The model is statistically significant

n The p-value for the F-test is 0.002.

n r2 = 0.6719 so the model explains about 67% of the variation in selling price (Y).

n But the F-test is for the entire model and we can’t tell if one or both of the independent variables are significant.

n By calculating the p-value of each variable, we can assess the significance of the individual variables.

n Since the p-value for X1 (square footage) and X2 (age) are both less than the significance level of 0.05, both null hypotheses can be rejected.

Binary or Dummy Variables

n Binary (or dummy or indicator) variables are special variables created for qualitative data.

n A dummy variable is assigned a value of 1 if a particular condition is met and a value of 0 otherwise.

n The number of dummy variables must equal one less than the number of categories of the qualitative variable.

Jenny Wilson Realty

n Jenny believes a better model can be developed if she includes information about the condition of the property.

X3 = 1 if house is in excellent condition = 0 otherwise

X4 = 1 if house is in mint condition = 0 otherwise

n Two dummy variables are used to describe the three categories of condition.

n No variable is needed for “good” condition since if both X3 and X4 = 0, the house must be in good condition.

Jenny Wilson Realty

Program 4.3A

Input Screen for the Jenny Wilson Realty Example with Dummy Variables

Jenny Wilson Realty

Program 4.3B

Output for the Jenny Wilson Realty Example with Dummy Variables

Model Building

n The best model is a statistically significant model with a high r2 and few variables.

n As more variables are added to the model, the r2-value usually increases.

n For this reason, the adjusted r2 value is often used to determine the usefulness of an additional variable.

n The adjusted r2 takes into account the number of independent variables in the model.

Model Building

SST

SSE

SST

SSR -== 1

2 r

n The formula for r2

n The formula for adjusted r2

)/(SST

)/(SSE

1 1 Adjusted

-- -=

kn r

n As the number of variables increases, the adjusted r2 gets smaller unless the increase due to the new variable is large enough to offset the change in k.

Model Building

n In general, if a new variable increases the adjusted r2, it should probably be included in the model.

n In some cases, variables contain duplicate information.

n When two independent variables are correlated, they are said to be collinear.

n When more than two independent variables are correlated, multicollinearity exists.

n When multicollinearity is present, hypothesis tests for the individual coefficients are not valid but the model may still be useful.

Nonlinear Regression

n In some situations, variables are not linear.

n Transformations may be used to turn a nonlinear model into a linear model.

* * **

** * * *

Linear relationship Nonlinear relationship

* *

** * *

* * **

Colonel Motors

n Engineers at Colonel Motors want to use regression analysis to improve fuel efficiency.

n They have been asked to study the impact of weight on miles per gallon (MPG).

MPG WEIGHT (1,000

LBS.) MPG WEIGHT (1,000

LBS.)

12 4.58 20 3.18

13 4.66 23 2.68

15 4.02 24 2.65

18 2.53 33 1.70

19 3.09 36 1.95

19 3.11 42 1.92

Table 4.6

Colonel Motors

Figure 4.6A

Linear Model for MPG Data

Colonel Motors

Program 4.4 This is a useful model with a small F-test for significance and a good r2 value.

Excel Output for Linear Regression Model with MPG Data

Colonel Motors

Figure 4.6B

Nonlinear Model for MPG Data

Colonel Motors

n The nonlinear model is a quadratic model.

n The easiest way to work with this model is to develop a new variable.

2 weight)(=X

n This gives us a model that can be solved with linear regression software:

22110 XbXbbY ++=ˆ

Colonel Motors

Program 4.5 A better model with a smaller F-test for significance and a larger adjusted r2 value

21 43230879 XXY ...ˆ +-=

Cautions and Pitfalls

n If the assumptions are not met, the statistical test may not be valid.

n Correlation does not necessarily mean causation.

n Multicollinearity makes interpreting coefficients problematic, but the model may still be good.

n Using a regression model beyond the range of X is questionable, as the relationship may not hold outside the sample data.

Cautions and Pitfalls

n A t-test for the intercept (b0) may be ignored as this point is often outside the range of the model.

n A linear relationship may not be the best relationship, even if the F-test returns an acceptable value.

n A nonlinear relationship can exist even if a linear relationship does not.

n Even though a relationship is statistically significant it may not have any practical value.

Chapter 13

To accompany Quantitative Analysis for Management, Eleventh Edition, by Render, Stair, and Hanna Power Point slides created by Brian Peterson

Waiting Lines and Queuing Theory Models

Learning Objectives

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

2. Understand the three parts of a queuing system: the calling population, the queue itself, and the service facility.

3. Describe the basic queuing system configurations.

4. Understand the assumptions of the common models dealt with in this chapter.

5. Analyze a variety of operating characteristics of waiting lines.

After completing this chapter, students will be able to:

Chapter Outline

13.1 Introduction

13.2 Waiting Line Costs

13.3 Characteristics of a Queuing System

13.4 Single-Channel Queuing Model with Poisson Arrivals and Exponential Service Times (M/M/1)

13.5 Multichannel Queuing Model with Poisson Arrivals and Exponential Service Times (M/M/m)

Chapter Outline

13.6 Constant Service Time Model (M/D/1)

13.7 Finite Population Model (M/M/1 with Finite Source)

13.8 Some General Operating Characteristic Relationships

13.9 More Complex Queuing Models and the Use of Simulation

Introduction

n Queuing theory is the study of waiting lines.

n It is one of the oldest and most widely used quantitative analysis techniques.

n The three basic components of a queuing process are arrivals, service facilities, and the actual waiting line.

n Analytical models of waiting lines can help managers evaluate the cost and effectiveness of service systems.

Waiting Line Costs

n Most waiting line problems are focused on finding the ideal level of service a firm should provide.

n In most cases, this service level is something management can control.

n When an organization does have control, they often try to find the balance between two extremes.

Waiting Line Costs

n There is generally a trade-off between cost of providing service and cost of waiting time. n A large staff and many service facilities generally results

in high levels of service but have high costs.

n Having the minimum number of service facilities keeps service cost down but may result in dissatisfied customers.

n Service facilities are evaluated on their total expected cost which is the sum of service costs and waiting costs.

n Organizations typically want to find the service level that minimizes the total expected cost.

Queuing Costs and Service Levels

Figure 13.1

Three Rivers Shipping Company

n Three Rivers Shipping operates a docking facility on the Ohio River.

n An average of 5 ships arrive to unload their cargos each shift.

n Idle ships are expensive.

n More staff can be hired to unload the ships, but that is expensive as well.

n 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.

Three Rivers Shipping Company Waiting Line Cost Analysis

NUMBER OF TEAMS OF STEVEDORES WORKING

1 2 3 4

(a) Average number of ships arriving per shift

5 5 5 5

(b) Average time each ship waits to be unloaded (hours)

7 4 3 2

35 20 15 10

(d) Estimated cost per hour of idle ship time

$1,000 $1,000 $1,000 $1,000

(e) Value of ship’s lost time or waiting cost (c x d)

$35,000 $20,000 $15,000 $10,000

(f) Stevedore team salary or service cost

$6,000 $12,000 $18,000 $24,000

(g) Total expected cost (e + f) $41,000 $32,000 $33,000 $34,000

Optimal cost

Table 13.1

Characteristics of a Queuing System

n 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.

n These components have their own characteristics that must be examined before mathematical models can be developed.

Characteristics of a Queuing System

Arrival Characteristics have three major characteristics: size, pattern, and behavior.

n The size of the calling population can be either unlimited (essentially infinite) or limited (finite).

n The pattern of arrivals can arrive according to a known pattern or can arrive randomly. n Random arrivals generally follow a Poisson

distribution.

Characteristics of a Queuing System

Behavior of arrivals

n Most queuing models assume customers are patient and will wait in the queue until they are served and do not switch lines.

n Balking refers to customers who refuse to join the queue.

n Reneging customers enter the queue but become impatient and leave without receiving their service.

n That these behaviors exist is a strong argument for the use of queuing theory to managing waiting lines.

Characteristics of a Queuing System

Waiting Line Characteristics n Waiting lines can be either limited or unlimited.

n Queue discipline refers to the rule by which customers in the line receive service. n The most common rule is first-in, first-out (FIFO).

n Other rules are possible and may be based on other important characteristics.

n Other rules can be applied to select which customers enter which queue, but may apply FIFO once they are in the queue.

Characteristics of a Queuing System

Service Facility Characteristics n Basic queuing system configurations:

n Service systems are classified in terms of the number of channels, or servers, and the number of phases, or service stops.

n A single-channel system with one server is quite common.

n Multichannel systems exist when multiple servers are fed by one common waiting line.

n In a single-phase system, the customer receives service form just one server.

n In a multiphase system, the customer has to go through more than one server.

Four basic queuing system

configurations

Figure 13.2

Characteristics of a Queuing System

Service time distribution

n Service patterns can be either constant or random.

n Constant service times are often machine controlled.

n More often, service times are randomly distributed according to a negative exponential probability distribution.

n Analysts should observe, collect, and plot service time data to ensure that the observations fit the assumed distributions when applying these models.

Identifying Models Using Kendall Notation

n D. G. Kendall developed a notation for queuing models that specifies the pattern of arrival, the service time distribution, and the number of channels.

n Notation takes the form:

n 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

Arrival distribution

Service time distribution

Number of service channels open

Identifying Models Using Kendall Notation

n A single-channel model with Poisson arrivals and exponential service times would be represented by:

M/M/1

n If a second channel is added the notation would read:

M/M/2

n A three-channel system with Poisson arrivals and constant service time would be

M/D/3

n A four-channel system with Poisson arrivals and normally distributed service times would be

M/G/4

Single-Channel Model, Poisson Arrivals, Exponential Service Times (M/M/1)

Assumptions of the model: n Arrivals are served on a FIFO basis.

n There is no balking or reneging.

n Arrivals are independent of each other but the arrival rate is constant over time.

n Arrivals follow a Poisson distribution.

n Service times are variable and independent but the average is known.

n Service times follow a negative exponential distribution.

n Average service rate is greater than the average arrival rate.

Single-Channel Model, Poisson Arrivals, Exponential Service Times (M/M/1)

n When these assumptions are met, we can develop a series of equations that define the queue’s operating characteristics.

n Queuing Equations: Let

l = mean number of arrivals per time period

µ = 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.

Single-Channel Model, Poisson Arrivals, Exponential Service Times (M/M/1)

1. The average number of customers or units in the system, L:

lµ

- =L

2. The average time a customer spends in the system, W:

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

lµ - =

1 W

)( lµµ

- =

q L

Single-Channel Model, Poisson Arrivals, Exponential Service Times (M/M/1)

4. The average time a customer spends waiting in the queue, Wq:

)( lµµ

- =

q W

5. The utilization factor for the system, r, the probability the service facility is being used:

l r =

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:

l -=1

0 P

7. The probability that the number of customers in the system is greater than k, Pn>k:

> ÷ ø

ö ç è

æ =

kn P

n Arnold’s mechanic can install mufflers at a rate of 3 per hour.

n Customers arrive at a rate of 2 per hour.

n So:

l = 2 cars arriving per hour

µ = 3 cars serviced per hour

Arnold’s Muffler Shop

- =

lµ W = 1 hour that an average car

spends in the system

2 =

- =

lµ

l L = 2 cars in the system

on average

Arnold’s Muffler Shop

hour 3

2 =

- =

)( lµµ

l q

W = 40 minutes average waiting time per car

)()()( 13

233

2 22

= -

= lµµ

l q L = 1.33 cars waiting in line

on average

670 3

2 .===

l r = percentage of time

mechanic is busy

330 3

2 11

0 .=-=-=

l P = probability that there

are 0 cars in the system

Arnold’s Muffler Shop

Probability of more than k cars in the system

k Pn>k = ( 2/3)

k+1

0 0.667 Note that this is equal to 1 – P0 = 1 – 0.33 = 0.667

1 0.444

2 0.296

3 0.198 Implies that there is a 19.8% chance that more than 3 cars are in the system

4 0.132

5 0.088

6 0.058

7 0.039

Excel QM Solution to Arnold’s Muffler Example

Program 13.1

Arnold’s Muffler Shop

n Introducing costs into the model: n Arnold wants to do an economic analysis of

the queuing system and determine the waiting cost and service cost.

n The total service cost is:

Total service cost

= (Number of channels)

x (Cost per channel)

Total service cost

= mCs

Arnold’s Muffler Shop

Waiting cost when the cost is based on time in the system:

Total waiting cost

= (lW)Cw

Total waiting cost

= (Total time spent waiting by all

arrivals) x (Cost of waiting)

= (Number of arrivals) x

(Average wait per arrival)Cw

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

Total waiting cost

= (lWq)Cw

Arnold’s Muffler Shop

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 + lWCw

And when based on time in the queue:

Total cost = mCs + lWqCw

Arnold’s Muffler Shop

n Arnold estimates the cost of customer waiting time in line is $50 per hour.

Total daily waiting cost

= (8 hours per day)lWqCw

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

n 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

n So the total cost of the system is:

Total daily cost of the queuing system

= $533.33 + $120 = $653.33

n Arnold is thinking about hiring a different mechanic who can install mufflers at a faster rate.

n The new operating characteristics would be:

l = 2 cars arriving per hour

µ = 4 cars serviced per hour

Arnold’s Muffler Shop

- =

lµ W = 1/2 hour that an average car

spends in the system

2 =

- =

lµ

l L = 1 car in the system

on the average

Arnold’s Muffler Shop

hour 4

1 =

- =

)( lµµ

l q

W = 15 minutes average waiting time per car

)()()( 18

244

2 22

= -

= lµµ

l q L = 1/2 car waiting in line

on the average

50 4

2 .===

l r = percentage of time

mechanic is busy

50 4

2 11

0 .=-=-=

l P = probability that there

are 0 cars in the system

Arnold’s Muffler Shop

Probability of more than k cars in the system

k Pn>k = ( 2/4)

k+1

0 0.500

1 0.250

2 0.125

3 0.062

4 0.031

5 0.016

6 0.008

7 0.004

Arnold’s Muffler Shop Case

n The customer waiting cost is the same $50 per hour:

Total daily waiting cost

= (8 hours per day)lWqCw

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

n The new mechanic is more expensive at $20 per hour:

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

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

n So the total cost of the system is:

Total daily cost of the queuing system

= $200 + $160 = $360

Arnold’s Muffler Shop

n 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

n It is now is now:

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

n The total daily system costs are less with the new mechanic resulting in significant savings:

$653.33 – $360 = $293.33

Enhancing the Queuing Environment

n Reducing waiting time is not the only way to reduce waiting cost.

n Reducing the unit waiting cost (Cw) will also reduce total waiting cost.

n This might be less expensive to achieve than reducing either W or Wq.

Multichannel Queuing Model with Poisson Arrivals and Exponential Service Times

(M/M/m)

Assumptions of the model: n Arrivals are served on a FIFO basis.

n There is no balking or reneging.

n Arrivals are independent of each other but the arrival rate is constant over time.

n Arrivals follow a Poisson distribution.

n Service times are variable and independent but the average is known.

n Service times follow a negative exponential distribution.

n The average service rate is greater than the average arrival rate.

Multichannel Queuing Model with Poisson Arrivals and Exponential Service Times (M/M/m)

n Equations for the multichannel queuing model:

n Let

m = number of channels open l = average arrival rate µ = average service rate at each channel

1. The probability that there are zero customers in the system is:

lµ

l >

- ÷ ø

ö ç è

æ + ú ú û

ê ê ë

é ÷ ø

ö ç è

æ =

å -=

P m

n for

Multichannel Model, Poisson Arrivals, Exponential Service Times (M/M/m)

2. The average number of customers or units in the system

lµ

µllµ +

-- =

02 1

P mm

)()!(

)/(

3. The average time a unit spends in the waiting line or being served, in the system

lµlµ

µlµ L P

mm W

=+ --

= 1

1 02

)()!(

)/(

Multichannel Model, Poisson Arrivals, Exponential Service Times (M/M/m)

4. The average number of customers or units in line waiting for service

l -= LL

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

lµ

L WW =-=

l r

m =

Arnold’s Muffler Shop Revisited

n Arnold wants to investigate opening a second garage bay.

n He would hire a second worker who works at the same rate as his first worker.

n The customer arrival rate remains the same.

0.5

2)3(2

)3(2

1 21

0 =

- ÷ ø

ö ç è

æ +

ú ú û

ê ê ë

é ÷ ø

ö ç è

æ =

å =n

= probability of 0 cars in the system

lµ

l >

-÷ ÷ ø

ö çç è

æ + ú ú û

ê ê ë

é ÷÷ ø

ö çç è

æ =

å -=

P m

n for

Arnold’s Muffler Shop Revisited

lµ

µllµ +

-- =

02 )()!1(

)/( P

mm L

minutes 2

122hour 8

3 ===

L W

n Average number of cars in the system

n Average time a car spends in the system

75.0 3

2 ) 2

1 (

]2)3(2[)!1(

)3/2)(3(2 2

=+ -

Arnold’s Muffler Shop Revisited

n Average number of cars in the queue

n Average time a car spends in the queue

0830 12

3 .==-=-=

l LL

minutes 2

12hour 0.0415 2

08301 ====-=

lµ

L WW

Arnold’s Muffler Shop Revisited

n 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)lWqCw

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

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

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

n So the total cost of the system is

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

n This is the cheapest option: open the second bay and hire a second worker at the same $15 rate.

Effect of Service Level on Arnold’s Operating Characteristics

LEVEL OF SERVICE

OPERATING CHARACTERISTIC

ONE MECHANIC µ = 3

TWO MECHANICS µ = 3 FOR BOTH

ONE FAST MECHANIC µ = 4

Probability that the system is empty (P0)

0.33 0.50 0.50

Average number of cars in the system (L)

2 cars 0.75 cars 1 car

Average time spent in the system (W)

60 minutes 22.5 minutes 30 minutes

Average number of cars in the queue (Lq)

1.33 cars 0.083 car 0.50 car

Average time spent in the queue (Wq)

40 minutes 2.5 minutes 15 minutes

Table 13.2

Excel QM Solution to Arnold’s Muffler Multichannel Example

Program 13.2

Constant Service Time Model (M/D/1)

n Constant service times are used when customers or units are processed according to a fixed cycle.

n The values for Lq, Wq, L, and W are always less than they would be for models with variable service time.

n In fact both average queue length and average waiting time are halved in constant service rate models.

Constant Service Time Model (M/D/1)

1. Average length of the queue

2. Average waiting time in the queue

)( lµµ

- = 2

q L

)( lµµ

- = 2

q W

Constant Service Time Model (M/D/1)

3. Average number of customers in the system

4. Average time in the system

l +=

q LL

1 +=

q WW

Garcia-Golding Recycling, Inc.

n The company collects and compacts aluminum cans and glass bottles.

n Trucks arrive at an average rate of 8 per hour (Poisson distribution).

n Truck drivers wait about 15 minutes before they empty their load.

n Drivers and trucks cost $60 per hour.

n A new automated machine can process truckloads at a constant rate of 12 per hour.

n A new compactor would be amortized at $3 per truck unloaded.

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: l = 8 trucks/hour arriving

µ = 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

Excel QM Solution for Constant Service Time Model with Garcia-Golding Recycling

Example

Program 13.3

Finite Population Model (M/M/1 with Finite Source)

n When the population of potential customers is limited, the models are different.

n There is now a dependent relationship between the length of the queue and the arrival rate.

n 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, first- served basis.

Finite Population Model (M/M/1 with Finite Source)

Equations for the finite population model: Using l = mean arrival rate, µ = mean service rate, and N = size of the population, the operating characteristics are:

1. Probability that the system is empty:

å =

÷ ø

ö ç è

= N

)!(

Finite Population Model (M/M/1 with Finite Source)

2. Average length of the queue:

( ) 0

1 PNL q

-÷ ø

ö ç è

æ + -=

µl

4. Average waiting time in the queue:

l)( LN

L W

q -

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

( ) 0

1 PLL q

-+=

Finite Population Model (M/M/1 with Finite Source)

5. Average time in the system:

1 +=

q WW

6. Probability of n units in the system:

( ) NnP

N P

n ,...,,

! 10 for

0 =÷

ö ç è

- =

Department of Commerce

n The Department of Commerce has five printers that each need repair after about 20 hours of work.

n Breakdowns will follow a Poisson distribution.

n The technician can service a printer in an average of about 2 hours, following an exponential distribution.

n Therefore:

l = 1/20 = 0.05 printer/hour

µ = 1/2 = 0.50 printer/hour

Department of Commerce Example

2. ( ) printer 0.21

050

50050 5

0 =-÷

ö ç è

æ + -= PL

q .

5640

050

0 .

)!(

! =

÷ ø

ö ç è

æ -

å =n

P 1.

3. ( ) printer 0.645640120 =-+= ..L

Department of Commerce Example

( ) hour 0.91

220

0506405

20 ==

- =

.).(

. q

5. hours 2.91 500

1 910 =+=

. .W

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

Excel QM For Finite Population Model with Department of Commerce Example

Program 13.4

Some General Operating Characteristic Relationships

n Certain relationships exist among specific operating characteristics for any queuing system in a steady state.

n A steady state condition exists when a system is in its normal stabilized condition, usually after an initial transient state.

n The first of these are referred to as Little’s Flow Equations:

L = lW (or W = L/l) Lq = lWq (or Wq = Lq/l)

n And

W = Wq + 1/µ

More Complex Queuing Models and the Use of Simulation

n In the real world there are often variations from basic queuing models.

n Computer simulation can be used to solve these more complex problems.

n Simulation allows the analysis of controllable factors.

n Simulation should be used when standard queuing models provide only a poor approximation of the actual service system.

Chapter 7

To accompany Quantitative Analysis for Management, Eleventh Edition, by Render, Stair, and Hanna Power Point slides created by Brian Peterson

Linear Programming Models: Graphical and Computer

Methods

Learning Objectives

1. Understand the basic assumptions and properties of linear programming (LP).

2. Graphically solve any LP problem that has only two variables by both the corner point and isoprofit line methods.

3. Understand special issues in LP such as infeasibility, unboundedness, redundancy, and alternative optimal solutions.

4. Understand the role of sensitivity analysis.

5. Use Excel spreadsheets to solve LP problems.

After completing this chapter, students will be able to:

Chapter Outline

7.1 Introduction

7.2 Requirements of a Linear Programming Problem

7.3 Formulating LP Problems

7.4 Graphical Solution to an LP Problem

7.5 Solving Flair Furniture’s LP Problem using QM for Windows and Excel

7.6 Solving Minimization Problems

7.7 Four Special Cases in LP

7.8 Sensitivity Analysis

Introduction

n Many management decisions involve trying to make the most effective use of limited resources.

n Linear programming (LP) is a widely used mathematical modeling technique designed to help managers in planning and decision making relative to resource allocation. n This belongs to the broader field of

mathematical programming.

n In this sense, programming refers to modeling and solving a problem mathematically.

Requirements of a Linear Programming Problem

n All LP problems have 4 properties in common: 1. All problems seek to maximize or minimize some

quantity (the objective function).

2. Restrictions or constraints that limit the degree to which we can pursue our objective are present.

3. There must be alternative courses of action from which to choose.

4. The objective and constraints in problems must be expressed in terms of linear equations or inequalities.

Basic Assumptions of LP

n We assume conditions of certainty exist and numbers in the objective and constraints are known with certainty and do not change during the period being studied.

n We assume proportionality exists in the objective and constraints.

n We assume additivity in that the total of all activities equals the sum of the individual activities.

n We assume divisibility in that solutions need not be whole numbers.

n All answers or variables are nonnegative.

LP Properties and Assumptions

PROPERTIES OF LINEAR PROGRAMS

1. One objective function

2. One or more constraints

3. Alternative courses of action

4. Objective function and constraints are linear – proportionality and divisibility

5. Certainty

6. Divisibility

7. Nonnegative variables

Table 7.1

Formulating LP Problems

n Formulating a linear program involves developing a mathematical model to represent the managerial problem.

n The steps in formulating a linear program are:

1. Completely understand the managerial problem being faced.

2. Identify the objective and the constraints.

3. Define the decision variables.

4. Use the decision variables to write mathematical expressions for the objective function and the constraints.

Formulating LP Problems

n One of the most common LP applications is the product mix problem.

n Two or more products are produced using limited resources such as personnel, machines, and raw materials.

n The profit that the firm seeks to maximize is based on the profit contribution per unit of each product.

n The company would like to determine how many units of each product it should produce so as to maximize overall profit given its limited resources.

Flair Furniture Company

n The Flair Furniture Company produces inexpensive tables and chairs.

n Processes are similar in that both require a certain amount of hours of carpentry work and in the painting and varnishing department.

n Each table takes 4 hours of carpentry and 2 hours of painting and varnishing.

n Each chair requires 3 of carpentry and 1 hour of painting and varnishing.

n There are 240 hours of carpentry time available and 100 hours of painting and varnishing.

n Each table yields a profit of $70 and each chair a profit of $50.

Flair Furniture Company Data

The company wants to determine the best combination of tables and chairs to produce to reach the maximum profit.

HOURS REQUIRED TO PRODUCE 1 UNIT

DEPARTMENT (T)

TABLES (C)

CHAIRS AVAILABLE HOURS THIS WEEK

Carpentry 4 3 240

Painting and varnishing 2 1 100

Profit per unit $70 $50

Table 7.2

Flair Furniture Company

n The objective is to:

Maximize profit

n The constraints are:

1. The hours of carpentry time used cannot exceed 240 hours per week.

2. The hours of painting and varnishing time used cannot exceed 100 hours per week.

n The decision variables representing the actual decisions we will make are:

T = number of tables to be produced per week.

C = number of chairs to be produced per week.

Flair Furniture Company

n We create the LP objective function in terms of T and C:

Maximize profit = $70T + $50C

n Develop mathematical relationships for the two constraints:

n For carpentry, total time used is: (4 hours per table)(Number of tables produced)

+ (3 hours per chair)(Number of chairs produced).

n We know that:

Carpentry time used ≤ Carpentry time available.

4T + 3C ≤ 240 (hours of carpentry time)

Flair Furniture Company

n Similarly,

Painting and varnishing time used ≤ Painting and varnishing time available.

2 T + 1C ≤ 100 (hours of painting and varnishing time)

This means that each table produced requires two hours of painting and varnishing time.

n Both of these constraints restrict production capacity and affect total profit.

Flair Furniture Company

The values for T and C must be nonnegative.

T ≥ 0 (number of tables produced is greater than or equal to 0)

C ≥ 0 (number of chairs produced is greater than or equal to 0)

The complete problem stated mathematically:

Maximize profit = $70T + $50C

subject to

4T + 3C ≤ 240 (carpentry constraint)

2T + 1C ≤ 100 (painting and varnishing constraint)

T, C ≥ 0 (nonnegativity constraint)

Graphical Solution to an LP Problem

n The easiest way to solve a small LP problems is graphically.

n The graphical method only works when there are just two decision variables.

n When there are more than two variables, a more complex approach is needed as it is not possible to plot the solution on a two- dimensional graph.

n The graphical method provides valuable insight into how other approaches work.

Graphical Representation of a Constraint

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100 T

N u

m b

e r

o f

C h

a ir

Number of Tables

This Axis Represents the Constraint T ≥ 0

This Axis Represents the Constraint C ≥ 0

Figure 7.1

Quadrant Containing All Positive Values

Graphical Representation of a Constraint

n The first step in solving the problem is to identify a set or region of feasible solutions.

n To do this we plot each constraint equation on a graph.

n We start by graphing the equality portion of the constraint equations:

4T + 3C = 240

n We solve for the axis intercepts and draw the line.

Graphical Representation of a Constraint

n When Flair produces no tables, the carpentry constraint is:

4(0) + 3C = 240 3C = 240 C = 80

n Similarly for no chairs: 4T + 3(0) = 240

4T = 240 T = 60

n This line is shown on the following graph:

Graphical Representation of a Constraint

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100 T

N u

m b

e r

o f

C h

a ir

Number of Tables

(T = 0, C = 80)

Figure 7.2

(T = 60, C = 0)

Graph of carpentry constraint equation

Graphical Representation of a Constraint

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100 T

N u

m b

e r

o f

C h

a ir

Number of Tables

Figure 7.3

n Any point on or below the constraint plot will not violate the restriction.

n Any point above the plot will violate the restriction.

(30, 40)

(30, 20)

(70, 40)

Region that Satisfies the Carpentry Constraint

Graphical Representation of a Constraint

n The point (30, 40) lies on the plot and exactly satisfies the constraint

4(30) + 3(40) = 240.

n The point (30, 20) lies below the plot and satisfies the constraint

4(30) + 3(20) = 180.

n The point (70, 40) lies above the plot and does not satisfy the constraint

4(70) + 3(40) = 400.

Graphical Representation of a Constraint

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100 T

N u

m b

e r

o f

C h

a ir

Number of Tables

(T = 0, C = 100)

Figure 7.4

(T = 50, C = 0)

Region that Satisfies the Painting and Varnishing Constraint

Graphical Representation of a Constraint

n To produce tables and chairs, both departments must be used.

n We need to find a solution that satisfies both constraints simultaneously.

n A new graph shows both constraint plots.

n The feasible region (or area of feasible solutions) is where all constraints are satisfied.

n Any point inside this region is a feasible solution.

n Any point outside the region is an infeasible solution.

Graphical Representation of a Constraint

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100 T

N u

m b

e r

o f

C h

a ir

Number of Tables

Figure 7.5

Feasible Solution Region for the Flair Furniture Company Problem

Painting/Varnishing Constraint

Carpentry ConstraintFeasible Region

Graphical Representation of a Constraint

n For the point (30, 20)

Carpentry constraint

4T + 3C ≤ 240 hours available

(4)(30) + (3)(20) = 180 hours used

Painting constraint

2T + 1C ≤ 100 hours available

(2)(30) + (1)(20) = 80 hours used

n For the point (70, 40)

Carpentry constraint

4T + 3C ≤ 240 hours available

(4)(70) + (3)(40) = 400 hours used

Painting constraint

2T + 1C ≤ 100 hours available

(2)(70) + (1)(40) = 180 hours used

Graphical Representation of a Constraint

n For the point (50, 5)

Carpentry constraint

4T + 3C ≤ 240 hours available

(4)(50) + (3)(5) = 215 hours used

Painting constraint

2T + 1C ≤ 100 hours available

(2)(50) + (1)(5) = 105 hours used

Isoprofit Line Solution Method

n Once the feasible region has been graphed, we need to find the optimal solution from the many possible solutions.

n The speediest way to do this is to use the isoprofit line method.

n Starting with a small but possible profit value, we graph the objective function.

n We move the objective function line in the direction of increasing profit while maintaining the slope.

n The last point it touches in the feasible region is the optimal solution.

Isoprofit Line Solution Method

n For Flair Furniture, choose a profit of $2,100.

n The objective function is then

$2,100 = 70T + 50C

n Solving for the axis intercepts, we can draw the graph.

n This is obviously not the best possible solution.

n Further graphs can be created using larger profits.

n The further we move from the origin, the larger the profit will be.

n The highest profit ($4,100) will be generated when the isoprofit line passes through the point (30, 40).

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100 T

N u

m b

e r

o f

C h

a ir

Number of Tables

Figure 7.6

Profit line of $2,100 Plotted for the Flair Furniture Company

$2,100 = $70T + $50C

(30, 0)

(0, 42)

Isoprofit Line Solution Method

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100 T

N u

m b

e r

o f

C h

a ir

Number of Tables

Figure 7.7

Four Isoprofit Lines Plotted for the Flair Furniture Company

$2,100 = $70T + $50C

$2,800 = $70T + $50C

$3,500 = $70T + $50C

$4,200 = $70T + $50C

Isoprofit Line Solution Method

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100 T

N u

m b

e r

o f

C h

a ir

Number of Tables

Figure 7.8

Optimal Solution to the Flair Furniture problem

Optimal Solution Point (T = 30, C = 40)

Maximum Profit Line

$4,100 = $70T + $50C

Isoprofit Line Solution Method

n A second approach to solving LP problems employs the corner point method.

n It involves looking at the profit at every corner point of the feasible region.

n The mathematical theory behind LP is that the optimal solution must lie at one of the corner points, or extreme point, in the feasible region.

n For Flair Furniture, the feasible region is a four-sided polygon with four corner points labeled 1, 2, 3, and 4 on the graph.

Corner Point Solution Method

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

| | | | | | | | | | | |

0 20 40 60 80 100 T

N u

m b

e r

o f

C h

a ir

Number of Tables

Figure 7.9

Four Corner Points of the Feasible Region

Corner Point Solution Method

n To find the coordinates for Point accurately we have to solve for the intersection of the two constraint lines.

n Using the simultaneous equations method, we multiply the painting equation by –2 and add it to the carpentry equation

4T + 3C = 240 (carpentry line) – 4T – 2C = –200 (painting line)

C = 40

n Substituting 40 for C in either of the original equations allows us to determine the value of T.

4T + (3)(40) = 240 (carpentry line) 4T + 120 = 240

T = 30

Corner Point Solution Method

Point : (T = 0, C = 0) Profit = $70(0) + $50(0) = $0

Point : (T = 0, C = 80) Profit = $70(0) + $50(80) = $4,000

Point : (T = 50, C = 0) Profit = $70(50) + $50(0) = $3,500

Point : (T = 30, C = 40) Profit = $70(30) + $50(40) = $4,100

Because Point returns the highest profit, this is the optimal solution.

Slack and Surplus

n Slack is the amount of a resource that is not used. For a less-than-or- equal constraint: n Slack = Amount of resource available –

amount of resource used.

n Surplus is used with a greater-than- or-equal constraint to indicate the amount by which the right hand side of the constraint is exceeded. n Surplus = Actual amount – minimum

amount.

Summary of Graphical Solution Methods

ISOPROFIT METHOD

1. Graph all constraints and find the feasible region.

2. Select a specific profit (or cost) line and graph it to find the slope.

3. Move the objective function line in the direction of increasing profit (or decreasing cost) while maintaining the slope. The last point it touches in the feasible region is the optimal solution.

4. Find the values of the decision variables at this last point and compute the profit (or cost).

CORNER POINT METHOD

1. Graph all constraints and find the feasible region.

2. Find the corner points of the feasible reason.

3. Compute the profit (or cost) at each of the feasible corner points.

4. Select the corner point with the best value of the objective function found in Step 3. This is the optimal solution.

Table 7.4

Solving Flair Furniture’s LP Problem Using QM for Windows and Excel

n Most organizations have access to software to solve big LP problems.

n While there are differences between software implementations, the approach each takes towards handling LP is basically the same.

n Once you are experienced in dealing with computerized LP algorithms, you can easily adjust to minor changes.

Using QM for Windows

n First select the Linear Programming module.

n Specify the number of constraints (non-negativity is assumed).

n Specify the number of decision variables.

n Specify whether the objective is to be maximized or minimized.

n For the Flair Furniture problem there are two constraints, two decision variables, and the objective is to maximize profit.

Using QM for Windows

QM for Windows Linear Programming Computer screen for Input of Data

Program 7.1A

Using QM for Windows

QM for Windows Data Input for Flair Furniture Problem

Program 7.1B

Using QM for Windows

QM for Windows Output for Flair Furniture Problem

Program 7.1C

Using QM for Windows

QM for Windows Graphical Output for Flair Furniture Problem

Program 7.1D

Using Excel’s Solver Command to Solve LP Problems

n The Solver tool in Excel can be used to find solutions to:

n LP problems.

n Integer programming problems.

n Noninteger programming problems.

n Solver is limited to 200 variables and 100 constraints.

Using Solver to Solve the Flair Furniture Problem

n Recall the model for Flair Furniture is:

Maximize profit = $70T + $50C Subject to 4T + 3C ≤ 240

2T + 1C ≤ 100

n To use Solver, it is necessary to enter formulas based on the initial model.

Using Solver to Solve the Flair Furniture Problem

1. Enter the variable names, the coefficients for the objective function and constraints, and the right- hand-side values for each of the constraints.

2. Designate specific cells for the values of the decision variables.

3. Write a formula to calculate the value of the objective function.

4. Write a formula to compute the left-hand sides of each of the constraints.

Using Solver to Solve the Flair Furniture Problem

Program 7.2A

Excel Data Input for the Flair Furniture Example

Using Solver to Solve the Flair Furniture Problem

Program 7.2B

Formulas for the Flair Furniture Example

Using Solver to Solve the Flair Furniture Problem

Program 7.2C

Excel Spreadsheet for the Flair Furniture Example

Using Solver to Solve the Flair Furniture Problem

n Once the model has been entered, the following steps can be used to solve the problem.

In Excel 2010, select Data – Solver.

If Solver does not appear in the indicated place, see Appendix F for instructions on how to activate this add-in.

1. In the Set Objective box, enter the cell address for the total profit.

2. In the By Changing Cells box, enter the cell addresses for the variable values.

3. Click Max for a maximization problem and Min for a minimization problem.

Using Solver to Solve the Flair Furniture Problem

4. Check the box for Make Unconstrained Variables Non-negative.

5. Click the Select Solving Method button and select Simplex LP from the menu that appears.

6. Click Add to add the constraints.

7. In the dialog box that appears, enter the cell references for the left-hand-side values, the type of equation, and the right-hand-side values.

8. Click Solve.

Using Solver to Solve the Flair Furniture Problem

Starting Solver

Figure 7.2D

Using Solver to Solve the Flair Furniture Problem

Figure 7.2E

Solver Parameters Dialog Box

Using Solver to Solve the Flair Furniture Problem

Figure 7.2F

Solver Add Constraint Dialog Box

Using Solver to Solve the Flair Furniture Problem

Figure 7.2G

Solver Results Dialog Box

Using Solver to Solve the Flair Furniture Problem

Figure 7.2H

Solution Found by Solver

Solving Minimization Problems

n Many LP problems involve minimizing an objective such as cost instead of maximizing a profit function.

n Minimization problems can be solved graphically by first setting up the feasible solution region and then using either the corner point method or an isocost line approach (which is analogous to the isoprofit approach in maximization problems) to find the values of the decision variables (e.g., X1 and X2) that yield the minimum cost.

The Holiday Meal Turkey Ranch is considering buying two different brands of turkey feed and blending them to provide a good, low-cost diet for its turkeys

Minimize cost (in cents) = 2X1 + 3X2 subject to:

5X1 + 10X2 ≥ 90 ounces (ingredient constraint A) 4X1 + 3X2 ≥ 48 ounces (ingredient constraint B)

0.5X1 ≥ 1.5 ounces (ingredient constraint C) X1 ≥ 0 (nonnegativity constraint)

X2 ≥ 0 (nonnegativity constraint)

Holiday Meal Turkey Ranch

X1 = number of pounds of brand 1 feed purchased

X2 = number of pounds of brand 2 feed purchased

Let

Holiday Meal Turkey Ranch

INGREDIENT

COMPOSITION OF EACH POUND OF FEED (OZ.) MINIMUM MONTHLY

REQUIREMENT PER TURKEY (OZ.)BRAND 1 FEED BRAND 2 FEED

A 5 10 90

B 4 3 48

C 0.5 0 1.5

Cost per pound 2 cents 3 cents

Holiday Meal Turkey Ranch data

Table 7.5

Holiday Meal Turkey Ranch

n Use the corner point method.

n First construct the feasible solution region.

n The optimal solution will lie at one of the corners as it would in a maximization problem.

Feasible Region for the Holiday Meal Turkey Ranch Problem

–

20 –

15 –

10 –

5 –

0 –

| | | | | |

5 10 15 20 25 X1

P o

u n

d s o

f B

ra n

d 2

Pounds of Brand 1

Ingredient C Constraint

Ingredient B Constraint

Ingredient A Constraint

Feasible Region

Figure 7.10

Holiday Meal Turkey Ranch

n Solve for the values of the three corner points.

n Point a is the intersection of ingredient constraints C and B.

4X1 + 3X2 = 48

X1 = 3

n Substituting 3 in the first equation, we find X2 = 12.

n Solving for point b with basic algebra we find X1 = 8.4 and X2 = 4.8.

n Solving for point c we find X1 = 18 and X2 = 0.

Substituting these value back into the objective function we find

Cost = 2X1 + 3X2 Cost at point a = 2(3) + 3(12) = 42

Cost at point b = 2(8.4) + 3(4.8) = 31.2

Cost at point c = 2(18) + 3(0) = 36

Holiday Meal Turkey Ranch

The lowest cost solution is to purchase 8.4 pounds of brand 1 feed and 4.8 pounds of brand 2 feed for a total cost of 31.2 cents per turkey.

Graphical Solution to the Holiday Meal Turkey Ranch Problem Using the Isocost Approach

Holiday Meal Turkey Ranch

–

20 –

15 –

10 –

5 –

0 –

| | | | | |

5 10 15 20 25 X1

P o

u n

d s o

f B

ra n

d 2

Pounds of Brand 1

Figure 7.11

Feasible Region

(X1 = 8.4, X2 = 4.8)

Solving the Holiday Meal Turkey Ranch Problem Using QM for Windows

Holiday Meal Turkey Ranch

Program 7.3

Holiday Meal Turkey Ranch

Program 7.4A

Excel 2010 Spreadsheet for the Holiday Meal Turkey Ranch problem

Holiday Meal Turkey Ranch

Program 7.4B

Excel 2010 Solution to the Holiday Meal Turkey Ranch Problem

Four Special Cases in LP

n Four special cases and difficulties arise at times when using the graphical approach to solving LP problems. n No feasible solution

n Unboundedness

n Redundancy

n Alternate Optimal Solutions

Four Special Cases in LP

No feasible solution n This exists when there is no solution to the

problem that satisfies all the constraint equations.

n No feasible solution region exists.

n This is a common occurrence in the real world.

n Generally one or more constraints are relaxed until a solution is found.

Four Special Cases in LP

A problem with no feasible solution

8 –

–

6 –

–

4 –

–

2 –

–

0 –

| | | | | | | | | |

2 4 6 8 X1

Region Satisfying First Two Constraints

Figure 7.12

Region Satisfying Third Constraint

Four Special Cases in LP

Unboundedness n Sometimes a linear program will not have a

finite solution.

n In a maximization problem, one or more solution variables, and the profit, can be made infinitely large without violating any constraints.

n In a graphical solution, the feasible region will be open ended.

n This usually means the problem has been formulated improperly.

Four Special Cases in LP

A Feasible Region That is Unbounded to the Right

15 –

10 –

5 –

0 –

| | | | |

5 10 15 X1 Figure 7.13

Feasible Region

X1 ≥ 5

X2 ≤ 10

X1 + 2X2 ≥ 15

Four Special Cases in LP

Redundancy n A redundant constraint is one that does not

affect the feasible solution region.

n One or more constraints may be binding.

n This is a very common occurrence in the real world.

n It causes no particular problems, but eliminating redundant constraints simplifies the model.

Four Special Cases in LP

Problem with a Redundant Constraint

30 –

25 –

20 –

15 –

10 –

5 –

0 –

| | | | | |

5 10 15 20 25 30 X1

Figure 7.14

Redundant Constraint

Feasible Region

X1 ≤ 25

2X1 + X2 ≤ 30

X1 + X2 ≤ 20

Four Special Cases in LP

Alternate Optimal Solutions n Occasionally two or more optimal solutions

may exist.

n Graphically this occurs when the objective function’s isoprofit or isocost line runs perfectly parallel to one of the constraints.

n This actually allows management great flexibility in deciding which combination to select as the profit is the same at each alternate solution.

Four Special Cases in LP

Example of Alternate Optimal Solutions

8 –

7 –

6 –

5 –

4 –

3 –

2 –

1 –

0 –

| | | | | | | |

1 2 3 4 5 6 7 8 X1

Figure 7.15 Feasible Region

Isoprofit Line for $8

Optimal Solution Consists of All Combinations of X1 and X2 Along the AB Segment

Isoprofit Line for $12 Overlays Line Segment AB

Sensitivity Analysis

n Optimal solutions to LP problems thus far have been found under what are called deterministic assumptions.

n This means that we assume complete certainty in the data and relationships of a problem.

n But in the real world, conditions are dynamic and changing.

n We can analyze how sensitive a deterministic solution is to changes in the assumptions of the model.

n This is called sensitivity analysis, postoptimality analysis, parametric programming, or optimality analysis.

Sensitivity Analysis

n Sensitivity analysis often involves a series of what-if? questions concerning constraints, variable coefficients, and the objective function.

n One way to do this is the trial-and-error method where values are changed and the entire model is resolved.

n The preferred way is to use an analytic postoptimality analysis.

n After a problem has been solved, we determine a range of changes in problem parameters that will not affect the optimal solution or change the variables in the solution.

n The High Note Sound Company manufactures quality CD players and stereo receivers.

n Products require a certain amount of skilled artisanship which is in limited supply.

n The firm has formulated the following product mix LP model.

High Note Sound Company

Maximize profit = $50X1 + $120X2 Subject to 2X1 + 4X2 ≤ 80 (hours of electrician’s

time available)

3X1 + 1X2 ≤ 60 (hours of audio technician’s time available)

X1, X2 ≥ 0

The High Note Sound Company Graphical Solution

High Note Sound Company

b = (16, 12)

Optimal Solution at Point a

X1 = 0 CD Players

X2 = 20 Receivers Profits = $2,400

a = (0, 20)

Isoprofit Line: $2,400 = 50X1 + 120X2

60 –

–

40 –

–

20 –

10 –

0 –

| | | | | |

10 20 30 40 50 60 X1

(receivers)

(CD players)c = (20, 0)

Figure 7.16

Changes in the Objective Function Coefficient

n In real-life problems, contribution rates in the objective functions fluctuate periodically.

n Graphically, this means that although the feasible solution region remains exactly the same, the slope of the isoprofit or isocost line will change.

n We can often make modest increases or decreases in the objective function coefficient of any variable without changing the current optimal corner point.

n We need to know how much an objective function coefficient can change before the optimal solution would be at a different corner point.

Changes in the Objective Function Coefficient

Changes in the Receiver Contribution Coefficients

Profit Line for 50X1 + 80X2 (Passes through Point b)

40 –

30 –

20 –

10 –

0 –

| | | | | |

10 20 30 40 50 60 X1

Figure 7.17

Old Profit Line for 50X1 + 120X2 (Passes through Point a)

Profit Line for 50X1 + 150X2 (Passes through Point a)

QM for Windows and Changes in Objective Function Coefficients

Input and Sensitivity Analysis for High Note Sound Data Using QM For Windows

Program 7.5B

Program 7.5A

Excel Solver and Changes in Objective Function Coefficients

Excel 2010 Spreadsheet for High Note Sound Company

Program 7.6A

Excel Solver and Changes in Objective Function Coefficients

Excel 2010 Solution and Solver Results Window for High Note Sound Company

Figure 7.6B

Excel Solver and Changes in Objective Function Coefficients

Excel 2010 Sensitivity Report for High Note Sound Company

Program 7.6C

Changes in the Technological Coefficients

n Changes in the technological coefficients often reflect changes in the state of technology.

n If the amount of resources needed to produce a product changes, coefficients in the constraint equations will change.

n This does not change the objective function, but it can produce a significant change in the shape of the feasible region.

n This may cause a change in the optimal solution.

Changes in the Technological Coefficients

Change in the Technological Coefficients for the High Note Sound Company

(a) Original Problem

3X1 + 1X2 ≤ 60

2X1 + 4X2 ≤ 80

Optimal Solution

60 –

40 –

20 –

– | | | 0 20 40 X1

S te

re o

R e c e iv

e rs

CD Players

(b) Change in Circled Coefficient

2 X1 + 1X2 ≤ 60

2X1 + 4X2 ≤ 80

Still Optimal

3X1 + 1X2 ≤ 60

2X1 + 5 X2 ≤ 80

Optimal Solutiona

60 –

40 –

20 –

– | | | 0 20 40

60 –

40 –

20 –

– | | | 0 20 40

f g

Figure 7.18

Changes in Resources or Right-Hand-Side Values

n The right-hand-side values of the constraints often represent resources available to the firm.

n If additional resources were available, a higher total profit could be realized.

n Sensitivity analysis about resources will help answer questions about how much should be paid for additional resources and how much more of a resource would be useful.

Changes in Resources or Right-Hand-Side Values

n If the right-hand side of a constraint is changed, the feasible region will change (unless the constraint is redundant).

n Often the optimal solution will change.

n The amount of change in the objective function value that results from a unit change in one of the resources available is called the dual price or dual value .

n The dual price for a constraint is the improvement in the objective function value that results from a one-unit increase in the right-hand side of the constraint.

Changes in Resources or Right-Hand-Side Values

n However, the amount of possible increase in the right-hand side of a resource is limited.

n If the number of hours increased beyond the upper bound, then the objective function would no longer increase by the dual price.

n There would simply be excess (slack) hours of a resource or the objective function may change by an amount different from the dual price.

n The dual price is relevant only within limits.

Changes in the Electricians’ Time Resource for the High Note Sound Company

60 –

40 –

20 –

–

25 –

| | |

0 20 40 60

50 X1

X2 (a)

Constraint Representing 60 Hours of Audio Technician’s Time Resource

Changed Constraint Representing 100 Hours of Electrician’s Time Resource

Figure 7.19

Changes in the Electricians’ Time Resource for the High Note Sound Company

60 –

40 –

20 –

–

15 –

| | |

0 20 40 60

30 X1

X2 (b)

Constraint Representing 60 Hours of Audio Technician’s Time Resource

Changed Constraint Representing 60 Hours of Electrician’s Time Resource

Figure 7.19

Changes in the Electricians’ Time Resource for the High Note Sound Company

60 –

40 –

20 –

– | | | | | |

0 20 40 60 80 100 120 X1

X2 (c)

Constraint Representing 60 Hours of Audio Technician’s Time Resource

Changed Constraint Representing 240 Hours of Electrician’s Time Resource

Figure 7.19

QM for Windows and Changes in Right-Hand-Side Values

Sensitivity Analysis for High Note Sound Company Using QM for Windows

Program 7.5B

Excel Solver and Changes in Right-Hand-Side Values

Excel 2010 Sensitivity Analysis for High Note Sound Company

Program 7.6C