rational decision making -...
TRANSCRIPT
Department of Decision Sciences
Rational Decision Making
Only study guide for
DSC2602
University of South AfricaPretoria
c©2010 University of South AfricaAll rights reserved.
Printed and published by theUniversity of South Africa,Muckleneuk, Pretoria.
DSC2602/1/2011
Cover: Eastern Transvaal, Lowveld (1928) J. H. Pierneef
J. H. Pierneef is one of South Africa’s best known artists.Permission for the use of this work was kindly grantedby the Schweickerdt family.
The tree structure is a recurring theme in various branchesof the decision sciences.
Preface
Everyday life is full of decisions. What should I wear today? What should I eat? Should I buythe red or blue shirt? Should I buy a specific house or buy a piece of land? What is the shortestroute from my house to work? . . . And many more.
Some of these decisions can be made without thinking or by guesswork. Some can be solved byreasoning or emotions. Some are a bit more difficult and may need additional information.
People have been using mathematical tools to aid decision making for decades. During WorldWar II many techniques were developed to assists the military in decision making. These devel-opments were so successful that after World War II many companies used similar techniques inmanagerial decision making and planning.
The decision making task of modern management is more demanding and more importantthan ever. Many organisations employ operations research or management science personnel orconsultants to apply the principles of scientific management to problems and decision making.
In this module we focus on a number of useful models and techniques that can be used in thedecision making process. Two important themes run through the study guide: data analysis anddecision making techniques.
Firstly we look at data analysis. This approach starts with data that are manipulated or processedinto information that is valuable to decision making. The processing and manipulation of rawdata into meaningful information are the heart of data analysis. Data analysis includes datadescription, data inference, the search for relationships in data and dealing with uncertaintywhich in turn includes measuring uncertainty and modelling uncertainty explicitly.
In addition to data analysis, other decision making techniques are discussed. These techniquesinclude decision analysis, project scheduling and network models.
Chapter 1 illustrates a number of ways to summarise the information in data sets, also known asdescriptive statistics. It includes graphical and tabular summaries, as well as summary measuressuch as means, medians and standard deviations.
Uncertainty is a key aspect of most business problems. To deal with uncertainty, we need a basicunderstanding of probability. Chapter 2 covers basic rules of probability and in Chapter 3 wediscuss the important concept of probability distributions in some generality.
In Chapter 4 we discuss statistical inference (estimation), where the basic problem is to estimateone or more characteristics of a population. Since it is too expensive to obtain the populationinformation, we instead select a sample from the population and then use the information in thesample to infer the characteristics of the population.
In Chapter 5 we look at the topic of regression analysis which is used to study relationshipsbetween variables.
In Chapter 6 we study another type of decision making called decision analysis where costs andprofits are considered to be important. The problem is not whether to accept or reject a statementbut to select the best alternative from a list of several possible decisions. Usually no statisticaldata are available. Decision analysis is the study of how people make decisions, particularlywhen faced with imperfect information or uncertainty.
Chapter 7 deals with project management. Project management consists of planning projects,acquiring resources, scheduling activities and evaluating complete projects. Managers are re-sponsible for project management. They must know how long a specific project will take tofinish, what the critical tasks are, and very often, what the probability is of completing the projectwithin a given time span.
In Chapter 8 the subject of network models is discussed. Network models consist of nodes andarcs. Many real-world problems have a network structure or can be modelled in network form.These include problems in areas such as production, distribution, project planning, facilitieslocation, resource management and financial planning. The graphical network representationof problems provides a powerful visual and conceptual aid to indicate the relationship betweencomponents of a system.
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Contents
1 Descriptive statistics 1
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Data collection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.3 Sampling methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3.1 Simple random sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3.2 Stratified random sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.3.3 Systematic sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4 Presentation of data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.4.1 Types of data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.4.2 The frequency table and histogram . . . . . . . . . . . . . . . . . . . . . . . . 7
1.4.3 The pie chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.4.4 The cumulative frequency polygon . . . . . . . . . . . . . . . . . . . . . . . 12
1.4.5 The stem-and-leaf diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.5 Descriptive measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.5.1 Measures of locality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.5.1.1 The mean . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
1.5.1.2 The median . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
1.5.1.3 The mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
1.5.2 Measures of dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
1.5.2.1 The variance of a data set . . . . . . . . . . . . . . . . . . . . . . . . 20
1.5.2.2 The standard deviation of a data set . . . . . . . . . . . . . . . . . . 22
1.5.2.3 The quartile deviation . . . . . . . . . . . . . . . . . . . . . . . . . . 22
1.5.2.4 The coefficient of variation . . . . . . . . . . . . . . . . . . . . . . . 24
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1.6 The box-and-whiskers diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
1.7 Summary of descriptive measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
1.7.1 Measures of locality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
1.7.2 Measures of dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
1.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2 Probability concepts 33
2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
2.2 Classical probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
2.3 Some rules in probability theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
2.3.1 Conditional probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
2.3.2 Joint probabilities: multiplication law . . . . . . . . . . . . . . . . . . . . . . 42
2.4 Summary of basic probability concepts . . . . . . . . . . . . . . . . . . . . . . . . . . 44
2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3 Probability distributions 49
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
3.2 Random variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
3.3 Discrete random variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.3.1 The probability distribution of a discrete random variable . . . . . . . . . . 50
3.3.2 Expected value of a discrete probability distribution . . . . . . . . . . . . . . 52
3.3.3 Variance of a discrete probability distribution . . . . . . . . . . . . . . . . . 53
3.3.4 Discrete distribution function . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
3.4 Discrete random distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
3.4.1 The binomial distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
3.4.2 The Poisson distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
3.5 Continuous random variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
3.5.1 The probability distribution of a continuous random variable . . . . . . . . 64
3.5.2 Continuous distribution function . . . . . . . . . . . . . . . . . . . . . . . . . 65
3.6 Continuous distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
3.6.1 The normal distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
3.6.1.1 The Standard Normal distribution . . . . . . . . . . . . . . . . . . . 67
3.6.2 The exponential distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
3.6.3 The uniform distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
3.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
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4 Estimation 83
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
4.2 Types of estimators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
4.3 Point estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
4.3.1 Estimating the mean . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
4.3.2 Estimating the variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
4.3.3 Estimating proportions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
4.4 Interval estimators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
4.4.1 The standard error . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
4.4.2 Confidence intervals for means . . . . . . . . . . . . . . . . . . . . . . . . . . 85
4.4.3 Confidence intervals for proportions . . . . . . . . . . . . . . . . . . . . . . . 90
4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
5 Correlation and regression 93
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
5.2 Correlation analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
5.2.1 The scatter diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
5.2.2 The correlation coefficient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94
5.2.2.1 Pearson’s correlation coefficient . . . . . . . . . . . . . . . . . . . . 96
5.2.2.2 Spearman’s rank correlation coefficient . . . . . . . . . . . . . . . . 97
5.3 Simple linear regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
5.3.1 The estimated regression line . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
5.3.2 The method of least squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
5.3.3 Rules and assumptions underlying regression analysis . . . . . . . . . . . . 100
5.3.4 Residual plot analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
5.3.5 The coefficient of determination . . . . . . . . . . . . . . . . . . . . . . . . . 101
5.3.6 The F-test for overall significance . . . . . . . . . . . . . . . . . . . . . . . . . 103
5.3.7 Forecasting accuracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
5.4 Fitting nonlinear relationships . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
5.5 The use of spreadsheets in simple linear regression . . . . . . . . . . . . . . . . . . . 111
5.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
6 Decision analysis 117
6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
6.2 Structuring decision problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
6.2.1 The basic steps in decision making . . . . . . . . . . . . . . . . . . . . . . . . 118
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6.2.2 Payoff tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
6.2.3 Decision trees . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
6.3 Decision making without probabilities or under uncertainty . . . . . . . . . . . . . 120
6.3.1 The optimistic approach - maximax criterion . . . . . . . . . . . . . . . . . . 121
6.3.2 The conservative approach - maximim criterion . . . . . . . . . . . . . . . . 122
6.3.3 Minimax regret approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
6.4 Decision making with probabilities or under risk . . . . . . . . . . . . . . . . . . . . 124
6.4.1 The expected value approach . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
6.4.2 Decision trees and the expected value approach . . . . . . . . . . . . . . . . 125
6.4.3 Sensitivity analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
6.4.4 Expected value of perfect information . . . . . . . . . . . . . . . . . . . . . . 132
6.4.5 Decision analysis with sample information . . . . . . . . . . . . . . . . . . . 134
6.5 Utility and decision making . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
6.5.1 Obtaining utility values for payoffs . . . . . . . . . . . . . . . . . . . . . . . 141
6.5.2 Utility curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
6.5.3 The expected utility approach . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
6.5.4 Utility functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
6.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147
7 Project management 153
7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
7.2 PERT/CPM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
7.3 Project scheduling with certain activity durations − single time estimates . . . . . . 155
7.3.1 Define the project . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155
7.3.2 Project network diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
7.3.2.1 Conventions for constructing network diagrams . . . . . . . . . . 156
7.3.2.2 Drawing network diagrams . . . . . . . . . . . . . . . . . . . . . . 160
7.3.3 Network calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
7.3.3.1 Calculating early and late event times . . . . . . . . . . . . . . . . 164
7.3.3.2 The duration of the project . . . . . . . . . . . . . . . . . . . . . . . 168
7.3.3.3 The critical path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
7.3.3.4 Total float . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
7.3.3.5 Other measures of float . . . . . . . . . . . . . . . . . . . . . . . . . 170
7.3.4 Using linear programming to find a critical path . . . . . . . . . . . . . . . . 174
7.3.4.1 Formulating an LP model . . . . . . . . . . . . . . . . . . . . . . . . 175
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7.3.4.2 Using LINDO or LINGO to solve the LP model . . . . . . . . . . . 176
7.4 Project scheduling with uncertain activity durations – multiple time estimates . . . 179
7.4.1 Probability of project completion time . . . . . . . . . . . . . . . . . . . . . . 182
7.5 Project scheduling with time-cost tradeoffs . . . . . . . . . . . . . . . . . . . . . . . 183
7.5.1 Formulating an LP model to crash a project . . . . . . . . . . . . . . . . . . . 186
7.5.2 Using LINGO to solve the LP model for crashing a project . . . . . . . . . . 188
7.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
8 Network models 193
8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
8.2 The shortest-route problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193
8.2.1 A shortest-route algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
8.2.1.1 The labelling phase . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
8.2.1.2 Backtracking phase . . . . . . . . . . . . . . . . . . . . . . . . . . . 199
8.2.2 Formulating the shortest-route problem as an LP model . . . . . . . . . . . 209
8.2.3 Solving shortest-path problems with LINGO . . . . . . . . . . . . . . . . . . 211
8.3 The maximum-flow problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213
8.4 The minimum-spanning tree problem . . . . . . . . . . . . . . . . . . . . . . . . . . 218
8.4.1 Minimum-spanning tree algorithm 1 . . . . . . . . . . . . . . . . . . . . . . . 218
8.4.2 Minimum-spanning tree algorithm 2 . . . . . . . . . . . . . . . . . . . . . . . 223
8.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227
A Solutions to exercises 231
A.1 Chapter 1: Descriptive statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
A.2 Chapter 2: Probability concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236
A.3 Chapter 3: Probability distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . 241
A.4 Chapter 4: Estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246
A.5 Chapter 5: Correlation and regression . . . . . . . . . . . . . . . . . . . . . . . . . . 248
A.6 Chapter 6: Decision analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250
A.7 Chapter 7: Project management . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260
A.8 Chapter 8: Network models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273
B Statistical tables 281
B.1 The cumulative Poisson distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . 281
B.2 The standard normal distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283
B.3 Student’s t-distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284
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B.4 The F-distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285
B.5 The cumulative binomial distribution . . . . . . . . . . . . . . . . . . . . . . . . . . 289
C Bibliography 295
CHAPTER 1
Descriptive statistics
1.1 Introduction
N business can exist without the information given by numbers. Managing numbers isan important part of understanding and solving problems. Numbers provide a universal
language that can easily be understood and supply a description of some aspects of most problems.
The collection of numbers and other facts such as names, addresses, opinions etc. provides data.The data only becomes information when it informs the user. Statistics is about changing data toinformation by analysing the data.
The Statistical analysis can be divided into two main branches namely descriptive statistics andinferential statistics.
• Descriptive statistics deals with methods of organising, summarising and representing datain a convenient and informative way by means of tabulation, graphical representation andcalculation of descriptive measures.
• Inferential statistics is a body of methods used to draw conclusions or inferences aboutcharacteristics of populations based on sample data by using the descriptive measurescalculated.
In this chapter we discuss methods for collecting data and some descriptive statistics.
1.2 Data collection
Data can come from existing sources or may need to be collected. Technology makes it possibleto collect huge amounts of data. For example, retailers collect point-of-sale data on products andcustomers and credit agencies have all sorts of data on people who have or would like to obtaincredit.
In the case where data must be collected, data can be collected from a census where everybody orevery item of interest is included or from a sample from the population of interest.
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First we look at an example to clarify some definitions.
A tyre company, Radial, advertises that its XXX tyres, generally known as “Triple X”, will com-plete at least 65 000 km before one of the four tyres will no longer meet the minimum safetyrequirements. Several complaints, however, have been received that the tyres completed only50 000 km before the minimum requirements were exceeded.
Radial sells directly to the public and it is company policy to keep a record of each customer.During the two years that XXX tyres have been manufactured, 2 600 sets have been sold. Radialfeels that they just do not have the time, personnel or money to locate and question all 2 600 oftheir customers. They feel that if they question 100 customers, they will get a good idea of theactual situation. In other words, they will take a sample of 100 from a population of 2 600.
Consider the following definitions:
V:Any property or characteristic that can be measured or observed, is called a variable. A variablecan take on a range of different values. For example, the distance completed on a set of tyresdiffers for each customer and the observations therefore vary continually.
In Radial’s case, distance completed is a variable.
S :The sample unit is the item that is measured or counted with regard to the variable being studied.
Radial’s sample unit is a set of tyres measured for minimum safety requirements.
P:A population is the set of all the elements or items being studied.
In Radial’s case, the 2 600 sets of XXX tyres that have been sold form the population.
S:A sample is a representative group or a subset of the population.
The 100 sets of tyres that Radial will investigate, form the sample.
Note:What is very important is that the sample must always be representative of the population. Itshould be designed and administrated in such a way as to minimise the chance of being biased(sample outcome does not represent the population of interest). If the sample is likely to leaveout certain people or there is a relatively high level of non-response, it would probably not berepresentative of the population.
1.3 Sampling methods
1.3.1 Simple random sampling
A good sample requires that every item in the population has an equal and independent chanceof being included in the sample.
A simple random sample of n elements is a sample that is chosen in such a way that every combi-nation of n elements has an equal chance of being the sample selected.
One method of drawing a simple random sample is to allocate a number to each item in the
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population. A computer is used to generate a sequence of random numbers. These numbers arethen used to identify items in the population to be included in the sample.
Example 1.3.1
Printapage, a printing company, has 30 clients with outstanding balances (in Rand) as shownin Table 1.1.
Account Accountnumber Balance number Balance
1 25 16 02 0 17 1023 605 18 2154 1 010 19 4295 527 20 1976 34 21 1597 245 22 2798 59 23 1159 667 24 27
10 403 25 2711 918 26 29112 801 27 1613 227 28 014 0 29 40215 47 30 570
Table 1.1: Outstanding balances (in rand)
The following random numbers are available:
22; 17; 83; 57; 27; 54; 19; 51; 39; 59; 84 and 20.
Use these numbers to draw a random sample of 5 from the 30 customer accounts.
Solution
Since the total number of elements in the population is 30, an account number larger than30 will be of no use.
The sample units are the numbers of the accounts to be drawn. These are
22; 17; 27; 19 and 20.
The corresponding outstanding balances are
279; 102; 16; 429 and 197.
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Example 1.3.2
A political candidate wishes to determine the opinions of the voters in his ward. He decideson a sample size of 20. Using random numbers, he chooses 20 telephone numbers from thedirectory for a telephonic survey. Is this procedure correct? Give a reason for your answer.
Solution
Not all residents will have telephones, and the numbers of those who do have telephonesmay not all be included in the telephone directory. Such a sample can therefore not beconsidered random.
1.3.2 Stratified random sampling
Simple random sampling requires no prior (a priori) knowledge of the population and cantherefore be done with relatively little effort. It could, however, happen that all the elementsdrawn for the sample, are nearly homogeneous or alike. This may cause the conclusions aboutthe population to be biased.
If, however, you have prior information about the population, you could rule out this problemto some degree and consider more correct information about the population by making use ofstratified random sampling.
The population is divided into mutually exclusive sets or strata. This means that a specific elementmay only belong to one group or stratum. The strata must be chosen in such a way that therewill be large differences between the strata, but small differences between the elements withinthe same stratum.
Now simple random samples are taken from each stratum. The number of elements taken fromeach is often proportional to the size of that stratum.
Example 1.3.3
Divide Printapage’s 30 customers into three strata as follows:
Stratum Balance1 < 2002 200 − 6003 > 600
A proportional sample of size 12 must be drawn from the population. How would you doit?
Solution
Printapage’s customers are divided into three strata, as shown in Table 1.2.
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Stratum 1 Stratum 2 Stratum 3
Account Balance Account Balance Account Balancenumber (< R200) number (R200 – R600) number (> R600)
1 25 5 527 3 6052 0 7 245 4 1 0106 34 10 403 9 6678 59 13 227 11 918
14 0 18 215 12 80115 47 19 42916 0 22 27917 102 26 29120 197 29 40221 159 30 57023 11524 2725 2727 1628 0
Frequency = 15 Frequency = 10 Frequency = 5
Table 1.2: Dividing Printapage’s data into 3 strata
To draw a proportional sample of size 12, the following number of items must be drawnfrom each stratum:
15
30× 12 = 6 elements from stratum 1,
10
30× 12 = 4 elements from stratum 2, and
5
30× 12 = 2 elements from stratum 3.
Lastly, a simple random sample as describe in the previous section is drawn from eachstratum.
1.3.3 Systematic sampling
Systematic sampling starts at a randomly selected starting point in the population. Each subsequentk-th element is then chosen.
Example 1.3.4
A political candidate wishes to determine the opinions of the voters in his ward. He has alist of voters available. One way of obtaining a systematic sample would be to start withvoter number 6 and then select every tenth voter to complete a questionnaire.
What are the advantages and disadvantages of such a method?
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Solution
Advantage(s): Systematic sampling is convenient, especially when the size of the populationis not known.
Disadvantage(s): If the variable being considered is periodic in nature, systematic samplingcould produce misleading results. For example, if we were to estimate a shop’s sales usinga 1-in-7 systematic sampling design, it could happen that only sales figures for Saturdayswere selected. Sales would then be overestimated.
1.4 Presentation of data
Once data has been collected either by you or by someone else, the initial task is to obtain someoverall impression of the findings. This can be done by visually representing the data usingfrequency tables, charts and diagrams. But before we try to visualise the data, let us first considerthe different types of data one might get.
1.4.1 Types of data
There are two main groupings of data—qualitative and quantitative.
Qualitative data is characterised by categorical answers such as yes or no, male or female, etc.Quantitative data is characterised by numerical values.
Quantitative data can further be divided into two groups, discrete data and continuous data.
• Discrete data include everything that can be considered as a separate unit because of its na-ture, for example, number of units sold, number of consumers, number of job opportunities,etc., that is everything that you can count on your fingers.
• Continuous data are usually the result of a measurement and do not consist of fixed, isolatedpoints. There can be a whole range of values between any two values. Examples are length,mass, time and temperature measurements.
Example 1.4.5
Classify the data in each of the following questions:
(a) Do you own a TV set? Yes No
(b) How many TV sets do you own?
(c) How many kilometres did you drive on your set of Radial tyres?
(d) What was your electricity bill last month?
7 DSC2602
Solution
(a) Qualitative
(b) Quantitative – discrete
(c) Quantitative – continuous
(d) Quantitative – continuous
Let’s look at Radial’s data again.
Radial, the company introduced in Section 1.2, has taken a sample of 100 and is happy that it isrepresentative. The sample elements (in thousands) are shown in Table 1.3.
61 42 59 98 61 82 45 51 14 32
38 24 72 64 77 16 62 32 54 46
19 77 46 64 29 78 45 86 51 22
58 70 50 72 26 34 58 40 54 30
66 46 37 59 62 70 90 62 67 61
64 69 78 88 80 50 86 70 64 74
72 45 66 75 22 69 62 40 69 74
66 46 75 67 83 54 50 67 51 62
64 59 66 45 53 78 56 80 48 64
75 16 67 61 51 77 58 66 72 75
Table 1.3: Sample elements for Radial
We identified Radial’s data as quantitative and continuous. Perhaps if we could picture the data,we would be able to form a better idea of what is going on.
1.4.2 The frequency table and histogram
The histogram is one of the most common ways of visually representing data. It is a graphicalrepresentation of a frequency table. A frequency table is a table in which the data are grouped intointervals. To draw a histogram we must first set up a frequency table. The steps needed to set upa frequency table are as follows:
Step 1 Find the range (R) of the data, where
R = maximum value of data set −minimum value of data set.
Step 2 Decide on the number of intervals.If the number of intervals used are too few or too many, one cannot get a good idea ofthe distribution of the data. It is not always easy to decide how many intervals to use.R10
is a good number if R is large, but any number between 5 and 8 is acceptable. Donot use fewer than 5.
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Step 3 Determine the width of the intervals as
R
number of intervals.
The width must be a whole number – this will make it easier to determine the limits ofthe intervals.
Step 4 Determine the interval limits.The limits should be such that there is no doubt into which interval a value falls. Forexample, when we are working with Radial’s data, we cannot choose intervals such as
55 – 6565 – 75
Why not? Well, where would you place a value of 65?
For the mathematical manipulations that we will be doing with grouped data, it is alsonecessary that we do not work with intervals such as
55 – just smaller than 6565 – just smaller than 75
What does “just smaller than 65” mean?
The rule that we will use is to take the lower limit of the first interval as half a unitless than the minimum value, so that there can be no confusion as to which interval avalue belongs. The lower limit of the first class must be a value which is smaller thanthe minimum data value and the upper limit of each interval is the same as the lowerlimit of the succeeding interval.
Step 5 Tabulate the data.
Example 1.4.6
(a) Set up a frequency table for Radial’s data shown in Table 1.3.
(b) What percentage of customers were able to do 80 000 km or more on a set of tyres?
(c) What percentage of customers were able to do 46 000 km or less on a set of tyres?
(d) Draw a histogram of the data using the frequency table.
9 DSC2602
Solution
(a) Set up the frequency table following these steps:
Step 1 The range of the data:
The minimum value is 14 and the maximum value is 98. The range is thereforeR = 98 − 14 = 84.
Step 2 The number of intervals:
The number of intervals is calculated asR
10= 8,4. Therefore, use 8 intervals.
Step 3 The interval width:
The interval width is calculated asR
8=
84
8= 10,5. Therefore, use a width of 11.
Step 4 The interval limits:
The interval limits are determined as follows:The minimum value is 14 so the lower limit of the first interval will start at half aunit less than the minimum, which is 13,5. The upper limit of the first interval isdetermined by adding the width to the lower limit, that is 13,5 + 11 = 24,5.The first interval is therefore 13,5 – 24,5.
The second interval starts at 24,5 and also has a width of 11. Its upper limit istherefore 24,5 + 11 = 35,5.
The last interval starts at 90,5 and its upper limit is 101,5, which is well above thelargest element in the sample.
The intervals are:=11
︷ ︸︸ ︷
13,5 – 24,5
24,5 – 35,5
35,5 – 46,546,5 – 57,557,5 – 68,568,5 – 79,579,5 – 90,590,5 – 101,5
Step 5 Tabulate the data:
The only remaining thing to do is to group the data into the intervals.Now go back to the data set and consider the first four sample elements in the firstrow, which are 61, 38, 19 and 58.Our aim is to find in which one of the following intervals they belong:
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Interval13,5−24,5 19← Fit in the first interval because it is greater
than 13,5 and less than 24,5.24,5−35,535,5−46,5 38← Fit in the third interval because it is greater
than 35,5 and less than 46,5.46,5−57,557,5−68,5 61 and 58← Fit in the fifth interval because they
are greater than 57,5 and less than 68,568,5−79,579,5−90,590,5−101,5
Instead of writing 19; 38; 61 and 58 in their corresponding intervals, we representthem with a line, |, as follows
Interval13,5−24,5 |24,5−35,535,5−46,5 |46,5−57,557,5−68,5 ‖68,5−79,579,5−90,590,5−101,5
The fifth element in a group of lines is indicated by a line drawn across the group:|||| represents a group of five.
Note:The total number of elements falling into an interval is called the frequency.The complete frequency table for Radial is given in Table 1.4.
Interval Frequency
3,5−24,5 |||| || 724,5−35,5 |||| | 635,5−46,5 |||| |||| ||| 1346,5−57,5 |||| |||| ||| 1357,5−68,5 |||| |||| |||| |||| |||| |||| 3068,5−79,5 |||| |||| |||| |||| || 2279,5−90,5 |||| ||| 8
90,5−101,5 | 1
Total 100
Table 1.4: The frequency table for Radial
It is clear that the highest frequency occurs in the interval 57,5 – 68,5. This showsthat most of the customers were able to do between 57,5 and 68,5 thousand kilo-metres on a set of tyres.
11 DSC2602
(b) The intervals 79,5 – 90,5 and 90,5 – 101,5 represent the number of customers who wereable to drive 80 000 km or more on a set of tyres.
The total number is thus the sum of the frequencies in these intervals, namely 8+1 = 9.
The percentage of customers who were able to do 80 000 km or more on a set of tyres is
9
100× 100% = 9%.
Note: The fraction9
100is called the relative frequency.
(c) The first three intervals account for the customers who were able to drive 46 000 km orless. The sum of the frequencies in these intervals, namely 7+ 6+ 13 = 26 is thus equalto the total number of customers. The percentage of customers who were only able todo 46 000 km or less on a set of tyres is
26
100× 100% = 26%.
(d) Now we can graphically represent the frequency table by drawing the interval lengthson a horizontal axis and the frequencies on a vertical axis. This is called a histogram.
The histogram for Radial is given in Figure 1.1.
0
5
10
15
20
25
30
35
13,5 24,5 35,5 46,5 57,5 68,5 79,5 90,5 101,5
Distance
Fre
qu
ency
Figure 1.1: Histogram for Radial
(Notice that the horizontal axis starts at 0 and that the zigzag line is there to break theline in order to prevent a huge space from appearing on the left of the actual graph.)
1.4.3 The pie chart
Another way of representing data is by means of a pie chart. A pie chart is drawn as a circle andthe “slices” of the circle represent the relative frequencies expressed as a percentage. It is oftendifficult to draw a pie chart by hand. In Radial’s case one needs to divide the circle into 100 equalslices – not an easy task! The pie chart for Radial is more or less as shown in Figure 1.2.
DSC2602 12
3 0 %
2 2 %
8 %7 %6 %
1 3 %
1 3 %
5 7 , 5 - 6 8 , 5
6 8 , 5 - 7 9 , 5
7 9 , 5 - 9 0 , 51 %
9 0 , 5 - 1 0 1 , 51 3 , 5 - 2 4 , 52 4 , 5 - 3 5 , 5
3 5 , 5 - 4 6 , 5
4 6 , 5 - 5 7 , 5
Figure 1.2: Pie chart representing Radial’s data
1.4.4 The cumulative frequency polygon
We calculated in Example 1.4.6 that 26% of the customers was only able to do 46 000 km or lesson a set of tyres. Such information can be presented graphically if we first obtain the “cumulativeless than” table. Such a table is set up from the frequency table, setting the upper limits to “lessthan . . .”.
The cumulative frequency table for Radial is given in Table 1.5.
Upper limit Frequency Cumulative frequency
< 24,5 7 7< 35,5 6 13 (7 + 6 = 13)< 46,5 13 26 (7 + 6 + 13 = 26)< 57,5 13 39 (7 + 6 + 13 + 13 = 39)< 68,5 30 69 (7 + 6 + 13 + 13 + 30 = 69)< 79,5 22 91 etc.< 90,5 8 99< 101,5 1 100
Table 1.5: The cumulative frequency table for Radial
You have probably realised that cumulative means “added up”.
This information can now be represented by a cumulative frequency polygon as shown in Figure 1.3.
13 DSC2602
b
b
b
b
b
b
b b
10
20
30
40
50
60
70
80
90
100
24,5 35,5 46,5 57,5 68,5 79,5 90,5 101,5
Cu
mu
lati
ve
freq
uen
cy
0
Distance
Figure 1.3: The cumulative frequency polygon for Radial
1.4.5 The stem-and-leaf diagram
The stem-and-leaf diagram is also a useful diagram and is easy to set up. The first step is to decidehow to separate each observation into two parts - the stem and the leaf.
Radial’s data can be separated in such a way that the first digit of each number is the stem andthe second digit is the leaf.
First we determine the biggest and smallest numbers in the data set and separate them into astem and a leaf.
The smallest number in the data, 14, has stem 1 and leaf 4.
The largest number, 98, has stem 9 and leaf 8.
Next we fill in the rest of the data.
All the other numbers lie between these two. We can therefore set up the stem from 1 to 9, withthe second digit of each number being written next to its stem, as shown in Table 1.6.
Stem Leaf Frequency
1 9 6 6 4 42 4 9 6 2 2 53 8 7 4 2 2 0 64 2 6 5 6 6 5 5 5 0 0 8 6 125 8 9 9 0 9 3 1 0 4 8 0 6 8 1 4 1 4 1 186 1 6 4 6 4 9 6 6 7 4 4 7 1 1 2 9 2 2 2 7 6 7 4 9 1 2 4 277 2 5 7 0 2 8 5 2 5 7 8 0 8 7 0 2 4 4 5 198 8 0 3 2 6 6 0 79 8 0 2
Table 1.6: Stem-and-leaf diagram for Radial (unsorted)
But every stem’s leaf MUST be in ascending order (from the smallest value to the largest). Radial’ssorted stem-and-leaf diagram is given in Table 1.7.
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Stem Leaf Frequency
1 4 6 6 9 42 2 2 4 6 9 53 0 2 2 4 7 8 64 0 0 2 5 5 5 5 6 6 6 6 8 125 0 0 0 1 1 1 1 3 4 4 4 6 8 8 8 9 9 9 186 1 1 1 1 2 2 2 2 2 4 4 4 4 4 4 6 6 6 6 6 7 7 7 7 9 9 9 277 0 0 0 2 2 2 2 4 4 5 5 5 5 7 7 7 8 8 8 198 0 0 2 3 6 6 8 79 0 8 2
Table 1.7: Stem-and-leaf diagram for Radial
Now turn the page on its side. It is easy to see that most of the customers drove between 60 000and 70 000 thousand kilometres on a set of tyres.
1.5 Descriptive measures
The presentation of charts and diagrams can be regarded as the first step in analysing data andis not sufficient for most purposes. They provide an overall picture of the data but give onlyan approximate indication of specific properties such as midpoint and spread of data. Properanalysis requires a summary of the data in the form of descriptive statistical measures.
Descriptive measures are single numerical values that indicate the shape or distribution of thedata set. There are descriptive measures of location, spread, symmetry and kurtosis.
1.5.1 Measures of locality
A measure of location or position gives an indication of the “midpoint” or general size of thedistribution. Examples are the mean, median and mode of a data set.
1.5.1.1 The mean
Radial advertises that its XXX tyres will last for at least 65 000 km before one of the four tyres willno longer meet the minimum safety requirements. What is the mean number of kilometres thatcan be driven on a set of XXX tyres? Radial has only the sample of 100 observations available forestimating the mean. If we consider the sample as being representative of the population, we canuse the sample mean as an estimator of the population mean.
To obtain the sample mean we add up all the observations and divide the result by the number ofobservations. (This is called the arithmetic mean.)
If we add up all the observations in Radial’s sample and divide the sum by the number ofobservations, we get
5 828
100= 58,28.
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We can therefore expect a set of tyres to last 58,28 × 1 000 = 58 280 km on average.
The formula for the mean is
x =1
n
n∑
i=1
xi
where
• x (read as x-bar) is the generally accepted symbol for the arithmetic mean,
• n is the number of observations,
•∑
is the Greek letter for S and means “sum”,
• xi represents the i-th observation, and
•n∑
i=1
xi is just another way of writing x1 + x2 + . . . + xn.
Note: You may enter the data into the statistics mode of your calculator and find the value ofthe mean by pressing a button. This is much faster than doing the calculation by hand. See themanual of your calculator.
When the “raw data”, that is the values in the original data set, are available, it is easy to calculatethe mean. Sometimes, however, the data is given in the form of a frequency table and the actualvalues are not known. Let’s look at Radial’s data again.
Assume that Radial’s sample data is available in the following form only:
(Distance in 1 000 km)
Interval Frequency ( fi)
13,5 – 24,5 724,5 – 35,5 635,5 – 46,5 1346,5 – 57,5 1357,5 – 68,5 3068,5 – 79,5 2279,5 – 90,5 890,5 – 101,5 1
—100
We do not know what the actual values in each interval are. For computational purposes wemake the following assumption:
All values in an interval are equal to the middle value of the interval.
The middle value is calculated by adding the lower and the upper limits of the interval anddividing the result by two.
The middle value of the first interval is:
13,5 + 24,5
2= 19.
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We thus assume that all the observations in the first interval are equal to 19.
The contribution of these seven observations to the grand total is therefore:
7 × 19 = 133.
The formula for the mean of a frequency distribution is
x =
∑ki=1 fixi
∑ki=1 fi
where
fi = the frequency for the i-th interval,xi = the middle value of the i-th interval, andk = number of intervals.
Note: Data in a frequency table are often referred to as grouped data.
Example 1.5.7
Calculate the mean distance travelled on a set of XXX tyres using the frequency distributionin Table 1.4.
Solution
In Table 1.8 the frequency, fi, the middle value, xi, and the product of these are given foreach interval.
Interval fi xi fixi
13,5 − 24,5 7 19 13324,5 − 35,5 6 30 18035,5 − 46,5 13 41 53346,5 − 57,5 13 52 67657,5 − 68,5 30 63 1 89068,5 − 79,5 22 74 1 62879,5 − 90,5 8 85 68090,5 − 101,5 1 96 96
100 5 816
Table 1.8: Calculating the mean from the frequency table
The mean is calculated as
x =
∑ki=1 fixi
∑ki=1 fi
=5 816
100= 58,16.
The mean distance is therefore 58,16 × 1 000 = 58 160 km.
Note: The intervals of the frequency distribution are all of equal width, that is 11. After youhave calculated the middle value of the first interval, the successive middle values can beobtained by adding 11 to the previous middle value.
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The calculation based on the above assumption resulted in a total that is different from the actualtotal. A mean calculated in this way will differ from the actual mean. For example, if theobservations in the interval 79,5 – 90,5 are:
88; 80; 83; 82; 90; 86; 86; 80,
their sum is 675. If we use the middle value and the frequency of each interval and then calculatethe contribution to the total of the observations in the interval 79,5 − 90,5, their sum is 680.
Interval Frequency Middle valuefi xi fixi
79,5 – 90,5 8 85 8 × 85 = 680
In our example the mean obtained using the original data values is 58 280 km, while the meanobtained using the frequency distribution is 58 160 km.
The mean is the measure of locality that is used most often. It can, however, be misleading. Forexample, if we calculate the mean of 2; 3; 5; 71, we see that the mean will be
x =2 + 3 + 5 + 71
4= 20,25.
When a data set has a mean of 20, one intuitively expects most of the values to lie in the vicinityof 20. In this instance, however, most of the values are less than 6, while one value is an outlierof 71!
The mean is rather sensitive to outliers and can often be misleading. On its own, without anyadditional information, it can often lead to incorrect conclusions.
Another disadvantage of the mean is that it is a difficult task to calculate the mean for an openfrequency distribution, as the following example will illustrate.
Example 1.5.8
Table 1.9 gives the property value distribution for 15 420 000 ratepayers in the SteelcityMetropolitan Substructure.
Property value Frequency(in Rands) (in thousands)
Less than 25 000 5 51825 000 – 50 000 3 37250 000 – 100 000 4 010
100 000 – 200 000 1 846200 000 – 500 000 576500 000 – 1 000 000 76
More than 1 000 000 22
Table 1.9: Property value distribution
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What assumptions will we have to make about the middle values of the first and the lastinterval? What is an acceptable lower limit for the first interval? What is an acceptableupper limit for the last interval?
If we have access to the original data, we may be able to make a good guess, but thisis not always possible. The use of the mean is therefore restricted when there are opendistributions.
A big advantage of the arithmetic mean is that it uses all the available data. Later we will see thatthis is not the case for the other measures of locality. Since the mean can be calculated exactly, itforms the basis for many advanced analyses and is not only descriptive in nature.
1.5.1.2 The median
Since the mean is sensitive to extreme values (outliers), and may often result in misleadingconclusions, the median is often preferred as a measure of locality. The median is the value thatdivides an ordered data set into two equal parts.
If the data set is sorted in ascending order, 50% of the data values will lie below, or to the left, ofthe median, and 50% will lie above, or to the right of the median.
The median is determined as follows:
If a data set of size n is sorted in ascending sequence, then the median (me) is the
n + 1
2-th value of the data set.
Example 1.5.9
Determine the median of the following data sets:
(a) 6, 9, 12, 12, 13, 15, 18, 24, 27
(b) 2, 3, 5, 71
Solution
(a) The data set is arranged in ascending order and n = 9.
The median (me) is the9 + 1
2= 5-th value, that is the median is 13.
(b) The data set is arranged in ascending order and n = 4.
The median (me) is the4 + 1
2= 2 1
2-th value.
The 2 12-th value is a value between the second and third values, that is halfway between
3 and 5 or3 + 5
2= 4. The median is therefore 4 and we can say that 50% of the data lie
to the left of 4 and 50% to the right of 4.
The median may also be calculated for a frequency table. The cumulative frequency table is usedto identify the median interval.
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Let’s use Radial’s data once again, as shown in Table 1.10.
Interval Frequency ( fi) Cumulative frequency
13,5 − 24,5 7 724,5 − 35,5 6 1335,5 − 46,5 13 2646,5 − 57,5 13 3957,5 − 68,5 30 6968,5 − 79,5 22 9179,5 − 90,5 8 9990,5 − 101,5 1 100
Table 1.10: Cumulative frequency table for Radial
To identify the median interval, find the interval within which the
n + 1
2=
100 + 1
2= 50,5th
value occurs. The median interval is therefore 57,5 − 68,5.
The biggest advantage of the median is that open intervals pose no problem and it is not affectedby extreme values. However, it ignores the largest part of the data and cannot be manipulatedmathematically.
1.5.1.3 The mode
The mode of a data set is the value that occurs most often.
Consider the following example: A survey was conducted amongst 1 000 married couples whowere married eight years ago. Table 1.11 gives the frequency table of the data obtained.
Number of children Number ofper couple couples
xi fi
0 364 Highest frequency1 3622 2263 444 4
Table 1.11: Frequency table for number of children per couple
It is clear that most couples have no children. The mode is therefore zero.
The mode, however, is not a good measure of locality. Sometimes there is no value that occursmore than any other value, or there is more than one value with the same maximum number ofoccurrences. In addition, the mode has the same drawbacks as the median. The only thing infavour of the mode is that it is easy to understand. For grouped data the modal interval is theinterval with the highest frequency.
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1.5.2 Measures of dispersion
1.5.2.1 The variance of a data set
In the previous section, we considered the problems that occur when working with a measureof locality only. Even though the arithmetic mean, uses all the values in a data set, it does notgive much information about what the data set really looks like. We also need information on thespread of the data around the mean.
Consider the data set5; 3; 8; 4; 1; 5; 0; 6
with a mean of
x =32
8= 4.
Let’s plot the data points around the mean:
. . . . . .. . . . . . . . . . . .
. . .. . . . . . . . .
. . . . . . . . . . . .. . .. . .
81
50
6
35
4
x = 4
Now calculate the distance from x = 4 to each value and then calculate the mean of these distances.
1 + 1 + 4 + 0 + 3 + 1 + 4 + 2 = 16 and16
8= 2.
When we work with Radial’s sample it may be quite difficult to calculate the mean distance –how long do you think it would take using 100 values?
An alternative is to calculate the deviation from the mean for each observation, that is x − x:
1; −1; 4; 0; −3; 1; −4; 2.
When these are added, you get 0, which tells you nothing! However, the number crunchers ofthe old days did not become discouraged, and came up with the clever idea of using (x − x), andsquaring it. The square of any value is always a positive number.
The mean of the squared deviations is called the variance. The positive square root of the varianceis called the standard deviation, and we will use this measurement to give an indication of thespread of data around the mean.
The variance of a sample is defined as
s2 =
∑ni=1(xi − x)2
n − 1.
Notice that we divide by n − 1 and not by n. The reason for this is that the sample variance (s2)is used to estimate the population variance (σ2). If we were to divide by n, it would give an
21 DSC2602
underestimation of the population variance. Division by n − 1 therefore gives a better estimator.Calculators normally can calculate s2 and σ2. Make sure which one you should use on yourcalculator. The squared deviations of our sample are calculated in Table 1.12.
xi (xi − x) (xi − x)2
5 1 13 −1 18 4 164 0 01 −3 95 1 10 −4 166 2 4
48
Table 1.12: Calculations to find the variance and the standard deviation
The variance s2 =48
(8 − 1)= 6,86 and the standard deviation s =
√6,86 = 2,62.
An alternative formula for computing the variance is
s2 =n(Σn
i=1x2
i) − (Σn
i=1xi)
2
n(n − 1)
Note: You may enter the data into the statistics mode of your calculator and find the value of thestandard deviation by pressing a button. This is much faster than doing the calculation by hand.See the manual of your calculator.
The variance can also be calculated for grouped data.
For a frequency table the sample variance is defined as follows:
s2 =
∑ki=1 fix
2i− nx2
n − 1where
x =
∑ki=1 fixi
∑ki=1 fi
=
∑ki=1 fixi
n;
xi = middle value of the i-th interval, and
n =
k∑
i=1
fi.
Let’s look at Radial’s data again.
The variance of Radial’s frequency table is calculated in Table 1.13. (Remember that we havealready calculated the mean as x = 58,16.)
The variance s2 =371 490 − 100 × 58,162
100 − 1=
33 231,44
99= 335,67.
DSC2602 22
Interval fi xi fix2i
13,5 − 24,5 7 19 2 52724,5 − 35,5 6 30 5 40035,5 − 46,5 13 41 21 85346,5 − 57,5 13 52 35 15257,5 − 68,5 30 63 119 07068,5 − 79,5 22 74 120 47279,5 − 90,5 8 85 57 80090,5 − 101,5 1 96 9 216
100 371 490
Table 1.13: Calculations for the variance of Radial’s frequency table
1.5.2.2 The standard deviation of a data set
The standard deviation is defined as the square root of the variance.
The standard deviation of a sample is defined as
s =√
s2
=
√∑ni=1(xi − x)2
n − 1.
But what does this tell us? It tells us how far away the observations are from the mean. Thelarger the standard deviation, the further away the data points are from the mean. The followingschematic representation shows how many of the data points lie between one standard deviationto the left and to the right of the mean, and between two standard deviations to the left and tothe right of the mean.
? ?? ?
−2s −s x +s +2s
68%
95%
The standard deviation plays a important role in inferential statistics – that is, the field wherethe problem of making scientifically based conclusions about populations, using sample data, isconsidered.
1.5.2.3 The quartile deviation
The median is that value which separates a sorted data set into two equal parts.
23 DSC2602
me
50% 50%
If we divide a sorted data set into four equal parts we get the following four quartiles:
meq2q1 q3
25% 25% 25% 25%
q1 represents the value that indicates the end of the first 25% of the data values; q2 representsthe value that indicates the end of the second 25% (or the value which divides the data set intotwo equal parts, that is the median); and q3 is the value indicating the end of the third 25%. Themiddle 50% of the data lies between q1 and q3.
The quartile deviation is
qD =q3 − q1
2
and is the measurement of the dispersion of the data around the median.
As with the median, the quartile deviation does not use all the observations. It ignores outlierssince the top 25% and the bottom 25% of the data values are not taken into account.
Example 1.5.10
The purchasing manager of a group of clothing shops has recorded the following 15 obser-vations on the number of days that pass between reordering items from a new range of children’sclothing.
Reordering intervals(in days)
17 18 26 15 1726 23 29 28 1822 5 12 23 22
Calculate and interpret the quartile deviation of the reordering intervals.
Solution
The value of the median or q2 is the value of the 12(n+ 1)th observation in a ranked data set.
Similarly, the value of q1 is the value of the 14(n + 1)th observation and the value of q3 is the
value of the 34(n + 1)th observation in an ordered data set.
(An ordered data set is a data set that is arranged in ascending order.)
The ordered data set is:
5; 12; 15; 17; 17; 18; 18; 22; 22; 23; 23; 26; 26; 28; 29.
DSC2602 24
The position of the first quartile q1 is:
1
4(15 + 1) = 4.
Therefore q1 is the 4th observation in the ordered data set and hence q1 = 17.
The position of the third quartile q3 is:
3
4(15 + 1) = 12.
Therefore q3 is the 12th observation in the ordered data set and hence q3 = 26.
The quartile deviation is:
qD =q3 − q1
2
=26 − 17
2= 4,5.
It can be expected that 50% of all observations will fall within 4,5 days on both sides of themedian, that is between 22 − 4,5 = 17,5 ≈ 18 and 22 + 4,5 = 26,5 ≈ 26 days.
It can also be said that 25% of the observations fall within 4,5 days to the left of the median,and 25% of the observations fall within 4,5 days to the right of the median.
1.5.2.4 The coefficient of variation
A stockbroker wishes to compare two funds, Fund A and Fund B. He has available the annualreturn rates of the two funds for the past 10 years, and calculates the mean and variance of eachas shown in Table 1.14.
Fund x s2
A 16 280,34B 12 99,37
Table 1.14: Mean and variance of funds A and B
Since the variance for Fund A is higher than that of Fund B, we can conclude that the riskassociated with Fund A is higher than that for Fund B.
Fund A, however, displayed a higher mean return than Fund B over the past ten years. Thisintuitively feels right - an investment having a higher associated risk should have a higher meanrate of return.
But what if Fund A had a mean rate of return of 21% with the same variance? Would we still beable to say that Fund A is subject to higher fluctuation than Fund B?
Variances can only be compared when the two means are close together.
25 DSC2602
To obtain a more reliable comparison we need a measure which displays the relative variability.The coefficient of variation is such a measure which cancels the differences between means.
The coefficient of variation is:
cv =s
x=
standard deviation
mean.
The coefficient of variation with the largest value has the highest relative variability.
For Fund A with a mean of 16 and a standard deviation of√
280,34 = 16,74,
cv(A) =16,74
16= 1,05.
For Fund B with mean 12 and a standard deviation of√
99,37 = 9,97,
cv(B) =9,97
12= 0,83.
This indicates that the observations of Fund A display a higher degree of variability than thoseof Fund B, which is the same conclusion as we reached before. However, if the mean of Fund Ais 21 and its standard deviation is 16,74, then
cv(A) =16,74
21= 0,80.
Fund A now displays a lower degree of variability, or variation, than Fund B.
Note: Sometimes the coefficient of variation is multiplied by 100 to present it as a percentage.
1.6 The box-and-whiskers diagram
A box-and-whiskers diagram is a graphical way to visualise the dispersion or spread of a dataset.
It uses only five observations from the dataset, namely the smallest value, the first quartile, themedian, the third quartile and the largest value.
Go back to the discussion on the quartile deviation in Example 1.5.10 for the information on thereordering intervals for children’s clothing. From the ranked data set we see that 5 is the smallestvalue and 29 is the largest. We also found that q1 = 17, me = 22 and q3 = 26.
To construct the diagram, draw a pair of coordinate axes with the vertical axis representing therange of observations. Draw a “box” from the first to the third quartile with the median in themiddle. Then draw lines from the box to the extreme points, as shown in Figure 1.4.
A box-and-whiskers diagram is especially useful when comparing two or more distributions.With the aid of a computer, even very large data sets can be analysed without too much effort.
DSC2602 26
-
6
. . . . . . . . .
. . . . . . . . .
. . . . . .
. . . . . .
. . . . . .
0
5
10
15
20
25
30
Smallest value = 5
q1 = 17
me = 22
q3 = 26
Largest value = 29
Reo
rder
ing
inte
rval
s
Figure 1.4: Box-and-whiskers diagram for reordering intervals of children’s clothing
Example 1.6.11
Rainfall measurements (in mm) in regions A, B and C yield the information given in Ta-ble 1.15.
A B C
Smallest value 0 200 0q1 50 300 200me 200 350 800q3 300 550 900Largest value 500 1 000 950
Table 1.15: Rainfall data
Draw a box-and-whiskers diagram of the rainfall measurements in regions A, B and C andinterpret the graphs.
27 DSC2602
Solution
The box-and-whiskers diagram is given in Figure 1.5.
-
6
0
100
200
300
400
500
600
700
800
900
1000
CA B
Rai
nfa
ll(i
nm
m)
Regions
Figure 1.5: Box-and-whisker diagram for the rainfall data
Keep in mind that the diagram is a visual representation of some information of a dataset.It is a quick way of examining one or more datasets graphically. You cannot make anysinificant statistical conclusions.
Looking at the diagram it is clear that region C has the highest median rainfall but also thelargest interquartile spread – 700 versus the 250 of A and B. The smaller interquartile spreadof A and B indicate a more reliable expectation of what the rainfall is going to be.
DSC2602 28
1.7 Summary of descriptive measures
1.7.1 Measures of locality
• Mean
Raw data x =1
n
n∑
i=1
xi
Frequency table data x =
∑ki=1 fixi
∑ki=1 fi
• Median me =n + 1
2th value in sorted data set.
• Mode Value in data set that occurs the most.
1.7.2 Measures of dispersion
• Variance (sample)
Raw data s2 =
∑ni=1(xi − x)2
n − 1
Frequency table data s2 =
∑ki=1 fix
2i− nx2
n − 1
• Variance (population) σ2 =
n∑
i=1
(xi − x)2
n
• Standard deviation s =√
s2
=
√∑ni=1(xi − x)2
n − 1
• First quartile q1 =14
(n + 1)th value in sorted data set
• Second quartile q2 = me
• Third quartile q3 =34(n + 1)th value in sorted data set
• Quartile deviation qD =q3 − q1
2
• Coefficient of variation cv =s
x=
standard deviation
mean
29 DSC2602
1.8 Exercises
1. As the manager of an insurance claims division, you have to set up performance levels.You have asked 30 of your experienced claims processing personnel to record the numberof claims they processed during a specific week.
They collected the data set given in Table 1.16.
31 30 28 33 35 3737 36 38 38 39 3638 34 39 31 30 3440 41 40 41 48 4546 44 39 34 40 42
Table 1.16: Claims processed by 30 claims processing personnel in a week
(a) Set up a frequency table for the data and draw a histogram of the claims processed.
(b) Set up a cumulative frequency table and a stem-and-leaf diagram for the claims data.
(c) During the past month, an experienced worker has processed only 26 claims per week.Do you sense a problem? Give reasons for your answer.
(d) Information obtained from a competitor indicates that for 50% of the time an experi-enced worker can process 36 to 39 claims per week. What is happening here?
(e) The manager decides to transfer some of the workers to other divisions if he finds thatless than 36 claims are processed per week for half of the time. Advise him.
2. Calculate the mean, median and the mode for the following data:
190; 104; 135; 314; 179; 175; 170; 146; 127; 131.
3. Consider the following frequency table:
Interval Frequency
27,5 – 31,5 531,5 – 35,5 535,5 – 39,5 1039,5 – 43,5 643,5 – 47,5 347,5 – 51,5 1
—30
(a) Calculate the mean.
(b) Identify the median interval.
(c) Identify the modal interval.
DSC2602 30
4. The selling price (in cents) of a specific share over 14 working days is as follows:
1 630; 1 550; 1 430; 1 440; 1 390; 1 400; 1 480; 1 490; 1 410; 1 905; 1 540; 1 890; 1 900; 1 900.
Calculate the standard deviation of the selling price.
5. A company that markets the seed of agricultural crops, tested three new wheat varieties, A,B and C, and obtained the following results: Maximum yield is not the company’s highest
A B C
Mean number of bushels per hectare 88 56 100Standard deviation 16 15 25
priority, but rather consistent yields. Which variety of seed would you recommend?
6. In one month the total costs (to the nearest ten rand) of the calls made by 23 male and 23female cellphone owners were:
Male owners:170 170 140 160 150 240 120 200 170 170 130 210150 140 140 200 210 90 150 220 190 270 140
Female owners:140 50 150 60 170 100 220 100 120 170 130 240
70 270 330 160 300 90 150 70 330 280 210
Draw a box-and-whiskers diagram to present the data graphically. Interpret the graph.
31 DSC2602
CHECKLISTYou should now be able to
⊲ describe how the following are done:
– simple random sampling,
– stratified sampling, and
– systematic sampling;
⊲ represent data by means of
– a frequency table,
– a histogram,
– a pie chart,
– a cumulative frequency table,
– a stem-and-leaf diagram, and
– a box-and-whisker diagram;
⊲ calculate the measures of locality for a data set, that is
– the mean,
– the median, and
– the mode;
⊲ calculate the measures of despersion for a data set, that is
– the variance,
– the standard deviation,
– the quartile deviation, and
– the coefficient of variation.
DSC2602 32
CHAPTER 2
Probability concepts
2.1 Introduction
E of us has some intuitive notion of what probability is. Everyday conversation is full ofreferences to it:
“There is a 60% chance that it will rain today.”“There is a 50-50 chance that the stock market will crash next month.”“The odds on winning the World Cup are high.”
What do these vague statements mean? What are the ideas behind them?
A probability is a numerical statement about the likelihood that an event will occur.
2.2 Classical probability
Since probability theory had its origin in games of chance, it is not surprising that the first methoddeveloped for measuring probability was used by gamblers.The so-called classical concept of probability was defined as the number of successes divided by thetotal number of possible outcomes:
P =number of successes
total number of possible outcomes.
Let me demonstrate this with the following example.
Example 2.2.1
A card is drawn at random from a pack of 52 playing cards. Determine the probability that:
33
DSC2602 34
(a) a red card is drawn;
(b) a spade is drawn;
(c) an ace is drawn;
(d) the jack of hearts is drawn.
Solution
Before determining the probabilities, let us use the example to illustrate some concepts.
A random experiment can be defined as the process of obtaining information through obser-vation or measurement. When we perform an experiment, we observe an outcome that iswell defined. On any single repetition of an experiment one, and only one, of the possibleoutcomes will occur.
In our example the experiment is to draw a card.
An outcome (sample point) is the most basic possible observation of an experiment. It cannotbe broken down into simpler occurrences. Any one of the 52 cards can be drawn. The samplespace for a random experiment consists of the entire set of possible outcomes. S denotes thesample space. The sample space for our sample is
S =
{
all 13 diamonds; all 13 hearts;
all 13 spades; all 13 clubs
}
S consists of 52 possible outcomes.
We define an event as a set of outcomes (or sample points) that possess a specific, commoncharacteristic. It is possible that an event could be satisfied by only one outcome; or anevent could be satisfied by more than one outcome; or it could be satisfied by no outcomes,in which case it is known as an impossible event.
For the first probability asked, that is a red card drawn, the event would be
A = {all 13 diamonds;all 13 hearts}
A consists of 26 possible outcomes.
The basic properties of probabilities are as follows:
• The number of successes cannot be less than zero. Therefore the minimum value ofthe probability P is 0.
• The number of successes cannot be more than the number of possible outcomes. There-fore the maximum value of the probability P is 1.
• The probability P can take on any value between 0 and 1. Therefore 0 ≤ P ≤ 1.
Let us now determine the probabilities asked in the example of the pack of playing cards.
35 DSC2602
A pack of cards consists of four suits of cards – diamonds, hearts, spades and clubs. In apack there are 13 diamonds, 13 hearts, 13 spades and 13 clubs – 52 cards in total. Of these26 are red and 26 are black cards – diamonds and hearts are red and spades and clubs areblack. In any one of the four suits, there are nine numbered cards (numbered 1; 2; 3; 4; 5; 6;7; 8; 9), a jack, a queen, a king and an ace.
If a card is drawn from the pack it can be any of the above.
(a) Using the classical definition we can write:
P(red card) =number of red cards
total number of cards=
26
52.
(b) There are 13 spades in a pack, therefore:
P(spade) =number of spades
total number of cards=
13
52.
(c) Each of the four suits contains one ace. There are four aces in a pack, therefore:
P(ace) =number of aces
total number of cards=
4
52.
(d) Each of the four types suits contains one jack. There is only one jack of hearts in a pack,therefore:
P(jack of hearts) =1
52.
Another example in which the classical definition works perfectly is the following:
Example 2.2.2
A survey of 350 households gave the following data on the number of children under 16living in each household:
Number of children under 16 Frequency
0 1851 512 90
3 or more 24—–350
Determine the probability that a household selected at random has:
(a) no children;
(b) three or more children;
(c) at least one child;
(d) no more than two children;
(e) at most one child.
DSC2602 36
Before we calculate the probabilities, let us first discuss the meanings of the words.
At least two: This means that two is the minimum value and if we say at leasttwo children, it means two or three or four or ... children.At most two: This means that two is the maximum value. At most two children means nochild or one child or two children.No more than two: This means that two is the maximum number, that is, two or one orzero children.Less than two: This means that two is not included and we are only interested in the valuessmaller than two, that is, zero or one.More than two: This means that two is not included and we are only interested in thevalues larger than two, that is, three, four, five, etc.
Solution
(a) The probability of no children is:
P(no children) = P(number of children = 0)
=frequency of 0 children
total number of children
=185
350.
(b) The probability of three or more children is:
P(three children or more) =24
350.
(c) The probability of at least one child is:
P(at least one child) = P(one child or more)
= P(one child or two children or
three children or more)
=51 + 90 + 24
350
=165
350.
(d) The probability of no more than two children is:
P(no more than two children) = P(two children or less)
= P(two children or one child or none)
=90 + 51 + 185
350
=326
350.
37 DSC2602
(e) The probability of at most one child is:
P(at most one child) = P(one child or less)
= P(one child or none)
=51 + 185
350
=236
350.
Example 2.2.3
A commercial bank categorised the loans that were made to corporate borrowers and thathad matured during the past year as excellent, good, fair, poor or bad with respect tocollection experience. The information collected is used as probabilities for the collection ofnext year’s loans. The following table lists the number of loans in each category:
Collection experience Number of loans
excellent 580good 418fair 140poor 42bad 20
(a) What is the probability that the collection experience of next year’s loan will be fair?
(b) If a profitable loan is defined as an excellent, good or fair loan, what is the probabilityof a profitable loan?
Solution
(a) The total number of loans is 1 200. Remember that probability is defined as:
P =number of successes
total number of possible outcomes.
The probability that collection experience is fair is:
P(collection experience is fair) =140
1 200
= 0,12.
(b) The probability of a profitable loan is:
P(profitable loan) = P(excellent or good or fair loan)
=580 + 418 + 140
1 200
= 0,95.
DSC2602 38
2.3 Some rules in probability theory
A graphic device that is often useful in providing a symbolic representation of operations involv-ing events, is a Venn diagram. In a Venn diagram, the sample space is usually represented as arectangle containing all the outcomes (sample points). The events of interest are represented ascircles or subsets of the rectangle containing those outcomes (sample points) that would cause theevent to occur. The following table illustrates the use of Venn diagrams for different outcomes:
Outcome indicated by grey Description Venn diagramarea in Venn diagram
• The complement of an event:
A Complement of event A. A
S
A
The complement of an event A is the collection of all outcomes in the sample spaceS that are not in A. The complement of A is denoted by the symbol A.
P(A) + P(A)= 1
The probability of the event A can be determined as follows:
P(A) = 1 − P(A).
• The union and intersection of an event:
A ∪ BEither event A or event B occurs, orboth events A and B occur. Alsoknown as union.
A BS
The union of events A and B is the event containing all outcomes belonging to eitherA or B or to both A and B. The union is denoted by A ∪ B
A ∩ BBoth events A and B occur. Alsoknown as intersection or joint occur-rence.
A BS
The intersection of A and B is the event containing all outcomes belonging to bothA and B. The intersection is denoted by A ∩ B. (Many authors use the notation AB.)The probability P(A ∩ B) is also referred to as the joint probability.
39 DSC2602
The addition law provides a way to compute the probability of event A or Bor both occurring. It is formally stated as
P(A ∪ B)=P(A)+P(B)−P(A ∩ B).
The first two terms in the addition law, P(A) + P(B), take all the outcomes in Aand all the outcomes in B into account. This means that the outcomes in A ∩ Bare counted twice. We can therefore not calculate P(A ∪ B) by simply adding P(A)and P(B). We have to correct for the double counting of the outcomes in A ∩ Bby subtracting the probability P(A ∩ B).
• Mutually exclusive:
A ∩ B = φ
It is impossible for events A andB to occur simultaneously. A andB are mutually exclusive events.
A BS
Two or more events are said to be mutually exclusive if the events do not haveany outcomes in common, that is if there are no outcomes in the intersection of theevents. For two events A and B to be mutually exclusive, P(A ∩ B) = 0. Formutually exclusive events the addition law becomes
P(A ∪ B)= P(A) + P(B).
To compute the probability of the union of two mutually exclusive events, wesimply add probabilities of the events.
Example 2.3.4
During a certain semester, 200 students enrolled for the modules OPS201 and OPS202.From these students, 160 students passed OPS201, 140 passed OPS202 and 124 passed bothOPS201 and OPS202.
The Department has decided that any student who passes at least one of the modules willreceive a note of congratulation. What is the probability of a student receiving a note ofcongratulation?
Solution
Let A = event of passing OPS201, andB = event of passing OPS202.
The relative frequency information leads to the following probabilities:
P(A) =160
200= 0,8;
DSC2602 40
P(B) =140
200= 0,7;
P(A ∩ B) =124
200= 0,62.
To receive a note of congratulation, a student must pass either OPS201, or OPS202, or bothOPS201 and OPS202. Thus, we need to find P(A ∪ B). Using the addition law for events Aand B, we have
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
= 0,8 + 0,7 − 0,62
= 0,88.
This result tells us that there is an 88% chance of a student receiving a note of congratulationbecause there is a 0,88 probability of students passing at least one of the modules, OPS201or OPS202.
The addition rule is applied when we are interested in the probability of any one of several eventsoccurring. On the other hand, we might be interested in the probability of two or more eventsoccurring either simultaneously or successively. We can compute the probability of obtaining twokings if two cards are drawn simultaneously from a well-shuffled pack of cards. Just as the pro-bability of any one of several events occurring is obtained by adding up appropriate probabilities,the probability of the joint occurrence of two or more events is obtained by multiplying appro-priate probabilities. When considering joint probabilities, the concept of conditional probabilitymust be introduced.
2.3.1 Conditional probability
The probability of one event occurring, given that another event has occurred, is a conditionalprobability. A conditional probability is defined in terms of the probabilities of the individualevents and combinations of them.
Let us consider the example of 20 persons classified by gender and colour of hair.
ColourGender Dark Blond Total
Male 2 6 8Female 9 3 12
Total 11 9 20
The probability of a person being male is:
P(person is male) =2 + 6
20
=8
20
=2
5.
41 DSC2602
The probability of a person being blond is:
P(person is blond) =6 + 3
20
=9
20.
If we want to determine the probability of a person being blond given that the person is male,then we have the situation where a prespecified condition holds.The condition is that we only consider the subsection “male”and we reduce the sample space toeight elements.
There are six blond males, therefore P(person is blond given person is male) =6
8.
This condition is indicated by the vertical line |.If you see the symbol |, then you know it is the mathematical shorthand for given.
Now let A1 represent the males and A2 represent the blond individuals. Then
P(person is blond given that the person is male) = P(A2|A1) =6
8.
If a person is selected at random from this group, the probability of the individual being bothmale and blond is given by
P(A1 and A2) = P(A1 ∩ A2) =6
20.
Here we note that this figure represents the probability of the individual being male multipliedby the probability of the individual being blond, given that he is male, that is,
P(A1) × P(A1 | A2) =8
20× 6
8
=6
20.
This gives us the equation used to calculate conditional probabilities. Suppose that we have anevent A with probability P(A) and we obtain new information or learn that another event, denotedB, has occurred. If A is related to B, we want to take advantage of this information in computinga new, or revised, probability for event A. This new probability of event A is written as P(A|B)where “|” can be read as “given”. The probability P(A|B) is read “the probability of A, given B”.
The conditional probability of A, given B, is
P(A|B) =P(A ∩ B)
P(B)if P(B) , 0.
The conditional probability of B, given A, is
P(B|A) =P(A ∩ B)
P(A)if P(A) , 0.
DSC2602 42
To identify the condition, always look out for the words “given” or “if” or “on condition that”.The words following given, or if, or on condition that, represent the condition.
2.3.2 Joint probabilities: multiplication law
Sometimes we want to determine the probability that two events intersect or occur at the sametime P(A ∩ B). In other words, we want to determine the probability of their joint occurrence orjoint probability.
The multiplication law can be used to find the probability of an intersection of two events. Themultiplication law is derived from the definition of the conditional probability, i.e
P(A ∩ B) = P(A|B)P(B) or
P(A ∩ B) = P(B|A)P(A).
If two events are statistically independent, then the multiplication law is reduced to:
P(A ∩ B) = P(A)P(B).
Example 2.3.5
Consider a random experiment of selecting 200 companies from the Johannesburg StockExchange. From the survey, the following information was gathered on the size of thecompanies and the type of industry in which the companies operate. The information issummarised in Table 2.1.
Company size(Number of employees)
Type Small Medium Largeof industry (< 20) (20 – 60) (> 60) TotalMining 0 0 70 70Finance 19 21 42 82Service 6 3 1 10Retail 14 18 6 38Total 39 42 119 200
Table 2.1: Information of 200 companies listed on the JSE
(a) What is the probability of selecting a large company from the JSE given that thecompany is known to be a retail company?
(b) What is the probability that a selected company from the JSE is a small company or aservice company or both?
(c) What is the probability that a selected company from the JSE is medium in size and afinance company?
(d) What is the probability that a selected company from the JSE is a small or a largecompany?
43 DSC2602
Solution
(a) Let
A = the event that the selected company is a large company, andB = the event that the selected company is a retail company.
There are 6 large companies out of 38 retail companies. Thus,
P(A|B) =6
38= 0,157.
Using the formula:
P(A|B) =P(A ∩ B)
P(B)=
6
20038
200
= 0,157.
(b) Let
A = the event that the selected company is a small company, andB = the event that the selected company is a service company.
The events are not mutually exclusive because they can occur simultaneously. FromTable 2.1 we can compute the following:
P(A) =39
200= 0,195
P(B) =10
200= 0,05 and
P(A ∩ B) =6
200= 0,03.
Then
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
= 0,195 + 0,05 − 0,03
= 0,215.
There is a 0,215 probability that a randomly selected company from the JSE will be asmall company or a service company or both.
(c) Let
A = the event that the selected company is a medium company, andB = the event that the selected company is a finance company.
Intuitively, from Table 2.1, it follows that
P(A ∩ B) =21
200= 0,105.
If we apply the multiplication rule we get
P(A ∩ B) = P(A|B)P(B)
=21
82× 82
200
=21
200= 0,105.
DSC2602 44
(d) Let
A = the event that the selected company is a small company, andB = the event that the selected company is a large company.
The events are mutually exclusive because both events cannot occur simultaneously.
From Table 2.1 it follows that
P(A) =39
200= 0,195, and
P(B) =119
200= 0,595.
Then
P(A ∪ B) = P(A) + P(B)
= 0,195 + 0,595
= 0,79.
There is a 0,79 probability that a randomly selected company from the JSE will be eithera small or a large company.
2.4 Summary of basic probability concepts
1. A random experiment is the process of obtaining information through the observation ormeasurement of a phenomenon, the outcome of which is subject to change.
2. An outcome (sample point) is a single result of an experiment.
3. An event is a collection of outcomes.
4. The sample space of an experiment is the collection of all possible outcomes.
5. The complement of event A indicated by (A) is the event containing all outcomes that are notin A, with
P(A) = 1 − P(A).
6. The Venn diagram is a graphic device for representing the sample space and relationshipsbetween events.
7. The intersection of events A and B (A∩B) is the event containing all outcomes that are in bothA and B. The product rule is used to calculate the joint probability, or the intersection oftwo events, where
P(A ∩ B) = P(A|B)P(B).
8. The union of events A and B (A ∪ B) is the event containing all outcomes that are in A, or inB, or in both. The addition law is a law to compute the probability of the union of events,that is
P(A ∪ B) = P(A) + P(B) − P(A ∩ B).
45 DSC2602
9. Two or more events are said to be mutually exclusive if they have no outcomes in common,that is P(A ∩ B) = 0.
10. If A and B are statistically independent, that is, they have no influence on each other, then
P(A|B) = P(A) and P(A ∩ B) = P(A)P(B).
11. The conditional probability is the probability of an event, given that another event has oc-curred.
2.5 Exercises
1. The Personnel Department has analysed the qualification profile of their 129 managersand the following contingency table was constructed. (The qualification level is the highestqualification achieved.)
Qualification Management level: Head of a
level Section Department Division Total
Matric 28 14 ? ?Diploma 20 24 6 ?Degree ? 10 14 ?
Total 53 ? 28 ?
(a) Complete the above contingency table.
(b) What is the probability that a randomly selected person
i. has only a matric?
ii. is head of a section and has a degree?
iii. is head of a department, given that he has a diploma?
iv. is head of a division?
v. is head of a division or of a section?
vi. has a matric, or a diploma, or a degree?
vii. has a matric, given that he is head of a department?
2. The data in the table refer to the employees of the Fairwaters Company and are classifiedaccording to gender and opinion. They concern a proposal to emphasise fringe benefitsrather than wage increases in an impending contract discussion.
OpinionGender In favour Neutral Opposed
Male 900 200 400Female 300 100 600
DSC2602 46
Calculate the probability that an employee selected from this group will be:
(a) a female opposed to the proposal;
(b) neutral;
(c) opposed to the proposal, given that the employee selected is a female;
(d) either a male or opposed to the proposal.
3. In a given firm, 50 of the 90 employees are union members. Of those who are unionmembers, 35 belong to union A and 15 belong to union B.
(a) What is the probability that any member selected at random is a union member?
(b) What is the probability that any member selected at random belongs to union B, giventhat he or she is a union member?
4. The number of cars owned by each of the families in a city is shown in the following table:
Cars owned 0 1 2 3 4 5Number of families 27 1 422 2 865 1 796 324 53
Calculate the probability that any family selected at random:
(a) owns at most three cars;
(b) owns at least one car;
(c) owns between two and four cars;
(d) will not own a car.
5. If there are 10 horses in a race, all with an equal chance of winning, what is the probabilitythat horse number 3 or 5 or 8 will win?
6. A securities analyst maintains that there is a chance of 0,3 that a certain company willbe purchased by Exon and a 0,3 chance that it will be purchased by Mobil. What is theprobability that it will be acquired by one or the other of these two companies?
7. What is the probability of drawing an ace or a spade from a pack of cards?
8. A company produces plastic toy elephants in two colours: blue and pink. Three machinesare used to produce these toys. Machine A is used for 10% of the production, Machine B for30% of the production and Machine C for the remainder. A production run consists in totalof 500 elephants.
Machine A’s production consists of 40% blue elephants and 60% pink elephants. Machine B’sproduction consists of 30% blue elephants and 70% pink elephants. Machine C’s productionconsists of 80% pink elephants with the remainder being blue.
(a) What proportion of the total production do blue elephants form?
(b) If a particular elephant is pink, what is the probability that it was produced by MachineB?
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CHECKLISTYou should now be able to
⊲ explain what probability is;
⊲ use Venn diagrams to represent events;
⊲ use the
– addition law,
– multiplication law
to calculate probabilities;
⊲ calculate the probability of the complement of an event;
⊲ calculate conditional probabilities.
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CHAPTER 3
Probability distributions
3.1 Introduction
I Cter 2 we discussed various ways of assigning probability values to the outcomes of anexperiment. In this section we continue the study of probability by introducing the concepts
of random variables and probability distributions.
3.2 Random variable
Let X be the number of pizzas sold by a certain Italian restaurant per day. X is called a randomvariable. A random variable assigns a real number to every possible outcome or event in anexperiment. It is usually presented by a capital letter such as X or Y.
We denote a random variable by a capital letter, X, while the value that X may assume is denotedby a small letter x. For example, if x denotes the possible outcome of a die, then x can take on thevalues x = 1; 2; 3; 4; 5; 6.
There are two types of random variables: discrete random variables and continuous random variables.
A random variable that may assume a finite (or infinite) sequence of discrete values, such as{1; 2; 3; 4; 5} (or {1; 2; 3; . . .}) is a discrete random variable. Random variables such as time, heightand temperature that may assume any value in a certain interval, are continuous random variables.
Examples of experiments and their associated random variables are given in Table 3.1.
49
DSC2602 50
Random PossibleExperiment variable (X) values (x)
Operate a Number of customers 0; 1; 2; . . .restaurant entering in one dayBuild a house Percentage of 0 ≤ x ≤ 100
house completed aftertwo months
Test the lifetime Time the bulb 0 ≤ x ≤ 100of a light bulb lastsToss a dice The number of dots 1; 2; 3; 4; 5; 6
on top of the diceResponse to a 3 if agreed 1; 2; 3questionnaire 2 if neutral
1 if disagree
Inspect a 0 if defective 0; 1machine 1 if not defective
Table 3.1: Examples of random variables
A probability model expresses the theoretical distribution of a random variable in a mathematicalformula. In the case of a discrete random variable the mathematical expression is known as theprobability distribution or probability mass function, and in the case of a continuous random variablewe refer to the density function.
3.3 Discrete random variables
3.3.1 The probability distribution of a discrete random variable
When we have a discrete random variable, a probability value is assigned to each event. Thesevalues must be between 0 and 1 and their sum must be 1. The probability mass function, denotedby p(x), provides the probability that the random variable, X, assumes a specific value.
The probability mass function of X is defined as
p(x) = P(X = x).
Furthermore, p(x) has the following characteristics: p(x) ≥ 0 for all x and∑
x∈Sp(x) = 1.
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Example 3.3.1
Consider a store that sells computers. The random variable under consideration is X, thenumber of computers sold on a specific day. The information on the number of computersthat have been sold over the past 120 days is summarised in Table 3.2.
Sales volume Number ofof computers daysNo sales 101 computer 202 computers 253 computers 304 computers 155 computers 20Total 120
Table 3.2: Computers sold
The probability mass function, p, provides the probability that the random variable Xassumes a specific value, x. Because the store did not sell any computers for 10 of the 120days, we can say that the probability of selling no computers, is
p(0) =10
120= 0,083.
Similar computations can be made for p(1), p(2), p(3), p(4) and p(5). Table 3.3 shows one wayto present the probability distribution for the number of computers sold per day.
x P(X = x)0 0,0831 0,1672 0,2083 0,2504 0,1255 0,167
Total 1
Table 3.3: Probability distribution for the number of computers sold per day
The probability distribution of X is graphically shown in Figure 3.1. The probability that Xassumes these values is shown on the vertical axis, while the values of the random variableare shown on the horizontal axis.
(Depending on the random variable, for example if the random variable is measured inintervals, it is also possible to draw a histogram from the information that follows from theprobability distribution.)
After identifying the random variable and its probability distribution, a variety of additionalprobability information can be determined. For instance:
DSC2602 52
0 , 0 50 , 1 00 , 1 50 , 2 00 , 2 50 , 3 0p ( x )
x0 1 2 3 4 5
Figure 3.1: Graphical representation of the probability distribution
• There is a 0,083 probability that no computers will be sold during a day.
• The most probable sales volume is three computers, with p(3) = 0,25.
• There is a 0,291 probability of a sales day with four or five computers being sold.
Using the probability information, the management of the store can better understand theuncertainties associated with the computer sales operation.
3.3.2 Expected value of a discrete probability distribution
Just as frequency distributions are described by measures of locality and spread, theoreticaldistributions of random variables can be described by similar measures of locality and spread.
Once the probability distribution of a random variable has been determined, the first characteristicthat is usually of interest is the “central tendency”, or average, of the distribution. The expected value,a measure of central tendency, is computed as a weighted average of the values of the randomvariable.
The mathematical formula for computing the expected value of a discrete random variable fora population of N observations
E(X) = µ =
N∑
i=1
xip(xi).
The calculation of the expected value for the sales of computers in Example 3.3.1 is shown inTable 3.4.
53 DSC2602
i xi p(xi) xip(xi)
1 0 0,083 0(0,083) = 02 1 0,167 1(0,167) = 0,1673 2 0,208 2(0,208) = 0,4164 3 0,250 3(0,250) = 0,755 4 0,125 4(0,125) = 0,56 5 0,167 5(0,167) = 0,835
E(X) = 2,67
Table 3.4: Expected value calculation for computer sales example
To compute the expected value, we must multiply each value of the random variable by itscorresponding probability and then add the resulting terms.
If the experiment is repeated a number of times, the expected value can be seen as the “long-run”average value for the random variable and not necessarily the outcome that we expect the randomvariable will assume the next time the experiment is conducted.
The expected value can be helpful to use as a tool for the management of the store in the planningof its inventory. For example, suppose the store is open for 90 days over the next four months.How many computers will be sold during that time? Although it is not possible to specify theexact number of computers that will be sold every day, the expected value of 2,67 computers perday provides an expected sales estimate of 90(2,67) � 240 computers over the next four months.
3.3.3 Variance of a discrete probability distribution
In addition to the central tendency of a probability distribution, we are often interested in thevariability, or the spread, of the distribution. If the variability is low, it is much more likely that theoutcome of an experiment will be close to the average or expected value. If the variability is high,the probability is spread over various random variable values. In such a case there is less chancethat the outcome of an experiment will be close to the expected value. The variance is a measurecommonly used to summarise the variability in the values of a random variable.
The mathematical expression for the variance of a discrete random variable in a population ofN observations is
Variance(X) = σ2 =
N∑
i=1
(xi − µ)2p(xi).
The variance can also be seen as a weighted average of the squared deviations.
The calculation of the expected value for Example 3.3.1 is shown in Table 3.5.
A related measure of variability is the standard deviation, σ. The standard deviation is defined asthe positive square root of the variance:
σ =√σ2.
DSC2602 54
i xi xi − µ (xi − µ)2 p(xi) (xi − µ)2p(xi)
1 0 0 − 2,67 = −2,67 7,13 0,083 7,13(0,083) = 0,592
2 1 1 − 2,67 = −1,67 2,79 0,167 2,79(0,167) = 0,466
3 2 2 − 2,67 = −0,67 0,45 0,208 0,45(0,208) = 0,094
4 3 3 − 2,67 = 0,33 0,11 0,250 0,11(0,250) = 0,027
5 4 4 − 2,67 = 1,33 1,77 0,125 1,77(0,125) = 0,221
6 5 5 − 2,67 = 2,33 5,43 0,167 5,43(0,167) = 0,907
σ2 = 2,307
Table 3.5: Variance calculation for computer sales example
For the computer sales example, the standard deviation of the number of computers sold per dayis
σ =√
2,307 = 1,52.
The standard deviation is preferred above the variance for interpretation purposes, because it ismeasured in the same units as the random variable (σ = 1,51 computers sold per day).
3.3.4 Discrete distribution function
A function, F, such that
F(x) = P(X ≤ x), with x any real number,
is known as a cumulative probability function or a distribution function.
The distribution function of any discrete random variable is a step function with step heightsequal to p(x), the probabilities of the various values.
A typical discrete distribution function is depicted in Figure 3.2.
F ( x )
xx 1 x 2 x 3 x 40
1
Figure 3.2: Discrete distribution function
The height of the jump at a discrete point, xi, gives the probability that the random variable, X,assumes the value xi. Let’s look at another example.
55 DSC2602
with F(x0) = P(X ≤ 0) = P(X = 0)
F(x1) = P(X ≤ 1) = P(X = 0) + P(X = 1)
F(x2) = P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)...
F(xi) = P(X ≤ xi) = P(X = 0) + P(X = 1) + · · · + P(X = xi)
The discrete distribution function can be written as follows:
xi P(X = xi) F(x) = P(X ≤ xi)
0 0,083 0,083 = 0,083
1 0,167 0,167 + 0,083 = 0,250
2 0,208 0,083 + 0,167 + 0,208 = 0,458
3 0,250 0,083 + 0,167 + 0,208 + 0,250 = 0,708
4 0,125... = 0,833
5 0,167... = 1,000
Table 3.6: Distribution function
The distribution function F(x) can be graphical depicted as follows:
0
1
1 2 3 4 5 x
F ( x )
0 , 0 8 30 , 2 5 00 , 4 5 80 , 7 0 80 , 8 3 3
Figure 3.3: Distribution function F(x)
An important feature of the distribution function F is that any probability of the stochastic variableconcerned can be determined by means of F. Thus, for example, it follows that:
P(X > xi) = 1 − P(x ≤ xi) = 1 − F(xi)
P(xi ≤ X ≤ x j) = F(x j) − F(xi)
P(X = xi) = P(X ≤ xi) − P(X < xi)
= F(xi) − F(xi−1)
DSC2602 56
3.4 Discrete random distributions
Data sets observed in practice can often be seen as realisations of some well-known randomvariable X. The exact probability distribution of X (discrete case) is often not known. However,the circumstances in which the random variable was observed, together with the characteristicsof the observed x values, usually identify an appropriate probability model.
In this section we discuss two widely used discrete distributions, namely the binomial distribution,and the Poisson distribution.
3.4.1 The binomial distribution
Suppose we randomly ask people whether they can drive a 10-ton truck. If an individual candrive such a truck, we label the outcome of our experiment a “success”. If the individual cannotdrive a 10-ton truck, we label the outcome of our experiment a “failure”.
We can describe this experiment by the following mathematical model, called a Bernoulli trial:
Let p be the probability of a “success” where p is a constant and 0 < p < 1. The random variableX is defined as
X =
{
1 if the result is a “success”,0 if the result is a “failure”.
It often happens that we repeat a Bernoulli trial n times.
If X = X1 + X2 + . . . + Xn, where X is the number of successes in the sample, then we say X has abinomial distribution with parameters n and p. Bi denotes two. In other words, there are only twopossible outcomes and we consider an outcome as a success or a failure. In short we say X ∼ b(n,p)where n =number of trails and p is the probability of a success.
The formula for a binomial distribution is developed from a specific set of assumptions involvingthe concept of a series of experimental trials. The trials take place under the following set ofassumptions:
1. Each trial has two mutually exclusive possible outcomes, which are referred to as “success”or “failure”.
2. The probability of a success, denoted by p, remains constant from trial to trial. The pro-bability of failure, denoted by q, is equal to 1 − p.
3. The trials are independent.The outcome of any given trial or sequence of trials does not affect the outcomes of subse-quent trials; the outcome of any specific trial is determined by chance.
The formula for a binomial distribution looks complicated but we will do a simple experiment tocalculate a probability and then generalise from it.
The experiment consists of tossing a coin and we are interested in the probability of turning upheads twice in five tosses. Assume that the appearance of heads on each toss is a success.
Suppose that the sequence of outcomes for five tosses of a coin is:
H T H T T(where H represents heads and T represents tails).
57 DSC2602
The probabilities of turning up heads once and tails once are: P(H) = p and P(T) = (1 − p) = q.
Remembering the rules of probabilities, we can write:
P(H T H T T) = P(H) × P(T) × P(H) × P(T) × P(T)
= p × q × p × q × q
= p2 × q3.
This is the probability of obtaining the specific sequence of successes and failures in the order inwhich they occur.We are not interested in any specific order of results but in obtaining a given number of successesin n trials.What is the probability of obtaining exactly two successes in five tosses? Nine other sequencessatisfy the condition of exactly two successes in five trials:
H H T T T
H T T H T
H T T T H
T H H T T
T H T H T
T H T T H
T T H H T
T T H T H
T T T H H
But each of these has the same probability of occurring, that is
p2 × q3.
(Do you believe me without testing it?)
Since there are 10 distinguishable sequences so we can write:
P(exactly two successes) = 10 × p2 × q3.
The coefficient 10 represents the total number of distinguishable arrangements that can be madeof the three Ts and two Hs.
A general symbol for the number of distinguishable arrangements is:
nCx or
(
n
x
)
where n is the number of trials and x is the number of successes. This is read as n combination x.
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The number of possible sequences of five trials that would contain exactly two successes is5C2 or
(52
). This can be written as:
(
5
2
)
= 5C2 =5!
(5 − 2)!2!.
In this case ! means factorial and 5! = 5 × 4 × 3 × 2 × 1.
Hence,
(
5
2
)
= 5C2 =5!
(5 − 2)!2!
=5!
3!2!
=5 × 4 × 3!
3!2!
=5 × 4
2!
=5 × 4
2 × 1
= 10.
Since the outcome of the coin can only be heads or tails,
p =1
2and q =
1
2.
The probability of heads occurring twice is:
P( heads occurring twice) = P(X = 2) = 5C2
(12
)2 (12
)3or
(52
) (12
)2 (12
)3
= 10 · 14· 1
8
=10
32
=5
16.
This result can now be generalised to
p(x) = P(X = x) = nCxpxqn−x or
(
n
x
)
px(1 − p)n−x
where
59 DSC2602
n = number of trails
p = probability of success in one trail (0 < p < 1)
x = number of successes in n trails (x = 0,1,2 · · · ,n).
Note: Most scientific calculators can calculate n! and(n
x
)directly. On your calculator
(nx
)is usually
probably indicated as nCr or nCx.
It can be shown that the expected value and the variance of the binomial distributionare
E(X) = np and Var(X) = np(1 − p).
The probability that the number of successes is less than, or equal to k, also known as thecumulative distribution, is calculated as follows:
P(X ≤ k) =
k∑
x=0
(
n
x
)
px(1 − p)n−x.
For example, if we have to calculate
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
=
(
n
0
)
p0(1 − p)n +
(
n
1
)
p1(1 − p)n−1 +
(
n
2
)
p2(1 − p)n−2
The above expression can be calculated by hand or you can make use of the cumulative discreteprobability tables in Appendix B.5. See example that follows.
Example 3.4.2
A certain public examination has a consistent past record of a 25% failure rate. For a groupof 15 people writing the examination this year, calculate the probability that:
(a) exactly two will fail;
(b) at least 12 will pass;
(c) all will pass.
Solution
In this case there are 15 people writing the examination. Since we can count them on ourfingers this is a discrete distribution.
A person can either pass or fail an examination and there are only two possible outcomes.We therefore have a binomial distribution.
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The correct way of attempting the problem is by writing down the information given. Thesample size, the probability of success (or failure) and that which we want to determinemust be written down in mathematical notation:
n = 15
p = 0,25 (in this case we can consider the probability of success being that of
failing the examination)
X = the number of successes, that is, the number of people failing.
(a) The probability that exactly two will fail is:
P(X = 2) = 15C2(0,25)2(0,75)13
=15!
13!2!(0,25)2(0,75)13
= 0,1559.
(b) The probability that at least 12 will pass is:
P(at least 12 will pass) = P(at most 3 will fail)
= P(X ≤ 3)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 15C0(0,25)0(0,75)15+15 C1(0,25)1(0,75)14
+
15C2(0,25)2(0,75)13+15 C3(0,25)3(0,75)12
= 0,0134 + 0,0668 + 0,1559 + 0,2252
= 0,4613.
We can also calculate P(X ≤ xi) for the binomial distribution using the cumulativebinomial probability tables in Appendix B.5. To use the table we need n,p and r (or xi).In the example n = 15 p = 0,25 and r or xi = 3. The answer from the table is 0,4613.
(c) The probability that all will pass is:
P(all will pass) = P(none will fail)
= P(X = 0)
= 15C0(0,25)0(0,75)15
= 0,0134.
Example 3.4.3
A sign on the petrol pumps of a service station encourages customers to have the oil checked,claiming that one out of every four cars should have its oil topped up.
(a) What is the probability that exactly three of the next 10 cars entering the station willhave the oil topped up?
(b) What is the probability that at least half of the next eight cars entering the station willhave the oil topped up?
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Solution
(a) n = 10.
p =1
4= 0,25.
X = the number of successes, that is, the number of cars having the oil topped up.
P(X = 3) = 10C3(0,25)3(0,75)7
=10!
7!3!(0,25)3(0,75)7
= 0,2503.
(b) “At least half of eight” means four or more.
P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6)+
P(X = 7) + P(X = 8)
= 8C4(0,25)4(0,75)4+8 C5(0,25)5(0,75)3
+
8C6(0,25)6(0,75)2+8 C7(0,25)7(0,75)1
+
8C8(0,25)8(0,75)0
= 0,0865 + 0,0231 + 0,0038 + 0,004 + 0,000015
= 0,1134.
or, working with the complement of the event,
P(X ≥ 4) = 1 − P(X < 4)
= 1 − P(X ≤ 3)
= 1 − P(X = 0) − P(X = 1) − P(X = 2) − P(X = 3).
Remember that the total probability is one. If we draw a line with the possible occurrences,we have
P = 1
P ( X < 4 ) P ( X > 4 )4
3.4.2 The Poisson distribution
There are many practical problems in which we may be interested in finding the probability thatx “successes” will occur over a given interval of time, or region of space. For example, the randomvariable of interest might be the number of cars arriving at a parking garage over a one-hour timeinterval, or the number of telephone calls received over a 20-minute interval.
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In such cases we will make use of the Poisson distribution to determine probabilities. The Poissondistribution was named after the Frenchman who developed it during the first half of the 19thcentury.
The Poisson distribution has been effectively applied to describe the probability functions ofphenomena such as product demand, demands for service, number of telephone calls transmitted througha switchboard, numbers of various types of accident, numbers of defects observed in products, et cetera.
What is important is that these things happen in an interval, that is, an interval of time or aninterval of space.
Examples in which time intervals are applicable are the number of telephone calls per hour or thedemand for an item per day.
Examples in which space intervals are applicable are the number of defects in a length of electricalcable or a count of the number of blemishes per square metre of sheet metal used for aircraft.
The Poisson distribution is concerned with occurrences that can be described by a discrete randomvariable. This variable X can take on the values x = 0; 1; 2; . . . , to infinity. The domain of theprobability function consists of all non-negative integers.
The probability distribution function of the Poisson random variable is given by
p(x) = P(X = x) =λxe−λ
x!for x = 0; 1; 2; . . .
where
λ = mean or average number of occurrences over an interval, or rate at which somethinghappens in the interval.
e = 2,17828, the base of the natural logarithms,x = number of observed occurrences over an interval, and
p(x) = the probability of x occurrences over the interval.
The mean and variance of the Poisson distribution are
Mean = λ and Variance = λ.
If the following three assumptions are satisfied, the Poisson probability distribution is applicable.
(a) The occurrence or non-occurrence of the event over any interval is independent of theoccurrence or non-occurrence over any other interval.
(b) The probability of the occurrence of an event is the same for any two intervals of the samelength.
(c) The occurrences are uniformly distributed throughout the interval of time. [See Section 3.6.3for the Uniform distribution.]
Example 3.4.4
From past experiences a cricket club manager knows that about six games, on average, willhave to be postponed during the season because of rain. What is the probability that threegames will have to be postponed this season?
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Solution
According to the manager’s experience, λ = 6. We are interested in the probability thatx = 3.
P(X = 3) = p(3) =63e−6
3!= 0,0892.
Thus, the probability that three games will have to be postponed because of rain is 0,0892.
Example 3.4.5
A tollbooth operator has observed that cars arrive randomly at an average rate of 360 carsper hour.
(a) What is the probability that only two cars will arrive during a specified one-minuteperiod?
(b) What will be the probability of at least four arrivals during a specified one-minuteperiod?
Solution
(a) Let X denote the number of cars arriving during a specified one-minute period. Themean value of X is:
λ =360
60= 6 cars per minute.
Notice that we have defined X and µ in terms of the same time interval of one minute.The probability that two cars will arrive, is
P(X = 2) =62e−6
2!
=36 × 0,00248
2 × 1
= 0,04464.
(b) The probability that at least four cars will arrive, isP(X ≥ 4) = 1 − P(X ≤ 3)
= 1 − P(X = 0) − P(X = 1) − P(X = 2) − P(X = 3)
= 1 − 60e−6
0!− 61e−6
1!− 62e−6
2!− 63e−6
3!
= 1 − 1 × 0,00248
1− 6 × 0,00248
1− 36 × 0,00248
2 × 1− 216 × 0,00248
3 × 2 × 1
= 1 − 0,1066
= 0,08934.
DSC2602 64
3.5 Continuous random variables
In many cases random variables can take on a seemingly unlimited number of values that include notonly integer values but also fractions. Such a random variable is said to be continuous.
Examples of continuous random variables are time measurements; interest rates; financial ratios;income levels; weight measurements; distance measures and volume measures.
In general, the value of a continuous random variable is found by measuring, whereas the valueof a discrete random variable is found by counting.
3.5.1 The probability distribution of a continuous random variable
The probability distribution of continuous random variables is represented by a probability densityfunction, whose graph approximates the relative frequency polygon for the population.
00 . 0 5
0 . 10 . 1 50 . 2
0 . 2 5
100
200
400
600
800
1000
1200
1300
1400
1600
1800
2000
2200
2400
2600
P r i c e s p e r s q u a r e m e t r e
R el a ti v e
n um b
e r o f
h ou s e
s
Figure 3.4: Histogram and frequency polygon of the relative frequency distribution of the priceper square metre of a population of 843 residences in Johannesburg
If we smooth the frequency polygon so that the area under the curve is equal to 1, the smoothcurve is the density function and the function f (x) describes the curve mathematically.
P r i c e p e r s q u a r e m e t r e
A r e a b e l o wd e n s i t y c u r v e = 1
f ( x )
x
65 DSC2602
A probability density function f(x) must satisfy the following two conditions:
1. f (x) is non–negative.
2. The total area under the curve representing f (x) equals one.
It is important to note that f (x) itself is not a probability, but a function.
In the case of continuous random variables, the probabilities are calculated by determining theareas of appropriate sections below the density function.
Given a probability density function f (x), the area under the graph of f (x) between two values aand b is the probability that X will take a value between a and b.
a b
The shaded area gives the probability of P(a < X < b).
The mathematical techniques of area determination, namely integral calculus are used to deter-mine probabilities for a continuous stochastic variable namely
P(a < X < b) =
∫ b
a
f (x)dx.
3.5.2 Continuous distribution function
The distribution function of the continuous random variable x with density function f (x) isdefined as
F(x) =
∫ a
−∞f (x)dx
for every real value of x.
Graphically the distribution function can be represented as
F ( x )
x
DSC2602 66
3.6 Continuous distributions
In this section we discuss a few continuous distributions that are commonly used, namely thenormal distribution, the exponential distribution and the uniform distribution.
3.6.1 The normal distribution
Although there are numerous continuous distributions, the normal distribution is one of the mostimportant distributions and plays a central role particularly in the area of statistical inference.
The normal distribution can also be regarded as the most useful continuous distribution. It pro-vides a good approximation of a number of other distributions such as the binomial distributionwhen the sample size is large.
If X is a continuous random variable for which the density function is given by
f (x) =1
σ√
2πe−
12 (x−µ)2/σ2
then X has a normal distribution with mean µ and variance σ2. This is denoted in short byX ∼ n(µ; σ2).
The graph of f (x) has a symmetrical bell-shaped appearance with the total area under the curveequal to one.
f ( x )
x
T o t a l a r e a u n d e r t h e c u r v ee q u a l t o o n e .
m
The value of µ moves the centre of the graph to the left or to the right on the x-axis, whereas σinfluences the size of the bell. If σ is relatively small, the bell will be narrow and high, and if σ isrelatively large, the bell will be wide and flat. Figure 3.5 is a graphic representation of the densityfunctions of a n(15; 9) and a n(30; 4) distribution.
1 5 2 0 2 5 3 0 3 5 4 0xFigure 3.5: The density functions of a n(25; 9) and a n(30; 4) distribution
67 DSC2602
The normal distribution has three distinct characteristics.
• If X has a n(µ; σ2) distribution, then the range of X is the set of real numbers
R = {x : −∞ < x < ∞}.
• If X has a n(µ; σ2) distribution, then
P(X ≤ µ − k) = P(X ≥ µ + k)
for all k. This means that X is symmetrically distributed around the mean µ.
k- m k
• If X has a n(µ; σ2) distribution, then
z =X − µσ
has a n(0; 1) distribution. We call this distribution the standard normal distribution.
3.6.1.1 The Standard Normal distribution
A normal distribution is completely specified by its mean and standard deviation. Thus, al-though all normal distributions are basically bell-shaped, different means and different standarddeviations will describe different bell-shaped curves.
To find the probability that a normally distributed variable, X, will lie between two specifiedvalues, say a and b, the probability density function (pdf) namely f (x) must be integrated overthe interval (a; b) to determine the area under the curve between a and b. It is quite difficult todetermine this integral.
Fortunately, it is possible to convert each of these different normal distributions into one stan-dardised form where the mean is 0 and the standard deviation is 1. These probability values aregiven in the table in Appendix B.2.
A random variable that has a normal distribution with a mean of 0 and a standard deviation of 1is said to have a standard normal distribution. The curve of a typical standard normal distributionis shown in Figure 3.6.
The letter z is used to designate this particular normal random variable. The units on the horizontalaxis (z) measure the number of standard deviations from the mean. To find the probabilitiesassociated with a continuous random variable, it is necessary to determine the probability of therandom variable having a value in a specific interval, [a; b]. Basically, we have to find the area
DSC2602 68
0 +1−1
Area or probability
σ = 1σ = 1
z
Figure 3.6: Areas or probabilities for the standard normal distribution
under the curve, between a and b. The mathematical technique for the calculation of these areasis beyond the scope of this module. We use the table for the standard normal distribution that isgiven in Appendix B.2.
To summarise: If X is a normal distributed variable, that is X ∼ n(µ; σ2), must be converted tothe standard normal distribution, z ∼ n(0; 1) by means of subtracting the mean and divide by thestandard deviation namely
z = (X − µ)/σ.
We can then use the standard normal table to calculate:
(1) the area of probability under the standard normal curve between zero and a given value z.See Figure 3.7.
0 z
Area or probability
Figure 3.7: Areas or probabilities for the standard normal distribution
or
(2) the z-value given the area of probability under the standard normal curve.
The standard normal table consists of three parts namely
(a) left-hand column: the z-value up to the first decimal.
(b) upper row: the second decimal of the z-value.
(c) inside of table: the area of probability between 0 and the z-value.
69 DSC2602
The standard normal table is graphically explained in the following figure:
z 0,00 0,01 0,02 0,03 0,04 0,05 0,06 0,07 0,08 0,09
0,0
0,1
0,2
0,3
0,4
"
"
"
"
4,0
(a)
z- va
lue
up to
firs
t dec
imal
(c) probabilities from 0 until 0,5
(b) second decimal of z- value
If the z-value is given and the probability has to be determined we read from the outside of thetable to the inside.
If the probability is given and we need to determine the z-value we read from the inside of the tableto the outside.
Note: The table is only defined for areas between 0 and a given positive z value that is smallerthan 0,5. If the area that you have to determine is not between 0 and a positive z-value you haveto change it to an area applicable to the table. See the following examples.
To illustrate the use of the table of the standard normal distribution, Appendix B.2, we look at thefollowing examples:
Suppose that the length of time, students take to write an entrance examination is known to benormally distributed with a mean of 60 minutes and a standard deviation of eight minutes. Whatis the probability of a particular student’s time being between 60 and 70 minutes?
The probability that we need to determine is P(60 < X < 70). Graphically we can depict it as thearea under the graph between 60 and 70:
6 0 7 0x
Before we can use the standard normal distribution table containing the values of z, we must firsttransform our normal random variable X into the standard normal random variable z.
The following illustrates what must happen:
µ = 60 and σ = 8↓ ↓µ = 0 and σ = 1
DSC2602 70
To do this we need to subtract the mean µ from x and divide by the standard deviation σ.
The formula that is used to accomplish this is:
z =X − µσ.
In our example we have z =X − 60
8.
But we want to determine P(60 < X < 70). If you recall your basic algebra, you will rememberthat you cannot change X without changing the values on either side of it.
P(60 < X < 70) = P(60 − 60
8<
X − 60
8<
70 − 60
8
)
= P(0 < z < 1,25).
It is always easier to draw a picture of the required area.
0 1 , 2 5 z
The table in Appendix B2, gives the area under the curve between 0 and z. We can now read offP(0 < z < 1,25) directly from the table. How?
In the first column the z-value up to one decimal is given. The other columns give the seconddecimal place. Now to determine P(0 < z < 1,25), find 1,2 in the z-column and move to thecolumn with 0,05 in the top row. The inside of the table gives the probability,
P(0 < z < 1,25) = 0,39435.
The probability that a student’s time being between 60 and 70 minutes is 0,39435.
Another important characteristic of the normal distribution is that the normal distribution issymmetrical around zero. The total probability under the curve is one (or 100%). Thus 0,5 (or 50%)lies to the left of zero and 0,5 (or 50%) to the right of zero.
71 DSC2602
0
0 , 50 , 5
z
The values on the horizontal line are the values that z can take on. They can be negative (to theleft of zero) or positive (to the right of zero). The values under the curve represent areas andare always positive. A probability can NEVER be negative! (Go back to the basic properties ofprobabilities in Chapter 2). The property of symmetry is extremely useful.
Let us look at some more examples:
a) Suppose we want to determine P(−2,4 < z < 0). The probability that z lies between −2,4and 0 is represented by the shaded area in Figure 3.8.
0−2,4 z
Figure 3.8: Finding P(−2,4 < z < 0)
The table only supplies values to the right of zero. Since the normal distribution is symme-trical, this probability (area) is equivalent to P(0 < z < 2,4) on the positive side. See Figure3.9
0 2,4 z
Figure 3.9: Finding P(0 < z < 2,4)
Read down the left-hand side of the table until you find z = 2,4. The probability isP(0 < z < 2,4) = 0,4918, which is the same as P(−2,4 < z < 0) = 0,4918.
b) Determine P(z > 1,92). The probability that z lies to the right of 1,92 is represented by theshaded area in Figure 3.10.
DSC2602 72
0 1,92 z
Figure 3.10: Finding P(z > 1,92)
The table consists only of values between 0 and z. We know that the total area to the rightof the midpoint at z = 0 is 0,5. P(z > 1,92) can thus be written as 0,5 − P(0 < z < 1,92).
Graphically we can explain it as:
0 0 01 , 9 2
= -
z z z
0 , 5
1 , 9 2
The probability that z will lie between 0 and 1,92 is found at z = 1,92 in the table, i.e.P(0 < z < 1,92) = 0,4726. The area to the right of 1,92 is found by subtracting the areabetween 0 < z < 1,92 from 0,5, i.e. P(z > 1,92) = 0,5 − 0,4726 = 0,0274.
c) Determine P(−2,5 < z < 1,32).
0 1,32-2,5 zz
Figure 3.11: Finding P(−2,5 < z < 1,32)
The area representing the required probability can be divided into two areas, one between−2,5 and 0 and the other between 0 and 1,32.
Due to the symmetry of the normal distribution, P(−2,5 < z < 0) = P(0 < z < 2,5) = 0,4938and P(0 < z < 1,32) = 0,4066. Therefore,
P(−2,5 < z < 1,32) = P(−2,5 < z < 0) + P(0 < z < 1,32) = 0,4938 + 0,4066 = 0,9004.
73 DSC2602
Graphically we can explain it as follows:
+=
z z z1 , 3 2
+=
z z z
1 , 3 2
- 2 , 5 0 1 , 3 2
- 2 , 5 - 2 , 50 0
0 2 , 5 0
0
1 , 3 2
Example 3.6.6
If X is normally distributed with µ = 25 and σ = 5, i.e. X ∼ n(25; 52), what is the probabilitythat
(a) X is greater than 30?
b) X is less than 15?
Solution
(a) The probability that X is greater than 30 is given by the area under the normal curve,to the right of x = 30, as shown in Figure 3.12(a).
The z-value corresponding to x = 30 is
z =x − µσ=
30 − 25
5= 1.
This is shown in figure 3.12(b).
25
σ = 5
30 x
(a) P(X > 30)0
σ = 1
1 z
(b) P(z > 1)
Figure 3.12: Finding the probability P(X > 30)
The table consists of values between 0 and z thus P(z > 1) = 0,5 − P(0 < z < 1)
Graphically we can explain it as follows:
DSC2602 74
0 0 0
= -
z z z0 , 5
1 1
From the table we find the area from z = 0 to z = 1 is 0,3413. The area to the right ofz = 1 is then 0,5 − 0,3413 = 0,1587.
Therefore, the probability of obtaining a value greater than 30 is 0,1587.
b) The probability that X is less than 15 is given by the area under the normal curve, tothe left of x = 15, as shown in Figure 3.13(a).
The z-value corresponding to x = 15 is
z =x − µσ=
15 − 25
5= −2.
(See Figure 3.13(b).)
2515
σ = 5
x
(a) P(X < 15)0−2
σ = 1
z
(b) P(z < −2)
Figure 3.13: Finding the probability P(X < 15)
P(z < −2) = P(z > 2) as the normal curve is symmetrical. From the z-tables we find thatthe area from z = 0 to z = −2 is the same as from z = 0 to z = 2 namely 0,4772. Thus,the area to the right of z = +2 or left of −2 is
0,5 − 0,4772 = 0,0228.
The probability of obtaining a value less than 15 is therefore 0,0228.We can explain this graphically as follows:
75 DSC2602
-=
=0 z2-
0 , 5
2 z0
2 z0z0
It often happens that we need to determine a specific value of a normally distributed randomvariable such that a given proportion of the area will be to the left or right of this value. In otherwords, we are given a cumulative probability and wish to determine the corresponding quantile value xof the random variable.
Example 3.6.7
Suppose it is known that the marks obtained for an assignment are adequately modelled bya normal distribution with mean 75 and standard deviation 10. What particular mark musta student receive so that only 10% of the students will have higher marks?
Solution
Remember that we can only use our table for values between 0 and z. Thus we have tochange the area to be suitable for the table. If 10% of the students’ marks exceed x, then 90%of the marks is less than or equal to x, i.e. P(X ≤ x) = 0,9.
µ = 75 X∗
10%
x
Figure 3.14: 10% of students’ marks exceed x
The problem is to find a z-value that has 90% of the area to the left of z. We know that thearea can be split into two areas as follows:
DSC2602 76
0 0
= +
0z z
9 0 % 4 0 %5 0 %
z
Therefore, we look in the probability portion inside the z-table to find a z-value that has 0,4of the area to its left. We use 0,4 (and not 0,9) since 0,5 of the 0,9 is to the left of z = 0. Theclosest entry is 0,3997, which corresponds to z = 1,28. We now find the value of x as follows:
z =X∗ − µσ
= 1,28
which can be simplified to
X∗ − µ = 1,28 × σ
giving
X∗ = (1,28 × σ) − µ.
We know that giving X∗ ∼ n(µ,σ2) ∼ n(75; 102), from which it follows that
X∗ = (1,28 × 10) − 75
= 87,8.
Approximately 90% of the assignment marks will therefore be less than 88%. If a particularstudent gets 88% for his or her assignment, only 10% of the students will have a better markfor the assignment.
3.6.2 The exponential distribution
A continuous probability distribution that is often useful in describing the time needed to completea task is the exponential probability distribution. The exponential random variable can describe thetime between telephone calls, the time required to load a ship, and so on. The exponentialdistribution is a continuous distribution. Its probability density function is
f (x) =1
λe−x/λ for x ≥ 0, λ > 0
wherex = random variable (service times, etc.),λ = average number of units the service facility can handle
in a specific period of time, ande = 2,7183 (the base of natural logarithms).
77 DSC2602
The general shape of the exponential distribution is shown in Figure 3.15.
x
f (x)
Figure 3.15: The exponential distribution
It can be shown that for the exponential distribution,
Expected value =1
λand Variance =
1
λ2.
For any probability distribution the area under the curve corresponding to some interval, providesthe probability that the random variable assumes a value in that interval.
For example, the probability that a truck is loaded in 20 minutes or less (x ≤ 20) is the area underthe curve from x = 0 to x = 20. If the mean, average time to load a truck is 18 minutes (λ = 18),the probability density function is
f (x) =1
18e−x/18.
The cumulative probability distribution is
F(x) = P(X ≤ x) = 1 − e−x/λ.
This formula provides the probability of obtaining a value for the exponential random variableof less than or equal to some specific value of X.
The probability that 20 minutes or less is needed to load a truck isF(20) = P(X ≤ 20)
= 1 − e−20/18
= 0,671.
3.6.3 The uniform distribution
The uniform probability density function, f , described by
f (x) =
1
b − afor a ≤ x ≤ b,
0 elsewhere,
DSC2602 78
is used to describe a continuous probability distribution that has a range of continuous valuesfrom point a to point b, inclusive.
The uniform distribution is often referred to as the rectangular distribution. It assumes that thereis an equally likely chance that any point along the range will occur. The probability densityfunction looks like a box, as shown in Figure 3.16.
a bx
f(x)
1b - a
Figure 3.16: The uniform probability density function
The graph of the probability density function, f , shows the height or value of the function at anyparticular value of x. Because we have a uniform pdf, the height, or value, of the function is thesame for each value of x between a and b.
It can be shown that the mean and variance of the uniform distribution are
Mean = µ =a + b
2and Variance = σ2 =
(b − a)2
12.
Suppose we want to find the probability that a chosen value will fall within an interval inside[a; b]. This is done by locating the interval and then finding the area enclosed by the x-axis,the pdf and the boundaries of the interval. Since the area is always rectangular, this is done bymultiplying the base of the rectangle, i.e. the interval, by the height which is always equal to1/(b − a).
Example 3.6.8
If a number is chosen at random from a set of numbers varying from 3 to 7, what is theprobability that the number will be between 5 and 6?
Solution
The numbers varying between 3 and 7 are uniformly distributed, seeing that each numberhas the same chance to be chosen. Since a = 3 and b = 7, the pdf is given by
f (x) =
1
7 − 3for 3 ≤ x ≤ 7,
0 elsewhere.
79 DSC2602
To find the probability P(5 < x < 6), we need to find the area of the shaded rectangle inFigure 3.17. Therefore,
P(5 < x < 6) = base × height
= (6 − 5) × 1
4= 0,25.
17 - 3
f ( x )
x3 4 5 6 7Figure 3.17: Uniform distribution for random numbers
Example 3.6.9
A project manager would like to determine the probability that a project will be between60% and 70% complete in six month’s time. At present, the project is 50% complete and themanager believes that project completion follows a uniform probability density functionthat ranges from 50% complete, the current state, to 100% complete.
Solution
Here a = 50 and b = 100. The pdf is therefore given by
f (x) =
1
50for 50 ≤ x ≤ 100,
0 elsewhere.
To find the probability P(60 < x < 70), we need to find the area of the shaded rectangle inFigure 3.18. Therefore,
P(60 < x < 70) = base × height
= (70 − 60) × 1
50= 0,20.
DSC2602 80
5 0 x
f ( x )
11 0 0 - 5 0
6 0 7 0 8 0 1 0 09 0
Figure 3.18: Uniform distribution for project
3.7 Exercises
1. A study of the memberships of boards of directors of major companies found that thedistribution in the table below provides an adequate approximation of the probabilitydistribution of x, the number of members on the board of a certain company who are alsomembers of at least one other major company. Find the mean and the standard deviationof x.
x f(x)
1 0,522 0,223 0,194 0,045 0,03
2. The telephone at Surety Insurance Agency is engaged 30% of the time. Assume that theengagement of the telephone is binomially distributed – the telephone is either engaged ornot!
(a) Determine the mean and the variance of the distribution, assuming that five calls aremade during the time under consideration.
(b) Suppose five calls are made to Surety, what is the probability that one of the five callerswill get an engaged signal?
(c) If five calls are made, what is the probability that less than two callers will get anengaged signal?
(d) If five calls are made, what is the probability that four or more callers will get anengaged signal?
(e) If five calls are made, what is the probability that at least two callers will get an engagedsignal?
3. On average six people per hour use a self-service banking facility in a department storeduring the prime shopping hours. What is the probability that
(a) three people will use the facility during the next hour?
(b) two people will use the facility during the next ten minutes?
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4. The ages of patients admitted to the intensive care unit of a hospital are normally distributedaround a mean of 60 years with a standard deviation of 12 years. What percentage of patientsare
(a) between 45 and 78 years old?
(b) older than 60 years?
(c) older than 78 years?
(d) younger than 45 years?
(e) between 30 and 45 years?
(f) What is the minimum age of the oldest 20% of the patients?
5. A venture capital company feels that the rate of return on a proposed investment is ap-proximately normally distributed, with a mean of 30% and a standard deviation of 10%.Determine the probability that
(a) the return will exceed 55%,
(b) the return will be less than 22%.
CHECKLISTYou should now be able to
⊲ calculate probabilities and cumulative probabilities for dis-crete variables by using
– the binomial distribution;
– the Poisson distribution;
⊲ name the characteristics of a continuous distribution function;
⊲ define the probability density function of a continuous distri-bution function;
⊲ calculate probabilities using
– the standard normal distribution,
– the exponential distribution, and
– the uniform distribution.
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CHAPTER 4
Estimation
4.1 Introduction
E is part of our everyday lives. For example, a housewife buying her monthlygroceries makes an estimate of how much of each product she has to buy.
It is therefore clear that any decision taken by a person or an institution is usually preceded byan estimate. To a large extent, the correctness of the decision is determined by the accuracy of theestimate. However, an estimate of an unknown quantity is impossible if no relevant informationis available.
Much information can be gained from a sample. As mentioned previously, the average of asample can be used as an estimate of the population average. We can also obtain an estimate ofthe population standard deviation on the basis of samples. Thus, one use of sample data is toestimate certain unknown quantities of the population. As mentioned, statistical inference is theprocess whereby statisticians make predictions about a population on the basis of samples.
In this chapter the estimation of a population mean (or average), population variance and popu-lation proportion will be discussed.
4.2 Types of estimators
There are two ways of estimating a population parameter: point estimation and interval estimation.
A point estimate is a single number, based on sample data, that is used to estimate the correspondingunknown population parameter.
Since even a good point estimate is only an approximation, an interval estimate or confidenceinterval gives us an interval of possible values that the unknown parameter can take on.
83
DSC2602 84
4.3 Point estimation
4.3.1 Estimating the mean
Suppose we need to know what the mean and the variance of a particular population attributeare. If we randomly sample the population and obtain n observations X1, X2, . . ., Xn then the
sample mean X is used as an point estimator for the population mean µ such that
µ = X =1
n
n∑
i=1
Xi.
(The hat indicates that it is an estimator.)
The sample mean x is called a point estimate for the population mean µ.
Note: An estimator for the mean is denoted by X, which is a formula, while an estimate of the meanis denoted by x, which is a value.
4.3.2 Estimating the variance
The population variance is usually unknown. The estimator for the population variance is usuallythe sample variance
σ2 = s2 =1
n − 1
n∑
i=1
(Xi − X)2.
4.3.3 Estimating proportions
Suppose we want to know what proportion of the population has some particular attribute, forexample, is male and smokes.
Then if we have drawn a random sample of size n from the population and X objects were foundto be males and smoking an estimator for the proportion of the population that has these attributesis
p =X
n.
From the previous chapter we know that X ∼ b(n; p) and that for a binomial random variable itholds that
E(X) = np and Var(X) = np(1 − p).
We deduce that for our estimator, p, it holds that
E(X
n) = p and Var(
X
n) =
p(1 − p)
n.
When a sample of size n is drawn from a binomial population with proportion parameter p, thenthe sample proportion has an approximately normal distribution with a mean p and standard
deviation
√
p(1 − p)
nas long as np > 5 and n(1 − p) > 5.
85 DSC2602
4.4 Interval estimators
4.4.1 The standard error
When estimating a population parameter such as the mean, the variance, or a proportion, it is alsoimportant to know how accurate the estimator is. This accuracy is usually expressed in terms ofthe estimator’s standard deviation and sample size and is called the standard error of the estimate.
If we use X to estimate µ and s2 to estimate σ2, then the standard error of X is s/√
n. Similarly, ifwe use p = X/n as an estimator for the population proportion p, then the standard error of p is
√
p(1 − p)
n.
4.4.2 Confidence intervals for means
Sometimes we hear statements like: “The average price of all new motorcars is between R90 000and R200 000 or “The average age of all South Africans is between 25 and 30”. These are examplesof interval estimates. Actually, in statistics it is customary to give not only an interval estimatefor a parameter, but also the probability that the method used to find this interval will lead toan interval that contains the parameter. This probability is called the level of confidence and theresulting interval is called a confidence interval.
A confidence interval is based on a confidence level. If we want to construct a 95% confidence interval,the level of confidence is 95% or 0,95.
This tells us that there is a probability of 0,95 that the population mean is in this interval and aprobability of 0,05 that it is not.
If 100 such intervals are constructed by taking random samples from the population, it is likelythat 95 of the intervals would include the population mean and five would not. The populationmean is either in the interval or not.
By the Central Limit Theorem the values of the sample mean x for random samples of size n havea distribution that is approximately normal with
µx = µ and σx = σ/√
n
when n is sufficiently large (n ≥ 30 will usually suffice.) Hence
z =x − µx
σx=
x − µσ/√
n
is approximately standard normal distributed.
If we want to determine a 95% confidence interval we want to determine upper and lower limitvalues (z-values) so that 95% of the values falls in the interval. Graphically we can denote it asfollows:
DSC2602 86
2 , 5 %2 , 5 % 9 5 %
2 , 5z- m 2 , 5z+
Mathematically we can write it as
P(−z0,025 ≤x − µσ/√
n≤ z0,025) = 0,95
The value 95% is known as the level of confidence and is also denoted by 1 − α. The 5%(2,5% × 2) not included in the interval or chance of error is denoted by α.
0 zα/2−zα/2
(1 − α)
α/2α/2
By using the standard normal table we can determine the z-values as follows:
First we convert the 0,95 z-area to one that is compatible with Appendix B2.
( s y m m e t r y )
+
0 0 0
=
0
0 , 9 5 0 , 4 7 5
0 , 4 7 5
0 , 4 7 5
0 , 0 2 5z- 0 , 0 2 5z- 0 , 0 2 5z+
0 , 0 2 5z+
=0 , 0 2 5z+
2 ×
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Next we determine the z-value that corresponds with an area or probability of 0,4750. Remember,to read from the inside of the table to the outside. A z-value of 1,96 is obtained.
P
(
−1,96 ≤x − µσ/√
n≤ 1,96
)
= 0,95
We want a confidence interval for the mean (µ) of the population. By algebraic manipulation wederive a 95% confidence interval for the population mean:
P
(
x − 1,96σ√
n≤ µ ≤ x + 1,96
σ√
n
)
= 0,95
In general:
P
(
x − zα/2σ√n≤ µ ≤ x + zα/2
σ√n
)
= 1 − α
The 100(1 − α)% confidence interval for µ is:
[
x − zα/2σ√
n; x + zα/2
σ√
n
]
Commonly used confidence coefficients are shown in Table 4.1.
Confidence Lower confidence Upper confidence
coefficient (1−α) α α/2 zα/2 limit limit
0,90 0,10 0,05 1,645 x − 1,645σ/√
n x + 1,645σ/√
n
0,95 0,05 0,025 1,96 x − 1,96σ/√
n x + 1,96σ/√
n
0,99 0,01 0,005 2,58 x − 2,58σ/√
n x + 2,58σ/√
n
Table 4.1: Confidence limits for µ
The above interval is suitable if the sample size is more than 30 and the population standarddeviation, σ, is known.
Sometimes, however, the sample size is more than 30 but we do not have the population standarddeviation. In this case we use the sample standard deviation as an estimator for the populationstandard deviation.
Example 4.4.1
A sample of 100 employees (n = 100) from a company was selected, and the annual salaryfor each was recorded. The mean and standard deviation of their salaries were found to be
x = R60 000 and s = R20 000.
DSC2602 88
Construct a 95% confidence interval for the population average salary, µ.
Solution
The 95% confidence limits arex ± 1,96
σ√
n.
Using s to estimate σ, we obtain
60 000 ± 1,9620 000√
100or 60 000 ± 3 920.
In repeated sampling 95% of the time, intervals constructed in such a manner will encloseµ. Hence, we are fairly confident that the average salary for employees of the company iscontained within the interval [R56 080; R63 920].
If we have a small sample (n < 30) and the standard deviation of the population is known wecan use the normal distribution.
Example 4.4.2
A company that produces snack foods uses a machine to package 454-gram bags of pretzels.
To check whether the machine is working properly the quality assurance department takes arandom sample of 24 bags. The mean weight is 450 grams. It is known from the specificationsof the machine that the standard deviation is 7,8 grams. Determine a 99% confidence intervalfor the true mean weight.
Solution
Givenn = 24
x = 450
σ = 7,8.
The confidence interval isx ± z α
2× σ√
n
= 450 ± 2,58 × 7,8√
24= 450 ± 4,1078.
The 99% confidence interval for the true mean weight is [445,8922; 454,1078]. The true meanlies between 445,8922 grams and 454,1078 grams.
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If we have a small sample (n < 30) and the standard deviation of the population is not known andmust be estimated by the standard deviation of the sample, we do not use the normal distribution.In such a situation the t-distribution must be used.
Instead of using z, we use t. In essence, we substitute zα/2 with tα/2;n−1 where n − 1 gives thedegrees of freedom within the sample; and we substitute σ with the known s. Thus, our final100(1 − α)% confidence interval for µ is
[
x − tα/2;n−1s√
n; x + tα/2;n−1
s√
n
]
.
The critical values of the t-distribution are dependent on the sample size and the table AppendixB.3 looks different from the normal table. In this case we work with the concept of degrees offreedom. The theory underlying the concept of degrees of freedom is beyond the scope of thismodule.
Example 4.4.3
A large insurance company is conducting a study to determine if a downward change inits automobile collision rates can be justified. As part of the study the company needs toestimate the average damage (in Rand) to a new car that hits a barricade at 40 km per hour.
Because of the high cost of this type of sampling the insurance company has decided tocrash only 10 cars.
The average repair cost is R9 364,5 with a standard deviation of R4 072,86. Determine a 95%confidence interval for the average repair cost.
Solution
Since n < 30 namely 10 and because the population standard deviation (σ) is not known wemake use of the t-distribution. The 95% confidence interval is
x ± t0,025;9s√n
x ± 2,262s√n
= 9 364,5 ± 2,2624 072,86√
10= 9 364,5 ± 2 913,35.
The interval is (6 451,15; 12 277,85).
The company’s actuaries are confident that the true average repair cost will be betweenR6 451,15 and R12 277,85.
The confidence interval calculation for the mean can be summarised as follows:
DSC2602 90
SConfidence interval
Large sample (n > 30)
σ known x ± z α2× σ√
n
Large sample (n > 30)
σ unknown x ± z α2× s√
n
Small sample (n < 30)
σ known x ± z α2× σ√
n
Small sample (n < 30)
σ unknown x ± t α2 ;n−1 ×
s√
n
Note: σ is the population standard deviation.
4.4.3 Confidence intervals for proportions
The proportion of individuals or objects in a population with a particular attribute is denoted byp. If, in a random sample of size n, we have X individuals or objects with the attribute underconsideration, we estimate p by
p =X
n
When probabilities involving the sample proportion have to be calculated, the value of p can bestandardised by using z:
Z =p − p
√
p(1 − p)
n
.
(Remember p is approximately normal distributed as long as np > 5 and n(1 − p) > 5.)
Similar to the confidence intervals for means, we can write 100(1 − α) % confidence interval for pas
p − zα/2
√
p(1 − p)
n; p + zα/2
√
p(1 − p)
n
.
91 DSC2602
Example 4.4.4
A new coin telephone has been installed in an airport baggage claim area. The first 300times the telephone was used a total of 45 users lost coins. Construct a 95% confidenceinterval for the true proportion of times that a user will lose a coin.
Solution
Here p =45
300= 0,15 and n = 300.
Lower confidence boundary:
p − 1,96
√
p(1 − p)
300= 0,15 − 1,96
√
0,15(1 − 0,15)
300= 0,15 − 1,96(0,02)
= 0,111.
Upper confidence boundary:
p + 1,96
√
p(1 − p)
300= 0,15 + 1,96
√
0,15(1 − 0,15)
300= 0,15 + 1,96(0,02)
= 0,189.
Thus, a 95% confidence interval for the true proportion of times that a user will lose a coinis [0,111; 0,189]. This means that there is a chance of between 11% and 19% that you willlose a coin if you use the new telephone in the airport baggage claim area.
DSC2602 92
4.5 Exercises
1. A clothing store analysed the value of purchases made on credit card by a sample of 25credit card customers. The sample mean was found to be R165,45 with a sample standarddeviation of R38,60. Construct a 95% confidence interval for the actual mean value of creditcard purchases at this store.
2. A recent survey amongst 120 street vendors in Johannesburg showed that 42 of them feltthat local by-laws still hampered their trading. Construct a 90% confidence interval for thetrue population proportion of street vendors who believe this.
3. The Pappi Paper Company wants to estimate the average time required for a new machineto produce a ream of paper. A sample of 36 reams required an average production time of1,5 minutes for each ream. Assuming that the population standard deviation is 0,3 minutes,construct an interval estimate with a confidence level of 95%.
4. The Milky Cove wants to estimate the average number of litres of ice cream sold per day.Twenty business days were monitored and an average of 32 litres was sold daily with astandard deviation of 12 litres. Calculate the confidence limits at the 95% confidence level.
5. The Tug-Bolt Company wants to estimate the percentage of credit customers who havesubmitted cheques in payment for bolts and whose cheques have bounced. A sample of150 accounts showed that 15 customers have passed bad cheques. Estimate at the 95%confidence level the true percentage of credit customers who have passed bad cheques.
CHECKLISTYou should now be able to
⊲ use sample data to estimate
– the mean
– the variance, and
– proportions
for the population from which the data come;
⊲ determine how accurate a population parameter is by meansof
– the standard error,
– confidence intervals for means, and
– confidence intervals for proportions.
CHAPTER 5
Correlation and regression
5.1 Introduction
I chapter we introduce methods for studying statistical relationships between variables.Establishing statistical relationships enables us to identify factors that contribute to variation
in the outcomes. Once the statistical relationship between variables has been determined, it canbe used to predict the value of certain variables.
Correlation analysis and regression analysis are two statistical methods that attempt to quantifyand describe the possible relationship between variables. Correlation analysis examines thestrength of this identified association between variables, while regression analysis is concernedwith quantifying the underlying structural relationship between variables. We will cover simplelinear and non-linear regression models in this chapter.
5.2 Correlation analysis
5.2.1 The scatter diagram
The first step to determining whether a relationship exists between two variables we could plotor graph the data available. A scatter diagram is drawn by plotting (x; y) observations on a graphwhere the independent variable, x, is recorded along the horizontal axis and the dependent variable,y, along the vertical axis.
A visual inspection of the possible relationship between the two variables x and y provides uswith initial insight into the likely regression and correlation results.
Example 5.2.1
Suppose that a sales manager has records containing data on annual sales volumes andyears of experience. The information is summarised in Table 5.1.
93
DSC2602 94
Sales Years of Annual salesperson experience (in R1 000s)
1 1 902 2 1073 3 1024 4 1125 6 1136 8 1217 10 1298 10 1339 11 134
10 13 136
Table 5.1: Historical sales figures and years of experience for 10 sales persons
Plot the data on a graph with years of experience on the horizontal axis and annual saleson the vertical axis. We now have a scatter diagram, so-called because the plotted pointsare “scattered” over the graph or diagram. The scatter diagram for these data is shown inFigure 5.1.
0 1 2 3 4 5 6 7 8 9 10 11 12 13 1480
90
100
110
120
130
140
An
nu
alsa
les
(R1
000)
Years of experience
•
••• •
•••• •
Figure 5.1: Scatter diagram of Annual sales versus Years of experience
5.2.2 The correlation coefficient
As we indicated in the introduction to this chapter, there are some situations in which the decisionmaker is not as concerned with the equation that relates two variables as with measuring theextent to which the two variables are related. In such cases, a statistical technique referred to ascorrelation analysis can be used to determine the strength of the relationship between the two pairsof variables.
The output of a correlation analysis is a number referred to as the correlation coefficient. Thecorrelation coefficient will always have a value between −1 and +1. A value of +1 means perfect
95 DSC2602
positive correlation and corresponds to the situation where all the dots on the scatter diagram lieexactly on a straight line with a positive slope coefficient. A value of −1 means perfect negativecorrelation and again corresponds to the situation where all the points on the scatter diagram lieon a straight line with a negative slope coefficient. Correlation is considered strong when thecorrelation coefficient is close to +1 or −1, and weak when it is close to 0. If the coefficient of linearcorrelation is zero, we say that there is no linear correlation, or that the two variables concerned arenot linearly related.
Note: If the correlation coefficient indicates that there is no linear association, it does not meanthat there is no relationship between the variables. The relation might be non-linear.
These possibilities are indicated in Figure 5.2.
1r = +
y valu
es
x v a l u e s(a) r = +1: Perfect positive linear correlation
y valu
es
x v a l u e s
1r = -
(b) r = −1: Perfect negative linear correlation
y valu
es
x v a l u e s
1r ® +
(c) 0 < r < +1: Positive linear correlation
y valu
es
x v a l u e s
1r ® -
(d) −1 < r < 0: Negative linear correlation
y valu
es
x v a l u e s
0r =
(e) r = 0: No linear correlation
Figure 5.2: Interpretation of correlation coefficients
There are two commonly used measures of correlation: Pearson’s correlation coefficient and Spear-man’s rank correlation coefficient.
DSC2602 96
5.2.2.1 Pearson’s correlation coefficient
Pearson’s correlation coefficient (r) computes the correlation between two quantitative variables.Pearson’s correlation coefficient is defined by
r =n∑n
i=1 xiyi −∑n
i=1 xi
∑ni=1 yi
√
n∑n
i=1 x2i− (
∑ni=1 xi)2
√
n∑n
i=1 y2i− (
∑ni=1 yi)2
wherer = the correlation coefficient,xi = the value of the independent variable for the ith observation,yi = the value of the dependent variable for the ith observation, andn = the number of paired data points.
From Figure 5.2 it is evident that the closer r is to −1 or +1, the stronger the association, and thecloser it is to zero, the weaker the relationship between the two pairs of variables.
Example 5.2.2
Calculate the correlation coefficient between years of experience and annual sales for Exam-ple 5.2.1.
Solution
The calculations required to compute the sample correlation coefficient are shown in Ta-ble 5.2.
Sales Years of Annualperson experience sales
i xi yi x2i
xiyi y2i
1 1 90 1 90 8 1002 2 107 4 214 11 4493 3 102 9 306 10 4044 4 112 16 448 12 5445 6 113 36 678 12 7696 8 121 64 968 14 6417 10 129 100 1 290 16 6418 10 133 100 1 330 17 6899 11 134 121 1 474 17 956
10 13 136 169 1 768 18 496
10 68 1 177 620 8 566 140 689
Table 5.2: Calculations required to compute the sample correlation coefficient
From the table we find that
r =10(8 566) − 68(1 177)
√
10(620) − (68)2√
10(140 689) − (1 177)2= 0,9648.
97 DSC2602
Because the correlation coefficient is close to +1, the association between years of experienceand annual sales is a very strong positive correlation. The years of experience can thereforeconfidently be used to estimate annual sales.
Note: You may enter the data pairs (xi; yi) into the statistical mode of your calculator and findthe value of r by pressing a button. You are not required to do the calculations by hand. Seethe manual of your calculator for details.
5.2.2.2 Spearman’s rank correlation coefficient
Spearman’s rank correlation coefficient (Rs) is used to find a measure of association betweentwo random variables if the data is generated by such measurements as rank, or position in asequence, or frequency of occurrence. Spearman’s rank correlation is defined by
Rs = 1 −6∑n
i=1 d2i
n(n2 − 1)
wheren = the number of paired ranks, anddi = the difference in ranks.
Example 5.2.3
At a cheese tasting function, two judges were asked to independently rank 10 of the cheesestasted from most desirable (rank = 1) to least desirable (rank = 10).
The cheeses were ranked as follows:
Cheese I II III IV V VI VII VIII IX X
Judge A 7 2 3 4 1 10 5 8 6 9Judge B 6 3 2 5 1 9 4 7 8 10
First we calculate the differences between the judges preferences:
Cheese I II III IV V VI VII VIII IX X
Judge A 7 2 3 4 1 10 5 8 6 9Judge B 6 3 2 5 1 9 4 7 8 10
Difference di 1 −1 1 −1 0 1 1 1 −2 −1
d2i
1 1 1 1 0 1 1 1 4 1
Now, since
10∑
i=1
d2i = 12 and n = 10 it follows that
Rs = 1 − 6(12)
10(100 − 1)= 0,927.
Spearman’s correlation coefficient is interpreted in the same way as a Pearson’s correlationcoefficient. In this example, the judges’ preferences coincide very closely, as shown by thehigh Rs value of 0,927.
DSC2602 98
5.3 Simple linear regression
5.3.1 The estimated regression line
In this section we limit ourselves to the task of finding the straight line that “best” fits the data ina scatter diagram. Simple linear regression aims to find a linear relationship between the valuesof two random variables only. Thus, we will be fitting a function that has the form
y = b0 + b1x
where
y = the estimated value for the dependent variable,x = the value of the independent variable,b0 = the intercept on the y axis (value of y when x = 0), andb1 = the change in the dependent variable corresponding to the
change in the independent variable (i.e. the slope of the line).
In regression analysis, the function that we use to describe the data is referred to as the estimatedregression function. Regression analysis dealing with a straight-line relationship involving onedependent and one independent variable is often referred to as simple linear regression. Where twoor more independent variables are used to predict values of the dependent variable, we (can) usemultiple regression. This, however, is beyond the scope of this module.
5.3.2 The method of least squares
When we draw an estimated regression line, the points on the scatter diagram will not all lieon the line. Some will lie above it and some below it. The difference between any point andthe corresponding point on the regression line is called an error, or residual, and is denoted by ei,where
ei = yi − yi or ei = yi − b0 − b1xi.
This represents the difference in value between what was predicted and what actually happened.It is shown graphically in Figure 5.3.
The least squares method seeks to estimate the slope and intercept parameters that minimise thesum of squared errors, or residuals, SSE, that is
Minimise
n∑
i=1
e2i =
n∑
i=1
(yi − b0 − b1xi)2.
It can mathematically be shown that, by taking partial derivatives of the sum of squared residualswith respect to b0 and b1, setting the partial derivatives equal to zero and solving the two equationssimultaneously,
b1 =n∑n
i=1 xiyi −∑n
i=1 xi
∑ni=1 yi
n∑n
i=1 x2i− (
∑ni=1 xi)2
and b0 = y − b1x
99 DSC2602
b
b b
b
b
b
b
b
b
b
b
b
x
y
Observed value
Predicted value
Difference (error or residual)
Estimatedregressionline
Figure 5.3: Differences between observed and predicted values
where xi = the value of the independent variable for the ith observation,yi = the value of the dependent variable for the ith observation,x = the mean value of the independent variable,y = the mean value of the dependent variable, andn = the total number of observations.
Example 5.3.4
Set up a regression line using the method of least squares to estimate annual sales, usingthe data in Example 5.2.1. (This is also called the least squares regression line.)
From Table 5.2 the following values are calculated:
x =
∑ni=1 xi
n=
68
10= 6,8 and y =
∑ni=1 yi
n=
1 177
10= 117,7.
The slope (b1):
b1 =n∑n
i=1 xiyi −∑n
i=1 xi
∑ni=1 yi
n∑n
i=1 x2i− (
∑ni=1 xi)2
=10(8 566) − 68(1 177)
10(620) − (68)2
= 3,57.
The intercept (b0):b0 = y − b1x
= 117,7 − 3,56(6,8)
= 93,43.
Thus, the estimated regression function found by using the method of least squares is
y = 93,43 + 3,56x.
DSC2602 100
Note that the slope (b1) is positive. This indicates that as years of experience (x) increase,so do annual sales (y). We can therefore say that there appears to be a positive relationshipbetween x and y in this problem.
Now compute the sum of the squares of the differences between the observed values ofthe annual sales and the estimated values based on the least-squares estimated regres-sion function y = 93,43 + 3,56x. The calculations are shown in Table 5.3 and we see that∑n
i=1(yi − yi)2 = 149,82.
Sales Years of Annualperson experience sales
i xi yi yi (yi − yi) (yi − yi)2
1 1 90 97,0 −7,0 49,002 2 107 100,5 6,5 42,253 3 102 104,1 −2,1 4,414 4 112 107,7 4,3 18,495 6 113 114,8 −1,8 3,246 8 121 121,9 −0,9 0,817 10 129 129,1 −0,1 0,018 10 133 129,1 3,9 15,219 11 134 132,6 1,4 1,96
10 13 136 139,8 −3,8 14,44
10 68 149,82
Table 5.3: Calculations of∑n
i=1(yi − yi)2 for the least-squares estimated regression line
The least-squares method guarantees that for this problem no other straight line exists witha sum of squares less than 149,82.
Note: You may once again enter the data pairs into the statistical mode of your calculator. Thevalues of b0 and b1 are readily available and it is not necessary to calculate them by hand. Theintercept and the slope of the regression line may not be b0 and b1, but may be a and b. See themanual of your calculator for these details.
5.3.3 Rules and assumptions underlying regression analysis
In order to conduct further statistical analysis involving the estimated regression function, thereare certain rules that have to be followed and assumptions that must be met.
A 1:The dependent and independent variables must have a linear relationship. Neither the regressionfunction nor the correlation coefficient accurately reflects the true relationship between observedvariables, unless the relationship is linear. (Alternative models for nonlinear relationships will bepresented in Section 5.4.)
A 2:The values of the dependent variable, y, are normally distributed for each value of the independentvariable, x.
101 DSC2602
A 3:If the scatter pattern around the regression line is consistent for all values of the independentvariable, x, it is homoscedastic (the variance is constant), as illustrated in Figure 5.4(a). If thevariation around some values of x is much greater than around others, it is heteroscedastic (thevariance varies), as shown in Figure 5.4(b). In this case the correlation coefficient exaggeratesthe relationship between the two variables for some values of x and underestimates it for others.The significance of the regression function and the accuracy of the estimates depend on howconsistent the scatter pattern is.
0
y
x
(a) Homoscedasticity
0
y
x(b) Heteroscedasticity
Figure 5.4: Examples of scatter patterns around the regression line
5.3.4 Residual plot analysis
For our purposes, we will accept that the assumptions discussed above are true. In practice,however, it is necessary to check the validity of these assumptions carefully. The differencesbetween the observed values of y and the estimated values of y play a major role in such analysis.As mentioned in Section 5.3.2, these differences are called residuals.
An approach that can be used to check the validity of the assumptions involved, is to simply plotthe residuals. The study of the residuals is important for deciding on the appropriateness of agiven forecasting model, especially in the case of multiple regression. (In multiple regression thefitted model is not only more complicated than simple linear regression, but you might end upwith more than one suitable model. In such instances the analysis of the residuals is essential forselecting the best model.)
If the residuals are essentially random, then the model may be a good one. If the errors show anykind of pattern, then the model is not taking care of all the systematic information in the data set.In such instances it may be necessary to try and transform the data in some way.
5.3.5 The coefficient of determination
The coefficient of determination, denoted by R2, is said to represent the proportion of variance in thedependent variable (y) that can be explained by the independent variable (x).
The coefficient of determination is defined as the ratio of these two sums of squares, and isdenoted by R2.
DSC2602 102
Thus
R2 =Sum of squares of explained deviations
Sum of squares of total deviations
=
∑ni=1(yi − y)2
∑ni=1(yi − y)2
.
R2 × 100 gives the percentage of the variation in y that can be explained by the variation in x. R2
is probably the most consulted statistic in regression analysis.
What is the difference between the coefficient of determination (R2) and the sample correlationcoefficient (r)?
In simple linear regression there is no difference between R2 and r2. However the coefficientof determination (R2) will always be positive while the sample correlation coefficient (r) can bepositive or negative. The sign of the sample correlation coefficient is determined by the sign ofthe slope (b1) of the estimated regression line. In regression other than linear regression R2 iscalculated and not r. r is only applicable to linear regression.
If r = 0,7 then R2 = 0,49. This implies that only 49% of the variation in y is explained by thespecific x in our model. It might be a good idea to investigate if the addition of more independentvariables will result in a better model. This is known as multiple regression (which are beyondthe scope of this module). It could be that you must try a non-linear relationship (see Section 5.4)
For simple linear regression:
r = ±√
R2 and R2 = r2.
(Note: For multiple regression the sample correlation coefficient and the coefficient of determi-nation are two different things. Multiple regression analysis is beyond the scope of this module.)
Example 5.3.5
Use the data contained in Example 5.2.1 to calculate the coefficient of determination.
Solution
The calculations for the coefficient of determination are shown in Table 5.4.
From the table we see that
10∑
i=1
(yi − y)2 = 2 005,68 and
10∑
i=1
(yi − y)2 = 2 156,1.
The coefficient of determination is
R2 =
∑ni=1(yi − y)2
∑ni=1(yi − y)2
=2 005,68
2 156,1= 0,93.
This means that 93% of the variation in annual sales can be explained by years of experience.R2 can also be calculated as R2 = 0,96482 = 0,93 from the correlation coefficient.
103 DSC2602
Sales Years ofperson experience Annual sales
i xi yi yi (yi − y)2 (yi − y)2
1 1 90 97,0 428,49 767,292 2 107 100,5 295,84 114,493 3 102 104,1 184,96 246,494 4 112 107,7 100,00 32,495 6 113 114,8 8,41 22,096 8 121 121,9 17,64 10,897 10 129 129,1 129,96 127,698 10 133 129,1 129,96 234,099 11 134 132,6 222,01 265,69
10 13 136 139,8 488,41 334,89
n = 10 y = 117,7 2 005,68 2 156,1
Table 5.4: Calculations for the coefficient of determination
5.3.6 The F-test for overall significance
Up to this point we have discussed the procedure for developing an estimated regression functionand the assumptions necessary to perform further statistical tests. We now want to determinewhether or not the relationship we have established between the x and y variables is statisticallysignificant. If we conclude that the relationship is not significant, we are saying that the variablesare not related. In this case the values of y would be independent of the values of x.
If there is a relationship between x and y, then the slope (b1) will not be equal to 0. The F-testallows us to test the significance of the overall regression model to be able to answer the statisticalquestion: Is there a significant relationship between y and x?
Without going into the mathematical detail, the F-test is defined as
F =
∑ni=1(yi − y)2/(k − 1)
∑ni=1(yi − yi)2/(n − k)
,
where
k = the number of parameters in the regression function,n = the number of observations,
(k − 1) = degrees of freedom, ν1, and(n − k) = degrees of freedom, ν2.
The F-test is therefore a ratio of two mean squares. The numerator refers to the variance explainedby the regression function, and the denominator refers to the variance of what is not explainedby the regression function, i.e. the residuals or errors.
For simple linear regression the number of parameters k = 2 (b0 and b1) and the degrees of freedomare ν1 = 2 − 1 = 1 and ν2 = n − 2.
The F-test is closely linked to the definition of the coefficient of determination. The following
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computational formula for F can be derived:
F =R2/ν1
(1 − R2)/ν2.
To determine whether a significant relationship exists between two variables, this value for F iscompared to the value obtained from the F-tables.
The shape of the F distribution is shown in Figure 5.5.
f(x)
x0Fν1,ν2;α
α
Figure 5.5: The density function of the F distribution
We have to decide beforehand on which level of significance to test. Say we decide on a 1% levelof significance, that is α = 0,01. Looking in the table for 1%, we find the F value corresponding tothe degrees of freedom, ν1 and ν2. If this value (Fν1 ,ν2;α) is less than the calculated F value, that isthe calculated value lies to the right of Fν1,ν2;0,01, or (F > Fν1.ν2;0,01) we can say that the relationshipis significant on a 1% level of significance.
Note: If the F value from the 1% table is not less than the calculated F value, but the F value fromthe 5% table is, we can say that although the relationship is not significant on a 1% level, it issignificant on a 5% level of significance.
Example 5.3.6
Determine whether the regression model fitted in Example 5.3.4 is statistically significant.
Solution
We choose to test the model on a 1% level of significance. From Example 5.3.5 it follows thatR2 = 0,93. Furthermore, k = 2 and n = 10 and the degrees of freedom for the F distributionare ν1 = 1 and ν2 = 10 − 2 = 8. Therefore,
F =R2/ν1
(1 − R2)/ν2=
0,93/1
(1 − 0,93)/8= 106,2.
From the F tables in Appendix B.4 for a 1% level of significance we find that Fν1,ν2;0,01 = 11,3.Since our calculated value of F = 106,2 > 11,3, we can say that the linear relationshipbetween annual sales and years of experience is highly significant on a 1% level.
Note: In the case of simple linear regression, the F-test is the same as testing the significance ofthe slope coefficient.
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5.3.7 Forecasting accuracy
A measure of the accuracy of predictions derived from regression analysis is given by the standarderror of the estimate, which is calculated as
se =
√∑ni=1(yi − yi)
2
n − 2=
√
SSE
n − 2
where
n = the number of observations, andSSE = the sum of squared errors or residuals.
It is usually true that approximately 68% of the values of y will be within se of the predicted valuey, and 95% of the values of y will be within 2se of the predicted value of y.
5.4 Fitting nonlinear relationships
A scatter diagram of two variables often shows that the relationship between them is not linearbut there is a nonlinear relationship of some sort. We can extend linear regression analysis to covermany nonlinear relationships by employing simple transformations of the nonlinear equationsto produce the familiar linear forms. The solutions to these linear cases can then be transformedback to describe the nonlinear relationship.
Let’s look at one example how to go about transforming one nonlinear model to a linear model,and use linear regression analysis to determine the unknown parameters. Table 5.5 summarisesother nonlinear models.
Suppose you plot the (x; y) data values and obtain a graph similar to Figure 5.6(a). Then theappropriate functional relationship may be y = β0 xβ1 .
Transform the function in the following way:
y = β0 xβ1
ln y = ln β0 xβ1
= ln β0 + ln xβ1
= ln β0 + β1 ln x
= β1lnx + lnβ0.
Plot the (ln x; ln y) values on a new graph. If these values are more or less on a straight line,calculate the slope (b1) and the intercept (b0) of the least squares regression line for the transformeddata values.
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0 i n t e r c e p tb =
1 g r a d i e n tb =
l n y
l n x
We obtain:
ln y = β1 × ln x + ln β0
↓ ↓ ↓ ↓y − axis slope x − axis intercept.
Thus β1 is equal to the slope (b1) of the regression line and ln β0 is equal to the intercept (b0) of theregression line. The estimate of the functional relationship is:
y = β0xβ1
= (eb0) × xb1 (since ln β0 = b0).
Table 5.5 and Figure 5.6 show examples of some nonlinear models with the relevant transformationand functional relationships1.
Appropriate Estimate offunctional Transform functional
Graph type relationship (x; y) into relationship
(a) or (b) y = β0xβ1 (ln x; ln y) y = exp(b0)xb1
(c) or (d) y = β0 exp(β1x) (x; ln y) y = exp(b0 + b1x)
(e) or (f) y = β0 + β1 ln x (ln x; y) y = b0 + b1 ln x
(g) or (h) y = xβ0x+β1
(1x; 1
y) y = x
b0x+b1
(i) y = exp(β0 +β1
x) (1
x; ln y) y = exp(b0 +
b1
x)
Table 5.5: Examples of nonlinear relationships
1 Table 5.5 and Figure 5.6 are copied, with permission, from the prescribed book Wayne L. Winston, Operations research,ITP(1994).
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0
y
x1
(b)
0
y
x1
(a)
0
y
x
(c)
0
y
x
(d)
.
0
y
x
(e)
0
y
x
.
(f)
0
y
x1
(g)
0
y
x
(h)
0
y
x
(i)
(β0, β1, x > 0)
β1 > 1
β1 = 1
0 < β1 < 1
β0
β0 β0
(β0 > 0, β1 > 0, x > 0)
−1 < β1 < 0
β1 < −1β1 < −1
(β1 > 0)
β0e
1β1
(β1 < 0)
(β1 < 0)(β1 < 0)
β0
e
− 1β1
β1 > 0
β1 > 0
1β0
1β0
− β1
β0
− β1
β0
Figure 5.6: Graphs of nonlinear functions
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To summarise we can use the following steps to estimate a nonlinear relationship:
Step 1 Draw a scatter diagram of the points (values of the two relevant variables) and findwhich graph in Figure 5.6 appears to fit the data best.
Step 2 Transform each data point according to the rules in the third column of Table 5.5.
Step 3 Estimate the least squares regression line for the transformed data. If b0 is the y-intercept of the least squares line (for the transformed data) and b1 is the slope of theleast squares line (for the transformed data), then we read the estimated relationshipfrom the fourth column of Table 5.5.
Note: Table 5.5 and Figure 5.6 are supplied on the formula sheet in the examination.
Example 5.4.7
An experiment was conducted to determine the relationship between sound intensity, x,and distance, y. The results of the experiment are shown in Table 5.6. Fit an appropriateregression function to describe the relationship between sound intensity and distance.
Distance Sound intensity(in metres) (in decibels)
i xi yi
1 0 95,02 2 82,03 4 77,04 6 73,55 8 71,06 10 68,57 12 66,38 14 65,09 16 63,7
10 18 62,5
Table 5.6: Sound intensity versus distance
Solution
Step 1 The scatter diagram of the data appears in Figure 5.7.
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6 06 57 07 58 08 59 09 5
1 0 0
0 2 4 6 8 1 0 1 2 1 4 1 6 1 8 2 0D i s t a n c e ( i n m e t e r s )
Soun
d inte
nsity
(in de
cibels
)
D i s t a n c e ( i n m e t r e s )
Figure 5.7: Scatter diagram of Sound intensity versus Distance
Step 2 The scatter diagram indicates that the relationship is not linear and has the same shapeas Figure 5.6(d). The relationship can, according to Table 5.5, be described by the
relationship y = β0 exp(β1x) and the estimate by y = e(b0 + b1x).
Step 3 The data need to be transformed by the transformation (xi; ln yi) as given in column 3of Table 5.5, that is (0; ln 95), (2; ln 82), . . . ,(18; ln 62,5).
Enter these transformed data pairs into the statistics mode of your calculator to obtainthe least squares coefficients b0 = 4,45 and b1 = −0,02 and the correlation coefficientr = −0,945. The estimated functional relationship is therefore
y = e4,45−0,02x.
Note: Without the transformation, that is when (x; y) is used to determine the regressionline, the correlation coefficient is −0,92. This seems good enough, but an analysis ofthe residuals will indicate that there is room for improvement in the simple linearregression model.
Example 5.4.8
Sales of VCRs for the last 24 months are given in Table 5.7. We want to predict future VCRsales.
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Month Sales Month Sales Month Salesxi yi xi yi xi yi
1 23 9 2 366 17 3 5422 156 10 2 942 18 3 3123 330 11 2 872 19 3 5474 482 12 2 937 20 3 3755 1 209 13 3 136 21 3 3756 1 756 14 3 241 22 3 4037 2 000 15 3 149 23 3 6978 2 512 16 3 524 24 3 495
Table 5.7: Monthly sales of VCRs
Solution
The first step is to draw a scatter diagram of the data, as shown in Figure 5.8.
05 0 01 0 0 01 5 0 02 0 0 02 5 0 03 0 0 03 5 0 04 0 0 0
0 2 4 6 8 1 0 1 2 1 4 1 6 1 8 2 0 2 2 2 4M o n t h
Sales
Figure 5.8: Scatter diagram of Sales versus Month
The scatter diagram indicates that the relationship is not linear and that the data needs to betransformed. The relationship appears to be S-shaped, like Figure 5.6(i), and an appropriate
functional relationship is y = exp(β0 +β1
x).
The data should be transformed to ( 1xi
; ln y). Enter the transformed data pairs(1/1; ln 23), (1/2; ln 156), . . . (1/24; ln 3 495) into the statistics mode of your calculator.The least squares regression coefficients for the the transformed data are b0 = 8,387 andb1 = −5,788. The estimated relationship between x and y is given by
y = exp(
8,387 − 5,788
x
)
.
Either one of these approaches can be used. Sometimes the one and sometimes the otherwill provide better results, especially if an in-depth analysis is done on the residuals of thefitted model.
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5.5 The use of spreadsheets in simple linear regression
Spreadsheets can be helpful in relieving the computational burden in regression analysis. Theuse of the data analysis tool in Excel is discussed here.
Excel 2003 was used for illustration. The steps might differ in other versions.
We illustrate the use of spreadsheets for the regression analysis with the following data on theassessment value of houses in a small town and the loans granted on them (both in R1 000s).
House Assessment value Loan grantedi xi yi
1 13,0 16,52 24,0 22,43 31,0 28,84 39,0 36,05 22,0 27,06 19,3 19,27 23,5 20,88 14,5 10,59 24,0 22,0
10 12,0 11,0
• Open a new file in Excel.
• Enter the data for the independent variable (x) in cells A2:A11 and the data for the dependentvariable (y) in the cells B2:B11. (See Table 5.8).
• Select the Tools pull-down menu.
• Select the Data Analysis option. (If this option is not shown on your menu, select Add-ins,mark Analysis Toolpak and click on OK.)
• When the Analysis Tools dialogue box appears, choose Regression.
• When the Regression dialogue box appears:
– Enter the range of the y values in the Input Y Range box by clicking on the first y value,typing : and then clicking on the last y value, giving $B$2:$B$11. Do the same for thex values in the Input X Range box to get $A$2:$A$11.
– Select the Output Range box and type in A12, the cell in which you want the analysisoutput to start.
– Click on OK to get the output.
From the output of the regression analysis, we get the following information:
• The coefficients of the regression line (cells B28 an B29) are b0 = 1,7332 and b1 = 0,8856. Theleast squares regression line is therefore y = 1,7332 + 0,8856x.
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1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
A B C D E F G
x y
13,0 16,5
24,0 22,4
31,0 28,8
39,0 36,0
22,0 27,0
19,3 19,2
23,5 20,8
14,5 10,5
24,0 22,0
12,0 11,0
SUMMARY OUTPUT
Regression Statistics
Multiple R 0,9377
R Square 0,8793
Adjusted R Square 0,8642
Standard Error 2,8987
Observations 10
ANOVA
df SS MS F Significance F
Regression 1 489,5973 489,5973 58,2692 0,0001
Residual 8 67,2187 8,4023
Total 9 556,8160
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%
Intercept 1,7332 2,7371 0,6332 0,5443 -4,5786 8,0449
X Variable 1 0,8856 0,1160 7,6334 0,0001 0,6181 1,1531
0,0
10,0
20,0
30,0
40,0
0,0 10,0 20,0 30,0 40,0 50,0
Assessment value
Lo
an
gra
nte
d
Table 5.8: Regression analysis using Excel
• The calculated F value is 58,2692 (cell E23). On a 5% level of significance the value obtainedfrom the F tables is F1,8;0,05 = 5,32. Since the calculated value is greater than the valueobtained from the tables, we can say that the regression line is significant.
The value of 0,0001 for Significance F (cell F23) is actually the same as the level of significance.Most statistical packages give a calculated level of significance (also referred to as a p value)that can be evaluated by the analyst to determine whether the model, or even the coefficients,are significant. In this case the level of significance is 0,0001 which is less than α = 0,01 andwe can say that the regression line is significant on a 1% level of significance, or even highlysignificant.
• The correlation coefficient r is equal to the value in cell B15, thus r = 0,9377.
• The coefficient of determination R2 is equal to the value in cell B16, thus R2 = 0,8793.
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5.6 Exercises
1. The marketing manager of a large supermarket chain would like to determine the effect ofshelf space (in m2) allocated to pet food on the sales of pet food. A random sample of 12equal-sized stores is selected, with the following results:
Store Shelf space, X Weekly sales, Y(m2) (thousands of rand)
1 5 1,62 5 2,23 5 1,44 10 1,95 10 2,46 10 2,67 15 2,38 15 2,79 15 2,8
10 20 2,611 20 2,912 20 3,1
(a) Assume a linear relationship and determine the weekly sales function by using regres-sion.
(b) Is the linear relationship assumption in (a) adequate?
(c) Use the F-test to determine whether the fitted line is statistically significant on a 5%level of significance.
(d) Predict the average weekly sales of pet food (in thousands of rand) for a new store with8m2 of shelf space for pet food.
2. The following data presented in Table 5.9 represent examination scores for two modules,QMG201-S and QMG203-U of 28 students in Decision Sciences.
This data was entered onto a worksheet and an output was obtained. See Figure 5.9
(a) Does a linear relationship exists between the examination scores of the two modules
(b) Determine the formula of the linear regression model using the computer output.
(c) How well does the model fit the data? Justify your answer.
(d) Is the linear regression model statistically significant for this data on a 5% level ofsignificance?
(e) How would you beter the fit of the data?
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Student QMG201-S QMG203-U Student QMG201-S QMG203-Ui xi yi i xi yi
1 47 33 15 48 392 54 49 16 49 373 48 40 17 53 424 47 44 18 53 405 50 48 19 49 406 45 36 20 52 477 50 35 21 40 378 56 50 22 48 349 54 46 23 43 21
10 48 37 24 50 4011 53 40 25 49 3412 47 39 26 52 3913 38 32 27 48 3814 47 42 28 42 34
Table 5.9: Examination scores
A B C D E F G 1 SUMMARY OUTPUT 2 3 Regression Statistics 4 Multiple R 0.672112581 5 R Square 0.451735322 6 Adjusted R Square 0.430648219 7 Standard Error 4.557416301 8 Observations 28 9
10 ANOVA 11 df SS MS F Significance F 12 Regression 1 444.943159 444.943159 21.42235 8.96361E-05 13 Residual 26 540.0211268 20.77004334 14 Total 27 984.9642857 15 16 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% 17 Intercept -7.39788732 10.06916283 -0.73470729 0.469093 -28.09534775 13.299573 18 X Variable 1 0.955985915 0.206546542 4.628428569 8.96E-05 0.531423422 1.3805484
0
20
40
60
0 20 40 60
QMG201-S scores
QM
G20
3-U
sco
res
Figure 5.9: Regression analysis using Excel
3. To determine how price influences sales, a company changed the price of a product over a20-week period. The price charged each week and the number of units sold is given in thefollowing table:
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Price Number of Price Number of(in R) units sold (in R) units sold
1 1 145 11 1792 788 12 2323 617 13 1834 394 14 1815 275 15 2226 319 16 2127 289 17 1868 241 18 1109 259 19 183
10 176 20 172
Fit an appropriate regression function on the data to describe the relationship between priceand number of units sold.
CHECKLISTYou should now be able to
⊲ understand the method of least squares;
⊲ calculate Spearman’s rank correlation coefficient;
⊲ use the statistics mode of a pocket calculator to find
– the least squares regression line and
– the correlation coefficient
for given data;
⊲ use the correlation coefficient to determine how well the esti-mated line fits the data;
⊲ use the F-test to determine the significance of the fitted regres-sion line;
⊲ identify a nonlinear relationship from a scatter diagram;
⊲ determine an appropriate nonlinear functional relationship forgiven data;
⊲ use the Data Analysis tool in Excel to do regression analysis ofgiven data and interpret the output.
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CHAPTER 6
Decision analysis
6.1 Introduction
I previous chapters we studied techniques for manipulating data in order to make decisionsabout population parameters and population characteristics.
In this chapter we will study another type of decision making where costs and profits are consid-ered to be important. The problem is not whether to accept or reject a statement but to select thebest alternative from a list of several possible decisions.
Example 6.1.1
Timberhut is a company that manufactures a range of wooden huts, but has not yet designedone specially for storing garden tools. Peter Mbeko is the financial manager who has todecide whether or not the company should manufacture and market the new product.Depending on whether the market is favourable or unfavourable, the company will decide toconstruct a large plant for production, a small plant or do nothing at all.
This example involves making a choice from various alternatives. Making such choices is calleddecision making. Decision making is difficult because the consequences associated with decisionalternatives are different, the consequences are not all equally valued and there is uncertaintyabout the consequences of each alternative.
Making decisions involves structuring the decision problem, assessing the possible impacts ofeach alternative, determining the preferences or values of the decision makers, and evaluatingthe alternatives.
Decision analysis is the study of how people make decisions, particularly when faced withimperfect or uncertain information, as well as a collection of techniques to support the analysisof decision problems.
Decision analysis differs from other modelling approaches by explicitly considering preferencesand attitudes, and modelling the decision process itself. Its purpose is to produce insight andpromote creativity to help people make better decisions.
117
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In this chapter we study the techniques formally known as decision analysis. We will discussboth deterministic and probabilistic approaches that explicitly address the uncertainty inherentin many decision-making problems. For example:
If you decide to invest your money in a savings account with a fixed interest rate for, say, a year,then you are dealing with decision making under certainty.
If you invest money and the interest rate varies according to economic conditions the decisionbecomes more complicated.
If we have an idea on what the chance or probability of a economic condition is, we have decisionmaking under risk.
When the probabilities are not known, we talk about decision making under uncertainty.
6.2 Structuring decision problems
6.2.1 The basic steps in decision making
Whether deciding about having a haircut today, building a multi-million rand plant, or buying anew camera, the steps in making a good decision are basically the same.
These basic steps are as follows:
1. Clearly define the problem at hand.
2. List the possible alternatives.
3. Identify the possible outcomes.
4. List the payoff or profit of each combination of alternatives and outcomes.
5. Select one of the mathematical decision analysis methods.
6. Apply the model and make a decision.
7. Do sensitivity analysis.
We use Maria Rocha’s decision problem to illustrate these basic steps. Maria is a freelance fashiondesigner living in Johannesburg.
1. The problem that Maria identified is what size dress shop, if any, to open in a shopping mallnear the university.
2. Maria has to generate the alternatives that are available to her. In decision analysis, analternative is defined as a course of action or a strategy that may be chosen by the decisionmaker. Maria decides that her alternatives are to open a small shop, a large shop or no shopat all.
One of the biggest mistakes made in decision making is to omit important alternatives.Although a particular alternative may seem inappropriate or of little value, it may turn outto be the best choice.
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3. The third step involves identifying the possible outcomes of the various alternatives. Thecriteria for action are established at this time. Maria determines that there are three possibleoutcomes: the market for a dress shop could be
• good, meaning that there is a high demand for her dresses,
• average, meaning that there is an average demand for her dresses, or
• bad, meaning that there is a low demand for her dresses.
A common mistake is to forget about some of the possible outcomes. Optimistic decisionmakers tend to ignore bad outcomes, while pessimistic managers may discount a favourableoutcome. If all possibilities are not considered, logical decisions cannot be made and theresults may be undesirable. Those outcomes over which the decision maker has no controlare called states of nature.
4. Maria’s next step is to express the payoff resulting from each possible combination ofalternatives and outcomes. Since she wants to maximise her profits, she can use profit toevaluate each consequence. Obviously every decision, of course, can be based on moneyalone — any appropriate means of measuring benefit is acceptable. In decision analysis,such payoffs or profits are called conditional values.
Maria has already evaluated the potential profits associated with the various outcomes. Ifthe market is good, she thinks a large shop would result in an annual net profit of R100 000.This R100 000 is a conditional value because her receiving the money is conditional uponboth her opening a large shop and having a good market. The conditional value of a largeshop if the market is average would be R35 000, and if the market is bad it would be a netloss of R60 000. A small shop would result in a net profit of R75 000 if the market is good, anet profit of R25 000 if the market is average and a net loss of R40 000 if the market is bad.Finally, doing nothing would result in R0 profit in any market.
The easiest way to present these values is by constructing a decision or payoff table or a decisiontree as discussed in the following section.
5 & 6. The following two steps are to select a decision analysis model and apply it to the datato help make the decision. Selecting the model depends on the environment in which thebusiness will operate and the amount of risk and uncertainty involved.
7. The last step is to apply sensitivity analysis to the model by considering the effect of changesin the probability estimates. If necessary, the decision making model can then be adjustedto include these effects.
6.2.2 Payoff tables
The payoff table is constructed by
• listing the different decision alternatives down the left side of the table,
• listing all the possible outcomes, or states of nature, across the top, and
• entering the payoff for each combination of the alternatives and the states of nature in thebody of the table.
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The payoff table for Maria’s decision problem is given in Table 6.1.
State of natureGood Average Bad
Decision market market marketalternative s1 s2 s3
Large shop (d1) 100 000 35 000 −60 000Small shop (d2) 75 000 25 000 −40 000No shop (d3) 0 0 0
Table 6.1: Payoff table for Maria’s problem (Payoffs in rand)
6.2.3 Decision trees
A decision tree provides a graphical representation of the decision-making process. Figure 6.1presents a decision tree for Maria’s problem. Note that the decision tree shows the natural orlogical progression that will occur over time. First, Maria must make a decision about the sizeof the shop. Then, after the decision has been implemented, one of the states of nature, good,average or bad market, will occur. The number at each end point of the tree indicates the payoffassociated with a particular sequence. For example, the top-most payoff of 100 000 indicates thata R100 000 profit is anticipated if Maria opens a large shop and the market turns out to be good.The next payoff of 35 000 indicates an anticipated profit of R35 000 if Maria opens a large shopand the market turns out to be average. Thus, the decision tree shows graphically the sequencesof decision alternatives and states of nature that provide the nine possible outcomes and payoffs.
An intersection, or junction point, of the decision tree is referred to as a node and the arc, orconnector, between the nodes as a branch. Figure 6.1 shows Maria’s decision tree with the nodesnumbered 1 to 4. When the branches leaving a node are decision branches, the node is referredto as a decision node and is presented by a square. Similarly, if the branches leaving a node arestate-of-nature branches, the node is referred to as a state-of-nature node and is presented by acircle. Hence, node 1 is a decision node, whereas nodes 2, 3 and 4 are state-of-nature nodes.
6.3 Decision making without probabilities or under uncertainty
In this section we consider approaches to decision making that do not require knowledge of theprobabilities of certain states of nature occurring. These approaches are appropriate in situa-tions where state-of-nature probabilities are not available or cannot be obtained easily, or whereconsidering best worst case analyses, independent of state-of-nature probabilities, is desirable.
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Good market (s1)
Average market (s2)
Good market (s1)
Good market (s1)
Average market (s2)
Average market (s2)
Bad market (s3)
Bad market (s3)
Bad market (s3)
Large shop (d1)
Small shop (d2)
No shop (d3)
100 000
35 000
-60 000
75 000
25 000
-40 000
0
0
0
1
2
3
4
Figure 6.1: Decision tree for Maria’s problem (Payoffs in rand)
6.3.1 The optimistic approach - maximax criterion
An optimistic person usually has an extremely optimistic view of the future and would tend tochoose the alternative which would generate the highest possible payoff. The optimistic approachtherefore evaluates each decision alternative in terms of the best payoff that could occur. Thisapproach is also called the maximax criterion since it selects the decision alternative that maximisesthe maximum payoff.
The optimistic approach is summarised in the following two steps:
• From the payoff table, identify the maximum payoff for each of the decision alternatives.
• Compare these payoffs and select the decision alternative that provides the maximum.
Table 6.2 illustrates how these steps are used to develop a recommendation for Maria’s problem.
Decision Maximumalternative payoff (in R)Large shop (d1) 100 000 MaximumSmall shop (d2) 75 000No shop (d3) 0
Table 6.2: Optimistic approach (Maximax criterion)
The optimistic approach recommends that Maria opens a large shop.
Note:
• For problems involving minimisation (for example the minimisation of cost) the optimisticapproach would lead the decision maker to choose the alternative with the smallest payoff.
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6.3.2 The conservative approach - maximim criterion
Not everybody, however, is optimistic about the future. The pessimist tends to be cautious aboutfuture outcomes and would like to be in the best position if the worst should happen. Theconservative approach evaluates each decision alternative in terms of the worst payoff that couldoccur and recommends the alternative that provides the best payoff of these.
This approach is also called the maximin criterion since it selects the alternative that maximises theminimum payoff.
The conservative approach is summarised in the following two steps:
• From the payoff table, identify the minimum payoff for each of the decision alternatives.
• Compare these payoffs and select the decision alternative that provides the maximum.
Table 6.3 illustrates how these steps are used to develop a recommendation for Maria’s problem.
Decision Minimumalternative payoff (in R)Large shop (d1) −60 000Small shop (d2) −40 000No shop (d3) 0 Maximum
Table 6.3: Conservative approach (Maximin criterion)
The conservative approach recommends that Maria should not open a shop.
Note:
• For problems involving minimisation the conservative approach identifies the alternativethat will minimise the maximum payoff.
6.3.3 Minimax regret approach
The minimax regret approach is an approach to decision-making that is neither purely optimisticnor purely conservative.
The minimax criterion differs from the previous criteria in that it does not work with the payofftable directly, but with a table that is constructed from the payoff table. This table is called theopportunity loss table or regret table.
First of all we need to know what regret is. Suppose Maria decides to open a small shop (d2) andthe market turns out to be good (s1). Table 6.1 shows that the resulting profit would be R75 000.However, we now know that the good market state of nature (s1) has occurred and the best decisionwould have been to open a large shop (d1), yielding a profit of R100 000. The difference betweenthe payoff for the best decision alternative (R100 000) and the payoff for the decision to open asmall shop (R75 000) is the opportunity loss, or regret, associated with decision alternative d2 whenstate of nature s1 occurs; in this case the opportunity loss or regret is R100 000−R75 000 = R25 000.Similarly, if Maria opens no store (d3) and the market turns out to be good (s1), the opportunityloss, or regret, associated with d3 would be R100 000 − R0 = R100 000.
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Opportunity loss thus refers to the difference between the optimal possible profit or return andthe actual return obtained. It is therefore the amount that is lost because the best alternative wasnot chosen.
In general, opportunity loss, or regret, is given by the following expression:
O L, R
Ri j = |V∗j − Vi j|where
Ri j = the regret associated with decision alternative di and state of nature s j,
V∗j = the payoff value corresponding to the best decision for state of nature s j, and
Vi j = the payoff value associated with decision alternative di and state of nature s j.
Note:
• In maximisation problems, V∗j
is the largest entry in column j of the payoff table.
• In minimisation problems, V∗j
is the smallest entry in column j of the payoff table.
• The absolute value of the difference between V∗j
and Vi j ensures that the regret is the
magnitude of the difference.
To calculate the opportunity loss, or regret, table for Maria’s problem, we simply subtract eachentry in a column of the payoff table, Table 6.1, from the largest entry in the column. Table 6.4shows Maria’s opportunity loss, or regret, table.
State of natureGood Average Bad
Decision market market marketalternative s1 s2 s3
Large shop (d1) 0 0 60 000Small shop (d2) 25 000 10 000 40 000No shop (d3) 100 000 35 000 0
Table 6.4: Opportunity loss (regret) table for Maria’s problem (Regret in R)
The minimax regret approach is summarised in the following three steps:
• Construct an opportunity loss (regret) table.
• From the opportunity loss (regret) table, list the maximum regret for each decision alterna-tive.
• Compare these regrets and select the decision alternative with the minimum regret.
Table 6.5 illustrates how these steps are used to develop a recommendation for Maria’s problem.
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Decision Maximumalternative regret (in rand)Large shop (d1) 60 000Small shop (d2) 40 000 MinimumNo shop (d3) 100 000
Table 6.5: Minimax regret approach
The minimax regret approach recommends that Maria opens a small shop, with a correspondingmaximum regret of R40 000.
Note:
• This approach is called the minimax regret approach since it selects the decision alternativethat minimises the maximum regret.
• The minimax regret approach is the same for both maximisation and minimisation problems.The best decision alternative for a minimisation problem is the smallest entry in a column.
Note that three different recommendations were made by the three approaches discussed, whichin itself is not bad. It simply reflects the difference in decision-making philosophies that underliesthe various approaches. In the end, the decision maker will have to choose the most appropriateapproach, and then make the final decision accordingly.
6.4 Decision making with probabilities or under risk
6.4.1 The expected value approach
In many decision-making situations, probability estimates for each of the states of nature canbe obtained. When such probabilities are available, the expected value approach, also called theexpected monetary approach, can be used to identify the best decision alternative.
We first define the expected value of a decision alternative and then apply it to Maria’s problem.
LetN = the number of states of nature, and
P(s j) = the probability that state of nature s j occurs.
Since one, and only one, of the N states of nature can occur, the probabilities must satisfy thefollowing conditions:
P(s j) ≥ 0 for all states of nature, andN∑
j=1
P(s j) = P(s1) + P(s2) + · · · + P(sN) = 1.
The expected value (EV) of decision alternative di is defined as follows:
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E V D A di
EV(di) =
N∑
j=1
P(s j)Vi j
where
Vi j = the payoff corresponding to decision alternative di and state of nature s j.
Saying this in words: The expected value of a decision alternative is the sum of the weighted payoffsfor the decision alternative. The weight of a payoff is the probability that the payoffwill occur, orthe probability of the associated state of nature. We look for a maximum expected value becausewe want to gain as much as possible.
To illustrate the application of the expected value approach, we return to Maria’s problem whereit is estimated that there is a 20% chance that the market will be good, a 50% chance that the marketwill be average and a 30% chance that the market will be bad. This means that P(s1) = 0,20,P(s2) = 0,50 and P(s3) = 0,30. Table 6.6 gives Maria’s payoff table with the probabilities included.
State of natureGood Average Bad
Decision market market marketalternative s1 s2 s3
Large shop (d1) 100 000 35 000 −60 000Small shop (d2) 75 000 25 000 −40 000No shop (d3) 0 0 0
Probability 0,20 0,50 0,30
Table 6.6: Payoff table with probabilities for Maria’s problem (Payoffs in R)
The expected value for each of the three decision alternatives is calculated as follows:
EV(d1) = 0,2(100 000)+ 0,5(35 000) + 0,3(−60 000) = 19 500 Maximum
EV(d2) = 0,2(75 000) + 0,5(25 000) + 0,3(−40 000) = 15 500
EV(d3) = 0,2(0) + 0,5(0) + 0,3(0) = 0.
As can be seen, the decision recommended by this approach is to open a large shop, with anexpected value of R19 500.
6.4.2 Decision trees and the expected value approach
It is often more convenient to use a decision tree to perform the calculations required to identifythe decision alternative with the best expected value. Figure 6.2 shows the decision tree forMaria’s problem with state-of-nature branch probabilities.
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Large shop (d1)
Small shop (d2)
No shop (d3)
Good market (s1)
Good market (s1)
Good market (s1)
Average market (s2)
Average market (s2)
Average market (s2)
Bad market (s3)
Bad market (s3)
Bad market (s3)
100 000
35 000
-60 000
75 000
25 000
-40 000
0
0
0
1
2
3
4
P(s1) = 0,20
P(s1) = 0,20
P(s1) = 0,20
P(s2) = 0,50
P(s2) = 0,50
P(s2) = 0,50
P(s3) = 0,30
P(s3) = 0,30
P(s3) = 0,30
Figure 6.2: Decision tree with probabilities for Maria’s problem (Payoffs in rand)
Working backward through the decision tree, we calculate the expected value at each state-of-nature node. That is, at each state-of-nature node we weigh each possible payoff by its chanceof occurrence. By doing so, we obtain the expected values for nodes 2, 3 and 4, as shown inFigure 6.3.
Large shop (d1)
Small shop (d2)
No shop (d3)
1
2
3
4
EV(d1) = 0,2(100 000) + 0,5(35 000) + 0,3(-60 000) = 19 500
EV(d3) = 0,2(0) + 0,5(0) + 0,3(0) = 0
EV(d2) = 0,2(75 000) + 0,5(25 000) + 0,3(-40 000) = 15 500
Figure 6.3: Applying the expected value approach using decision trees
At node 1, a decision node, the decision maker has to make a decision about which action totake. Maria’s objective is to maximise profit. According to the expected values calculated onthe decision tree, the best decision branch is d1. The decision tree analysis therefore leads to arecommendation to open a large shop, with an expected profit of R19 500. Note that this is thesame recommendation as obtained by the expected value approach applied to the payoff table.
Decision problems may be substantially more complex than Maria’s problem, but if there is areasonable number of decision alternatives and states of nature, the decision tree approach canbe used.
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The decision tree approach can be summarised as follows:
1. Draw a decision tree consisting of decision and state-of-nature nodes, and branches thatdescribe the sequential nature of the problem.
2. For the expected value approach, determine the probabilities for each of the state-of-naturebranches.
3. Calculate the expected value at each state-of-nature node.
4. Select the decision branch leading to the state-of-nature node with the best expected value.The decision alternative associated with this branch is the recommended decision.
Example 6.4.2
Highveld Development Corporation (HDC) has purchased land for a luxury security com-plex. The individual units will be priced from R500 000 to R1,8 million, depending on thearea of the unit and optional features such as fireplaces and swimming pools.
HDC has had preliminary architectural drawings drawn up for three different project sizes:30 units, 60 units or 90 units. When asked about the possible market acceptance of the project,management took the view that it was an all-or-nothing situation. That is, they believe thatthe market acceptance would be one of two possibilities: high market acceptance of theproject and hence a substantial demand for the units, or low market acceptance of theproject and hence a limited demand for the units. Although management could exercisesome influence over market acceptance with advertising, the high prices asked for the unitsmean that demand will probably depend on a variety of other factors over which HDC hasno control.
HDC is optimistic about the potential of the complex and an initial subjective probabilityassessment of 0,8 that the market acceptance will be high, is provided. This leads to acorresponding probability of 0,2 that market acceptance will be low. Thus, P(s1) = 0,8 andP(s2) = 0,2.
Using the best information available, management have estimated the payoffs for the secu-rity complex project as presented in the payoff table, Table 6.7.
State of natureHigh Low
Decision acceptance acceptancealternative s1 s2
Small complex (d1) 8 7Medium complex (d2) 14 5Large complex (d3) 20 −9
Probability 0,8 0,2
Table 6.7: Payoff table for the HDC security complex (Payoffs in millions of rand)
(a) Construct a decision tree for HDC’s problem.
(b) What would HDC’s decision be according to
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i. the optimistic approach;
ii. the conservative approach;
iii. the minimax regret criterion;
iv. the expected value approach.
Solution
(a) The decision tree for HDC’s problem is given in Figure 6.4.
High acceptance (s1)
Low acceptance (s2)
Small complex (d1)
Medium complex (d2)
Large complex (d3)
Low acceptance (s2)
Low acceptance (s2)
High acceptance (s1)
High acceptance (s1)
8
1
2
3
4
7
14
5
20
-9
P(s1) = 0,8
P(s1) = 0,8
P(s1) = 0,8
P(s2) = 0,2
P(s2) = 0,2
P(s2) = 0,2
Figure 6.4: Decision tree for HDC’s problem (Payoffs in millions of rand)
(b) i. If HDC’s management follow an optimistic approach, the decision will be madeaccording to the maximax decision criterion. Here HDC would choose to build alarge complex for maximum profit of R20 000 000 as illustrated in Table 6.8.
Decision Maximumalternative payoff (in millions of rand)Small complex (d1) 8Medium complex (d2) 14Large complex (d3) 20 Maximum
Table 6.8: Optimistic approach (maximax criterion)
ii. If management, on the other hand, follow a conservative approach, they will chooseaccording to the maximin criterion where the best of the worst possible payoffs ischosen. According to the conservative approach, HDC will choose to build a smallcomplex, as illustrated in Table 6.9.
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Decision Minimumalternative payoff (in millions of rand)Small complex (d1) 7 MaximumMedium complex (d2) 5Large complex (d3) −9
Table 6.9: Conservative approach (maximin criterion)
iii. To apply the minimax regret approach, we need to construct the opportunity loss(regret) table as given in Table 6.10.
State of natureHigh Low
Decision acceptance acceptancealternative s1 s2
Small complex (d1) 12 0Medium complex (d2) 6 2Large complex (d3) 0 16
Table 6.10: Opportunity loss (regret) table
The minimax regret criterion will choose the decision alternative that minimises themaximum regret. According to this criterion, HDC will choose to build a mediumcomplex with maximum regret of R6 000 000. This is illustrated in Table 6.11.
Decision MaximumAlternative Regret (in millions of rand)Small complex (d1) 12Medium complex (d2) 6 MinimumLarge complex (d3) 16
Table 6.11: Minimax regret approach
iv. The expected value approach will choose the decision alternative with maximumexpected payoff. Using the payoffs in Table 6.7 the expected value for each decisionalternative is calculated and the maximum is chosen.
EV(d1) = 0,8(8) + 0,2(7) = 7,8
EV(d2) = 0,8(14) + 0,2(5) = 12,2
EV(d3) = 0,8(20) + 0,2(−9) = 14,2. Maximum
Thus, using the expected value approach, the large complex with an expectedvalue of R14 200 000 is chosen.
6.4.3 Sensitivity analysis
In this section we consider how changes in the probability estimates for states of nature affect, oralter, the recommended decision alternative. The study of the effect of such changes is referred
DSC2602 130
to as sensitivity analysis.
When doing sensitivity analysis, different probabilities for the states of nature are considered andthe expected value for each decision alternative is computed. By repeating this computation forseveral different probabilities, we can begin to learn how changes in the probabilities for statesof nature affect the recommended decision. Suppose, for example, that the probability of highacceptance in the HDC problem is reduced to 0,2 and that the probability of low acceptance isincreased to 0,8. Now the expected values for the different decision alternatives become
EV(d1) = 0,2( 8) + 0,8( 7) = 7,2 Maximum
EV(d2) = 0,2(14) + 0,8( 5) = 6,8
EV(d3) = 0,2(20) + 0,8(−9) = −3,2.
With these new probabilities the recommended decision alternative is to build a small complex,with an expected value of R7 200 000. Since the probability of high acceptance is only 0,2, sobuilding a large complex is the least preferred alternative, with an expected loss of R3 200 000.
Thus, when the probability of high acceptance is large, HDC should build the large complex;when the probability of high acceptance is small, HDC should build the small complex. We couldcontinue to modify the probabilities of the states of nature and learn more about how changesin the probabilities affect the recommended decision alternative. The only drawback to thisapproach is the numerous calculations required to evaluate the effect of several possible changesin the state-of-nature probabilities.
For the special case of two states of nature, as in the HDC problem, a graphical proceduresimplifies the process of sensitivity analysis. To demonstrate this procedure, we let p denote theprobability of state of nature s1, that is P(s1) = p. With only two states of nature the probability ofstate of nature s2 is
P(s2) = 1 − P(s1) = 1 − p.
Using the payoff values in Table 6.7, the expected value for decision alternative d1 is determinedas follows:
EV(d1) = P(s1)(8) + P(s2)(7)
= p(8) + (1 − p)(7)
= 8p + 7 − 7p
= p + 7.
Repeating the expected value computations, the following expressions are obtained for the ex-pected values of decision alternatives d2 and d3:
EV(d2) = 9p + 5 and
EV(d3) = 29p − 9.
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We now have three equations that show the expected value of the three decision alternatives asa function of the probability of state of nature s1. These equations can be presented on a graphwith values of p on the horizontal axis and the associated EVs on the vertical axis. Since allthree equations are linear equations, the graph of each is a straight line. Each line is drawn byidentifying two points satisfying the equation and drawing a line through the points.
For line EV(d1) = p + 7 : If p = 0 then EV(d1) = 7 and if p = 1 then EV(d1) = 8.
For line EV(d2) = 9p + 5 : If p = 0 then EV(d2) = 5 and if p = 1 then EV(d2) = 14.
For line EV(d3) = 29p − 9 : If p = 0 then EV(d3) = −9 and if p = 1 then EV(d3) = 20.
By connecting these points the lines EV(d1), EV(d2) and EV(d3) in Figure 6.5 are obtained.
0
5
10
15
20
-5
-10
0,2 0,4 0,6 0,8 1,0
p
Exp
ecte
d v
alu
e (
EV
)
EV(d1)
EV(d2)
EV(d3)d1 provides
highest EV
d2 provides
highest EV
d3 provides
highest EV
.
Figure 6.5: Expected value for HDC decision alternatives as a function of p
The graph shows how the recommended decision changes as the probability of the high accep-tance state of nature, p, changes. For small values of p, decision alternative d1 (small complex)provides the largest expected value and is the recommended decision. When the value of p in-creases to a certain point, decision alternative d2 (medium complex) provides the largest expectedvalue and is recommended. For large values of p, decision alternative d3 (large complex) becomesthe recommended decision.
The value of p for which the expected values of d1 and d2 are equal, is the value of p correspondingto the intersection of the EV(d1) and EV(d2) lines. To determine this point, set EV(d1) =EV(d2) andsolve for p.
p + 7 = 9p + 5
8p = 2
p = 0,25
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Whenever p = 0,25, decision alternatives d1 and d2 provide the same expected value. Repeatingthis calculation we find the value of p corresponding to the intersection of lines EV(d2) and EV(d3)as p = 0,70.
From Figure 6.5 we can conclude that decision alternative d1 provides the largest expected valuefor p < 0,25, decision alternative d2 provides the largest expected value for 0,25 < p < 0,70 anddecision alternative d3 provides the largest expected value for p > 0,70.
The benefit of performing sensitivity analysis is that it can provide a better perspective on manage-ment’s original judgement regarding the state-of-nature probabilities. Management originallyestimated the probability of high customer acceptance as P(s1) = 0,8. As a result, decision alterna-tive d3 was recommended. After conducting the sensitivity analysis, we can now tell managementthat, as long as P(s1) > 0,70, the d3 decision alternative remains optimal.
Note: The graphical sensitivity analysis procedure is only applicable to decision analysis problemswith two states of nature. It is, however, also important in problems with more than two statesof nature, like Maria’s problem which we discussed earlier. In these cases a spreadsheet can beused to assist with the computations.
6.4.4 Expected value of perfect information
Suppose Highveld Development Company (HDC) has the opportunity to conduct a marketresearch study that would help evaluate buyer interest in the security complex project andprovide information that management could use to improve the probability assessments for thestates of nature. To determine the potential value of this information, we begin by assumingthat the study could provide perfect information on the states of nature. By perfect informationwe mean knowing in advance what states of nature will occur. Although we never have perfectinformation in practice, it is worth knowing how much the value of our decision can be improvedif such information was available. This is called the expected value of perfect information (EVPI).
To show how perfect information can be used to develop a decision strategy1 that HDC shouldfollow, once it is known which state of nature will occur, we return to HDC’s payoff table whichis reproduced in Table 6.12.
State of natureHigh Low
Decision acceptance acceptancealternative s1 s2
Small complex (d1) 8 7Medium complex (d2) 14 5Large complex (d3) 20 −9Probability 0,8 0,2
Table 6.12: Payoff table for the HDC security complex (Payoffs in millions of rand)
If HDC knew for sure that state of nature s1 would occur, the best decision would be d3, with apayoff of R20 million. Similarly, if it knew for sure that state of nature s2 would occur, the best
1 A decision strategy is a decision rule that specifies the decision alternative to be selected after the new informationbecomes available.
133 DSC2602
decision alternative would be d1, with a payoff of R7 million. HDC’s optimal decision strategybased on perfect information is therefore as follows: If state of nature s1 occurs, select d3, and ifstate of nature s2 occurs, select d1.
To calculate the expected value for this decision strategy, we return to the original probabilityestimates for the states of nature: P(s1) = 0,8 and P(s2) = 0,2. Based on these probabilities, there isa 0,8 probability that the perfect information will indicate state of nature s1 and a 0,2 probabilityfor state of nature s2. Therefore, the expected value of the decision strategy that uses perfectinformation is
0,8(20) + 0,2(7) = 17,4.
This expected value of R17,4 million is called the expected value with perfect information (EVwPI).
According to the expected value approach (see p129) the recommended decision alternative is d3
with an expected value of R14,2 million. Because this decision recommendation and expectedvalue computation were made without the benefit of perfect information, it is called expected valuewithout perfect information (EVwoPI).
Because the expected value with perfect information is R17,4 million and the expected valuewithout perfect information is R14,2 million, the additional expected value that could be obtainedif perfect information about the state of nature were available, is R17,4−R14,2 = R3,2 million andthis is called the expected value of perfect information.
Generally speaking, a market research study will not provide “perfect” information; however,if the market research study is a good one, the information gathered might be worth a sizableportion of the R3,2 million. HDC should seriously consider the market survey as a way to obtainmore information about the states of nature, but only if the survey will cost less than R3,2 million.
In general, the expected value of perfect information is given by the following expression:
E V P I
EVPI = |EVwPI − EVwoPI|where
EVPI = expected value of perfect information,
EVwPI = expected value with perfect information about the states of nature, and
EVwoPI = expected value without perfect information about the states of nature.
Note:
• For minimising problems the expected value with perfect information is always less than,or equal to, the expected value without perfect information. The absolute value of thedifference gives the EVPI as the magnitude of the difference between EVwPI and EVwoPI.
• The opportunity loss (regret) table for the HDC problem is restated in Table 6.13:
The expected opportunity loss (EOL) for each decision alternative is
EOL(d1) = 0,8(12) + 0,2( 0) = 9,6
EOL(d2) = 0,8( 6) + 0,2( 2) = 5,6
EOL(d3) = 0,8( 0) + 0,2(16) = 3,2.
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State of natureHigh Low
Decision acceptance acceptancealternative s1 s2
Small complex (d1) 12 0Medium complex (d2) 6 2Large complex (d3) 0 16Probability 0,8 0,2
Table 6.13: Opportunity loss (regret) table for HDC’s problem (Regret in millions of rand)
Now, regardless of whether the decision analysis involves maximisation or minimisation,the minimum expected opportunity loss always provides the best decision alternative. De-cision alternative d3, with EOL(d3) = 3,2, is therefore the recommended decision.
In addition, the minimum expected opportunity loss is always equal to the expected value ofperfect information. That is
EOL(best decision) = EVPI.
For the HDC problem this value is R3,2 million.
6.4.5 Decision analysis with sample information
In applying the expected value approach, we saw how probability information about the statesof nature affects the expected value calculations and therefore also the decision recommendation.Frequently, decision makers have preliminary, or prior, probability estimates for the states ofnature that are the best probability values available. To make the best possible decision, however,the decision maker may want to seek additional information about the states of nature. The newinformation can be used to revise, or update, the prior probabilities so that the final decision isbased on more accurate probability estimates for the states of nature.
Additional information is usually obtained through experiments designed to provide sampleinformation about states of nature. Raw material sampling, product testing and market researchstudies are examples of experiments that may enable management to revise or update the state-of-nature probabilities.
Next we examine a problem in which we apply our knowledge of conditional probabilities toobtain the relevant probabilities.
Jack Smith owns the oil prospecting and development rights for a plot of land. A propertydevelopment company has offered him R105 000 for the plot. Jack must decide whether to sellthe plot or sink an oil well.
If he sinks the well and it is dry, he stands to lose R20 000. If oil is found in commercial quantities– a so-called “wet well” – he stands to earn R100 000. If the well is really flowing – a so-called“soaking well” – he could earn R500 000.
From past experience he knows that there is a
• 70% chance that the well is dry − P(dry) = 0,7;
135 DSC2602
• a 20% chance that the well is wet − P(wet) = 0,2 and
• a 10% chance that the well is soaking − P(soaking) = 0,1.
The decision tree of Jack’s problem is as follows:
S e l l
D r i l l1
D r y
W e t
S o a k i n g
( 0 , 7 )
( 0 , 2 )
( 0 , 1 )
1 0 5 0 0 0
1 0 0 0 0 0
5 0 0 0 0 0
2 0 0 0 0-
The expected values of his two alternatives are:
EV (sell) = 105 000 ←− Maximum
EV(drill) = 0,7 × −20 000 + 0,2 × 100 000 + 0,1 × 500 000 = 56 000.If Jack has to make a decision now he would choose to sell.
Suppose Jack wants additional information on the state of the land to help him to make a betterdecision. He can contract a geologist to undertake a seismic survey of the land at an additionalcost of R30 000.
The survey report can be bad or good. The probability of a good report is 0,5 and the probabilityof a bad one is 0,5.
The geologist supplies the following information as on the reliability of such surveys (all figuresare probabilities).
State of Survey reportwell bad good Total
dry 0,44 0,26 0,7wet 0,05 0,15 0,2
soaking 0,01 0,09 0,1Total 0,5 0,5 1,0
Jack must now make two decisions:
• Should he contract the geologist?
• Should he drill the well or sell?
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There are also two sets of states of nature facing Jack:
• The result of the survey (is it good or bad?)
• The true state of the well (is it dry, wet or soaking?).
We can now draw a decision tree for Jack’s problem. Remember that Jack must pay R30 000 forthe survey, an amount that must be deducted from the amount he stands to lose or gain.
S e l l
D r i l l3
D r y
W e t
S o a k i n g
1 0 5 0 0 0
1 0 0 0 0 05 0 0 0 0 0
- 2 0 0 0 0
S e l l
D r i l l4
D r y
W e t
S o a k i n g
7 5 0 0 0
S e l l
D r i l l5
D r y
W e t
S o a k i n g 7 0 0 0 04 7 0 0 0 0
-
1
2
N o s u r v e y
S u r v e y
R e t u r n
( 1 0 5 0 0 0 3 0 0 0 0 ) -
5 0 0 0 0-( 2 0 0 0 0 3 0 0 0 0 ) -7 0 0 0 0
4 7 0 0 0 0
7 5 0 0 0
5 0 0 0 0
-
Figure 6.6: Decision tree of Jack’s problem
From the decision tree we see that there are five decision points facing Jack. Before he can decideon point 1 he must decide on point 2 and 3 and before he can decide on 2 he must decide on 4 and5. We therefore have to analyse the tree from right to left – a method called the rollback principle.
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First we analyse point 3:
S e l l
D r i l l3
D r y
W e t
S o a k i n g
( 0 , 7 )
( 0 , 2 )
( 0 , 1 )
1 0 5 0 0 0
1 0 0 0 0 0
5 0 0 0 0 0
- 2 0 0 0 0
The expected values at point 3 are:
EV (sell) = 105 000←−maximum
EV(drill)= 0,7 × −20 000 + 0,2 × 100 000 + 0,1 × 500 000 = 56 000.
If decision point 3 is reached, Jack will choose to sell.
Next we analyse point 4:
If we now consider decision point 4 we need some conditional probabilities, that is, the probabilitythat the well is dry given a good report, the probability that the well is wet given a good reportand the probability that the well is soaking given a good report.
To calculate these we use the table of probabilities supplied by the geologist above, as well as theformula
P(A | B) =P(A ∩ B)
P(B)(do you remember?)
P(dry well | good report) =0,26
0,5= 0,52
P(wet well | good report) =0,15
0,5= 0,3
P(soaking well | good report) =0,09
0,5= 0,18
The decision tree at point 4 is:
S e l l
D r i l l4
D r y
W e t
S o a k i n g
( 0 , 5 2 )
( 0 , 3 )
( 0 , 1 8 )
7 5 0 0 0
7 0 0 0 04 7 0 0 0 0
- 5 0 0 0 0
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The expected values at point 4 are:
EV(sell) = 75 000
EV(drill) = 0,52 × −50 000 + 0,3 × 70 000 + 0,18 × 470 000 = 79 600 ←− Maximum
So if decision point 4 is reached Jack should drill.
Next we look at point 5:
The conditional probabilities calculated using the geologist’s table are:
P(dry well | bad report) =0,44
0,5= 0,88
P(wet well | bad report) =0,05
0,5= 0,1
P(soaking well | bad report) =0,01
0,5= 0,02
The decision tree at point 5 is:
S e l l
D r i l l5
D r y
W e t
S o a k i n g
( 0 , 8 8 )
( 0 , 1 )
( 0 , 0 2 )
7 5 0 0 0
7 0 0 0 04 7 0 0 0 0
- 5 0 0 0 0
The expected values at point 5 are:
EV(sell) = 75 000←−Maximum
EV(drill) = 0,88 × −50 000 + 0,1 × 70 000 + 0,02 × 470 000 = −27 600
So if decision point 5 is reached he should sell.We can now evaluate decision point 2.
First write down the outcomes at points 4 and 5:
Decision point 4
EV(drill) = 79 600 best option←−Maximum
Decision point 5
EV(sell) = 75 000 best option
Lastly we look at point 1:
For decision point 1 we now decide between the best value at decision point 3 and the best valueat decision point 2.
Remember EV (point 3) = EV (no survey) and EV (point 2) = EV (survey)
EV(no survey) = 105 000←−Maximum
EV(survey) = 77 600
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No survey is the best option and at decision point 2 the best option is to sell the plot.
To conclude, Jack should decide not to undertake a survey but to sell the plot.
S e l l
D r i l l3
D r y
W e t
S o a k i n g
( 0 , 7 )
( 0 , 2 )
( 0 , 1 )
1 0 5 0 0 0
1 0 0 0 0 05 0 0 0 0 0
- 2 0 0 0 0
S e l l
D r i l l4
D r y
W e t
S o a k i n g
( 0 , 5 2 )
( 0 , 3 )
( 0 , 1 8 )
7 5 0 0 0
-
S e l l
D r i l l5
D r y
W e t
S o a k i n g
( 0 , 8 8 )
( 0 , 1 )
( 0 , 0 2 ) 7 0 0 0 04 7 0 0 0 0
-
1
2
N o s u r v e y
S u r v e y
R e t u r n
( 1 0 5 0 0 0 3 0 0 0 0 ) -
5 0 0 0 0-( 2 0 0 0 0 3 0 0 0 0 ) -7 0 0 0 0
4 7 0 0 0 0
7 5 0 0 0
5 0 0 0 0
[ 1 0 5 0 0 0 ]
[ 1 0 5 0 0 0 ][ 5 6 0 0 0 ]
[ 7 5 0 0 0 ]
[ 7 9 6 0 0 ][ 7 9 6 0 0 ]
[ 7 5 0 0 0 ]
[ 7 5 0 0 0 ]
[ 7 7 3 0 0 ]
[ 2 7 6 0 0 ]
( 0 , 5 )
( 0 , 5 )
[ 1 0 5 0 0 0 ]1
-
Figure 6.7: Decision tree with branch probabilities and expected values.
6.5 Utility and decision making
Up to now payoffs were expressed in terms of monetary values. When probability informationwas available, the decision alternative with the best expected monetary value was recommended.However, in some situations this approach does not provide the most desirable decision. Themost desirable decision is the decision preferred by the decision maker after taking into accountnot only monetary value, but also other factors such as the risk associated with the outcomes.
One example of a situation in which selecting the decision alternative with the best expectedmonetary value may not lead to the most preferred decision, is the decision to buy house insurance.Clearly, buying such insurance does not provide a higher expected monetary value than notbuying it. Similarly, people gamble even though the expected monetary value of such a decisionis negative. In these cases monetary value is not the sole measure of the true worth of the outcometo the decision maker.
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The value (or worth) of an outcome in cases where expected monetary value does not lead to themost preferred decision alternative is expressed in terms of its utility and the expected utility isused to identify the most favourable decision.
Utility is a measure of the total worth of a particular outcome. It reflects the decision maker’sattitude towards a collection of factors such as profit, loss and risk. Let’s look at an example:
Consider the problem faced by Mitchell Inc., a relatively small real estate investment company.Mitchell currently has two investment opportunities which require approximately the same cashoutlay. Owing to limited cashflow, they are not able to make more than one investment now.Consequently, the following three possible decision alternatives, d1, d2 and d3, may be considered:
d1 = make investment A;
d2 = make investment B;
d3 = do not invest.
The monetary payoffs associated with the investment opportunities depend largely on whathappens to the real estate market during the next six months. The states of nature are as follows:
s1 = real estate prices go up;
s2 = real estate prices remain stable;
s3 = real estate prices go down.
Using the best information available, Mitchell has estimated the payoffs associated with eachdecision alternative and state-of-nature combination. The resulting payoff table is shown inTable 6.14.
State of naturePrices up Prices stable Prices down
Decision alternative s1 s2 s3
Investment A (d1) 30 000 20 000 −50 000Investment B (d2) 50 000 −20 000 −30 000Do not invest (d3) 0 0 0
Probability 0,3 0,5 0,2
Table 6.14: Payoff table for Mitchell Inc. (Payoffs in R)
The expected values for the three decision alternatives, using the estimates of the probabilitiesfor each state of nature as given in the payoff table, are
EV(d1) = 0,3(30 000) + 0,5(20 000)+ 0,2(−50 000) = 9 000 Maximum
EV(d2) = 0,3(50 000) + 0,5(−20 000)+ 0,2(−30 000) = −1 000
EV(d3) = 0,3(0) + 0,5(0) + 0,2(0) = 0.
Using the expected value approach, the optimal decision is to select investment A, with anexpected monetary value of R9 000. But, is this really the best decision alternative? Will Mitchellbe able to absorb the R50 000 loss if it invests in investment A and real estate prices go down?
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Mitchell’s ability to currently undertake only one investment suggests that its financial position isvery weak. Furthermore, the company’s chief executive officer (CEO) feels that if this investmentresults in substantial losses, Mitchell’s future will be in jeopardy. Although the expected valueapproach selects d1, either d2 or d3 will probably be selected to avoid the possibility of losingR50 000. In fact, if a loss as great as even R30 000 could drive Mitchell out of business, the CEOwould select d3, feeling that both investment A and investment B are too risky for Mitchell’scurrent financial position. This decision problem can be resolved by incorporating his ownattitude towards risk using utility theory.
Next we look at how we obtain utility values and then, how we use them in decision making.
6.5.1 Obtaining utility values for payoffs
Utility assessment begins by assigning the worst outcome a utility of 0 and the best outcome autility of 1. All other outcomes will have a utility value between 0 and 1. Assessing the utility ofany other outcome involves determining the probability, p, that makes you indifferent betweenalternative 1, which is the gamble between the best and worst outcomes, and alternative 2, whichis obtaining the other outcome for sure. In setting the probability, you should be aware thatutility assessment is completely subjective. It is a value set by the decision maker that can not bemeasured on an objective scale.
Consider Mitchell Inc’s investment problem again. It is faced with choosing between alternativesthat could result in a variety of payoffs, from a profit of R50 000 to a loss of R50 000. The first stepin determining their utility function is to arrange the payoffs in rank order from highest to lowest.Then arbitrarily assign a utility of 1,0 to the highest payoff and a utility of zero to the lowest.
For each payoff between the highest and the lowest, we then present the company with thefollowing situation:
Suppose you have the opportunity of achieving a guaranteed return of Q, or taking a chance ofreceiving R50 000 with probability p or losing R50 000 with probability (1 − p). What value of pwould make you indifferent to these choices? This is illustrated in the decision tree in Figure 6.8.
pR 5 0 0 0 0
Q
1 p- - R 5 0 0 0 0
Figure 6.8: The lottery for determining utility of payoff Q
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Let us start off with Q = 30 000. Because this is a relatively high value, the company decidesthat for p = 0,95 it would make no difference to it whether it chose a guaranteed R30 000 or thechance. This process is repeated for each payoff.
Using the above procedure, utility values for the rest of the payoffs in Mitchell’s problem weredeveloped. The results are presented in Table 6.15.
Monetary Utilityvalue (in rand) value
50 000 130 000 0,9520 000 0,90
0 0,75−20 000 0,55−30 000 0,4−50 000 0
Table 6.15: Utility of monetary values for Mitchell Inc.
Note: You will not be expected to do these calculations in this course. Utility tables will be given.
With the utility values for all monetary values determined, the original payoff table for Mitchell’sproblem can be written in terms of utility values, as shown in Table 6.16, and is called a utilitytable.
State of naturePrices up Prices stable Prices down
Decision alternative s1 s2 s3
Investment A (d1) 0,95 0,90 0,0Investment B (d2) 1,00 0,55 0,40Do not invest (d3) 0,75 0,75 0,75
Probability 0,3 0,5 0,2
Table 6.16: Utility table for Mitchell Inc.
6.5.2 Utility curve
A graph that plots utility value versus monetary value is called a utility curve. We can graphicallyrepresent Mitchell Inc’s utility function Figure 6.9.
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0 , 10 , 20 , 30 , 40 , 50 , 60 , 70 , 80 , 91
4 0 0 0 0 3 0 0 0 0 2 0 0 0 0 1 0 0 0 0 1 0 0 0 0 2 0 0 0 0 3 0 0 0 0 4 0 0 0 0 5 0 0 0 0- - - -5 0 0 0 0-Utilit
yP a y o f f ( i n R )
Figure 6.9: Michell Inc’s utility function.
Mitchell’s utility function is concave (the function opens downwards). This type of curve ischaracteristic of risk averse individuals. Such decision makers avoid risk, and prefer the expectedvalue of the gamble for certain and do not want to participate in the gamble. Thus the ratio choosesthe guaranteed value.
Other individuals might be more optimistic and therefore willing to take risks. These individualswould take a gamble that offers higher rewards even if the expected value is less than a certainpayoff.
The utility function of a risk-taking individual would be generally convex (function opens up-wards), as shown in Figure 6.10.
- R 1 0 0 0 - R 5 0 0 0 R 5 0 0 R 1 5 0 0R 1 0 0 0P a y o f f
0 , 20 , 40 , 60 , 81 , 0
U ti l i t
y
R i s k t a k i n g
Figure 6.10: A risk-taking utility function
Finally, some individuals are risk neutral — they prefer not to take risks, or to avoid them. Theirutility function would be linear, as shown in Figure 6.11. For any payoff, the probability ofwinning the gamble is such that the expected utility is the same as the expected payoff.
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- R 1 0 0 0 - R 5 0 0 0 R 5 0 0 R 1 5 0 0R 1 0 0 0P a y o f f
0 , 20 , 40 , 60 , 81 , 0
U ti l i t
y
R i s k n e u t r a l
Figure 6.11: A risk-neutral utility function
6.5.3 The expected utility approach
In the same way as expected monetary values were used to select an optimal decision alternativebefore, expected utilities (EU) can now be calculated and used to select an optimal decision alter-native. If there are N possible states of nature, the expected utility of a decision alternative di canbe calculated.
Let
N = number of states of nature, and
P(s j) = probability that state of nature s j occurs.
E U D A di
EU(di) =
N∑
j=1
P(s j)Ui j
where
Ui j = the expected utility corresponding to
decision alternative di and state of nature s j.
The expected utility for each of the decision alternatives in the Mitchell problem is calculated asfollows:
EU(d1) = 0,3(0,95) + 0,5(0,9) + 0,2(0) = 0,735
EU(d2) = 0,3(1,0) + 0,5(0,55) + 0,2(0,4) = 0,655
EU(d3) = 0,3(0,75) + 0,5(0,75) + 0,2(0,75) = 0,75. Maximum
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The optimal decision using the expected utility approach is therefore not to invest. When sum-marising the utility assignments and the associated monetary values (Table 6.17) we see that eventhough investment A has the highest expected monetary value of R9 000, Mitchell should declinethe investment since the CEO considered the 0,2 probability of a R50 000 loss a serious risk. Theseriousness of this risk and its associated impact on the company were not adequately reflectedby the expected monetary value of investment A. It was necessary to assess the utility for eachpayoff to adequately take this risk into account.
Decision Expected Expected monetaryalternatives utility value (in rand)
Investment A 7,35 9 000Investment B 6,55 −1 000Do not invest 7,50 0
Table 6.17: Summary of expected utility and expected monetary values for Mitchell Inc.
Unfortunately, determining the appropriate utilities is not a trivial task. It requires a degree ofsubjectivity on the part of the decision maker, and different decision makers will have differentutility functions. This aspect of utility often causes decision makers to feel uncomfortable aboutusing the expected utility approach. However, in a decision situation in which we are convincedthat monetary value is not the only relevant measure of performance, utility analysis should beconsidered.
6.5.4 Utility functions
Up to now we have seen that utility function can be specified in terms of a graph, as in Figure 6.10,or given as a table like in Table 6.16. A third form is a given mathematical expression, likeU(x) = log(x), U(x) = 1 − e−x/R or U(x) =
√x.
The utility values contained in an utility table can be graphed against their corresponding mone-tary values as was shown earlier. Likewise, mathematical expressions can be graphed. ConsiderFigure 6.12 in which utility functions of a risk-seeking, a risk-averting and a risk-neutral individ-ual are shown graphically.
Wealth
Utility
RISK AVERSE
Wealth
Utility
RISK NEUTRAL
Wealth
Utility
RISK SEEKING
Figure 6.12: Graphs of utility functions
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Example 6.5.3
A certain shopkeeper, Noah, values the changes in his assets according to the utility function
U(M) =√
M + 40 000 − 200
where M expresses the change in cash position associated with each outcome. Table 6.18shows the payoff table for Noah’s decision to buy or not to buy fire insurance.
State of natureDecision alternative Fire No fire
Buy insurance (d1) −100 −100Do not buy insurance (d2) −40 000 0Probability 0,002 0,998
Table 6.18: Noah’s payoff table (Payoffs in rand)
According to the table, if no insurance is bought and there is a fire, Noah will lose R40 000.Using the utility function with M = 40 000, the utility is calculated as
U(−40 000) =√−40 000 + 40 000 − 200
= −200.
Calculating the utilities for the other outcomes in a similar manner, we find that U(0) = 0and U(−100) = −0,25. Noah’s utility table is given in Table 6.19 with the expected utility ofeach decision alternative in the last column.
State of natureDecision alternative Fire No fire Expected utility
Buy insurance (d1) −0,25 −0,25 −0,25Do not buy insurance (d2) −200 0 −0,40Probability 0,002 0,998
Table 6.19: Noah’s utility table
Noah’s expected utilities were calculated as follows:
EU(d1) = (0,002)(−0,25) + (0,998)(−0,25) = −0,25 ←− Maximum
EU(d2) = (0,002)(−200) + (0)(0,998) = −0,40
According to the expected utility criterion, Noah should buy fire insurance.
Noah’s utility function is concave downwards as shown in Figure 6.13 which indicates thathe is a risk averter.
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U t i l i t y
- 2 0 0 0 0- 4 0 0 0 0 2 0 0 0 0
1 0 0
- 2 0 0
- 1 0 0
Figure 6.13: Noah’s utility function
6.6 Exercises
1. Carl Bender and Betty Atherton have known each other since high school and are now intheir second year BCom. In an attempt to make extra money and use some of the knowledgegained from their business courses, Carl and Betty have decided to look into the possibilityof starting a small company that would provide typing services to students who neededterm papers or reports typed in a professional manner. They have identified three strategies.Strategy 1 is to invest in a fairly expensive microcomputer system with a high-quality laserprinter. In a favourable market, they should be able to obtain a net profit of R10 000 overthe next two years. If the market is unfavourable, they could lose R8 000. Strategy 2 is topurchase a less expensive system. With a favourable market, they could get a return ofR8 000 during the next two years. With an unfavourable market, they would incur a loss ofR4 000. Their final strategy, strategy 3, is to do nothing. Carl is basically a risk taker, whileBetty tries to avoid risk.
(a) Construct a payoff table for this problem.
(b) What decision criterion would Carl use? What would his decision be?
(c) What decision criterion would Betty use? What would her decision be?
(d) If Carl and Betty decide to put their differences aside and go for the strategy that willnot make them regret going into the business, what decision criterion should they use?What would their decision be?
2. Flora, a florist in Bloemfontein, sells long stem roses for R35 per dozen. She buys the rosesat the local flower market for R20 per dozen. At the end of the day, all unsold roses are soldto local restaurants and hotels for R10 per dozen. If demand is not met, Flora will incure acost of R12 per dozen. The daily demand for the roses (in dozens) and the probabilities areas follows:
Daily demand 20 30 40 50
Probability 0,1 0,3 0,4 0,2
Flora’s problem is to decide how many dozen roses to buy at the market each morning.
DSC2602 148
(a) Construct the payoff table for her problem. (Remember pro f it = revenue − cost).
(b) In each of the following questions, state which decision criterion is used, set out thedecision rule clearly and give the appropriate decision.
i. How many dozen roses should she order if she follows a risk-seeking approach?
ii. How many dozen roses should she order if she follows a risk-averting approach?
iii. If she does not want to take unnecessary risks while avoiding regrets because ofnot buying enough, how many dozen roses should she buy?
(c) If Flora is risk neutral, how many dozen roses should she buy?
(d) How much should Flora be willing to pay for information that will tell her what thedaily demand for roses will be?
3. The Burger Inn is contemplating opening a new restaurant on Church Street. Managementhave three possible models for the restaurant, each with a different seating capacity. Theyestimate that the average number of customers per hour will be 80, 100 or 120. The payofftable for the three models is given in the following table:
State of natureDecision 80 customers 100 customers 120 customers
alternative per hour per hour per hour
Model A 10 000 15 000 14 000Model B 8 000 18 000 12 000Model C 6 000 16 000 21 000
Management estimates that the probability of 80 customers per hour is the same as theprobability of 120 customers per hour and twice as much as the probability of 100 customersper hour.
(a) Find the optimal decision by using a decision tree.
(b) What is the expected value of perfect information?
4. A vendor at a football stadium must decide whether to sell ice cream or cold drinks attoday’s game. The vendor believes that the profit made will depend on the day’s weather.If he sells cold drinks and it is a cool day, he estimates that he will make a profit of R500. Ifhe sells cold drinks and it is a hot day, he estimates that he will make a profit of R600. If hesells ice cream and it is a hot day he estimates that he will make a profit of R900 against aprofit of R300 if it is a cool day. Based on his past experience of this time of year, the vendorestimates the probability of hot weather to be 0,60.
(a) Construct a payoff table for his problem.
(b) What would his decision be if he is an optimistic person? State the criterion that isused.
(c) What would his decision be if he did not want to take unnecessary risks and did notwant to regret his decision? Show all calculations and indicate the criterion you used.
(d) Draw a decision tree for the problem. What should his decision be according to thedecision tree?
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(e) Use sensitivity analysis to determine graphically the effect of a change in the probabil-ities of the weather conditions on his choice.
(f) If perfect information on the weather was available, what would he be willing to payfor this information?
5. Consider the payoff table in Table 6.20.
State of natureDecision alternative s1 s2 s3
d1 100 000 40 000 −60 000d2 50 000 20 000 −30 000d3 20 000 20 000 −10 000d4 40 000 20 000 −60 000
Probability 0,1 0,3 0,6
Table 6.20: Payoff table (Payoffs in rand)
(a) If the decision maker is risk neutral, what is the optimal decision? (Use the expectedvalue approach)
(b) Suppose two decision makers have utility values as given in Table 6.21. Graph theutility curves for the two decision makers and classify each as being risk neutral, a riskseeker or a risk taker.
UtilityPayoff (in rand) Decision maker 1 Decision maker 2
100 000 1,00 1,0050 000 0,94 0,5840 000 0,90 0,5020 000 0,80 0,35−10 000 0,60 0,18−30 000 0,40 0,10−60 000 0 0
Table 6.21: Utilities of two decision makers
(c) Find the optimal decision for each of the decision makers, if the following utility tablesfor each decision maker were given:
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State of natureDecision alternative s1 s2 s3
d1 1,00 0,90 0d2 0,94 0,80 0,40d3 0,80 0,80 0,60
Probability 0,1 0,3 0,6
Table 6.22: Utility table for Decision Maker 1.
State of natureDecision alternative s1 s2 s3
d1 1,00 0,50 0d2 0,58 0,35 0,10d3 0,35 0,35 0,18
Probability 0,1 0,3 0,6
Table 6.23: Utility table for Decision Maker 2.
6. A small advertising agency wants to implement one of three possible advertising campaignsfor a client who is introducing a new product. The return of the campaign depends on thedemand for the related product. The estimated return (in thousand rand units) is given inTable 6.24:
Campaign type High demand Medium demand Low demand
Aggressive (d1) 30 18 −2Strong (d2) 25 20 2Cautious (d3) 10 8 5Probability 0,25 0,5 0,25
Table 6.24: Agency’s return table.
As a talented decision analyst you have helped the agency assess their utility function,which turned out to be
U(x) = ln(x + 3)
Where x is in thousands of rands.
Which campaign should they take according to the expected utility criterion?
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CHECKLISTYou should now be able to
⊲ construct a payoff table for a given decision problem;
⊲ construct an opportunity loss table for a given problem;
⊲ use the
– optimistic approach (maximax criterion),
– pessimistic approach (maximin criterion),
– minimax opportunity loss approach, and
– expected values
to make a decision.
⊲ construct a decision tree for a given decision problem;
⊲ use decision tree analysis to make a decision;
⊲ do sensitivity analysis;
⊲ calculate and interpret the expected value of perfect infor-mation (EVPI) and the expected value of sample information(EVSI);
⊲ use conditional probabilities and how to obtain the utilityvalues;
⊲ understand what utility means;
⊲ use the expected utility approach to make a decision.
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CHAPTER 7
Project management
7.1 Introduction
Almost every industry worries about how to manage projects effectively. A project can be definedas a collection of activities or tasks (with a specific goal in mind) that has to be completed over aperiod of time. A builder constructing an office building, for example, must complete thousandsof activities costing thousands or millions of rand in time and within a certain budget. NASA,for example, must inspect countless components before it launches a rocket.
In any project, the three factors of concern are time, cost and resource availability. Unnecessarydelays and cost overruns lead to millions of rand being wasted. This normally occurs because ofpoor planning and scheduling. How can such problems be solved?
Project management can be used to manage complex projects. Project management usually con-sists of planning of projects, acquiring resources, scheduling activities and evaluating completeprojects.
Managers are responsible for project management. They must coordinate the various activities sothat the entire project is completed in time. They must know how long a specific project will taketo finish, what the critical tasks are, and often, what the probability is of completing the projectwithin a given time span. In addition, it is often important to know the effect on the total projectof delays at individual stages.
To help managers effectively managing their projects they can make use of critical path schedulingtechniques. Critical path scheduling refers to a set of graphical techniques used in planning andcontrolling projects. These graphic forms relate a project’s activities in such a way that theactivities that are critical to the project’s completion time can easily be identified.
In this chapter, two popular critical path scheduling techniques will be discussed to help managersplan, schedule, monitor and control large and complex projects, namely the program evaluationand review technique (PERT) and the critical path method (CPM).
153
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7.2 PERT/CPM
Both PERT and CPM techniques were developed during the late 1950s. Although PERT and CPMhave the same general purpose and utilise much of the same terminology, the techniques weredeveloped independently. PERT was developed to handle uncertain activity times and CPM forindustrial projects for which activity times were general known.
Although PERT and CPM are similar in their basic approach, they do differ in the way activitytimes are estimated. For every PERT activity, three time estimates are combined to determine theexpected activity completion time and its variance. Thus PERT is a probabilistic technique; it allowsus to find the probability that the entire project will be completed by any given date. CPM, onthe other hand, is called a deterministic approach. It uses two time estimates, the normal time andthe crash time, for each activity. The normal completion time is the time we estimate it will take tocomplete the activity under normal conditions. The crash completion time is the shortest time itwould take to finish an activity if additional funds and resources were allocated to the task.
Today’s computerised versions of PERT and CPM have combined the best features of bothapproaches. It is therefore no longer necessary to distinguish between the two techniques andwe will refer to the project scheduling procedures as PERT/CPM.
There are six steps common to both PERT/CPM, namely:
1. Define the project and all of its significant activities or tasks.
2. Develop the relationships between the activities. Decide which activities must precede andfollow others.
3. Draw the network connecting all of the activities.
4. Assign time and/or cost estimates to each activity.
5. Compute the longest time path through the network; this is called the critical path.
6. Use the network to help plan, schedule, monitor and control the project.
To perform network calculations, we need an activity network representing the project and theduration of all the activities in that network. Network analysis is only a tool — its value dependsentirely on the way in which it is used and the information on which it is based. Consequently,ascertaining the duration of activities from records or estimating durations is an important partof the exercise.
If the activities have been performed previously and it can be assumed that the same resources andprocedures will be used, their durations may be obtained from records. If records are unavailable,some form of estimation is necessary.
In the following sections, we will discuss the PERT/CPM technique using
• project scheduling with certain activity durations − single time estimates
• project scheduling with uncertain activity durations −multiple time estimates
• project scheduling with time-cost tradeoffs
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7.3 Project scheduling with certain activity durations −single time estimates
In this section we will explain the basic PERT/CTM methodology and procedure using certainactivity durations or single time estimates.
7.3.1 Define the project
PERT/CTM provide a planned approach to project management. The first step is to define theproject and all project activities. To be effective, PERT/CTM first requires a clear definition of allthe tasks or activities that make up the project and time estimates of each activity.
Secondly, when the activities involved in a project have been identified, the project manager mustset up a immediate predecessor list. This will show which activities depend on the completion ofother activities before they can commence.
Let us look at an example.
Example 7.3.1
The flight manager of World Airline has identified the following tasks that need to beperformed between the arrival and departure of an aeroplane:
Activity Description
A Unload the passengersB Unload the luggageC Refuel the enginesD Clean the interiorE Load the mealsF Load the luggageG Board the passengersH Perform the safety check
The meals cannot be loaded or the interior cleaned until the passengers have been unloaded.The departing luggage cannot be loaded until the arriving luggage has been unloaded. Thepassengers cannot board until the interior has been cleaned. The safety check can beperformed only after the engines have been fuelled and the meals, luggage and passengershave been loaded. Identify the immediate predecessor(s) of each task.
Solution
The meals cannot be loaded nor the interior cleaned until the passengers have been un-loaded. This means that the activity, unload the passengers (A), is a predecessor of theactivities, load meals (E) and clean the interior (D).
The departing luggage cannot be loaded until the arriving luggage has been unloadedmeans the activity, unload the luggage (B), is a predecessor of the activity, load the luggage(F).
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The passengers cannot board until the interior has been cleaned means the activity, cleanthe interior (D), is a predecessor of the activity, board the passengers (G).
The safety check can be performed only after the engines have been fuelled and the meals,luggage and passengers have been loaded. This means that the activities, refuel the engines(C), load the meals (E), load the luggage (F) and board the passengers (G) are all predecessorsof the activity, perform the safety check (H).
The immediate predecessor(s) of each task can be summarised as follows:
ImmediateActivity Description predecessors
A Unload the passengers -B Unload the luggage -C Refuel the engines -D Clean the interior AE Load the meals AF Load the luggage BG Board the passengers DH Perform the safety check C,E,F,G
After completing the immediate predecessor(s), a list a network can be drawn of the project.
7.3.2 Project network diagrams
A network diagram is a graphical representation of an entire project. In this module one particularapproach to network construction, namely the activity on arc method is discussed. This is themost widely used method. An alternative, namely the activity on node method, can be found inthe literature, but will not be discussed here.
7.3.2.1 Conventions for constructing network diagrams
Network diagrams consist of two basic elements, namely activities and events (or nodes).
• An activity is a task or job that requires time and resource, such as counting the number ofdefective items or writing a report. An activity is represented by an unbroken arrow calledan arc.
The arrow shows where the activity starts and where it is completed. The name of theactivity and the time taken to complete it, are usually indicated on the arc. The name ofthe method of constructing a network is called the activity on the arc method. Note that thelength of the arrow is not related to the duration of an activity.
• An event is a point in time and we represent it by a circle (or a node). It indicates when aparticular activity starts or ends. We usually number events, as shown in Figure 7.1, so thatwe can easily describe paths within the network.
157 DSC2602
Activity A1 2
Event Event
Figure 7.1: Representation of activities and events
The network diagram is constructed by showing all activities in a logical order. For example, thenetworks below relate to a decorating job. No activity may begin until all the activities leading toit have been completed.
• Figure 7.2: The walls can be painted only after they have been cleaned.
AClean the walls
1 2 3
BPaint the walls
Figure 7.2: Logical order of activities
• Figure 7.3: The walls can be papered only after the old paper has been removed and thenew paper is available.
1
2
3 4
LStrip old paper off walls
M
Obtain new paper
NPaper the walls
Figure 7.3: Logical order of activities
Activities occurring on the same path are sequential and are directly dependent on one another.Parallel activities on different paths are independent of one another. See Figure 7.4.
Parallel activitiesSequential activities
A B
C
D
E
F
G
H
I1 2 3
4
5
6
7
8 9
Figure 7.4: Sequential and parallel activities
DSC2602 158
When drawing networks, one of the conventions is to allow time to flow from left to right. Eventsare numbered so that those on the left of the diagram have smaller numbers than, events on theright of the diagram. Events with smaller numbers occur before events with bigger numbers.
It is not possible to use network diagrams in which “loops” or “dangles” occur.
• A loop like the one in Figure 7.5 may be a perfectly legitimate sequence of operations where,for example, a certain amount of ratification or reprocessing of materials takes place but,because of the calculations that must be performed on the diagram, it cannot be accepted innetwork diagrams.
4 5 6
7
8P QR
S
T
Figure 7.5: Loops are unacceptable in network diagrams
• A network diagram must start in a single node and finish in a single node. It is not acceptableto leave events “dangling,” as shown in Figure 7.6.
2 3 4 6E F H
5
G
Figure 7.6: Dangling events are unacceptable in network diagrams
An additional element of a network is a dummy activity. Dummy activities are sometimes used innetworks for the convenience in network constructions. Dummy activities take up no time – theyare of zero duration. Dummy variables are represented by a dotted line with an arrow head.
Dummy variables are necessary on the following occasions:
• To make the diagram logical. If, for example, the following were given:
159 DSC2602
ImmediateActivity predecessor
A -B -C AD BE C, DF DG EH F
then Figure 7.7 is correct,
A
B
C
D
E
F
G
H
1
2
3
4
5
6
7
Figure 7.7: Only activity E depends on both activities C and D
and Figure 7.8 is incorrect.
A
B
C
D
E
F
G
H
1
2
3
4
5
6
Figure 7.8: Activities E and F depend on both activities C and D
Figure 7.8 is incorrect because activities C and D have to be completed before either E or F maybegin. In Figure 7.7, activity E depends on the completion of both activities C and D, andactivity F depends on D alone. A dummy activity is required to represent the logic.
• To avoid having more than one activity with the same beginning and end event nodes. If,for example, the following were given:
ImmediateActivity predecessor
A -B AC AD B, C
DSC2602 160
Figure 7.9 is now unacceptable
AB
CD3 4 5
Figure 7.9: Unacceptable way of representing activities B and C
and Figure 7.10 is correct.
A
B
C D3
4
5 A
B
C D3
4
5
o rFigure 7.10: Using dummy activities to represent activities B and C
Figure 7.9 is incorrect because it is not acceptable to represent activities in this manner, sinceactivities B and C would both be described as being the activity from event node 3 to 4,and neither would be described uniquely. A dummy activity, as shown in Figure 7.10, istherefore necessary.
It may be necessary to use dummy activities when initially constructing networks to avoidcomplicated and untidy diagrams. However, since the amount of subsequent analysisdepends on the number of activities in a diagram, redundant dummies should be eliminatedin order to save calculation time.
When drawing large networks for projects involving many activities, we have often foundthat it is easier to begin at the end of the project and work backwards. It is often helpfulto consider large projects in separate parts, that is, certain sections of the project or, if moreappropriate, certain periods during the project, and then to piece together several smallernetworks rather than try to construct the complete network from scratch.
7.3.2.2 Drawing network diagrams
Drawing a network is a matter of practice and experience. A number of computer packages willconstruct the network diagram, but activities and precedence need to be fully specified. Networkconstruction is an iterative process and several attempts may be needed to achieve a correct re-presentation. Every network starts and finishes with a node. A good practice would be to avoidarrows that cross and arrows that point backwards.
Consider the following immediate predecessor list for a project:
161 DSC2602
Activity Immediate predecessor(s)
A -B -C A, BD BE C, DF C, D
The network must begin in a single node. Both activities A and B have no immediate predecessorand stem from the first node.
A
B
1
2
3
Figure 7.11: Activities A and B both start in node 1
Activity C has both activities A and B as immediate predecessors, while activity D has onlyactivity B as its immediate predecessor. A dummy activity is therefore necessary between nodes2 and 3. This is shown in Figure 7.12.
A
B
1
2
3
C
D
4
5
Figure 7.12: Dummy activity needed
Activities E and F both have activities C and D as their immediate predecessors. A dummyactivity is necessary between nodes 4 and 5 to accommodate this. See Figure 7.13.
DSC2602 162
A
B
1
2
3
C
D
4
5
6
7
E
F
Figure 7.13: Activities E and F have the same immediate predecessors
The network can now be completed. The diagram must finish with a single node and, sinceactivities E and F both start in one node, a dummy activity is necessary to be able to identifythem individually. The dummy activity used to show activity C as an immediate predecessor ofactivities E and F is unnecessary and can be eliminated. The complete project network is shownin Figure 7.14.
A
B
1
2
3
C
D
4
5
6
E
F
Figure 7.14: The complete project network
Let us look at an example.
Example 7.3.2
Draw a network (activity on arc) for the following project.
Project Immediateactivity predecessor(s)
A -B -C -D BE A, DF CG E, F
163 DSC2602
Solution
The network must begin in a single node. Activities A, B and C have no predecessor andstem from the first node.
A
1 BC
2
3
4
Activity D has B as its predecessor and activity E, A and D. A dummy node is thereforenecessary between nodes 2 and 3.
A
1 BC
D E
2
3
4
5 6
Activity F has C as its predecessor, while activity G has E and F as its predecessors. Thesolution is one of the following diagrams:
A
1 B
C
D E GF
765
4
3
2
(or drawn more symmetrically)
A
1 B D E G
FC765
4
2
3
DSC2602 164
7.3.3 Network calculations
The objective of initial network calculations is to determine the overall duration and critical path ofthe project. The critical path is defined by those activities that must be completed on time for theproject to be completed on time.
To find the duration and critical path, we need to determine the earliest and latest times that anactivity can begin and end.
7.3.3.1 Calculating early and late event times
For the purpose of network calculation, each node in the network diagram is divided into three,as shown below:
L a t e s t e v e n t t i m e ( L T )
E v e n t n u m b e r
E a r l y e v e n t t i m e ( E T )
( n o d e )
• Early event time (ET)
The early event time (ET) is the earliest time that an activity could begin, assuming all thepreceding activities are completed as soon as possible.
Consider the following network:
AB
C
D
E
F
G
H
I
24 5
1
3 5
623
Steps to determine the early event times:
We start at the first node and move forward through the network (left to right).
1. To start, the convention is always that the first node’s ET is equal to zero.
2. The ETs for the nodes are determined from the beginning of the network by movingforward (from left to right) through the network.
3. Where activities occur sequentially (one after the other), the ETs are found by addingthe duration of all preceding activities. For example
ET(5) = ET(3) + 4 = 9
165 DSC2602
1 2
4
5
6
7
8
AB
C
D
E
F
G
H
I
24 5
1
3 5
623
9
3
0 3
5 9 1 4
1 274
5 + 4 = 9
Figure 7.15: Calculating early event times
4. If two (or more) activities lead to one event or node, the next activity cannot beginuntil all the preceding activities have been completed. Consequently, the last of theseactivities to finish determines the starting time for the subsequent activity. The earlyevent time is therefore given by the maximum of all the possible early event times.
In Figure 7.15, the early event time at event 9 is day 20 (assuming that the project startsat day 0), since activity I can only start when the later of activities I and H, which is H,has been completed. The mathematical representation for this is
ET(9) = max{ET(7) + 6 = 20; ET(8) + 1 = 13} = 20.
5. Repeat steps 2 and 3 until the last node’s ET has been calculated. The ET of the lastnode is equal to the overall duration of the project.
1 2
4
5
6
7
8
AB
C
D
E
F
G
H
I
24 5
1
3 5
6
23
9
3
0 3
5 9 1 4
2 0
1 274Figure 7.16: Calculating early event times
• Late event time (LT)
The late event time is the latest time an activity can begin without causing a delay in theoverall duration of the project. We use the overall duration given in the last node and workbackwards.
Steps to calculate the late event times
We start at the back of the network in Figure 7.16 and move forward (right to left).
1. The late event time (LT) of the last event is equal to its early time.
2. Now working backwards through the network, the late event times are calculatedbackwards for activities occurring one after the other, by successively subtracting theduration of the activities from the preceding late event times. For example
DSC2602 166
LT(8) = LT(9) − 2 = 20 − 2 = 18
LT(3) = LT(5) − 4 = 9 − 4 = 5.
1 2
4
5
6
7
8
AB
C
D
E
F
G
H
I
24 5
1
3 5
6
23
9
3
0 3
5 9 1 4
2 0
1 274
5 9 1 4
2 0
1 81 31 0Figure 7.17: Calculating late event times
3. If two or more activities lead backwards to a node, then the minimum of the two lateevent time is used. In Figure 7.18, the late event time at node 2 is 3, since activity B muststart on day 3 if the project is not to be delayed. Mathematically, this is represented asfollows:
LT(2) = min{LT(3) − 2 = 3; LT(4) − 1 = 9} = 3.
1 2
3
4
5
6
7
8
A
B
C
D
E
F
G
H
I
2
4 5
1
3 5
6
20
5 9 14
4 7 12
203 3 20
1495
181310
309
Figure 7.18: Calculating late event times
Example 7.3.3
Consider the following project network in Figure 7.19 where the activities and their dura-tions, in days, are represented on the arcs. Determine the early event time and late eventtime for each activity.
167 DSC2602
A
C
1
2
3
4
5
66
2
C
D
E
4
3
5
F2
G3
Figure 7.19: Project network
Solution
First we determine the earliest event times:
Early time node 1: ET(1) = 0
Early time node 2: ET(2) = ET(1) + 6 = 0 + 6 = 6
Early time node 3: ET(3) = ET(1) + 2 = 0 + 2 = 2
Early time node 4: ET(4) = ET(2) + 4 = 6 + 4 = 10
Early time node 5: If more than one node leads to node 5, we have to determine the
maximum of all the possible early event times.
ET(5) = max {ET(2) + 3 = 6 + 3 = 9; ET(3) + 5 = 2 + 5 = 7;
ET(4) + 0 = 10 + 0 = 10; } = 10
Early time node 6: We use the same method as node 5 because G and F lead to node 6.
ET(6) = max {ET(4) + 2 = 10 + 2 = 12;
ET(5) + 3 = 10 + 3 = 13; } = 13
Graphically the early times can be represented as in Figure 7.20:
A
C
1
2
3
4
5
66
2
C
D
E
4
3
5
F2
G3
0
6
2
1 0
1 0
1 3
Figure 7.20: Early event times calculated
Secondly, we move backwards through the network and determine the late event times.
Late time node 6: LT(6) = ET(6) = 13
Late time node 5: LT(5) = LT(6) − 3 = 13 − 3 = 10
Late time node 4: More than one node leads to node 4. We have to determine
the minimum of all the possible late event times.
LT(4) = min{LT(6) − 2 = 13 − 2 = 11;
LT(5) − 0 = 10 − 0 = 10} = 10
DSC2602 168
Late time node 3: LT(3) = LT(5) − 5 = 10 − 5 = 5
Late time node 2: We use the same method as node 4 because D and C lead to node 2.
LT(2) = min{LT(5) − 3 = 10 − 3 = 7;
LT(4) − 4 = 10 − 4 = 6} = 6
Late time node 1: same as node 2.
LT(1) = min{LT(3) − 2 = 5 − 2 = 3;
LT(2) − 6 = 6 − 6 = 0} = 0
The late and early times for the project can be presented graphically as follows:
A
C
1
2
3
4
5
66
2
C
D
E
4
3
5
F2
G3
0
6
2
1 0
1 0
1 30
6
5
1 0
1 0
1 3
Figure 7.21: Late and early times of project
7.3.3.2 The duration of the project
After calculating the ETs and LTs of the network the overall duration of the project can be deter-mined. This is done so that either a delivery date can be given to the customer, or the alterationnecessary for the project to be completed by a certain date, can be considered.
The overall duration of the project can be read from the network diagram. The project’s durationis equal to the early time of the last event or node.
The duration of our example project as seen in Figure 7.18 is thus equal to 20.
7.3.3.3 The critical path
The critical path consists of all the activities that are critical for the completion of the project intime. This means those activities that can not be delayed at all, without delaying the completionof the entire project. In general, the critical path will pass through all the nodes where the ET isequal to the LT. This provides an easy way of scanning the network for the critical path. Thecritical path of the project representing in Figure 7.18 is therefore A→ B→ D→ F→ H.
However, in some circumstances, this approach may suggest that a noncritical activity is critical.To be sure that an activity is included in the critical path you need to check that a measure called,the total float, is equal to zero.
169 DSC2602
7.3.3.4 Total float
The total float of an activity is the amount by which the duration of an activity can be increasedwithout delaying the project. Total float is therefore a measure of the flexibility in the duration ofan activity.
The total float for an activity is the difference between the maximum time available for that activityand the duration of that activity. The total float for an activity with a start node of i and a finishnode of j is given as:
total float (TF) = LT for j − ET for i − duration of activity(ti j)
For our example in Figure 7.18, the total floats are calculated as follows:
Activity A: TF(1; 2) = LT(2) − ET(1) − 3 = 3 − 0 − 3 = 0
Activity B: TF(2; 3) = LT(3) − ET(2) − 2 = 5 − 3 − 2 = 0
Activity C: TF(2; 4) = LT(4) − ET(2) − 1 = 10 − 3 − 1 = 6
Activity D: TF(3; 5) = LT(5) − ET(3) − 4 = 9 − 5 − 4 = 0
Activity E: TF(4; 6) = LT(6) − ET(4) − 3 = 13 − 4 − 3 = 6
Activity F: TF(5; 7) = LT(7) − ET(5) − 5 = 14 − 9 − 5 = 0
Activity G: TF(6; 8) = LT(8) − ET(6) − 5 = 18 − 7 − 5 = 6
Activity H: TF(7; 9) = LT(9) − ET(7) − 6 = 20 − 14 − 6 = 0
Activity I: TF(8; 9) = LT(9) − ET(8) − 2 = 20 − 12 − 2 = 6
We see that activities A, B, D, F and H have zero total float, while activities C, E, G and I each hastotal float of 6 time units. This means that each event on the path C→E→G→I can be delayed (orthe duration of the activities on this path can each be increased) by 6 time units, without delayingthe completion of the project.
Next an example in which not all the nodes with ET = LT is on the critical path. If we look at thefollowing network, the ETs and LTs make the critical path obvious: A, D, H, K.
ABC
DG
I
H
J
EF
K
1
2 22
3
3
34
4
11 00 021 0
31 3 1 4
61 4 1 6
71 5 1 6
41 4 1 5
51 4 1 4
81 7 1 7
91 7 1 8
1 02 0 2 01 0
2
Figure 7.22: Critical path indicated by ‖
DSC2602 170
However, suppose that activities C and G were combined into a new activity L, taking six days.This part of the network is shown in Figure 7.23
DH3
4
1 021 0
51 4 1 4
81 7 1 7
L6
AB
K
Figure 7.23: New activity L
The ETs and LTs are equal on nodes 2 and 8, but activity L is not a critical activity. The total floatfor activity L is
= LT for j(node 8) – ET for i(node 2) – duration of activity L= 17 − 10 − 6= 1
The total float is 0 on activities D and H and they remain critical.
7.3.3.5 Other measures of float
There are two other important measures of float namely, free float and independent float.
• Free float
Another measure of flexibility in the duration of an activity is free float.
Free float is the time that an activity could be delayed without affecting any of the activitiesthat follow. However, free float does assume that previous activities run to time.
Free float = ET for j − ET for i − duration of activity
The free floats for our example project in Figure 7.18 are calculated as follows:
Activity A: FF(1; 2) = ET(2) − ET(1) − 3 = 3 − 0 − 3 = 0
Activity B: FF(2; 3) = ET(3) − ET(2) − 2 = 5 − 3 − 2 = 0
Activity C: FF(2; 4) = ET(4) − ET(2) − 1 = 4 − 3 − 1 = 0
Activity D: FF(3; 5) = ET(5) − ET(3) − 4 = 9 − 5 − 4 = 0
Activity E: FF(4; 6) = ET(6) − ET(4) − 3 = 7 − 4 − 3 = 0
Activity F: FF(5; 7) = ET(7) − ET(5) − 5 = 14 − 9 − 5 = 0
Activity G: FF(6; 8) = ET(8) − ET(6) − 5 = 12 − 7 − 5 = 0
Activity H: FF(7; 9) = ET(9) − ET(7) − 6 = 20 − 14 − 6 = 0
Activity I: FF(8; 9) = ET(9) − ET(8) − 2 = 20 − 12 − 2 = 6
171 DSC2602
Only activity I has free float > 0, meaning that I is the only activity that can be delayed (by6 time units) without delaying the start of any other activity beyond its earliest possiblestarting time.
• Independent float
The independent float gives the time that an activity could be delayed if all the previousactivities were completed as late as possible and all the following activities were to start asearly as possible.
Independent float = ET for j − LT for i − duration of activity
1 21 21 5
1 32 2 2 4
C o n s i d e r t h e f o l l o w i n g a c t i v i t y M :
M5
The determination of the total, free and independent float of activity M is illustrated inFigure 7.24.
D A Y1 2 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0 2 1 2 2 2 3 2 4
M( 5 d a y s ) ( 2 4 - 1 2 - 5 )
M( 5 d a y s ) ( 2 2 - 1 2 - 5 )
F R E E F L O A T
T O T A L F L O A T
M A X I M U M T I M E A V A I L A B L E
M I N I M U M T I M E A V A I L A B L E
M( 5 d a y s )
F L O A T( 2 2 - 1 5 - 5 )
I N D E P E N D E N T
Figure 7.24: Measures of float
It can be seen that total float is seven days, free float is five days, and independent float twodays.
DSC2602 172
Lastly, we look at an example.
Example 7.3.4
Consider the project in Table 7.1.
Immediate DurationActivity predecessor(s) (weeks)
A – 3B – 5C B 3D A, C 4E D 8F C 2G F 4H F 2I B 5J H, E, G 3
Table 7.1: Project with immediate predecessors and activity durations
Draw the project network and determine the critical path and the duration of the project.Also calculate the free float of each activity.
Solution
Figure 7.25 depicts the project network:
1
2
3
4 5
6
7
8
9
A
B C
D E
F G
H
I
J
3
5 3
2
4 8
2
4 3
5
Figure 7.25: The project network
173 DSC2602
Note the following important factors to consider when drawing the network:
• Both activities A and B have no immediate predecessors and both stem from node 1.
• Both activities C and I have activity B as their immediate predecessor, and begin innode 2.
• Activity D has both activities A and C as its immediate predecessors, but activity F hasonly C. A dummy activity is therefore necessary to split nodes 3 and 4.
• Both activities H and G have activity F as their immediate predecessor, and begin innode 6.
• Activity J has activities E, H and G as its immediate predecessors and all three mustend in node 7. Since both activities H and G go from node 6 to node 7, a dummyactivity is needed to be able to identify them uniquely.
• Since the network must end in one node, activities I and J end in node 9.
Once the network has been constructed, it is necessary, to check whether it is correct. Beginin the last node and check whether each activity has the immediate predecessors, as givenin the immediate predecessor list.
In order to determine the critical path and the duration of the project, the early and lateevent times must be calculated. This is shown in Figure 7.26.
1
2
3
4 5
6
7
8
9
A
B C
D E
F G
H
I
J
23 23
20
2016
128
8
5
3
4 8
5
2
3
2
4 3
5
0 0
5
8
8 12
10 14
20
Figure 7.26: Calculating early and late event times
The duration of the project is 23 weeks. The not-so-obvious event times are calculated asfollows:
Node 4: ET(4) = max{0 + 3; 8 + 0} = 8
Node 7: ET(7) = max{12 + 8; 10 + 2; 14 + 0} = 20
Node 9: ET(9) = max{5 + 5; 20 + 3} = 23
Node 6: LT(6) = min{20 − 2; 20 − 4} = 16
Node 3: LT(3) = min{8 − 0; 16 − 2} = 8
Node 2: LT(2) = min{8 − 3; 23 − 5} = 5
Node 1: LT(1) = min{5 − 5; 8 − 3} = 0
According to the definition, those activities with zero total float are on the critical path. Wecalculate the total float of each activity, where TF(i; j) = LT( j) − ET(i) − ti j. Remember ti j isthe duration of activity i j.
DSC2602 174
Activity A: TF(1; 4) = 8 − 0 − 3 = 5
Activity B: TF(1; 2) = 5 − 0 − 5 = 0
Activity C: TF(2; 3) = 8 − 5 − 3 = 0
Activity D: TF(4; 5) = 12 − 8 − 4 = 0
Activity E: TF(5; 7) = 20 − 12 − 8 = 0
Activity F: TF(3; 6) = 16 − 8 − 2 = 6
Activity G: TF(6; 8) = 20 − 10 − 4 = 6
Activity H: TF(6; 7) = 20 − 10 − 2 = 8
Activity I: TF(2; 9) = 23 − 5 − 5 = 13
Activity J: TF(7; 9) = 23 − 20 − 3 = 0
The critical activities are B, C, D, E and J. The critical path is therefore B→C→D→E→J. Thecritical path can also be described in terms of the nodes: 1-2-3-4-5-7-9.
Note: The critical path can also be identified from the network itself. Those events for whichthe early and late event times are equal, are on the critical path. From Figure 7.26 we seethat the early and late times on nodes 1, 2, 3, 4, 5, 7 and 9 are equal. The duration of activityA, however, does not correspond to the early and late times of nodes 1 and 4 and is thereforenot on the critical path.
The free float of each activity is calculated, where FF(i; j) = ET( j) − ET(i) − ti j (duration ofactivity i j)
Activity A: FF(1; 4) = 8 − 0 − 3 = 5
Activity B: FF(1; 2) = 5 − 0 − 5 = 0
Activity C: FF(2; 3) = 8 − 5 − 3 = 0
Activity D: FF(4; 5) = 12 − 8 − 4 = 0
Activity E: FF(5; 7) = 20 − 12 − 8 = 0
Activity F: FF(3; 6) = 10 − 8 − 2 = 0
Activity G: FF(6; 8) = 14 − 10 − 4 = 0
Activity H: FF(6; 7) = 20 − 10 − 2 = 8
Activity I: FF(2; 9) = 23 − 5 − 5 = 13
Activity J: FF(7; 9) = 23 − 20 − 3 = 0.
Activity A can be delayed by 5 weeks, activity H by 8 weeks and activity I by 13 weeks,without delaying any other activity beyond its earliest starting time.
7.3.4 Using linear programming to find a critical path
It is quite easy to write a program in order to find the critical path in a project network. Linearprogramming can be used to determine the critical path of a project through programming. Linearprogramming (LP) is a tool for solving optimisation problems. Such an optimisation problem isformulated as a mathematical model consisting of a linear objection function and linear constraints.It is therefore called linear programming.
We first need to write the project diagram as an LP program and then we can solve it using asoftware package.
175 DSC2602
7.3.4.1 Formulating an LP model
Consider the project network in Example 7.3.4. The network diagram is shown once again foryour convenience.
1
2
3
4 5
6
7
8
9
A
B C
D E
F G
H
I
J
3
5 3
2
4 8
2
4 3
5
Figure 7.27: The project network of Example 7.3.4
First a mathematical model has to be formulated and the variables to be used in the model,namely decision variables, defined. For our example the decision variables are defined as
x j = the time that the event corresponding to node j occurs, where j = 1; 2; . . . , 9.
This mean that
x1 = time event corresponding to node 1 occurs
x2 = time event corresponding to node 2 occurs, etc.
Secondly, the objective (or goal) of formulating the LP model has to be defined. In our example,the objective is to minimise the time required to complete the project. This is given by thedifference between the time the last event occurs and the time the first event occurs. The objectivefunction is therefore
Minimise z = x9 − x1.
For each activity (i; j), we know that before node j occurs, node i must occur and activity (i; j)must be completed. This implies that for each arc (i; j) in the project network, x j ≥ xi + ti j, whereti j = duration of activity (i; j).
Consider, for instance, activity F (or activity (3; 6)). Before node 6 can occur, node 3 must haveoccurred and activity F, with duration t36 = 2, must be completed. Therefore x6 ≥ x3+2. This is theconstraint for activity F. A constraint for each arc in the network must appear in the mathematicalmodel.
The complete LP model is given in Model 7.3.1.
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Minimise z = x9 − x1
subject to x4 ≥ x1 + 3 [Activity A]
x2 ≥ x1 + 5 [Activity B]
x3 ≥ x2 + 3 [Activity C]
x5 ≥ x4 + 4 [Activity D]
x7 ≥ x5 + 8 [Activity E]
x6 ≥ x3 + 2 [Activity F]
x8 ≥ x6 + 4 [Activity G]
x7 ≥ x6 + 2 [Activity H]
x9 ≥ x2 + 5 [Activity I]
x9 ≥ x7 + 3 [Activity J]
x4 ≥ x3 [Dummy activity (3; 4)]
x7 ≥ x8 [Dummy activity (8; 7)]
All the variables unrestricted.
Model 7.3.1: The LP model for finding the critical path
7.3.4.2 Using LINDO or LINGO to solve the LP model
The software accompanying the prescribed book by Wayne L Winston may be used to solve theLP model. The User’s guide explains in detail how to install the package and use it to solve LPmodels.
After installing the package an input data file must be created. The following points are importantwhen creating the input file for LINDO:
• There is no equal sign in the objective function.
• Mark the beginning of the constraints with the statement Subject to, s.t. or st.
• Enter the constraints with all the variables on the left-hand side of the inequality sign.
• Label the constraints as Act A, Act B, etc. These labels may not be longer than 7 charactersand may not contain spaces, therefore Act A and not Activity A. Each label ends with around right bracket. (These labels are used to enable us to interpret the solution.)
The input file for LINDO is as follows:MIN X9-X1
Subject to
Act_A) X4-X1>3
Act_B) X2-X1>5
Act_C) X3-X2>3
Act_D) X5-X4>4
Act_E) X7-X5>8
Act_F) X6-X3>2
Act_G) X8-X6>4
Act_H) X7-X6>2
Act_I) X9-X2>5
Act_J) X9-X7>3
Dummy34) X4-X3>0
Dummy87) X7-X8>0
177 DSC2602
LINDO provides the following solution:
L P O P T I M U M F O U N D A T S T E P 0 O B J E C T I V E F U N C T I O N V A L U E 1 ) 2 3 . 0 0 0 0 0 V A R I A B L E V A L U E R E D U C E D C O S T X 9 2 3 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 X 1 0 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 X 4 8 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 X 2 5 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 X 3 8 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 X 5 1 2 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 X 7 2 0 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 X 6 1 6 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 X 8 2 0 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 R O W S L A C K O R S U R P L U S D U A L P R I C E S A C T _ A ) 5 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 A C T _ B ) 0 . 0 0 0 0 0 0 1 . 0 0 0 0 0 0 A C T _ C ) 0 . 0 0 0 0 0 0 1 . 0 0 0 0 0 0 A C T _ D ) 0 . 0 0 0 0 0 0 1 . 0 0 0 0 0 0 A C T _ E ) 0 . 0 0 0 0 0 0 1 . 0 0 0 0 0 0 A C T _ F ) 6 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 A C T _ G ) 0 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 A C T _ H ) 2 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 A C T _ I ) 1 3 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 A C T _ J ) 0 . 0 0 0 0 0 0 1 . 0 0 0 0 0 0 D U M M Y 3 4 ) 0 . 0 0 0 0 0 0 1 . 0 0 0 0 0 0 D U M M Y 8 7 ) 0 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 N O . I T E R A T I O N S = 0
P r o j e c t d u r a t i o n
A c t i v i t i e s w i t h n e g a t i v ed u a l p r i c e s a r e c r i t i c a l .
----
--
The solution gives a project duration of 23 weeks. The critical activities can be identified as thosewith negative dual prices. Activities B, C, D, E and J all have dual prices of −1. This means thatif any one of these activities is delayed by 1 week, the duration of the project will increase by 1week. The critical path is therefore B→C→D→E→J. The dummy activity (3, 4), with a dual priceof −1, lies on the critical path.
LINGO can also be used to solve LP models. The input data file for LINGO differs from the inputfile for LINDO in the following respects:
• The LINGO statements end with semicolons.
• The objective function has an equal sign.
• The Subject to statement is omitted.
• The constraints can be given in any format — it is not necessary to give all the variables onthe left-hand side of the inequality sign.
DSC2602 178
• The identifying labels appear between square brackets and there is no restriction on thenumber of characters in the label. (Spaces in the labels are still not allowed.)
MIN = X9-X1;
[Activity_A] X4>X1+3;
[Activity_B] X2>X1+5;
[Activity_C] X3>X2+3;
[Activity_D] X5>X4+4;
[Activity_E] X7>X5+8;
[Activity_F] X6>X3+2;
[Activity_G] X8>X6+4;
[Activity_H] X7>X6+2;
[Activity_I] X9>X2+5;
[Activity_J] X9>X7+3;
[Dummy34] X4>X3;
[Dummy87] X7>X8;
LINGO gives the following solution, which is exactly the same as the solution given by LINDO.
O p t i m a l s o l u t i o n f o u n d a t s t e p : 0 O b j e c t i v e v a l u e : 2 3 . 0 0 0 0 0 V a r i a b l e V a l u e R e d u c e d C o s t X 9 2 3 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 1 0 . 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 4 8 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 2 5 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 3 8 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 5 1 2 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 7 2 0 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 6 1 0 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 8 1 4 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 R o w S l a c k o r S u r p l u s D u a l P r i c e 1 2 3 . 0 0 0 0 0 1 . 0 0 0 0 0 0 A C T I V I T Y _ A 5 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 A C T I V I T Y _ B 0 . 0 0 0 0 0 0 1 . 0 0 0 0 0 0 A C T I V I T Y _ C 0 . 0 0 0 0 0 0 E + 0 0 1 . 0 0 0 0 0 0 A C T I V I T Y _ D 0 . 0 0 0 0 0 0 E + 0 0 1 . 0 0 0 0 0 0 A C T I V I T Y _ E 0 . 0 0 0 0 0 0 E + 0 0 1 . 0 0 0 0 0 0 A C T I V I T Y _ F 0 . 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 A C T I V I T Y _ G 0 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 A C T I V I T Y _ H 8 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 A C T I V I T Y _ I 1 3 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 A C T I V I T Y _ J 0 . 0 0 0 0 0 0 1 . 0 0 0 0 0 0 D U M M Y 3 4 0 . 0 0 0 0 0 0 1 . 0 0 0 0 0 0 D U M M Y 8 7 6 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0
P r o j e c t d u r a t i o n
C r i t i c a l a c t i v i t i e s h a v e n e g a t i v e d u a l v a l u e s .
----
--
179 DSC2602
7.4 Project scheduling with uncertain activity durations – mul-
tiple time estimates
Thus far, we have assumed that a single certain time can be given for the duration of everyactivity. For one-of-a-kind projects or for new jobs, providing activity time estimates is not alwaysan easy task. It is known that many activities are dependent upon the weather, and the elementof chance is to be expected, say, when building a wall or completing a long sea journey. Withoutsolid historical data, managers are often uncertain about activity times. Uncertain activity timesare best described by a range of possible values rather than by one specific time estimate.
For this reason, the developers of PERT employed a probability distribution based on three timeestimates for each activity:
Optimistic time (a) = the time an activity will take if everything goes as well as possible.
There should be only a small probability (say, 1100
) of
this occurring.
Pessimistic time (b) = the time an activity would take assuming very unfavorable conditions.
There should also be only a small probability that the activity will
really take this long.
Most likely time (m) = the most realistic time estimate to complete the activity.
Let us look at an example.
Consider a project that was undertaken to determine how long a feasibility study for the PVPcompany would take. The necessary activities were identified and the immediate predecessorsof each determined. This information is given in Table 7.2.
ImmediateActivity Description predecessor(s)
A Develop product design –B Plan market research –C Prepare routing (manufacturing engineering) AD Build prototype model AE Prepare marketing brochure AF Prepare cost estimates (industrial engineering) CG Do preliminary product testing DH Complete market survey B, EI Prepare pricing and forecast report HJ Prepare final report I
Table 7.2: Activity and immediate predecessor list for PVP
The network for the PVP project is given in Figure 7.28.
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A
B
C
D
E
F
G
H I
1
2
3
4
5
6 7
8
J
Figure 7.28: Network for the PVP project
The time estimates below were obtained for the activities in the PVP project. These are shown inTable 7.3.
DurationOptimistic Most probable Pessimistic
Activity a m b
A 4 5 12B 1 1,5 5C 2 3 4D 3 4 11E 2 3 4F 1,5 2 2,5G 1,5 3 4,5H 2,5 3,5 7,5I 1,5 2 2,5J 1 2 3
Table 7.3: Time estimates for the PVP project
The procedure to calculate the critical path can be summarised as follow:
1. First, the average time for each activity is calculated. Using activity A as an example, we seethat the most probable time is 5 weeks with a range from 4 weeks (optimistic) to 12 weeks(pessimistic). If the activity could be repeated many times the average time for the activitycould be calculated. This average time, or expected duration, (t) is given by the formula
t =a + 4m + b
6. (7.1)
For activity A, the average time or expected duration is
tA =4 + 4(5) + 12
6= 6.
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2. Secondly, we need to determine the variation of the activity time values.
With uncertain activity times, the variance can be used to describe the dispersion or variationin the activity time values. The variance of the activity time is given by the formula
σ2 =
(
b − a
6
)2
. (7.2)
The difference between the pessimistic (b) and optimistic (a) time estimates greatly affectsthe value of the variance. Large differences in these two values reflect a high degree ofuncertainty in the duration of activities. The measure of uncertainty, that is, the variance,of activity A is
σ2A =
(12 − 4
6
)2
= 1,78.
Equations 7.1 and 7.2 are based on the assumption that the activity time distribution can bedescribed by a beta probability distribution (which need not concern us to much here).
The expected durations and variances of all the activities can be calculated in the same way.The results are summarised in Table 7.4.
Expected durationActivity (weeks) Variance
A 6 1,78B 2 0,44C 3 0,11D 5 1,78E 3 0,11F 2 0,03G 3 0,25H 4 0,69I 2 0,03J 2 0,11
Table 7.4: Expected durations and variances for the PVP activities
3. We can now proceed with the network calculations to determine the critical path and theexpected duration of the project. The expected durations of the activities are treated as ifthey were the fixed length or known duration of each activity. The early and late eventtimes were determined, as discussed in section 7.3.3.1, and are represented graphically inFigure 7.29.
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A
B
C
D
E
F
G
H I
1
2
3
4
5
6 7
J2
8
2
6
2
3
3
5
4
2
30 0
6
9 13 15
11
9
17 17
15139
14
156
Figure 7.29: Early and late event times for the PVP project
4. Lastly the duration and critical path is calculated. The expected completion time for theproject is 17 weeks. The critical path for the project is A→E→H→I→J. Those events withequal early and late times are on the critical path. If you are unsure whether A or B isincluded remember to calculate the total float. The activity with a zero float is included.The total float of A:6 − 0 − 6 = 0 and the total float of B:9 − 0 − 2 = 7. A is thus included.
7.4.1 Probability of project completion time
In the previous section, three time estimates were used to determine the critical path and expectedduration. The time estimates are thus not fixed. Variation may exist, and can have an impact onthe project completion.
Variation in the duration of critical activities can impact on overall project completion time –possible delaying it. Variation in the duration of noncritical activities ordinarily has no effect onthe project completion time because of the total float time associated with them. However, if anoncritical activity is delayed long enough to extend its total float time, it becomes part of a newcritical path and may affect the project completion time.
PERT uses the variance in the critical path activities to determine the variance in the projectcompletion time. The variance in the project completion time is computed by summing thevariance of the critical activities.
project variance =∑
variance of activities on the critical path
Let us say we want to determine the probability that the PVC project will be completed in 20weeks. In the PVC example, the critical path was determined as A, E, H, I, J. The variance of theproject completion time can now be calculated using the variances of each activity as calculatedin Table 7.4.
σ2 = σ2A + σ
2E + σ
2H + σ
2I + σ
2J
= 1,78 + 0,11 + 0,69 + 0,03 + 0,11
= 2,72.
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Knowing that the standard deviation is the square root of the variance, the standard deviationfor the PVP project completion time is
σ =√σ2 =
√
2,72 = 1,65.
How can this information be used to help answer questions about the probability of finishing theproject on time? PERT makes two more assumptions: (1) total project completion times follow anormal probability distribution; and (2) activity times are statistically independent.
To find the probability that this project will be finished on or before the 20-week deadline, weneed to determine the appropriate area under the normal curve. The standard normal equationcan be applied as follows:
z =due date − expected date of completion
σ
where z is the number of standard deviations the due date or target date lies from the mean orexpected date. The expected value is given by the expected duration of the critical activities. Forinstance, for our example
E(T) = tA + tE + tH + tI + tJ
= 6 + 3 + 4 + 2 + 2
= 17.
The z-value for the normal probability distribution at T = 20 is
z =20 − 17
1,65= 1,82.
µ = 17 20 T
σ = 1,65 weeks
Figure 7.30: Probability of completing the PVP project within 20 weeks
Looking up z = 1,82 in the table for the standard normal distribution, we find that the probabilityof the project being completed in 20 weeks is 0,4656+ 0,5000 = 0,9656. Thus, even though activitytime variability may cause the completion time to exceed 17 weeks, there is a 96,56% chance thatthe project will be completed before the 20-week deadline.
7.5 Project scheduling with time-cost tradeoffs
Consider the following table, which gives the immediate predecessor list and the expected activitydurations for a project involving the maintenance of two machines used in production.
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Immediate Expected durationActivity Description predecessor(s) (days)
A Overhaul machine I - 7B Adjust machine I A 3C Overhaul machine II - 6D Adjust machine II C, A 3E Test system B, D 2
Table 7.5: Immediate predecessor list for machine maintenance project
The project network and the network calculations are shown in Figure 7.31.
1
2
3
54
A B
DC
E7 3
6 3
20 0
7
7
10 12 1210
7
7
Figure 7.31: Project network with early and late event times
From the network we see that all events have equal early and late times. If we compare theactivity durations to the early and late times, we see that activities A, B, D and E, have no totalfloat, while activity C has a total float of 1. We can therefore say that only activity C is not on thecritical path.
There are thus two critical paths: A→B→E and A→D→E and the duration of the project is 12days.
Suppose the production manager has urgently requested that owing to the current productionlevels the maintenance of the machines must be completed within 10 days. What can we do? Itis clear that, unless selected activity durations can be shortened, it is impossible to complete theproject in 10 days. This shortening of activity durations in order to meet a certain completiondate, is referred to as crashing. Crashing is usually achieved by adding extra resources (such asequipment or people) to an activity. Naturally, crashing costs more money, and managers areusually interested in speeding up a project at the least additional cost.
But how do we decide which activities to crash and by how much? In order to determine whichactivities to crash and by how much, two sets of time and cost estimates for the activities must beavailable:
1. The normal time estimate is the same as the expected time.
2. The normal cost is an estimate of how much money it will take to complete an activity in itsnormal time.
3. The crash time is the shortest possible activity time.
4. Crash cost is the price of completing the activity on a crash or deadline basis.
The normal and crash times and normal and crash cost for the maintenance project are shown inTable 7.6.
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Duration (days) Total cost (in rand)Activity Normal(ti) Crash(t′
i) Normal(Ci) Crash(C′
i)
A 7 4 500 800B 3 2 200 350C 6 4 500 900D 3 1 200 500E 2 1 300 550
1 700 3 100
Table 7.6: Crashing data for machine maintenance project
Note for example that activity A’s normal time is 7 days and its crash time is 4 days. This meansthat the activity can be shortened by 3 days (7-4) if extra resources are provided. The normal costis R500 while the crash cost is R800. This implies that crashing activity A will cost an additionalR300 (800 − 500).
The crash cost per day for activity A is thus equal to 300 ÷ 3 which is R100.
In general terms, we can say that the crashing cost per unit of time (Ki) is given by
Ki =crash cost − normal cost
normal time − crash time
=C′
i− Ci
ti − t′i
Similarly the crash costs per day for all the other activities can be computed. Table 7.7 consists ofthe Ki values of all the activities.
Max reduction Crash costDuration (days) Total cost (in rand) in time per day
Activity Normal(ti) Crash(t′i) Normal(Ci) Crash(C′
i) (Mi = ti − t′
i)
(
Ki =C′
i−Ci
Mi
)
A 7 4 500 800 3 100B 3 2 200 350 1 150C 6 4 500 900 2 200D 3 1 200 500 2 150E 2 1 300 550 1 250
1 700 3 100
Table 7.7: Crashing data for machine maintenance project
The crashing cost per unit of time can also be calculated using the assumption that the time-costrelationship for an activity is linear. This assumption is common in practice and helps us derivethe cost per day (Ki) to crash an activity. This value may be found directly by using the formulato take the slope of the line.
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Slope = crash cost per time unit =crash cost − normal cost
normal time − crash time
Figure 7.32 shows the graph of the time-cost relationship for activity A.
1 2 3 4 5 6 7
R200
R400
R600
R800
R1 000
Act
ivit
y c
ost
Activity duration (days)
Crash
Normal
Figure 7.32: Time-cost relationship for activity A
Now the question is: Which activities should be crashed, and by how much, to meet the 10-day projectcompletion deadline at minimum cost? It makes sense to crash the activities on the critical path thathas the lowest cost activity per time unit. The activities that are common to both the critical pathsin our example are A or E. Activity A has the lowest crashing cost and crashing this activity by 2days will reduce the project duration to the desired 10 days. Keep in mind, however, that as youcrash the current critical path activities, other paths may become critical. You will need to checkthe critical path in the revised network and perhaps either identify additional activities to crashor modify your initial crashing decision. Repeat this procedure until the time of completion issatisfactory, or until there can be no further reduction in the project completion time.
For a small network this trial-and-error approach can be used to make crashing decisions. In largernetworks, however, a mathematical procedure is required to determine the optimal crashingdecisions. The following discussion shows how linear programming can be used to solve thenetwork crashing problem.
7.5.1 Formulating an LP model to crash a project
The LP model for finding the critical path for this network is given in Model 7.5.1. The decisionvariables are defined as
x j = the time that the event corresponding to node j occurs, where j = 1, 2, 3, 4, 5.
187 DSC2602
Minimise z = x5 − x1
subject to x2 ≥ x1 + 7 [Activity A]
x4 ≥ x2 + 3 [Activity B]
x3 ≥ x1 + 6 [Activity C]
x4 ≥ x3 + 3 [Activity D]
x5 ≥ x4 + 2 [Activity E]
x3 ≥ x2 [Dummy activity (2; 3)]
All variables unrestricted
Model 7.5.1: The LP model for finding the critical path
This model can be extended to determine which activities to crash, and by how much, to achievethe 10-day deadline at minimum cost.
First we need decision variables for the number of days to crash each activity, and define them asfollows:
A = the number of days activity A must be crashedB = the number of days activity B must be crashed...
...E = the number of days activity E must be crashed
Secondly, we need to determine the objective function. In Table 7.7, the column Crash cost per daygives the cost of reducing the duration of each activity by one day. It will, for instance, costR100 to reduce activity A by one day, R150 to reduce activity B by one day, etc. The objective offormulating this LP model is to minimise the cost of reducing the project duration to 10 days. Theobjective function is therefore:
Minimise COST = 100A + 150B + 200C + 150D + 250E.
Thirdly, we need to determine the crash time constraints. In Table 7.7, the column Max reduction intime, gives the maximum number of days by which each activity can be crashed. Activity A can,for instance, be crashed by at most 3 days. This constraint is entered into the model as A ≤ 3.
Fourthly, we need to determine the constraints describing the network. The constraint dealingwith the starting time of activity B has to be altered for the crashing. The constraint x2 ≥ x1 + 7must change to x2 ≥ x1 + 7 −A to accommodate the possible shortening of activity A.
Fifthly, we need to determine the project completion constraint. In our example, the project shouldbe completed in 10 days x5−x1 ≤ 10. The LP model for crashing the project is given in Model 7.5.2.
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Minimise COST = 100A + 150B + 200C + 150D + 250E
subject to A ≤ 3 [Crash time activity A]
B ≤ 1 [Crash time activity B]
C ≤ 2 [Crash time activity C]
D ≤ 2 [Crash time activity D]
E ≤ 1 [Crash time activity E]
x5 − x1 ≤ 10 [Project completion time]
x2 ≥ x1 + 7 − A [Activity A] [Network constraints]
x4 ≥ x2 + 3 − B [Activity B] [Network constraints]
x3 ≥ x1 + 6 − C [Activity C] [Network constraints]
x4 ≥ x3 + 3 −D [Activity D] [Network constraints]
x5 ≥ x4 + 2 − E [Activity E] [Network constraints]
x3 ≥ x2 [Dummy activity (2; 3)]
A,B,C,D,E ≥ 0;
x j unrestricted, j = 1; . . . ; 5.
Model 7.5.2: The LP model for crashing the project
7.5.2 Using LINGO to solve the LP model for crashing a project
The input data file for LINGO: (Note that an asterisk (*) is used for multiplication.)
[Objective] min = 100*A+150*B+200*C+150*D+250*E;
[A] A<3;
[B] B<1;
[C] C<2;
[D] D<2;
[E] E<1;
[Deadline] X5-X1<10;
[Act_A] X2>X1+7-A;
[Act_B] X4>X2+3-B;
[Act_C] X3>X1+6-C;
[Act_D] X4>X3+3-D;
[Act_E] X5>X4+2-E;
[Dummy23] x3>x2;
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LINGO gives the following solution:
O p t i m a l s o l u t i o n f o u n d a t s t e p : 5 O b j e c t i v e v a l u e : 3 5 0 . 0 0 0 0 V a r i a b l e V a l u e R e d u c e d C o s t A 2 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 B 0 . 0 0 0 0 0 0 0 E + 0 0 5 0 . 0 0 0 0 0 C 0 . 0 0 0 0 0 0 0 E + 0 0 5 0 . 0 0 0 0 0 D 1 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 E 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 5 1 0 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 1 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 2 5 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 4 8 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 3 6 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 R o w S l a c k o r S u r p l u s D u a l P r i c e O B J E C T I V E 3 5 0 . 0 0 0 0 1 . 0 0 0 0 0 0 A 1 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 B 1 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 C 2 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 D 1 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 E 1 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 D E A D L I N E 0 . 0 0 0 0 0 0 0 E + 0 0 2 5 0 . 0 0 0 0 A C T _ A 0 . 0 0 0 0 0 0 0 E + 0 0 1 0 0 . 0 0 0 0 A C T _ B 0 . 0 0 0 0 0 0 0 E + 0 0 1 0 0 . 0 0 0 0 A C T _ C 0 . 0 0 0 0 0 0 0 E + 0 0 1 5 0 . 0 0 0 0 A C T _ D 0 . 0 0 0 0 0 0 0 E + 0 0 1 5 0 . 0 0 0 0 A C T _ E 0 . 0 0 0 0 0 0 0 E + 0 0 2 5 0 . 0 0 0 0 D U M M Y 2 3 1 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0
-----
M i n i m u m c o s t
N u m b e r o f d a y sa c t i v i t i e s s h o u l db e s h o r t e n e d o rc r a s h e d .
N e g a t i v e v a l u e si n d i c a t e c r i t i c a lp a t h a c t i v i t i e s .
The following information is read from the LINGO solution:
• The minimum cost of crashing the project to 10 days is R350.
• In the Value column, the values of the variables A, B, C, D and E indicate the number of daysby which each activity should be shortened. Activity A should be shortened by 2 days andactivity D by 1 day.
• From the Dual price column, we see that each of the activities A, B, C, D and E has a negativedual price. This indicates that all the activities are critical. The critical paths are A→ B→ Eand C→ D→ E. (The path A→ D→ E is not a critical path since dummy activity (2, 3) hasa zero dual price.)
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7.6 Exercises
1. Cabinets Unlimited is a company in which all types of cabinets are manufactured andassembled. A new model is to be manufactured, which requires the following tasks:
Task Description Task time (min)
A Prepare the wheels 10B Mount the wheels 5C Assemble the sides 15D Attach the top 11E Attach the base 10F Insert the brackets 5G Insert the shelves 5H Attach the doors 10I Attach the back panel 10J Paint the unit 15
The wheels are mounted after they have been prepared. The base cannot be attached untilthe sides are assembled and the wheels mounted. The top cannot be attached and thebrackets cannot be inserted until the sides have been assembled. The shelves are insertedafter the brackets have been installed. The back panel is attached after the base and top havebeen attached. The doors are attached after the shelves have been inserted and the top andbase attached. The unit is painted after the back and doors have been attached. Identify theimmediate predecessors of each task.
2. Draw a project network for each of the following three projects.
Immediate predecessorsProject Project Project
Activity IPSO FACTO DELTA
a - - -b - - -c - a, b a, bd a, b, c a, b ae d c, d c
3. Draw a project network for each of the following projects. Determine the duration and thecritical path of each project. Also calculate the total float and the free float for each activity.
Immediate predecessor(s)Duration Project Project Project
Activity (weeks) ALPHA BETA GAMMA
A 2 - - -B 4 - A -C 1 A A -D 3 A, B B, C A, B, CE 2 C B, C BF 3 C, D D, E D, E
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4. Consider the project presented in Table 7.8.
Duration (in hours)Immediate Optimistic Most probable Pessimistic
Activity predecessor(s) a m b
A - 4 6 8B - 1 4,5 5C A 3 3 3D A 4 5 6E A 0,5 1 1,5F B, C 3 4 5G B, C 1 1,5 5H E, F 5 6 7I E, F 2 5 8J D, H 2,5 2,75 4,5K G, I 3 5 7
Table 7.8: Immediate predecessors and multiple time estimates
(a) Construct the network diagram for this project.
(b) Identify the critical path and calculate the expected duration of the project.
(c) What is the probability of the project being completed within one day (24 hours)?
5. Consider the information in Table 7.9. The activities have to be performed in order toassemble a product.
Immediate Duration (minutes) Crashing costActivity predecessor(s) Normal(ti) Crash(t′
i) per minute (in R)
A - 10 5 1B A 5 4 2C - 15 10 1D C 11 6 1E B, C 10 7 1F C 5 4 2G F 5 4 2H D, E, G 10 8 3I D, E 10 8 3J I, H 15 10 4
Table 7.9: Crashing data for product assembly project
(a) Construct a network diagram for the project.
(b) Determine the critical path and project duration by using linear programming.
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(c) Use linear programming to answer the following questions:
i. How much would it cost to achieve an assembly time of 48 minutes? Which tasksshould be shortened and by how much should they be shortened to achieve thisgoal? Identify the critical path(s) after crashing.
ii. What is the shortest assembly time that can be achieved by spending an additionalR19?
CHECKLISTYou should now be able to
⊲ for a given project
– set up an immediate predecessor list,
– construct a network diagram,
– calculate early and late event times,
– calculate total float and free float,
– determine the critical path and project duration;
⊲ determine the expected durations and the variance of activitiesusing time estimates for a project;
⊲ use the normal distribution to determine the probability of aproject being completed within a certain time;
⊲ formulate linear programming models
– to find the critical path of a project and
– to crash a project;
⊲ use LINGO to solve the LP models;
⊲ interpret the LINGO outputs.
CHAPTER 8
Network models
8.1 Introduction
IChapter 7 we looked at a network approach to project scheduling. Network models consistingof nodes and arcs and network analysis techniques arise frequently in managerial science
applications. Many real-world problems have a network structure or can be modelled in networkform. These include problems in areas such as production, distribution, project planning, facilitieslocation, resource management and financial planning. The graphical network representation ofproblems provides a powerful visual and conceptual aid to indicate the relationship betweencomponents of a system.
This chapter presents three network models: the shortest-route problem, the minimum-spanningtree and the maximum-flow problem. In each case we will discuss an algorithm for solving theproblem by hand, as well as formulate a mathematical model to be solved on a computer.
8.2 The shortest-route problem
The first model we consider is the shortest-route problem. The primary objective of shortest-routeproblems is to determine the shortest route or path from one node to a series of nodes in a network.The arcs usually represent distances, time or costs.
To illustrate the problem and an algorithm for solving it let us consider the problem facing theJack Russell Construction Company. Jack has several construction projects located throughoutthe province with some of the sites located as far as 100 kilometres from the depot where his mainoffice is situated. With multiple daily trips carrying personnel, equipment and supplies to andfrom the construction sites, the costs associated with transportation activities are substantial. Forany given construction site the alternative routes between the site and the depot can be describedby a network of roads.
The network shown in Figure 8.1 describes the travel alternatives to and from six of Jack’sconstruction sites.
193
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1
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5
6
73 0
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1 21 01 2
3 4
D e p o t
Figure 8.1: Alternative routes to and from the depot
The circles or nodes of the network correspond to the site locations, while the roads appear as thearcs in the network. The distances between the sites are shown on the corresponding arcs. Jackwould like to determine the routes or paths that will minimise the total distance from the depotto each site.
Note: The length of an arc is not necessarily proportional to the travel distance it represents.
8.2.1 A shortest-route algorithm
The shortest-route algorithm can be used to minimise the total distance from any starting node toa final node. In our example, in order to minimise transportation costs, Jack needs to determinethe shortest route from the depot to each of the construction sites.
The shortest-route algorithm is an iterative process. At each step, the shortest distance to onenode is determined. The procedure consists of two parts, the labelling phase to determine theshortest distance from node 1 to every other node, and the backtracking phase to determine theactual route from node 1 to any other node.
8.2.1.1 The labelling phase
The labels used consist of two numbers enclosed in brackets like the one in the figure below. Thefirst number in the label for a particular node indicates the distance from node 1 to the nodeunder consideration, while the second number indicates which node precedes this one on theroute from node 1 to the node under consideration. The label for each node is shown directlyabove or below the node in the network.
[ 1 5 ; 3 ]N o d e
T h e d i s t a n c e f r o m n o d e 1 t o t h i s n o d e i s 1 5 .
O n t h e r o u t e f r o m n o d e 1 , n o d e 3 p r e c e d e s t h i s n o d e .
N o d e l a b e l
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There are two kinds of labels:
• If the distance from node 1 to the particular node may still be shorter along another routethan the route described by the label, we say that the node has a tentative label.
• When we have determined the shortest distance from node 1 to the labelled node, we saythat the node has a permanent label. Such a node is shaded to indicate that it has a permanentlabel.
Now, let us solve Jack’s problem.
I (see Figure 8.2)
Begin by giving node 1 the permanent label [0; S]. The 0 indicates that the distance from node 1to itself is zero and the S identifies node 1 as the starting node.
Shade node 1 to show that it is permanently labelled. Mark node 1 with an arrow to show that ithas most recently been permanently labelled.
Consider every node that can be reached directly from node 1:
• Node 2: The direct distance from node 1 to node 2 is 30 km and node 2 can be tentativelylabelled [30; 1].
• Node 3: The direct distance from node 1 to node 3 is 20 km. The tentative label for node 3is [20; 1].
In Figure 8.2 the initial identification of the depot as the starting node is shown with a permanentlabel and nodes that can be reached directly from node 1 are shown with tentative labels.
1
2
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6
73 0
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1 21 01 2
3 4
D e p o t[ 0 ; B ]
[ 3 0 ; 1 ]
[ 2 0 ; 1 ]
Figure 8.2: Initial labelling for Jack’s shortest-route problem
I 1 (see Figure 8.3)
Consider all the nodes with tentative labels, that is node 2 and node 3. Identify the node with thesmallest distance value in its label. The travelling distance in the label of node 2 is 30 km, while thetravelling distance in the label of node 3 is 20 km. We therefore select node 3.
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Since there is no shorter route to reach node 3 from node 1, we label node 3 permanently byshading it and marking it with an arrow.
Consider all nodes that can be reached directly from node 3:
• Node 2: The direct distance from node 3 to node 2 is 6 km. If this is added to the 20 kmindicated in the permanent label of node 3, the shortest distance from node 1, we see thatnode 2 can be reached in 20 + 6 = 26 km. The tentative label is revised to [26; 3] to indicatethat a shorter distance of 26 km from node 1 to node 2, via node 3, has been found.
• Node 5: The direct distance from node 3 to node 5 is 8 km. By adding this to the 20 kmindicated in the permanent label of node 3, we see that the shortest distance from node 1 tonode 5, via node 3, is 28 km. Node 5 is tentatively labelled [28; 3].
1
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73 0
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1 21 01 2
3 4[ 3 0 ; 1 ]
[ 2 0 ; 1 ]
D e p o t[ 0 ; B ]
[ 2 6 ; 3 ]
[ 2 8 ; 3 ]
Figure 8.3: Iteration 1
I 2 (see Figure 8.4)
Consider all tentatively labelled nodes (nodes 2 and 5). Select the tentative label with the smallestdistance value, that is, [26; 3]. Since there is no shorter path to reach node 2 from node 1, node 2is permanently labelled and marked with an arrow.
Consider all nodes without permanent labels that can be reached directly from node 2:
• Node 4: Add the direct distance from node 2 to node 4 (12 km) to the distance value in thepermanent label of node 2. Node 4 can be reached in 26+ 12 = 38 km. Node 4 is tentativelylabelled [38; 2].
• Node 6: Node 6 can be reached in 26 + 34 = 60 km and is tentatively labelled [60; 2].
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1 21 01 2
3 4
[ 2 0 ; 1 ]
D e p o t[ 0 ; B ]
[ 2 6 ; 3 ]
[ 2 8 ; 3 ]
[ 6 0 ; 2 ]
[ 3 8 ; 2 ]
Figure 8.4: Iteration 2
I 3 (see Figure 8.5)
Consider all tentatively labelled nodes, that is, nodes 4, 5 and 6. Select the node with the smallestdistance value in its label, node 5 with the label [28; 3]. Node 5 is permanently labelled andmarked with an arrow.
Consider all nodes without permanent labels that can be reached directly from node 5:
• Node 4: Add the direct distance from node 5 to node 4, that is, 8 km, to the distance valuein the permanent label of node 5. Node 4 can be reached in 28 + 8 = 36 km which is lessthan the previous distance of 38 km. Revise the tentative label at node 4 to [36; 5].• Node 7: Node 7 can be reached in 28 + 4 = 32 km and is tentatively labelled [32; 5].
1
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1 21 01 2
3 4
[ 2 0 ; 1 ]
D e p o t[ 0 ; B ]
[ 2 6 ; 3 ]
[ 2 8 ; 3 ]
[ 6 0 ; 2 ]
[ 3 8 ; 2 ][ 3 6 ; 5 ]
[ 3 2 ; 5 ]
Figure 8.5: Iteration 3
I 4 (see Figure 8.6)
Consider all tentatively labelled nodes, that is, nodes 4, 6 and 7. Node 7 with the label [32; 5]has the smallest distance value in its label. Node 7 is therefore permanently labelled and markedwith an arrow.
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Consider all nodes without permanent labels that can be reached directly from node 7:
• Node 6: Add the direct distance from node 7 to node 6, that is, 12 km, to the distance valuein the permanent label of node 7. Node 6 can be reached in 32 + 12 = 44 km, which is lessthan the previous distance of 60 km. The tentative label is therefore altered to [44; 7].
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1 21 01 2
3 4
[ 2 0 ; 1 ]
D e p o t[ 0 ; B ]
[ 2 6 ; 3 ]
[ 2 8 ; 3 ]
[ 6 0 ; 2 ]
[ 3 6 ; 5 ]
[ 3 2 ; 5 ]
[ 4 4 ; 7 ]
Figure 8.6: Iteration 4
I 5 (see Figure 8.7)
Consider all tentatively labelled nodes, that is, nodes 4 and 6. Node 4 with the label [36; 5] has thesmallest distance value in its label. Node 4 is therefore permanently labelled and marked withan arrow.
Node 6 is the only remaining node without a permanent label. Add the direct distance from node4 to node 6, that is, 10 km, to the distance value in the permanent label of node 4. Node 6 can bereached in 36 + 10 = 46 km. This is more than the previous distance, 44 km. The tentative labeltherefore remains unchanged.
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1 21 01 2
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[ 2 0 ; 1 ]
D e p o t[ 0 ; B ]
Figure 8.7: Iteration 5
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I 6
Since node 6 is the only remaining node with a tentative label, it is now permanently labelled.Figure 8.8 shows the final network with all the nodes permanently labelled.
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1 21 01 2
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[ 2 0 ; 1 ]
D e p o t[ 0 ; B ]
[ 2 6 ; 3 ]
[ 2 8 ; 3 ]
[ 3 6 ; 5 ]
[ 3 2 ; 5 ]
[ 4 4 ; 7 ]
Figure 8.8: Final network with all nodes permanently labelled
8.2.1.2 Backtracking phase
To find the shortest route, we need to backtrack through the network. The information givenby the permanent labels can be used to find the shortest route from node 1 to each node in thenetwork. For example, the shortest route to travel from node 1 to node 6 is identified as follows:
Start at node 6 and work back to node 1:
Node 6’s permanent label [44; 7] gives the next direct link. The 7 in the label indicates that thenode that comes just before node 6 on the route from node 1, is node 7.
Node 7’s permanent label [32; 5] gives the next direct link. The 5 in the label indicates that node5 comes just before node 7.
Node 5’s permanent label [28; 3] gives the next direct link. Node 3 precedes node 5.
Node 3’s permanent label [20; 1] gives the next direct link. Node 1 precedes node 3.
The shortest route from node 1 to node 6 is therefore 1-3-5-7-6, and from node 6’s label we seethat the shortest distance from node 1 to node 6 is 44 km.
Node 6’s label tells us that the shortest distance from node 1 to node 6 is 44 km. The shortestroute is 1-3-5-7-6.
The shortest routes identified for Jack’s transportation network are given in Table 8.1.
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Shortest route from DistanceNode node 1 (in kilometres)
2 1-3-2 263 1-3 204 1-3-5-4 365 1-3-5 286 1-3-5-7-6 447 1-3-5-7 32
Table 8.1: Shortest routes for Jack’s transportation network
For a relatively small problem such as Jack’s, you probably could have found the shortest routesjust as quickly, if not more quickly, by inspection. However, for problems with 15 to 20 nodes, asystematic procedure such as the preceding labelling procedure is required.
The algorithm is summarised as follows:
S- A
Consider a network with N nodes. Use the following procedure to find the shortest route fromnode 1 to each of the other nodes in the network:
Step 1 Assign the permanent label [0; S] to node 1.The 0 indicates the distance from node 1 to itself.The S indicates that node 1 is the starting node.
Step 2 Find tentative labels for the nodes that can be reached directly from node 1.The first number in the label (the distance value) is the direct distance from node 1 tothe node in question.The second number in the label (the precedingnode value) indicates the preceding nodeon the route from node 1 to the node in question.
Step 3 Identify the tentative label with the smallest distance value and mark it as permanentlylabelled.If all nodes are permanently labelled, go to Step 5.
Step 4 Consider all nodes without permanent labels that can be reached directly from the newpermanently labelled node identified in Step 3. Compute new tentative labels for thesenodes by adding the direct distance between the new permanently labelled node to thenode in question to the distance value of the new permanent label.
∗ If the sum is less than the distance value of a tentative label, change the tentativelabel as follows:
· The distance value becomes the shorter distance,
· The preceding-node value becomes the new permanent node’s number.
∗ If the sum is more than the distance value of a tentative label, leave the tentativelabel unchanged.
∗ Return to Step 3.
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Step 5 The shortest route to a given node can be found by starting at the given node andmoving to its preceding node. Working backwards through the network will providethe shortest route from node 1 to the node in question.
Steps 1 to 4 are the labelling phase and step 5 the backtracking phase.
Note:
• N − 1 iterations of the algorithm are required to find the shortest distance from node 1 toall other nodes. If the shortest distance to every node is not needed, the algorithm can bestopped when the nodes of interest have been permanently labelled.
• The algorithm can easily be modified to find the shortest distance from any node, say nodek, to all other nodes in the network. This is done by merely labelling node k with thepermanent label [0; S] and then applying the algorithm in the usual way.
Example 8.2.1
Find the shortest path from node 1 to node 6 in Figure 8.9, using the shortest-route algorithm.
2 0 0
1
2
3
4
5
61 5 05 0
4 0
Figure 8.9: Network for example problem
Solution
We first start with the labelling phase.
I (see Figure 8.10)
Step 1 Assign the permanent label [0; S] to node 1.
Step 2 Compute tentative labels for nodes that can be reached directly from node 1.
2 0 0
1
2
3
4
5
61 0 0
1 0 02 0 0
1 5 0
1 0 0
1 0 05 0
4 0
[ 1 0 0 ; 1 ]
[ 2 0 0 ; 1 ]
[ 0 ; B ]
Figure 8.10: Initial labelling
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I 1 (see Figure 8.11)
Step 3 Node 2 has the smallest distance value in its tentative label, that is, 100, and it ispermanently labelled. Mark it with an arrow.
Step 4 New tentative labels are computed for all nodes that can be reached directly fromnode 2:
∗ Node 3: The distance value of node 2 (= 100) plus c23 (= 50) is smaller than thedistance value of node 3 (= 200). The label therefore changes to [150; 2].
∗ Node 4: 100 + 200 = 300. The label assigned to node 4 is [300; 2].
∗ Node 5: 100 + 100 = 200. The label assigned to node 5 is [200; 2].
2 0 0
1
2
3
4
5
61 0 0
1 0 0
2 0 0
1 5 0
1 0 0
1 0 05 0
4 0
[ 1 0 0 ; 1 ]
[ 2 0 0 ; 1 ][ 1 5 0 ; 2 ]
[ 2 0 0 ; 2 ]
[ 3 0 0 ; 2 ]
[ 0 ; B ]
Figure 8.11: Iteration 1
I 2 (see Figure 8.12)
Step 3 Node 3 has the smallest distance value (= 150) in its tentative label and is perma-nently labelled. Mark it with an arrow.
Step 4 New tentative labels are computed for all nodes that can be reached directly fromnode 3:
∗ Node 5: 150 + 40 = 190 < 200. The label changes to [190; 3].
2 0 0
1
2
3
4
5
61 0 0
1 0 0
2 0 0
1 5 0
1 0 0
1 0 05 0
4 0
[ 1 0 0 ; 1 ]
[ 2 0 0 ; 1 ][ 1 5 0 ; 2 ]
[ 2 0 0 ; 2 ]
[ 3 0 0 ; 2 ]
[ 0 ; B ]
[ 1 9 0 ; 3 ]
Figure 8.12: Iteration 2
I 3 (see Figure 8.13)
Step 3 Node 5 has the smallest distance value (= 190) in its tentative label and is perma-nently labelled. Mark it with an arrow.
Step 4 New tentative labels are computed for all nodes that can be reached directly fromnode 5:
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∗ Node 4: 190 + 150 = 340 > 300. The label stays the same.
∗ Node 6: 190 + 100 = 290. The label assigned to node 6 is [290; 5].
2 0 0
1
2
3
4
5
61 0 0
1 0 0
2 0 0
1 5 0
1 0 0
1 0 05 0
4 0
[ 1 0 0 ; 1 ]
[ 2 0 0 ; 1 ][ 1 5 0 ; 2 ]
[ 2 0 0 ; 2 ]
[ 3 0 0 ; 2 ]
[ 0 ; B ]
[ 1 9 0 ; 3 ]
[ 2 9 0 ; 5 ]
Figure 8.13: Iteration 3
I 4 (see Figure 8.14)
Step 3 Node 6 has the smallest distance value (= 290) in its tentative label and is perma-nently labelled. Mark it with an arrow.
Step 4 New tentative labels are computed for all nodes that can be reached directly fromnode 6:
∗ Node 4: 290 + 100 = 390 > 300. The label stays the same.
2 0 0
1
2
3
4
5
61 0 0
1 0 0
2 0 0
1 5 0
1 0 0
1 0 05 0
4 0
[ 1 0 0 ; 1 ]
[ 2 0 0 ; 1 ][ 1 5 0 ; 2 ]
[ 2 0 0 ; 2 ]
[ 3 0 0 ; 2 ]
[ 0 ; B ]
[ 1 9 0 ; 3 ]
[ 2 9 0 ; 5 ]
Figure 8.14: Iteration 4
I 5 (see Figure 8.15)
Node 4 is the only node not permanently labelled. There are no other nodes thatcan be reached directly from node 4, and it is permanently labelled. Figure 8.15shows the final network with all nodes permanently labelled.
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2 0 0
1
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5
61 0 0
1 0 0
2 0 0
1 5 0
1 0 0
1 0 05 0
4 0
[ 1 0 0 ; 1 ]
[ 2 0 0 ; 1 ][ 1 5 0 ; 2 ]
[ 2 0 0 ; 2 ]
[ 3 0 0 ; 2 ]
[ 0 ; S ]
[ 1 9 0 ; 3 ]
[ 2 9 0 ; 5 ]
Figure 8.15: Iteration 5
Next we look at the backtracking phase.
Step 5 The shortest route from node 1 to node 6 is identified by working backward fromnode 6. The shortest routes, with a distance of 290 units can be identified, that is,1-2-3-5-6.
Example 8.2.2
Mary has just (at time 0) purchased a new car for R120 000. The cost of maintaininga car for a year depends on the age of the car at the beginning of the year, as given inTable 8.2(a). To avoid the high maintenance costs associated with an older car, Mary maydecide to trade in her car and purchase a new one. The price she receives on a trade-in will depend on the age of the car at the time of the trade-in (see Table 8.2(b)). Tosimplify the computations we assume that, at any time, it costs R120 000 to purchase anew car. What strategy should Mary follow if she wants to minimise the total net costs(ie purchasing costs+maintenance costs−money received for trade-ins) incurred over thenext five years? Solve this problem by using the shortest-route algorithm.
(a) Maintenance costs
Age of car Annual maintenance(in years) cost (in rand)
0 20 0001 40 0002 50 0003 90 0004 120 000
(b) Trade-in prices
Age of car Trade-in price(in years) (in rand)
1 70 0002 60 0003 20 0004 10 0005 0
Table 8.2: Data for Mary’s problem
Solution
This problem can be represented by a network consisting of 6 nodes, where node i is thethe beginning of year i. Arc (i; j) represents a time span of one year. For i < j, an arc (i; j)corresponds to purchasing a new car at the beginning of year i and keeping it until thebeginning of year j. The total length of arc (i; j), call it ci j, is the total net cost incurred in
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owning and maintaining a car from the beginning of year i to the beginning of year j, if anew car is purchased at the beginning of year i and this car is traded in for a new car at thebeginning of year j. Thus
ci j = maintenance cost incurred during years i; i + 1; . . . ; j − 1
+ cost of purchasing car at beginning of year i
− trade-in value received at beginning of year j.
Using the information in Table 8.2 in this formula, the total net costs (in R10 000s) can becalculated as follows:
c12 = maintenance cost incurred during year 1 (age of car = 0 years)
+ cost of purchasing car at beginning of year 1
− trade-in value received at beginning of year 2 (age of car = 1 year)
= 2 + 12 − 7 = 7.
c13 = maintenance cost incurred during years 1 and 2 (age of car = 0 and 1 years)
+ cost of purchasing car at beginning of year 1
− trade-in value received at beginning of year 3 (age of car = 2 years)
= 2 + 4 + 12 − 6 = 12
c26 = maintenance cost incurred during years 2, 3, 4 and 5 (age of car = 0, 1, 2, 3 years)
+ cost of purchasing car at beginning of year 2
− trade-in value received at beginning of year 6 (age of car = 4 years)
= 2 + 4 + 5 + 9 + 12 − 1 = 31
By calculating the total net costs for all possible purchasing and trade-in transactions,Table 8.3 can be constructed:
Purchase car at Trade car in atbeginning of beginning of year j
year i 1 2 3 4 5 6
1 - 7 12 21 31 442 - - 7 12 21 313 - - - 7 12 214 - - - - 7 125 - - - - - 7
Table 8.3: Total net costs for Mary’s problem
The network for Mary’s problem is given in Figure 8.16. Note that the arcs are directedforward since we are considering time — a car bought today cannot be sold yesterday!
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1 2 3 4 5 67 7 7 7 7
12 12
12 12
21 21
21
31 31
44
Figure 8.16: Network for Mary’s problem
To find the optimal strategy, we have to find the shortest path from node 1 to node 6 inFigure 8.16, using the shortest-route algorithm.
I (see Figure 8.17)
Step 1 Assign the permanent label [0; S] to node 1.
Step 2 Compute tentative labels for nodes that can be reached directly from node 1.
1 2 3 4 5 67 7 7 7 71 2 1 2
1 2 1 2
2 1 2 1
2 1
3 1 3 1
4 4
[ 0 ; B ] [ 7 ; 1 ] [ 1 2 ; 1 ]
[ 2 1 ; 1 ] [ 3 1 ; 1 ] [ 4 4 ; 1 ]
Figure 8.17: Initial labelling
I 1 (see Figure 8.18)
Step 3 Node 2 has the smallest distance value in its tentative label, that is, 7, and it ispermanently labelled. Mark it with an arrow.
Step 4 New tentative labels are computed for all nodes that can be reached directly fromnode 2:
∗ Node 3: The distance value of node 2 (= 7) plus c23 (= 7) is greater than thedistance value of node 3 (= 12). The label therefore stays the same.
∗ Node 4: 7 + 12 = 19 < 21. The label changes to [19; 2].
∗ Node 5: 7 + 21 = 28 < 31. The label changes to [28; 2].
∗ Node 6: 7 + 31 = 38 < 44. The label changes to [38; 2].
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1 2 3 4 5 67 7 7 7 71 2 1 2
1 2 1 2
2 1 2 1
2 1
3 1 3 14 4
[ 0 ; B ] [ 7 ; 1 ] [ 1 2 ; 1 ]
[ 2 1 ; 1 ] [ 3 1 ; 1 ] [ 4 4 ; 1 ][ 1 9 ; 2 ] [ 2 8 ; 2 ] [ 3 8 ; 2 ]
Figure 8.18: Iteration 1
I 2(see Figure 8.19)
Step 3 Node 3 has the smallest distance value (= 12) in its tentative label and is perma-nently labelled. Mark it with an arrow.
Step 4 New tentative labels are computed for all nodes that can be reached directly fromnode 3:
∗ Node 4: 12 + 7 = 19, which is the same as the distance value of the previouslabel. This means that there are two paths by which node 4 can be reached in19 units and these are indicated as such on the network.
∗ Node 5: 12 + 12 = 24 < 28. The label changes to [24; 3].
∗ Node 6: 12 + 21 = 33 < 38. The label changes to [38; 3].
1 2 3 4 5 67 7 7 7 71 2 1 2
1 2 1 2
2 1 2 1
2 1
3 1 3 14 4
[ 0 ; B ] [ 7 ; 1 ] [ 1 2 ; 1 ]
[ 1 9 ; 2 ] [ 2 8 ; 2 ] [ 3 8 ; 2 ][ 2 4 ; 3 ] [ 3 3 ; 3 ][ 1 9 ; 3 ]
Figure 8.19: Iteration 2
I 3 (see Figure 8.20)
Step 3 Node 4 has the smallest distance value (= 19) in its tentative label and is perma-nently labelled. Mark it with an arrow.
Step 4 New tentative labels are computed for all nodes that can be reached directly fromnode 4:
∗ Node 5: 19 + 7 = 26 > 24. The label stays the same.
∗ Node 6: 19 + 12 = 31 < 33. The label changes to [31; 4].
I 4 (see Figure 8.21)
Step 3 Node 5 has the smallest distance value (= 24) in its tentative label and is perma-nently labelled. Mark it with an arrow.
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1 2 3 4 5 67 7 7 7 71 2 1 2
1 2 1 2
2 1 2 1
2 1
3 1 3 14 4
[ 0 ; B ] [ 7 ; 1 ] [ 1 2 ; 1 ]
[ 1 9 ; 2 ] [ 2 4 ; 3 ] [ 3 3 ; 3 ][ 3 1 ; 4 ]
Figure 8.20: Iteration 3
Step 4 New tentative labels are computed for all nodes that can be reached directly fromnode 5:
∗ Node 6: 24 + 7 = 31. This is the same as the distance value of the previouslabel. This means that there are two paths by which node 6 can be reached in31 units and these are indicated as such on the network.
1 2 3 4 5 67 7 7 7 71 2 1 2
1 2 1 2
2 1 2 1
2 1
3 1 3 14 4
[ 0 ; B ] [ 7 ; 1 ] [ 1 2 ; 1 ]
[ 1 9 ; 2 ] [ 2 4 ; 3 ] [ 3 1 ; 4 ][ 3 1 ; 5 ][ 1 9 ; 3 ]
Figure 8.21: Iteration 4
I 5 (see Figure 8.22)Node 6 is the only node not permanently labelled. There are no other nodes thatcan be reached directly from node 6 and it is permanently labelled. Figure 8.22shows the final network with all nodes permanently labelled.
Step 5 The shortest route from node 1 to node 6 is identified by working backwards fromnode 6. Three shortest routes, each with a distance of 31 units can be identified,namely 1-3-5-6; 1-3-4-6 and 1-2-4-6.
The results can be interpreted as follows:
• 1-3-5-6: Mary should keep the car she bought at time 0 for 2 years until the beginningof year 3 when she should trade it in for a new car. This new car should then be tradedin after another 2 years at the beginning of year 5. At the end of the 5-year period,Mary’s car strategy will have cost her R310 000. She will need to find a new strategyfor deciding when to trade in the new car she bought at the beginning of year 5.
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1 2 3 4 5 67 7 7 7 71 2 1 2
1 2 1 2
2 1 2 1
2 1
3 1 3 14 4
[ 0 ; B ] [ 7 ; 1 ] [ 1 2 ; 1 ]
[ 1 9 ; 2 ] [ 2 4 ; 3 ] [ 3 1 ; 4 ][ 3 1 ; 5 ][ 1 9 ; 3 ]
Figure 8.22: Iteration 5
• 1-3-4-6: Keep the new car for 2 years and trade it in at the beginning of year 3.Keep this car for 1 year and trade it in at the beginning of year 4.Keep this car for 2 years and trade it in at the beginning of year 6.
• 1-2-4-6: Keep the new car for 1 year and trade it in at the beginning of year 2.Keep this car for 2 years and trade it in at the beginning of year 4.Keep this car for 2 years and trade it in at the beginning of year 6.
8.2.2 Formulating the shortest-route problem as an LP model
For large problems with many nodes, the algorithm is necessary and can be easily implementedon a computer.
A shortest-route problem can be formulated as an LP model with the following characteristics:
• The objective is to minimise the cost of sending one unit from the supply node (the source)to the demand node (the sink).
• All other nodes in the network are nodes through which units can be sent.
• The cost of sending one unit from node i to node j is the length of arc (i; j).
Consider the network in Figure 8.23 which represents the transportation of power from a powerplant to a city. When power is sent from the plant (node 1) to the city (node 6), it must passthrough relay substations (nodes 2, 3, 4 and 5). The distances to possible connections are givenon the arcs. The objective is to minimise the distance that the power must travel. If the cost ofshipping power were proportional to the distance the power travels, the shortest route wouldalso indicate the minimum shipping costs.
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Figure 8.23: The network for sending power from the plant to the city
We begin by defining the decision variables:
The following decision variables are defined.
Let xi j =
{
0 if no unit is shipped from i to j,
1 if one unit is shipped from i to j.
For example
x12 = units shipped from 1 to 2
Secondly, we define the objective function:
The objective is to minimise the distance the power has to travel (or the shipping cost). Theobjective function is therefore
Minimise DISTANCE = 4x12 + 3x13 + 3x24 + 2x25 + 3x35 + 2x46 + 2x56.
Thirdly, we define the constraints:
The constraints are formulated in such a way that the shortest path will be identified when themodel has been solved. The decision variables x22, x33, etc. are dummy variables used to ensurethat an arc that is not on the shortest route will have a value of zero. x33 = 1 states that the unitthat is sent through node 3 stays at node 3. The arc following node 3 therefore has a value of zero,and is not on the shortest route. Because one unit must leave node 1 and one unit must arrive atnode 6, dummy variables, x11 and x66, are not used in the model.
• Constraints on the supply at nodes are formulated as follows:
– Node 1: One unit can be shipped from node 1 to either node 2 or node 3. The constraintis x12 + x13 = 1.
– Node 2: One unit can be shipped from node 2 to either node 4 or node 5 or it canremain at node 2. The constraint is x22 + x24 + x25 = 1.
– Etc.
• Constraints on the demand at the nodes are formulated as follows:
– Node 2: One unit can arrive at node 2 from either node 1 or from node 2 itself. Theconstraint is x12 + x22 = 1.
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– Node 5: One unit can arrive at node 5 from node 2, node 3 or from node 5 itself. Theconstraint is x25 + x35 + x55 = 1.
– Etc.
Lastly, variables must be non-negative.
The complete LP model is given in Model 8.2.1.
Minimise DISTANCE = 4x12 + 3x13 + 3x24 + 2x25 + 3x35 + 2x46 + 2x56
Supply constraints:
x12 + x13 = 1 (Unit sent from 1 to 2 or 3.)
x22 + x24 + x25 = 1 (Unit sent from 2 to 2, 4 or 5.)
x33 + x35 = 1 (Unit sent from 3 to 3 or 5.)
x44 + x46 = 1 (Unit sent from 4 to 4 or 6.)
x55 + x56 = 1 (Unit sent from 5 to 5 or 6.)
Demand constraints:
x12 + x22 = 1 (Unit arrives at 2 from 1 or 2.)
x13 + x33 = 1 (Unit arrives at 3 from 1 or 3.)
x24 + x44 = 1 (Unit arrives at 4 from 2 or 4.)
x25 + x35 + x55 = 1 (Unit arrives at 5 from 2, 3 or 5.)
x46 + x56 = 1 (Unit arrives at 6 from 4 or 5.)
xi j ≥ 0 for i = 1; 2; 3; 4; 5 and j = 2; 3; 4; 5; 6.
Model 8.2.1: The LP model for finding the shortest route
8.2.3 Solving shortest-path problems with LINGO
The input data file for LINGO:
[Objective] min=4*x12+3*x13+3*x24+2*x25+3*x35+2*x46+2*x56;
[From_1] x12+x13=1;
[From_2] x22+x24+x25=1;
[From_3] x33+x35=1;
[From_4] x44+x46=1;
[From_5] x55+x56=1;
[To_2] x12+x22=1;
[To_3] x13+x33=1;
[To_4] x24+x44=1;
[To_5] x25+x35+x55=1;
[To_6] x46+x56=1;
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The solution obtained by LINGO:
Optimal solution found at step: 2
Objective value: 8.000000
Variable Value Reduced Cost
X12 0.0000000E+00 0.0000000E+00
X13 1.000000 0.0000000E+00
X24 0.0000000E+00 1.000000
X25 0.0000000E+00 0.0000000E+00
X35 1.000000 0.0000000E+00
X46 0.0000000E+00 0.0000000E+00
X56 1.000000 0.0000000E+00
X22 1.000000 0.0000000E+00
X33 0.0000000E+00 0.0000000E+00
X44 1.000000 0.0000000E+00
X55 0.0000000E+00 0.0000000E+00
Row Slack or Surplus Dual Price
1 8.000000 1.000000
FROM_1 0.0000000E+00 -3.000000
FROM_2 0.0000000E+00 1.000000
FROM_3 0.0000000E+00 0.0000000E+00
FROM_4 0.0000000E+00 3.000000
FROM_5 0.0000000E+00 3.000000
TO_2 0.0000000E+00 -1.000000
TO_3 0.0000000E+00 0.0000000E+00
TO_4 0.0000000E+00 -3.000000
TO_5 0.0000000E+00 -3.000000
TO_6 0.0000000E+00 -5.000000
Interpretation of the solution:
According to the solution, variables x13, x35, x56, x22 and x44 all have values of 1. The shortest routeis therefore 1-3-5-6. x22 = 1 and x44 = 1 indicate that these nodes are not on the shortest route.
Note: When this problem is solved by hand, the solution in Figure 8.24 is obtained.
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33
2
32
2[ 0 ; B ]
[ 3 ; 1 ] [ 6 ; 3 ][ 6 ; 2 ]
[ 8 ; 5 ]
[ 7 ; 2 ][ 4 ; 1 ]
Figure 8.24: Solution using the algorithm
From this solution it is clear that there are two shortest routes, each with a length of 8 units. Thesepaths are 1-3-5-6 and 1-2-5-6. LINGO gives only one of the possible shortest routes. Dependingon the software you are using, you may obtain either one of them.
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8.3 The maximum-flow problem
In a maximum-flow problem, the objective is to determine the maximum number of units, suchas vehicles, messages, petrol, oil, etc, that can enter a network at one end, move through thenetwork, and exit at the other end of the network. The node where units can enter the network iscalled the input (or source) node and the node where the units can exit the network is called theoutput or (sink) node. In such problems we attempt to send as many units through all arcs of thenetwork as possible.
The number of units that can flow through a network is limited, because the connections maybe able to carry only a certain amount of flow. Such limitations may be things like pipe sizes inan oil distribution system or roads with only one lane in a road network. We refer to the upperlimit on the flow on an arc as the flow capacity of the arc. The flow capacity of an arc in a certaindirection appears on the arc, close to the node from which the flow comes. In the figure belowthe flow capacity in the direction from node 1 to node 2 is 3 units, while the flow capacity in thedirection from node 2 to node 1 is 1 unit. This means that the arc can carry a maximum of 3 unitsof flow from node 1 to node 2 and a maximum of 1 unit from node 2 to node 1.
1 23 1
M- A
Step 1 Find any path from the input (source) node to the output (sink) node that has flowcapacities in the direction of flow greater than zero on all arcs of the path. If no path isavailable, stop, because the optimal solution has been reached.
Step 2 Find the smallest arc capacity, P f , on the path selected in Step 1. Increase the flowthrough the network by sending an amount of P f along this path.
Step 3 For the path selected in Step 1, reduce all arc flow capacities in the direction of flow byP f and increase all arc flow capacities in the reverse direction by P f . Go to Step 1.
Let us look at an example.
Consider the problem facing the city planners of Dunbar, a town in the Eastern Province. Theyare in the process of developing a road system to accommodate the heavy traffic expected overthe holiday season. They would like to determine the maximum number of cars that can flowthrough the town from west to east. The road network, with flow capacities in hundreds of cars,is shown in Figure 8.25.
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7West East
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Figure 8.25: Dunbar’s road network
The flow capacities are based on the direction of flow. For example, the section of road betweennodes 1 and 3, or arc 1-3, shows a capacity of 600 vehicles per hour in the 1-3 direction. However,a zero capacity exists in the 3-1 direction. This can be interpreted as being a one-way street orthat it is undesirable to direct traffic towards node 1, which is the input and a potential traffic jamlocation. The road section between nodes 2 and 3, however, shows capacities of 200 vehicles inboth directions.
I 1 (see Figure 8.26)
Start by selecting any path through the network. Let us start by selecting the path 1-2-5-7. Theflow capacity on arc 1-2 in the West-East direction is 5 units, on arc 2-5, 3 units and on arc 5-7, 8units.
A maximum of 3 units can flow from node 1 to node 7 along this route since arc 2-5 cannot handlemore. The minimum arc capacity on this path is therefore P f = 3.
To determine the remaining capacities along this path, the capacities in the direction of flow arereduced by 3. To permit flows assigned along this path to take an alternative route, should it benecessary for optimality, the flow in the reverse direction is increased by 3 units.
The total flow through the network after one iteration is 3 units.
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Total flow afteriteration 1
Figure 8.26: Iteration 1 (3 units sent along path 1-2-5-7)
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I 2 (see Figure 8.27)
Select path 1-3-6-7. P f = 6, as determined by arc 1-3.
The total flow after iteration 2 is 3 + 6 = 9.
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Total flow afteriteration 2
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Figure 8.27: Iteration 2 (P f = 6 along path 1-3-6-7)
I 3 (see Figure 8.28)
Select path 1-2-3-5-7. P f = 2, as determined by arc 1-2 and arc 2-3.
The total flow after iteration 3 is 9 + 2 = 11.
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Total flow afteriteration 3
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Figure 8.28: Iteration 3 (P f = 2 along path 1-2-3-5-7)
I 4 (see Figure 8.29)
Select path 1-4-6-7. P f = 1, as determined by arc 6-7.
The total flow after iteration 4 is 11 + 1 = 12.
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12 12
0 3
Total flow afteriteration 4
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Figure 8.29: Iteration 4 (P f = 1 along path 1-4-6-7)
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I 5 (see Figure 8.30)
Select path 1-4-6-5-7. P f = 1, as determined by arc 6-5.
The total flow after iteration 5 is 12 + 1 = 13.
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0 1
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13 13
0 3
Total flow afteriteration 5
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Figure 8.30: Iteration 5 (P f = 1 along path 1-4-6-5-7.)
I 6 (see Figure 8.31)
Select path 1-4-6-3-5-7. P f = 1, as determined by arc 3-5.
The total flow after iteration 6 is 13 + 1 = 14.
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0 3
Total flow afteriteration 6
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Figure 8.31: Iteration 6 (P f = 1 along path 1-4-6-3-5-7.)
I 7
An inspection of Figure 8.31 indicates that only arc 1 to 4 has a positive capacity leading out ofthe source node. No more units can flow from 1 to 2 or 1 to 3 because their capacities are zero.Although the capacity of arc 1 to 4 is 2 no way exists to send flow along any route of the networkto the sink node, without encountering a zero capacity along the way. The optimal solution hastherefore been reached.
The maximum flow is thus reached, if it is not possible to send any additional flow along anyroute from the source node to the sink node; thus if no path from arc to sink exists with a positive(> 0) flow.
In our example there are no more paths from node 1 to node 7 with flow capacities greater thanzero on all the arcs of the path in the west-east direction. The flow of 1 400 vehicles per hour istherefore optimal. The maximum flow through the network is 1 400 vehicles per hour.
The iterative process can be summarised in the following table:
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Iteration Path Flow
1 1-2-5-7 32 1-3-6-7 63 1-2-3-5-7 24 1-4-6-7 15 1-4-6-5-7 16 1-4-6-3-5-7 1
Total flow: 14
Note: In iteration 6, a flow of 1 unit (100 vehicles per hour) was permitted in the 6-3 direction.
From the initial network, however, we know that the flow capacity in the 6-3 direction is zero.The flow of 1 unit in the 6-3 direction represents a fictitious flow. The real effect of this flow is todivert 1 unit of flow originally committed to the 3-6 arc (iteration 2), along the 3-5 arc to enableus to get 1 more unit of flow through the network.
Lastly, to summarise, we can determine the final flow in each arc. The flows on each arc in thenetwork can be found by keeping track of the flow assignments or by comparing the final arcflow capacities with the initial arc flow capacities; the difference equals the flow. Consider, forinstance, arc 4-6 with initial and final capacities as follows:
5 04 4 2 36 6
d i f f e r e n c e = 3
d i f f e r e n c e = 3
I N I T I A L F I N A L
The flow over arc 4-6 is therefore 3 units. By comparing final and initial flow capacities for allarcs in the network, the final flow pattern can be determined, as shown in Figure 8.32.
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7 W e s t E a s t
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Figure 8.32: The final flow pattern
The results of the maximum flow analysis indicate that the road network will be able to carry amaximum of 1 400 vehicles per hour from west to east through Dunbar.
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8.4 The minimum-spanning tree problem
In a minimum-spanning tree problem the objective is to identify the arcs in a network that connectall nodes of the network to each other in such a way that the total length of all the arcs used,is minimised. A spanning tree for a N-node network is a set of N − 1 arcs that connects everynode to every other node. A minimum spanning tree is the set of arcs that does so at minimumtotal arc cost, distance, etc. The minimum-spanning tree is applicable to transportation systemsin which the nodes are terminals and the arcs highways, airways, etc, as well as communicationand teleprocessing systems.
Two algorithms that can be used to solve minimum-spanning tree problems are discussed in thismodule, namely the minimum-spanning tree algorithms 1 and 2.
8.4.1 Minimum-spanning tree algorithm 1
We can summarise the algorithm 1 as follows:
M-S T A 1
Step 1 Arbitrarily begin at any node and connect it to the closest node in terms of the criterionbeing used (eg time, cost, distance, etc). The two nodes are referred to as connectednodes, and the remaining nodes are referred to as unconnected nodes.
Step 2 Identify the unconnected node that is closest to any of the connected nodes. Break tiesarbitrarily if two or more nodes qualify as the closest node. Add this new node to theset of connected nodes. Repeat this step until all nodes have been connected to at leastone other node in the network.
Example 8.4.3
Sage Construction is currently developing a housing project at Pebble Beach. They wantto determine the least expensive way to provide each house with water and power. Thenetwork of houses and the possible connections (in metres) are shown in Figure 8.33.
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30
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1060
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3050
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Figure 8.33: The network for Sage Construction
219 DSC2602
Solution
I 1 (see Figure 8.34(a))
Arbitrarily select node 1 to begin at and mark it as connected (Step 1).
Node 1 is connected to nodes 2, 3 and 4. Node 2 is 30 metres away from 1, node 3 is 20metres away from 1 and node 4 is 50 metres away from 1. Of these three nodes, the nodenearest to node 1 is node 3, which is 20 metres away. Mark node 3 as connected and showthe connection as a thick line on the network (Step 2).
Connected Connectednode to Distance
1 2 30
3 20 ←−minimum
4 50
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(a) Iteration 1
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(b) Iteration 2
Figure 8.34: Iterations 1 and 2
I 2 (see Figure 8.34(b))
Consider all the connected nodes, namely nodes 1 and 3. Determine which of the nodesthat are connected to 1 and 3, namely nodes 2, 4, 5, 6 and 7, is the closest.
Node 4, 20 metres from node 3, is identified as being the closest to any of the connectednodes. Mark node 4 as connected and show the connection.
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Connected Connectednodes to Distance
1 2 304 50
3 2 30
4 20 ←−minimum
5 506 307 70
I 3 (see Figure 8.35(a))
Consider the connected nodes 1, 3 and 4 and their connections:
Connected Connectednodes to Distance
1 2 30 ←−minimum
3 2 30 ←−minimum5 506 30 ←−minimum7 70
4 6 60
Node 2, 30 metres from node 3, is arbitrarily identified as being the closest to any of theconnected nodes. Mark node 2 as connected and show the connection.
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(a) Iteration 3
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(b) Iteration 4
Figure 8.35: Iterations 3 and 4
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I 4 (See Figure 8.35(b))
Consider nodes 1, 3, 4 and 2 and their connections:
Connected Connectednodes to node Distance
2 5 30 ←−minimum
3 5 506 30 ←−minimum7 70
4 6 60
Nodes 5 and 6 are both 30 metres from a connected node. Select node 5 arbitrarily, mark itas connected and show the connection.
I 5 (see Figure 8.36(a))
Consider nodes 1, 3, 4, 2 and 5 and their connections:
Connected Connectednodes to node Distance
3 6 30 ←−minimum
7 70
4 6 60
5 7 40
Node 6, 30 metres from node 3, is identified as being the closest to any of the connectednodes. Mark node 6 as connected and show the connection.
I 6 (see Figure 8.36(b))
Consider nodes 1, 3, 4, 2, 5 and 6 and their connections:
Connected Connectednodes to node Distance
3 7 70
5 7 40
6 8 10 ←−minimum
Node 8, 10 metres from node 6, is identified as being the closest to any of the connectednodes. Mark node 8 as connected and show the connection.
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(a) Iteration 5
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(b) Iteration 6
Figure 8.36: Iterations 5 and 6
I 7 (see Figure 8.37)
Consider nodes 1, 3, 4, 2, 5, 6 and 8 and their connections:
Connected Connectednodes to node Distance
3 7 708 7 20 ←−minimum
Node 7, 20 metres from node 8, is identified as being the closest to any of the connectednodes. Mark node 7 as connected and show the connection.
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Figure 8.37: Iteration 7
All the nodes are now connected and we can stop. The minimum-spanning tree is givenby the thick connections on the network, as shown in the seventh and final iteration. Theminimum total length of the connections is 20 + 20 + 30 + 30 + 30 + 10 + 20 = 160 metres.
When considering a large number of nodes, it is not easy to keep track of all the possible con-nections between connected and unconnected nodes. The shortest distance may be overlookedand the final solution may not be optimal. The following algorithm gives a systematic method forfinding the minimum-spanning tree.
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8.4.2 Minimum-spanning tree algorithm 2
This method can be summarised as:
M-S T A 2
Step 1 Set up a table with all the node numbers written down the left-hand column and againacross the top row. Enter the direct distance between nodes i and j in the correctposition in row i and column j. If two nodes are not directly connected, leave the spaceblank or enter a “-”. These spaces are not taken into consideration when the algorithmis carried out. Add a column on the extreme left-hand side of the table and label it“connected”.
Step 2 Arbitrarily select any node and consider it connected. Place a X in the left-handcolumn beside the chosen node and cross out the column corresponding to this node.This is done to show that the node is connected and should not be considered further.
Step 3 Consider all rows corresponding to connected nodes (ie all theX rows) and all columnsthat are not crossed out. Find the smallest value (i; j) and draw a circle around it. Thismeans that i and j are connected, with j being the new connected node.
Step 4 Place a X in the left-hand column beside the new connected node and cross out thecolumn corresponding to the new connected node. Return to Step 3 until all nodeshave been connected.
Step 5 The minimum spanning tree is given by the connections identified by the circled values.The total length of the connections is given by the sum of the circled values.
Example 8.4.4
Find the minimum-spanning tree of the network in Figure 8.38 using the minimum spanningtree algorithm 2.
1
3
4
2
5
6
7
8
9
10
45
30 2535
50
15
40
45
50 20
60
4030
35
20
2530
Figure 8.38: The network
DSC2602 224
Solution
I (See Table 8.4)
Construct the table and arbitrarily select node 6 to begin with. Mark it as connected in theleft-hand column and cross out column 6.
Connected Node 1 2 3 4 5 6 7 8 9 10
1 - 30 45 - - - - - - -2 30 - - - 25 - 35 50 - -3 45 - - 50 - 20 - - 60 -4 - - 50 - - 45 - - - -5 - 25 - - - 40 15 - - -
X 6 - - 20 45 40 - - - 30 357 - 35 - - 15 - - 30 - 208 - 50 - - - - 30 - - 259 - - 60 - - 30 - - - 40
10 - - - - - 35 20 25 40 -
Table 8.4: Initial table
I 1
Consider Table 8.4. The smallest value in the connected row (row 6) is 20, which appears inposition (6; 3) in the table. Circle this value. Nodes 6 and 3 are now connected with node3, the new connected node. Place a X next to row 3 in the left-hand column and cross outcolumn 3. See Table 8.5.
Connected Node 1 2 3 4 5 6 7 8 9 10
1 - 30 45 - - - - - - -2 30 - - - 25 - 35 50 - -
X 3 45 - - 50 - 20 - - 60 -4 - - 50 - - 45 - - - -5 - 25 - - - 40 15 - - -
X 6 - - 20 45 40 - - - 30 35
7 - 35 - - 15 - - 30 - 208 - 50 - - - - 30 - - 259 - - 60 - - 30 - - - 40
10 - - - - - 35 20 25 40 -
Table 8.5: Iteration 1
225 DSC2602
I 2
Consider the values in the connected rows (rows 6 and 3) in Table 8.4 that are not crossedout. Select the smallest value — in this case 30 which appears in position (6; 9). Node 9is the new connected node. Place a X next to row 9 in the left-hand column. Cross outcolumn 9. See Table 8.6.
Connected Node 1 2 3 4 5 6 7 8 9 10
1 - 30 45 - - - - - - -2 30 - - - 25 - 35 50 - -
X 3 45 - - 50 - 20 - - 60 -4 - - 50 - - 45 - - - -5 - 25 - - - 40 15 - - -
X 6 - - 20 45 40 - - - 30 35
7 - 35 - - 15 - - 30 - 208 - 50 - - - - 30 - - 25
X 9 - - 60 - - 30 - - - 4010 - - - - - 35 20 25 40 -
Table 8.6: Iteration 2
I 3
Consider the values in the connected rows (rows 6, 3, and 9) that are not crossed out inTable 8.6. Select the smallest value — in this case 35, which appears in position (6; 10).Node 10 is the new connected node. Place a X next to row 10 in the left-hand column andcross out column 10. See Table 8.7.
Connected Node 1 2 3 4 5 6 7 8 9 10
1 - 30 45 - - - - - - -2 30 - - - 25 - 35 50 - -
X 3 45 - - 50 - 20 - - 60 -4 - - 50 - - 45 - - - -5 - 25 - - - 40 15 - - -
X 6 - - 20 45 40 - - - 30 35
7 - 35 - - 15 - - 30 - 208 - 50 - - - - 30 - - 25
X 9 - - 60 - - 30 - - - 40X 10 - - - - - 35 20 25 40 -
Table 8.7: Iteration 3
DSC2602 226
I 4
Consider the values in the connected rows (rows 6, 3, 9 and 10) that are not yet crossed outin Table 8.7. Select the smallest value — in this case 20, which appears in position (10; 7).Node 7 is the new connected node. Place a X next to row 7 in the left-hand column. Crossout column 7. See Table 8.8.
Connected Node 1 2 3 4 5 6 7 8 9 10
1 - 30 45 - - - - - - -2 30 - - - 25 - 35 50 - -
X 3 45 - - 50 - 20 - - 60 -4 - - 50 - - 45 - - - -5 - 25 - - - 40 15 - - -
X 6 - - 20 45 40 - - - 30 35
X 7 - 35 - - 15 - - 30 - 208 - 50 - - - - 30 - - 25
X 9 - - 60 - - 30 - - - 40
X 10 - - - - - 35 20 25 40 -
Table 8.8: Iteration 4
Repeat Steps 3 and 4 until all the nodes are connected, as shown in Table 8.9.
Connected Node 1 2 3 4 5 6 7 8 9 10
X 1 - 30 45 - - - - - - -
X 2 30 - - - 25 - 35 50 - -
X 3 45 - - 50 - 20 - - 60 -X 4 - - 50 - - 45 - - - -
X 5 - 25 - - - 40 15 - - -
X 6 - - 20 45 40 - - - 30 35
X 7 - 35 - - 15 - - 30 - 20
X 8 - 50 - - - - 30 - - 25X 9 - - 60 - - 30 - - - 40
X 10 - - - - - 35 20 25 40 -
Table 8.9: Final table
227 DSC2602
The minimum-spanning tree is identified by the positions of the circled values in the finaltable. The connections are: (2; 1), (5; 2), (6; 3), (6; 4), (6; 9), (6; 10), (7; 5), (10; 7) and (10; 8).The minimum-spanning tree is shown in Figure 8.39. The minimum length of the spanningtree is 30 + 25 + 15 + 20 + 25 + 35 + 45 + 20 + 30 = 245.
1
3
4
2
5
6
7
8
9
10
45
30 2535
50
15
40
45
50 20
60
4030
35
20
2530
Figure 8.39: The minimum-spanning tree
8.5 Exercises
1. Use the shortest-route algorithm to find the shortest route from node 1 to each of the othernodes in the following transportation network:
1
2
3
4
5
6
7
7
9
18
3
5
4
3
3
6
2
2. Brooklyn Taxi Association has identified 10 primary pickup and drop locations in Pretoria.In an effort to minimise travel time, improve customer service and improve the utilisationof their fleet of taxis, management would like the taxi drivers to take the shortest routebetween locations whenever possible. The following network shows roads and streets withflow in both directions. (Distances are indicated on the arcs.) Formulate an LP model todetermine the route a driver should take from location 1 to location 10. Use LINGO to solvethe LP model.
DSC2602 228
1
2
3
4
5
6
7
8
9
10
8
13
15
10
5
56
3
94
15
4
4 4
2
5 5
7
5
12
3. Petrochem, an oil refinery, is designing a new plant to produce diesel fuel. The network inFigure 8.40 shows the main processing centres, represented by the nodes, and the possibleflows (in thousands of litres of fuel) on the arcs.
1
24
5
6
7485
3
0
00
0
11
1
11
22
4
4
50
6
33
33
3
S i n kS o u r c e
Figure 8.40: The network for Petrochem’s new plant
Determine the maximum amount of fuel that can flow through the plant, from node 1 tonode 7. Show your workings on the network and give the path and the flow in each iteration.Gee die finale vloeipatroon deur elke boog.
4. Southern College is installing an electronic mail system. The following network shows thepossible electronic connections between the offices. Distances between the offices are inhundreds of metres. Design an office communication system that will enable all offices tohave access to the electronic mail service. What is the minimum length of the connectionsthat will be needed to install the mail system?
1
2
3
4
5
6
7
8
2
3
2
4
30,5
1
11,5
1,2
11,6
4
3 2,5
0,5
229 DSC2602
5. The distances (in kilometres) between the cities Alpha, Beta, Gamma, Delta and Epsilonare shown in the following table. A road system has to be built to connect these cities.For political reasons, no road can be built connecting Gamma and Epsilon and no road canbe built between Alpha and Beta. Use minimum-spanning tree algorithm 2 to determine theconnections between the cities that will minimise the total length of roads to be built. Showthe connections and give the total length.
City Alpha Beta Gamma Delta Epsilon
Alpha - 132 217 164 58Beta 132 - 290 201 79
Gamma 217 290 - 113 303Delta 164 201 113 - 196
Epsilon 58 79 303 196 -
CHECKLISTYou should now be able to
⊲ solve network models by using
– the shortest-route algorithm,
– the maximum-flow algorithm,
– minimum-spanning tree algorithm 1, and
– minimum-spanning tree algorithm 2;
⊲ LINGO-uitvoere interpreteer
⊲ interpret the solutions found.
DSC2602 230
APPENDIX A
Solutions to exercises
A.1 Chapter 1: Descriptive statistics
� 1(a) The smallest value is 28 and the largest is 48. The range is R = 48 − 28 = 20.
If we decide to use 5 intervals, the width of each interval will be20
5= 4. The first interval will
then be 27,5 − 31,5.
A quick calculation (5 (intervals) ×4 (width) = 20 and 20 + 27,5 = 47,5) gives an upper limit of47,5 for the last interval, which poses a problem since it does not contain the largest value.
We will use 6 intervals. The width of the intervals is20
6= 3
1
3, hence we choose a width of 4. Now
the first interval is 27,5 − 31,5. The limit of the last interval is now 6 × 4 + 27,5 = 51,5, which isperfect!
The frequency distribution of the claims data is given in Table A.1.
Interval Frequency
27,5 − 31,5 |||| 531,5 − 35,5 |||| 535,5 − 39,5 |||| |||| 1039,5 − 43,5 |||| | 643,5 − 47,5 ||| 347,5 − 51,5 | 1
30
Table A.1: Frequency distribution of the claims data
The histogram of the insurance claims data is given in Figure A.1.
� 1(b) The cumulative frequency table of the claims data is given in Table A.2.
231
DSC2602 232
-
6
∧∨0
2
4
6
8
10
27,5 31,5 35,5 39,5 43,5 47,5 51,5Claims
Fre
qu
ency
Figure A.1: Histogram of the claims data
Upper limit Cumulative frequency
≤ 31,5 5≤ 35,5 10≤ 39,5 20≤ 43,5 26≤ 47,5 29≤ 51,5 30
Table A.2: Cumulative frequency distribution of the claims data
(In the cumulative frequency table we can use either “smaller than” or “smaller than or equal to”(< or ≤), since we always work with an interval with an upper limit that is halfway between twosubsequent values.)
The stem-and-leaf diagram is given in Table A.3.
Stem Leaf Frequency
2 8 13 0 0 1 1 3 4 4 4 5 6 6 7 7 8 8 8 9 9 9 194 0 0 0 1 1 2 4 5 6 8 10
Table A.3: The stem-and-leaf diagram of the claims data
The values (or leaves) in a stem may be separated so that the values from 0 to 4 are separated fromthe values from 5 to 9. The new diagram gives more information and is provided in Table A.4.
Stem Leaf Frequency
2⋄ 8 13⋆ 0 0 1 1 3 4 4 4 83⋄ 5 6 6 7 7 8 8 8 9 9 9 114⋆ 0 0 0 1 1 2 4 74⋄ 5 6 7 3
Table A.4: Stem-and-leaf diagram (stem separated)
233 DSC2602
� 1(c) The minimum number of claims processed per week is 28. If a worker processes only 26claims per week for a whole month, the chances are good that there is a problem.
� 1(d) In the interval 35,5 − 39,5, the frequency is 10. Therefore, for
10
30× 100% = 33
1
3%
of the time, 36 to 39 claims are processed per week. Is the competitor more productive? Thisdepends on whether the claims being processed are similar. More information should be obtainedbefore the manager takes action.
� 1(e) Looking at the cumulative frequency table, it is easy to see that, for 33 13% of the time, fewer
than 36 claims are processed per week. I would advise the manager to transfer some of hispersonnel to other divisions.
� 2 The mean is given by
x =
∑ni=1 xi
n
=1 671
10
= 167,1.
The data must be arranged in ascending order before the median can be determined. Hence:
104; 127; 131; 135; 146; 170; 175; 179; 190; 314.
The median is the observation in the(n + 1)
2th position, that is, the
10 + 1
2= 5,5th observation.
Because there is no such value, we calculate the mean of the fifth and sixth values:
me =146 + 170
2= 158.
The mode is the value that occurs most often. There is no value that occurs more than once andthere is no mode.
� 3(a) The mean is given by
Interval Frequency Cumulativefi xi fixi frequency
27,5 – 31,5 5 29,5 147,5 531,5 – 35,5 5 33,5 167,5 1035,5 – 39,5 10 37,5 375 2039,5 – 43,5 6 41,5 249 2643,5 – 47,5 3 45,5 136,5 2947,5 – 51,5 1 49,5 49,5 30
— ——-30 1 125
DSC2602 234
x =
∑ki=1 fixi
∑ki=1 fi
=1 125
30
= 37,5.
� 3(b) The median is the observation in the(n + 1)
2th position, that is, the
30 + 1
2= 15,5th observation.
This observation falls in the interval 35,5 – 39,5.
The median interval is 35,5 – 39,5.
� 3(c) The modal interval is the interval with the highest frequency.
The modal interval is 35,5 – 39,5.
� 4 The mean is calculated as
x =
∑
xi
n=
22 355
14= 1 596,79.
The calculations for the mean and standard deviation are shown in Table A.5.
i xi (xi − x) (xi−x)2
1 1 630 33,21 1 102,902 1 550 −46,79 2 189,303 1 430 −166,79 27 818,904 1 440 −156,79 24 583,105 1 390 −206,79 42 762,106 1 400 −196,79 38 726,307 1 480 −116,79 13 639,908 1 490 −106,79 11 404,109 1 410 −186,79 34 890,50
10 1 905 308,21 94 993,4011 1 540 −56,79 3 225,1012 1 890 293,21 85 972,1013 1 900 303,21 91 936,3014 1 900 303,21 91 936,30
565 180,30
Table A.5: Calculations for the variance and standard deviation
The variance s2 =565 180,30
(14 − 1)= 43 475,41
and the standard deviation s =√
43 475,41 = 208,51.
235 DSC2602
� 5 The coefficient of variation for each variety is
cv(A) =16
88= 0,1818 = 18,18%
cv(B) =15
56= 0,2679 = 26,79%
cv(C) =25
100= 0,25 = 25,00%.
Variety A displays the least variation in the yield size.
� 6 In the diagram representing the costs of calls for females is larger than the diagram represent-ing the costs of calls for males, emphasizing the greater variation in costs for females. The factthat the median line in the costs of calls for females box is positioned low down within the boxsuggests that the middle half of the distribution is skewed. The quarter of observations betweenthe median and the upper quartile are more widely spread than the quarter of observations be-tween the median and the lower quartile. In contrast, the median line in the box that representsthe costs of calls for males is midway between the top and the bottom of the box and indicatesthat the spread of values in the middle half of the costs for males is symmetrical.
05 0
1 0 01 5 0
2 0 0
2 5 0
3 0 0
3 5 0
F e m a l e s M a l e sG e n d e r
Cost o
f calls (R)
DSC2602 236
A.2 Chapter 2: Probability concepts
� 1(a) The completed table follows:
Qualification Management level: Head of a
level Section Department Division Total
Matric 28 14 8 50Diploma 20 24 6 50Degree 5 10 14 29
Total 53 48 28 129
� 1(b)(i) First define the possible events: Let A = the event that the selected person has a matric.
There are 50 persons out of the 129 with a matric. Thus
P(A) =50
129= 0,387.
� 1(b)(ii) Define the possible events:
Let A = the event that the selected person is head of a section, andB = the event that the selected person has a degree.
Intuitively, from the table it follows that
P(A ∩ B) =5
129= 0,038.
If we apply the multiplication rule, we get
P(A ∩ B) = P(A|B)P(B) =5
29× 29
129= 0,038.
� 1(b)(iii) Define the possible events:
Let A = the event that the selected person is head of a department, andB = the event that the selected person has a diploma.
There are 24 out of the 50 heads of departments who have diplomas. Thus
P(A|B) =24
50= 0,48.
By using the formula:
P(A|B) =P(A ∩ B)
P(B)=
24129
50129
= 0,48
� 1(b)(iv) Let A = the event that the selected person is head of a division.
There are 28 out of the 129 who are heads of divisions. Thus
P(A) =28
129= 0,22.
237 DSC2602
� 1(b)(v) Define the possible events:
Let A = the event that the selected person is head of a division, andB = the event that the selected person is head of a section.
The events are mutually exclusive since they cannot occur simultaneously.
NowP(A ∪ B) = P(A) + P(B)
=28
129+
53
129= 0,22 + 0,41
= 0,63.
� 1(b)(vi) Define the possible events:
Let A = the event that the selected person has a matric,B = the event that the selected person has a diploma, andC = the event that the selected person has a degree.
The events are mutually exclusive since the highest qualification is considered.
From the table we find that
P(A) =50
129= 0,387 P(B) =
50
129= 0,387 P(C) =
29
129= 0,224.
Now,P(A ∪ B ∪ C) = P(A) + P(B) + P(C) = 0,387 + 0,387 + 0,224 = 1.
� 1(b)(vii) Define the possible events:
Let A = the event that the selected person has a matric, andB = the event that the selected person is head of a department.
P(A|B) =P(A ∩ B)
P(B)=
14129
48129
= 0,292.
� 2(a)
OpinionGender In favour Neutral Opposed Total
Male 900 200 400 1 500Female 300 100 600 1 000
Total 1 200 300 1 000 2 500
P(a female opposed to the proposal) =600
2 500.
� 2(b) P(neutral) =300
2 500.
DSC2602 238
� 2(c) P(opposed to the proposal, given a female)
=P(opposed and a female)
P(a female)
where
P(opposed and a female) =600
2 500,
and
P(a female) =1 000
2 500.
Hence P(opposed given a female)
=P(opposed and a female)
P(a female)
=
600
2 5001 000
2 500
=600
1 000.
� 2(d) P(either a male or opposed to the proposal)
= P(a male ∪ opposed)
= P(a male) + P(opposed) − P(a male and opposed)
=1 500
2 500+
1 000
2 500− 400
2 500
=2 100
2 500.
� 3(a)
Let U be the event that a member is a union member.
Let A be the event that a member belongs to union A.
Let B be the event that a member belongs to union B.
P(any member selected at random is a union member)
= P(U)
=50
90.
239 DSC2602
� 3(b) P(any member selected at random belongs to union B given that he is a union member)
= P(B | U)
=P(B ∩U)
P(U)
=
15
5050
90
=15
50× 90
50
=135
250= 0,54.
� 4(a) Let X denote the number of cars owned. P(family owns at most three cars)
= P(X ≤ 3)
= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
=27 + 1 422 + 2 865 + 1 796
6 487
=6 110
6 487
= 0,9419.
� 4(b) P(family owns at least one car)
= P(X ≥ 1)
= P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
= 1 − P(X = 0)
= 1 − 27
6 487
= 0,9958.
� 4(c) P(family owns between two and four cars)
= P(2 ≤ X ≤ 4)
= P(X = 2) + P(X = 3) + P(X = 4)
=4 985
6 487
= 0,7685.
DSC2602 240
� 4(d)
P(family owns no car) = P(X = 0)
=27
6 487
= 0,0042.
� 5 Only one of the horses can win. Therefore the events are mutually exclusive.
P(A or B or C) = P(A ∪ B ∪ C)
=1
10+
1
10+
1
10
=3
10= 0,3.
� 6 Once again the events are mutually exclusive. Therefore,
P(Exon or Mobil) = P(Exon ∪Mobil)
= 0,3 + 0,3 = 0,6.
� 7 Here the possibility of drawing the ace of spades (i.e. an ace and a spade) must be taken intoaccount.
P(ace or spade) = P(ace ∪ spade)
= P(ace) + P(spade) − P(ace ∩ spade)
=4
52+
13
52− 1
52
=16
52= 0,31.
� 8
Machine Blue Pink TotalA 50 × 0,40 = 20 50 × 0,60 = 30 500 × 0,10 = 50B 150 × 0,30 = 45 150 × 0,70 = 105 500 × 0,30 = 150C 300 × 0,20 = 60 300 × 0,80 = 240 500 × 0,60 = 300
125 375 500
� 8(a) The proportion of the total that blue elephants form is
P(X = blue) =125
500= 0,25.
The percentage blue elephants is 25%.
241 DSC2602
� 8(b) The probability that an elephant was produced by machine B, given that it is pink is
P(Machine B | pink) =105
375= 0,28
orP(Machine B ∩ pink)
P(pink)
=
105
500375
500
= 0,28.
A.3 Chapter 3: Probability distributions
� 1 The probability distribution is shown in the following figure:
0
0,1
0,2
0,3
0,4
0,5
0,6
f (x )
1 2 3 4 5
Number of board members
Probability distribution
x
The mean and variance are calculated in the following table.
x f (x ) xf (x ) x -E (x ) (x -E (x ))2
(x -E (x ))2f (x )
1 0,52 0,52 -0,8400 0,7056 0,3669
2 0,22 0,44 0,1600 0,0256 0,0056
3 0,19 0,57 1,1600 1,3456 0,2557
4 0,04 0,16 2,1600 4,6656 0,1866
5 0,03 0,15 3,1600 9,9856 0,2996
Total 1 1,84 5,8000 16,7280 1,1144
We know that the mean is given by
E(X) = µ =
n∑
i=1
xi f (xi).
DSC2602 242
From the table we find that E(x) = µ =∑5
i=1 x f (x) = 1,84.
We also know that the variance is given by
Variance(X) = σ2 =
n∑
i=1
(xi − µ)2 f (xi).
From the table we see that σ2 = 1,1144.
The standard deviation is therefore σ =√
1,1144 = 1,056.
� 2(a) If we assume that the event of getting an engaged signal is a “success”, p = P(success) = 0,3.
The mean: E(X) = np = 5 × 0,3 = 1,5.
The variance: Var(X) = np(1 − p) = 5 × 0,3 × 0,7 = 1,05.
� 2(b) Here p = 0,3 and n = 5 and we want to determine the probability of one successful trial.
p(1) = P(X = 1) =
(
5
1
)
0,31(1 − 0,3)5−1
=5!
1!(5 − 1)!0,310,74
= 0,3602.
� 2(c) P(less than 2 engaged signals) = P(X = 0 or X = 1)
= P(X = 0) + P(X = 1)
=
(
5
0
)
0,300,75 + 0,3602
= 0,1681 + 0,3602
= 0,5283.
� 2(d) P(4 or more engaged signals) = P(X = 4 or X = 5)
= P(X = 4) + P(X = 5)
=
(
5
4
)
0,340,71 +
(
5
5
)
0,350,70
= 0,0284 + 0,0024
= 0,0308.
� 2(e) P(at least 2 engaged signals) = P(X = 2 or X = 3 or X = 4 or X = 5)
= 1 − P(less than 2 engaged signals)
= 1 − P(X = 0 or X = 1)
= 1 − 0,5283
= 0,4717.
243 DSC2602
� 3(a) This is a Poisson distribution with λ = 6 and x = 3.
P(X = x) =λx · e−λ
x!
P(X = 3) =63 · e−6
3!= 0,089.
� 3(b) If the average per hour is 6, the average per 10-minute period is 1. Therefore
P(X = 2) =12 · e−1
2!= 0,1839.
� 4(a)
µ = 60 7845 x
σ = 12
µz = 0 1,5-1,25 z
σz = 1
P(45 < X < 78) = P(45 − 60
12<
X − µσ<
78 − 60
12)
= P(−1,25 < z < 1,5)
= P(−1,25 < z < 0) + P(0 < z < 1,5)
= 0,39435 + 0,43319
= 0,82754.
Of the patients admitted to the ICU, 82,75% are between 45 and 78 years old.
� 4(b)
60 x
σ = 12
0 z
σz = 1
P(X > 60) = P(
z >60 − 60
12
)
= P(z > 0) = 0,50
Therefore 50% of the patients are older than 60 years.
DSC2602 244
� 4(c)
60 78 x
σ = 12
0 1,5 z
σz = 1
P(X > 78) = P(
z >78 − 60
12
)
= P(z > 1,5)
= P(z > 0) − P(z < 1,5)
= 0,5 − 0,43319
= 0,06681.
Therefore 6,68% of the patients are older than 78 years.
� 4(d)
6045 x
σ = 12
0−1,25 z
σz = 1
The area from z = −1,25 to z = 0 is equal to the area from z = 0 to z = 1,25 which is 0,39435 (fromthe standard normal table). The area to the left of −1,25 is therefore
0,5 − 0,39435 = 0,10565.
Therefore 10,565% of the patients are younger than 45 years.
� 4(e)
604530 x
σ = 12
µz = 0−2,5 −1,25 z
σz = 1
P(−2,5 < z < −1,25) = P(1,25 < z < 2,5) (due to symmetry)
= P(z < 2,5) − P(z < 1,25)
= 0,49379 − 0,39435 (from standard normal table)
= 0,09944.
Therefore 9,944% of the patients are between 30 and 45 years old.
245 DSC2602
� 4(f)
µ = 60 X∗
0,20
The oldest 20% will fall on the right-hand side of the distribution. The area between the meanand the cut-off point of the oldest 20% is 30% or 0,30. In the body of the standard normal table,look up 0,30 or the nearest value to 0,30 and read off the z-value, that is, 0,84.
We know that z =X∗ − µσ
=X∗ − 60
12and we set this equal to the z-value:
X∗ − 60
12= 0,84
which gives X∗ = 70,08 years. The oldest 20% of patients will be 70,08 years or older.
� 5(a) The probability that the return will exceed 55% is
P (return exceeds 55%) = P(X > 0,55)
= P
(
X − µσ>
0,55 − 0,3
0,1
)
= P(z > 2,5)
= 0,5 − 0,49379
= 0,00621.
� 5(b) The probability that the return will be less than 22% is
P (return less than 22%) = P(X < 0,22)
= P(
z <0,22 − 0,3
0,1
)
= P(z < −0,8)
= P(z > 0,8)
= 0,5 − 0,28814
= 0,2119.
DSC2602 246
A.4 Chapter 4: Estimation
� 1 Here x = 165,45, s = 38,6 and n = 25 credit card customers. Since the sample size is small, thatis, fewer than 35, we use the t-distribution to find the confidence interval. The 95% confidencelimits are
x ± t α2 ; n−1
s√
n.
By using s to estimate σ, we obtain
165,45 ± 2,06438,6√
25or 165,45 ± 15,93
which gives the 95% confidence interval as [R149,52; R181,38] .
It is therefore 95% certain that the actual mean value of credit card purchases at this store liesbetween R149,52 and R181,38.
� 2 From the given information it follows that
n = 120 x = 42 p =42
120= 0,35.
Lower confidence boundary:
p − 1,645
√
p(1 − p)
n= 0,35 − 1,645
√
0,35(1 − 0,35)
120= 0,35 − 1,645(0,043)
= 0,279.
Upper confidence boundary:
p + 1,645
√
p(1 − p)
n= 0,35 + 1,645
√
0,35(1 − 0,35)
120= 0,35 + 1,645(0,043)
= 0,421.
It is 90% certain that the true proportion of the population of street vendors who believe that localby-laws still hamper their trading, falls in the interval [0,279; 0,421].
� 3 Here we have to determine a 95% confidence interval for the population mean µ. Thefollowing is given: The sample size n = 36, the sample mean x = 1,5 and the population standarddeviation σ = 0,3. The 95% confidence interval for the mean is
x ± 1,96σ√
n.
Upper confidence boundary:
x + 1,96σ√
n= 1,5 + 1,96
0,3√
36= 1,5 + 1,96(0,05)
= 1,598.
247 DSC2602
Lower confidence boundary:
x − 1,96σ√n= 1,5 − 1,96
0,3√36
= 1,5 − 1,96(0,05)
= 1,402.
It is 95% certain that the new machine will, on average, require between 1,402 and 1,598 minutesto produce a ream of paper.
� 4 Here x = 32, s = 12 and n = 20. Since the sample size is small, that is, less than 30, we use thet-distribution to find the confidence interval. The 95% confidence limits are
x ± t α2 ; n−1
s√
n.
Upper confidence boundary:
x + t0,025; 20−1s√
n= 32 + 2,093
12√
20= 32 + 5,616
= 37,616.
Lower confidence boundary:
x − t0,025; 20−1s√
n= 32 − 2,093
12√
20= 32 − 5,616
= 26,384.
� 5 Here n = 150 and p = 15150= 0,1. The 95% confidence interval for the percentage of customers
who have passed bad cheques is
p − z0,025
√
p(1 − p)
n; p + z0,025
√
p(1 − p)
n
.
Lower confidence boundary:
p − z0,025
√
p(1 − p)
n= 0,1 − 1,96
√
0,1(0,9)
150= 0,1 − 0,048
= 0,052.
Upper confidence boundary:
p + z0,025
√
p(1 − p)
n= 0,1 + 1,96
√
0,1(0,9)
150= 0,1 + 0,048
= 0,148.
It can be said, with 95% certainty, that between 5,2% and 14,8% of all credit customers have passedbad cheques.
DSC2602 248
A.5 Chapter 5: Correlation and regression
� 1(a) y = a + bx.Enter the data points into your calculator’s statistic mode and obtain
a = 1,45
b = 0,074.
Thus
y = 1,45 + 0,074x
or
Sales = 1,45 + 0,074x.
� 1(b) Using the statistic mode of your calculator, r = 0,8270 is obtained. Yes, a strong positivelinear relationship exists between the weekly sales and shelf space.
� 1(c) r = 0,8270 α = 0,01 n = 12
R2 = 0,683929 k = 2 r1 = k − 1 = 2 − 1 = 1
r2 = n − k = 12 − 2 = 10
F =R2/r1
(1 − R2)/r2=
0,683929/1
(1 − 0683929)/10= 21,64
F-table value F1;10;0,05 = 4,96
Since F calculated = 21,64 > 4,96 the linear relationship is significant on a 5% significance level.
� 1(d) Given y = 1,45 + 0,074x. Determine y if x = 8. y = 1,45 + 0,074(8) = 2,042
The average weekly sales are equal to R2 042,00.
� 2(a) We have to determine how strong the linear relationship between the two modules is. Thecorrelation coefficient (r) gives an indication of how strong the linear relationship between them is.From the output cell B4 we can get the r-value, namely 0,67211. There exists a linear relationshipbetween the two courses, although it is not very strong.
� 2(b) We have to determine y = b0 + b1x. From the output we can get the coefficients of theregression line. The intercept (b0) in cell B17 is equal to −7,3979 and the slope (b1) in cell B18 isequal to 0,9560. The regression line is therefore
y = −7,3979 + 0,9560x
� 2(c) To determine how well a fitted model fits the data we use the R2 value. The R square value(cell B5) of 0,4517 indicates that only 45% of the variation in the y-values can be declared by thex-values. The model does not fit the data very well.
249 DSC2602
� 2(d) The F-value is already calculated in cell E12, namely 21,42235. On a 5% level of significancethe value obtained from the F-tables is
F1;26;0,05 ≈ 4,24
Since the calculated value is greater than the value obtained from the tables, we can say that theregression line is significant on a 5% level of significance.
� 2(e) Draw a scatter diagram of the data and determine if a nonlinear relationship of some sortexists. Fit an appropriate nonlinear model.
� 3 A scatter diagram of the data is given in the Figure A.2.
0
500
1000
1500
0 5 10 15 20 25
Price (in R)
Un
its s
old
Figure A.2: Scatter diagram of sales versus price
From the scatter diagram it is clear that the relationship between price and number of units soldis not linear, and the shape is similar to Figure 5.6(b). The relationship can, according to Table 5.5,be described by y = β0xβ1 .
The data need to be transformed by the transformation (ln xi; ln yi) as given in column 3 ofTable 5.5, that is (ln 1; ln 1 145),((ln 2; ln 788), . . . (ln 20; ln 172). Enter these transformed data pairsinto the statistics mode of your calculator to obtain the least squares coefficients b0 = 6,984 andb1 = −0,666. The estimated functional relationship is therefore
y = e6,984x−0,666.
You may also use a spreadsheet to verify your answer.
DSC2602 250
A.6 Chapter 6: Decision analysis
� 1(a) The payoff table is given in Table A.6.
State of natureDecision alternative Favourable market Unfavourable market
Strategy 1 10 000 −8 000Strategy 2 8 000 −4 000Strategy 3 0 0
Table A.6: Payoff table (payoffs in rand)
� 1(b) Since Carl is a risk seeker, he should use the maximax decision criterion. This is illustratedin Table A.7.
Decision Maximumalternative payoff (in rand)Strategy 1 10 000 MaximumStrategy 2 8 000Strategy 3 0
Table A.7: Maximax decision criterion
Carl’s decision is to select strategy 1, which is an optimistic decision approach.
� 1(c) Betty should use the maximin decision criterion as illustrated in Table A.8.
Decision Minimumalternative payoff (in rand)Strategy 1 −8 000Strategy 2 −4 000Strategy 3 0 Maximum
Table A.8: Maximin decision criterion
Betty would select strategy 3, which reflects a pessimistic approach.
� 1(d) Carl and Betty would use the minimax regret decision criterion. Table A.9 is the opportunityloss or regret table for Carl and Betty’s problem.
The minimax regret criterion is illustrated in Table A.10.
They would select strategy 2.
251 DSC2602
State of natureDecision alternative Favourable market Unfavourable market
Strategy 1 0 8 000Strategy 2 2 000 4 000Strategy 3 10 000 0
Table A.9: Regret table (in rand)
Decision Maximumalternative regret (in rand)Strategy 1 8 000Strategy 2 4 000 MinimumStrategy 3 10 000
Table A.10: Minimax regret criterion
� 2(a) The payoff table is constructed by listing the possible actions, that is, how many dozen rosesto buy, down the left of the table and the states of nature, that is, how many dozen roses will bedemanded, across the top of the table. The payoff for different combinations of actions and statesof nature is calculated as follows:
• If 20 dozen roses are bought and 20 dozen are demanded (supply = demand) the payoff is
Pro f it = revenue − cost
= 20(35) − 20(20)
= 300.
• If 20 dozen roses are bought and 50 dozen are demanded (supply < demand) the payoff is
Pro f it = revenue − cost − ill will
= 20(35) − 20(20) − 30(12)
= −60.
• If 50 dozen roses are bought and 20 dozen are demanded (supply > demand) the payoff is
Pro f it = revenue − cost
= [revenue from regular sales + revenue from sales to hotels] − cost
= [20(35) + 30(10)] − 50(20)
= 0.
The complete payoff table is given in Table A.11.
DSC2602 252
State of nature: DemandDecision 20 dozen 30 dozen 40 dozen 50 dozen
Alternative: Buy s1 s2 s3 s4
20 dozen (d1) 300 180 60 −6030 dozen (d2) 200 450 330 21040 dozen (d3) 100 350 600 48050 dozen (d4) 0 250 500 750
Probability 0,10 0,30 0,40 0,20
Table A.11: Payoff table for Flora’s problem (payoffs in rand)
� 2(b) i. If Flora follows a risk-seeking approach, she will use the maximax decision criterion, asset out in Table A.12.
Decision Maximumalternative:Buy payoff (in rand)
20 dozen (d1) 30030 dozen (d2) 45040 dozen (d3) 60050 dozen (d4) 750 Maximum
Table A.12: Maximax criterion (optimistic approach)
She should buy 50 dozen roses each morning.
� 2(b) ii. If Flora follows a risk-averting approach, she will use the maximin decision criterion, asset out in Table A.13.
Decision Minimumalternative: Buy payoff (in rand)
20 dozen (d1) −6030 dozen (d2) 200 Maximum40 dozen (d3) 10050 dozen (d4) 0
Table A.13: Maximin criterion (conservative approach)
She should buy 30 dozen roses each morning.
� 2(b) iii. If Flora does not want to take unnecessary risks while avoiding regrets, she will makeher decision according to the minimax regret criterion. The opportunity loss, or regret, table isgiven in Table A.14.
The minimax regret decision criterion is set out in Table A.15.
Flora should buy 40 dozen roses each morning.
253 DSC2602
State of nature: DemandDecision 20 dozen 30 dozen 40 dozen 50 dozen
alternative: Buy s1 s2 s3 s4
20 dozen (d1) 0 270 540 81030 dozen (d2) 100 0 270 54040 dozen (d3) 200 100 0 27050 dozen (d4) 300 200 100 0
Probability 0,10 0,30 0,40 0,20
Table A.14: Regret table for Flora’s problem (payoffs in rand)
Decision Maximumalternative:Buy regret (in rand)
20 dozen (d1) 81030 dozen (d2) 54040 dozen (d3) 270 Minimum50 dozen (d4) 300
Table A.15: Minimax regret criterion
� 2(c) If Flora is risk neutral, she will use the expected value approach to make a decision.
EV(Buy 20 dozen) = 96
EV(Buy 30 dozen) = 329
EV(Buy 40 dozen) = 451 Maximum
EV(Buy 50 dozen) = 425
She will buy 40 dozen roses each morning.
� 2(d) Flora should not pay more for information than the expected value of perfect information(EVPI). We know that EVPI is equal to the minimum expected opportunity loss. From the regrettable, we find the expected opportunity loss for each decision alternative:
EOL(Buy 20 dozen) = 459
EOL(Buy 30 dozen) = 266
EOL(Buy 40 dozen) = 104 Minimum
EOL(Buy 50 dozen) = 130
She should therefore not pay more than R104 for information.
DSC2602 254
� 3(a) We must first of all determine the probabilities for 80, 100 and 120 customers.
It is given that P(80) = P(120) and P(80) = 2P(100) and it is known thatP(80) + P(100) + P(120) = 1.
From this follows that P(80) + 0,5P(80) + P(80) = 1
from which we can say that 2,5P(80) = 1.
Therefore P(80) = 0,4; P(100) = 0,2 and P(120) = 0,4.
The decision tree is given in Figure A.3.
10 000
12 000
18 000
8 000
14 000
15 000
1
Model A
Model C
Model B
2
80 customers/hour
120 customers/hour
100 customers/hour
P(80) = 0,4
P(100) = 0,2
P(120) = 0,4
3
80 customers/hour
120 customers/hour
100 customers/hour
P(80) = 0,4
P(100) = 0,2
P(120) = 0,4
4
80 customers/hour
120 customers/hour
100 customers/hour
P(80) = 0,4
P(100) = 0,2
P(120) = 0,4
6 000
16 000
21 000
Figure A.3: Decision tree for the Burger Inn
In Figure A.4 the expected value for each decision is calculated.
1
2
3
4
Model A
Model C
Model B
EV = 0,4(10 000) + 0,2(15 000) + 0,4(14 000) = 12 600
EV = 0,4(8 000) + 0,2(18 000) + 0,4(12 000) = 11 600
EV = 0,4(6 000) + 0,2(16 000) + 0,4(21 000) = 14 000
Figure A.4: Expected values for the Burger Inn
Select the model with the largest expected value — Model C.
255 DSC2602
� 3(b) If perfect information were available, the optimum decision under each state of nature wouldbe selected. The expected value with perfect information is thereforeEVwPI = 0,4(10 000) + 0,2(18 000) + 0,4(21 000) = 16 000.
The expected value without perfect information has been calculated in (a) as EVwoPI = 14 000.
Therefore EVPI = EVwPI − EVwoPI = R16 000 − R14 000 = R2 000.
� 4(a) The payoff table is
State of natureDecision
alternative Cool day Hot day
Ice cream 300 900Cold drinks 500 600
Probabilities 0,4 0,6
� 4(b) He would use the maximax criteria if he is an optimistic person.
Maximumpayoff (in rand)
Ice cream 900 MaximumCold drinks 600
He should sell ice cream.
� 4(c) He should use the minimax criteria.Regret table in R is:
Cool day Hot dayIce cream 200 0
Cold drinks 0 300
Maximumregret (in rand)
Ice cream 200 MinimumCold drinks 300
He should sell ice cream.
DSC2602 256
� 4(d) The decision tree is
5 0 0( 0 , 4 )
1
2
3
i c e c r e am
h o t d a yc o l d d r i n k s
c o o l d a y
c o o l d a y
h o t d a y 9 0 0( 0 , 6 )
3 0 0( 0 , 4 )
6 0 0( 0 , 6 )
5 6 0V W =
6 6 0V W =
The expected value for each decision is calculated as
EV(Ice cream) = (300) × (0,4) + (900) × (0,6) = 660 ←− Maximum
EV(Cold drink) = (500) × (0,4) + (600) × (0,6) = 560
Select the option with the largest expected value, namely sell ice cream.
� 4(e) Let
P(Cool day) = p
P(Hot day) = 1 − p
Using the payoff values, the expected values for the decision alternatives are as follows:
EV (Ice cream) = 300p + (1 − p)900 (A.1)
= 900 − 600p (A.2)
EV (Cold drinks) = 500p + (1 − p)600 (A.3)
= 500p + 600 − 600p (A.4)
= 600 − 100p (A.5)
Equations A.2 and A.5 can graphically be presented as
257 DSC2602
1 2 3 4 5 61 4 4 4 4 2 4 4 4 4 3
**
3 0 0
6 0 0
9 0 0
* S e l l i c e c r e a m* * S e l l c o l d d r i n k s
p
E xp ec t e
d va l u
e (E V
)
*{0 . 6 1 2 3 4 5 6
The value of p for which the expected values of d1 and d2 are equal, is the value of p where EV(d1)and EV(d2) lines intersect. The value of p can be determined as follows:
EV(d1) = EV(d2)
900 − 600p = 600 − 100p
−500p = −300
p = 35= 0,6
If p < 0,6 he must sell ice cream and if p > 0,6 he must sell cold drinks.
Note: If you choose P (cool day) = (1 − p) and P (hot day) = p the two equations are:
EV (cool day) = 300 + 600p
EV (hot day) = 500 + 100p
and the point of intersect 0,4.
� 4(f) He should not pay more for information than the expected value of perfect information(EVPI). The EVPI is equal to the minimum expected opportunity loss. From the regret table(question (c)) we find that the expected opportunity loss for each decision is:
EOL (Ice cream) = (200)(0,4) + (0)(0,6) = 80 ←− Minimum
EOL (Cold drinks) = (0)(0,4) + (300)(0,6) = 180
He should therefore not pay more than R80 for information.
� 5(a) If the decision maker is risk neutral, the expected value approach is applicable.
EV(d1) = 0,1(100 000) + 0,3(40 000) + 0,6(−60 000) = −14 000
EV(d2) = 0,1( 50 000) + 0,3(20 000) + 0,6(−30 000) = − 7 000
EV(d3) = 0,1( 20 000) + 0,3(20 000) + 0,6(−10 000) = 2 000 Maximum
DSC2602 258
The optimal decision is d3 with an EV of R2 000.
Note: We do not need to calculate the EV for d4 because d4 is dominated by d2. When one decisionis better than another under all states of nature, we say that it dominates the other. If one decisionis dominated by another, its EV will always be less than the EV of the dominating decision.
� 5(b) The graph of the decision makers’ utility functions is shown in Figure A.5.
20
40
60
80
100
-60 -40 -20 0 20 40 60 80 100
Monetary value (in R1 000s)
Uti
lity
Decision maker 1Decision maker 2
Figure A.5: Utility functions for the two decision makers
Decision maker 1 has a concave utility function — a risk averter.
Decision maker 2 has a convex utility function — a risk taker.
� 5(c) Note again that d4 is dominated by d2 and hence is not considered. The expected utility foreach decision maker is:
State of natureDecision alternative s1 s2 s3 Expected utility
d1 1,00 0,90 0 0,37d2 0,94 0,80 0,40 0,57d3 0,80 0,80 0,60 0,68 ←− Maximum
Probability 0,1 0,3 0,6
Table A.16: Utility table for decision maker 1
Decision maker 1 should make decision d3 since it has the largest expected utility.
259 DSC2602
State of natureDecision alternative s1 s2 s3 Expected utility
d1 100 50 0 25,0 ←− Maximumd2 58 35 10 22,3d3 35 35 18 24,8
Probability 0,1 0,3 0,6
Table A.17: Utility table for decision maker 2
Decision maker 2 should make decision d1 since it has the largest expected utility.
� 6 First we have to find the utilities for the monetary values using the given utility function,U(x) = ln(x + 3), and the payoff table. For example
U(30) = ln(30 + 3) = 3,50
U(18) = ln(18 + 3) = 3,045
U(−2) = ln(−2 + 3) = ln(1) = 0
Next we construct the agency’s utility table namely:
Campaign type High demand Medium demand Low demand
Aggressive (d1) 3,5 3,045 0Strong (d2) 3,33 3,14 1,61Cautious (d3) 2,56 2,40 2,08Probability 0,25 0,5 0,25
Table A.18: Agency’s utility table
Now we can use the utilities to determine the expected utility for each venture:
EU(d1) = (3,5)(0,25) + (3,045)(0,5) + (0)(0,25) = 2,40
EU(d2) = (3,33)(0,25) + (3,14)(0,5) + (1,61)(0,25) = 2,81 Maximum
EU(d3) = (2,56)(0,25) + (2,40)(0,5) + (2,08)(0,25) = 2,36
The agency should invest in venture d2.
DSC2602 260
A.7 Chapter 7: Project management
� 1 The list of immediate predecessors is as follows:
ImmediateTask Description predecessors
A Prepare the wheels NoneB Mount the wheels AC Assemble sides NoneD Attach the top CE Attach base B, CF Insert brackets CG Insert shelves FH Attach doors D, E, GI Attach back panel D, EJ Paint the unit H, I
� 2
• The network for project IPSO is given in Figure A.6.
1
2
3
4
5 6
a
b
c
d e
Figure A.6: Project IPSO
• The network for project FACTO is given in Figure A.7.
1
2
3
4
5 6
a
b
c
d e
Figure A.7: Project FACTO
• The network for project DELTA is given in Figure A.8.
261 DSC2602
1
2
3
5
4
a
b
c
d
e
Figure A.8: Project DELTA
� 3
• Project ALPHA:
The network for project ALPHA is given in Figure A.9.
1
2
3
5
4
A
B
D
C
6
E
F
2
4
1
3
2
3
Figure A.9: Project network for project ALPHA
The early and late event times are calculated on the diagram in Figure A.10.
1
2
3
5
4
A
B
D
C
6
E
F
2
4
1
3
2
30
2 4
4 4
3 7
7 7
10 10
min{2; 0}
0
max{2; 4} max{3; 7}
max{5; 10}
min{7;8}
min{4; 6}
Figure A.10: Early and late event times for project ALPHA
The duration of project ALPHA is 10 weeks.
The critical path is B→D→F (or 1-3-4-6 in terms of nodes).
DSC2602 262
Activity Total float* Free float**A 2 0B 0 0C 4 0D 0 0E 5 5F 0 0
* TF(i; j) = LT( j) − ET(i) − ti j ** FF(i; j) = ET( j) − ET(i) − ti j
• Project BETA:
The network for project BETA is given in Figure A.11.
1 2
3
4
5
6
C
B
E
D F7
A
Figure A.11: Project network for project BETA
The early and late event times are calculated on the diagram in Figure A.12.
1 2
3
4
5
6
C
B
E
D F 7A0 0 2 4
1
3
2
32 2
3 6
6 6
8 9
9 9 12 12
min{2; 5} max{3; 6}
min{6; 7}
max{8; 9}
Figure A.12: Early and late event times for project BETA
The duration of project BETA is 12 weeks.
The critical path is A→B→D→F (or 1-2-4-6-7 in terms of nodes).
Activity Total float Free floatA 0 0B 0 0C 3 0D 0 0E 1 0F 0 0
263 DSC2602
Note: The two dummy activities have free float (which could be considered to be the freefloat of activities C and E):
Dummy (3; 4): FF(3; 4) = 6 − 3 − 0 = 3Dummy (5; 6): FF(5; 6) = 9 − 8 − 0 = 1
• Project GAMMA:
The network for project GAMMA is given in Figure A.13.
1
2
3
4
5 6
B
A
C
F
D
E
Figure A.13: Project network for project GAMMA
The early and late event times are calculated on the diagram in Figure A.14.
1
2
3
4
5 6
B
A
C
F
D
E
2
4
1 3
2
30 0
4 4
4 4
4 4
7 7 10 10
min{4; 5}
max{7; 6}
Figure A.14: Early and late event times for project GAMMA
The duration of project GAMMA is 10 weeks.
Here, the critical path cannot be easily identified from the diagram. By calculating the totalfloat, however, we find that the critical path is B→D→F (or 1-2-3-4-5-6 in terms of nodes).
Activity Total float Free floatA 2 2B 0 0C 3 3D 0 0E 1 1F 0 0
DSC2602 264
� 4(a) The project network is given in Figure A.15.
1
2
3
4
5
6
7
A
B
C
D
E
F
G
H
I
J
K
Figure A.15: The project network
� 4(b) The expected time for each activity must be calculated before the critical path and projectduration can be determined. In Table A.19 the expected duration and variance of each activityare calculated.
Expected duration (hrs) VarianceActivity t = (a + 4m + b)/6 σ2 = ((b − a)/6)2
A 6 0,44B 4 0,44C 3 0,00D 5 0,11E 1 0,03F 4 0,11G 2 0,44H 6 0,11I 5 1,00J 3 0,11K 5 0,44
Table A.19: Expected durations and variances for activities
We use the expected durations of the activities and do the network calculations as shown inFigure A.16.
The critical path is A→C→F→I→K. The expected project duration is 23 hours.
265 DSC2602
1
2
3
4
5
6
7
A
B
C
D
E
F
G
H
I
J
K
6
4
3
5
1
4
2
6
5
3
5
0 0
6
9
13
18
19
23 23
20
18
13
6
9
Figure A.16: The network calculations
� 4(c) The expected project duration is 23 hours.
The variance of the critical path is σ2 = σ2A+ σ2
C+ σ2
F + σ2I + σ
2K = 1,99 and the standard deviation
is σ =√
1,99 = 1,411.
The probability of completing the project within 24 hours is given by the shaded area in Fi-gure A.17.
µ = 23 24 T
σ = 1,99
Figure A.17: Probability of completing the project within 24 hours
The probability of completing the project within 24 hours is
P(T ≤ 24) = P(z ≤ 24 − 23
1,411)
= P(z ≤ 0,71)
= 0,5 + 0,2612 (from table)
= 0,7612.
There is a 76,12% chance that the project will be completed within one day.
� 5(a) The project can be presented by the network diagram in Figure A.18.
DSC2602 266
1
2
3
4
5
6
7
8 9
A
B
C
D
E
F G
H
I
J10
15
5 10
11
5 5
10
10
15
Figure A.18: The project network
� 5(b) The LP model, Model A.7.1, can be formulated to determine the critical path and projectduration. The decision variables are defined as follows:
Let x j = the time that the event corresponding to node j occurs.
Minimise z = x9 − x1
subject to x2 ≥ x1 + 10 [Activity A]
x4 ≥ x2 + 5 [Activity B]
x3 ≥ x1 + 15 [Activity C]
x6 ≥ x3 + 11 [Activity D]
x6 ≥ x4 + 10 [Activity E]
x5 ≥ x3 + 5 [Activity F]
x7 ≥ x5 + 5 [Activity G]
x8 ≥ x7 + 10 [Activity H]
x8 ≥ x6 + 10 [Activity I]
x9 ≥ x8 + 15 [Activity J]
x4 ≥ x3 [Dummy activity (3; 4)]
x7 ≥ x6 [Dummy activity (6; 7)]
All variables are unrestricted.
Model A.7.1: LP model for finding the critical path
The following data are entered into LINGO:
min = x9-x1;
[A] x2>x1+10;
[B] x4>x2+5;
[C] x3>x1+15;
[D] x6>x3+11;
[E] x6>x4+10;
[F] x5>x3+5;
[G] x7>x5+5;
[H] x8>x7+10;
[I] x8>x6+10;
[J] x9>x8+15;
[Dummy34] x4>x3;
[Dummy67] x7>x6;
267 DSC2602
The solution obtained by LINGO:
Optimal solution found at step: 0
Objective value: 51.00000
Variable Value Reduced Cost
X9 51.00000 0.0000000E+00
X1 0.0000000E+00 0.0000000E+00
X2 10.00000 0.0000000E+00
X4 15.00000 0.0000000E+00
X3 15.00000 0.0000000E+00
X6 26.00000 0.0000000E+00
X5 20.00000 0.0000000E+00
X7 26.00000 0.0000000E+00
X8 36.00000 0.0000000E+00
Row Slack or Surplus Dual Price
1 51.00000 1.000000
A 0.0000000E+00 0.0000000E+00
B 0.0000000E+00 0.0000000E+00
C 0.0000000E+00 -1.000000
D 0.0000000E+00 -1.000000
E 1.000000 0.0000000E+00
F 0.0000000E+00 0.0000000E+00
G 1.000000 0.0000000E+00
H 0.0000000E+00 0.0000000E+00
I 0.0000000E+00 -1.000000
J 0.0000000E+00 -1.000000
DUMMY34 0.0000000E+00 0.0000000E+00
DUMMY67 0.0000000E+00 0.0000000E+00
From the solution we see that the minimum duration of the project is 51 minutes and the criticalpath is C→ D→ I→ J (dual prices are negative).
� 5(c)(i) The LP model for crashing the project to 48 minutes is given in Model A.7.2. (The numberof minutes each activity can be crashed is given by ti − t′
i.)
Define the decision variables:
Let x j = the time that the event corresponding to node j occurs ( j = 1; 2; . . . ; 9);A = the number of minutes activity A must be crashed;B = the number of minutes activity B must be crashed;...
...J = the number of minutes activity J must be crashed.
DSC2602 268
Minimise COST = A + 2B + C +D + E + 2F + 2G + 3H + 3I + 4J
subject to x9 ≤ x1 + 48 [Deadline]
x2 ≥ x1 + 10 − A [Activity A]
x4 ≥ x2 + 5 − B [Activity B]
x3 ≥ x1 + 15 − C [Activity C]
x6 ≥ x3 + 11 −D [Activity D]
x6 ≥ x4 + 10 − E [Activity E]
x5 ≥ x3 + 5 − F [Activity F]
x7 ≥ x5 + 5 − G [Activity G]
x8 ≥ x7 + 10 −H [Activity H]
x8 ≥ x6 + 10 − I [Activity I]
x9 ≥ x8 + 15 − J [Activity J]
x4 ≥ x3 [Dummy (3; 4)]
x7 ≥ x6 [Dummy (6; 7)]
A,B,C,D,E, F,G,H, I, J ≥ 0;
x j are unrestricted, j = 1; . . . , 9.
Model A.7.2: LP model for crashing the project to 48 minutes
The input data for LINGO:
[ O b j e c t i v e ] m i n = A + 2 * B + C + D + E + 2 * F + 2 * G + 3 * H + 3 * I + 4 * J ; [ D e a d l i n e ] x 9 - x 1 < 4 8 ; A < 5 ; B < 1 ; C < 5 ; D < 5 ; E < 3 ; F < 1 ; G < 1 ; H < 2 ; I < 2 ; J < 5 ; [ A ] x 2 > x 1 + 1 0 - A ; [ B ] x 4 > x 2 + 5 - B ; [ C ] x 3 > x 1 + 1 5 - C ; [ D ] x 6 > x 3 + 1 1 - D ; [ E ] x 6 > x 4 + 1 0 - E ; [ F ] x 5 > x 3 + 5 - F ; [ G ] x 7 > x 5 + 5 - G ; [ H ] x 8 > x 7 + 1 0 - H ; [ I ] x 8 > x 6 + 1 0 - I ; [ J ] x 9 > x 8 + 1 5 - J ; [ D u m m y 3 4 ] x 4 > x 3 ; [ D u m m y 6 7 ] x 7 > x 6 ;
269 DSC2602
The solution obtained by LINGO:
O p t i m a l s o l u t i o n f o u n d a t s t e p : 1 6O b j e c t i v e v a l u e : 5 . 0 0 0 0 0 0 V a r i a b l e V a l u e R e d u c e d C o s t A 2 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 B 0 . 0 0 0 0 0 0 0 E + 0 0 1 . 0 0 0 0 0 0 C 2 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 D 1 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 E 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 F 0 . 0 0 0 0 0 0 0 E + 0 0 2 . 0 0 0 0 0 0 G 0 . 0 0 0 0 0 0 0 E + 0 0 2 . 0 0 0 0 0 0 H 0 . 0 0 0 0 0 0 0 E + 0 0 1 . 0 0 0 0 0 0 I 0 . 0 0 0 0 0 0 0 E + 0 0 3 . 0 0 0 0 0 0 J 0 . 0 0 0 0 0 0 0 E + 0 0 2 . 0 0 0 0 0 0 X 9 4 8 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 1 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 2 8 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 4 1 3 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 3 1 3 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 6 2 3 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 5 1 8 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 7 2 3 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 8 3 3 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 R o w S l a c k o r S u r p l u s D u a l P r i c e O B J E C T I V E 5 . 0 0 0 0 0 0 1 . 0 0 0 0 0 0 3 3 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 4 1 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 5 3 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 6 4 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 7 3 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 8 1 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 9 1 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 1 0 2 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 1 1 2 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 1 2 5 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 A 0 . 0 0 0 0 0 0 0 E + 0 0 - 1 . 0 0 0 0 0 0 B 0 . 0 0 0 0 0 0 0 E + 0 0 - 1 . 0 0 0 0 0 0 C 0 . 0 0 0 0 0 0 0 E + 0 0 - 1 . 0 0 0 0 0 0 D 0 . 0 0 0 0 0 0 0 E + 0 0 - 1 . 0 0 0 0 0 0 E 0 . 0 0 0 0 0 0 0 E + 0 0 - 1 . 0 0 0 0 0 0 F 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 G 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 H 0 . 0 0 0 0 0 0 0 E + 0 0 - 2 . 0 0 0 0 0 0 I 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 J 0 . 0 0 0 0 0 0 0 E + 0 0 - 2 . 0 0 0 0 0 0 D U M M Y 3 4 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 D U M M Y 6 7 0 . 0 0 0 0 0 0 0 E + 0 0 - 2 . 0 0 0 0 0 0
DSC2602 270
Interpretation of the results:
• It will cost R5,00 to crash the project to 48 minutes.
• Shorten activity A by 2 minutes, activity C by 2 minutes and activity D by 1 minute (valuecolumn of output).
• Activities A, B, C, D, E, H and J are critical activities. There are two critical paths, namelyA→ B→ E→H→ J and C→D→H→ J. (Dummy activity (6; 7) is a critical activity.)(Dualprices are negative.)
� 5(c)(ii) There is R19 available to crash the project and we must determine which activities couldbe shortened and by how much. The objective is to minimise the duration of the project.
The LP model is the same as the model for crashing the project to 48 minutes, except for thefollowing changes:
• The objective function changes to minimise z = x9 − x1.
• Replace the Deadline constraint x9 ≤ x1 + 48 with the Extracost constraintA + 2B + C +D + E + 2F + 2G + 3H + 3I + 4J = 19.
The input data for LINGO:
[Objective] min = x9-x1;
[Extracost] A+2*B+C+D+E+2*F+2*G+3*H+3*I+4*J=19;
A<5; B<1; C<5; D<5; E<3;
F<1; G<1; H<2; I<2; J<5;
[A] x2>x1+10-A;
[B] x4>x2+5-B;
[C] x3>x1+15-C;
[D] x6>x3+11-D;
[E] x6>x4+10-E;
[F] x5>x3+5-F;
[G] x7>x5+5-G;
[H] x8>x7+10-H;
[I] x8>x6+10-I;
[J] x9>x8+15-J;
[Dummy34] x4>x3;
[Dummy67] x7>x6;
271 DSC2602
The solution obtained by LINGO:
O p t i m a l s o l u t i o n f o u n d a t s t e p : 7 O b j e c t i v e v a l u e : 4 3 . 0 0 0 0 0 V a r i a b l e V a l u e R e d u c e d C o s t X 9 4 3 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 1 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 A 5 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 B 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 2 5 0 0 0 0 0 C 5 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 D 3 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 E 2 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 F 1 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 G 1 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 H 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 2 5 0 0 0 0 0 I 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 2 5 0 0 0 0 0 J 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 2 5 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 4 1 0 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 3 1 0 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 6 1 8 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 5 1 4 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 7 1 8 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 X 8 2 8 . 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 R o w S l a c k o r S u r p l u s D u a l P r i c e O B J E C T I V E 4 3 . 0 0 0 0 0 1 . 0 0 0 0 0 0 E X T R A C O S T 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 2 5 0 0 0 0 0 3 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 4 1 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 5 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 5 0 0 0 0 0 0 6 2 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 7 1 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 8 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 9 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 1 0 2 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 1 1 2 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 1 2 5 . 0 0 0 0 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 A 0 . 0 0 0 0 0 0 0 E + 0 0 - 0 . 2 5 0 0 0 0 0 B 0 . 0 0 0 0 0 0 0 E + 0 0 - 0 . 2 5 0 0 0 0 0 C 0 . 0 0 0 0 0 0 0 E + 0 0 - 0 . 7 5 0 0 0 0 0 D 0 . 0 0 0 0 0 0 0 E + 0 0 - 0 . 2 5 0 0 0 0 0 E 0 . 0 0 0 0 0 0 0 E + 0 0 - 0 . 2 5 0 0 0 0 0 F 0 . 0 0 0 0 0 0 0 E + 0 0 - 0 . 5 0 0 0 0 0 0 G 0 . 0 0 0 0 0 0 0 E + 0 0 - 0 . 5 0 0 0 0 0 0 H 0 . 0 0 0 0 0 0 0 E + 0 0 - 0 . 5 0 0 0 0 0 0 I 0 . 0 0 0 0 0 0 0 E + 0 0 - 0 . 5 0 0 0 0 0 0 J 0 . 0 0 0 0 0 0 0 E + 0 0 - 1 . 0 0 0 0 0 0 D U M M Y 3 4 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0 D U M M Y 6 7 0 . 0 0 0 0 0 0 0 E + 0 0 0 . 0 0 0 0 0 0 0 E + 0 0
DSC2602 272
Interpretation of the solution:
• The minimum assembly time is 43 minutes.
• This is achieved by shortening activity A by 5 minutes, activity C by 5 minutes, activity Dby 3 minutes, activity E by 2 minutes, activity F by 1 minute and activity G by 1 minute.
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A.8 Chapter 8: Network models
� 1 The final network after 6 iterations is given in Figure A.19.
1
2
3
4
5
6
7
7
9
18
3
5
4
3
3
6
2
[0; S]
[7; 1][12; 2]
[18; 5][17; 6]
[14; 5][18; 1]
[17; 6]
[9; 1]
Figure A.19: Final network
The shortest route from node 1 to each of the other nodes is given in Table A.20.
Shortest route from DistanceNode node 1 (in kilometres)
2 1-2 73 1-3 94 1-2-5-6-4 175 1-2-5 126 1-2-5-6 147 1-2-5-6-7 17
Table A.20: Shortest routes
� 2 The decision variables are defined as
xi j =
{
0 if no unit is shipped from i to j
1 if one unit is shipped from i to j.
The LP model is given in Model A.8.1.
DSC2602 274
Minimise DISTANCE = 8x12 + 13x13 + 15x14 + 10x15 + 15x27 + 5x23
+5x32 + 6x34 + 5x36 + 6x43 + 4x45 + 3x46
+4x54 + 9x56 + 12x59 + 5x63 + 3x64 + 9x65
+4x67 + 2x68 + 5x69 + 15x72 + 4x76 + 4x78
+4x710 + 4x87 + 2x86 + 5x89 + 7x810
+5x98 + 5x96 + 12x95 + 5x910
subject to
x12 + x13 + x14 + x15 = 1 (One unit sent from node 1.)
x22 + x23 + x27 = 1 (One unit sent from node 2.)
x32 + x33 + x34 + x36 = 1 (One unit sent from node 3.)
x43 + x44 + x45 + x46 = 1 (One unit sent from node 4.)
x54 + x55 + x56 + x59 = 1 (One unit sent from node 5.)
x63 + x64 + x65 + x66 + x67 + x68 + x69 = 1 (One unit sent from node 6.)
x72 + x76 + x77 + x78 + x710 = 1 (One unit sent from node 7.)
x86 + x87 + x88 + x89 + x810 = 1 (One unit sent from node 8.)
x95 + x96 + x98 + x99 + x910 = 1 (One unit sent from node 9.)
x12 + x22 + x32 + x72 = 1 (One unit arrives at node 2.)
x13 + x23 + x33 + x43 + x63 = 1 (One unit arrives at node 3.)
x14 + x34 + x44 + x54 + x64 = 1 (One unit arrives at node 4.)
x15 + x45 + x55 + x65 + x95 = 1 (One unit arrives at node 5.)
x36 + x46 + x56 + x66 + x76 + x86 + x96 = 1 (One unit arrives at node 6.)
x27 + x67 + x77 + x87 = 1 (One unit arrives at node 7.)
x68 + x78 + x88 + x98 = 1 (One unit arrives at node 8.)
x59 + x69 + x89 + x99 = 1 (One unit arrives at node 9.)
x710 + x810 + x910 = 1 (One unit arrives at node 10.)
xi j ≥ 0 for i = 1; 2; . . . ,9 and j = 2; . . . ,10.
Model A.8.1: The LP model for finding the shortest route
275 DSC2602
The input data file for LINGO:[Objective] min = 8*x12+B*x13+15*x14+10*x15+5*x23+15*x27+5*x32+6*x31+5*x36+
6 * x 4 3 + 4 * x 4 5 + 3 * x 4 6 + 4 * x 5 4 + 9 * x 5 6 + 1 2 * x 5 9 + 5 * x 6 3 + 3 * x 6 4 + 9 * x 6 5 + 4 * x 6 7 + 2 * x 6 8 + 5 * x 6 9 + 1 5 * x 7 2 + 4 * x 7 6 + 4 * x 7 8 + 4 * x 7 1 0 + 4 * x 8 7 + 2 * x 8 6 + 5 * x 8 9 + 7 * x 8 1 0 + 5 * x 9 8 + 5 * x 9 6 + 1 2 * x 9 5 + 5 * x 9 1 0 ; [ F r o m _ 1 ] x 1 2 + x 1 3 + x 1 4 + x 1 5 = 1 ; [ F r o m _ 2 ] x 2 2 + x 2 3 + x 2 7 = 1 ; [ F r o m _ 3 ] x 3 2 + x 3 3 + x 3 4 + x 3 6 = 1 ; [ F r o m _ 4 ] x 4 3 + x 4 4 + x 4 5 + x 4 6 = 1 ; [ F r o m _ 5 ] x 5 4 + x 5 5 + x 5 6 + x 5 9 = 1 ; [ F r o m _ 6 ] x 6 3 + x 6 4 + x 6 5 + x 6 6 + x 6 7 + x 6 8 + x 6 9 = 1 ; [ F r o m _ 7 ] x 7 2 + x 7 6 + x 7 7 + x 7 8 + x 7 1 0 = 1 ; [ F r o m _ 8 ] x 8 6 + x 8 7 + x 8 8 + x 8 9 + x 8 1 0 = 1 ; [ F r o m _ 9 ] x 9 5 + x 9 6 + x 9 8 + x 9 9 + x 9 1 0 = 1 ; [ T o _ 2 ] x 1 2 + x 2 2 + x 3 2 + x 7 2 = 1 ; [ T o _ 3 ] x 1 3 + x 2 3 + x 3 3 + x 4 3 + x 6 3 = 1 ; [ T o _ 4 ] x 1 4 + x 3 4 + x 4 4 + x 5 4 + x 6 4 = 1 ; [ T 0 _ 5 ] x 1 5 + x 4 5 + x 5 5 + x 6 5 + x 9 5 = 1 ; [ T 0 _ 6 ] x 3 6 + x 4 6 + x 5 6 + x 6 6 + x 7 6 + x 8 6 + x 9 6 = 1 ; [ T o _ 7 ] x 2 7 + x 6 7 + x 7 7 + x 8 7 = 1 ; [ T o _ 8 ] x 6 8 + x 7 8 + x 8 8 + x 9 8 = 1 ; [ T 0 _ 9 ] x 6 8 + x 7 8 + x 8 8 + x 9 8 = 1 ; [ T o _ 1 0 ] x 7 1 0 + x 8 1 0 + x 9 1 0 = 1 ;
[To_2] x12+x22+x32+x72=1;
[To_3] x13+x23+x33+x43+x63=1;
[To_4] x14+x34+x44+x54+x64=1;
[T0_5] x15+x45+x55+x65+x95=1;
[T0_6] x36+x46+x56+x66+x76+x86+x96=1;
[To_7] x27+x67+x77+x87=1;
[To_8] x68+x78+x88+x98=1;
[T0_9] x59+x69+x89+x99=1;
[To_10] x710+x810+x910=1;
DSC2602 276
The solution obtained by LINGO:
Optimal solution found at step: 7
Objective value: 25.00000
Variable Value Reduced Cost
X12 0.0000000E+00 0.0000000E+00
X13 0.0000000E+00 0.0000000E+00
X14 0.0000000E+00 0.0000000E+00
X15 1.000000 0.0000000E+00
X23 0.0000000E+00 0.0000000E+00
X27 0.0000000E+00 1.000000
Optimal solution found at step: 13
Objective value: 25.00000
Variable Value Reduced Cost
X12 0.0000000E+00 0.0000000E+00
X13 0.0000000E+00 0.0000000E+00
X14 0.0000000E+00 0.0000000E+00
X15 1.000000 0.0000000E+00
X23 0.0000000E+00 0.0000000E+00
X27 0.0000000E+00 1.000000
X32 0.0000000E+00 10.00000
X34 0.0000000E+00 4.000000
X36 0.0000000E+00 0.0000000E+00
X43 0.0000000E+00 8.000000
X45 0.0000000E+00 9.000000
X46 1.000000 0.0000000E+00
X63 0.0000000E+00 10.00000
X64 0.0000000E+00 6.000000
X65 0.0000000E+00 17.00000
X67 1.000000 0.0000000E+00
X68 0.0000000E+00 0.0000000E+00
X69 0.0000000E+00 2.000000
X72 0.0000000E+00 29.00000
X76 0.0000000E+00 8.000000
X78 0.0000000E+00 6.000000
X710 1.000000 0.0000000E+00
X87 0.0000000E+00 2.000000
X86 0.0000000E+00 4.000000
X89 0.0000000E+00 4.000000
X810 0.0000000E+00 1.000000
X98 0.0000000E+00 6.000000
X96 0.0000000E+00 8.000000
X95 0.0000000E+00 23.00000
X910 0.0000000E+00 0.0000000E+00
X22 1.000000 0.0000000E+00
X33 1.000000 0.0000000E+00
X44 0.0000000E+00 0.0000000E+00
X55 0.0000000E+00 1.000000
X66 0.0000000E+00 0.0000000E+00
X77 0.0000000E+00 0.0000000E+00
X88 1.000000 0.0000000E+00
X99 1.000000 0.0000000E+00
277 DSC2602
X22 1.000000 0.0000000E+00
X33 1.000000 0.0000000E+00
X44 0.0000000E+00 0.0000000E+00
X55 0.0000000E+00 1.000000
X66 0.0000000E+00 0.0000000E+00
X77 0.0000000E+00 0.0000000E+00
X88 1.000000 0.0000000E+00
X99 1.000000 0.0000000E+00
TO_2 0.0000000E+00 15.00000
TO_3 0.0000000E+00 10.00000
TO_4 0.0000000E+00 8.000000
T0_5 0.0000000E+00 13.00000
T0_6 0.0000000E+00 5.000000
TO_7 0.0000000E+00 1.000000
TO_8 0.0000000E+00 4.000000
T0_9 0.0000000E+00 0.0000000E+00
TO_10 0.0000000E+00 -3.000000
To save space, the part of the solution dealing with slack and dual prices has been omitted, sincewe do not need it to find the shortest route.
The shortest route is 1-5-4-6-7-10. The distance is 25 kilometres.
Note: The purpose of using computers is to save time and to make life easier for us. If a shortest-path problem is small enough to solve by hand in less time than it would take to formulate amathematical model and solve it with LINGO (or some other computer package), it makes senseto solve it by hand. However, the more one uses the computer to solve problems, the easier itbecomes and the quicker one can find solutions!
� 3 After 6 iterations of the maximum-flow algorithm we see that a maximum of 10 000 litres ofoil can flow through the network. See Figure A.20.
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00
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654
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2 4
3
3
0
23
3
4
40
61
5
6
125
2
0
0
35
1
0
0
212
76
3 1 10
Figure A.20: The maximum-flow algorithm applied
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The iterative process is summarised in Table A.21.
Iteration Path Flow
1 1→ 2→ 5→ 7 32 1→ 2→ 4→ 7 13 1→ 5→ 7 24 1→ 5→ 6→ 7 15 1→ 3→ 6→ 7 16 1→ 5→ 2→ 4→ 7 2
Total flow: 10
Table A.21: The iterations of the maximum-flow algorithm
The flow through each arc is shown in Figure A.21.
1
24
5
6
7
3
M a x i m u m f l o w :1 0 u n i t s
44
3
3 0 3
5
21
12 1
Figure A.21: The flow through each arc
The maximum amount of fuel that can flow through the plant is 10 000 litres.
279 DSC2602
� 4 The minimum-spanning tree is given in Figure A.22.
1
2
3
4
5
6
7
8
2
3
2
4
30,5
11
1,5
1,2
11,6
4
3 2,5
0,5
Figure A.22: The minimum-spanning tree
They will need 800 metres of connections.
� 5 Table A.22 is constructed for the execution of the algorithm.
Connected City Alpha Beta Gamma Delta Epsilon
Alpha - - 217 164 58Beta - - 290 201 79
Gamma 217 290 - 113 -Delta 164 201 113 - 196
Epsilon 58 79 - 196 -
Table A.22: Table for execution of minimum spanning tree algorithm 2
Arbitrarily select Alpha to begin with. Mark it as connected and cross out the Alpha column asshown in Table A.23.
Connected City Alpha Beta Gamma Delta Epsilon
X Alpha - - 217 164 58Beta - - 290 201 79
Gamma 217 290 - 113 -Delta 164 201 113 - 196
Epsilon 58 79 - 196 -
Table A.23: Select Alpha to begin with
The smallest value in row Alpha is 58 in position (Alpha; Epsilon). Circle the value, mark Epsilonas connected and cross out column Epsilon.
Continue the algorithm until all cities are connected, as shown in Table A.24.
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Connected City Alpha Beta Gamma Delta Epsilon
X Alpha - - 217 164 58
X Beta - - 290 201 79X Gamma 217 290 - 113 -
X Delta 164 201 113 - 196
X Epsilon 58 79 - 196 -
Table A.24: Final table
The roads must be built between Alpha and Delta, between Alpha and Epsilon, between Deltaand Gamma and between Epsilon and Beta. The minimum distance is 414 kilometres. Theminimum-spanning tree is shown in Figure A.23.
Alpha
Beta
Delta
Epsilon
Gamma
164
11358
79
Figure A.23: The minimum-spanning tree
APPENDIX B
Statistical tables
B.1 The cumulative Poisson distribution
The entries in the table are the probabilities
P(X ≤ r) =
r∑
x=0
f (x)
wheref (x) =
λxe−λ
x!,
λ = the mean, or average, number of occurrences over an interval, and
x = the number of observed occurrences over the interval.
λ-values
0,10 0,2 0,3 0,4 0,5 0,6 0,7
0 0,9048 0,8187 0,7408 0,6703 0,6065 0,5488 0,49661 0,9953 0,9825 0,9631 0,9384 0,9098 0,8781 0,84422 0,9998 0,9989 0,9964 0,9921 0,9856 0,9769 0,96593 1,0000 0,9999 0,9997 0,9992 0,9982 0,9966 0,99424 1,0000 1,0000 0,9999 0,9998 0,9996 0,99925 1,0000 1,0000 1,0000 0,99996 1,0000
r-va
lues
Continued on the next page. . .
281
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T P (continued)
λ-values
0,8 0,9 1 1,5 2 2,5 3
0 0,4493 0,4066 0,3679 0,2231 0,1353 0,0821 0,04981 0,8088 0,7725 0,7358 0,5578 0,4060 0,2873 0,19912 0,9526 0,9371 0,9197 0,8088 0,6767 0,5438 0,42323 0,9909 0,9865 0,9810 0,9344 0,8571 0,7576 0,64724 0,9986 0,9977 0,9963 0,9814 0,9473 0,8912 0,81535 0,9998 0,9997 0,9994 0,9955 0,9834 0,9580 0,91616 1,0000 1,0000 0,9999 0,9991 0,9955 0,9858 0,96657 1,0000 0,9998 0,9989 0,9958 0,98818 1,0000 0,9998 0,9989 0,99629 1,0000 0,9997 0,9989
10 0,9999 0,999711 1,0000 0,999912 1,0000
r-va
lues
λ-values
4 5 6 7 8 9 10
0 0,0183 0,0067 0,0025 0,0009 0,0003 0,0001 0,00001 0,0916 0,0404 0,0174 0,0073 0,0030 0,0012 0,00052 0,2381 0,1247 0,0620 0,0296 0,0138 0,0062 0,00283 0,4335 0,2650 0,1512 0,0818 0,0424 0,0212 0,01034 0,6288 0,4405 0,2851 0,1730 0,0996 0,0550 0,02935 0,7851 0,6160 0,4457 0,3007 0,1912 0,1157 0,06716 0,8893 0,7622 0,6063 0,4497 0,3134 0,2068 0,13017 0,9489 0,8666 0,7440 0,5987 0,4530 0,3239 0,22028 0,9786 0,9319 0,8472 0,7291 0,5925 0,4557 0,33289 0,9919 0,9682 0,9161 0,8305 0,7166 0,5874 0,4579
10 0,9972 0,9863 0,9574 0,9015 0,8159 0,7060 0,583011 0,9991 0,9945 0,9799 0,9467 0,8881 0,8030 0,696812 0,9997 0,9980 0,9912 0,9730 0,9362 0,8758 0,791613 0,9999 0,9993 0,9964 0,9872 0,9658 0,9261 0,864514 1,0000 0,9998 0,9986 0,9943 0,9827 0,9585 0,916515 0,9999 0,9995 0,9976 0,9918 0,9780 0,951316 1,0000 0,9998 0,9990 0,9963 0,9889 0,973017 0,9999 0,9996 0,9984 0,9947 0,985718 1,0000 0,9999 0,9993 0,9976 0,992819 1,0000 0,9997 0,9989 0,996520 0,9999 0,9996 0,998421 1,0000 0,9998 0,999322 0,9999 0,999723 1,0000 0,999924 1,0000
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B.2 The standard normal distribution
The table entries give the probabilities P(0 ≤ Z ≤ z), where Z ∼ n(0; 1), denoted by the shadedarea in the following figure.
0 z
Second decimal place of z
z 0,00 0,01 0,02 0,03 0,04 0,05 0,06 0,07 0,08 0,09
0,0 0,00000 0,00399 0,00798 0,01197 0,01595 0,01994 0,02392 0,02790 0,03188 0,035860,1 0,03983 0,04380 0,04776 0,05172 0,05567 0,05962 0,06356 0,06749 0,07142 0,075350,2 0,07926 0,08317 0,08706 0,09095 0,09483 0,09871 0,10257 0,10642 0,11026 0,114090,3 0,11791 0,12172 0,12552 0,12930 0,13307 0,13683 0,14058 0,14431 0,14803 0,151730,4 0,15542 0,15910 0,16276 0,16640 0,17003 0,17364 0,17724 0,18082 0,18439 0,18793
0,5 0,19146 0,19497 0,19847 0,20194 0,20540 0,20884 0,21226 0,21566 0,21904 0,222400,6 0,22575 0,22907 0,23237 0,23565 0,23891 0,24215 0,24537 0,24857 0,25175 0,254900,7 0,25804 0,26115 0,26424 0,26730 0,27035 0,27337 0,27637 0,27935 0,28230 0,285240,8 0,28814 0,29103 0,29389 0,29673 0,29955 0,30234 0,30511 0,30785 0,31057 0,313270,9 0,31594 0,31859 0,32121 0,32381 0,32639 0,32894 0,33147 0,33398 0,33646 0,33891
1,0 0,34134 0,34375 0,34614 0,34849 0,35083 0,35314 0,35543 0,35769 0,35993 0,362141,1 0,36433 0,36650 0,36864 0,37076 0,37286 0,37493 0,37698 0,37900 0,38100 0,382981,2 0,38493 0,38686 0,38877 0,39065 0,39251 0,39435 0,39617 0,39796 0,39973 0,401471,3 0,40320 0,40490 0,40658 0,40824 0,40988 0,41149 0,41308 0,41466 0,41621 0,417741,4 0,41924 0,42073 0,42220 0,42364 0,42507 0,42647 0,42785 0,42922 0,43056 0,43189
1,5 0,43319 0,43448 0,43574 0,43699 0,43822 0,43943 0,44062 0,44179 0,44295 0,444081,6 0,44520 0,44630 0,44738 0,44845 0,44950 0,45053 0,45154 0,45254 0,45352 0,454491,7 0,45543 0,45637 0,45728 0,45818 0,45907 0,45994 0,46080 0,46164 0,46246 0,463271,8 0,46407 0,46485 0,46562 0,46638 0,46712 0,46784 0,46856 0,46926 0,46995 0,470621,9 0,47128 0,47193 0,47257 0,47320 0,47381 0,47441 0,47500 0,47558 0,47615 0,47670
2,0 0,47725 0,47778 0,47831 0,47882 0,47932 0,47982 0,48030 0,48077 0,48124 0,481692,1 0,48214 0,48257 0,48300 0,48341 0,48382 0,48422 0,48461 0,48500 0,48537 0,485742,2 0,48610 0,48645 0,48679 0,48713 0,48745 0,48778 0,48809 0,48840 0,48870 0,488992,3 0,48928 0,48956 0,48983 0,49010 0,49036 0,49061 0,49086 0,49111 0,49134 0,491582,4 0,49180 0,49202 0,49224 0,49245 0,49266 0,49286 0,49305 0,49324 0,49343 0,49361
2,5 0,49379 0,49396 0,49413 0,49430 0,49446 0,49461 0,49477 0,49492 0,49506 0,495202,6 0,49534 0,49547 0,49560 0,49573 0,49585 0,49598 0,49609 0,49621 0,49632 0,496432,7 0,49653 0,49664 0,49674 0,49683 0,49693 0,49702 0,49711 0,49720 0,49728 0,497362,8 0,49744 0,49752 0,49760 0,49767 0,49774 0,49781 0,49788 0,49795 0,49801 0,498072,9 0,49813 0,49819 0,49825 0,49831 0,49836 0,49841 0,49846 0,49851 0,49856 0,49861
3,0 0,49865 0,49869 0,49874 0,49878 0,49882 0,49886 0,49889 0,49893 0,49896 0,499003,1 0,49903 0,49906 0,49910 0,49913 0,49916 0,49918 0,49921 0,49924 0,49926 0,499293,2 0,49931 0,49934 0,49936 0,49938 0,49940 0,49942 0,49944 0,49946 0,49948 0,499503,3 0,49952 0,49953 0,49955 0,49957 0,49958 0,49960 0,49961 0,49962 0,49964 0,499653,4 0,49966 0,49968 0,49969 0,49970 0,49971 0,49972 0,49973 0,49974 0,49975 0,49976
3,5 0,49977 0,49978 0,49978 0,49979 0,49980 0,49981 0,49981 0,49982 0,49983 0,499833,6 0,49984 0,49985 0,49985 0,49986 0,49986 0,49987 0,49987 0,49988 0,49988 0,499893,7 0,49989 0,49990 0,49990 0,49990 0,49991 0,49991 0,49992 0,49992 0,49992 0,499923,8 0,49993 0,49993 0,49993 0,49994 0,49994 0,49994 0,49994 0,49995 0,49995 0,499953,9 0,49995 0,49995 0,49996 0,49996 0,49996 0,49996 0,49996 0,49996 0,49997 0,499974,0 0,49997 0,49997 0,49997 0,49997 0,49997 0,49997 0,49998 0,49998 0,49998 0,49998
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B.3 Student’s t-distribution
Entries in the table are the values tν;α for various degrees of freedom (ν) and one-tailed proba-bilities α, where P(Tν ≥ tν,α) = α.
Tail probability α
0,25 0,1 0,05 0,025 0,01 0,005
1 1,0000 3,0777 6,3138 12,7060 31,8197 63,65672 0,8165 1,8856 2,9200 4,3026 6,9645 9,92473 0,7649 1,6377 2,3534 3,1824 4,5407 5,84094 0,7407 1,5332 2,1318 2,7764 3,7469 4,60415 0,7267 1,4759 2,0150 2,5706 3,3649 4,03206 0,7176 1,4398 1,9432 2,4469 3,1427 3,70747 0,7111 1,4149 1,8946 2,3646 2,9979 3,49948 0,7064 1,3968 1,8595 2,3060 2,8965 3,35549 0,7027 1,3830 1,8331 2,2622 2,8214 3,2498
10 0,6998 1,3722 1,8125 2,2281 2,7638 3,169211 0,6974 1,3634 1,7959 2,2010 2,7181 3,105812 0,6955 1,3562 1,7823 2,1788 2,6810 3,054513 0,6938 1,3502 1,7709 2,1604 2,6503 3,012314 0,6924 1,3450 1,7613 2,1448 2,6245 2,976915 0,6912 1,3406 1,7531 2,1314 2,6025 2,946716 0,6901 1,3368 1,7459 2,1199 2,5835 2,920817 0,6892 1,3334 1,7396 2,1098 2,5669 2,898218 0,6884 1,3304 1,7341 2,1009 2,5524 2,878419 0,6876 1,3277 1,7291 2,0930 2,5395 2,860920 0,6870 1,3253 1,7247 2,0860 2,5280 2,845321 0,6864 1,3232 1,7207 2,0796 2,5176 2,831422 0,6858 1,3212 1,7171 2,0739 2,5083 2,818823 0,6853 1,3195 1,7139 2,0687 2,4999 2,807324 0,6848 1,3178 1,7109 2,0639 2,4922 2,796925 0,6844 1,3163 1,7081 2,0595 2,4851 2,787426 0,6840 1,3150 1,7056 2,0555 2,4786 2,778727 0,6837 1,3137 1,7033 2,0518 2,4727 2,770728 0,6834 1,3125 1,7011 2,0484 2,4671 2,763329 0,6830 1,3114 1,6991 2,0452 2,4620 2,756430 0,6828 1,3104 1,6973 2,0423 2,4573 2,750035 0,6816 1,3062 1,6896 2,0301 2,4377 2,723840 0,6807 1,3031 1,6839 2,0211 2,4233 2,704560 0,6786 1,2958 1,6706 2,0003 2,3901 2,6603
100 0,6770 1,2901 1,6602 1,9840 2,3642 2,6259∞ 0,6745 1,2816 1,6449 1,9600 2,3264 2,5759
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B.4 The F-distribution
Table entries give the values Fν1 ,ν2;α for the given values of ν1, ν2 and αwhere P(Fν1,ν2≥ Fν1 ,ν2;α) = α.
U 5%
Degrees of freedom in the numerator, ν1
1 2 3 4 5 6 7 8 9 10
1 161 199 216 225 230 234 237 239 241 2422 18,5 19,0 19,2 19,2 19,3 19,3 19,4 19,4 19,4 19,43 10,1 9,55 9,28 9,12 9,01 8,94 8,89 8,85 8,81 8,794 7,71 6,94 6,59 6,39 6,26 6,16 6,09 6,04 6,00 5,965 6,61 5,79 5,41 5,19 5,05 4,95 4,88 4,82 4,77 4,74
6 5,99 5,14 4,76 4,53 4,39 4,28 4,21 4,15 4,10 4,067 5,59 4,74 4,35 4,12 3,97 3,87 3,79 3,73 3,68 3,648 5,32 4,46 4,07 3,84 3,69 3,58 3,50 3,44 3,39 3,359 5,12 4,26 3,86 3,63 3,48 3,37 3,29 3,23 3,18 3,14
10 4,96 4,10 3,71 3,48 3,33 3,22 3,14 3,07 3,02 2,98
11 4,84 3,98 3,59 3,36 3,20 3,09 3,01 2,95 2,90 2,8512 4,75 3,89 3,49 3,26 3,11 3,00 2,91 2,85 2,80 2,7513 4,67 3,81 3,41 3,18 3,03 2,92 2,83 2,77 2,71 2,6714 4,60 3,74 3,34 3,11 2,96 2,85 2,76 2,70 2,65 2,6015 4,54 3,68 3,29 3,06 2,90 2,79 2,71 2,64 2,59 2,54
16 4,49 3,63 3,24 3,01 2,85 2,74 2,66 2,59 2,54 2,4917 4,45 3,59 3,20 2,96 2,81 2,70 2,61 2,55 2,49 2,4518 4,41 3,55 3,16 2,93 2,77 2,66 2,58 2,51 2,46 2,4119 4,38 3,52 3,13 2,90 2,74 2,63 2,54 2,48 2,42 2,3820 4,35 3,49 3,10 2,87 2,71 2,60 2,51 2,45 2,39 2,35
21 4,32 3,47 3,07 2,84 2,68 2,57 2,49 2,42 2,37 2,3222 4,30 3,44 3,05 2,82 2,66 2,55 2,46 2,40 2,34 2,3023 4,28 3,42 3,03 2,80 2,64 2,53 2,44 2,37 2,32 2,2724 4,26 3,40 3,01 2,78 2,62 2,51 2,42 2,36 2,30 2,2525 4,24 3,39 2,99 2,76 2,60 2,49 2,40 2,34 2,28 2,24
28 4,20 3,34 2,95 2,71 2,56 2,45 2,36 2,29 2,24 2,1930 4,17 3,32 2,92 2,69 2,53 2,42 2,33 2,27 2,21 2,1634 4,13 3,28 2,88 2,65 2,49 2,38 2,29 2,23 2,17 2,1240 4,08 3,23 2,84 2,61 2,45 2,34 2,25 2,18 2,12 2,0848 4,04 3,19 2,80 2,57 2,41 2,29 2,21 2,14 2,08 2,03
60 4,00 3,15 2,76 2,53 2,37 2,25 2,17 2,10 2,04 1,9980 3,96 3,11 2,72 2,49 2,33 2,21 2,13 2,06 2,00 1,95
120 3,92 3,07 2,68 2,45 2,29 2,18 2,09 2,02 1,96 1,91∞ 3,84 3,00 2,60 2,37 2,21 2,10 2,01 1,94 1,88 1,83
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,ν2
Continued on the next page. . .
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F- (continued)
U 5%
Degrees of freedom in the numerator, ν1
0,05 12 15 20 24 30 40 60 120 ∞1 244 246 248 249 250 251 252 253 2542 19,4 19,4 19,4 19,5 19,5 19,5 19,5 19,5 19,53 8,74 8,70 8,66 8,64 8,62 8,59 8,57 8,55 8,534 5,91 5,86 5,80 5,77 5,75 5,72 5,69 5,66 5,635 4,68 4,62 4,56 4,53 4,50 4,46 4,43 4,40 4,37
6 4,00 3,94 3,87 3,84 3,81 3,77 3,74 3,70 3,677 3,57 3,51 3,44 3,41 3,38 3,34 3,30 3,27 3,238 3,28 3,22 3,15 3,12 3,08 3,04 3,01 2,97 2,939 3,07 3,01 2,94 2,90 2,86 2,83 2,79 2,75 2,71
10 2,91 2,85 2,77 2,74 2,70 2,66 2,62 2,58 2,54
11 2,79 2,72 2,65 2,61 2,57 2,53 2,49 2,45 2,4012 2,69 2,62 2,54 2,51 2,47 2,43 2,38 2,34 2,3013 2,60 2,53 2,46 2,42 2,38 2,34 2,30 2,25 2,2114 2,53 2,46 2,39 2,35 2,31 2,27 2,22 2,18 2,1315 2,48 2,40 2,33 2,29 2,25 2,20 2,16 2,11 2,07
16 2,42 2,35 2,28 2,24 2,19 2,15 2,11 2,06 2,0117 2,38 2,31 2,23 2,19 2,15 2,10 2,06 2,01 1,9618 2,34 2,27 2,19 2,15 2,11 2,06 2,02 1,97 1,9219 2,31 2,23 2,16 2,11 2,07 2,03 1,98 1,93 1,8820 2,28 2,20 2,12 2,08 2,04 1,99 1,95 1,90 1,84
21 2,25 2,18 2,10 2,05 2,01 1,96 1,92 1,87 1,8122 2,23 2,15 2,07 2,03 1,98 1,94 1,89 1,84 1,7823 2,20 2,13 2,05 2,01 1,96 1,91 1,86 1,81 1,7624 2,18 2,11 2,03 1,98 1,94 1,89 1,84 1,79 1,7325 2,16 2,09 2,01 1,96 1,92 1,87 1,82 1,77 1,71
28 2,12 2,04 1,96 1,91 1,87 1,82 1,77 1,71 1,6530 2,09 2,01 1,93 1,89 1,84 1,79 1,74 1,68 1,6234 2,05 1,97 1,89 1,84 1,80 1,75 1,69 1,63 1,5740 2,00 1,92 1,84 1,79 1,74 1,69 1,64 1,58 1,5148 1,96 1,88 1,79 1,75 1,70 1,64 1,59 1,52 1,45
60 1,92 1,84 1,75 1,70 1,65 1,59 1,53 1,47 1,3980 1,88 1,79 1,70 1,65 1,60 1,54 1,48 1,41 1,32
120 1,83 1,75 1,66 1,61 1,55 1,50 1,43 1,35 1,25∞ 1,75 1,67 1,57 1,52 1,46 1,39 1,32 1,22 1,00
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Continued on the next page. . .
287 DSC2602
F- (continued)
U 1%
Degrees of freedom in numerator, ν1
1 2 3 4 5 6 7 8 9 10
1 4052 4999 5404 5624 5764 5859 5928 5981 6022 60562 98,5 99,0 99,2 99,3 99,3 99,3 99,4 99,4 99,4 99,43 34,1 30,8 29,5 28,7 28,2 27,9 27,7 27,5 27,3 27,24 21,2 18,0 16,7 16,0 15,5 15,2 15,0 14,8 14,7 14,55 16,3 13,3 12,1 11,4 11,0 10,7 10,5 10,3 10,2 10,1
6 13,7 10,9 9,78 9,15 8,75 8,47 8,26 8,10 7,98 7,877 12,2 9,55 8,45 7,85 7,46 7,19 6,99 6,84 6,72 6,628 11,3 8,65 7,59 7,01 6,63 6,37 6,18 6,03 5,91 5,819 10,6 8,02 6,99 6,42 6,06 5,80 5,61 5,47 5,35 5,26
10 10,0 7,56 6,55 5,99 5,64 5,39 5,20 5,06 4,94 4,85
11 9,65 7,21 6,22 5,67 5,32 5,07 4,89 4,74 4,63 4,5412 9,33 6,93 5,95 5,41 5,06 4,82 4,64 4,50 4,39 4,3013 9,07 6,70 5,74 5,21 4,86 4,62 4,44 4,30 4,19 4,1014 8,86 6,51 5,56 5,04 4,69 4,46 4,28 4,14 4,03 3,9415 8,68 6,36 5,42 4,89 4,56 4,32 4,14 4,00 3,89 3,80
16 8,53 6,23 5,29 4,77 4,44 4,20 4,03 3,89 3,78 3,6917 8,40 6,11 5,19 4,67 4,34 4,10 3,93 3,79 3,68 3,5918 8,29 6,01 5,09 4,58 4,25 4,01 3,84 3,71 3,60 3,5119 8,18 5,93 5,01 4,50 4,17 3,94 3,77 3,63 3,52 3,4320 8,10 5,85 4,94 4,43 4,10 3,87 3,70 3,56 3,46 3,37
21 8,02 5,78 4,87 4,37 4,04 3,81 3,64 3,51 3,40 3,3122 7,95 5,72 4,82 4,31 3,99 3,76 3,59 3,45 3,35 3,2623 7,88 5,66 4,76 4,26 3,94 3,71 3,54 3,41 3,30 3,2124 7,82 5,61 4,72 4,22 3,90 3,67 3,50 3,36 3,26 3,1725 7,77 5,57 4,68 4,18 3,85 3,63 3,46 3,32 3,22 3,13
28 7,64 5,45 4,57 4,07 3,75 3,53 3,36 3,23 3,12 3,0330 7,56 5,39 4,51 4,02 3,70 3,47 3,30 3,17 3,07 2,9834 7,44 5,29 4,42 3,93 3,61 3,39 3,22 3,09 2,98 2,8940 7,31 5,18 4,31 3,83 3,51 3,29 3,12 2,99 2,89 2,8048 7,19 5,08 4,22 3,74 3,43 3,20 3,04 2,91 2,80 2,71
60 7,08 4,98 4,13 3,65 3,34 3,12 2,95 2,82 2,72 2,6380 6,96 4,88 4,04 3,56 3,26 3,04 2,87 2,74 2,64 2,55
120 6,85 4,79 3,95 3,48 3,17 2,96 2,79 2,66 2,56 2,47∞ 6,63 4,61 3,78 3,32 3,02 2,80 2,64 2,51 2,41 2,32
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omin
ator
,ν2
Continued on the next page. . .
DSC2602 288
F- (continued)
U 1%
Degrees of freedom in numerator, g1
0,05 12 15 20 24 30 40 60 120 ∞1 6107 6157 6209 6234 6260 6286 6313 6340 63662 99,4 99,4 99,4 99,5 99,5 99,5 99,5 99,5 99,53 27,1 26,9 26,7 26,6 26,5 26,4 26,3 26,2 26,14 14,4 14,2 14,0 13,9 13,8 13,7 13,7 13,6 13,55 9,89 9,72 9,55 9,47 9,38 9,29 9,20 9,11 9,02
6 7,72 7,56 7,40 7,31 7,23 7,14 7,06 6,97 6,887 6,47 6,31 6,16 6,07 5,99 5,91 5,82 5,74 5,658 5,67 5,52 5,36 5,28 5,20 5,12 5,03 4,95 4,869 5,11 4,96 4,81 4,73 4,65 4,57 4,48 4,40 4,31
10 4,71 4,56 4,41 4,33 4,25 4,17 4,08 4,00 3,91
11 4,40 4,25 4,10 4,02 3,94 3,86 3,78 3,69 3,6012 4,16 4,01 3,86 3,78 3,70 3,62 3,54 3,45 3,3613 3,96 3,82 3,66 3,59 3,51 3,43 3,34 3,25 3,1714 3,80 3,66 3,51 3,43 3,35 3,27 3,18 3,09 3,0015 3,67 3,52 3,37 3,29 3,21 3,13 3,05 2,96 2,87
16 3,55 3,41 3,26 3,18 3,10 3,02 2,93 2,84 2,7517 3,46 3,31 3,16 3,08 3,00 2,92 2,83 2,75 2,6518 3,37 3,23 3,08 3,00 2,92 2,84 2,75 2,66 2,5719 3,30 3,15 3,00 2,92 2,84 2,76 2,67 2,58 2,4920 3,23 3,09 2,94 2,86 2,78 2,69 2,61 2,52 2,42
21 3,17 3,03 2,88 2,80 2,72 2,64 2,55 2,46 2,3622 3,12 2,98 2,83 2,75 2,67 2,58 2,50 2,40 2,3123 3,07 2,93 2,78 2,70 2,62 2,54 2,45 2,35 2,2624 3,03 2,89 2,74 2,66 2,58 2,49 2,40 2,31 2,2125 2,99 2,85 2,70 2,62 2,54 2,45 2,36 2,27 2,17
28 2,90 2,75 2,60 2,52 2,44 2,35 2,26 2,17 2,0630 2,84 2,70 2,55 2,47 2,39 2,30 2,21 2,11 2,0134 2,76 2,61 2,46 2,38 2,30 2,21 2,12 2,02 1,9140 2,66 2,52 2,37 2,29 2,20 2,11 2,02 1,92 1,8048 2,58 2,44 2,28 2,20 2,12 2,02 1,93 1,82 1,70
60 2,50 2,35 2,20 2,12 2,03 1,94 1,84 1,73 1,6080 2,42 2,27 2,12 2,03 1,94 1,85 1,75 1,63 1,49
120 2,34 2,19 2,03 1,95 1,86 1,76 1,66 1,53 1,38∞ 2,18 2,04 1,88 1,79 1,70 1,59 1,47 1,32 1,00
Deg
rees
offr
eedo
min
the
den
omin
ator
,ν2
289 DSC2602
B.5 The cumulative binomial distribution
The entries in the table are the probabilities
P(X ≤ r) =
r∑
x=0
P(X = x)
whereP(X = x) =
(
n
x
)
px(1 − p)n−p,
n = the number of trials,
p = the probability of a success, and
x = the number of successes in n trials.
The probability of a success, p
n r 0,05 0,10 0,20 0,25 0,30 0,40 0,50
5 0 0,7738 0,5905 0,3277 0,2373 0,1681 0,0778 0,03131 0,9774 0,9185 0,7373 0,6328 0,5282 0,3370 0,18752 0,9988 0,9914 0,9421 0,8965 0,8369 0,6826 0,50003 1,0000 0,9995 0,9933 0,9844 0,9692 0,9130 0,81254 1,0000 0,9997 0,9990 0,9976 0,9898 0,96885 1,0000 1,0000 1,0000 1,0000 1,0000
8 0 0,6634 0,4305 0,1678 0,1001 0,0576 0,0168 0,00391 0,9428 0,8131 0,5033 0,3671 0,2553 0,1064 0,03522 0,9942 0,9619 0,7969 0,6785 0,5518 0,3154 0,14453 0,9996 0,9950 0,9437 0,8862 0,8059 0,5941 0,36334 1,0000 0,9996 0,9896 0,9727 0,9420 0,8263 0,63675 1,0000 0,9988 0,9958 0,9887 0,9502 0,85556 0,9999 0,9996 0,9987 0,9915 0,96487 1,0000 1,0000 0,9999 0,9993 0,99618 1,0000 1,0000 1,0000
10 0 0,5987 0,3487 0,1074 0,0563 0,0282 0,0060 0,00101 0,9139 0,7361 0,3758 0,2440 0,1493 0,0464 0,01072 0,9885 0,9298 0,6778 0,5256 0,3828 0,1673 0,05473 0,9990 0,9872 0,8791 0,7759 0,6496 0,3823 0,17194 0,9999 0,9984 0,9672 0,9219 0,8497 0,6331 0,37705 1,0000 0,9999 0,9936 0,9803 0,9527 0,8338 0,62306 1,0000 0,9991 0,9965 0,9894 0,9452 0,82817 0,9999 0,9996 0,9984 0,9877 0,94538 1,0000 1,0000 0,9999 0,9983 0,98939 1,0000 0,9999 0,9990
10 1,0000 1,0000
Continued on the next page. . .
DSC2602 290
C (continued)
The probability of a success, p
n r 0,05 0,10 0,20 0,25 0,30 0,40 0,50
12 0 0,5404 0,2824 0,0687 0,0317 0,0138 0,0022 0,00021 0,8816 0,6590 0,2749 0,1584 0,0850 0,0196 0,00322 0,9804 0,8891 0,5583 0,3907 0,2528 0,0834 0,01933 0,9978 0,9744 0,7946 0,6488 0,4925 0,2253 0,07304 0,9998 0,9957 0,9274 0,8424 0,7237 0,4382 0,19385 1,0000 0,9995 0,9806 0,9456 0,8822 0,6652 0,38726 0,9999 0,9961 0,9857 0,9614 0,8418 0,61287 1,0000 0,9994 0,9972 0,9905 0,9427 0,80628 0,9999 0,9996 0,9983 0,9847 0,92709 1,0000 1,0000 0,9998 0,9972 0,9807
10 1,0000 0,9997 0,996811 1,0000 0,999812 1,0000
15 0 0,4633 0,2059 0,0352 0,0134 0,0047 0,0005 0,00001 0,8290 0,5490 0,1671 0,0802 0,0353 0,0052 0,00052 0,9638 0,8159 0,3980 0,2361 0,1268 0,0271 0,00373 0,9945 0,9444 0,6482 0,4613 0,2969 0,0905 0,01764 0,9994 0,9873 0,8358 0,6865 0,5155 0,2173 0,05925 0,9999 0,9978 0,9389 0,8516 0,7216 0,4032 0,15096 1,0000 0,9997 0,9819 0,9434 0,8689 0,6098 0,30367 1,0000 0,9958 0,9827 0,9500 0,7869 0,50008 0,9992 0,9958 0,9848 0,9050 0,69649 0,9999 0,9992 0,9963 0,9662 0,8491
10 1,0000 0,9999 0,9993 0,9907 0,940811 1,0000 0,9999 0,9981 0,982412 1,0000 0,9997 0,996313 1,0000 0,999514 1,0000
Continued on the next page. . .
291 DSC2602
C (continued)
The probability of a success, p
n r 0,05 0,10 0,20 0,25 0,30 0,40 0,50
18 0 0,3972 0,1501 0,0180 0,0056 0,0016 0,0001 0,00001 0,7735 0,4503 0,0991 0,0395 0,0142 0,0013 0,00012 0,9419 0,7338 0,2713 0,1353 0,0600 0,0082 0,00073 0,9891 0,9018 0,5010 0,3057 0,1646 0,0328 0,00384 0,9985 0,9718 0,7164 0,5187 0,3327 0,0942 0,01545 0,9998 0,9936 0,8671 0,7175 0,5344 0,2088 0,04816 1,0000 0,9988 0,9487 0,8610 0,7217 0,3743 0,11897 0,9998 0,9837 0,9431 0,8593 0,5634 0,24038 1,0000 0,9957 0,9807 0,9404 0,7368 0,40739 0,9991 0,9946 0,9790 0,8653 0,5927
10 0,9998 0,9988 0,9939 0,9424 0,759711 1,0000 0,9998 0,9986 0,9797 0,881112 1,0000 0,9997 0,9942 0,951913 1,0000 0,9987 0,984614 0,9998 0,996215 1,0000 0,999316 0,999917 1,0000
20 0 0,3585 0,1216 0,0115 0,0032 0,0008 0,0000 0,00001 0,7358 0,3917 0,0692 0,0243 0,0076 0,0005 0,00002 0,9245 0,6769 0,2061 0,0913 0,0355 0,0036 0,00023 0,9841 0,8670 0,4114 0,2252 0,1071 0,0160 0,00134 0,9974 0,9568 0,6296 0,4148 0,2375 0,0510 0,00595 0,9997 0,9887 0,8042 0,6172 0,4164 0,1256 0,02076 1,0000 0,9976 0,9133 0,7858 0,6080 0,2500 0,05777 0,9996 0,9679 0,8982 0,7723 0,4159 0,13168 0,9999 0,9900 0,9591 0,8867 0,5956 0,25179 1,0000 0,9974 0,9861 0,9520 0,7553 0,4119
10 0,9994 0,9961 0,9829 0,8725 0,588111 0,9999 0,9991 0,9949 0,9435 0,748312 1,0000 0,9998 0,9987 0,9790 0,868413 1,0000 0,9997 0,9935 0,942314 1,0000 0,9984 0,979315 0,9997 0,994116 1,0000 0,998717 0,999818 1,0000
Continued on the next page. . .
DSC2602 292
C (continued)
The probability of a success, p
n r 0,05 0,10 0,20 0,25 0,30 0,40 0,50
25 0 0,2774 0,0718 0,0038 0,0008 0,0001 0,0000 0,00001 0,6424 0,2712 0,0274 0,0070 0,0016 0,0001 0,00002 0,8729 0,5371 0,0982 0,0321 0,0090 0,0004 0,00003 0,9659 0,7636 0,2340 0,0962 0,0332 0,0024 0,00014 0,9928 0,9020 0,4207 0,2137 0,0905 0,0095 0,00055 0,9988 0,9666 0,6167 0,3783 0,1935 0,0294 0,00206 0,9998 0,9905 0,7800 0,5611 0,3407 0,0736 0,00737 1,0000 0,9977 0,8909 0,7265 0,5118 0,1536 0,02168 0,9995 0,9532 0,8506 0,6769 0,2735 0,05399 0,9999 0,9827 0,9287 0,8106 0,4246 0,1148
10 1,0000 0,9944 0,9703 0,9022 0,5858 0,212211 0,9985 0,9893 0,9558 0,7323 0,345012 0,9996 0,9966 0,9825 0,8462 0,500013 0,9999 0,9991 0,9940 0,9222 0,655014 1,0000 0,9998 0,9982 0,9656 0,787815 1,0000 0,9995 0,9868 0,885216 0,9999 0,9957 0,946117 1,0000 0,9988 0,978418 0,9997 0,992719 0,9999 0,998020 1,0000 0,999521 0,999922 1,0000
Continued on the next page. . .
293 DSC2602
C (continued)
The probability of a success, p
n r 0,05 0,10 0,20 0,25 0,30 0,40 0,50
32 0 0,1937 0,0343 0,0008 0,0001 0,0000 0,0000 0,00001 0,5200 0,1564 0,0071 0,0012 0,0002 0,0000 0,00002 0,7861 0,3667 0,0317 0,0067 0,0012 0,0000 0,00003 0,9262 0,6003 0,0931 0,0252 0,0055 0,0001 0,00004 0,9796 0,7885 0,2044 0,0698 0,0189 0,0007 0,00005 0,9954 0,9056 0,3602 0,1530 0,0510 0,0028 0,00016 0,9991 0,9642 0,5355 0,2779 0,1131 0,0091 0,00037 0,9999 0,9883 0,6982 0,4325 0,2118 0,0248 0,00118 1,0000 0,9967 0,8254 0,5935 0,3440 0,0575 0,00359 0,9992 0,9102 0,7367 0,4951 0,1156 0,0100
10 0,9998 0,9589 0,8464 0,6440 0,2046 0,025111 1,0000 0,9833 0,9196 0,7717 0,3233 0,055112 0,9939 0,9622 0,8674 0,4618 0,107713 0,9980 0,9841 0,9306 0,6039 0,188514 0,9994 0,9940 0,9673 0,7324 0,298315 0,9999 0,9980 0,9862 0,8352 0,430016 1,0000 0,9994 0,9948 0,9080 0,570017 0,9998 0,9982 0,9537 0,701718 1,0000 0,9995 0,9791 0,811519 0,9999 0,9916 0,892320 1,0000 0,9970 0,944921 0,9991 0,974922 0,9997 0,990023 0,9999 0,996524 1,0000 0,998925 0,999726 0,999927 1,0000
DSC2602 294
APPENDIX C
Bibliography
1 Thomas M. Cook and Robert A. Russell. Introduction to Management Sciences. Prentice-HallInc, fourth edition, 1989.
2 Barry Render, Ralph M. Stair Jr., Michael E. Hanna. Quantitative Analysis for Management.Prentice Hall, eighth edition, 2003.
3 Kamlesh, Mathur, Daniel Solow. Management Science - The Art of Decision Making. PrenticeHall, 1994.
4 Jon Curwin and Roger Slater. Quantitative Methods for Business Decisions. Thomson, fifthedition, 2002.
5 Steyn, Smit, Du Toit, Strasheim. Modern Statistics in Practice. Van Schaik Publishers, 1994.
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