MATHEMATICAL

OPTIMISATION'
AND
PROGRAN/I MING
t
TECHNIQUES FOR
ECONOMIC ANALYSIS
Differential Equations
Linear Prograrhming
Difference Equations

, Routh Theorem
Venkatesh Seshamani Pontryagin
Principle Hamiltonian
Obrian Ndhlovu
Functions
Theresa Mwale
 ,!
i

IVIATHENIATICAL
OPTIMISATION
AND
PROGRANIMING
TECHNIQUES FOR
ECONOMIC
ANALYSIS

Venkatesh Seshamani

Obrian l dhlovu Theresa Mwale

Department of Economics,
University of Zambia

9 e:5,2/er-r
'r^aLhbo
/

DEDICATION

To the memory of my parents-in-law,

To my mother
and
To rny wife Lalitha, my fellow-pilgrim in the Journey

- Venkatesh Seshamani

ALrthor & Publislrcr Venkatesh Seshamani

Universigy of Zambia, Professor of To the memory of myAunt HappyNkosi
Economics, Great East Road Carnpus, To my daughter Victoria
HSS Building, P.O.Box 32379
Lusaka -Obrian Ndhlovu
Zambia
Tel -t260 2tt 29047 5
Email sclash4@gmail.com

o Vcnkatesh Seshamani
To the memory of my father
ISBN 978-9982-70-056-6
-Theresa Mwale
First Edition April2014

Printed in India by Unnegh Printers

Address Fort Mumbai - 400 001. India
Tel. +9t 22 2266 0024
Web. www. urme ghprinters. com
 I
:

CONTENTS
4 12 Rank of a matrix 79
LIST OF TABTES 411 Eigen Values and Eigen vectors 80
414 Diagonalisation of a matrix 81
LIST OF FIGURES.,,.....
4 1.5 Positive and non-negative square matrlces 83
PREFACE 416 Cayley-Hamilton Theorem. 84
411 lnput-output modelsl... 84
PART l: PROTEGOMENON 1 414 Decomposable and indecomposable matrices a7
1 ECONOMTCS A5 A QUANTTTATTVE SCtENCE. .. ....... 3 479 Perron-Frobenius theorem 90
1 1 Background 3 4 20 Stochastic matrices 92
1 2 Quantification in sciences 4 4 21 quadratic forms 94
13 quantification in economics
,,5 4 )) Test ForSign Definitene\\ ..,,,...,... 95
L4 The utility of mathematical methods ,,5
15 The inutility of mathematical methods 10 DIFFERENTIAL CALCULUS 99
1 6 Conclusion L) 51 I ntroduction oo
52 The Concept of Derivative: 99
ECONOMTCS AS A SCTENCE OF OpTtMtSAT|ON......:.... 13 Non-differentiability
2l lntroduction 13 54
)) Components of optimal decision making .15
Differentiation
Economic Applications
23 Types of Optimisation under Certainty 15 56 Higher Order Derivatives 109
24 Summary of the optimisation process L9 57 Partial Derivatives and their Applications: , ,
58 Extensions to Two & More Variables : Total Differention 13
PART ll: PRELIMINARY CONCEPTS AND PRTNCTPLES 2L

SETS, RELATIONS AND FUNCTIONS 23

V,, INTEGRAL CATCUTUS
lntroduction rfi
3.1 Introduction 62 lnverse differentiation and the lndefinite lntegral
23
3.2 What is a set?... 63 Fundamental Theorem of Calculus
23
3.3 Representation of Sets 24
6.4 Rules of integration 118
34 Types of sets 65 lnitial conditions and boundary values 125
25
3.5 Algebra of Sets 66 Economic Applications 726
29
36 Ordered Pairs and Cartesian Products 61 Definite integration \27
.34
3.7 What is a Relation? 68 Properties of the definite integral 131
35
3.8 Mapping of Sets 69 Multiple inteBrals 1,32
31
39 What is a Function? 610 lmproper integrals 133
39
3 10 Types of Functions 6 11 L'H6pital's Rule 134
42
3 11 Sequences 6 12 Economic applications 135
47
3 12 Limits of a Function... . ..41
3.13 More on Sequences and Limits .. 49 PART lll: CAICUtUS TECHNIQUES. 139
51

MATRIX ALGEBRA STATIC OPTIMISATION: UNCONSTRAINED OPTIMISATION 147

55 1
4 1, lntroduction 7 lntroduction. .747
55
42 What is a vector? 72 What is optimisation? ..747
55
43 Vector Operations 13 Meaning of the signs of first and second derivatives: ..142
57 14 Unconstrained Optimisation in a Single Choice Variable: 145
44 What is a matrix? 61
45 Types of Matrices 15 Economic Application 148
62
46 Matrix Operations 16 A note on points of inflexion 150
64 17 Hessian and .lacobian determinants
41 What is a determinant? .752
70 7.4 Unconstrained Optimisation in two choice variables 754
48 Minors and cofactors
49 Properties of Determinants 79 Hessian and Jacobian Determinants in Optimisation 159
13
4 10 lnverse of a matrix , ,
110 Unconstrained optimisation in n-choice variables 161
74 762
4 11 Cramer's rule for solving simultaneous equations 7 1,1,
 STATIC OPTIMISA'rlON: CONS'TRAINED OPTIMISATION WITH EqUALITY CONSTRAINTS . .. .....................165 PART lV: PROGRAMMING TECHNIQUES 26L
8 1 lntroduction 165
8.2 optimisation of a function in two variables subject to a single constraint. . 1,67 13 I.INEAR PROGRAMMING 263
8 3 The Lagrangean Multiplier Method .. - l7-1. 13 1 lntroduction 263
8 4 Economic lnterpretation of the Lagrangean multiplier L76 13 2 What ls Linear Programming? 263
B 5 Optinrisation of a function in n-variables subject to a single constraint L78 I3 3 AdvantaBes and Limitations of Linear Programming
86 Optimisationofafunctioninnvariablessubjecttomconstraints(m < n) ..,,, 180 13 4 Formulation of Linear Pro8ramming Problems
A7 EconomicApplicalions .. .. l8'l 1l 5 Duality
13 5 Graphical Solution of Linear Programmes
STATIC OPTIMISATION: CONSTRAINED OPTIMISAIION WITH INEQUAI-ITY CONSTRA1NTS............,..........183 73 7 Nature of Linear Programming Solutions
9I Constrdints
lnequality Constrdints 183 13 8 The simplex Method (Algebraic)
92 Binding and Non binding {slack) Constraints
Constraints 183 13 9 Comparison between the Simplex and Graphical Methods
3 Conditions 10
* 9
9.4
Karush-Kuhn Tucker
Sufficien( Conditions. .
185
1A7
13 Recent developments in Solving Linear Programming Problems

9.5 Optimisation with Mixed Constraints... - 1A7 74 soMC EXTENSTONS 0F LTNEAR PROGRAMM1NG.........................
74 7 lntroduction,
10 ECONOMIC DYNAMICS lN CONTINUOUS TIME: DIFFERENTIAL EQUATIONS . 191 1,4 2 Transportation model
10 1 lntroduction 191 143 Test for Optimality
10 2 What is a differential equation? 191 L44 AssrBnment Ir4odel
10 3 First order linear differential equations 192 1,4 5 Transhipment model
L0-4 Economicapplications 199
10 5 Non-linear differential equations of the first order and first degree .. 207 INTRODUCTION TO GAME THEORY
10 6 Economic applications 205
70 7 Higher-order differential equations 201 15 2 What is a game?
10 8 Complex Numbers 215 15 3 The terminology of game theory,.
10 9 Economic applicatrons....... ....,.. ......... . 226 75 4 Types of Games
10 10 Simultaneous differential equations 227 15 5 Mixed strategies
10 11 Economic applications 231, 15 6 Game theory as a linear programme

11 ECONOMIC DYNAMICS lN DISCREIE TlMt: DIFFERENCE EQUATIONS 231

11 1 lntroduction 233
LL2 What is a difference equation? 233 16 AN INTRODUCTION TO DYNAMIC PROGRAMMING 329
11 3 First-order linear difference equations 234 16.1 A heuristic problem _129
L.l,4 Economicapplicatjons 234 LG 2 Steps in dynamic programming problem .330
11 5 Second order difference equations 24L 16 3 The classic stagecoach problem 330
11 6 Economic applications )52 1.6 4 Solving a dVnamic programming problem 332
11,1 A note on higher-order difference equations 2s3 15 5 Bellman's Principle of optimality 333

T2 DYNAMIC OPTIMISATION: AN INTRODUCTION TO OPTIMATCONTROLTHEORY 255

12,1, lntroduction 255 PART V: POSTSCRIPT 335
1,22 An lllustrative Example 255
337
72 3 Concepts RelatinB to Optimal Control I hcory 251 SOME KEY MESSAGES,
337
7) 4 Sufficient Conditions 258 I he Main Takeaway
337
12 5 Economic Application: A Mining Problem 259 l lcxibilitv in Problem Solving
Versatility of Techniques 338
Efficiency 339

\ELECTED REFERENCES. 34t

CONCEPTS AND THEOREMS NAMED AFTER MATHEMATICIANS 343

 346
I iBure 4 5 Five Sector Decomposable Model 89
EXAMINATION TYPE QUESTIONS
PAPER ONE QUESTIONS 346 I igure 5 1 Non differentiable functions 103

PAPER TWO QUESTIONS -- ... ....344 Figure 6.1. Area under a Curye 128
PAPER THREE QUESTIONS 350 I rLure 6 2: Riemann Sum 124
PAPER FOUR QUTSTIONS. 352 I icure 6 3: Possibility for a Negative Area 130
PAPER FIVE QUESTIONS 354 I rBUre 6 4: Consumer and Producer Surplus 137
PAPER SIX QUESTIONS .. ...... ............ ....356
Figure 7 1: A depiction of changing direction 143
PAPER SEVEN QUESTIONS ..358
Figure 7.2. Graphs showing different combinations of first- and second-order derivatives. L44
PAPER EIGHT QUESTIONS 360
PAPER NINE QUESTIONS 362 ligure 7 3: Maximum and Minimum Points 746
PAPER IEN QUESTIONS 364 l\Nte7 4 Total vs marginal functions r52
I igure 7 5: Three dimensional graphs for optimisation 155
INDEX 367 Figure 7 6: A diagram showing a saddle point 156
Figure 8.1, Constraint in one choice case 166
LIST OF TABLES 168
FiBUre 8 2: Constrained Optimisation
Table 11 1 The Significance of b
Figure 9.L. Binding and Slack constraints 184
Table L3 1 The Simplex Tableau
Table: problem..........
Lion pr,
le 14.1.Simple TransportaLion
Figure 10 1 Case of non-oscillatory convergence 201,

Table.
le 15.1.Prisoners' Dilemmaa lllustr
lllustration........... Figure 10.2. ArBand diagram 216
Table:
le 15.2. Game of Trade liberalisal
eralisation.............. FiBure 10.3: Sine and Cosine functions.. 218
le 15-3- Decision to Advertise
Table: tise or Not.............. Figure 10.4. A detailed Argand diagram .220
le 15.4.A general
Table: payoff matrix..
ratrix ...................... Figure 1.0 5: An oscillatory time path plot 224
Figure 1L.1: An oscillatory time path 231
Figure 11 2 Cobweb model: Case of convergence 240
usTT (OF FIGURES Figure 11 3 Cobweb model: Non conver8ence and non divergence .240
-relationships between mathematic
rre 1.1: lllustration of interrelatio Figure 11.4. Cobweb model: Case of divergence .241
Figure rtics, statist and ecor
:istics and economrcs.
Figure 11 5: Oscillatory time path 245
rre 2.1:
FiBure smooth and Noisy Cost Functions
Fl

rre 2.2: Steps in

FiBure
ll
Optimisation......
rre 3.1: Set and Subset................
Figure
rre 3.2. Disjoint 5ets....................
Figure
::
Figure 11.6: Argand diagram (Difference equations)
Figure 13 1 lllustration offeasible solution
Figure 13.2 lllustration of irrelevant constraint
241
272
274

;et ----------...---.....--.-.
rre 3.3. Complement of a Set...... Figure 13.3. case of multiple optimal solutions 275
Figure
rre 3.4. Convex and Non-convex:
Figure Figure 14.1 Source and Destination nodes 2a6
onvex Sets..............
Figure 14 2: A transhipment model 308
rre 3.5. A convex lndifference
Figure Curve...............
nce Cu
FiBure 16.1i Cost of moving from one state to another 331
rre 3.6. Non convex indifference
Figure )rence curve,,...,,.,,.
rre 3.7. Set intersection...............
Figure Figure L6.2; Route Network with associated costs available to the traveler 331

'ibutive law --...-------

rre 3.8. lllustration of Distributiv(
Figure
rre 3.9.lllustration of number
Figure ber of telements in a Union set ........
Figu rre
re 3.10. All outcomes of a throw of a pair of dice....................
Figu rre
re 3.11. lllustration of functiona
ctional forms......... 44
Figu rre
re 3.l2.lllustrations of pecial
a special form of rational functionj
I Recta
Rel ; ,,
ngula /per bola;
;;;";;; 45
rre 3.13. Graphs of exponential
Figure :ntial aand logarithmic functions ...,.. 45
rre 3.l4.lllustration of functions
Figure :tions \with and without 1imits......... 49
rre 3.15. Continuous and discontl
Figure liscontinuous functions................... 51
rre 3.16. Non-smooth and smoot
Figure smooth functions.. 52
Figu rre
re 4.1: DiaBramatical representa
esentation of Vectors...................... 57
lre
FiBure 4.2. tt4ultiplication of vectors
ectors .........,,,,,,..... 61,
rre 4.3. Multiplication of Matrice
Figure 4atrices -........,........ 66
Figure 4 4 Seven sectorcompletely decompasable model
r X

PREFACE
)

Although there are numerous textbooks on the application of mathematical methods to

economic analysis, our motivation for writing this book is to provide a treatment of the subject
from a perspective that would enable a better understanding and appreciation by potential
users in Zambian universities and other universities in this region. The various techniques are
illustrated with applications to situations with a local flavour. Many textbooks contain
illustrations drawn from North American, European and other contexts that students in this
part of the world may fail to adequately understand or appreciate And understanding the
contexts of problems is very critical for students to appreciate the value of the techniques used
in solving the problems. The relevance of the techniques appears more meaningful and this in
turn enhances the students'desire to grasp the techniques

Mathematics for economics is a required course that is tauBht at various levels in universities. lt
is taught at undergraduate freshman, sophomore and higher levels in economics major
programs; at masters level economics programs; in other masters programs such as programs
in Economic Policy Management or Defence Studies that draw students with first degrees not
only in economics but in a variety of social science disciplines; in economics courses that are
taught in other schools such as Natural Resource Economics in the Schools of Natural Sciences,
Engineering Economics in Engineering Schools, Health Economics in Schools of Medicine,
Agricultural Economics in Schools of Agriculture; and so on. Our book is designed and
sequenced in a way in which it can cater to the requirements of all these courses

The three authors of this book have a variety of experiences Venkatesh Seshamani is a
Professor of Economics who has received specialised training in all areas of quantitative
methods -mathematics, statistics, econometrlcs, programming and operations research - from
the University of Mumbai and Stanford University. He has 45 years of teaching experience, 36
of which have been in African universities. Obrian Ndhlovu has an outstanding scholastic record
from the University of Zambia and Oxford University and is currently a lecturer in the University
of Zambia where he teaches economics courses, notably mathematics and other quantitative
methods. Theresa Mwale is a very promising graduate student who is currently at an advanced
stage of completion of her dissertation

The combined insights of the authors garnered over the years have resulted in the production
of this book Feedback from various users of this book - students, teachers and practitioners -
would be very welcome in improving later editions of the book

Authors
 I

Malhematical Optimisation and Programming Techniques for Economic Analysis

Chapter 1

I ECONOMICS AS A QUANTITATIVE SCIENCE

1.1 Background
ln the course of its history since its birth as a formal discipline in 1776 with the publication of
Adam Smith's magnum opus1, economics has been subjected to a variety of concepts and
definitions. Smith himself referred to economics as political economy and defined it as "an
rnquiry into the nature and causes of the wealth of nations". This definition in fact was the title
of his work which earned him the epithet of "father of modern economics".

It was in the nineteenth century that the term economics came to be used to refer to what was
nevertheless recognised as a science. John Stuart Mill (1344)2 defined economics as "the
science that traces the laws of such phenomena of society as arise from the combined
operations of mankind in the production of wealth, in so far as those phenomena are not
modified by the pursuit of any other object". Even when Thomas Carlyle (1849)3 cynically
described economics as a "dismal science", there was an implicit acknowledgement of the
subject being a science.

Alfred Marshall (1890)a emphasised that economics "on the one side was a study of wealth and,
on the other and more important side, a part of the study of man". Lionel Robbins (1932)s
described economics as "the science that studies human behaviour as a relationship between
ends and scarce means which have alternative uses". Thus both Marshall and Robbins
underscored the nature of economics as a social behavioural science.

Economics has indeed travelled a long distance since its initial conception as political economy
and its modern-day status as a science But this has not sidelined the importance of political
economy. ln so far as it is recognised that institutional and legal frameworks, socio-cultural
of economic activity, etc. impinge on economic policy
processes, environmental impacts
outcomes, political economy that subsumes all these factors within its canvas is a
multidisciplinary subject in comparison to economics that is largely a more unified discipline.
And in an age where economic and development policy analysts repeatedly discuss issues of

t A Smith (1776): An lnquiry into the Nature and Causes of the Wealth of Nations, Methuen and Company,
London
'l.S uill 1ta++1, On the Definition oJ Politicol Economy, ond on the Method of lnvestiqotion Proper to tt, Essay V in
Essays on Some Unsettled Questions of Political Economy, John W Parker, London
r
T Carlyle 17849): Occasional Discourse on the Negro Question, Fraser's Magazine, Vol XL
)A Varshalt llSg}l Principtes of Politicol Economy, Macmillan, London.
'L. Robbins (1932): An Essdy on the Nature and Significance of Economic Science, Macmillan, London
 l

N4rtltetnitttca! Oplinrisirliorr lrncl Progranrminr'Iecltrticltrcs lor I i ottrrtttir r\tltlt,sj:

I ECONOMTCS AS A QUANTTTATIVE SCTENCE

democracy, governance, human rights, transparency, accountability and other similar issues, l..l Qu:rrrlilication in ecortonrics
such issues are more appropriately encompassed under political economy than under the more tr r.,not difficUlt to understand why economics is the most quantitative of the social sciences
limited scope of economics. ln other words, we regard political economy as a multidisciplinary lVlo.,l of economic analysis deals with Ihree variables that are directly measurable, namely,
field covering the interface between economics, politics, law, ethics, etc. and as a subject in its I rr rr cs, outputs and values
(which are products of prices and outputs)
own right as distinct from economics as a social science. lndeed, today we even witness
disagreements between economic scientists and political economists on a number of issues! rrrlnotallvariablesrelatetoprices,outputsandvalues Someofthecriticalvariablessuchas
l,r,l('s and preferences and expectations cannot be exactly quantified Also, many economic
Economics is one of several social sciences that include, among others, anthropology, lroices and decisions are of a qualitative nature As Maddala and Nelson (1974)6 state,
education, history, human geography, political science, psychology, sociology and social work. ,lr,risions such as to buy or not to buy a car or a house, the mode of travel to use or the
Social sciences differ from the natural sciences such as physics, chemistry, biology, astronomy ,), ( upation one must choose are all of a qualitative nature Economists do use a number of
and earth sciences in that while social sciences study human behaviour in a societal context, Llrr,rntitative techniques using proxy and dummy variablesT to measure such variables indirectly
natural sciences study the natural world, natural behaviour and natural condition. t.r i:xample, logit and multinomial logit models and random walk models are employed to
rr,rlyze and prcdict outcomes But the results will always be subject to errors and can never be
Subjects like logic, mathematics and statistics are also sciences but they are neither social nor
natural. They belong to a class of their own.
, \,r( I ,rs when one uses, for example, Einstein's famous equatlon e - tnc2 (energy equals
rr r.,s trrTt(.s the speed of light squared), or when electrolysis of water invariably yields for every
7.2 Quantification in sciences rrolr, ol water, a mole of hydrogen gas and a half-mole of oxygen gas in their diatomic form

Every science contains a catalogue of concepts, laws and theories. Consider some typical t.lorwiLhstanding the limitation described above, economics today is characterised by extensive
sample illustrations of these. Economics has concepts such as production, distribution and r ,,,oI quantitative methods. The application of quantitative methods in economics takes three
consumption of goods and services; theories such as theory of consumer behaviour, theory of ,rrlrrr forms: mathematical economics, statistical economics/economic statistics and
the firm and theory of distribution; and laws such as law of equi-marginal utility, law of , ,rrrometrics
demand, and law of supply. Political science has concepts such as democracy, freedom,
citizenship, state, government; laws such as Greene's 48 laws of power and Duwerger's law; lvt,rllr0matical economics ls the application of mathematical methods to represent economic
and theories such as Hobbes's theory of social contract, Rawls's theory of justice and t lr, rrr ies and analyze economic problems

Kropotkin's theory of anarcho-communism. Physics has concepts such as gravity, temperature 'L rtr,,tical economlcs is the statistical analysis of economic relationships lt is more commonly
and entropy; laws such as Newton's laws of motion, Coulomb's lnverse-square law, Kepler's
1, r(.d as economic statistics since it involves the collection, processing, compilation,
to
laws of planetary motion and laws of thermodynamics; and theorles such as Einstein's theory of
llr relativity, big bang theory and string theory.
lr ,,,nrrnation and analysis of economic data.

I , ,,1r)ltctrics is a method that combines economic theory, mathematics and statistics in the
All theories (and laws) in every science postulate relationships between concepts (specified as
il rlrlrlrr,rlrorl to the analysis of economic phenomena lt is concerned with the empirical
variables) that can be tested empirically using quantitative and qualitative techniques. To the
, 1,,r,'rrltnation of economic relationships
extent the concepts are directly measurable, they can be analyzed using quantitative
il techniques. And to the extent they can be more occurotely measured, the more exact will be
the resu lts.
, Ivt,rrlrlrla & F D Nelsoo 1L974): Analysis of qualitotive voriables, Working Paper No 70, National Bureau of
Natural sciences deal with concepts that are accurately measurable and hence are known as , rrr !{(.!carch
r,
exact sciences. Social sciences can never be exact sciences since they deal with human ,, , ,r I quantifiable substrtutes for qualitative variables that cannot be quantified For example, household
, ,, ,, ir ry he used as a proxy variable for household's standard of livinB Again, the residential status of
behaviour. Nor can all key concepts be directly measured. The degree to which concepts can be
, ,lrr rl, rrr,rV be oftwo categoriesi citizen5 and expatriates A dummy variable may be used, with value 1 for
quantified will vary from one social science to another. I rrrr value 0 for expatriates Thus, the dummy variable is assigned numerical values to capture the
, ,r' , r v,rrLrble relatingto an individual's legal status in a.ountry
 I

ECONOMICS AS A QUANTITATIVE SCIENCE N4athcnrltrctl Optinrisirlion irntJ Programming'Iechniques firr lleonomic Analysis I

democracy, Bovernance, human rights, transparency, accountability and other similar issues, 1.3 QuantificaLion in economics
such issues are more appropriately encompassed under political economy than under the more It is not difficult to understand why economics is the most quantitative of the social sciences.
limited scope of economics. ln other words, we regard political economy as a multidisciplinary Most of economic analysis deals wlth three variables that are directly measurable, namely,
field covering the interface between economics, politics, law, ethics, etc. and as a subject in its prices, outputs and values (which are products of prices and outputs).
own right as distinct from economics as a social science. lndeed, today we even witness
disagreements between economic scientists and political economists on a number of issues! to prices, outputs and values. Some of the critical variables such as
But not all variables relate
tastes and preferences and expectations cannot be exactly quantified Also, many economic
Economics is one of several social sciences that include, among others, anthropology, choices and decisions are of a qualitative nature. As Maddala and Nelson (1974)b state,
education, history, human geography, political science, psychology, sociology and social work. decisions such as to buy or not to buy a car or a house, the mode of travel to use or the
Social sciences differ from the natural sciences such as physics, chemistry, biology, astronomy occupation one must choose are all of a qualitative nature. Economists do use a number of
and earth sciences in that while social sciences study human behaviour in a societal context, quantitative techniques using proxy and dummy variables'to measure such variables indirectly.
natural sciences study the natural world, natural behaviour and natural condition. For example, logit and multinomial logit models and random walk models are employed to
Subjects like logic, mathematics and statistics are also sciences but they are neither social nor analyze and predict outcomes. But the results will always be subject to errors and can never be

natural. They belong to a class of their own. exact as when one uses, for example, Einstein's famous equation e = mc2 (energy equals
mass times the speed of light squared), or when electrolysis of water invariably yields for every
1.2 Quantification in sciences mole of water, a mole of hydrogen gas and a half-mole of oxygen gas in their diatomic form.

Every science contains a catalogue of concepts, laws and theorles. Consider some typical Notwithstanding the limitation described above, economics today is characterised by extensive
sample illustrations of these. Economics has concepts such as production, distribution and use of quantitative methods. The application of quantitative methods in economics takes three
consumption of goods and services; theories such as theory of consumer behaviour, theory of major forms: mathematical economics, statistical economics/economlc statistics and
the firm and theory of distribution; and laws such as law of equl-marginal utility, law of econometrics-
demand, and law of supply. Political science has concepts such as democracy, freedom,
citizenship, state, government; laws such as Greene's 48 laws of power and Duwerger's taw; Mathematical economics is the application of mathematical methods to represent economic
and theories such as Hobbes's theory of social contract, Rawls's theory of justice and theories and analyze economic problems.
Kropotkin's theory of anarcho-communism. Physics has concepts such as gravity, temperature
Statistical economics is the statistical analysis of economic relationships. lt is more commonly
and entropy; laws such as Newton's laws of motion, Coulomb's lnverse-square law, Kepler's
referred to as economic statistics since it involves the collection, processing, compilation,
laws of planetary motion and laws of thermodynamics; and theorles such as Einstein's theory of
dissemination and analysis of economic data.
relativity, big bang theory and string theory.
Econometrics is a method that combines economic theory, mathematics and statistics in the
All theories (and laws) in every science postulate relationships between concepts (specified as
application to the analysis of economic phenomena. lt is concerned with the empirical
variables) that can be tested empirically uslnS quantitative and qualltative techniques. To the
determination of economic relationships.
extent the concepts are directly measurable, they can be analyzed using quantitative
techniques. And to the extent they can be more occurotely measured, the more exact will be
the results.
u
G s Maddala & F D Nelson 1974): Anolysis of quolitotive voriobles, Working Paper No 70, National Bureau of
Natural sciences deal with concepts that arc accuratoly measurablc and hence are known as
Economic Research
exact sciences. Social sciences can never bo oxact iclences slnce they deal with human ' These are quantifiable substitutes for qualatative variables that cannot be quantified For example, household
behaviour. Nor can all key concepts be dlrectly moasurod, Thc degree to which concepts can be income may be used as a proxy variable for household's standard of living Again, the residential status of
individuals may be of two categories: citizens and expatriates A dummy variable may be used, with value 1 For
quantified will vary from one social sclence to another, citizens and value 0 for expatriates Thus, the dummy variable is assigned numericai values to capture the
categorical variable relating to an individual's leBal status in a country
 1

I ECONOMICS AS A QUANTTTATTVE SCTENCE lvllrlherlaticll Optinrlsittrotr rnd Programming Tcchniqucs lor lrroilorrric Arrrlysis

It may be noted that statistics itself can be studied using the tools of mathematics and such Turkish economist Dani Rodrik who aiso lectures at Harvard posed the above question in a
e
study of statistics from a mathematical standpoint is called mathematical statistics dramatic way and also gave his answer

Figure 1.1: lllustration of interrelationships between mathematics, statistics and economics Question: Why do students of economics have to know about quasi-concavity and all that in
order to improve the lives of the poor?

Answer: lf you are smart enough to be a Nobel prize-winning ecor]omist maybe you can do
without the math, but the rest of us mere mortals cannot We need the math to make sure we
thlnk straight - to ensure that our conclusions follow fronr our premises and that we haven't
left loose ends hanging in our argument. ln other words, we use math nrt because we are
smart but because we are not smort enough We are just smart enough to recognise we are not
smart enough And this recognition, I tell my students, will set them apart from a lot of people
out there with very strong opinions about what to do about poverty and development

Similar arguments are also found in Calzi & Basille (2004)10 who wrote: "Apparently, therefore,
there is (by far!) more mathematics in economics than in any of the other social sciences and
even than in more traditional scientific dlsciplines. lt is therefore all the more important for
economists to have a solid mathematical background, so as to avoid suffering from any
llri
inferiority complexes and to be able to distinguish between good and bad economics
autonomously".

But what is the utility of mathematics that makes it seem such an indispensable tool for
In this book, we deal with the application of mathematical methods. economists? The utility can be summed up in three words: clarity, brevity and efficiency. These
three aspects are not independent but interlinked.
1.4 The utility of mathematical methods
On the aspect of clarity, a very graphic statement was provided by lrving Fisher:11 "The effort of
Mathematics today has virtually become a compulsory prerequisite language that anyone
the economist is to see, to picture the interplay of economic elements The more clearly cut
wanting to be a proficient economist has to learn.
these elements appear ln his vision, the better; the more elements he can grasp and hold in his
Harvard economics professor Gregory Mankiw wrote in his biog8 in 200G the reasons why an mind, the better The economic world is a misty region. The first explorers used unaided vision
aspiring economist today needs mathematics Among the several reasons he cited were the Mathematlcs is a lantern by which what was dimly visible now looms up in firm, bold outlines.
following: The old phantasmagorias disappear. We see better We also see further"

Every economist needs to have a solid foundation in the basics of economic theory and Economics is one of those fields of knowledge where interactions and interrelationships among
econometrics. You cannot get this solid foundation without understanding the language different elements are highly complicated and hence difficult to comprehend. According to
of mathematics that these fields use, Greek economlst Stephan Valavanis, there are two disciplines economics and astronomy -
As a policy economist, you need to be able to read the academic literature to figure out
what research ideas have policy relevance That literature uses a lot of mathematics, so
1'
you need to be equipped with mathematicaltools to understand it intelligently Dani Rodrik l2OO7): Why We LJse Moth in Eeconomics, Dani Rodrik's WebloB, September 4
'o M L Calzi & A Basille \2OO4l: Economists and Mothemotics in M Emmer (Ed): Mathematics and Culture l,
Mathematics is good training for the mind lt makes you a more rigorous thinker.
Springer
tt I Firh", the theory of volue ond p,.e5, Section 10, Appendix 3, Yale
8 \tg26): Mothemqticql lnvestigotions in
Greg Mankiw's BloB (2006): Why Aspiring Fconomists Need Moth, September 15 University Press

\
 I

I ECONOMTCS AS A QUANTITATIVE SCTENCE Mrthcrniltictl Optirttisrrlion and Programming Technitlucs lirr llconornic Analysi:

which deal with phenomena where "everything depends on eveMhing erse,, 12

To give a simpre One is the well known "adding up problem". Total product produced at any time is distributed
illustration of economic interrelationships, the price of labour (wages) depends
on the demand among the factors of production What would constitute a fair distribution of the product
for labour and the suppry of rabour The demand for rabour wiil depend on how
much of a among the factors? The answer was, in order to ensure a fair distribution, each factor should be
commodity, say X, labour herps to produce. This wiil depend on the demand
for commodity x. paid a price equal to the value of its marginal product But the question, as posed by Joan
This demand in turn will depend on the incomes of the buyers. The
incomes of the buyers who Robinson (1934)14, was: "How do we know that if each factor is paid its marginal product, the
are mostly labourers depend on their wages. so we are back to wages again.
such relationships total product is disposed of without residue, positive or negative?" The proof could be provided
can be more clearly, more simpry and more briefry expressed through mathematics
than only by using the eighteenth century Swiss mathematician Euler's theorem relating to
through tediously long verbal explanations.
homogenous functlons Euler's theorem is explained in a later chapter of this book
The combination of clarity and brevity of expression makes for efficiency.
Now, it is often said, Our second illustration relates to the proof of the existence of a general equilibrium French
and quite rightly, that there is very little in economics that mathematics
can explain which economist Leon Walras 1L874fs developed the theory of general competitive equilibrium and it
cannot also be explained through verbal exposition. Then why bother to
learn and use was proved that the equilibrlum is efficient. But even though the equilibrium may be efficient,
mathematics? The answer can be provided by means of an anarogy
There is nothing that a one needs to prove the existence of the equilibrium in the first place. This proof came through
tractor can do which cannot also be done by a plough or even strong bare
hands Then why are a series of developments. Nicola Giocoli (2003)16 summarises the sequence as follows:
tractors used? lt is because for the same amount of effort, a tractor can
furrow the rand much
faster, deeper and wider than a prough or bare hands. rn other words, the "The technique most frequently used in economics for establishing the existence of solutions to
tractor is a more
efficient tool of farming than a plough or bare hands. Likewise, mathematics
is a more efficient
an equilibrium system of equations is that of searching for a fixed point of a suitably
method of economic analysis than verbal expositions. constructed function or correspondence

samuelson (1952)13 brought out this idea using another anarogy. He wrote: ,,To if we manage to transform our equilibrium problem into a fixed point problem we can easily
get to some
destinations it matters a great dear whether you go afoot or ride by a train. No wise man derive existence results for economic equilibrium.
studying the motion of a top wourd voruntariry confine himserf to words,
I
forswearing arl The application of fixed-point techniques in economics is one of the legacies of the game-
symbols. Similarly, no sensible person who had at his command both
the techniques of literary theoretic revolution brought forth by mathematicians such as John von Neumann and.John F.
argumentation and mathematical manipulation would tackle by words alone problem
a like the Nash. A well-defined sequence of contributions led to that outcome The sequence started with
following: Given that you must confine ail taxes to excises on goods or factors,
what pattern of von Neumann's first paper on game theory... , passed through his 1937 general equilibrium
excises is optimal for a Robinson crusoe or for a community subject
to prescribed norms? |
paper.. . and the 1944 Theory of Games and Economic Behaviour paper, written in
could go on and enumerate other probrems. But that is not necessary. All you
have to do is to collaboration with Oscar Morgenstern..., then featured Nash's new equilibrium concept. ..,
pick up a copy of any economic journal and turn to the articles
on literary economic theory, and and culminated in Arrow and Debreu's existence proof, which relied on game theory and Nash's
you will prove the point a hundred times over,,
equilibrium This sequence is arguably one of the most important in the whole history of
Mathematics has also had a more fundamentar rore to play in economics cconomics Besides the fixed-point techniques, the above-mentioned works offered the
Not onry has it
contributed to a more efficient analysis of economic problems; the
development of solutions to cconomists' community a score of other tools and methods that became the backbone of
some problems in economics had to wait for development/application modern mathematical economics: convex analysis, linear algebra, duality theory, etc."
of mathematics. Many
illustrations can be provided of such contributions of mathematics to economics. Here, we
mention two instances.

'' J Robinson (1934],t Euler's Theorem ond the Problem ot' Distributior, The Economic Journal, Vol.44, No 175, pp
398- 474
" L Walras 11874)t Elements of Pure Economics or the Theory of Sociol Weolth, Routlege edition, Taylor and
Francis, October 2010.
l]:.Y:::ll::11:::It@nometrics: An tntroduction ro Maximum Liketihood Methods, McGraw Hiil and company
p. 5a muerson
\L95) )t Economic Theory ond Mothematics _ An Approisol, American Economic Review, 42.
"'N Giocoli(2003):FixingthePoint:theContributionofEarlyGameTheorytotheToolboxofModernEconomics,
lournal of Economic Methodology, 10:1, pp. l.
- 39
 I

10 I ECONOMTCS AS A QUANTTTATTVE SCTENCE

/ N4tthematical Oplirnisalion and Programming Techniqucs lirr llconomic Analysis

The developments delineated in the second illustration above involve tairly high-level
mathematics and hence their detailed expose is beyond the scope of this book.
lcst he be misunderstood that he was attacking the use of mathematics in economics, Krugman
lollowed his write-up some ten days later with a clarifying note.2o He said that mathematics in
1.5 The inutility of mathematical methods |conomics could be extremely useful but economists should have it as their servant and not as
ll)eir master.
while the utility of mathematical methods in economics has been well appreciated,
disillusionment with what has probably been an over-zealous application of the methods has Krugman rightly pointed out that good mathematics does not imply good economics. Equally,
also crept among economists (many of whom were themserves prominent appriers of you can have great work in economics with little or no mathematics. As an illustration of the
mathematics in economics!) leading often to strident commentaries on the risks and limitations lormer, he stated that the mathematics of real business cycle (RBC) models is much more
of doing so. As the saying goes, one can have too much of a good thing And this seems to be olegant than New Keynesian models, but it does not make them less silly. As an example of the
the case with the use of mathematics in economics. over the past decades, there has been a latter, he cited Ackerlof's market for lemons" that had virtually no explicit mathematics but
trend of increasing 'mathematicisation' of economic theory and, ironically, in tandem a rising was nevertheless transformative in its insight.22
trend in the criticism of this trend! The basic fear seems to be that there is a tendency for
overuse, tantamount to misuse, of mathematics leading to a poorer instead of a heightened ls it then possible to titrate the right dosage of mathematics in the context of any economic
understanding of economic realities. ln a scathing attack, Robert Kuttner (19g5)17 wrote: theorizing and analysis? Can one suggest a certain quantum of mathematics use as the most
"Departments of economics are graduating a generation of idiot savants, brilliant ,rppropriate? ln our view, it may not be easy to offer even rules of thumb here apart from
at esoteric
mathematics, yet innocent of actual economic life.,, 18 providing a heuristic description of the dilemma that poses this challenge. And this dilemma is
the age-old conflict between rigor and realism.
ln a candid critiquels he wrote in the september 2,2009 issue of the New york Times, noted
Nobel prize-winning economist paul Krugman said that just when economists had begun In a well-known article,23 D.G. Champernowne wrote: "Unfortunately for the cautious
to
believe that in the real world they had things under control and that the central problem economist, his economic models will be judged according to the degree in which they appearto
of
depression-prevention had been solved, the global crisis, unforeseen by anybody, occurred in be relevant to the real world; so that in avoiding the appearance of being wrong, he may yet
2008 and everwhing fell apart. Krugman attributed this to the economic profession gorng appearto be silly in publishing a long article whose relevance to any practical issue seems to be
astray "because economists as a group mistook beauty, clad in impressive-looking mathematics superficial This danger of manufacturing mere "toys" is especially great since the assumptions
for truth" and "the central cause of the profession's failure was the desire for an ail-
which are most convenient for model-building are seldom those which are appropriate to the
encompassing, intellectually elegant approach that also gave economists a chance to real world" Champernowne further went on to emphasise that: "The ability to judge the
show off
their mathematical prowess". relevance of an economic theory and its conclusions to the real world is but rarely associated
with the ability to understand advanced mathematics".
Krugman elaborated this point further. He wrote: "unfortunately, this romanticized and
sanitized vision of the economy led most economists to ignore all the things that can go lhe use of mathematics tends to warrant assumptions that are convenient for analysis but may
wrong.
They turned a blind eye to the limitations of rationality that often led to bubbles and not be appropriate to the real world. One can find 'learned' articles on say the theory of the
busts; to
the problems of institutions that run amok; to the imper-fections of markets especaally firm which begin with the assumption of a continuum of firms or of a number of firms
-
financial markets - that can cause the economy's operating system to undergo sudden,
unpredictable crashes; and to the dangers created when regulators don't believe in regulation,,.
'" P Krugman l2)Ogal Mothemotics ond economics, New York Times, September 11, 2009,
'' George Ackerlof (1970);Ihe Morket for lemons: Quolity lJncertointy ond the Morket Mechonism, Quarterly
lournal of Economics,84(3), pp488 - 500 The paper discusses information asymmetry and is a seminal
(ontribution to the economics of information. ln 2001, Ackerlof shared the Nobel prize in economics for his
p.l4 44. contribution, with Michael Spence and Joseph Stiglitz
applicable to top American " Debates and disagreements over several substantive contents of Krugman's first article continue. See, for
large a dreaded subject and instance, J H Cochrane l2O1-1-): How did Poul Krugmon get it so wrong?, Economic Affairs, pp,36 - 40 However, in
our view, the observations Krugman makes on the use of mathematics are largely valid
'' D G Champernowne (1954): The lJse ond Misuse ol Mothemotics in Presentinq Economic Theory, Review of
l.conomics and Statistics, Vol. 35(a), pp.369 - 372-
r
\

L2 ECONOMICS AS A QUANTITATIVE SCIENCE Vlalhcmilticrl C)Plirrisirtiorr lrrrd Pro-qrrmrninq fcchuiqucs lrrt lronotttic i\ttal-fsis 1-l

corresponding to points on a real line! These theories will undoubtedly score high on rigor and
generality but what can one say about their practical relevance? Chapter 2
The conflict between rigour and realism was best brought out by Nobel laureate Tjalling
Koopmans (19571'z4: "As we strive for greater rigor and precision in the formulation of 2 ECONOMICS AS A SCIENCE OF OPTIMISATION
postulates and propositions, the inadequacies and lack of realism of these postulates are
2.t Introduction
thereby made to stand out in stronger relief. As we succeed in incorporating one aspect ofthe
reality in our models, our failure to incorporate other aspects becomes more apparent". ln Chapter 1, we referred to Robbins's definition of economics as the science that studies
human behaviour as a relationship between ends and scarce means that have alternative uses.
ln sum, there is an inevitable trade-off between rigor and realism, between convenience of As we would be already familiar from ECON 100 series, the two critical terms in this definition
analysis and correspondence to reality and between the beauty of the mathematical models are 'scarce' and 'alternative uses'.
and the truth content of the models. The ingenuity of an economist lies in deciding on what All relationships between ends and means need not in principle be characterised by scarcity.
constitutes the desired trade-off. Conceptually, there could be four possible relationships:
1Many ends, many resources;
7.6 Conclusion 2Few ends, few resources;
Whether one likes lt or not, the use of mathematics is here to stay. lt is a language in which 3.
Few ends, many resources;

every student of economics must gain at least a modicum of proficiency if he/she has to 4.
Many ends, few resources
It should be obvious that it is only in the last case that one is faced with the problem of scarcity.
communicate with fellow economists and share ideas.
Scarcity then simply means that there are just not enough resources to meet all the ends, the
Some of the most advanced mathematical methods are used in economics today. However, ends here being goods and services that satisfy human wants. ln such a case one has to make a
choice regarding what ends to meet and to what extent, with the limited resources that are
such highly advanced mathematics may not be necessary for understanding the basic principles
availa ble.
of economics. Most of basic economic theory can be well-grasped by a student who has also
grasped well the basic principles of set theory, calculus and matrix algebra which are covered in The above kind of choice, however, assumes that such a choice is possible to make. Suppose
the ends that we want to meet are various goods such as cooking oil, sugar, clothes, etc. Now
this book
imagine that the resources available are only in the form of ice boxes. Obviously, ice boxes can
Although economics is a science that is amenable to a high level of quantification, it is a subject be used only for one purpose - to store ice. They cannot be used for anything else. ln other
words, they have only one use; they cannot be put to alternative uses. Hence we have to
in which a quote attributed to Einstein would well hold: "Not everything that can be counted
assume that resources have alternative uses. Money is one such resource. So are land, labour,
counts; and not everything that counts can be counted". Never sideline the qualitative and and other factors of production.
eccentric aspects of human behaviour.
The choice of putting the resources that one has to alternative uses is made with some
Finally, while learning mathematics, one can draw inspiration from another famous quote of objective in mind. The objective depends on the agent making the choice lf the agent is a
Einstein: "Politics are only a matter of present concern. A mathematical equation stands for consumer, the objective is to maximise satisfaction or utility from the consumption of goods
and services. lf the agent is a producer, the objective will be to maximise revenue or profit from
ever".
or to minimise cost of the production of output of some good(s). lf the agent is the
government, the objective is to maximise the overall welfare of the society or citlzens.

ln brief, the objective of every choice-making agent is to maximise "profit" or minimise "cost" in
a larger sense of the terms. Anything that is consldered as benefit (utility, revenue, social
welfare) is profit and anything that is considered as suffering (disutility, expenditure incurred in
producing a good, labour expended to earn an income) is cost.

The maximisation of profit or minimisation of cost is called optimisation. ln simple language,

'oTC Koopmans(1951}:ThreeEssqysonthestoteofEconomicscience,Wrley,rntlsor,,,N(,wyork optimisation is making the most of given resources.
/'

L4 I ECONOMICS AS A SCIENCE OF OP rMr\AilON l\4rtherrrtical Oplinrisrrtiorr rrrtl Progrrnrnring'l'echniqrres lirr l,tlurrtric Anrlvsis 15

optimisation thus implies making the best choices possible in given situations. And making .! l {.onrponents of optimal decision-making
decisions is tantamount to makanB, decisions on what actions to take A set of actions that can
tlr,, lr,,iic,,r'ing are the components of optimal decision-making in a situation of certainty
be taken is called a programme On could decide to choose among several programmes and the
process of making the decision is called programming- The obvious aim is to choose the optimal l\ r.l1r.,l.rr1;maker with eoals :a consumer who wants to maximise utillty; a
programme, that is, one that optimises (maximises or minimises as required) the objective on | ,,' i r, , ,,.,ho wants to maximise output or minimise costs; a :eller who wants to maximise
hand. This is the reason that another expression used for optimisation is mothemoticol ,.rlr,', ,r,r irlvestor who wants to maximise returns; a worker who wants to maximise wages or
progromrntng rrinirr ,.l {lisutility from labour, etc
Economics in its quintessential nature involves optimisation ln other words, it can be regarded A lcn.,,',q .ret of alternative actions: a consumer has to decide what goods to consume and in
as a science of optimisation. whar c;uantities; an investor has to decide in what projects to invest and how much; etc
Now, decision-making can be done in one of three possible situations: certainty, risk and A i!!-.r,i-p_essibleoutcomes: different sets of quantities of goods and services consumed will
uncertainty. Situations of certainty are known as deterministic situations while situations of risk viclC dilicrent levels of utilities to the consumer; different portfolios of investment in various
are also called stochostlc situations. proiecis will yield different aggregate returns to the investor; etc
Deterministic situations are those in which the outcomes of alternative courses of action are A1 qi,ifit
knovrn with certainty. lt is then possible for the decision-maker to evaluate these outcomes and ,rr-c1qre1!r 1!91!gp!9111q[91: a utility function which is a function that shows how utilities are
choose that course of action to which the outcome with the highest valuation corresponds This lien.iaii.rt by the quaftities of goods and servlces consumed; an earnings function which shows
of course is the optimal outcome, and when a decision-maker selects such an outcome, he can lloflr i:di'irings are Benerated by the levels of investment in various projects; etc
be considered to have made an optimal decision
/\dg{tri nl/yhich is the action : that decision is the best which optimises the value
Where the number of alternative actions and outcomes is pretty l.rrgc, there are special of th:.rtrjr:ctive function: a consumer will decide to maximise his utility function; a producer
techniques available for evaluating them. One such technique is linear programming which is lvill 11.,,:ici: to minimise his cost function; an investor will decide to maximise his earnings
explained in Chapter 13. Itlrri:iirll. r'1'c
When the outcome in any situation is not certain, we can have risk or uncert.rinty The technical '1.3 l l pr"s of Optimisation under Certainty
distinction between risk and uncertainty was made by the American cconornist Frank Knight,
best known for his book Risk, Uncertointy ond Profit. A situation whc.ro thc outcome is not /\ ij:: iii[r,r-i']raker may be required to tal(e decisions under various circumstances ln many
known but the probabilities of alternative outcomes are known is a risk or slochastic situation. !i1i:i.'. r:r\ '.rire results of the decision taken by a decision-mal<er are affected by the actions of
Where neither the outcomes nor the probabilities of alternative outcorncs are known, the rrtlr:r, r i iri! is the crux of Gdme Theory which deals with optimisation as a game involving two
sitltation is described as one of uncertainty Knight explained that the distinction between risk 1)r i,l.ii i irlayers An individual player or decision-maker has to take into his reckoning the
and uncertainty was a significant one- ln the context of the theory oI llre firm, for instance, ni,.r,, ir.: ::ctionsof otherplayerswill haveonhisdecisions Agameishostileif Ihe decision
uncertainty could give rise to excess profits for a firm that could not be eliminated by perfect rrr,rli:: i: c,rnfronted by rivals who possess both the motive and the capacity to take such
competition. r.1ir,r:, ;ri',r,'ill minlmise the gains or payoffs to the decision-maker But this need not be always
tho r-,s,. ii'r iJ'tany situations, the game may be non hostile in the sense that the declsion-maker
The gist of the above distinction is that while risk is measurable, uncertainty is not measurable.
or i;i:rvr- is confronted by an indifferent opponent ln such a situation, the game effectively
And yet, economic decision makers are occasionally faced with satuations of uncertainty. How
rrrvr;lr.r: r,r,i,r one player, namely, the decision-maker in question
can one make optimal decisions in such situations? The standard techniques provided by
mathematics and statistics cannot readily apply to such decision making, A number of S1111-r'1,1.ii,r:re are two major producers of a certain product in the market. Each producer will
alternative rules for decision-making have been proposed and the choice among these rules lry .i;1r, ihe maximum space for himself/herself in the market The rival producer will try to
i1-r

would depend on the decision-maker's attitude and psychological state of mind. Such rules do ilrr'l;:,:r',re Thus the two players are rivals and the actions of one player will be hostile
include, among many, Maximin Rule, Maximax Rule, Hurwicz Rule, Minimax Regret Rule, and La tow,ricls I ie other player.
Place Rule, The discussion of these rules is beyond the scope of this book which will deal with
SupllD!1r I
riudent has been given a certain monthly allowance to spend on various goods and
optimal decision making only in deterministic decisions that can be carried out with the help of
rerviils llr: will
decide to do so in a manner that will maximise his utility function But this will
mathematical techniques Decision-making in situations of risk and uncertainty require
not be,tf any concern to his fellow students who consequently will not do anything to
knowledge of statistics, in pafticular the theory of probability.
undern-rine his actions aimed at his personal utility maximisation This is an example of a non
hostile qaile which de facto involves only one player.
 \l

16 ECONOMICS AS A SCIENCE OI OI'IIMI5N IION Malhematical Optimisation and Programming Techniques for Economic Analysis ll

We deal with game Theory in Chapter 15 All the earlier chapters deal with optimisation by a ln other optimisation problems involving minimisation, the constraint imposed on the
sole decision-maker concerned variable may be of the (>) type. This imposes a lower bound on the value of the
va ria ble
Now, the techniques of optimisation available to a single decision mal(er themselves are of
dlfferent kinds, depending on a number of factors These are: ln all the above type of cases, the techniques used to solve optimisation problems with equality
constraints may not be the most efficient ones to solve such problems involving inequality
Choice variables: Suppose a flrm produces a single product and wants to maximise its profits
constraints.
from the production and sale of that product Here, the profit maximisation will be determined
by the firm's choice of the level of output of that single product lf, however, the firm produces Time: ln a paper that he published in 1954 in Oxt'ord Economic Popers, Nobel laureate Sir iohn
two or more products and wants to maximise its overall profit from the production and sale of Hicks distinguished between two categories of firms, stickers and sndtchers The snatchers were
those products, then the profit maximisation will be determined by the firm's choice of the those interested in maximising immediate expected profits by snatching whatever was

levels ofoutput ofthose products The output ofthe product is thc choice or decision variable immediately available, irrespective of consequences. The stickers, on the other hand, would be
and the technique of optimisation (maximisation in these examplcs) will depend on the number wary of the longer-term possible adverse consequences of snatching now by losing consumer
of choice or decision variables The number of choice variables involved can be one, two, three goodwill, and hence would opt for less than maximum profits now in order to maximise long-
and, in general, n term expected profits. Reflecting on the number of banks that came to grief in Zambia in the
1990s in the post-liberalisation period, one may feel that they probably reaped the desserts of
Constraints: Suppose the firm in the above examples wants to maxinrise its profits but with the
behaving like rapacious snatchers.
condition that it has a fixed amount of money to spend during thl pr.riod of production. Then,
the firm is not free to choose any level of output(s) but only such lovc:l as is feasible to produce The distinction between stickers and snatchers is based on their respective perspective on time.
under that condition Such a condition is known as a consltoit)l or side relation lt is also The snatchers were interested in instantaneous profit maximisation; time did not enter into
referred to as a restraint or subsidiory condition An optimisalion problcrn can have one, two, their model of optimisation. Such a model is called a stotic model and the corresponding
three, and, in general, m constraints An essential requircnr('nt {or solving a constrained optimisation static optimisation. lt is a model without time. The stickers, on the other hand,
optimisation problem is that the number of constraints must br, llwor than the number of were those who realised that their optimising decisions today would affect their decisions
choice variables ln other words, m < n tomorrow. Time entered the model and the objective was to maximise profits over a period of
time rather than at any given point in time. Such a model involving time is called a dynomic
An optimisation problem which has no constraints is called /ree o( un( onslroined optimisation,
model and the corresponding optimisation, dynamic optimisation.
while a problem involving constraints is called constrained oplrnrs,rlron The distinction
between the two types of optimisation has been humorously broLryilrt otrL by one writer in the It is, however, important to understand that, while a static model is a timeless model, a model
following words: "The essence of an optimisation problem is 'c.rtr lrirrli ,r black cat in a dark in which time enters need not be dynamic. As Silberberg and suen (2001) in their book Ihe
room in minimat time'. A constrained optimisation problem corrr',,1roncls to a room full of Structure of Economics: A Mothemoticol Anolysis stote.'"The fundamental property of dynamic
furniture " The complexity of the optimisation technique will depcrrrl orr the number of choice models is that decisions made in the present affect decisions in the future". They further go on
variables and the number of constraints The sirnplest case is unconslr,rinod optimisation of an to explain that dynamic optimisation problems are those "where decisions are linked, that is,
objective function in one choice variable At the other end of the rp(,ctrum will be the case of where a decision in one time period affects the level of some relevant variable in the future ln
optimisation of an objective function in n choice variables sub1ect to rn ronstraints (m < n) that case, simple replication of past decisions will not be optimal; each decision imposes an
'externality' on the future. lt is only then that a problem becomes truly dynamic".
Further, a constrained optimisation problem may involve constraint., in thc Iorm of equations (

(> or < ) The above concepts will be elucidated later especially in chapter 12 on Dynamic optimisation.
- )or inequalities Solving constrained optimisation problems with inequality
constraints requires techniques different from those required tor solving constrained Given a static optimal solution, one may analyze how a change in any choice variable may affect
optimisation problems with equality constraints the solution. Such analysis once again is not dynamic but is referred to as comporotive stotics. lt
ln order to understand the difference between equality and inequality constraints, suppose a is comparative because we are comparing the initial static solution with a new solution arising
firm has to produce output of its product subject to the constraint of a fixed amount of man from the change. But it is static because one does not analyze the time path ofthe change: how
hours of labour available durlng the production period. Then, such a constrainl places an upper exactly the change took place The well-known Keynesian 45" degree diagram showing shifts in
bound or ceiling on the quantity of labour that can be used to maximise profit from the equilibrium resulting from shifts in aggregate demand, covered in any intermediate
productionandsaleoftheproduct lnsuchacase,alloftheavailablr:labourneednotbeused macroeconomic course, is an example of comparative statics.
in maximising the profit function lf it is used, labour becomes a binding constrainU else it is Dynamic models of optimisation are of two kinds, depending on how time is treated. For
s/ock at the optimum example, suppose the decision variable is price. Then price may be viewed as changing over
 Nlrrthenratical OPlilrislrtiorr il[rl Progrenrrnirtg Tcchnic[rcs l()r I 1 ()ilr)ilrc Anitl_Vsis 19
18 ] t.ONOMICS AS A SCIENCE OI- OP I IMI\A I I( )N

i ,l,t, ( lrv(,\' l ( ()ilomists may at times wish to optimise more than onc objective function
time in two ways. The price may bo .djusted continuously over time or it may be adjusted over rrrrrll,rrrr.ou,,ly optimising multiple objectives is very difficult, if not impossible The main
successive periods of time, whcrc cach period may be a year, a quarter, etc ln the first case, ,, r ( )il loT lhrs is that the multiple objectives may be conflicting and it may not be possible to
time is a continuous variable and the optimisation wjll involve dit't'erentiol equations. ln the ,rl,rrrvcthevalueof oneobjectivefunctionwithoutdiminishingthevalueof another Students

I second case, time is a discrete variable and the optimisation will involve dit'ference equations
Linearitv: lf, in a given optimisation problem, the oblective function to be optimised and allthe
,,,,ronon'ricswouldimmediatelyrecalltheconceptof Poretooptimalitywhichdefinesprecisely
, lr ,r situation A solution would be deemed Pareto optimal if none of the individual objective
l,rr( lrons can be improved without impairment of some other function
constraints are linear, then one can adopt the techniques of linear optimisation Linear
programmlng and some of its extensions dealt with in later chapters of this book are examples ,\ r)llrer prominent illustration is the achievement of Nosh equilibrium (after the economics
of linear optimisation t,,lr(,1 laureate John Nash who is basically a mathematician) through a process of bargaining

On the other hand, if the objective function and/or some or all of the constraints are non, ,rrrong n decision-makers that makes all ofthem equally happy or unhappy
linear, then you have a problem of non linear optimisation Besidcs the standard calculus tItrcatmentof multipleobjectiveoptimisationsisbeyondthescopeofthisbook
techniques, a number of seorch or iterotive methods are avail.rble to solve non-linear
optimisation problems. These include Conjugate Gradient, Doublc Dogleg, Neldor-Mead, .'. ,l Summary of the optimisation process
Newton-Raphson, Quadratic and Trust Region methods among sovr-ral others Discussion of
these methods is beyond the scope of this book I t ,,rrc 2 2: Steps in Optimisation on next page

Smoothness: Most of the optimisation techniques that are discrrs:r,rl rs well as those that are
not discussed in this book assume that the objective function to l;r,optimised is smooth ln
other words, it has no discontinuities or irregularities 2s Consequr,nt ly, I lrt.sc techniques cannot
be applied if the objective function is non-smooth. ln Figurr' ,r t, wt. provide a simple
iii illustration of a cost function that is smooth and non smooth ot noi\y
lr Figure 2 1: smooth and Noisy Cost Functions

rii

Smooth cost function "Noisy" cost function

Techniques based on Perturbotion Theory and Pottern Seorch methods are used to handle
non-smooth functions. Their treatment again is beyond the coverage of this book.

25
Technicaliy, as will be explained in the chapter on Differentiation, the function mu!t hc twrce differentiable
V

ECONOMICS AS A SCIENCE OF OPIIMISATION

Optim isation
(maximisation or _/
minimisation

,/

Formulation of
objective function
./

Stipulation of constraints

Stlpulation of variable //'

bounds (upper and lower)

,/
 Mathematical Optimisation and Programming Techniques lor IJcortorlie Analysis

Cha pter 3

I SETS, REI,ATIONS AND FUNCTIONS

I I lntroduction
llrr, principal objective of this chapter is to introduce the basic set-theoretical taxonomy that
wrll be adopted throughout the text. We therefore start with an intuitive discussion of the
rriltron of sets, covering its various types as well as representations, the basic algebra of sets,
,'rr[,red palrs and Cartesian products, and relations After the quick excursion to set theory,
lilt( tions are introduced extensively Since constraints in economic studies of optimisation are
rr,rr,rlly represented by sets and the critical issues of efficiency and economic growth are usually
!.Ir r:sented by functions which take on various graphical representations, the understanding of
tlr,,,c aforementioned concepts is imperative to the full grasp of the idioms used in subsequent
toprr:s of this book This chapter will therefore provide the necessary tools required for
lxllnsive understanding of the subsequent chapters. We assume here that the reader is

l,rrrriliar with the elementary properties of the real numbers

t /. What is a Set?
lil ilricroeconomic studies of production theory for instance, we use concepts such as the
lrrorluction set and input requirement set. The input requirement set ls defined as thatsetof
rnprts required to produce a given Ievel of outputs. The production set is therefore a set of all
tlrr, ;rossible outputs that can be produced using the input requirement set. ln analyzing policies
ril tlr(' world today, we are concerned with alternative policies that will help us achieve a certain
rlt ol goals, for instance in monetary policy strategies such as inflation targeting, we are
I r,n( (,rned with the set of instruments to use to achieve our objective. What then is a set?

',tnrl)ly put, a set is a well deflned collection of objects26. These objects are called the elements
rl llrr, set ln the words of Georg Cantor, the great founder of abstract set theory in the 1870s,
'',r ,,r'l is a Many which allows itself
to be thought of as a One " Therefore, in many texts for easy
r,rrr;rrehension, sets are usually denoted by capital letters such as A,B,C....X,Y Z,the
rlIrronts are usually denoted by small letters such as a,b,c .... x,y,z. The symhol € is used to
rh,urte "belongs to" or "is a member of" while the symbol G is used to denote "does not belong

'llr, rrotion ol "object'is leti undefined, thaL is, it can bc givcn rny rncrning Howcvcr, lhe ''objects" must hc
l,,1,r,,rllyr/irtirr,r;rri,rhableThaLis,if aqnd,b arctwoobjects.a=banda+bcanrotholdsimultaneously,andtllc
r,rr, [! [l "cithcr a: b or a+ b''ist tautology
 /-

24 | ,r-r, *rro.,ors AND FUNcToNS Mathematical Optimisation and Programming Techniques for licottotuic Ana!ysis I
25

Example 3.1 \ [r:r isaneuensingLedigitinflotionnumberevera.ttoinedinZum.hia], where x

r',lrrcsents the element, in this case the even inflation number. The colon symbol ":" is read as
Suppose,4 is the input requirement set of a certain production technology with only two
,,rrr h that".
inputs, capital, denoted by k and labour denoted by l. Show using set notation that I and
k are elements in,4. I I .l Venn-Euler diagrams:

t,k e A ; read as landk are elements that belong to the setA lf this statement is I lrr,elements are represented by points in a closed figure such as a circle, a square or an ellipse.
wrong,thiscansimplybewrittenasL,k€. A,readasLandkdonotbelongtoA. lf only A V0nn diagram must contain all the possible zones of overlap between its curves, representing
I belongs to A and k does not belong to A or is not in A, this can be written as, I e A, 'll
(ombinations of inclusion/exclusion of its constituent sets, while in an Euler diagram, some
ke A r,,rres might be missing. Thus, a Venn diagram is a restrictive form of an Euler diagram. Since
rlr(, essence is to show the overlapping (or non-overlapping) of all the sets, involving both the
tr The above example can be generalised to say that if ,4 is an input requirement set with n
Vr,rrn and Euler diagram concepts provide a more explicit representation of sets. An example of
elements, that is, xr, xz. x3 ... ... . xn, then this is denoted in set notation as x7,xz,x3 ... ... . xn e A.
,r Vcnn-Euler diagram consisting of 3 overlapping sets is shown in Figure 3.1 below:
It means the elements listed on the left are members of the set given on the right.

I r;itrre 3.1: Venn-Euler diagram for three sets

3.3 RepresentationofSets
Sets can generally be represented by three methods: the tabular form or Roster method, rule
method and the Venn-Euler diagrams method.

3.3.1 Tabular form or Roster method:

With this kind of representation, all the elements of the set are explicitly listed within the
braces. This implies that if you describe a set, for instance, let M be a set of all the goals of
monetary policy in Zambia, this means all the goals have to be listed within the braces
regardless of their number. Assuming low and stable prices(p), high employment(e), economic C
growth(g), stability of financial markets(f), interest rate stability (i) and stability in foreign
exchange markets(m) are the main monetary policy goals, this can be represented in tabular
lormas:M : {p,e,g,f ,i,m}.
Similarly, if your set involves numerical values, such as a set X representing all single even digit
values of inflation ever attained in Zambia since independence, this can be represented
'.1
4 Types of sets
y: {2,4,6,8} llnite set: This is a set with a finite number of elements; otherwise, the set is called an infinite
^r. r
3.3.2 Rule method:
'.r.1 For instance, lA = {x:x ts anaturalnumber} is an infinite set
wlrile B : {Sunday,Monday,Tuesdoy} is a finite set.
This is more of an implicit way of representing sets and usually more convenient compared to
the tabular form. A set is indicated by stating the rule satisfied by its elements and written in t mpty set or null set: This is a set which has no elements. lt is denoted by { } or @. An empty set

braces. Consider the examples in section 3.3.1 above, the sets M and X using the rule method rk,cs not mean it does not exist, it represents emptiness.
can be represented as follows: \ilbset: Set,4 is said to be a subset of I if every element of / is an element of B. We denote
M ={mi m is o monetory policy goal}, where the sym bol m represents the ele me nt, that is, a ny tlrr relationship by the symbol c and write.4 c B. Note that a null set is a subset of every set'
element that was earlier listed in the braces. Propersubset: Set.4 is a proper subset ofanother set B ifevery element ofz4 is an element of
ll and the set B contains at least one element which is not an element of .4' The tyrnbg1i

\
)
SI:'[S, RELAIIONS AND FUNCTIONS
I\l,rtlr(rrrirticat Optirnisrtiou and Progrrurming Tcchniques tir lrcotrrtttic Attitlysis 27

i,s=rrl I ' l,t lrrilll \(.tS

usedtodenotetherelationshipofpropersubset.Forinstance,/:{2,4,6} isapropersubset

)O
B = 12,4,6,8,j. Subsets are represented using a Venn diagram as shown in Figure 3.1

Figure 3.1: Set and Subset

ltnlvrt\ul or porent set: this set consists of all the possible elements. ln a Venn-Euler diagram, it
.r,rr r r,, uI ,rll the elements within and outside the sets (circles) but within the box. lt is a super
r r ,,,r,,r,,tr11,, of all the sets under consideration The universal set is normally denoted by the
tr,r rr,l I I 0r [ (See Figure 3.21. For instance, when we discuss the set of letters in some words in
I rr11lr.,lr, llrr, universal set is the set of 26 English alphabets.

t tnt,l.'nrcnt ol a set: lf the set A is a subset of the universal set U, then the complement of A
The empty set is a proper subset of every set
@ willr r.,,1)('ct to U ls the set of all those elements of U which do not belong to A, denoted as A'
',r ,l' Allornatively, Ac = {x: x e U and x G,4. }.Therefore the complement of
Equol sets: Sets ,4 and B are said to be equal if they contain the same elements, denoted any set is the
A= B. -' I , ,l .rll I hose elements that do not belong to the particular operation of the set but belong to
rlr,,rrrrrvlrsal set Figure 3 3 below shows,4'
Equivolent sets: The number of elements in one set must be equal to the number of
in another set. For example, sets /4 = {1,2,3} and g = {A, B, C] are equivalent sets Equal sets
t tilil,, l l Complernent of a Set
are equivalent sets but equivalent sets are not equal sets.

Ordered sets: A set is an ordered set if the order of elements is considered. This implies that if a
set is S: {1,2,3,4,5}, then it is different from S'= t2,t,3,4,5} though containing the -same
elements.

Similor sets: Sets are said to be similar if there is one-to-one correspondence established
between ordered sets
r rrvtx set: lf we let x € S and y € S and 2 € [0,1], and if [rlr + (l - l)yl € S, then we say

s1: [r, z r' +] rlr rr ,,r'l S is a convex set This implies that for a set to be convex, any tw l ,Ull"
,t rrrrrt lie entir-e_ly_within that set. --'-
$++{}
s2={4'3'2' 7}
I t11rrrr,3 4 Convex and Non-convex Sets
A simllar set is an equivalent set but a stronger version of an equivalent set. ln a similar set, the
elements are ordered, so the first element in S, must correspond to the first element in S, and
so on, but in an equivalent set, there is just a one to one correspondence regardless of the
order. That is to say, similarity is stronger than equivalence.

Disjoint sets: Two sets,4 and B are called disjoint sets if they have no common elements. For
example, A = tl,2,3,4j and I : {6,7,8,91are disjoint. They are depicted in Figure 3.2
r
I

2a I sL rs, RELATTONS AND FUNCTTONS Mrtllemilticrl Optimisation and Prtrgramming Techniques lirr l,)rorrortic Artaly:is 29

closed and open sets: ln Figurc 3 4 above, point z is a boundory point where as x and y which I rgure 3.5. Non-convex indifference curve
lie strictly within the set arc interior points A set which includes the boundary points in

I
addition to the interior points is called a closed set A set which includes only the interior points
and not the boundary points is an open set.

It is important to note that the concepts of being open and closed in the context of sets do not
correspond to their being physically open and closed. As we shall see in Chapter 13 dealing with
Linear Programming, the feasible region of a linear programme may appear physically open but
is always a closed convex set.

ln economlcs, the concept of convexity is used in consumer theory when drawing the
indifference curve, that is, it must be drawn in such a way that it conforms to the notion of the
convex set This is because the principle of diminishing marginal rate of substitution uses the
mathematical notion of a convex set.
lhis function does not obey the assumption of convexity and cannot be used to depict
Figure 3.5. A convex lndifference Curve rrrdifference curve. The counterpart of indifference curves is isoquants used in production
theory which also must obey convexity for the various assumptions to hold.
i

il.5 Algebra ofSets

li
l-lnion of sets: take two sets.A and B, the union of the two sets denoted by A U B , consists of
,rll elements in,4 and all elements in B and all elements comrnon in A and 8.This is represented
lry the shaded area in the diagram below.

The straight line lies completely in the shaded area. point A contains more of y and less of X.
llrt: following are some standard relations on the union of sets:
Point B contains more of X and less of y. Therefore, more balanced bundles are preferred
extreme ones. As one moves down along the curve AB, the marginal rate of substitution of I AUA: ,4, idempotent property
good X for Y diminishes, this means you are willing to give up less and less of y as we move tl AcB,thenAUB :B andifBc A,thenAUB =A
down the curve. By contrast, consider Figure 3.6 below: iii) AUA' : u
iv) AU@ = 4, identity ProPerty
ul A and B are both subsets of,4UB
vi) lf Ac B andB c C,lhenAUB c C,Transivity.
vii) Commutative properly, AUB : BUA.
httersection oI o set: This is denoted by A n B and includes all the elements common to both
/ ,rnd B as shown by the shaded area in the diagram.

L
 Mathematical Optimisation and Programtning Techniques tir Iicorrotnic Analysir I
31
30 | ,rrr, *rro,o*sAND FUNcroNs

Figure 3.7. Set intersection TheorderinwhichthesetsinaUnionarearrangedisimmaterialTheUnionsimply

meanselementsbelongingtosomeofthesetsorallofthem.ltmattersnotwhichis
written first.
AUB=BUA

5. Cumulative law of two sets:

Theintersectionoftwosetsgivenanyorderareequal,thlsissimilartotheruleof
numbers ob = ba. The law is expressed as:
AoB=B^A
The following are some standard relations on intersection sets: 5. Distributive proPerty of union and intersection:
similar to the rule a(b'l c) = q6 * bc' this can be illustrated as follows:
I An A= A, idempotentproperty
The union of sets is distributive over the intersection:
ii) tf,4 c B,AoB:A
.4 u (8 n C) = (Au B) n (,4 u C)
iii) AnA'--0
iv) An0=l,identitylaw The intersection of sets is distributive over the union of sets:
v) A n B is a subset of both,4 and B. /4n(Buc)=(AnB)u(Anc)
vi) lf C is a subset of both.4 and B, then C c A n B.
vii) AnB = B nl,commutative property I rliure 3.8. lllustration of Distributive law
Dilference of two sets: A - B is a set of all elements of .4 that do not belong to B.

A_B={x:xeA,xeB}
It A - B =,4 and B - A = B, then/4 and B are disjoint.

A-B=B-AOnlywhenA=B
3.5.1 Summary of Set Theorems

The following theorems of sets will prove helpful in dealing with set problems.

X. Equality of two sets: i ,4n(8uc) ii AU(B1C)

Two sets A and B are equal if and only if A is a subset of B and B is a subset of A 7. Associative proPerty of sets
as: (,4 U B) U C = AU (B U C) =C U (A U B) = (C U.A) U B
A=BtffAcBandBcA (,4 n B) n c = An(B n c) = c n (A n p) = (c n1] n B
2. ldempotent law:
section of their
(o)AuA=Aand(b)AnA=A
3. ldentity law: complements. That is--
lf.4 is any set, @ is the null set and U, the universal set, ('4 u 8)' = A'n B'
Av@= A The complement of an intersection of two sets A and B is the union of the
AnLl =A complements.
(AnB)'=A'vB'
4. Commutative law of union of two sets
r 32 SETS, RELATIONS AND FUNCTIONS Mathemzrtical Optimisation and Programming Techniques for licorronric Analysis I

9. Theorem: The solution is as follows:

al A n(ts - C) = (An 8) - (,4 n C)
b) (,4 - B) u (B - A) = (Au B) - (A n B) a To findn(A n B n C),since we know the number of persons who did not prefer any of
the three products(2520) and also know the total number of persons interviewed
3.5.2 Number of elements in sets: (14520),we can find the number in the union set of the three products.
This is simply denoted by n(A) and represents the total number of elements in a particular set. Therefore, the number of persons who liked at least one of the products isj$Zq-
The total number in the union of two sets can be shown as follows: 2520 = t2OO0. Using the formula: - >

n(A v B) = n(A - B) + n(B - A) - n(A B) n(A u B v c) : n14-S + n(B) + n(c) -Jr{A

I B.) - n(8. c) - n(c e A) + n(A B
^ ^ ^
= n(A) + n(B) - n(A B) ^ c),
= n(A) + n(8) if A and^ B are disjoint :
12000 8150 + 5540 +4010 - 3150 - 1820 - 1030 + n(A n B n C),

illl Figure 3.9 lllustration of number of elements in a Union set Therefore: n(A i
B n C) = 300

b. To get the number that liked A but not B, we remind of the following set
representations. A n B' = A - B, B n A' = B - /. Therefore, the number is derived
usrng

n(AnB')= n(A)-n(AnB)
= 8150 - 3150
= 5000

c. To find number of persons who liked A only, n(A n B' n C'), a glance at the graph
below makes it clear:
ln a case of 3 sets, A, B and C, the number of elements in the union set can be expressed as
follows:

n(A u B v c) = n1a\ + n(B) + n(c) - n(A n B) - n(B n c) - n(c n A) + n(A n B n c)

AaBnC,
Example 3.2
A marketing survey of a company wishes to find consumer preferences for three of its
best selling products A, B and C. The following results were obtained: ANB NC

The total number of consumers interviewed was 14,520. Of these, product A was
preferred by 8150 consumers, product B was preferred by 5540 and product C by 4010.
Products A and B were preferred by 3150 persons. Products B and C were preferred by
1820 persons and consumers who preferred A and C were 1030. The persons who did
not prefer any of the three products were 2520.

Find:

a. The number of persons who liked all the three products.

b. The number of persons who liked A but not B.
We proceed asfollows: n(A n B) = n(A n B n C) + n(A n B n C')
c. The number of persons who liked A only. 3150 = 300 + n(A^B nC')
 Analysis
I

34 I SETS, RELATTONS AND FUNCTTONS Mathematical Optimisation and Programming Techniques lirr Ireorrornie I
35

n(AnBnc')=2959 .,,t z1 has n elements and set B has m elements, we can form m X n ordered pairs. This leads

And rr,, to the concept of Cartesian products.

n(A n C) = n(A iB n C) + n(A n B' n C) I r, .l l'he Cartesian Droduct
1030 = 300 +n(A^B'nC) llrr', is the product set of ,4 and B, it consists of the set of all ordered pairs (x,y) where r €
n(A B' n C) -- 730 I ttil(l e B. For instance,if = {a,b,c},/g = {1,2}, the Cartesian between the sets A and B
y A
^ l, r* rted bV ,4 x 8ls shown as follows: t /
But '

n(e) = n11 n B' n c) + n(A n B n c') + n(A n B' A c) + n(A n Bn c) . AxB:[(x,y):xeA,ye8], this is a rule method of representation earlier
discussed.
8150 = n(.4 n B' C')+ 2850 + 730 + 300
^ . AxB:{(a,L),(a,2),(b,L),(b,z),(c,t),(c,2)} , this is a tabular wav of
n(AnB'nC')=4270 representation. The product was obtained by "multiplying" every element in / with all
Therefore, the number of persons who liked only.4 is 4270.Ihe obtained solution can be the elements in B.This has formed "3 x 2" ordered pairs.
represented as follows in a Venn diagram:
Nrrrr, that the product A x B + B x A : {(1, a), (2, a), (1,b), (2, b), (l' c)'(2, c)} as earlier
rrrr.Ilionedbecausethelattercaseisa"2x3" lnthesameway,theproductBxBisgivenby
11 ,, 11 = {(1,1)(t,2),(2,r),(2,2)}

l7 What is a Relation?

the concept of ordered pairs and Cartesian products, you will notice that these terms are
I l,,rr11i

Irrr.rrclated and that a function is a special type of a relation. Simply defined, a.relation from-
, r,, y is a set of ordered pairs(.x,y), such that to each x € X, (eech element that belongs to X)
tlrr.rr. corresponds at least oney eY. Note that x and y need not include all the elements in
\ ,[r(/ y To shed more light, let x and Y be two nonempty sets. A subset R of a cartesian
y,thatis,if Risarelationfromxto
lrrrrrlrrctxxYiscalledalbinorylrelotionfromXtoY.lf X=
r wr, simply say that il is a relotion on X. Put differently, R is a relation on X iff (if and only if)
/i , X).lf (x,y) e R, then we think of R as associating the object x with y, and if
Yt(X x
lrr,v),(y,x)] o R = @, we understand that there is no connection between x and y as
3.6 Ordered Pairs and Cartesian Products
Irrvr,,,rBed by R ln concert with this interpretation, we adopt the convention of writing x Ry

trnt,,,rrl of (x, y) e R throughout this text.

Given the extensive introduction to sets, we are now in a positlon to introduce the concepts of
ordered pairs and Cartesian products. lt should be recalled that capital letters represents sets lrrr rlIrrtration, consider all the possible outcomes when a die is tossed twice. Define set.A to
while small letters represent elements. l[,,|rt(omes from the first throw and set B to be outcomes from the second throw. Let the
rrrrllrr,rl pairs be denoted bV@,y), where x e A,y e B The ordered pair.4 x B is a Cartesian
3 6.1 An Ordered pair
1rl,,rlrrr I which contains 36 ordered pairs as shown below.

An ordered pair is a set of elements for instance (x,y). where x is the first element and y is
the second element. However, (x,y) + (y,x), that is, the two sets are not commutative if the
set is an ordered pair. The two elements nevertheless need not be distinct: this simply means
that (x,x) ot (y,y) is possible where as if it is a set and not a pai,{x,x}: {x} ln general, if
7
I

35 SI I\, RELATIONS AND FUNC IION\ Mathematical Optimisation and Programing Technic;ues ftr-l-icorrorrric Analysts 37

Figure 3 10. All outcomes of a throw of a pair of dice

lr r r' noles of relations

)a).)a i A Relation R in a set ,4 is called reflexive if aRa,thot is,(a,a) € R for every
5 a])aO] oeR
ii A relation R in a set,4 is called symmetric if whenever aRb,then bRa, that is if
4 )sa))t (o,b)eR)(b,a)eR
iii A relation R in a set A is called transitive if when aRb andbRo, then aRcthat
)SO]O] is,(a,b) e R,(a,b) € R + (a,c) € R
a
O])aa]
1 ]att)t r il Mapping of Sets
irrl(,r(,d pairs obtained from functions (special type of relation) and relations can be
1r"^ r6
r , 1,r ,,,,r.nted by Mappings.

Let R be the relation such that the sum of the 1't and 2nd throw is greater than 8. The solutions : t1,2,3j,Y = {1,2,3,4,5,6}
lstance, Considertwo setsX andY;X
are the ordered pairs:
p = [(3,0), (4,s),(4,6), (s,4), (s,s), (s,6), (6,3), (6,4),6,s), (6,6)J XY
Alternatively, this can be represented in rule form ,f,/I\
as:
\i\
P : {(x,y): x + y > B,(x,y) e A x B}
,#,
1+1

Therefore, given sets A and B, a relation R is a subset of the Cartesian product,4 x B. The
elementsofthesubsetRareorderedwhichareorderedpairs(r,y)
ll4
It
lirl suchthatxEA,yeBx
and y need not include all elements of ,4 and B ln the above example for instance, x takes on
/\s
/1
l\h/
values 3, 4,5 and 6, y too takes on the same values, therefore, the domain and range are / "\/
identical in this case.

3.7.1, Domain, Range and inversc ol ;r r cl:ttion Itom the above mapping, it can clearly be seen that all elements of X have a corresponding
lfument of Y, while only three elements of Y are paired with X. ln such a correspondence where
Assuming two sets A and B as in the above example, where x € A, and y € B. The domoin and
ronge of relation are defined as follows The domoin of the relation is defined as a set of
lll elements of Y are not paired with the elements of X, we say that the set X has been mapped
lnto the set y. The Cartesian product gives a total of 18 ordered pairs, givenX = t1,2,3j,
elements x paired with y in (x, y) which belongs to R. This can be represented as:
Y e {1,2,3,4,5,6}.However; the above mapping shows only the ordered pair (1,1), (2,2),(3,3).
X - lx: lot some y,(x,y) e R) It lsthus, a relation (also a function as will be seen later)
It the domain of the relation R The ronge of the relation
is called is defined as a subset of y in B
Consider anothe r exa mple, X = t1,2,3,4,5,6j, Y = {1,2,3,4,5,6}
over which y varies: Like the domairr, it is represented by:

Y ly: lot som.e y,(x,y) e Rj

it is read as the range of R or range /? Therefore, the first element (x) in an ordered pair forms
the domain while the second element (y) in an ordered pair forms the range

lnverseofarelotion:suppose /iisarelationfromAtoB,theinverseofRdenotedbyR-1is
simply a relation from ll lrt A ln other words, the domain and range are reversed.
Jd | ,rrr, orro,o*s AND FUNCToNS Mathematical Optimisation and Programming Techniques for lrconornic Analysis 39

,,1 \, there is associated only one unique element of Y.This moves us to the introduction of
lrlt( Iions

I ') What is a Function?

I rrilr lions are important in the economic study ofthe relationshlps between various cost curves
rr l! total cost and marginal cost Functions can be used to show
,rs average variable cost,
, " I the
rlrr rtlyrelationships between these costs and also shows us that if one of the cost curves
I r r , ,r particular shape, each of the others has another specific shape. The positions at which the

rv,'r.r1,,e and marginal curves cross one another can also be exactly determined ln production

llri,ry, functions can be used to show the relationship between total, average and marglnal
lf every element in X is matched wi h every corresponding element in )/ and vice versa, then X I r, rrlrrcts This section thus gives insight as to what functions entail.
has been mopped ontoY. For X and I, there are 36 ordered pairs when their Cartesian product ',rrrrlrly defined, a function is a relation / (i.e. a subset of ordered pairs) such that to each
is obtained. The six ordered pairs cor rstitute the above mapping which is a relation (this is also a
, l, rr('ntr € X, there is a unique elementy € y. It is a subset of ordered pairs characterised by
function as will be seen later).
,,,rl,rin given conditions, Denoted as f: X + Y
Consider another mapping: y
tt, rrl ,rs: f is a function that maps X onto
x=11,2,3,4,s,6,7,8,9,1o,I71, =17,2,3,4,s,6j
Y
r) The domain of f is equal to X
rr) To each X, there corresponds a unique y e Y
, / (r) is an often-used expression to show the association between the element x and the
,,rrlrponding unique element ofthe ordered pair (x,y) that is an element ofthe function /. n
rrrrctimes called the orqument or independent vorioble of the function and y is the entry
tn. y . f(x), called the dependentvariable. The correspondences can be one ofthe following:
2 I = x * 2, for every x in X, there will correspond one and only one
One-to-one: e 9., y
y in }/, there will correspond one and only one x in X.
y in Y,and for every
3
.' Many-to-one: e.g y: x2 +3x - 2, IgrylSZ-q y,there will correspond more than
---.-
onerin X
I one-to-many: e.g y2 = x * 4, for one x in X, there will correspond more than one y in
Y

4 Many-to many: e.g.

;-#= 1, for more than one x in X, there will correspond more
than one y in Y.
rlr, rr,lore, the values taken by :r form the domain while the values of y obtained from the
,,rrr1r'Considerthefollowingexampleof mappingstotestyourunderstandingoffunctions:
Since to every element in X there corresponds an element in Y and every element in y is
associated with at least one element in X, in this case, we also say that X has been mopped
ontoY.Theabovemappingshows12orderedpairsoutofthe12x6 orderedpairsofthe
Cartesian productX x y.
From the examples shown above, what is common in each case is
that regardless as to whether the set X was mapped onto or mapped into \, for every element
40 | ,rrr, *rro,o*sAND FUNCToNs Mathematical Optimisation and Progrming Techniques for Economic Analysis 4L

' ilomain range

This is a function. You can tell by tracing from ,olrrtions:
-6 each x to each y. There is only one y for each x; there
-I
0
,3
t5
is only one arrow coming from each x. This is a one
to one correspondence.
'[-l 'l ,*-.
l\ I ti\
domain range
This isa function. There is only one arrow coming
from each x; there is only one y for each x. lt just so
happens that it's always the same y for each x, but it _r \lill
is only that one y. {a} (b)

1,

ilomain
_3
range
This one is not a function: there are two arrows
coming from the

Each element
number 1; the number 1 is
associated with two different range elements. So this
is a relation, but it is not a function.

of the domain that has a pair in the

lrit lr*r
'l

{d} {.1
tt
tt
(0;
_2____+-t range is nicely well-behaved. But what about
-----+_60
_l--______+ that 16? lt is in the domain, but it has no range al y = -2x + 6 is a function because for each value of the independent variable x there
o----------+ 3 element that corresponds to it! So then this is not a is one and only one value of the dependent variable y. For example' if x = t' y :
1-------------+ 15
function. lt is not even a relation. -2(1) + 5 = 4. The graph would be similar to (a).This is called a linear function used to
I6 represents some constraints in utility and profit maximisation.

Example 3.3 bl y2 : x, which is equivalent lo y : 1r[*, is not a function because for each value of x,
there are two values of y' For example, if y2 = 9, y = +3' The graph would be similar to
This example will help exercise your graphing skill of equations. Thus, if you are given
that of (c) illustrating that a parabola whose axis is parallel to the r axis cannot be a
data, you should be able to produce total revenue curves, costs or any other curves in
fu nction.
economics and various sciences.
c\ y = x2 is a function. For each value of / there is only one value of y' For instance, if
Graph the following equations and determine which among them are functions.
x = -5,y: 25. While it is also true that y = 25 when x = 5, it is irrelevant. The
a. Y=-2x+6 definition of a function simply demands that for each value of x, there be one value of y.
b. y-.2 -^ The graph would be like (d).

c. !=x2 dl y : -x2 + 6x + 14 is a function. For each value of r there is a unique value ofy. The
d. !=-xz*6x*1,4 graph would be like (bf,r
,{J
e. x2+y2=64 e1 x2 + yz = 54 is not a function. lt x = 0,y2 = 64, andy = +8. The graph would be a
f. x:5 circle, similar to (e).
7-
I Mathematical Optimisation anrJ Programming Techniques tbr Ecorromie Analysis I
43
42 I SETS, RELATTONS AND FUNCTTONS

fl x=5 is not a function The graph of x=5 is a vertical line. This means that at
1 rr,, \-rf * en-1xn-l + """ + arx2 + arxl + a6xo,
x = 5, y has many values. The graph would be like (f). llJlq.rr,, (11,G1,....c, are real numbers with au + 0 and n is a positive integer This implies that
rlr, l.r'.1 two expressions can simply drop to a1r ond. ao respectively using the property of
3.10 Types ofFunctions , ,t,,,u,'rtts27. Thus the constant function described above is actually a special case of polynomial
3. 1.0. L Constant Functions: Irinlrrrrrs,withn=0.Thiscanbederivedasfollows,substitutingn=0 intheaboveequation,
,rll lll.'tcrms drop except !: oo- ln the same line, various functions common to us can be
A function that maps every element of the domain to a single element of the co-domain is
called a constant function .That is, the range of a constant function is always a singleton sel, rllr rvr.rl by simply substituting n.Thus depending on the value of the integer n (which specifies
tlr,, lrrlilrest power of x), we have several subclasses of polynomial function:
Generally, we write a constant function as /(x) = k, where k is a fixed number. For instance if
k = 3.5, this can be shown graphically as:
Constant function
=0 f=oo
1 y=atx+ao Linear function

2 !=azx2+o'tx+oo Quadratic function

(r) = 3.s Iy=or*=+or*'+o.*+oo\ Cubic function

WlrIrr lrlotted in the coordinate plane, the aforementioned functions appear as follows. (a)
tllr,,ir,rtes the case of ar> 0, involving a positive slope and thus an upward-sloping line; if
,i, 0,thelinewill bedownwardsloping Aquadraticfunctionplotsaparabola,acurvewitha
trr1,l. built-in bump or wiggle The particular illustration in figure (b) implies a negative 42, a

l,rrr, trrrn with a maximum. ln the case of a2 ) 0, the curve will open the other way, displaying a
lrrrrr tron with a minimum. The graph of a cubic function will, in general, manifest wiggles, as

rllrr,.tr,rled in figure (c), attaining both maximum and minimum values at itsturning points.

As shown above, in the coordinate plane, the constant function will appear as a horizontal
straight line. ln economics, this graph is mainly used to depict the average revenue curve ofthe
hypothetical competitive market which takes on a horizontal shape, implying that the prices,
marginal revenue and average revenue are all equal, prices are plotted on the vertical axis are
fixed. Likewise, in national income models, when investment / is exogenuously determined, we
may have an investment of the form I : K70 million, or I = /0, which typifies the constant
fu nction.

3. 10.2 Polynomial Functions:

A polynomial takes many different forms; it can be a constant function (as shown above), linear,
{- cubic or even a quadratic functlon. Polynomials are widely used to model different lrr rrrhscript power (n) of the x is calted the exponent Any tem raised to the power zero is one by property of
I
, s anrl any term raised to the power one is jtself. The value of n, is often called the degree of the polynomial
.1 r,rrL,r rl
maximisation problems of production as well as in consumer theory using the various cost,
trrrr rrrrr: lirr instance, a quadratjc function is a second-degree polynomial, and a cubic lunction is a third-degree
revenue, consumption, demand as well as supply functions. The word polynomial means l,,tlrrrrrriirl Theorderinwhichtheseveraltermsappeartotherightoftheequalsignisinconsequential;theymay
"multiterm". A polynomial is an expression of one variable for instance y in the form l, rrr,rrrgcd in descending order of power instead- Also, even though we put the symbel y on the leti. it is also
,,,,, t,rirl)lc to write f(x) in its place. (see Chiang & Wainwright,20l3 )
7 I

Mathematical Optimisation and Programming Techniques for Economic Analysis

I

44 I sr rs. RELAT|oNS AND FUNCTTONS

Figure 3.11. lllustration of functional forms I the positive degree. A special rational function that has interesting applications in
ls the one depicted below. Since the product of two variables is a fixed constant,
I impliesxy = a, this can be used to depict the demand curve with prices (p) on
Lrnear
Quadretic 'x =
r0 )
"1* y= aO+ a1x + a2 x rtlcal axis and quantities (Q) on the horizontal axis. Similarly, the average fixed cost curve
) I also rectangular-hyperbolic because AFCx Q (total fixed cost) is a constant.

a1

{ case of a, (o )
3 ll I tionssofaa special f ormnof
ustrati(
:'"tional function: Rectangu
I ar hyperbo a

0 tangu
,ular-- hyperbol
Rectangular-
Recta |hyperbohc
IC

3
(b)
x

\
laa >ool
Y= "O* '1* * ', '2 * r, n3

I
x

I xpoOnential rd log
al anc nlcc furrnctiions
rgarithmtr ls

13. Graphs
( sofeexpoonentiallan
a nd loga
lr rrithhmic fu
I r n ctions

tial
:xponentia
Ex
3.10.3 Rational functions .xx
Y=D
A rotionol function is any function which can be written as the ratio of two polynomial
functions. Neither the coefficients of the polynomials nor the values taken by the function are
necessarily rational numbers. According to this definition, any polynomial function must itself
be a rational function because it can always be expressed as a ratio of 1, and 1 is a constant.
Therefore, a constant function is a rational function. 1l
{b}1)
ln the case of onevariable r,a function is called a rational function if and only if it can be
I

written in the form diag m1

iaram
P(x\
f@ =
a@)
rl rlr,rnr function of the form y = bx .fhis is called an exponential function where b
L depicts a
tlr l),rse (positive real number) andx is the exponent The prominent feature of this function
Where P and Q are polynomial functions in x and Q is not the zero polynomial.
rlr,rr lhe values of the function will always be positive for all real values of a(--, {-),. the
Thedomainof/isthesetofall pointsofxforwhichthedenominatorQ(x) isnotzero,where
l,,irr,rrrr is the set of real numbers while the range is (0, -). Exponential functions with the base
one assumes that the fraction is written in its lower degree terms, that is, P and Q have several
t
I

46 I SETS, RELATTONS AND FUNCTTONS Mathematical Optimisation and Prograrnming Techniques lbr Economic Analysis 47

e is greatly used in economics, where, e is a real number, whose value is approximately equals Io solve an exponent such as t in S: P(1 * i)', use the logarithmic transformation.
to2.7L82.Fot instance, the income y is an exponential function of time period t. That is, if ys is ',ilbstituting the given values for the terms in the equations, we have
the initial income and r is the growth rate of income, then ,y : yoe't. 1,2 = s.6(1+ 0.15)t
Diagram 2 depicts a logarithmic function of the form f (x) -- log6x. Here b is a positive real l,rking natural logs on both sides
number (b> 1) and called the base of the logarithmic function. Logarithmic functions are the ln12:In56*tIn1.15
inverse to the exponentional functions as clearly shown by the diagrams. Thus, its domain is
2.4849t : 1.72277 + 0.1397 6t
(0,-) and the range is (--,1-;. When we use the number e as the base, we get natural
logarithmic functions. When there is no ambiguity about the base, it is common to write the 0.I3976t = 0.76214, thus, r - 5.45 Years
logarithmic function as: f(x) = logx. These are used in estimating growth rates from data
points and also in compounding of interest rates in economics.
III Scquences
I ilil,,rrl(,r the following expression,
Example 3.4
lrt r, t-r,/4--+x ___
Assume for a given principal P compounded annually at an interest rate i for a given .
number of years t will have a value S at the end of that time given by the exponential Wlt,'tt t =0,y:2,
fu nction, r l, y=*V5,etc.
s:p(1 +i)t
I lrl,, 1,, ,r sequence of values of y corresponding to integer values of x. A sequence is defined by
Example 3.5 I lrp v,rltrcs of the function as x takes on positive values.
A small firm with current sales of $10,000 projects a 12 percent growth in sales annually. I
11, , -, where n:7,2,3,........, then !y!2,!2, ........!n will form a sequence.
Its projected sales in 4 years are calculated in terms of an ordinary exponential function.
ll r, / (x), what then will be the value of as x tends to some number? This brings us to the
y
s = 10,000(1 +0.t2)+ r rtlr r'pls of limits.
= 10,000(1.s735) = 1s,73s
1 I2 I,iulits of aFunction
Example 3.6
lurrrl,rrnental question asked in limits is if you have any function, be it a consumption,
A S-year development plan calls for boosting investment from 2.6 million a year to 4.2 Ir rrrhrr lion or demand function, what is the value of that function as it tends to a certain value?
I million. What average annual increase in investment is needed each year? ll tlrr, lunctional values f(x) of a function / draws closer to one and only one finite real
:2.6(t + i)s
4.2 rrrrrulrtr /,for all values of x as x draws closer to a certain value "c" from both sides but does
rrl r,rlual "a" , I is called the limit of the function as x tends to a. This can be represented by,
1615=(1+i)s
!r1;f @ =
t
1+,=V1515 =1.10061
I 1r.rr,lrrre, for a limit to exist, limr-o- / @) = limr-"+ f (x)
i:0.10061=10% f

I r,rrrrple 3.8
Example 3.7
gx?42
A developing country wishes to increase savings from a present level of 5 5 million to 12 Iind/(x)=j;:rasx+-
million. How long will it take if it can increase savings by 15 percent a year? 2.
5ubstituting tne x = ; into the function gives an indeterminote formf(x) = 9. The task is
to change this toJEt-erminate form by some algebraic operators. -
r
48 SI I\, RTLATIONS AND FUNCTIONS Mathe matical Optiirisation and Programrning Techriqucs firr Ircorrorrric Analysis 49

9x) 4 ('3x +'2X3x - 2) I4 illustration of functions with and without limits

312 3x-2 =3x*2
So,
9x2-4 .a'
lim-=4
''43x-2
Example 3.9

Given the function f (i - rina lim,-2 /(x)

=;if,function, an indeterminate form results. The function gtves
When 2 is substituted into the
r,r ,lr,rtirarr 1, as the value of x approaches 4 from either side, the value of /(x)
it is clear that
zero both in the numerator and denominator. This means there is a common factor in the
,1,;,r,r,rrhcs2 Thismeansthe!imitof/(x)asxapproaches4isthenumber2,writtenas:
numerator and denominator which need to be factored out.
tinf U) =z
lim \'-, -' -x')-\: - ?) + x(-x - z) + lx - z)
x,2Jf(x) ,-) y2(y _ 2) _ x(x _ 2) + (x _ Z) ,i r ,r1;lrroaches 8 from either side in diagram 1, where the open circle in the graph of f(x')
.. (x'+ Y1t)(x-2) rl,rrrlrls there is a gap in the function at that point, the value of f(x) approaches 3 even though
ii(xz-x-+1)(x-2) t1,,, lurction is not defined atthat point.
ix2+x+l)
=lim---j l,rllr,rlir,rm2,asxapproaches5fromtheleft(fromvalueslessthan5),writtenasx)5,9(x)
x-z (xz - x -t l)
,l,l,r(),rches 3, called a one-sided limit, as x approaches 5 from the right (from vaiues greater
-7t
- /3 written as x 5+, g(;t) approaches 4, the limit does not exist, therefore, since g(.t)
tlr rrr ',), -
,1, r 111;f approach a single number as;r approaches 5 from both sides.

Example 3 10 I " rrrrplr-.3 11

For the function f (i :

gI:1, \uppose the cost function of a firm is c(q) = Vq + K, where q is output, 7 is variable
find lim,.-u / (x)
r ost per unit and /( is fixed cost What will be the average cost as output becomes very
Like in Example 3 9 above, the solution for this question cannot be found b merely
l, rrge?
substituting zero in place of h" This will result in an indeterminate form To avoid this, a
way has to be found that will negate the indeterminate form. We set the solution as \o ution:
I

follows lhe average cost is simply the cost function dlvided by output; therefore, trying to find
x3 +3x2h+3xh2 +h3 -x3 out the limit as q tends to infinity is as follows:
f;1X/(") = h r Kr
3x2h + 3xh2 + h3 tim
q+*\ I
Iz +-lq/ :V
h
:3x2 +3xh*h2 t Lt l\4ore on Sequences and Limits
: 3x2
,ul )l )o5e you have the sequence
Consider the following two functions, f (x) and g(x) graphically depicted in Figure 3 14:
1
X,,--
n
7-

50 SITS, RELATIONS AND FUNCTIONS lVlathcnraticirl Optimisation and Programnring Techniqucs lirr l:eorrortric Analysis 51

111 I I ( oill inuity ofa Function

- 2'3'4
analysis, we are concerned with the continuity offunctions as well as the existence
As ?1 n@, 1+0 lrilrrt., (ll those functions to forecast and predlct certain outcomes in our economic Suppose
n
,rr .rr,, ,rrralyzinga discontinuous production function, what are the implications for the
lf xn=r *2,asn+-,thesequencetendsto2. lnbothcases,thesequencehasalimit,itisa
',n
' r rrl'. .rrd marginal product curves? Simply defined, a continuous function is defined as a
convergent sequence. A sequence that has no limit is called a divergent sequence. r, ltrrl which has no breaks in its curve. Simply put, it can be drawn without lifting the pencil

€.8. xn = n * 3; 4,5,6,7, ... ... ... . rr tlt',l),lper

3.13.1 Sumnrary of properties for limits l,rllrwing graphs give a clear picture ofthe concept of continuity and discontinuity

Assuming thal limr-o P(x) and lim*-"Q(x) both exist, in a nutshell, the rules of limits are rrrI I I5 Continuous and discontinuousfunctions
given below;

1-. limx-ah : hwhere hisa constant

2. limr-oxn = on wheren is a positive integer
3. lim,-o hP(x) = h limx-a P(r) where h is a constant
4. lim*-olP(x) t Q(x)l = lim,-o P(r) * lim,-o Q(x)
5. lim,-o[P(x).0(x)] = lim,-o P(x).lim,-"Q@)
5. lim,-o[P(r) = 0(x)] = lim,-o P(x) + lim,-o 0(x) where [lim"*" Q(x) + 0]
V. lim*-olP(x)1" = Iim*-oI P(r)]a Where rz > 0
3 13.2 Key points about limits of functions i, t'1,,!ir'r, Oler t.iervnl aasdb

The following are the key points for limits of function:

7. lim*nr,,means the limit as x approaches z, from the right hand side and limr-rr- ts
the limit from the left hand side.
2. For a rational function (quotient of polynomials):
a) lf the degree of the numerator is less than the degree of the denominator, then the
limits at - and -- are both zero.
b) lf the degree of the numerator is the same as the degree of the denominator, then
the limits at - and -- are both the quotient of the coefficients of those of the
highest degree
c) If the degree of the numerator is greater than the degree of the denominator, then
as x 4 @ or x - --, the function approaches either - or -- according to the
signs of the numerator and denominator
(d
3 A function /(x) is continuous at 11 in its domain if limr-r, /(x) exists and is /(xr). A Srtak in do,m*|fl
continuous function is one that is continuous at every point in its domain.
4 Sums, products and quotients of continuous functions are again continuous functions.
(Quotients are not defined where denominators become zero). llr, rl,,rvr,liraphs are quite self explanatory. Graph (A) shows continuity over an interval a and
5 A differentiable function is continuous. l, tl,r ,,rr tlearly be drawn without lifting a pencil. Graph (B) shows a jump between a and b,
llrr rl I rlr,,continuous. Graph (C) shows infinite discontinuity as both curves will never meet
'1, ;,rr,,',llriping convergence. Graph (D) shows a break in domain, the domain is from a to c,
1,,,,l,,ll, lhcn takes off from d to b B and C are discontinuous in the range.
r
52 i \r r\, Rf LATIoNS AND FUNCTTONS Mathematical Optirristrtion and Prograntnting Techniques fbr L,corrrrrrric Anrlvsis 53

:l l4 1 Pojnt continuity
A function /(r) is continuous at a point x = u if for any posltive value E-, however small, there
exists a positive number d such that: ----------{

I f(x) - f(a) l< E, Provided I x - u l< r)

In other words, l(x) + /(a) + a, where x + a either from the left or from the right. lf
as x -{
the two one sided limits are unequal, the function is discontinuous alx: a. Therefore, a
function / is continous at a point x : n if the following hold:

t. f(x) is detined, that is, it exits at x = d

2. lim*-" f (x) exists and 5
3. lim,-"f (x) - f (a) Duratton {minutEs}

Therefore, a function is continuous over the interval (o,b) if it is continuous at every point in limr-r* c(t) : 4,lim1-3_ c(t) = 3

= s, )1p_c(t) :
that interval. The graph between x = a and x = b can be drawn without lifting pen from paper a,
as earlier stated
,\?r.(O
3.14 2 Types of continuous functions ,\T_.(O:6,lim c(t): s,

llrr' liocs on and on, showing discontinuity in the function.

Figure 3 16. Non-smooth and smooth functions
,'rr',rrlcr the following example to show that a limit may exist in a discontinuous function:

c{x}
4

Z
3

b)

From the above graphs, (b) is a smooth continuous function whereas (a) is not completely ilr,' (rn opcn circle means a gap in the function and a continuous function can be sketched
smooth, it has a sharp points. ,drtl)orl ever removing pencil from paper, it is clear that /(x) is discontinuous at x = 5 and
/ ( r ) as discontinuous at x : 4 Diagram (a) shows that the limit does not exist
Consider the following discontinuous function: if international calls cost K3 up to 3 minutes,
rlrrrr, . /(x) -Z,linr-q*f (l) = 3).On the other hand, diagram (b) shows that the limit
then a K1 every additional minute, the discontinuous function is obtained.
,"r,t ,sincelimr-og(:,r) is2approachingfrombothsides Thetypeof discontlnuitydepictedin
I I t ' a jump discontinuity Discontinuity depicted in (b) is called a removoble discontinuity,
,,,, rrrse if the function is redefined at x=4 such that /(a) -- limr-" f (4), it becomes
, rrIruous. Thus, a function may be discontinuous but a limit may exist. All polynomial
r,rrr( lrons are continuous, as are all rational functions, except where undefined, i e., where that
l, r rorninator equal zero
 Mirlhcnratical Optirnisrtion rrtd Prograrrnring lcchniilues lirr lrcorronrir Arrirlysis 55

Chapter 4

I MA'I'IIIX ALGEBRA

II lnl rrrduction
trr rlrr,world of Economics, many characteristics or attributes are used to compare the
I'r r l,,rrrr.rrrce or relative position of different countries, regions or even firms in a country. For a
t,rr',rt, lrrnforexample,onemighthavetolool<atitsmarl<etshare,itsizeofsalesrevenue,the
-trr rl llro work form and ultimately the level of profits either in absolute or in proportion to
",rh , lrr look at how two or more firms compare requires looking at this array of variables or
rrl,,rr,rlron For each firm, there must be a measure for market share, sales revenue, and the
lt | 1,,,r.', on
rrrv,,rr tlris information, the summation or subtraction for aggregation purpose must be
Ir rrrrrtl|rl lf information is available for all the sectors in the economy, that is total output, the
dv' r,11(.wage rate, and so on, then it should be possible to compute, for the same variable,
rr,rlrrrrr,rl statistics. lf prices in different sectors change, ceteris poribus, should require getting
rr.w rrrlrrrmation on the new value of output? There should be no need to redo this because the
r lr,rrrltr,rrr prices can easily be factored in through multipllcation.
llrr,, rlr,rpter is thus devoted to look at matrices (and their special forms, vectors) and their
,r111',lrr,r We explore with relevant examples the different operations of matrices and how
rr,rtr( be used to aid understanding of economic phenomena. As Chiang and Wainwright
r,,, can

put it, matrices provide a compact way of writing an equation system, a way of testing
I 'rtrt',)'rr
tlr, ,.xr,,lence of solution by evaluation of a determinant and ultimately providing a method of
Itrrlr rli lhat solution (if it exists). Before endeavouring to lool< at the different operations of
rr,rtrr (.\, it is important to first consolidate the understanding of the matrix itself and the
rltlllrlrrt lorms it can manifest.

|) What is a vector?

',rr1r1rrr,,r'that for each variable like GDP, there are observations for several countries, or for
r ,r, lr r orrntry, observations are made on several variables. ln the latter case, this gives an array
r,l rrrrrrrlreTs representing the population size, level of GDP or per copito GDP and many other
v rrl,rlrlt,s The order in which these numbers are presented now matters as opposed to sets

r lrr.rrrg & K Wainwright, 2013.

 r
I
56 I MArRrx ALGEBRA Mf,lhenluticrl Optirnisrtiolt alrd Programrring'fechniqucs lbr [icorrornie Anrrlvsis 57

studied in the preceding chapter The first number now stands for population size, the second 4. 1: Diagramatical representation of Vectors
for level ofGDP and so on This illustration gives rise to the vector

Formally, a vector is deflned as an ordered set of numbers and is denoted by X Simply put, a

vector is an ordered n-tuple With n-elements, the vector is said to be of order n The actual
arrangement can lal(e either a column way or a row way ln the case of the former, it is known
as a column vector and the latter referred to as a row vector. AnV given vector X with elements
xt, xz, , and x, can be represented as

rx1 I
lr,
x,=l,l
I

1,,]
if it is a column vector and if a row vector, its appearance changes to
Xr=(xr,r2 " ,,r,,)
The elements in the latter form are not separated by commas and this is not as a result of an
error lt is deliberately so. ln vectors, elements are not separated by commas but by spaces
Note that the natural order of a vector is taken to be a column Changing from one form to
another, from column to row or vice visa is called trdnsposing ln the above two vectors, since tt ru rrli ikrfined the vector and its forms, two special vectors deserve attention. These are nul/
theelementsaswell astheorderremainthesame,then oneisatronsposeof theother.Thisis tt I tutl vectors. lf all the n elements in a vector are identically zero, then the vector is referred
denoted by a primed symbol or more unequivocally, by a superscript T lt is needless to say that t , r ,r rrull vector ln the Euclidean plane, this is represented by a point at the origin. Stnce it is
a transpose of a transpose is a vector itself. Thus tlrr rrrigin,itlacksbothdirectionandmagnitude,twocritical requirementsofavector
Xz:Xt or X2=XT lr rry1'y1.1 one and only one of the elements is unity and the rest remain identically zero, such
Since vectors are simply an ordered n tuple, they can be sometimes be interpreted as a point ri,,, tor is call a unit vector. The term unit in this context emanates from its magnitude or
rrl,tlr which, as will be discussed later in the chapter, is unit
on an n-dimensional space civen two vectors A and B represent"d by (;t)
""n (;:)
respectively, A and B can be represented as a points on the two dimensional Euclidian plane II V(.('tor Operations
rrlrlrrr',i.t[g;g is a regional block of three countries, each with three production sectors or
il,lr ,ril0s These can be the Manufacturing, Agriculture and the Service sector and from each
',t,)r, output measures in physical units are obtained. ln general, this is represented in a
tMl
,,,, r,,r for m X :lrll where the letters represent output from respective sectors Subscripts
lsl
,rr lr,,added to represent countries For a specific country, the output vector is written as
rl0l
r I tf the Manufacturing
l,lI which means Country one
produced 10 units from sector,

,tlrrrrll frofir the Agriculture and 7 r:nits from the service sector. From the above vectors, many
i rr r ,t r( s can be generated. We can add country specific vectors to get output from the regional
ll,,l rs a whole Alternatively, for each country, we may multiply each vector by a vector of
7T

59
I Mathematical Optimisation and Programing Techniques for Economic Analysis
58 MATRIX ALGFBRA

r10 + Bl
prices for each sector's output. This gives a measure equivalent to the Gross Domestic Product.
t t tt = I o + s I
Thus vector operation is a look at how to add (subtract) vectors and multiplication of vectors. lz + rrl
We begin with the former. ro that the final answer turns to be

Since by definition vectors are ordered n-tuple, the summation then must also recognise this
order. ln the particular example under consideration, the first number represents output from ,,
the manufacturing sector. To add for different countries then, first-elements must add by
ll{l
themselves, the second-elements added and the same for the third The resulting vector will
show, for the regional block, how many units came from the Manufacturing, Agriculture and hl llIul
the Service sectors.
tlrr, procedure remains the same as in (a) above'

x= , t A=l:l.t'r'l
[[l. lul.lu) hrl lt l
lMt l- M2+Ml ' llr|n add the individual elements in the two sets
x= A.l 42+
Itl I +s3 l s2
t8 + 101

',t:ls+ol
li, lrr+zl
t r.lt the final answer turns to be
I18l
x= ,r,rl=lsl
lr, A,,
hel
would have been expected the answer is the same as in (a) above. Adding country A to

lI', ll or country B to A should not change the results. This outcome leads us to perhaps one
;rrrlrular mathematical rule, that addition of
vectors is commutotive'

For each sector, the sum is made across the different countries so that the Manufacturing
sector for example, its output in the three countries is added. The sum is the output from the
sector in the regional block.

Example 4.1:

dua e ements n the three sets

Given three vectors.4 = , = .,r r =
['f] [ri] f],r ".
al A+8,
I10t I8r e
A+B=[;].t,ll t-201
A+B+c=[r+] rr,i' answer too enables us discover another rule related to
Then add the individual elements in the two sets
( ommutat ve rule above. Notice that
 I

60 ] va rrrrx ALGTBRA lvlathcmatical Optintisation ancl Progmrnnting Tcchniques tirr lreottotttit Attrl-Vsili 61

, = *,^,-li
{A+B)+c=
il.Iil -lill
which is a necessary condition to show that addition of vectors is also ossociotive
r, , lrri(lues of multiplication as shown in
H]
figure below

Ll' ri, ,l I Multiplication of Vectors

Two ingredients are needed for two or rrore vectors to be able to add. The first as noticed from
the above example is that vectors should be ofthe same order, they should all have the same
number of elements In some cases however, one or more sectors are missing from one country
such as the Aviation industry in Zambia. These can be shown with zeros against them as l,P =
opposed to leaving the variahle out. The second is that the vectors nrust be of the same type.
Thc're is need for all vectors in summation to be of one type, either column or row vectors.

The next and perhaps more intriguing operation of vectors is the multiplication of vectors
vectors, like matrices, multiply in a rather special way. Elements of one rour frorr the first llI lrr',1 vector, transposed, is still the output vector developed earller. The second is a vector
vector correspond with elements of one column from the second vector The multiplication is rrl ;rr rlr, giving prices for the respective sectors. The product of the two vectors is given by
thr:s a scalar product of a row from the first vector and a column from the second A necessary Ar.P:a.M+F.A+y.S
condition is that the two vectors be of the same order. llr ', r rrrvariably the value of total output, the GDP, because it is a summation of the value of
,rlrllrill lrom all thethree sectors.
l'bt1
lr"l Ir,rtrrlr r.4-2:
To ,llustratc this, take two vectors, = l.rr e2 an) and g =l ,'I Thu forr",. ,, u
lt
.4

A newly opened retail outlet in Woodlands market sells three brands of maize meal,
Lb" l N,)tional milling's Mama's pride, Simba milling's No1; and Superior rnilling's Mealile with
row vector while the latter is column but of the same order since both have the same number
tlrcrespectivepricesof K465O; K4800; andK4450 lf thevolumeof forthemonthof
(n) of elcments. Since A is row and B is column, the requrremerrt for multiplication is already
met. This precludes the need to transpose any of thc, two The multiplication thus proceeds as r,rnuary is given by the vector , = : t..r the rotal sales Revenue (rR) for the
follows [H] [i?]
rrronth of January.
'l'll : P . Q where P is the row vector of respective prices

A' B -AB- (at Ltz ",,|?,tl t tt (46 s 48

t45t
+4 s)1621
Ls rl
Iu,] 't
-
tt 469 * 45 * 48 * 62 * 44500 + 51
AB = atbr + azb2 + .. ...+ a.rb, 't'R
- 2,092.50 +2,976.00 +2,269.50
This is a scalar product that results frorn the nrultiplication of the two vectors. t'R = K7334.00
ln the above example of a three country regional block, multiplying any of the country-vectors .l ,l What is a matrix?
is devoid of any economic meaning Nonetheless, another vector can be defined so that its
product with any of the above vectors carries economic meaning, Given two vectors x and p preceding subchapter of vectors, consideration was on a single economy with multiple
as
follows: , lnformation however might need to be presented for a number of countries, each with
r

62 I varnrx ALGEBRA
Mtthematical Optirnisation and Programming Techniqucs for Economic Analysis I Uf

respective sectors This kind pf information cannot be presented in a single vector but needs an
Wlr.rr rrr and n are equal, the number of row equal the number of columns, the matrix takes a
array, with rows and columns Wlth rows representing countries, columns will be for sector.
,1rr,rr,. .,hape and is called a squore motrix.fhe elements o11,o22,---,o- form the principal
Each element in the array will be linked to a sector or a particular country.
,lt,r11,rrr.rl of the square matrix. For instance, the Leontief lnput-output analysis considered the
ln the case of a retail outlet, we considered only one outlet selling multiple maize meal brands. ltrl,,rtir,,, among many sectors of the economy. lt shows how much each sector is producing and
Suppose now that the outlet is just but part of a family of outlets This must not be difficult to lrrrwllr,rIoutputisabsorbedasinputinothersectorsaswellasthesectoritself Therowsand
conceive. lmagine the same company operating the outlet at woodlands market also has ','lilrlrs represent the input and output of the same sectors. As such, the input-output matrix
several others in different shopping malls around Lusaka. To present sales data, there must be a r ,rlw,rys a square matrix.
column for each brand of maize meal sold and a row for each outlet
l,rl,rrrli the principal diagonal as a dividing line, the matrix is divided into two triangles wlth the
Since each element points to a sector of a particular country or a brand sold from a specific ,rr,, number of elements. These can be loosely called the upper and lower triangles. lf all
outlet, the arrangement of these elements is of critical importance Each element has a specific r'l|trr|nts in one of these two triangles are zeros, then the matrix is called a triongulor motrix
position tied to a country and a sector and cannot be freely repositioned. This results in ',1r|r rlir:ally, the matrix is upper triangular if the non-zero elements are only in the upper
an
ordered array or arrangement of numbers into columns and row. ln mathematical language, trt,rrrpilc. When the non-zero elements are in the lower triangle, the matrix is called lower
such an array is called a matrix Formally, a matrix is defined as an array of numbers, Irr,rrrlitrlar matrix. The matrix A below is an example of a Lower triangular matrix while B
parameters or variables. Thus r,,1rrtscnts the uppertriangular case.

f atL anl

is a matrix
A = laz, arrl
[o.r arr)
with three rows and two columns. lt is consequently referred to as a three by two
^:li i, rl '=ffi fi 3l
matrix commonly written in its short form 3 * 2 and the matrix written as,43*, The subscripts
llrl what if both triangles are zero? lf all the elements off the principal diagonal are equal to
on the elements represent the row and column. As a shorthand form, the above matrix can be
,'r'ro, the matrix is known as a diogonol motrix. This is a matrix with bothtriongles equal to zero.
written as O - aU,i = 7,2,3 and j = 1,,2. The matrix a has elements a1; where ai1 is an
element in the ith row and the fth column ln general, we deal with matrices of dimensions tlr,'matrix
l2 0 0l
l0 -6 0l is a diagonal matrix of order 3 because non-zero elements are only
m * n where both any are any positive integers. lf m or n but not both is unit, then the matrix is 10041
called a vector. Thus vector considered earlier are just a special type of matrices. ,rIrng the principal diagonal and off the principal diagonal, all elements identically zero. ln
,rl(lition to being diagonal if the elements along the principal diagonal are not only non-zero,
4.5 Types of Matrices lrrtl are strictly unit, then the matrix is called a unit ot identity motrx denoted by I The identity

Matrices are of different dimensions or shape and can be written in other forms Alternatively, 110001
rrr,rlrixisoftheforml!
they can be written as J ? !lwithzerosoff theprincipal diagonal andunitsatong Thisis

drt dtz atn [ooor]

azt ,r specialtype of matrix which in matrix multiplication, behaves like a unit in scalar algebra.
,:( azz a2n

Ilrc other pair of matrices deserving attention is lhe symmetricol and skew-symmetricol
am1 dmz o^n
rrr.rtrices. These relate tothe mirror effect of the principal diagonal. lf for a squzre matrix the
Matrices can take different forms and called by different names depending on the nature and rrrirror is placed along the principal diagonal, are elements in one triangle a reflection of
arrangement of elements. lf m and n remain unequal, the matrix is said to be rectangular r'lements in the other? lf yes, then the matrix is symmetrical. The matrix M below is an example
because it takes a rectangular shape when all the elements, regardless of the shape are ol a symmetrical matrix The elements in the upper triangle are a reflection of the element in
identically zerosla4 = 0,V i, j), the matrix is called a null matrix denoted by O. tlre lower triangle and vice-versa
7

Mathcmaticai Optinlisatjon and Programnring Techniques lirr

Economic Analysis I
64 ] MATRIX ALGEBRA

A has an element in B
13 1 71 llre matrices are both of the order 4 * 3 and each element in matrix
u=lr e _zl torrespondingtoitsposition.ThisisalsotrueabouteachelementinBToaddthetwo
b z +l rnatrices, we sum the corresponding elements as follows'
ln some cases however, there could still be a reflection but the elements are changing signs, bn br:l
ratr an dr.rl lbl
from positive to negative and vice-versa Such a matrix is known as a skew symmetrical motrix.
For the latter matrix, principal diagonal elements are equal to zero. Thus a square matrix e * e =l1l', Z1:, Z:=\l.lil', i:: l',1
l=["iilissymmetricolif aij =ajiforall iand-7 andisskewsymmetricolif oi1 =-oi,1.,.11 i
loo, dqz o+zl ln^, boz borl
and 7. Since the condition ati= -ari cannot hold for non-zero elements, elements along the l0it + b1t an+ bt2 a13 + br3l
la21 + b2, o2 I b2 a2.
-l bzzl
principal diagonal are equal to zero for a skew symmetrical matrix as stated a priori. An
| 43, + b31 ar2 * b=, av * fu31
example of this type of matrix is given in tV below.
laal + ba1 an2 I ba2 a4 * ba3)

,: [+ i, Asacorollary,subtractionbetweenanytwomatricesfollowsthesamerulesasaddition.For
the two matrices A and B defined above, then
3]
The different operations and use ofthese matrices are considered in subsequent subsections.

4.6 Matrix Operations ^'),i',,,iii:Nklri::iA

Matrices have been defined as an array of data, representing a continuum of scenarios and
latt - btt atz- brz ar: - brsl
contexts There can be matrices representing output from various sectors ofthe economy, sales lor, - b7
= f rr, b,
ozz- bzz az: - bzzl
of different commodities, matrices of prices, and many other variables. Depending, on the type - asz - bsz an - brrl
of data represented, there may be need to add (or subtract) or multiply matrices. Division of lao, - bo, a+z - bqz aa3 - ba3)

matrices is also an important part of matrix algebra as it enables equation solving. This section of indivldual
Both the sum and the difference are the sum and difference, respectively,
will deal with the addition and multiplication of matrices but postpone the discussion of the
correspondingelementsTheorderoftheresu|tantmatrixalsoremainsunchanged,inthis
division aspect. Like under vectors, as special matrices, the subtraction of matrices can be
partlcularcase,4*3.
imbedded in the addition and the discussion of the latter must be construed to include the
former. txample 4 3

4.6.1 Addition of Matrices Given two matrices aefined uv.a = [! 3 -rt] .ro , :l-; I ll rrr'a
ln the addition of matrices, corresponding elements are added This requires that for any two
matrices to add, they must be of the same shape so that each element in one matrix has an
a) A+B

element corresponding to lts position in the other matrix. The position in matrices is defined by
the row and column that the element is in the sum will also be a matrix of the same order with
/+B:[?
t-2
3 ;'1-'
73r
t7 i ll
elementsequal tothesumofthecorrespondingelementsofthetwomatricesadding Fortwo =ls l0 Bl
matrices A and B given by

tQtL An Qrz b| bD brzl

b) B-A
n _lo^ azz azs b^ bn br=l
" - lazt Qsz o:r: andB= bT b, b..
loo, a+z aqz
I B-A=l-; ; il-t? ?,
-11
5l
bo, bo, bor]
:r: =^ :,1
r
68 | ,o.*,* ALGEBRA l\larhcnraticul C)ptinrisation rntl l'rogrrnrnringTechnir;ues lirr licorrrrrrir Arrrllsis 69

,,v,, ,,laLement has implications on certain type of matrices. We defined above that
is interested in the total number of each animal. This is given UV tt" ,urn ot [7 rical matrices have elements above the principal diagonal more of a reflection of
Irs s below. Transposing such a matrix will leave it unchanged. Thus given a symmetrical

the two now have to decide urhether to so to Livinsstone, the tourist capi A, then,4' = A. This result also extends to more special symmetrical matrices. The first is
=
[;] tji] l,,rr,rl matrices which has zero above and below the principal diagonal For a diagonal
or Lusaka, the capital city Both markets are of fairly the same distance hence , l), rlron D' - D Ihe second and perhaps more special is the identity matrix I which also
transportation cost A rational farmer will prefer where one's total revenue is higher to the condition /' : 1

Prices in Lusaka for the three animal are given as follows: Goat: l( 105, Pig: 320, Chicke I lrr0c propcrtics oltransposes are vvorth stating:
K 25, which can be expressed in a row vector (105 320 25) For Livingstone, l: When transposing a matrix, what are rows are written as columns and what are
are slightly different and are represented by another row vector (130 300 25).
columns become rows. lf this procedure is repeated, swapping again, then the result is
price matrix can be defined with rows representing the two markets and columns for
the original matrix. The rows which had become columns are again rewritten as rows.
different animals
Thus transposing a matrix twice leaves the matrix unchanged. This is the general result
To get an idea of how farmers will decide, multiply the price matrix and a quantity when the number of transposing is even. (A')' : 4
I

in which rows represent animals and columns for the two f

' ll: Given two matrices A and B. lf the product AB exists, then its transpose is equal to
I r os 3zo z5rl:
e.
| : pazs 27ssl the product of transposed A and B taken in reverse order. That is (AB)' : B'A' .
-
it
I rao 3oo zsl
[,i, ln] Izz:s Zszol ple45
ln the product rnatrix, columns represent farmers while rows aTe markets. lf they go
ii
Lusal<a, farmer 1 will make l< 2, 825 while farmer 2 makes K 2, V95 Clearly, farmer 1
better off with Lusaka while farmer 2 would prefer Livingstone " ,,, --
r) -.3 6r
(1 ,. i)."0
, /3
(;, ,l)
5r
,n""
Since they are using
choose
the same transport, they must choose one marl<et. Which one t
will depend on each farmer's bargaining power. In the extreme, farmer 2
even consider compensating the other for the opportunity loss and still remaln better-off
ma'
.' i )C; )l
l(1
4.6.3 Transpose
i
=l=; ru lil
i lf columns of a matrix are written as rows, or vice verso, the rectangular shape of the iil:rz
changes. lf it was an m x n, then after changing columns to rows and vice versa, the A'B' does not even exist since the two matrices are
Notice that in this particular example,
becomes n xmin dimensions. The new matrix is known as a transpose of the former and vi not conformable. However, swapping the position in the multiplication evades the
versa. Like in the vectors discussed above, the transpose of a matrix is denoted by a prim, problem. The product now is the same as above, proving the law.
symbol or a superscript 7 Given a matrix A, then its transpose is denoted l:y A' or Ar

^:l\, ?]
,,". e'ore'=[ln i J
e a,=ll,
l+ zlb -s 4
lli
-,' ?l

An element a1; which is in the ith row andTth column in matrix A changes to a;;, it is now in
j'h column and ith row in the transposed matrix. ln simple terms, to transpose is to mirror
t =ri , itl; i il l;r ,i, lit
matrix along the principal diagonal Thus, transposing does not affect elements along r ,,i,lV lll: The transpose of a sum of matrices is a sum of the transposes Algebraically,
principal diagonal 31
(A + B)' = A' + B' . Given two matrices A and B defined by ,4 = 54 and
2-4
| ,or*,, ornrr*o Mathematical Optimisation and Programming Techniques for Economic Analysis

lrrlllrr,(l l-o clearly show this, the matrix is written with letters for columns and numerical
, :l:; jl,*"" in the equation above we solve the two sides of the equation to rlr,,r rrpls for rows.

show that the equation holds: lat bt cr I

rrr rlrl rlrrterminBntlAl =la, b2 crl, repeat the first two columns after the last one. The
(A+ B)' : A'+ B' lo= b. crl
l, lilrr.0l the expanded matrix is of the form.
:
rc !_). (:; )l [; !-l.l:; ')

I i] : r:, i :t.t:^ ,t il
t: ? 8l:t: i 2t

4.7 What is a determinant?

A determinant is a uniquely defined scalar or single number associated with a square matrix.
For a matrix A, the determinant is denoted by 1.41. This need not be confused with its use for
, lr ,rr(rw represents a product of the three elements it joins. With the two pairs of arrows,
absolute numbers. ln scalar algebra, such type of brackets indicate that the algebraic sign of a
,rrrr tlrc upward sloped and the downward sloped separately. The determinant is given by
number be ignored. ln matrices however, this is just a way of showing a determinant which can
,,l,tr r( ting the upward arrow sum from the downward arrow sum Thus
unquestionably be negative. Care should also be exercised to distinguish this from the double
DetA = (arb2ca I brcro= * crarbr) - (a3b2c1 * b3cra, I crarbr)
bracket introduced earlier in the chapter. The determinant is an integral part of matrix algebra
hr tlr,, lull equation, there are nine different elements, given a 3 x 3, and each element appear
and will prove useful in latter concepts.
lwtr r, ,rrrd only twice.

tn a 2 x 2 matrix given o, e = lZ)', Z::), ,n.determinant is of second order and is defined bv lrrrrrrrkrrstand the second method, we need to introduce some additional concepts which we
rlr ilr llrc next subsection.
oetn=1f,":):l
,l ll l\,linors and cofactors
- AttAzz - OZtQtZ

which is a scalar since it is a sum of products of scalars. For instance, the determinant of a lllr' r'r;uation for the determinant given above can be factorised to make it look attractive. The
matrix such as i, oU,"in"a through cross multiplication as follows: il,lr.rtion of elernents to factor has to be strategic however The strategy is to factorise
[! l] rr[,rrr,nts from one row or one column. For illustration purpose, we select the first column

13 1l:r" !-sxz llr.nr,nts, the a1 a2 and a3.

=3-10:-7 D etA = at(bzcs - brcr) + ar(b=c, - brca) + a3(brc, - brcr)

To obtain the determinant of higher order matrices, we have to resort to methods that will = a1(b2cs - btcr) - a2(brca - b.rcr) + a3(brc2 - brcr)
o'
eventually reduce the calculation process to cross multiplication of second order determinants.
= ,, l: ;, ,,1 - '.'l * ,.1'. ll, ::)
Two methods stand out. The first is loosely referred lo as exponsion by columns The second is l' b' c.l ",1.
l. t= crl l* .- :l
t},e Loploce Exponsion. We deferthe latter and deal with the former now. I lrr' lrarts in bracket or determinant bracket are determinants of 2 x 2 matrices. These are sub

To get the determinant, for instance of a third order, write the matrix and repeat the first two 'l''lcrminants of the matrix A which are multiplying with elements from the first column. Their
r rrrrr;rosition is however not arbitrary For the first sub determinant in the equation, it is as
columns after the third. This should result in five columns and the order should remain
tlr)rgh the row and column of the element multiplying is deleted That rule applies for the
r

72 I MATRIX ALGEBRA Mtthcnrzrtical Optirrisation and Programnring Teclrniques for Economic Analysis 73

other two. The sub determinants are called minors denoted by Mil.fhe minor Milis associated ll'rlng the Laplace expansion, select the first row and expand using cofactor
with the element in the ith row and 7'h column. lt is gotten by ,deleting, the lih row and
column and evaluating the determinant of the remaining sub matrix
7th t;l='l::l-,li ll.,13 il
= 1(18) -2(4) +3(20)
A concept more related with minors is that of Cot'octors denoted by c1;. A cofactor :
is a mi -50
with an algebraic sign added. The rule for attaching a sign is as follows: if the sum of i and 7 for
a specific minor is even, the minor takes a positive sign; otherwise it takes on a
negative | ,t l'r 0l)crties of Determinants
can be condensed in the equatio n Ct j - (--1)i+) Mi7 where I and are as defined
7 ex onte. Wttlr tlrr, (loterminant at hand, a couple of insights of the determinant are quite useful These
Armed with this knowledge, we are now ready to evaluate a third-order determinant such ,rrr ,,rl[,rl properties of determinants. lt is essential to know how the determinants changes
as:
lorr dtz orrl *ltlr tlr matrix changing form. This is simply an assessment of the behaviour of the
lAl = la1 ezz arrl rlr,l,'l lr lll l.lnt
lo., atz a=rl
The method is as follows: the value of the determinant is obtained by the following I'r rpr,tly 1: lf any two rows or columns are interchanged, the value of the determinant of the
equations
where M1; and C17 are the minors and cofactors of the determinant matrix only changes in sign. Swapping another pair of rows or columns (as the case
may be) will again change the sign only Since the change in sign is binary, negative
DetA = arrMr., - aztMzt* arrMy or positive, an even number of row or column interchange will leave the
= atrCtt * a2rCr, + a3)C3t determinant unchanged

= r,,,r:,, fatr arz atsl

2" "
This method is what is referred to as the Laplace Expansion mention before. lt is
Considera matrix I =lan Qzz nrrl frorthe
Lo' atz arrl
matrixA, ittwo matrices B and C

named after a
19th century French mathematician and astronomer pierre-simon Loploce. we put are defined as
it as the
second and more formal method of evaluating a third order determinant. The above
expression loil or.i ntrl 1,,r, or: o',1
does not put any restriction on which column (or row) to use for expansion. The formulae B-latt Qtr azzlaodC=lozt att atrl
works with any column or any row, provided the cofactor are adjusted accordingry. The [rr, ezz arz] 1o.,, ett ur,)
Matrix B is gotten by interchanging the second and third columns. The property
expansion can still be used for higher order determinants but the procedure will
be states that the determinant of matrix B will be negative of the determinant of
multistage2e. ln general, the determinant of a kth order matrix A is given by matrix.A.
k

ltl =\oaci1 lBl: -lll

k
The matrix C is derived by interchanging the first and third columns of matrix B. This
:Y.
/J4ttvLt "
means moving from
Therefore, the determinant of
A to C involves and even number of column
will equal the determinant of A.
swapping.
j=r C

Example 4.5 lct = l/l

l)roperty 2: ln a matrix, if any row or column can be expressed as a linear function of the
remaining columns or row, then the determinant is zero. This embraces a more
/
tzi t
Find the determinant of a matrix =
extreme case where two columns or rows are identical Consider a matrix
il 6 2t
e:h 1 sl
2e
At every use of Laplace Expansion, the cofactors are of order one less than the parent
matrix Thus given a fourth
order determinant, the first application will produce third order cofactors, which will in
turn require the same
lezrcl
procedure to reduce to second order.
r
I

74 I MATRIX ALGEBRA
Nlrtlrerraticrl Optirnisation :Lnd Prograrrming Techniques tbr Econonric Arrrlysis

where the third row is twice the second one. We should then expect that
determinant is zero. 'l rtvt'tse of a matrix is the answer. Given a non singular matrix A, a matrix that pre-
or post-multiplies by A and gives an identity matrix is called an inverse of A it is

^- l),'.-,li ,11
,.1 i by ,4-1 and may be simply referred to as a reciprocal of A. For a singular matrix, the

-
1(0) 3(s6) + 6(28) rloes not exist and this can be revealed at determinant stage. With the problem at
-0 lhF, solution can now be put simply as
This confirms our expectation that if any two columns or rows are linearly relat
the determinant is zero. Caution should however be exercised so as not to co A-1Ax - A-18
a linear and non-linear relationship. This property only applies to linear relations
lx = A-18
and does not include a case where one row or column is a square ofanother.
x : A-78
Property 3: lf any row or column in a matrix is multiplied by a constant k, the value of i
determinant also multiplies by the same constant k. lf another row or column ,' rtrrx r'1 - [o,r], aeiine a matrix of cofactors where each element in A is replaced by its
multiplied by the same constant, the determinant will increase by a square of The new matrix will te C : lcql, where C;; is a cofactor of a;;. Then take the product
constant. ln a more general case, the determinant will increase byk! wherep is "rllr/'commonlyreferredtoasodjointof A ltissymbolised6yadjA.Thetwomatrices
, ol, )ilnable since they are square matrices of the same order Thus:
number of columns or rows scaled up by k. Given , ,.trl* 1=
[3 i]
determinant can be calculated with ease
for instance, the new matrix is,41 =
lll = -l
[3 ]] rn"
tf the first row is

aet"rminant of the new matrix

multiplied by
is ( '":^', ,',')(?,t 'j,t, ?,')
13 ll - -
o1+1 s1ol /ai atz
o:'
Czt crr\
c,,,
Czz
=-zt=?t2 3r
=1": I
'ls 4l
\d nr art2
.

':)(?,," crn C nn/

I
Property 4: if any row is changed by subtracting or adding a multiple of another row or co
as the case may be, the value of the determinant is unaltered. Consider a

lZ j]. lt, a"t.rrin.nt l"

uol=
"o- bc. From the first cotumn, subtract
(; atjCtj
n
Yo,
L''''
r"
t a rtcnt

second column twice so

calculate the new determinant.
that the new matrix . li _3, j] .rd nro.""a ozjCtj
L
4
\-) (12
tL2 t
t az)cnt

uol
_3i = or" - 2b) - b(c - 2d)
17
:ad-2bd-bc*2bd
=ad-bc
Il orrjCt j

:lz ,,1 thls expression, each element is a Laplace expansion for the determinant of A. for elements
the principal diagonal, elements from a particular column are being multiplied by their
tive cofactors and should equal the determinant of A denoted by lAl. ln off diagonal
4.10 Inverse of a matrix
however, column elements are multiplied by what was referred to as olien cofactors
ln matrix algebra, the division of matrices is inconceivable. Matrices cannot be divided (the row value I defers between the element and multiplying cofactor). We already know
scalars and thus dividing two matrices such as 4 is not permissible But then how can one I r, ruld equal to zero

an equation of the form ,4r : B other than by writing , : B le as in scalar algebra?

7-

76 I MATRIX AI GFBRA MltthcnlLticitl Optirrisation rnrl [)rLrglrnrnrinl I cchnirlLrcs ]irr [rcorrorrrir '\rlLllsis

This produces a diagonal matrix with l,4l along the principal diagonal and zeros elsew lr, 1,1.,1 .,"0 to get the inverse of a matrix is to find its determinant 1hu,; using thc
/A o .. 0\ rt 0 0r ,1,l,rr c method, which the reader should now lre conversant with, the delurnrrnant of
I
ec:( g
""-\;; I oi_,/o r o\ rrr trrx A is lAl :43. The determinant is non zero, so the inverse is assured NexL i:; to
l,rrl llro CofactoT matrix and the adioint, then .4 | can be derived
A)-''\;o ,)
AC, : IAII
1, ')
ll :,: l:., _"ll
Pre-multiplying both sides of the equation by
formulae yields,4C' = l,4l
A the inverse of A and mal<ing it the subject
,-llo,:l
" -1,: l:, -orl
till
I

4 t46' : lAlA 1
L r; :t t:: ;t
c, = lAlfi_a
'" t '-!c'
lAl
= | '.1
L t,,
;i1 4l ',1

Note that l,4l is not a matrix but a scalar and so can be divided like an ordinary scalar I t4 -10 ,A
determinant plays a critical role in the inverse of a matrix In the equation, if the determinan c-l-s 2?
l/ I - 0, then the inverse is undeflned, it does not exist Such a rnatrix, with no inverse, is I t,, r

114
AtiA=l n -s
a singular motrix The existence of an inverse depends on having a non-zeTo determinant whl
also depends on linear independence of rows and columns.

Since we defined C' = Adj A,the inverse of A can also bewritten as

22
t'r ') l']
tl14 ))s
,4r:-l-10 -
-10
A t:AAdiA
I
431 -; 1.

ln summary, gettlng the inverse of .4 requires the following three steps t I (lramer's rule for solving simultaneous equations

1 Find the determinant of the matrix, l,4l ',,rre commonly used in linear algebra With a linear algebra given in matrix form
2 Replace all the elements in the matrix by their respective cofactors to get the matrix ,/ with A a coefficient rnatrix, x is a vector of variables and d is a vector of constants As
cofactors, C y known by now, vectors are just a special kind of matrices and multiplying them should
3. Transpose the new matrix of cofactors C to get the adjoint of A (,4d1 A) ,,rrrrusual The solution is found by pre multiplying by the inverse of A on both sides to
r / td This is a tedious method as it requires finding the inverse of A
Then, the inverse of A is defined
r, r,)u: theorem called Cromer's rule proves a lot easier it is named after an eighteenth
A t--AdiA
1

r y Swiss mathematician Gabriel Cramer, whose worl<s brought it into the world lt is a lot
tAt
,rnd convenient to start with a system of two equations with two unknowns
a..xr + o,2x2 = tl,
eztxt+o.22r2=d2
can be expressed in matrix fornt Ax = tl. as follows
Example 4.7
lz:: z:ltI'):lx:l
tor the matrix
12
II
0
, ., nverse.4 1 rr rrrlillreeliminationmethodof solvingsimultaneousequations,thesolutionsforxrisgivenby
.4 = 2
l-z -r i] "o
(at t azz - arrorr)x, = dtozz - dzatz,
r

18 ,or^,* ALGEBRA l\4rtltet)latictl Optintistrlion and Prograrnrning lcchnitlues lirr []torrorrrrr t\tlrl\sis 19
|

dtazz /2 -l l\/rr\ /4'

rr , =- - dzotz
O11422 - 421 Q17 (i :',;;Xi) (il)
I
Before movingto x2, a closer look at this one will be helpful. A closer examination of Ll ,rng Cramer's rule (with the help of the Laplace expansion for determinant. First B{rt thc
numerator and the denominator will reveal that they are actually determinants of 2 x ,lllcrminant ofA and then A1, A: and A:.
matrices. A careful arrangement willgive
21. 1
lAl = 342
3-2 4
x7
:,1:, il- =l_; 7l . , 17 -:,1
Corollary, x2 will also be given by - 2(r2) 3( 6) + 3(6) : 60

t4
-t ll
lr,l-lrr 4 -zl
lr z 4l
ln both solutions, the denominator is a determinant of A, the coefficient matrix. The numera
forx, is still a determinantthough of a new matrixA, formed by replacingthefirst column of -ol:, -;l ,,1 ) ;11+,rl i -)l
with the vector of constants d. Similarly for rz, the numerator is a determinant of a new - 4(12) 11( 6) + 11(6) = 1ss
42formed by replacingthe second column with the vector d. With the two new determi Thus

x,' , -- lA,l
the solutions can thus be written as

lAl
xt" = 1,4, I

: 1Bo =
rAr :.t
b(l
_, _lAzl
^' - CorollarY, xr' = f, a.' - |
lll
though this is illustrated using a smaller matrix, the method works for a coefficient matrix A
ll'll:illkofamatrix
any order. The method does not require calculating the inverse of A. lt only relies
determinants This is the method referred to as Cramer's rule above ln a more generalform ,n X n matrix is a matrix with m rows and n columns. lf all columns and rows are linearly
1

a coefficient of any order, the rule is written as

,,,,l,, ndent, then whichever is lower between m and n, is the rank of suclle-Ila!rix. The
rlx is said to have a full rank. However, in some matrices, not all rows or columns will be

xi. - lal EiT-wtren there is linear dependence among coiumni-ffirows,then the rank
rtr ,,irisiders the number of linearlyln?&'b-ii#ftfa-otr"rriin;aiid ro,//s. Thua m and n need to
where -47 is formed by replacing theTth column in A with the vector of constants d number of linearly dependent rows and columns
d, by reducing each by the
Example 4.8
ly Then whichever remains lowest is the rank. Formally, the rank of a matrix,
t, rr,,1r,rl bf r(.4), is defined as the number of independent rows, or columns, and where the
Given the following system of equations, use the Cramer's rule to find the equili ', rlrllc-'r, the lower value counts Linking this with the earlier discussion on the inverse of a
values x1-, x2* and xa* ,, r rx, we conclude that only matrices with full rank will be non-singular

2x, x2-x3=4
|
13 z tl
m -n= 3
3x, I 4x.. 2x, : 11 .r ilr,rlTixgivenby,4 ll 4 3lisasquarematrixwiLh Thereshouldnotbe
3xr-Zx.r*4xr=11 Ls ro 7l
the equations are arranged in a matrix form Ax = d as
r, irt)tation to think that the rank (/) : 3 A closer look should reveal that the third row is a
 I

80 I MATRIX ALGEBRA Mathematicar optimisation and programming Techniques for Economic Analysis
81

linear combination of the first two rz -= rt * 2rr.fhe number of linearly independent rows (or
tA_^rt=ln;^ ,r_^l=o
columns) is not three but two. The rank of matrlx A is /(A) : 3
+@-t)(7-1)-(z)(z)=o
4.13 Eigen Values and Eigen vectors -72-ttl+24=0
The whole idea behind matrices is the search for a better and convenient way of handling
- 8) = o
=?('1
mathematical expressions. We continue to transform and explore different types of matrices ^,:, ";of,
llhlch are eigen values of the matrix A.
and operations so as to equip the reader with mathematical prowess necessary !o handle
different economic problem. The next step is to consider what are known as eigen volues and
lor,tr = 3, we have (A - 3t)x : O
'
eigen vectors. The word eigen is a Germany word translatin1 to own. So we are finding a
matrix's own values and vectors. These are sometimes referred lo aschoracteristic roots ti itfr:t='
lhr coefficient matrix is singular as can be noticed from the rows or columns . x1
* 2x2:
Consider a set of equation: Ax : 7x where A is a square matrix of dimensions n; .t is a non null 0 ..o x, = -2x2 andthe normalisation equation
column vector with n elements and tr is a scalar called eigen volue or chorocteristic root.
x12 lx22=1
Ax=7x ) (A-ll)x=0 lhr unique solution forl., : 3 is
Th,e identity matrix / is added for conformability, that is, to enable the two subtract For a non-
trivial solution to exist for this set of equations, the new coefficient matrix ,4 -,i/ must be
singular. This implies that its determlnant lA - lll must be zero. ',=l-i1"1
I t,,ts l
lor ), = B, we have (A - Bl)x = o
lA - lrl
t;'!,l,11)l='
An expansion of this determinant will produce a polynomial in ,tr often referred to
lx1 - x' = | 1 1, : Zxrwiththe normalisationxrz + xr2 = 1
characteristic polynomial. The roots of the polynomial are called eigen volues or chorocteris
roots will prove useful later. A vector x associated with each eigen value is called elgen
or chorocteristic vector. With a singular coefficient matrix, the consequence is that a un
solution cannot be found. There will be linear dependence in the equations One is a multiple
",=llfl
I J-sl
r
the other. The resulting vector will merely be a ratio of the two elements, which is
Dlrgonalisation of a matrix
indicative of direction Eigen vectors lack the second attribute of vectors, that is, magnitude.
impose this attribute, eigen vectors are normalised to a unit magnitude. Thus, the rlg.rr values and eigen vectors at finger tips, diagonalisation of a matrix is now within
l)r'rrionalisation, also known as spectrol decomposition, is the transformation
equation of an
txr2 =1 v rrr,rlrix into a diogonol form, with off-diagonal elements all equal to zero. The
x12 drive for
r'rli ,,rrr,. is to find a matrix that is easy to manipulate. Diagonar
is used to get a unique eigen vector. The r, are not vectors but elements ofa single vector. rnatrices play a very
l,rrrl rrrl' in simplifying the mathematical operations of economics.
elucidate this distinction, the letter u is used to denote a vector with elements denoted by xr. For instance, diagonal
' rrov(' much easy to multiply than ordinary matrices. ln addition, they provide a short
Example 4.9 1,,,1'l('rminingthesigndefinitenessof quadraticformsthatcomelaterinthechapter

r ,'lri,'. v,rlues are distinct, the matrix A can be transformed into a diagonal
Given a square matrix A --l: l],
fina tf,u eigen values and the respective eigen nratrix by using
Ir rrration matrix T. This is a matrix of eigen vectors
of A. That is
We require that
T: (\ uz vn)
 ALGEBRA Matheuratical Optinrisation rnd Prograrrming 1-cchniques lirr licorrrrric Arrirlysis
I MATRIX

forasquarenxnmatrixA.Wehavenointentionsofdevelopingtheprocedure,ratherwe
give the results. The procedure is well developed by chiang and wainwright
(2005).30 For t-r'-'r] t;l]
=
'
given
xt+x2=0=xr=x,
I matrix A, its diagonal equivalent D is by31

D:T-1AT
ThediagonalmatrixDwillhaveeigenvaluesalongtheprincipaldiagonalintheorderWhi(
correspondstotheorderinwhichtheeigenVectorsappearinthecolumnsofT.Thatis,Dw lrr 11 |

be given by rtr(,refore, the matrix of eigen vectors T is given by, = I iY'o v.,ri, t i, of
-/vm ,'.V2
'/al 1

|
rt :" 3 3l two, and its inverse must be pretty easy to find, The solution is left to the reader to
'=[ s' ;il
r rr deT

;,r ove32 For the equation

D=
thei'aretherespectiveeigenvaluesorcharacteristicrootsfortheeigenvectorsusln
matrix T.

:A
Example 4.10
17;
Given the matrix /4 = [t tina tft" eigen values and vectors and the corre:
]1,
diagonal matrix D.

tA-^rt=lo;^ ,t_^l=o t?;

(4 - 1)(2 - 1) -
-+72-67+5=o
(3)(1) = 0
[3r
Given this characteristic polynomial, the eigen characteristic
roots are
'lr uld have been expected, that once eigen values are known, the diagonal matrix is implied
L= I and 7z- 5
rr,sult has just proved that the eigen values are the elements of the diagonal matrix. With
For )', = l, rli,rgonal matrix known, there is no more need to go through that lengthy process Once

t3 il tlll :, v,rlues are known, so is the diagonal matrix In the diagonal equation, we can still make A
',ul)ject of formulae and make use of the properties of a diagonal matrix. Simply pre
3x1* x2 = O =+ xz = -3xz ly with T and post multiply by the inverse of T on both sides. Then
and the normalisation equation x12 + x22 = 1 the unique solution for tr' = 1 ;5
A : TDT-1

'; l'ositive and non-negative square matrices

and non-negative matrices must not be perceived as a new concept. Even for a'novice'

: rr.rnes should give a clue. A positive motrix is one all of whose elements are positive. lf all
And for /1, 5
nLs in a matrix are positive, then the matrix is said to be a positive matrix. On a number
the zero divides it into two potions, the negative and the positive side. The zero itself does
ro
chiang & Wainwright (2013)
tt Fo, oihogon.l eijen vectors (rr), it is permissible to use the
transpose as opposed to inverse of the matrix '$/ rt ) I I rc principal diagonal elements, cha nge the signs of off-diagonal elements a nd divide the whole matrix by
the equation nant
 Milthematical Optimisation and Progriurming Techniques tbr Econorrric Analysis
84 MATRIX ALGEBRA

not fall on either side. As a consequence, a positive matrix does not include zero, because
l ilr instance, the agriculture sector uses output from the industry in form of farming
as well as its own output as seed.
number is not positive.

When a matrix has a zero in it but all other numbers are positive, this would imply that
r,,rr h sector, there are two sources of demand. The first is from the same sectors as

matrix has no negative number in it. lt only has zero and positive numbers. lt is thus e inputs in the continuation of production. The second is the final demand. When
r,, used to brew beer, then it is said to be used as an intermediate input into the
non-negotive motrix. lt is defined as a matrix all of whose elements are non-negative. A
rrr lron of the final commodity beer The total maize needed for domestic consumption is
related matrix is a semi-positive matrix. This is a non-negative matrix with each row and
r', roferred to as final demand. For equilibrium to occur, each sector must satisfy its total
has at least one positive element.
'1, rr,ilr(1, that is, it must produce only enough to meet the input requirements of all the sectors
4,16 Cayley-Hamilton Theorem r w,'ll ,15 the total demand. This is also called the Leontief lnput-Output model, m;r-twentieth
r,rtrry Russian-American economist Wassily Leontief Let us now develop the algebra.
For any square matrix A, the characteristic equation is a polynomial in lambda (.1). The equat '

is denoted bV f(1) = 0. Cayley-Hamilton Theorem says any such matrix satisfies its I t,,lrrrr, a coefficient aij as the total amount of output i needed to produce a unit ofl. But the
characteristic equation. lf the matrix A is substituted in place of tr in the equation, the ptorlttction of7 will not be restricted to a unit. As such the total of an intermediate input I that

still holds.Thatisif/(i):0,then f(A)=O.Foranillustration,takea matrixA=lSZ wrll lrc required to produce xj of output/ will be a1;x7 For instance, if it takes 20kgs of maize
3]
(lrr ,rrldition to other inputs) to produce a unit of beef, the it will require 100k9 to produce five
characteristic polynomial or equation is given by
rrrrrls of beef. Since each sector must satisfy its total demand,
f(^): lA -.r11 = 6
x7: * ... * I
:l'-r^ ,1,,1=o ar7x1 + a72x2 a1nxr.

ttrr the left hand side is the total of commodity 1 produced. The first term on the right is how
d,

:(2 - - 1)- 2(4) = o rrtttchofcommodity1isneededtoproducecommodity1-@

^)(8
=12-107+24:0 r.rluirements into secto, and so on. The final demand for commodity 1 is depicted by d1. lt
the equation should also hold if the rnatrix A is substituted for.1. The terms, in the
nrrst suffice to state that nothing prevents any ofthe a;7s or dls from being zero. Sectors will
multiplying with the matrix will now be matrices themselves while the constant term remains
rrrrL require inputs of every sector, making some ai1 zerc.ln addition, some commodities may
scalar. This will defy the rules of addition of matrices. To circumvent this, the constant
multiplied by the identity matrix I of the same order as the matrix A. Thus
lrl --
purely for intermediate use only As such, their final demands are unequivocally zero.
I lowever, none of these elements can be negative.
f(A):42-roA+Z4t=0 or the other sectors, the equations are as follows:
=ll, tll:, Zl-*ll, []+z+l[ ?]='
I

: *'.. t *
:l-i, t:;l-l::, ;31.t"- !rl:'
x2 e2axa + ozzxz a2nxn d2

t0 9t:o
=19 0l xn= dn1x1+ en2xz+...1- annxn+ dn

ln matrix form, the model of the form.

4.17 Input-output models
d7
ln an economy with many sectors, there is always some interdependence among all or
d2
sectors. ln addition to the primary input of labour (and land), output from various sectors
used as intermediate inputs in the other sectors. For instance, the agriculture sector depe ;^
on the industry for its implements; the industry on the other hand will rely on the agricultu x:Axtd"
for its raw material and so on. ln some cases, a sector may use it own output as an inter
86 I ,o.*,* o,nrr*o Mathematical Optimisation and Programing Techniques fbr Economic Analysis 87

The matrix.4 is the coefficient matrix rt is square

and nonnegative. rt dimensions ar( r,r r,l,.r lor example a commodity like Lime. This is an industrlal output. lt is used in the
determined by the number of sectors or industries and
the prevairing production techniquer iErt, ullur,',r.ctorandinthemanufactureofcement ltisneveruseddomesticallyimplyingthat
determine the actual elements Given the input output
coefficient and the final demand ue.tor, rt- lrr,rl rlr,n)and is zero. This should not mean that Iime should not be produced since doing so
the task is to find output vector.r The process is purely
matrix algebra. wlll rrr,rl,r, irrrpossible or inefficient the production of cement as well as cement which depend
x=Ax*d lr, llrrr,,

(t-A)x=d { lll l)r'('ornposable and indecomposable matrices

Then pre-multiply by the inverse of / -,4, also known as the Leontief inverse I rr',r(lr,r ,r square matrix A with dimensions n x n. The matrix is said to be decomposable if it is
x:(t-A)-Ld 1,r,,,,,t1r[, lo interchange or rearrange its rows and columns is such a way as to obtain a matrix of
llrr,lrrrtl
x=8d", whereB=(t_A)-1
, lAl Arrf
Example 4.11
^= lA^ Arrl
Lrr lr r,[,rnent in the above matrix is a matrix. Thus decomposing is breaking the matrix into four
A three sector economy has the input output matrix "rrlr rrr.rlrices, with some pattern however. This pattern is what will distinguish a decomposable
, = [.i 1 : and the final
[.0 .4 2 rrr,rlrrx lrom an ordinary. lt is required that the sub matrices take to following form. Matrices
t3'l. 1,, ,rrrd.422 must be square matrices with dimensions kxk and (n-k)x(n-k)
demand for the s ctors' outputs ,, ,: I *nr, reu"r of output from a, the three r,.1rr,rlively. Thiswill leave A' and A21 with dimensions kx(rz-k) and (n-k)xk
Irooo]
sectors will ensur that the final demand ir satisfied? ,,...1,,,r tively33.

using the formura iilustrated above, the first step llr'. l)cacon for decomposability lies in the matrix .4rr. ls it possible to rearrange so that this
is to find what was referred to as the
Leontief inverse. This requires getting the inverse
of the matrix (/ -,4). We assume the
rrr.rlrrx is null? A matrix is decomposoble if after this rearrangement to form sub matrices, the
reader is arready famiriar with inverting a matrix ,l ,, will be a null matrix, with zeros. lf the above is not achievable, the matrix is said to be
discussed under section 4 L0. I _ A
I .7 -.2 = tutlt'romposoble. As a rule of thumb, a matrix will be indecomposable if there isa zero in every
l- r .; -.;l
-.41
[.0 .8]
-.4
and thererore u\! - a/
41-, = lfi ZZ l|l
*l r.zo thererore , rrlurnn or row
.zl.s3l
x=(l-A)-1d lrr r,( onomics, a group
of industries or sectors as said to be indecomposable if every industry in

1 [ 64 .32 .401 l 500 I llrr.cconomy is linked, directly or indirectly through others, to all other sectors. A sector is
=,,*l:3 (lrr.ctly linked to another sector if one sector directly depends on the other for inputs or
i3 3111,,$,J rrr,rrket for its output. lf the linkage is only through the two sectors' linkage to a common third
..= [ ?;gl ] ,,r.r lor, it is called indirect linkage.
lzfit.sl Wlren some sectors directly or indirectly sell their output to others but do not buy from them,
The above example is v )ry interesting and worth llrc economy is said to be decomposable. Assume a seven sector economy represented
a comment ln the fi al demand vector, the by
final demand for commodity 2 is zero. Nonetheress, rliure 4.4 below.
the moder stiil says 187s
units of the
I
commodity must be pr rduced This shourd echo
the point that some commodities may be
producedpurerytomeettheinputneedsofarthesectors
Thewholeoutputofcommodity2is
used as an input into producing the same commodity
2 and in producing the other two
commodities

' the matrices would as well be vectors or scalars These are just special types of matrices and form part of
possible sub matrices
r

88 I MATRIX ALGEBRA Mathcmatical Optinrisation and Programming Techniques for Economic Analysis 89

Figure 4.4. Seven sector completely decompasable model I ', I rve Sector Decomposable Model

Though the economy has seven-sector represented by the numbered nodes, the sectors are
all inter-linked. Sector 1 is only linked to sector 2 while sector 3 is linked only to sector 4.
thc above figure, all the sectors are linked, some directly in either a two-way fashion or one-
same applies for the last three sectors.
l,rshion and others indirectly. The technological matrix is given by matrix B below
The technological matrix representing the economy in Figure 4.4 is given by the matrix A be
('6= o6) (il:)l
(i1:H (,:,)l
(o o) (b.r)l
Bn Brsl
B, Br.l
lA, 0 0l o B33l
=lo A2 ol
Io , ,.] The economy is divided into several subgroups of industries suchthat when the industries are
properly numbered, the technological matrix is block triongulor. The sub-matrices B,i are
The above economy is divided into three independent major sectors lt can be viewed as
lquare, each of them representing an indecomposable sub-matrix. Below the diagonals are only
three separate economies lts input-output model (/ - A)x = d can as well be broken i
teros and above the diagonals aie blocks of non-negative elements with at least one positive
three separate input output models given by:
alement in each column. There may be some zeros above the main diagonal, but not all can be
(1 A)x7 = da zero. lf so, the system will be completely decomposable as in Figure 4.4.
(l -A)x2=d2
The industries in the above graph fall into three categories:
I (l-A,)x..=7,
o Group 1: lndustries 1 and 2 which sell to each other and to lndustries 3 and 4 in
When, through appropriate interchange of rows and columns, the technological matrix of
group 2 as well as sector 5 in group 3.
economy can be represented by,4 above, the economy is said to be completely decomposoble. . Group 2: lndustries 3 and 4 which sell to each other and to sector 5;
This applies, as demonstrated above, when the economy comprises of sectors that can be o Group 3: Only industry 5 is in this group and does not sell to any sector.
categorised in completely independent groups. One group of sectors in an economy has The matrices 81; (forming the principle diagonal) are square, each of them representing an
nothing to do with sectors in other groups. indecomposable matrix. Below the diagonal matrices are only zeros while above the diagonal
Consider another economy where the graph is as in Figure 4.5. are blocks of non-negative elements with at least one positive element in each column. There
may be some zeros in these matrices but they cannot all be zero, else the system will be
completely decomposable.

Decomposable economies with block triangular technology matrices are somewhat different
from indecomposable economies. Suppose that final demand d1 is increased for some industry
 I

90 I MATRIX ALGEBRA Mathematical Optimisation and Programing Techniques for Economic Analysis 91

i which falls in group k. Since every sector in group k sells directly of indirectly to all industries
llrlrnrrlrr,, root should be less than one (2- < 1). Further, if the technological matrix A is
in the group, every output rj in group k will increase. But since industries in group k do not buy lttrler rrrr;rosable, then with 2' ( 1, then the matrix (l - 41-t )) 0 which implies that r )> 0.
from industries in groups indexed above k, x7 in such industries will remain unchanged We say
industries in group k do not have bockword linkoges with industries from such groups. trrrrrlh,4 12

The x; in some group r with an index smaller than k may or may not increase depending on (,rv(.n the following input output -':|atrix A =
to 21

whether one or more industries in group buy from one or more industries in group ll/3 al
r. Backward
linkages may exist with such groups. l)r,tcrmine the Frobenius root
Considerthe matrix equation of theform Ax = dgivenby. I lrrd the matrix (l - A)-'
I lrrding characteristic roots was introduced and discussed earlier in the chapter and the

t:it 4' ziltil hl

To decompose coefficient matrix A, interchange row 1 and 2; then column 1 and
t

r
t,ader is now familiar with the steps. Therefore, we just state the roots and leave it to the
cader to verify.
2 The
resulting matrix is

l0zz Qzt aztl

EE
7.=-l-and7-=l-
'
;=lo drr o,=l .j3 .J3
Ihe root of interest, the Frobenius root, must be positive and no less than another root
I o ast a..]
From the above matrix, it is possible to define the four sub matrices required for in absolute value. Therefor )' = P-.The associated characteristic vector (without
These are A, =[azz), Arr=fan azz), Arr=
" 13
decomposition. ."0 Orr=[Z:: :::)
[3] r.lGl
normalisation) is u' =
The coefficient matrix /4 is thus decomposable. [vij
11 -21
4.19 Perron-Frobenius theorem r_A=l_tl3 Ll
lf z4 is a semi-positive matrix and among lts characteristic roots, there is one particular root 2*
called the dominont root or Frobeniusroot with an associated characteristic vector u*, such
We now need to find its inverse Q - n)-t. For a Z x 2 matrix, the inverse is much simpler

that: using the shortened cofactor method. The determinant is

l. fhen in the matrix, swap the
principal diagonal elements and change the signs ofthe off-diagonal elements. Thus
a) 2- is a real number and non-negative
b) No other root exceeds,l.* in absolute value _ 111
(l - A)-' =
21
c) ,l- is non-negative vector --ltt
t/31,13 ll I

d) For all p > 7', ttl - / is non singular and its inverse (pl - A)-t is a semi positive matrix.
lf in addition to being semi positive, matrix,4 is also indecomposable, then (a), (c), and (d) can _t3
-lr 61
3J
be strengthened to:
Note that the (l - A)-1 >> A
a) ,tr. is positive and is not a repeated root
c) u' is a strictly positive vector Wo conclude that for every sector in an indecomposable system, the total requirement of every
dl fut -.4) 1 is strictly positive rrrput exceeds its direct requirement.
matrix
This theorem is quite helpful in dealingwith solutions involving linear system of equations and Metzler's Theorem: if A is a non-negative matrix and ,1- is its Frobenius root, then a necessary
matrices. ln the Leontief solution, x- = (/ -;4) 1d where the final demand vector d is non ,rnd sufficient condition for 2- is that allthe principal minors of (/ A) are positive -
negative, all that is needed to guarantee a strictly positive output vector (x* >> 0) is that the
 93
Mathematical Optimisation and Programnring Techniques for Economic Analysis
92 MATRiX ALGFBRA

I 0 (.)
= 0.1 and for the unemployed person, the probability of being unemployed in
4.20 Stochastic matrices
tlr, next period is I- O.4 -- 0.5. Thus
A good farmer ought to have a good focused of weather. S/he must know the chances of havi
0.41
a particular weather condition the following day given the current day's weather. ln a , = [3:? 0.61
a trader must make careful judgment ofthe price tomorrow, given today's, so as to decide
For the equilibrium rate of employment, we need to find the vector
the optimal quantity to supply. Or one may ask, what is the probability that a
trrl I emPloYedworkers I
x = lrrl=
unemployed in one time period will have a job in the next period? Will still be lunemployed workersl
even in the next period? The probability of getting into state i from state 7 is of rrr /'t = .x
importance, both for policy and decision making Government only needs to intervene in
+(P-l)x=0
market if a distortion persists. But what is the probability that a distortion today will also
there tomorrow, ceteris paribus? This is best tackled using stochastic matrices. , l/'- 1l = 0 since lambda the eigen value is unit (tr = 1) in this equation' then (P - /)
r,,,,rngular.
A stochastic matrix is a non-negative square matrix whose column sums are unit. lt is a
of probability or stochastic movements The element aii gives the probability of having I [-0 1 0.4 ltrrl
the previous period was in state
Iol -o+llx'l=u
l. Define a vector of employment status as
-xr=4xz
employed llrc equation give the employed unemployed ration of 4 to 1' The unemployment then
"* _{1,
12, unemployed of employed
rrrust be one-fifth, equivalent to 2O%. Alternatively, since the total number
Then the stochastic matrix is of the form
,rnd unemployed is always equal to 100,000, then using the equation x, * x2 = 100'000'
:lo" o"f = 200,000 Again unemployment rate is 20%'
',
we have xr = 800,000 and x2
Lazt azzl
stochastic For a
The element the probability that someone who had a job in period t (j = ll will have
a, is A .,tochastic matrix can be column stochastic or row stochastic or doubly
,,,/r,,)rn stochastic, columns add to unit and for row stochastic, it is rows that add to unit' lf both
job in period 2 This need not be the same job because the interest here is that
add to unit, then the stochastic matrix is said to be doubly stochastic.
employed, whether by the same company or a different one now. The element aI is rrrw., and columns

probability that one who was employed (/ : 1) will have lost employment by period 2. Si ( rn,,ider a simple weather model.
can be
the state of being employed and being unemployed are mutually exclusive and exhaustive, llr,,probability of a weather condition, given the weather on the preceding day
their sum must be unit. Thus arr+a21:1. This confirms the earlier statement that the rr.prt.sented by a tronsition matrix
column sums must be unit. The corollary applies to the second column, for someone
unemployed in the initial period. ',=[09 0.51
to.1 0.51
to
A toiny year is 90% likely to be followed by a rainy year and a drought year is 50% likely be
Example 4.13
l,rlkrwed by another drought year. suppose the rainfall condition is known
for the initial year to
Suppose there are 100,000 individuals participating in a particular labour market. lf an ,rainy,. Then the vector x(0) : probabilities of rainy and drought for known initial
individual is employed in the current period, there is a probability of 0.9 that the person
t)r. [l] t,*t
and the
yr,,tr. For the following year, focused have to be made based on current condition
will be employed in the next period. lf the individual is unemployed in the current period,
lr,rnsition matrix.
there is a probability of 0.4, the they will have found employment in the next period.

Write out the stochastic matrix.

xo) - Pr(o)
r0.9
What is the equilibrium rate of employment in the market? t0.1 8:il tll
To write the matrix, first compute the two remaining probabilities. For a person employed = r3?l
in the current period, the probability that they will be unemployed in the next period is
 Mathematical Oprimisation and Programming Techniques lbr Economic Analysis 95
94 | ,or*,, ALGEBRA

I rrr r vr,1 y vl.ctor X other than the null vector, the function X'AX is positive definite it X'AX >
and for year 2
ll llrr',r',,ilrassurancethatthefunctionisthroughoutitsdomainabovezero.ltnevertouches
xQ) = tlrr rr,r,r I or any values of xr,x, ...,rk not all zero, the function is always positive. lf however
llr', lrrr( lron does touch zero but never cross into the negative, then the function is said to be
1", lrlv'. ,,r,rni-definite. lt is defined by X'AX > 0. Alternatively, the function can be said to be
trr'tt ttr'rlulive

A lur, lron is negotive definite if it is negative for all values of xyx2 -.',xp, not all zero.
The probability that year 2 will be rainy is 86% and chances of a drought is 14%. This focused is a function is negative definite if X'AX < 0. lf zero is permitted into the range,
Al11,,lrr,rrtally,
based on the initial year and will change as we become certain on year 1.Once year 1 is known, tlrlrr Ilrr,function becomes negative semi-definite This is when X'AX < 0. For instance, if
year 2 wlll no longer have to be based on the focused on year 1 but on actual outcome. But the
,,,,,,,,, n *h"," 1 = [1 tne rorm wirr ue
adjustment is not effected in this manner once year t has passed and is known, then it l],
becomes the present or initial period, and what was year two now becomes year 1 and so on. x'AX=txt,,lll llt;jl
ln general, with the latest period known with certainty taken as the initial period, the focus for
the nth period made in the initial period is given by ' lxl + x2+ xrll:tl
xt

,(n)=p*(n-t):pn*(:o) = xtz * xrx, * x.rx, * xrz

The matrix p raised to power n need not be forbidding. Diagonal matrices discussed earlier in = xrz * 2xrx2 * x22
the chapter can be of service here. The stochastic matrix P can be transformed into a diagonal - (x, I xr)2
lur whatever values ofx, and 12, not both zero, the function returns a positive number. This
matrix and raislng it to any power should not be a problem.
r,rr be seen from the square which only produces a positive number. The quadratic form is

4.21 Quadratic forms prr',rtive definite. A slight modification of A will however alter the sign definiteness. Suppose

Consider a polynomial expression in several variables. lf the sum of the exponents in each term r,,w the matrix A is given by o =ll, then the form will be
l'],
of the polynomial is the same, the polynomial has a form. Alternatively, a form can be defined
as a polynomial expression in which each term has a uniform degree. The uniform degree
x'AX = lxt *,1
I], l'l l;ll
defines the order of the form. For example = lx, - x2 -xt + xrl[:l]
f (*,y,2) -- 4z - 9y * z = ,r' - xlxz - xzxt + xz2
is a linear form because the degree is one The expression = xr2 2xrx, * x22
: (x, - x2)z
f(x,Y,') = 4xz * 9xY -l 3Yz I \xz * 722
hktr the previous function, this one too has a square which will prevent a negative number
is a form of the second degree. lt is therefore known as a quodrotic t'orm in three variables-
r rr curring. The function will therefore be non negative. However, the difference x, - x, can be

suppose two matrices X (a vector) and A are defined as , = l:1rl o = l::: Z;11tn" rtro (and therefore its square) even if both variables are not zero. This kind offunction can be
""0 rk'scribed as positive semi definite.
x' AX = lx, -nll":" z::ll:I
+ a2x21 4 22 Test for Sign Definiteness
- tx,
t r ,"1farxr
"lorrxr+a22x2l I he question yet to be answered is, 'under what conditions will the quadratic form be positive
= dttxt2 + azx7x2 + a21x2x1 * o22x22 r)r negative definite?' What are the diagnostic tests for positive or negative definiteness? Two
: errxt2 * (arr'l a21)xrx2 * d22x22
rnethods are used to determine the sign definiteness of a quadratic form. The first is known as
which proves to be a quadratic form.
r
I

96 I MATRIX ALGEBRA Mathematical Optimisation and programming Techniques for Economic Analysis
91

the determlnantal test, the name invariably coming from its use of the determinant in the test. Frerrrple 4 14
The second is the Eigen value test, which uses Eigen values.
Determine the sign definiteness of the following quadratic forms

Assume a quadratic form described by the coefficient matrix / : Lh

[f flbt so that a. Q = x? + 6xl + 3x! - 2xrx, - 4xrx=
Q = oxl2 *Zhxrx2+bxrz,then b. Q =2x? +3xj - x! + 6xrxr-Bxrx, -2x2x3
The first step is to write out the coefficient matrix. For principal diagonal elements, these
definire,, andab - h2 > 0
o
'' [x"T:'il:] ffi : 3] are coefficients of squared variables. For the off diagonal elements, halve each
coefficient
Now ab - lt2 is the value of the determinant of the coefficient matrix A. That is, to create a symmetric matrix. For part (a)

o0-n'=lh la ht l1
-1 0llxrl
bl Q =lx, xz xrll-r 6 -?llxrl
The determinant is known as lhe discriminont of the quadratic form A discriminant in other lo -z el[,,]
words is a determinant formed by the coefficient of the terms in the quadratic form. And since t1 -7 0t
Discriminant lol = l-r 6
a quadratic form is a symmetric matrix, ai; = aii. Hence the coefficient of the cross product -zl
term in the function is equally divided between oij and aji
lo -z 3l
The principal minors will be given by
The condition for sign definiteness can be restated in a more general way. The function Q is
lDrl =r>0,
positive definite if both principal minors, lal ana l[ fl ,re Ootfr positive. For a negative
definite function Q the first principal minor lal must be negative and the secona
lf; il
rD,r= l_11
l1 -1 0r
7l:,, o'

positive. Notice that no specific condition is place on b. This however should not mean that the lD.l = l-r 6 -2]1= 11 > 0
value of b is immaterial in the determination of sign definiteness. lnstead the condition on b is
lo -z 3l
All the principal minors are positive. Therefore, the quadratic form is positive definite.
hidden in ob - h2 > O + ab > /r2. Since h2, by virtue of the square, will always be positive,
then ab can only exceed h2 if and only if both c and b are of the same sign, positive or For part (b), Q = 2x? + 3xl - x! + 6xrx, - Bx1x3 _ 2x2x3
negative. Thus a condition on o is in fact a condition on b.
12 3 4 1thl
For instance, in the quadratic form Q = 5X2 +3XY +2Y2, the coefficient matrix is .4 = Q=1x., x2 " x, ll: S
''L; :t -rllx, I

tf, ') and the discriminant lDl = l,.ss 'rtl rhe principal minors are: t2 3 -ilt;;l
4t
: l: 3
lal :5>0,
Discriminant lOl
lDl=7.7s>o l+ -1 -11
Therefore the quadratic form Q is positive definite. -11
The principal minors will be given by
ln a more general case, with n-variables, a quadratic form is positive definite if all the principal
minors are positive. That is lDtl > O,lDzl > 0, lDrl > 0, ...,1D"1 > 0. For a negative definite lDrl=2>o,
quadratic form, the principal minors must alternate in sign Particularly, all odd-numbered tD,t =|tr 3l
: -' . ,,
principal minor must be negative and all even-numbered principal minors must be positive.
ThuslDrl <0,lDzl >0, lDgl <0, ID+l >0,....|f principal minorsof anyquadraticformfail to p,t =13
l+ -1
', :rl =
-23 < o
adhere to any of the two patterns, then such a form is indefinite. lt is positive for some part of -11
Both positive and negative definite require that the second principal minor is positive,
the domain and negative for some other part.
being even numbered. The case at hand is however opposite. This is enough evidence
to
suggest the quadratic form is indefinite. Nonetheless, even the third, by being
different
from the first (allodd numbered) confirms the results.
 I Mathematical Optimisation and Programming Techniques for Economic Analysis 99
98 I MATRTX ALGEBRA

The second method involves the use of characteristic roots or eigen values. A concise
eigen vectors. Given a uadratic form Q =
discussion
x' AX , the
Cha pter 5
of eigen values is under eigen volues ond q

linked to the eigen values of the coefficient matrix A. We summarise

form has sign definiteness
the sign definiteness in the three statements below: 5 DIFFERENTIALCALCUTUS
. Q is definite if and only if every eigen value i l; I Introduction
. Q is semi definite if and only if every eigen, ] llrc subject of differential calculus is essentially concerned with the rate of change of a

r Q is some eigen values are positive and some ns the rlr,pendent variable with respect to an independent variable. ln the study of many
form Q is negative for some part of the domain and positlve for another part' ;rhenomena, we are concerned with changes in quantities or the rate of change. For instance, in
romparative static analysis and the concept of margins in economics, the speed of a rocket and
llrc study of voltage of an electrical signal in physics all involve the important underlying
rrrncept of "rate of change" of a variable. The concepts of limits and continuity discussed in
rlrapter 3 are basic to the study of calculus. Furthermore, the discussion of variables and
lunctions in the same chapter showed that as a variable changes, the values of all functions
rhpendent on that variable also change. Thus, if a variable quantity xchanges by an
tr)crementw,insteadof x,wewritex+w.Thenfunctionsof xsuch asx2, x3,ji1 t.k"on
rrow values. For instance, x2 becomes, x2+2xw+w2.

5.2 TheConceptofDerivative:
t onsider a continuous function y = f(x).lf x changes by a certain quantity, denoted as Ax,
tlren y will change bya certain quantity, Ay.

Av
he ratio may be regarded as the overoge rote of chonge of the function with respect to .r
I
f
over the intervalAx As Ax becomes smaller and smaller, Ay will also change and the limit of
if tnit limit exists, as Ax tends to zero, may be called the instontoneous rote oJ chonge of
fI,
lhe function with respect to /. Mathematically, it is called the derivative of the function y with
rcspect to x and is denoted asdfi or f' (x)ory'. ln other words,

f o'Y' = lim6,-o^2'

I xample 5.1

An apple grower agrees to supply crates of apples, with a dozen apples in each crate,
according to the supply function S(r) = lgrz where x is the price per crate. As the price
goes up, the supplier naturally supplies more apples.

i. What is the average rate of change in supply when price changes from K25 to
K50 per crate?
1OO DIFFTRENTIALCALCULUS Mrthernatical C)primisation and Programming Techniques lbr Economic Analysis 101

ii. What is the rate of change in supply from K25 per crate to (25 + L,x) per crate? : 10(50 + Ax) : 500 + 10Ax

iii What value does in (ii) approach as Ax tends to zero? I rlrrrli the limit as Ax + 0 ln (ii), we get lim4r-6fl = Soo crates of apples per Kwacha
fr
l rr,,rrc in the price ofthe crate
Solution
I ,t r\ now see the graphical meaning of the derivative Refer to the diagram below.
Now at K25 per crate, the number of crates of apples supplied is S(25) = 10(25),
6,250 and at K50 per crate, the numbers of crates are.t(50) = 10(50), : 25,000.

{x+ trx,Y+}Y}

__l

We first have the point(r,y). Then, when I increases to x+Lx, y also changes to
PritP per [rate in dollars y + Ly. Frcm elementary geometry, we know that the ratio f is the slope of the chord
joining the two points. As Ar becomes smaller and smaller, the point (x + Lx'y + Ay)
The average rate in supply from K25 per crate to K50 per crate is given fyff in the
approaches more and more to the point (x,y) and the chord also shrinks. ln the limit
figure. That is, when Ar tends to zero, the chord becomes the tangent to the curve at the point (r,y)
As _ .r(so) -.s(2s) and the slope of the tangent is then given by the limit of the ratio f when Ar tends to
Lx 50-25 zero. Thus the derivative at the point (x, y) given by
25,000 - 6,250
-25
fr = tt^or-oo]: li^or-o[9#@ is nothing but the slope of the tangent to a curve
18,750
at that point.
-25
= 750
that is, 750 crates of apples for the increase of the price of a crate by every kwacha.
lct us now discuss the meaning of the derivative in economics and business. The concept of
a.r s(zs + ax) - s(25) moryin in economics corresponds exactly to the mathematical notion of the derivative. Margin
Ar- Ax lt simply the derivative of the 'total' function. For instance, suppose you have the total cost
10(25 -t Lx)z - L0(2r2 lunction C = f Q),where C denotes cost and 0, output. Then the derivative ff is the marginal
_
L,x cost or the rate of change in total cost at some value ofthe output Q. Let c =10+2Q+
1.0252 + 50Ar * (Ax)z - ZSz 0.5Q2 .then
L,x
IO2 I

DIFFERENTIALCALCULUS Mathematical Optimisation and Programming Techniques ibr Economic Analysis 103

dc .. I(a+LQ)-fQ) lEIr r, ', I Non differentiable functions

= o'do".
da Ao
+ LQ) + 0.5(Q +AQ)2 -rc-2Q
=l ,_
10 + 2(Q -0.5Q'
L0
LQ
^Qi0
-
.
li*
z\e + eLQ + 0.SAQ2
LQ
^Q+o
lim(2+Q+0.540)
= A0-o'
=(2+Q)
lfQ:26, 11,"n,

dCr
- @lo=ro=2+2o=22
trr lrrrtlr the diagrams,derivative does not exist at the value x6 of x. A function, in order to be
The +l means the derivative 4 evaluated at O = 20 rlrll',rr.ltiable must therefore be continuous and posses no kinks. ln other words, it must be
dQto 20 dQ
rrrrr rrllt
A few other common marginal functions together with their mathematical counterparts in the
form of derivatives are mentioned below. h 4 l)ilferentiation
Fu nction Derivative Itllr,il,ntiation refers to the process of obtaining the derivative of a function. Though the
Utility function U(x) r|,rrv,rtive of any function can be obtained directly in a manner in which we obtained the
# = marginal utility \r/ rrr,rr;lrrral cost from the total cost in the previous section, there are rules which can often be
Revenue function R = r(x) ,rlrllrr,d mechanically to differentiate different types of functions. These rules are explained
H = marginal revenue
lr,,lrrw
Production functionO f(l)
=
# = marginal Product
\,tl constant functions: Suppose we have a function / = o where o is a constant
Consumption function C : C(/)
fi = ,.rr,n., propensity to consume thenAy = O sothat# : 0 and hencethe derivativeQ = 0.
Savingsfunction S = S(y)
= marginal propensity to save (bl Powetfunctions.'Thegeneral formofapowerfunctionisy=qra.Whereaandbare
,,q
constants. The derivative of such a function is given by
dv
5.3 Non-differentiability bax'-'
f,r--
It is possible that a function y = f (x) may possess derivatives at some values of x but not at
I x,rmple 5.2
some other values. When the derivative exists at every point (x,y),the function is said to be
differentiable; else it is non-differentiable. lt is obvious that a discontinuous function is non- Lety=516.Then 9=30xs
differentiable. At the point of discontinuity the derivative does not exist. Graphically, it is not
possible to draw a tangent to the curve at this point. Verify as a special case that it y : I 1lzn9{ = 1

Note: Part (iii) of Example 5.1 could have just been solved by usinS the power function
However, to be differentiable, it is not enough for a function to be continuous. lt must also not
rule evaluated at K25 as follows:
contain any kink or sharp points lf a kink exists, then again no unique tangent can be drawn to
S(x) = 1g'z
the curve at this point. There will be an infinite number of lines that can be made at that point.
ds
The above two instances of non-differentiability are graphed below.
--2Ox
dx
7O4 | DTFFERENTTAL CALCULUS Mathematical Optimisation and Programming Techniques for Economic Analysis 105

= 20(2s) d.v
: 500 E = fr(x). fr(x).fs'(x) + f1@)ft(x)f2'@) + fr(x)f=(x)f,'(x)
Crates of apples per Kwacha increase in the price ofthe crate. ln general, if y is a product of n functions of x, that is,
(c) SumoIequotions:Supposeafunctiony=f(x) isinfactasumoftwoormoresepar
functions ofx such as fr(x),f2Q) etc. that is,
!=fte)+fr(x)+...
,:fir,at
Then the derivative of y with respect to x is the sum of the derivative of each of whlre each of n functions is differentiable, then
separate functions with respect to.r. That is,
dv
^( , )
fior f '(x) = h'Q) + f2'@) +...
#:Zlro,1-1ro,I
,=' ,,=*l
Example 5.3

LetY = 16 * !2Ox 2r3' Then

[ )
- lel Quotient ol two functions: Let the variable y be equal to the quotient of two separate
dvddd functions /r(x) and /r(r), that is,
:'= dx
ax . (10)+.dx(120x)+-(-Zx3)
=0-l 720-6x2
dx
t- f{x)
f(n
= 720 - 6x2 Then
dy fr(x). fr'(x) - fr(x). fr'(x)
dx IfzQ)12
ldl Product ol two functions: Suppose J, is the product of two separate functions f.(x) and
This is called Quotient rule
/r(x). That is,
l.xample 5.5
Y: h@).fz@)
Then 1g1 : :_ then
'5
dy ...df,(x) -dfr(x) dy (xs)(0) - 5(sa)
4*= ft@;+ fr(r):i dx (xslz
This is referred to as product rule -25x4
- x]o
Example 5.4 -25
x6
SuPPose Y = x2 (3x * 4). Then
(fl Function ol o function: Suppose y -- fr,(u) and u = f2(x) , then the derivative of y
T
--L ^dBx+4\ d(xz\
--:------------:- + (3x + 4) --:----: respect to r is given by
= xz
dxaxx dy dy du
= x2(3) + Gx + a)(zx)
:3x2 + 6x2 +Bx dr'd" d*
d"d
=9x2 +Bx = *fr(u). *f,(x).
= x(gx + B) This is called choin rule. This is because it involves differentiation of a chain of functlons.
The rule can be extended but the extension becomes complicated. lf Example 5.5

Y : ft@). fr(x).f=(x),then LetY:Ai-T

105 I DTFFERENTTAL CALCULUS
,0,
Mathematical Optimisation and Programming Techniques for Economic Analysis |
i,: t!
Here, we may put 2x2 - 1,: u, so that we have y = rli = ri, th.n I rrrexample, let y = sx,then9 = ex anflif y = Ae',then*= Ae'
dy dy du lhis is the only case of a function being equal to its own derivative.
d-: d"'d,
1. lll Logorithmic fundions:
=_u 2.4x i y=to1x
: zx(zxz _
H=:
t)-l ii. Y = lsgY
2x
- ,,1'zx,
where u : /r(x) "* =:#
--r
(gl iii. y=logar *=!bgo"
lnverse function: rf y = /(x), then the derivative ofx with respect to y is Biven by
dxL iv. Y = logou
= *-, ,where the function / is assumed one - to - one whereu=/(x)
* *=!t"c,"#
ai Note that each succeeding function is a more generalised case ofthe preceding one.
Example 5.7
I x,rmple 5.8
lety - 4x2, where the function is defined only for positive values of x.
Lety =lqg/x2 -L Then, we can see that this corresponds to the function form in (ii)
rhen, above. Hence,
fl=r. dv11
What is we have x =,[y 1 +, tnen,
-1 =--3=G!.'- t)-zZx
_112
ay ax .Jxz-lt '

i
dx 1 uyr-] 1
x
ay=z\il '4 - xr-l
) (jl lmplicit function: When we have a functional relationship between r and y in the
='\
=!(+*,\
n,) implicit form as f (x,y) = 0, we can differentiate each term on the tHS of the equation

1... , treating y as a function of x and then solve for

f.
= glx")-2
1.7 1
Example 5.9
= 8G:4_ Lelxy2-2x2+y3:g
dx
(hl Exponentiol function: lf y Differentiating with respect to x, we get
- an, where u=
/(r), then
dy dv
-!- + vz- - 4x *
du -dv
ii= o" loga*
2xv 3vz -1-
ax =
o
ax
For example, let y = 5x'-r. tn"n, -dv 4x
(zxY + 3Y'z)-fi= - !2
dv-
;;
dx'
(s,'-')(tog s)(zx) dy_ 4x-y2
= 2x'5"-1logs dx 2xy t3y2
As a special case, if we puto : esothaqy : e, where u = f(x),then
dy .n. du
dx-dx
\.,i,
t' .,
\.
'\-.

."1,
+;
 I Mathematical Optimisation and Programming Techniques for Economic Analysis 109
108 I DTFFERENTIAL CALCULUS

5.5 EconomicApPlications
t,6 Higher Order Derivatives
Ar has been pointed out, the derivative of a function y = f (x) is also a function of x and hence
The demand function for a commodity is given as 0 : 100 - P, where P is the price per u
and Q is the number of units. Find the marginal revenue when Q = 20 units. we noted
Grn be differentiated with respect to r.
The derivative of the original function y = f(x) is
hlnce often known as the rrst derivotive. The derivative of the first derivative is the second
marginal revenue is the derivative of the total revenue. The total revenue R = P0. we
dcrlvative;the derivative ofthe second derivative isthethird derivotlue; and so on.
P: 100 - Q hence R : 100Q - 92. Then, marginal revenue

iidRd
dQ= -fdQ.
rooQ - Q')
Althefirstderivativeisdenoted asfforf'(x)ory',sothenotationforthesecondderivative

:L00-ZQ hdfi or f " (x) or y", for the third derivative

fi or f "'(x) or y"' . tngeneral, the nth derivative

Note that the derivative of a function is itself a function and hence can be evaluated lr denoted asffior f"(x) or f .

particularvaluesof thevariable Thus,whenQ=20,marginal revenueisequal to60'Furth

For example, let y - sxa. Then
the marginal revenue reaches zero when Q = 50.
dv
Lety = f (x) bea differentiablefunction of x.The elosticity of thefunction isgiven bythe -!- = 2Ox3
dx
of the proportional change in y to proportional change in r. That is, elasticity of y with res d2v d rdvt
to x is given by: 7F= ildx):6ox
Lv/v d3v d / d,zv\
rzox
- L,x/x d.'= d-\aF)=
xay dav d /d3v\
rzo
y'Lx a,r = a,\ap)=
This is in fact the orc e/osticity of the function over the range Ax. we can get the point e dsv d /dav\
of the function at the point r as a limit of the arc elasticity as Ax + 0. Point elasticity thus aF= a.\a.r)=o
xdv
' to -:.
be equal yax Andthen,# , d,-{l and all the subsequent derivatives are equal to zero.
Suppose we have a demand function 5x + By = 12 where r the quantity is demanded and y
We shall study in the subsequent chapters; second order derivatives have useful applications. lf
the price We have the first derivative tells us about the slope of a curve at a point or the rate of change in a
12B lunction at a given value of its argument, the second derivative tells us about the slope of the
'=T- sY rlope of the curve at a point or the rate of change of the rate of change, in other words, the
dx8
-;- = --
then,'dv occeleration of the function. The first derivative tells us whether the function is increasing or
5
decreasing. The additional knowledge about the second derivative value informs us whether
Therefore, point elasticity of demand is given by
the function is changing (increasing or decreasing) at an increasing rate or decreasing rate.
y dx _8Y These interpretations ofthe first and second derivatives will be discussed elaborately later.
xdy 5x
'8v
:-:7r-6
___ 6,7 Partial Derivatives and their Applications:
s
\--s-l So far we have considered only functions of the form y: f(x) involving one dependent
variable and one independent variable. More generally, we may have a function in several
_8y
8Y-72 v.rrraLrles of the form

_Zy Y= f(xvxz, xn)

2v-3
 Mathematical Optimisation md Programming Techniques for Economic Analysis 111
I
110 IDTFFERENTTALCALCULUS

d
With only one dependent variable, the derivative of the function y = f (x) tells us the wa
changes when.r changes, x being assumed to be the sole influencing variable ony.
^ (4x-3y)
= ox
=4
instance, if you take the production function O = f(l). The derivative
f, is ttre value of 0 r0z:, 022 A2f
marginal product ofthe input/. Now suppose the production function it6: f(L,K) = zw = fit
the two inputs land /( are labour and capital respectively. We wish to study the
i'z=
separ
"1il '7=
0z
influences of labour and capital on output. To do this, we may assume one of the inputs to - -(-3x
oy
+ ISY")
kept constant at a certain level (i.e. treat it as a constant) and differentiate output with
= 3Oy
to the other factor. This would tell us the rate of change in output when that factor
varies. lf say, capltal is kept constant, then the derivative of output with respect to labour 0 t|zt, 022 A2 f zY* = ftx
give us the marginal product of labour; likewise, keeping labour constant, we can different *lrt)= 6;6'= di6=
0z
the function with respect to capital and obtain the marginal output of capital. But in - -(-3x + l\y")
ox
case, the function is differentiated portiolly with respect to one independent variable, _, -J
the other variable constant. ln other words, the marginal products of labour and capital -

obtained as portiol derivotives of output with respect to labour and capital respectively. d t|z:, d2z Azf
general, if you have the variable y as a function of n independent variables, one can get ,rl;) = dr*= 6iil= z*Y = f'Y

partial derivative for each of the n variables, treating the remaining n - 1 variables a

consta nts.
^ (4x-3y)
= o\,,
=-s
Example 5.10
lhcfirsttwoderivativesaresimilartooursimplederivatives oltheformfi.Thelattertwo
letz:2x2-3xy+t4**t
rlr.rlvatives are called cross or mixed portiol derivotives. ln our example, we note that the
values
We will have two first partial derivatives, one with respect to x and the other with ol the two cross partial derivatives are equal. This is generally the case. fn, and f*, will be equal

to y. To distinguish a partial derivative from the ordinary derivative th" for."r. y

ftlr all values of r and for which they are continuous. This has been proved by a theorem
f,
denoted by a curled d. Thus, known as Young's Theorem.

0z 5,7,1 Economic Applicotions of Portiol Derivotives

-=4x-3y
,,;
fxample 5.11
i'= -'* * "t' Suppose you have

ffcan also be written as{since z : f (x,y),2*or f, and s : f(r,K)

-U- can also be written as{.,,2, or J" Where o is output, L is labour .na r liiipliE156-ry-,oduction function is said to reflect
dy dy'
constant returns to scale if when labour and capital are
One can differentiate each of these first partial derivatives to obtain second order
derivatives. But since the first derivatives are themselves functions of both r and y, each !t turow and cap-Gft6lbGs the output, and a sgljgqsg!9ilnj!- of labour
them may therefore be differentiated with respect to x or y. We shall thus have four and capital reduces output too
order partial derivatives. Let us calculate them for Example 5.10 above.
f(11,LK)=Lf(L,x)=1o
0 Irlztl- 022 a2f, F
dr\dr) - d-7 -- jj - zxx -- txx
--
trz l r,rrr*rr,o.aALCULUS
Mathematical Optimisation and Programming Techniques tbr Economic Analysis 113

Such a to be homogenous of degree one or

production function is said . +u+ = partial elasticiqr of demand Qrwith respect to p,
at dPt
homogenous. A theorem, known as Euler's Theorem, states that for a linear homogenou
. ?*:dqz partial elasticity of demand Q2with respecr ro p2
productionfunction O = lU,,K) Qz

a0 ao . ?* = partial elasticity of clemand Qrwith respect to p,

LV+K7=0 Qt dPt
. +* = partial elasticity of demand Qrwith respect to p,
Qz dP,
This means that if the prices of labour and capital are set equal to their re
Ihe last two elasticities are called cross elosticities of demand and they will have the same
marginal products, then the total payments made to labour and capital will together
rigns as those of their respective marginal and demand functions.
equal to the value of output produced. This in fact is a solution to the so called odding
problem ot product exhoustion problem. l.rrrr1tle513
A popular example of a linear production function is the Cobb-Douglos product ll two commodities are jointly produced, their cost function will be
fu nction:
c = f (e,,er)
= AL"K7-", where A and a are constants
_O The partial derivatives $ ana $ are the marginal cost functions and are usually
oo o dQ' dQ2

OL
= oorn-'y' positive
ao
(t - a)ALo K-o For example, let the joint cost function be given by
i=
do do C=log(10*Qr)Qr.then,
:. L.- + K :dK = L(ttALo rK1-")+ /([(1 - a)AL"K-"1 AC Q,
dL
6Q, 10 + 01
Example 5.12 AC
Ios(10 * O,')
dQz
Let the demand functions for two goods be given as -:
Qt= f (Pt,q), Qz= tt(Pz,qz) ', 8 Extensions to Two or More Variables: Total Differentials
t# #, ,ff afi
The four partial derivative una are called marginal demand functions. ltr y: f (xr,x2). This means the variable y is a function of not only one variable but two
A.,suming independence in the two explanatory variables, then x, and x2 will vary
lf the demand functions are well-behaved in the sense of adhering to the law of i
rrrdependently. These variations will each have an effect on the dependent variable y Since
relationship between the price of the good and lts demand, then {! and will be
ffi ,,,rch x is causing a change in y, the total change in y will then be the sum ofthe two changes
negative. The signs of ffi and witl however depend on the nature of relationship I rcing caused by x, and x, respectively This is given by:
ff
between the two commodities. lf the two goods are complementory, afall in the price af
dl=^dxt+*dxz
af
either commodity will raise demand for both commodities so that both ff ^"a !^
It is called the totol differentiol of y. lt differs from the partial derivatives discussed in earlier
be negative. By similar reasoning, if both ffi anaafi are positive, the two goods
lcctions in that the partial derivative assume variation in one variable at a time and look at the
competitive. But if one of the partial derivatives is positive and the other negative, the cffect caused on / ln total derlvatives however, all variables vary independently and our
goods are neither complementary nor competitive. interest is the total change in y caused by independent changes in all the explanatory variables.

Once the marginal demand functions are obtained, one can also obtain the expression for
the partial elasticities of demand by multiplying bythe relevant price quantity ratio.
 115
Mathematical Optimisation and Programming Techniques for Economic Analysis
r1.4 | otrrrRentrnr cALCULUS

dllferentiation. Then the second order total derivative would simplify to

ln a more general case, with many explanatory variable5 the total derivative is the sum of
dzy = flrd.xl * Zfrrdxrdx2 + f22dxl
changes caused by each explanatory variable. With n explanatory variables, the total deriva
is given below. Ar we shall see later, the second order total derivative has an important role
to play tn
af df af optlmisation.
dY = 1- dx1 * dx, * "' * -:-dx,
=L
dxr'dxz'dxn
The total derivative above has two terms that show the change caused by the first variable and
the second variable As in the case of second order derivatives, one can also get the second
order total derivative by differentiating the first order total derivative that we have discussed.
To get to the second order total derivative, we must differentiate this with respect to the two
variables in the first order total derivative again. lt is simple to comprehend if we think of the
first derivative as the original function which must be totally differentiated. Thus we must get
the change (in the first total derivative) caused by the two variables and then sum them.

dzY : d(dY)
=u9"
0x, 0,, *d(dY)
}xz
0,,
a1ffa,, +
frax,1 d*, +
d(9Y dx. * 9Y ar.t
dx, ' -!L--;--9!!:dr,
dr-

At this stage, it is demonstrated that the second order total derivative is the derivative of the
first. ln the first total derivative, the two terms will each be differentiated with respect to each
of the independent variables.

a,y- = (2ar,
\dxi
2a*,)
* dxtdxz - r*, *
/ \dx2dxl
(j
.0.^, *2a,,\a*,
dxi - /
: 02v ^ 02v 02v 02v
+ + +
a*idri aifr;dxzdxr u;;fi;dxtdxz 5fidxi
d2v - 02v dzv
-u4d
+2 xtd xz + xl
=
#,0'l ffid
ln the last line, the expression for the second order derivative can be shortened by using the
symbol fi; inOlaceof
ffi.
d2y
l,i = artat
The left hand side is second order derivative ofthe function with respect to variable i and then
7 By now we know using Young's theorem that the order in which the function is differentiated
does not matter. This is the reason we stick to the order of starting with variable i in the
 Mathematical Optimisation and Programnring Techniques for Economic Analysis 777

Cha pter 5

rr lN'l'IiGRALCALCULUS

h I lillt'oduction
tr rl!,rt)l('r 5, we dealt with moving from a function to a derivative, and how this applies in
1,,,n,)rrics ln other economic problems however, this process has to be done in reverse. Given
tlrl r[,rivative, what is the original function? This process of reverse differentiation is known as
Itlt tltttlrcn-

Jt ;, lDverse differentiation and the Indefinite Integral

r ,rr.,r(l('r a function y whose derlvative

#: rr.The derivative is known but the function itself
r, r, rt known. Obviously one clue is that there must be a function which when differentiated,
grvr... the derivative in question, 2x.fhe easiest way to get round this is to differentiate a
lrrrrr tron and see how it transforms into a derivative. Take a simple quadratic polynomial
Y:x2+2x-4
I l..rn1i the techniques of differentiation covered in the preceding chapter, we find the derivative
,.',1,', -2, ]' 2.For everytermof apolynomial orbwhereaisthecoefficient,xisthevariable
,rrrrl /r is the exponent, the derivative baxb-7 is arrived at in two stages:

r Multiply the term by its exponent

rr Reduce the exponent by unit or subtract one from the exponent.
1,, (Jrry out inverse differentiation, simply execute the above two processes in reverse. lt is
wrIth stating here that execution in reverse also means reversing the order_ o!executing the
two stages. Since differentiation starts with the first to end with the second, integration-h-as 6
,,l,rrt with the second and end with the first step. lt is also important to be mindful that the
rr.vcrse of multiplication is-division, while t!9I9-y"Fg of addit!o11 is, s_u,b!r:-e,S!iSn.

lr r cverse, the steps are therefore rearranged as

r lncrease the exponent by a unit or add one to the exponent

ii Divide the term with the resulting exponent
(,rven the derivative baxb-7, the new procedure should bring backthe original function. The
rh,rivative above can then be reversed to become.

Iox+zldx=x2+zx
J'
llrough this function looks similar to the original function above, one component is still missing.
llreconstantisabsent lntheoriginalfunctiontheconstantequals4andonemayjustopttofix
118
I

TNTEGRALCALCULUS Mathematical Optimisation and Programming Techniques for Fronomic Analysis J 119
|
I
it in the above equation. However, it will not be known o priori lhal the value of the constant Pttwcr Rule:
Jt
4. Hence in order to provide for the unknown constant, the right hand side of the naa. may suggest, this rule of integration applies to functions with powers. These are
Ilha
equation must include an arbitrary constant C. Thus we shall have referred to as polynomial functions and are discussed in chapter 3. Since polynomial
Jimnnty
ler+z)dx=x2-t2xtC ftly m..nt multi-term, polynomials are functions of several terms. ln the integration, terms
(2005) when they
J J ht.gr.t"a separately. This is exactly as put by Chiang and Wainwright
Generally, an indefinite integral is of the form.
J|l tnot 'the integral of the sum of finite number of functions is the sum of the integrals of
lfgtar=F(x\+c Jlr functions'. lt must be cleared here that the term 'sum' as used in the statement also
polynomial can be taken as a function
J
It is an indefinite integral because it lacks a definite value. lt will be a function of x and hencc
]UOll 'difference'. ln this regard, each term in a
hhnf , polynomial a sum of power functions.fhus the statement of the power rule refers to a
vary with it. The opposite case, definite integral is discussed later in the chapter ln the integral,
rtrrlllr. lr,r rn of any function.
the symbol / which looks like an elongated s is called the integral sign. lt is an instruction to
integrote. The function /(x) is what must be integrated. lt is called the integrand. As a function
in general, we expect that it will take on various form or types of functions The last part dr
Itrp.tr.t,tl,forthefunctiongivenby f(x):ax" whereaisthecoefficientandnisthepower,
t I lr, | | rw('r ru les is expressed as follows
specifies the variable of integration. lt is similar to the dr under differentiation which specifies
the variable of differentiation fhe dx in integration means integration is to be executed with r1
respect to variable r. lx"dx=:xn*r]-C (n+-l)
J n-ll
The C on the right side of the integral is an arbitrary constant. lt is called the constont ol tlr| orrly restriction on the rule is that the power must be non-negative-one. This restriction is
integrotion. lt represents the constant that disappears when a function is differentiated. Since vr,ry ilnportant for if ignored, the function may collapse. Other than that, n can take on any
integration reverses differentiation, the constant of integration is a place holder for the
v,rlrrr.r ln particular, since n : 0 is admissible, the rule can be used to integrate functions of
unknown constant, which can also be zero. This constant also serves to indicate the multiple
parentage of the integrand. That is, a given derivative will result from multiple functions, v,1rlrrs power including constant functions without requiring any special treatment. This might
differing in the vertical intercept or constant. lr,. rrrlriguing but it is easier to execute lf a constant function f(x) = a can be rewritten as

llt) -axo Evenifthevariablexappearsinthefunction,itspowermakesislogicallyabsent.

6.3 FundamentalTheoremofCalculus
ln view of the discussion so far, applying differentiation and integration on a function, one after I x.rtttple 6.1:
the other, should leave the function unchanged. This leads lo the fundomentol theorem of lt can be rewritten oxodx.
Find J adx.This is a constant function explained above. as I
colculus. The theorem states that if /(r) is a continuous function, then the derivative of the
With this the integral can be found with ease as follows.
indefinite integral of /(x) is the function /(r) itself. Algebraically, this can be shown as follows

al l"a*=[o*oa,
dxJI\x)=llx) 1
:a-xetL+L
0+L
6.4 Rules of integration
=ax*C
The rules of integration show how integration of different functions is to be executed. As will be
noticed, these rules are similar or related to the rules of differentiations studied earlier. lt is Example 6.2

assumed here that the reader is already familiar with the latter. ln this section, we provide for Find J1/'. Thistype of function looks a little strange. lt can however be transformed to a
I
each type of function its integral. ln general, the rules are classified into four. These apply to the
more familiar form, the power form. Notice that JI : rI. Thus
four classes of functions.

I*= I,+
12O INTEGRAL CALCULUS Malhcntatical Optimisation nnd Progrlntrning Tcchniques lbr Econorric
Anrlysis
121

1 --l*, tL, .
: l lrt, Logarithmic Rule
rr,, t,,l,,tlthmic function is a special type of function with a special integral. As
2+l rnenti
21 ,,,,,rr llrc rules of integration are derived from the rules of differentiation. U,nder,;ilH;
: -xz * C
r rl, r, rtidtion, the derivative of a special function f (xl = ! is given by llnx _ t ,,
642 The Exponential Rule , ., ., o[ rhis differential will provide a clue tor the integrati;n ot a togarit;mic
;[l' ]j"
As the name implies, the rule applies to exponential functions. Though exponential is I rr t lrrnic rule states as

taken as synonymous to the constant e = 2.7t828, the book takes a more general case.
general form of an exponential function is /(x) = b', where b is the base. For such a funct
the integral is given by ,
Ii* =rnx * c
rellef because it provides a way out with a function that was not
, rUle comes as a
adrnissible
,rrrt.r rhe power rule. The function =, ' ir a term with n = -1 and was not
i
1
I t'ar =!1*,
tnb ,rrt,,r the power rule above. With the logarithmic rule on the menu, such a function
uOrni..iUl"
-t'"tt will no
in the special case where b = e = 2.71828, the rule reducesto , rr1i, t be inadmissible.

le*dx:e'+C
J
slnce ln e = 1 ln some cases, the power in the function may not be a single variable x The
power may be a function of x Such need not come as a strange concept lt is provided for by
the variant of the exponential rule stated as

r".",
[ 4* = , ,
Jf,G)
",,^,
Example 6.3
t,lr] o*
J llx) -
function with base e. The power is specified and I tn r(x)+ L
Find I ex'+2dx. This is an exponential
integral should not be mystifying. Before dealing with the whole function, it is helpful ^'" t,' f'(x) in the numeraLor is the derivative of [(x) in the denominator. Given a quot.
someone who acknowledges ineptitude for algebra to begin with the power alone | ,, rrons for integration, it is a necessary condition that the numerator ,, . a"r,uali""t'unt of
the power alone and differentiate it since the integral required the differential of ,r,,r.minator if the rule is to be applied. The method and procedure
power. For a connoisseur, this may not be necessary. , l, rv.rtive relationship is left to the reader.
"t 0",;;:;tll:
The power is f(x) : x2 + 2 and the derivative is f'(x) = 2x. The integral can then
!<
worKeo out as | ,,rrrrple 6 4

| [
JJ"r,rr4*- "r,,t4,
(x)
sf
l,
--f
I'lx)
+2

--+c
"x2
2x between the terms of the quotient The derivative of h(x) - [n.r is q(x) = 1 ,n,,ih
th"
722
I

| TNTEGRALCALCULUS
Mathematical Optimisation and programming Techniques for Economic Analysis L23

new revelation, the integral takes the torm l!)!ax which now points to the llr. nrethod allows the simplification of complex integration problem. As the name suggest, it
'h(x)
rrrv,lvcs substituting one variable for another to reduce a more complex function to a simple
logarithmic rule. lts execution has already been illustrated above.
rrrt| A new variable u replaces x time it is realised that the integrand is a chain rule derivative.
I lrl rnethod of substitution is of critical importance as the outcome hinges on how this is done.
I r rr I lre above integral problem, the new variable is u = h(r). Then dz : h' (x)dx and the new
Irrlr,lirand is

= c(n@)) + c
Is@la,=G(u)+c
These rules of integration need not be used in isolation. As will be discovered is course, I r,rrrrple 6.5

functions may require the use of more than one rule. There is no harm using one rule for a
Find[,(2',!1) dx
ofthe function and another rule for other parts. This does not vitiate the answer. "
lx-z)(x+3)

Some functions however may be combined or expressed ln more complex forms. For exa Sometimes the way questions are presented, it may not be clear at first site which rule of
'Gt
the methods covered so far have not provided for the integration of a function which is
product of independent functions. The term independent is used in a more restricted man numerator and the denominator and check whether one is a derivative of the other. The
It is used to mean the functions do not satisfy the relationship required for the logarithmic rule of thumb for polynomial functions is that the derivative is always of a lower order.
exponential rules of integration to apply. The integrals of such complex functions can be combine this with the fact that the derivative, as a derived function cannot be in the
by way of substitution method or integration by parts. These are discussed below.
denominator. This leaves only one possibility to be checked; whether the numerator is a
derivative of the denominator, which we confirm.
6.4.4 Substitutionmethod
Then introduce a new variable as follows:
Since integration is the reverse of differentiation, we will often get back to differentiation
reload on the concepts applied in differentiation. This should make the work in integration a u=(x-2)(x+3)
easier as will be demonstrated for more complex functions. One concept prominent =x2+x-6
differentiation is the chain rule, which applies when one function is a function of another
fly=(2x+L)dx
ls, the function C(h(x)) has the differential
Then substitute into the integral
d
.h'(x) (2x+1.)
dxc(h@) = s(h(x)) t fdu
The reverse ofthe above will be the integration ofthe right hand side ofthe equation to get Jt._zx.l:ro*= J ,
left hand side of it. Formally, this can be stated as =lnu*C
I
= ln[(x - 2)(x + 3)l + c
I s(h(x)) h'(x) = G(h(x)) + c
. (r,4.5 lntegration byparts
J
This provides a way of integrating a product of functions. However, caution must be \ome functions however may not possess the relationship required for the substitution method
before applying the formulae. The method requires that the derivative relationship be L apply. suppose now we have to deal with the integral of a product of two completely
the two factors of a function is accurately identified This is known as the substitution lndependent functions. The substation method will not offer anv clue, nor will anv of the
of integration. lt is used to integrate a function which is a product offunctions which possess n|cthodsdiscussed",,ti"ieu5Withcompletelyno
special relationship. lt is a requirement that this derivative relationship exist if the method is nlternative.
be used
 Mathenratical Optimisation and Programming Techniques for Economic AnaJysis 1,25

tz4 | ,n.ro*ot .ALCULUS lrlr ,ruse with successive differentiation, the differentiable function must be getting to a
mind that theprocess of
itmust always be in one's ' "n\tant Thus the method of integration by
parts will have to be performed as many times as it
Though this may be over emphasising' there is no harm to ever
* *" of differentiation' As such' l,rhos to differentiate the function u to a constant. Since both functions are likely to be
integration i'
'r'.|" '"u]|.J "ot"" 'lrll(,rentiable, the strategy is differentiating one which ultimately becomes constant if
u functions of x' its differential
trekbacktocon'iaeraiferentiationconceptst;tt";ofuseinintegrationThistime'the
Uotf' u
eiu"n'tl'naion f (x) = " *ltt' rr,lrcatedly differentiated. For instance, the function 1as simple as may look cannot become
'product "na
'uf"' is given bY rorrstant with repeated differentiation. A function of the form ax' can be reduced to a
using the Product rule
rrrrr.,tant with repeated differentiation. This should guide the selection of which function is
'rlrllerentiable' and which one to 'integrate'.
*
with this derivative at
hand' ,":,t:l,;t#:;# #'",! n,esra,e o*n *0*
'd tr.5 Initial conditions and boundary values
above differential' A', rnentioned earlier in the chapter, the lost constant in differentiation cannot be retrieved
lrorn the differential alone. With the constant unknown, the best was to include an arbitrary C

i; (,, ar = ir"'+uu'\dx
'ur=lrr'dx+luu'dx
iil rls place. But the integral that results is a function which can be plotted on a Euclidean plane,
lrrrt for the unknown vertical intercept as is sometimes called.

int"grJt ot a differentlal
of a function is a function Ilrc shape and slope of the function is known but the exact location hinges on the unknown
,...ur"',n" sum of the integral of rrrrstant. But what is the constant? And why does it remain so critical? To the first question,
sum of functions the
r
is
httd oit
itself3a. on the right 'id;';;n'"g'ur then needs to be rearranged llr('constant in the function is the values of the function when the explanatory variable
nl' nolnt"g"tion' rh" l"tion
the functions' The left
hand not matter ,r,,,,umes a null value. With a constant C, then the function passes through the point (0, c). For
'''i" hand side' Here it does
to have the intesration ;;;;;; ;r'no]on' :' ::" llr.second question, the constant is not as critical as is normally put. lt is only critical to the
" ' '-"n
whichofthetwofromthe..'gf'"t"n't"ototheleft'Nonetheless'thechoicewillhaveabearini r.xlont that it is simpler to locate or work with the point (0, c) for any function.
identification of the functions' Irr cconomics, the point (0,c) is often referred to as the initiol condition Since most economic
expect that
formally and the authors
by parts cjn be stated v,rriables are restricted to the positive side of the plane, it is befitting to refer to x = 0 as the
At this stage, the integration
The functio&is-/--\. ',l,rrting point. lt gives the starting or initial value of the function Any information about it
reader will not have any problem
the precise point (0, c) from which the constant of an indefinite integral can be
r,,,,,c.ntially gives
rlllinitised. From the foregoing, the arbitrary C in the indefinite integral is now definitised to a
lr,rrticular C = c, where c is the initial condition.

Allornatively, a point along the function may be given. This is equivalent to stating the value of
llrt function for particular value(s) of the regressors. For instance, in a time series, information
rrrr the function may be available for a particular year. Thus in the function Y(t) : I f(t)dt + C
,r grarticular or definite value of C can be found by substituting the values of (t,Y) of a known
l)oint Once the value of C is definitised, the integral ceases to be indefinite. lt becomes a
rlr'[inite integral and gives a particular function with the integrand as its derivative
the function r..--+i^^r This
ofr functions' Tt-'
integration of a product I x,rmple 6.6
But the result side of
the equation also has an by parts will also re
t^"t the outcome of integration : 19 pind the
not be intimidating nlthougfL;;i"*" pc The marginal cost of producing a given level of output is given by MC
former' This is made *+
tn" *'' not be as complex as the
integration by parts' total cost function given that overhead costs total 15.
'ttt"'
of Calculus at 6-3
\efer to Fundamental Theorem
 Mathematicar optimisation anrl programming Techniques for Economic
L76
I

I TNTEGRAL CALCULUS Analvsis 727

It is already clear from microeconomics theory that the marginal cost is the derlvative of TRr=o = Q

the total cost function. Therefore, the total cost function is an integral of the marginal - C =O
cost function. So, proceed to integrate
TR=25x-3x2
lx 10
TC |
= J10 tr.7 Definiteintegration
-+-dxx
x2 All the integrals discussed so far are indefinite integrals.
we explained already what makes
=-* l0lnx*C llr.m indefinite; it is the lack of a definite numericar varue. Even when
the constant of
This is an indefinite integral. lt has two parts, the variable part and the constant part. The lrl.gration is know, there is stilr no specific value ofthe integral because it
depends on the x_
variable part is the Total Variable Cost (TVC) and the constant is the Total Fixed Cost v, rlu e.

(TFC) Since the latter is given, then the value of the arbitrary constant is known,
[ 161a* = F(x) + c
C = T FC = 15. To total cost function is therefore given as J
ll we identify two values of x, a and b such that a < b, we can substitute
x2 into the indefinite
TC = n* 10lnx+ 15 lItcgral to get a definite numerical value.

6.6 EconomicApplications IF(b) + cl_lr@) + ct =_gJL

ll''
constant will be common in both F(a) and F(b) and wiil disappear
on differencrng. The
lntegration is a useful concept in the field of economics in general. With a widespread use of
,",rrlt is a definite value, free of the variable x and the arbitrary constant.
lhe morginol concept, integration provides a critical link from the marginal function to the total This is what is
rr'fcrred to as
function. For instance, ifthe marginal revenue curve is known to us, integration provides a way le
llre second b is the up
of getting the total revenue function which can further be used to get the average function, definite
limits below and above the integral sign. The lower limit will be below
These are critical functions in microeconomics. the
u pper
Consider the revenue example. Suppose the marginal revenue function is given by
t lrrs is provided by the Jundomentor theorem
of corcurus, which says the definite integrar can be
MR=25-6x r()rnputed by subtracting the endpoints-of an integrar. For
a function f(x) continuous in the
where x is the quantity of the commodity produced and we wish to find the total revenue ,rI.rval (a,b), the definite int"graiis*-'.--
function. Like the total cost function presented in Example 6.6Example 6.6 above, the total
b - ..=.-
revenue function is the integral of the marginal revenue function.
f
f | [(x) dx : F(b) - F(d )
TR:125-6xdx J'
a ,_.---'
J
Io put this in context, suppose ;" ;;i;"d
=25x-3x2+C
in the area bound by the function
horizontal axis and two rimits, rower and upper. This is shown in
f (x), the
the figure berow.
The marginal revenue function depicts the behaviour oftotal revenue as quantity changes. lts
information is insufficient to determine the exact position (determined by the vertical intercept
C = cl of the revenue function. So, an additional piece of information is required. When no
such information is explicitly provided, then we must resort to theory.

In the above case, the initial condition is not explicitly provided. lt can however be implied from
theory. The guidance from theory is that revenue must be zero when no output is produced
and sold.
t28 l ,rrrn*o,-.or.rr* 729

Figure 6.1. Area under a Curve

f(x)
f (")

For such a shape, geometry does not provide any method of precisely getting its
Nonetheless, it should still be possible to use geometric methods to estimate the area.
starting point isthe Riemonn sum, named after a nineteenth Germany mathematician Be
Riemann For the above area, partition it into rectangles whose area is a product of the width
(w) and heieht (h).

Figure 6 2: Riemann Sum

fhf lntegration takes care of infinitesimal division and sets the limit of summation as n
f(x) larger. The precise area can thus be given as
flcomet
b
I
I
a= | [(x)dx=F(b)-F(a)
J
I
a
I

I
llr lliemann integral approaches integration from an area of a shape point of view. lt
I

I
, rlr ttlates the area between the function and the x-axis. Since the function, as illustrated in
I

I
I ty,rrt: 6 2, both the height and width are positive and so will the area. When the function falls

lrllrrw the axis, the height of each partition becomes negative. With a positive width, the area
wrll be negative Consider Figure 6.3.

ln Figure 6.2 above, the area is partitioned into seven rectangles. The width ofeach rectangle is
the difference between two successive values of x. That is w : xi+t - xi = Ax The number of
partitions we can form will depend on the width of each partition. The guiding formula is
'# = o, The partitions will be small the more they are or the number will increase with a

reduction the average width. For the rectangle bordered by x1 on the left and x; * Ax on the
is
right, the height is the value of the function evaluated at xi. We now state the area of each
rectangle as:
 Mathemrtical OptimisatioD and Programing Tcchniqucs tbr Economic Analysis I f af
130 | cALcuLUS
'*.ro*o.
l,,ll Properties ofthe definite integral
Figure 6.3: Possibility for a Negative Area
lrr tlre preceding discussion on integration, some properties emerge. Though it will be too much
1,, l)ighlight all of them, the book looks at a few key ones. The reader can infer from the few
Ilvr,n to generate the rest.

l'r)l)erty l: Given a function with a constant coefficient, the constant coefficient can be factored
out of the integral Formally this is shown as follows
bb
rr
1tf(x)ax=kJfQ\dx
a4
this property is also ideal for dealing with negative in an integral such as

It" -f fOa*. Perhaps the word negative mentioned may be a source of worry.
The simple way to view the negative is that it is a constant coefficient and thus
can be treated like k in the property above. ln fact, most cases will present the
negative as part of the constant, that is, the constant coefficient k is negative.
the left
the function and the axis bound by a on
Suppose we are interested in the area between
two limit Riemonn integrol l'roperty il: Since
f /(r) d.x = F(b) - F(a), then interchanging the limits of the integration
and d on the right Finding this area by simple
only changes the sign of the definite integral.
f ab
) f(,t
a* (t
J
fQ)ax = F(a) - F(b) = - ) fjta*
wouldbeincorrect.ltwouldbeincorrectbecausewithinthetwobounds'thefunctionisabove ba
for some other(s). As a consequence, the area . The same property can also be used to prove that when the two limits are
the horizontal axis for some part(s) and below it
equal, the definite integral equal zero.
wouldbepositiveforsomepart(s)(thoseabovethehorizontalaxis)andnegativeforother(s)
graph, the totar area under the curve
(those below the horizontar axis). Thus in the above I'roperty lll: The third property emanates from the fact that a given area can still be calculated
between a and d would be given bY: as a sum of its partitions Given an area under the function f(x) bound by a and

TotalArea: A-B+C d on the left and right respectively, the area remains unaltered if partitioned into
as manypartsas possible. Definetootherlimits c and b such thata < c <b <
bcd
d. The sum of the three partitions defined by the four limits equals the area
= l tr-,ar - | frrtar+ [ frnax bound by the lowest and highest limits. This is stated as follows.
ab
dbcd

Sinceareaofafixedshapeisadefinitenumber,definiteintegralswillalwaysresultindefin
given, the marginal utility function' we are able
to determine the total uti I I@ar
J'
-- | tatar + [ f<,)ar + s61a*
J' J' J
|
numbers Thus, aabc
units of a good' Given the marginal cost
function' we are able to determ This property permits the evaluation of area defined by discontinuous function
from consuming 7t

total cost of producing Q amount of output' since the partitions can be made to follow the points of discontinuity lt is often
referred lo as odditivity property of integrotion.

I'roperty lV: The last of the four properties involves a sum of functions. This also includes the
difference since a sum and difference apply the same principles, mutotis
 Mathe matical Optimisation and Programming Techniques lbr Ecrlnorric Analysis | ,r,

r32 \ ,rrtu*ot
cALCULus
variable so that the resulting integral is of one variable only. lntegrate the part in square
then the propertY
a sum of functions /(r) = g(x) + h(r)' brackets, treating the other variable as if a constant.
mutondis Given
states that
tlrl
i ,,.rror= Jf
J";
i''t h(x)ldx = I n'*1 x t )nt)ax ll*'.v2)dvdx
the rule of
of function to-'annlV
separating a polynomial
This propertv allows the same
orten'':l':h:*":,1i[]}"i'[il:: :"'
'"'"'l'o"'"''"o'raterv the use
form and thus will requlre

6'9 Multipte integrals r:s^-^^+i'ta a frtnction with respect to more

than ono

n* : ::r ::,iiil, i i ?i" :l : :: :JiT::x": i:,": r,'"'ffi ;:t* [*: :i J :: i.H I

{, l0 lmproper integrals
pec'i'io
i'il1riii'll' ;" ;;;;' c" n w'ih re s
ru n
'i,rr('times in integration, one of the limits or both may be infinite. For example in the
fl""1ltr i*:;*.:l'J;:ni::#Jl
a tim€
"
invorves integrati-ng
with respect to one at
,.ro0,", This rrrrr,liratlon
E tfOd.t, the upper limit is infinite. Such a problem cannot be evaluated because
more than one independen,
as constants tlr,infinite on the right literary means its open on the right. Moreover, area can only be
while treating the others may first be integratot
z
to integrate a function = f @'y)"The unction , rI rrlated for a closed space This crumples the integral.
Suppose one wants given limits The res rlting function
x, and *"';; t;;;;e 1,, rlcal with such a problem, it is necessary to assume a finite upper limit which can be
with respect to limits' The process is
called
to y and ""t''"t*'"""' other
r,,.,rrned to approach infinite This takes us back to the concept of limits studied in earlier
with respect or more
one variable' one
'lven lr,rl)ters The solution to the above example is found using limits as follows.
because, white integratin-;;;;;,;o
I
with
with the order of integrating *b
constant. The same '""n"t"'o"l*n
integrols written as tr tx)ax
interchanged This is shown usingdouble
b h(y)
Jf
aa
- l\t) rota,
l rlrr' limit exists, the integral is said to be convergent. The term convergent is used with a very
| | roaardY and this refers to a space bounded on all the sides except the right, the space
rrrrlrle meaning. Since
L oiy't
b h(x) ',rrr still be considered a closed spoce if the function converges to zero as the explanatory
v,rrr,rble approaches infinite. This is the case with asymptotic functions which virtually converge
t,, /(,ro as the explanatory variable approaches infinite- This permits calculating the area even
rlrorrgh technically there is no upper llmit. When the limit does not exist, then the function ls

,lrvlrgent. This refers to functions that don't close-up with the axis

I r,rrnple 5.8
ExamPle 6.7
Eva I uate
0

* **,*-,,, J":i: il il,*il ;:" r';11'J;

Jilr'l n= | e*dx
l
il##fil=
134 INTEGRAL CALCULUS Mathematical Optimisrtion and Programming Techniques lbr Econonic Analysis 135

This integral defines the area of a shape to the left ofthe y-axis bound by the function tl rrrnt,rining the point a, then the limit can be evaluated more easily by replacing each function in
the x-axis. Since the function in questlon never becomes zero, the shape is open and a tlrr,rluotient by the respective derivatives. Formally,
cannot be found. Nonetheless, since the function becomes asymptotic I + -e,
uPWhhet:,,'ITsdened 65 it r
a'(
limf(r) = i,-
x+o n \X)
x)
iff tim
x-a s(x) : x-o
limh(x) = 0 or-

ll tlrc first derivative still give the same output, l'H6pital's rule allows differentiating the
rltlllrentials. This leads to second order derivatives. The process of differentiating can continue
rrrrtrl a determinate outcome is obtained. Caution must be exercised when to use the rule. lt is
rrillrortant to bear in mind that the rule only applied in limited circumstances. lt requires that
tlr,'rrndifferentiated functions are simultaneously zero or infinite as approaches.

,:: ,-
I r.rtnple 6.9

ln evaluating limits, it is possible to get outcomes of the form 9 :

I Fi nd

lq,
e'I
f, = -. nnswersl
O and
these quotients are non-trivial. The former is guided by the fact that dividing any number ir1f,
Substituting the zero into the function gives the indeterminate case of ! The quotient
zero gives zero. The latter is based on the fact that any number divided by zero is infinite. Thl!
also does not have any common factor that may be factorised and eliminated. Thus
forms, though involving zero, are determinate.
f, l'H6pital's rule remains the only option.
In some instances, it is possible to get expressions ofthe form Dlfferentiate the numerator and denominator separately to get
I ex
-l e'
lim_=linr-
r+0 X x-O '1,

such expressions have no determinate ,",3,,"J^, such, any such expression i, ,uia to lim er
= l+0
I
indeterminate. The limit cannot be determined. other indeterminate forms may involve,f,
=eo
difference between two infinite values. As a remedy for such problems, we factorise dtf
eliminate the common factor that drives both the numerator and denominator to zero. fhoul
this sounds common and is used in many instances, it is not always possible to find thl r'. l2 Economic applications
common factor. Some functions may lead to this scenario is though there is no common faclil
4,, noted already, the concept of integration, both indefinite and definite can be applied to the
between the numerator and denominator. il ',ludy of economics in many ways. lt provides a link between many econometrics problems. ln
The l'H6pital's rule, named after a seventeenth century French mathematician euillaume t ,rrrother way, integration, like differentiation, provides an additional tool for working with
I Hop.ital, provides a more general way of dealing with such functions. Given a function of tl lurctions. Given marginal utility, integration provides a method of finding the total utility or

r@:#,
f-irvcn the net investment rate, integration makes it possible to derive the level of capital as a
lrrrrction of time. This section gives practical economics examples in which integration is used.

llr| first scenario is where integration provides a link between marginal and total function. This
tf utility, revenue and cost functions. For illustration, we use the

then the rimit or the runction ,ril:,:'";J:2.1T].=r0.,.,,.,,.r varue a is ina"t".,inatl

I l', ,r broader area encompassing
r r rrt function but the reader must be able to apply the concept to other areas Suppose

MC=4*6x1-1-5x2
L'H6pital's rule states that if the functions g(x) and h(x) are differentiable in the intervil
136
I

| TNTEGRAL CALCULUS Mathematical optimisation and programming Techniques for Economic Analysis
737

, what is the total cost function? lt is known from theory that total cost function is the integf{ Fllttro 6.4: Consumer and Producer Surplus
of marginal cost function Therefore, proceed with integration
r
rC = J| MC(x)dx
f
= Jf (l5xr*6x-t4)dx
=5.x3+3x2+4x+C
This integration provides an opportunity to once again consider the arbitrary constant C. the
constant C was not in the marginal cost function but only emerges in the total cost function. ln
addition, it does not depend on the level of output x. Microeconomic theory defines this typo
of cost as fixed cost, a type of cost related to overheads. Recall that the marginal cost measures
the incremental costs, and must have nothing to do with overheads. lt would however be
misleading to ignore this type of cost in the total cost function because the firm incurs it.

The second application of integration as a measure of area under a curve is in the measurement
of consumer and producer surpluses. The former is the area between the demand curve and
the price line while the latter is measured by the area between the supply curve and the price
line For illustration again, there is no need to consider the two cases but the reader is left to supply curve, D is demand curve, p' and Q- are equilibrium price and
S is

corollory apply the concept to the other The example used here is the consumer surplus. The quantity, CS is consumer surplus and pS is producer surplus

figure below depicts the consumer surplus in the price-quantity plane.

llr('integration studied referred to an area bound by the axis on the lower side. on the
r 0ntrary, consumer surplus in not bordered by the axis but by the price line, which
is above the
,rxis since price is eternally positive for a good. This should not prove problematic as some
',rrrrple algebra is sufficient to get round the problem. The limits are not explicitly stated and it
rrr,ry be mistakenly taken as an
indefinite integrotion. The area of interest is running from Q = Q
to the last quantity. For the last quantity, microeconomics comes into play lt is the quantity
rrP

where the demand intersects the given price line The rationale is that a consumer will continue
lrttying successive unit for as long as his/her marginal utility (demand) remains above the price.
llris point will define the upper limit of integration.

lor the area below the price line, one must recognise that this is always a rectangle. lt is
rlofineci by four straiSht lines meeting a right ongle. The formula for its area is defined in
|lementory mothemotics as a product of its two orthogonal sides. To sum it, consumer surplus
rs given by the area under the curve bound by the zero and equilibrium quantities
less the
r cctangle.
o'
cs= [ prO; de-p.Q.
t"

lhe rectangle being subtracted perhaps deserves attention and it may not suffice to just glossy
 I

138 I INTEGRAL CALCULUS

over it. Recall that consumer surplus is defined as utility in excess of what the consumer pays
for, This is the difference between total utllity from consuming a good (area under the curve)
and the value of money exchanged for the same goods (the rectangle)

The third area of application of integration is in the analysis of investment and the behaviour of
capital stock, a functlon of time. ln Corporote Finonce, net investment /(t) is the measure of
net addition to capital K(t). lt is the rate of capital formation. Algebraically, thls is shown as

o: IG)
dt
The stock of capital increases if net investment is positive and declines with a negative net
investment. Thus capital stock is the sum of all net investments if time is discrete. For a
continuous time case, capital stock is the integral of net investment. For a particular example,
suppose net investment is given by /(t) = 31tt s and that initial capital stock ((0) is zero. What
is the time path of capital stock?

The solution lies in integrating the investment function. Though no limits are given, the initial
condition provided suffices to definitise the integral. Proceed as follows.

t
K(t) = | /(r)dr
J

= J[ ,rot 4,
:2t1s + C
this is an indefinite integral because there is still an arbitrary constant in the function. Using the
initialcondition,

K(0)=Zx(0)rs+C
:C
Since K(0) : 0, given in the condition, then C = 0. The time path for capital stock is thus

K(t) : 2st s

The rate of capital formation is a function of time only. The accumulated capital also follows as
a function of time only. This example must however not be taken to describe the general
behaviour of capital. lts interpretation must be restricted to the above given scenario, that net
investment is only a function of time.
 Mathematical Optimisation and Programing Techniques for Economic Analysis 1,41

Chapter 7
7 STATICOPTIMISATION: UNCONSTRAINEDOPTIMISATION
7,1 Introduction
ln a perfectly competitive market, each player is too small to influence the market prices. Take
an example of a producer who hires ,L amount of labour and K worth of capital to produce
output. The quantity produced is denoted by Q. The prices of output, labour and capital are
glven as P,w and r respectively. The sole objective ofthe producer is presumably profit
maximisation. The producer optimlses the profit given by
tt=PQ-rK-wL
ln this optimisation, the producer is not limited or restricted either by output or how much
lnputs can be utilised. We know that both capital and labour can be hired on assurance that
they will be paid from the resulting sales. As such, the producer can hire as much capital and
labour needed limit in order to maximise profits. He continues to engage more units of inputs
as long as the Marginal Revenue Product (MRP) remains above the unit price of the particular
lnput. No budget constraint exists since the producer can hire all the capital and labour that is
needed on assurance that such will be paid from the resulting sales of output.
Because ofthe nature ofthe market, the producer is also not bound on how much to produce
sincethe producer is too small to affect total supply. The same exists in the factor market
where a single employer can employ as much as he is able without affecting the price or
exhausting the supplies. Because there are no restrictions to the optimisation, the optimisation
is referred to as Unconstroined optimisotion.

7,2 What is optimisation?

As already explained in Chapter 2, optimisation is synonymous with the 'search for the best'.
But searching can only take place where there are many alternatives. Optimisation involves
choosing, from amongst many alternatives, a combination of factors that gives the best of what
one is interested in So in optimisation, there has to be choice.
The word best also needs to be understood in a dual way. lf one is availed many bundles of
goods from which to get only one, the interest from these goods is to generate benefit called
utility in economic terminology. Utility or happiness is a desired characteristic or state lt is
something one wants. ln this context, best means getting the highest possible level of utility or
satisfaction. Optimisation will therefore be synonymous to maximising This entails searching
for a bundle that gives the maximum utility.
lf the commodity under consideration was pollution, the choice would be different because the
objective would have changed. Pollution is something undesirable. No one wants to have it,
only that nature imposes it on us. Given many bundles with different levels of pollution, ceterE
poribus, the best option or bundle ls one with the lowest level of pollution. Thus, while the
't47 STATICOPTIMISATION: UNCONSTRAINED OPTIMISATION Mathematical Optimsation and Programrning Techniques fbr Economic Analysis | ,0,

terminologies best or optimum remain the same, the direction of optimisation changes
llrr' 1;oint , = -* is the demarcation At this point, the derivative is neither positive nor
dependlng on the nature of the variable. For a desirable variable, optimisation means
maximisation. When the variable is undesirable such as pollution, payment, etc, optimisation r,'[,rtive. lt is zero. lt is a point at which the function changes direction, from decreaslng to
rrr r[asing and vice-versa. lt is called a turning point.The turning point therefore is a point at
implies minimisation.
wlrtr lt the function is changing direction or simply turning. At this point, the first derivative is
Choice is not only restricted to bundles ln many cases, choice involves choosing the level of r,,ro, separating the negative and positive portions.
some given variable. lt may be choosing how much to produce so that profits are maximised
(optimised). lt may also be choosing the quantity to produce so that the per unit cost is at its llrrr', when the first derivative is zero, the function is turning. such a point is also the maximum
,,r ilrinimum of the function since the function is no longer going any further, either downwards
lowest More precisely, optimisation refers to the choice of values of certain variables which
could maximise or minimise the value of a function. For example, given a profit function ,rt ttpwards. Both the maximum and minimum require that the first derivative is zero. The
,,;rlrttsite will however not be definite. Getting a zero first derivative will not be specific on
fl = f Q) where Q is the level of output, the objective is to choose that level of output which
will maximise the profit. Alternatively, given an average cost function 6 = f (xr,x) where x1 wlrllrer the point is maximum or minimum. For this reason, it is safer to simply call the point
and x2 are the outputs of two products, the objective would be to choose the values of outputs
.rr,ptimum point since the word optimum encompasses both maximum and minimum. we
which minimise average cost. These problems are examples of unconstrained optimisation. wrll discover later however that, under rare circumstances, a zero first derivative may neither
lrc rrrinimum nor maximum.
7 .3 Meaning of the signs of first and second derivatives:
I lpirrrc 7 1: A depiction of changing direction
As noted in chapter 5, the first derivative measures the rate of change of the dependent
variable as the independent variable changes by a unit. Consider a function in one variable,
y = f(x) The first derivative ,u.rrr", the rate at which / changes with respect to r. The
fl
function is changing positively or increasing when the first derivative is positive lt is changinB
negatively or decreasing when the first derivative is negative.
When the first derivative is zero, it means the function is not changing or, more commonly,
constant. This rate of change may be constant as in the case of a linear function. Take for
instance the function
y=ax+b
where a and b are constants The derivative of this function isL = a Since a was defined as a

constant, the derivative is constant. lt is the same for the entire domain of the function. Save
for a constant function, linear functions have no finite maximum or minimum.
For a nonlinear function, the rate of change is not constant. lt can only be specified for a
specific point or value of the domain. Take an example of a quadratic function I r1;nre 7 1, above shows two cases of a turning point. The first (a) is a maximum and the second

y=ox2+bx+c {l') r\ a minimum. Points along the graph show points of different slopes. At point a1, the slope
r'. t)()sitive and the function is increasing; At point a2, the slope is zero. The function is constant
The derivative of the above (non linear) function is
,rrrrl on the graph, it is represented by a horizontal line. point a, shows a negatively sloped part

!=zo'+t'
dx
,rl llrc function. At this point, the function is decreasing. Similar comments can be made on part
(lr) ofthe figure.
The derivative contains the variable x. This means it also varies. As the independent variable
llrr"'econdderivative'ssignshedslightonthecurvatureofthefunction.Atthepoint x=a,if
changes, so does the derivative which measures the slope When x < -ra, th" derivative is
tlrl second derivative,fi at a or f"(a) is positive, the first derivative will be increasing and
negative and the function is decreasing. When x > -!, tn" derivative is positive and the
llr. lunction will be concove upwords. A function is concave downwards when its first derivative
function is increasing. Thus the function has two portions, one for which the function is
t', il('creasing so that the second derivative is negative, that is/"(a) < 0. lf the function,s
decreasing and another for which it is increasing. These are separated by the point, = - * ,r, r cleration is momentarily nil and the function is changing its curvature at that point (from
 AN
Mathematical Optinisation and Programming Techniques for Economic Analysis ,0,
uNcoNSTRAINED oPTlMlsATloN
|
r44 | ,aor,a oPTlMlsATloN:

concavitydownwardtoconcavityupwardorviceversa),thenthereisapointofinflexionand
f" (a) = 0 at that Point'
Note however the following:
i. concavity upward at x: a implies f "(a)> 0 and f"(a) > 0 implies concavity

and f"(o)<0 implies concavitv

:::/J:ii, downward at x=a implies /"(a)(0
', downwards;
0 does not Case (iv) f'(x) = o, case (v) /'(x) = 0, f" (x) = 0 case lvil f' (x) = o, f" (x) < o
iii point of inflexion at x = aimplies f "(a) = Q [u1 /"(a) i"*:-t:l'I T:T
-= t"@)>0
thatx=oi"pointofinflexionunlessweknowthatf"(o)changessignatr=a't'e'
derivative is a necessary but not a
zero value ofthe second
f"'(o) +0. i" Jan"t *"'as,
point of inflexion'
sufficient condition for a

Sincethefirstasweltasthesecondderivativecantakeonpositive,zeroornegativevalues,we
have a total of nine Possible cases:

(i) f'(x)<0; f"(x)>0;

i,il'f'(i<o; f"(x)=o;f"'(x)+o;

+0 case (vii) f'(x) > 0, Case (iixl f'(x) > 0, f" (x): 0Case (ix) /'(x) > 0, f,'(x) < O
t"@) > 0

+0
lr can be seen that in case (v), the first derivative was decreasing up to zero but instead of
the interplay of
d scenarios to help in understanding , rossing into the negative, it suddenly starts to increase Such a point, though having
f,(x) =
the shape ofthe graph' {), is not a turning point since the function continues in the same direction lt is instead referred
the first and second derivatives in determining lo as a point of inflexion. cases (iv) and (vi) show minimum and maximum points respectively.
\ince, in optimisation, we are interested chiefly in such points we shall discuss them below.
FigureT2'Graphsshowingdifferentcombinationsoffirst-andsecond.orderderivatives
7.4 Unconstrained Optimisation in a Single Choice VariaL!9q___
'-----.\--
tl y: f(x) is a twice differentiable function, that is, /(x) and f'(x) are both smooth and
r ontinuous at a particular point x : a, then the function has

(i) Aminimumvalue atz: a if

\ \ \ f'(") = o, and 7'"(x) > o;
(ii) A maximumvalueatx: aif
/'(x) < 0, f" (x) = o case (iii) f '(x) < 0, /"(x) < 0 f'(") = o, and f"(x) < o
case (i) f'(x) < O, case (ii)
(iii) The condition in (i) and liil is a sufficient but not a necessory condition for a
f"(x)>o minimum or maximum. The necessary and sut'ficient condition for a minimum
maximum) is that /'(x) = 0 and /(") is positive (negative) where /(n) is the first
non-zero higher-order derivative and n is an even integer

lhe terms 'maximum' and 'minimum' in defining these points is not used in an absolute sense
,rr they need not be unique for a function A function may have more than one minimum or
 Mathematical Optimisation and Programming Techniques for Economic Analysis | ,0,
t46 uNcoNSTRAINED oPTlMlsATloN
| ,roa,a oPTlMlsATloN:
dv

maximum in the sense of fulfilling conditions

(i) and (ii) stated above. Consider the followinS *=tz-t"
graph: l or an extreme value, we need I = O.
Figure 7.3: Maximum and Minimum Points 12- 3x2'= 0
.------=-
x = rl+
y=i(x) x:-12
The above solution shows that the function has extremes at two points. Therefore,
obviously, one is a minimum and the other is a maximum. We therefore use the second
order derivative to determine which one is maximum and minimum.
dry
-:--= = -t''r
d.x'
f "(z) = -12 < 0
Therefore, atx = 2, we have a maximum. The value of y is

A'C'EandGandfourpointsof minimum;&4F f (z) : 1.00 + 1'2(z) - Q)3

Thefiguredepictsfour(4) pointsof maximum the function since :100+24-8=1.16
sense of it being the highest value of
and H. Point A is not a maxlmu' in tf''"
thereareseveralothervalueswhicharehigherthanA.itisamaximumonlyinarelativesense a + Ax
that the function has a highest value at x =
a than those at all neighbouring points However, f ' (-Z) = 12 > 0, thereforc,at x = -2, we have a minimum.The value of y is

in x ln other words' /(a) f > (a+ Ax)' Similarly'

where Ax is an arbitrarily s"ll '"r.iation f (-2) = 1.00 + 1.2(-2) - (-D3
<f(b+Ax)'PointssuchasAandBare :100-24+B:a4
pointBtooisaminimuminarelativesensethatf(b)
thereforecalledpointsofrelotivemoximumandrelotiveminimumbecausetheyaremaximum
points'
and minimum relative to the neighbouring
I x,rmple 7.2
PointcisthehigheStofallmaximumpoints.ltshouldthenbethehighestofallvaluesofthe
j; y : f(x) =
runction. rr this were .*", ,"-,"i . *J,'o
o" Find the extreme values of the function 9xa
;:l'",1,," ::^i::'#tri;:il;l"ilii',lilil'il:l
C is not The first step is always to take the first derivative and locate a point where it is zero
Point
Point C on the Idx = 32r'
is unequivocal
F is the lowest Clearly, the first derivative is zero at.x = 0. Next, we get the second derivative to check
global minimum' for possible maximum or minimum. But the second derivative is also zero at this point
of all minimums. lt can be loosely call a This leads to no decision as yet.
global maximum or global minimum from the
first
There is no automatic way of locating the It must now be clear from section 7.3 that the necessory and sufJicient condition for a
andsecondorderconditionsonthederivatives.ltisnecessarytocomparealltherelative minimum (maximum) is that /'(x):0 and /(") is positive (negative) where /(n) ;t 11,"
one/s is arc global'
maxima or relative minima to check which first non-zero higher-order derivative and n is an even integer. Since the second order
derivative is zero when evaluated at r = 0, we proceed to find a non-zero higher order
ExamPle 7.1
derivative, therefore;
Find the extreme values of the function
Y:100*l2x-x3
f"'(x) = 192x, f"'(o) = o
point where it is zero' f"" (x) = 192
The first step is to find the first derivative and locate a
 I

L48 | Srnrrc oPTlMlsATloN: UNCONSTRAINED OPTIMISATION Mathematicar optilnisation and programming Techniques
fbr Economic Anaiysis 1,49
I

The forth derivative turns out to be a non-zero constant. When evaluated at any point 5ince the average cost function,-c, given
above has a single minimum at r: 1, it is u_shaped.
including the point of interest x : 0, it is positive Thus the first non-zero value of the lhe minimum average cost is 1. At n
= 1., the marginal cos"t
derivative alx=0,occursat/(a)whichisanevenorder. Also, 1""(o):192> 0'Hence
MC = 4 _ 7z(1.) + 9(7)2 = 1,3 _ 12 7
the function has a minimum at x = 0. The minimum value of Ihis proves the equarity of the average .ort ina =
marginar cost when the former is at its
rrrinimum. The other two prepositions about
l=Bxa the reratioribetween average ano ..[in c.n ue
= B(o)a: o "imilarly verified. The reader can arso check that the marginar cost reaches its minimum even
r',trlier at output level x= 3.
'3
7.5 EconomicAPPlication
I xample 7.4
Example 7.3
Let a monoporists'demand function be given as p:15
The total cost function for a commodity ls given by C = 4x - 6xz + 3r3 Find the value c = x2 + 4x' Let a tax of t kwacha per
-unit -zx and his cost function as
of out-put be imposed on the monoporist.
of the output x for which the average cost is lowest. we wish to find the maximum profit obtainabre by the
monoporist and the tax revenue
The average cost is a cost per unit of output. lt is gotten by dividing the total cost by the obtainable by the government.
level of output. Let C denote average cost. Then The total profit for the producer is the totar
revenue ress the production cost. Since the
_C
F---L-6x+3x2
producer must arso pay some tax, the totar
tax payment or obrigation is treated as a cost h/.
since it eats on the profits. Ll
Onemustalwaysbecareful nottogofortheoptimisationofthecostfunction Moreover,
such a point seldom exists since the total cost function is normally a monotonically
increasing function As a custom now, the first order derivative is.

*=-u*u,
ax
To maximise profits, we need

dlt
The first order derivative is zero at I = 1. From economic theory, we know this point -:-=0
dx
represents a minimum. However, this should never preclude the need for a test using the
dlt
second order condition.
E=tt-6x-t
d2e
--:-=:6 : 11,-6x-t=O
d.x"
d2c
> o
. 11 _ f
17or,=r:6 6

Thus average cost is minimum at output .x = 1. The minimum average cost is

The second order derivative giues
e:+-o(r)+s(r)'Z ff = -6 < 0. This is negative for a, varues of x.
Therefore, profit is maximised at x* :
=4-6+3=1-. #. ao get the rever of profit at the optimar rever
of output, substitute the optimal z. into ihe profit
ln the above example, the marginal cost function would have been given by taking the first function-
derivative of the cost function, that is, frmax=(1L_t)x_3x2
MC={=4-ryx-t9x2 (r1-r)f-')-3111-t"
t, ,/ \
dr 6-)
The relationship between the average cost and marginal cost curves (for U-shaped cost curves)
(ll - t)z-\ (11 - t)2
is as follows; when average cost is minimum, marginal cost is equal to average cost; when
(7r
6-n
-n-
marginal cost is below average cost, the average cost is falling; and when marginal cost is highef t)2
than average cost, average cost is rising.
 150 I

| STATIC OPTIMISATION: UNCONSTRAINED OPTIMISATION N4attrcnratical OptimisiLtion and Progr:iltuting Tcchniqucs lirr Econolric r\ntrli,sis 151

The government's total tax revenue : tr, substituting .r, we get the following;

_ r(11 - f)
6
11 1 " 13
V=--{'-r-+ -x+
'(t2 tt)
- 6" 6'
dy 1^ 3
?x'I'-
dxz2-xt
The profit level in Example 7.4 above is the maximum for the producer. For as long as
rate, price and the production technology remain the same; profits cannot be increa --
further. For the government however, the tax revenue is for a given tax rate. lt does dtv t')
represent the optimum tax level for the government. There still remains room for ds2
government to increase tax revenue, either by increasing or reducing the rate. ln either d, y,

the tax level cannot be increased indefinitely. Though an increase in tax rate should i --:0atx-2
ax'
the tax collection, at a certain point, the fall in output resulting from increased taxation
,13 v
outweigh the increase in the rate.
,tar-l*o
But how can such a point be identified? This scenario is similar to the duopoly setup Wr' have a point of inflexion al x= 2
each market player takes into account the reaction of the rival. ln this case, government
tax rate that when the producer will maximise his profits, tax collection will also be maxi LI rl ( o\t curves and
total productivity curves often display points of inflexion lf the marginal
l

The question is: Given the tax function, find the tax rate for which tax is optimised. We pt , I ( rrve is U shaped, it rneans that over initlal ranges of output it is falling, and then reaches
as follows: , r rrrllllum and thereafter it starts rising Correspondingly it means that the total cost curve
I I I r,,cs at a decreasing rate when the firm is enjoying economies of scale Subsequently, it
To maximise tax revenue, we set ,r,",,)t an increasing rate as diseconomies of scale set in At the output level, when the
dT d,rll 1 r ,, rrllrll,ll cost is minimum, we have a point of inflexion on the total cost curve The curve at that
ar:alzt -at')= 0
I ,,rrt rcasesto be concave downward and becomes concave upward
tt -;t
1. r r\Nilrll to the laws of returns, the marginal product curve is generally inverted U shaped. This
-
bJ = 0
rr, ,rtrs that the total product curve first increases at an increasing rate and thereafter increases
11 rt ,r rlccreasing rate. At the input level where the marginal product is at a maximum, we have a
*L:T:c.c
- .- - lr,rrl of inflexion on the total product curve. The curve at that, point ceases to be concave
,,1,w,rrd and becomes concave downward
The second order condition
fiftO -i-- a O,for all t. This means the set tax rate gives
I 1,,, lypical total product and total cost curves are graphed below
maximum tax level, given the producer maximises profits. Thus the maximum level of
revenue given all the above conditions is.
211 - 5.5r
r=s.sl-l
\12 1
t2t
- 48'
7,6 A note on points of inflexion
As already noted, a point of inflexion occurs when the concavity ofthe curve changes. While
point of maximum or minimum is necessarily a stationary point, a point of inflexion may or
not be stationary.
The geometric feature of an inflexional point is that the tangent at the point crosses the curve.
L52
I

| STATIC OPTIMISATION: UNCoNSTRAINED OPTIMISATION Mathematical Optimisation and Programming Techniques for Economic Analysis 153

Figure 7.4. Total vs marginal functions

a column vector X and let t be a scalar parameter on which each of the.ri,s

dxt /
/dt
dx, /
dX t d.t
s. The full derivative is then given by because it is a derivative with
dt

clxn/
tdt
toonlyt. Now lety -- f (x): f (xr,xr,xs.......xn),t the gradient vector is given as

af!) yafQ) af@) af@) af@).r

0x I dxr ' dx, ' 0x, """ 0x. I
Conslder now the derivative of the gradient vector with respect to (w.r.t) an ?z x1 column
Vfctor X.
a,f@)_apfQ)1
ar, = o.\ a. )
7,7 Hessian and f acobian determinants
Hessian and Jacobian determinants are also widely used in optimisation. The Hessian
was developed in the 19th century by the German mathematician Ludwig Otto Hesse and
named after him, while the Jacobian matrix is named after another 19th century
mathematician Carl Gustav Jacob Jacobi.
A Jacobian is a matrix of first order partial derivatives while a hessian is a matrix of seco
order partial derivatives. Jacobian determinants l/l are used to test for functional depen
for both linear and non linear functions while a Hessian determinant is used to check for
second order conditions in optimisation.

lfu frz " frrl

fr, " fr^l
--lf^
[r, ;, r,,)
n x n matrix called a Hession mdtrx. Notice that the hessian matrix is ;-nade up of all
This is an
the second-order partial derivatives, with the second-order direct partial derivatives on the
principal diagonal and the second-order cross partials off the principal diagonal.

Next, consider a column vector of m functions.

 Mathematical optimisation and programing Techniques for Econornic Anarysis
r54 | ,ror,. oprMrsATroN: uNcoNSTRAtNE, 155
""
,::];f I hr 11,nr,ra I form of such fu n ctions is give n by

z: t'@,y)
,t"=[:[,] wlt|t r' function of x and y ln total, the equation has three variables and therefore drawn
,Z is a
Irr llrr.r'dimensions, one for each variable. This will form a shape with
volume. The actual
rh,r;r.will bedeterminedbythespecificfunction. ThetwobasicformsareshowninFigureT.5
Where x is an n-dimensional column vector Then the derivative of g w.r.t x i Il,lr rw
differentiating each function in g w.r.t to each variable in .r. There are m functions in g and
variables in x This gives an ?n x 7I matrlx with rows representing functions and columt tlqrrrr, / 5: Three dimensional graphs for optimlsation
representing variables, given by:
og, og, og,
0x, ixz )xn
dg
09, 09, 0gr.

0x
0x, 0x, 0x,
::'.:
0g^ 0g^ 0g^
0x, ixz )xn
This m x n matrix is called a Jocobion Motrix. Notice that the elements of each row are
partial derivatives of one function 91 with respect to each of the in
variables x1, x2, ... ... xp and the elements of each column Ere the partial derivatives of each
the functions g, (x)SrG) ... ... g^(x) with respect to one of the independent variables x;.
It can be noted from the above definitions of the Hessian and the Jacobian that there are
points of distinction between them:
. A Hessian is a matrix of second-order partial derivatives while a Jacobian is a matrix
first order partial derivatives.
. A Hessian is always square while a Jacobian need not always be square.

7.A Unconstrained Optimisation in two choice variables lrr the figure above, part a. shows something like a dome or a hill which
has a summit. lt has a
rrr,rximum point. Part b. can be described as a trough, something one can use
So far we have looked at unconstrained optimisation in one choice variable, denoted by r in to fetch water. lt
lr'rs a minimum point But the minimum or maximum points may not
our previous section This is where a dependent variable such as utility or profit depends on arways exist. They may
lroth be absent in various functions
entirely one variable. lf a person consumes only one commodity, then utility will only depend
on how much of that commodity is consumed. For a single-product firm, its profit will only lor a two variable function such asZ = f(x,y), the necessary conditions for the function to
depend on how much of that product it produces. All these were cases of single choice variable, Irave an extremum at the value z : o and y = b is that at this pair of values, we must have the
Decision is only made on one variable. lollowing three conditions met:

An extension of the single-choice variable is a two-choice variable. We recognise that in reality, (i) u*
= O ^nO fi = O, rn" first order partial derivatives must equal zero
utility depends on quantities of many goods that an individual consumes. But we are
conservative enough not to rush to multiple-choice variable. lt is important to tackle this in a
simultaneously. This meansat this given point (a, b), the function is neither
increasing nor decreasing.
gradual manner Therefore, as a step towards reality, we first look at two-choice variable case.
(ii) The second-order direct partial derivatives, when evaluated at the critical point
ln this case, utility does not depend on a singly quantity because it is now naive to imagine this.
(a,b), must be both negative for a rerative maximum and positive for a rerative
On the other hand, it does not depend on many quantities, because it would be too much to
minimum. This ensures that from a relative plateau at (a, b), the function is concave
analyse. lnstead, two quantities determine utility
and moving downward in relation to the principal axes in the case of a maximum
155
I

J STATIC OPTIMISATION: UNCONSTRAINED OPTIMISATION Mathematical Optimisation and programing Techniques for
Economic Analysis 1,57

and convex and moving upward in relation to the principal axes in the case of a aJ)-(#)':0,
\ n(* then further investigation is needed before arriving at any
minimum. This condition is summarised (4aO andd^3" ( 0; for maximum
", dxz dy' r 111r ls5i6n. The test is therefore inconclusive.
"' , o .nd 0". 2 g, for a minimum. Note that these second-order
andl7' direct
I x,rrrrple 7.6
-9!a
partial derivatives are found along the principle diagonal of the Hessian matrix.
(iii) The product of the second-order direct partial derivatives evaluated at the critical Given Z : xz + xy + y2 - 6x + 5, find its extremum point.
point must exceed the product of the cross partial derivatives also evaluated at tha To locate an extreme value, we need
critical point. This added condition is needed to preclude an inflection point or r)z az
saddle point condition (iii) is summarise d by, (r*t1.#) - (#,)' , o. -=-=0.
tty dy

NOTE So we have

1) AccordingtoYoung'stheorem,thecrosspartialderivativesareequal,tnatis:!1=!1 az
Aady dy.lx -:-=2x*v-6
ox
2) lf we have condition (i), (ii) met and instead of condition (iii), we havc
a3=*+zu
I/()22 A2z\r-l-l/ Azz\z <0
\a*,' ay, 1 \axay) This results into two simultaneous
"Or3J"nr.
this means we have a soddte point. This occurs when *dxz ^na* have different signs, 2x+y-6=0
dy'
This point is called saddle because the function is at a maximum when viewed from ona
x*2Y=g
axis but at a minimum when viewed from the other axis. See Figure 7.5 below.
Figure 7.6: A diagram showing a saddle point

Given this point, the step th

dition to determine
the nature of the point, whe
the two. Recall that
the second order condition g the second order
direct partial derivatives and get the following;
a2z
0x2 -'
a2z
--1
)
oy'- '
--
drz _.
-r"",r,;::::',','::"*"o,,",, the x-axis, it appears to be a maximu,, o* *n"",,"r1 0x0y-'
from the y-axis, it appears to be a minimum.
I Atthe point x = 4 andy : -2,
3) However, lf we have condition (i), (ii) met and instead of condition (iii), we a[
a2z azz
;;) - (=;'< 0, but
# *o#have the same sisn, then we h"'"
'n '''r'{ ;-->0,
ox' dY"
i:;:# and -)0
/azz o2z\ / azz\2
= Q)Q) - (1)'z : 3 >
\aP ur'- )- \aA) o
 Mathematical Optimisation and Programming Techniques lbr Econonric Analysis 159
oPTlMlsATloN
1s8 I ,aor,a orr,r'SATloN: uNCoNSTRAINED
look dC,
the func ion' The whole shape must f +x1 f'=-::axt (1)
The pointx = 4 andy = -2 isa minimum for
likeaninverteddomeThelowestvaluecanbefoundbyevaluatingthefunctionatthe Corollary, for the second duopolist, we will get
turning Point
(-2)'-
z(4,_2) = (4)2 -r (4)(-2) +
5(4) + s dc-
6-8+4-24+5 1+xr.f'=7i Q)
1
The assumption of fixed x2for the first aropolir, and fixed .r1 for the second duopolist
means that each duopolist assumes that there will be no reaction from his rival to his own
ExamPle 7 7 action. Equation (1) can be used to solve for.r, in terms of x2 and equation (2) can be
same market' Though they decide
on output
Two producers (duopolists) face the. used to solve for xz in terms of .rr. These solutions can be written as
independently,eachiswellinformedthatthemarketpricewilldependontotalquantity
their individual outputs' Let
precisely' on the sum of
x1: R1(x2), x2 = R2(x)
supplied on ti" rn"ft"t or more
These two functions are called reaction functions and give the output of each duopolist as
x1 = suPP|Y of the first duoPolist a function of the other. lf the equations corresponding to the two reaction functions are
12 : suPPlY of the second duoPolist solved simultaneously, they will give the equilibrium outputs of both the duopolists,
o ana #'o rhe cost which may be denoted as irand x2.This is shown in thefigure below.
The r,arket price P, is given by P: f (xr+x2) with $<
given as
functions for the two duopolists are respectively
h=e
Cz=
The duopolists' profits n, and n2 are
Er= P
frz: P

whic
lf each duopolist determines the output
conditions are:
Ant
oxt
=
I tl Hessian and facobian Determinants in Optimisation
For the first duoPolist,
obian determinants l,[l are used to test for functional dependence for both linear and non
alt l,rr

u*' !er,-
= c,) Irrrr,,rr functions. lf Vl = 0, the equations are not all functionally independent. ff Vl + O, the
r.rlrr,rtions are functionally independent. By functional dependence, we mean that a function
--lL.
oxr
,,, ,r', ,,rrr be obtained by the linear combination or any other non linear operation of another
lun( lion. Consider the following example
= f(xr+ xr) + x
When firm one is deciding on output'
it doe
I x,rrnple 7.8
the output of the other as a c
So it takes
derivative is equated to zero'
SuPPose: gt=Sxtl3xz
f (xt+ x) + 4f '( gz-25xt2 *3oxrx2*9x22
First, we take the first- order partials,
f(xr+x.-)+x''
f +x ogt
ox
ln summarY, ',
160
I

| STATIC OPTIMISATION: UNCONSTRAINED OPTIMISATION Mathematical Optimisation and Programming Techniques for Economic Analysis 151

a_9.!
a2z
0x, =
3
0
iJy'
0o" -:2)
#=50xr]-30xz a2z
dxt
a*ar=
t
do"
3o'' + rcr'
fi= ',',lling up the Hessian, we have

Setting up the Jacobian, we have fl = 156r, i :Or, SOr, I fArrl

, =li il
Evaluating the determinant,

l/l : 5(30rr r 1.8x) - 3(50r, * 30.tr) n \ shown above, both values of the principle minor are clearly greater than zero. Note
: 150x1 I 90x2 - t5Ox, - 9Ox, lrat they need not be equal in value.
: (150 - 150)x, + (90 - 90)r, I

=u on the other hand, (ii) above is the same as lHrl > 0, where lHzl is the determinant of
Since l/l : O, there is functional dependence between the equations. Though the llre second principal minor, in this case, the determinant of the Hessian above. Solving
does not specify the specific relationship, it is easy to tell that, rrsing our example,

(5x, * 3xr) 2 : 25x12 + 3oxrx, * 9xr2 lH2l = Q)(2)- (1)(1) = 4- 1= 3> 0

That is, 92 = @r)2 lherefore, just as in Example 7 6, using the Hessian, we arrive at the same conclusion that
Recall the 2nd order conditions for optimisation, that is; the function is a rninimum at the optimal points.

(i) #.ornaff<0;foramaximunana ff> oandfn)0;foraminimum ln general, a Hessian matrix must be negative definite for a maximum. ln a similar
f fashion, for a minimum, the Hessian must be positive definite. ln the above example, this
(ii) wnere(u*fr.ff) - (#)' ) 0 must hotd for both cases in (i). > 0 and lflzl > 0This is also consistent with the second order
Now consider a second order Hessian matrix given as follows:
il means lHrl = 022

'T,*Wl I
conditions for optimisation in (i) and (ii).
I
7. I 0 Unconstrained Optimisation in n-choice variables
I r,l

y = f (xrxz... .... .,xn)

0y = frdx, t f2dx, I ... f,dxn
Example 7.9 l|
I or a critical value, we require all first order partial derivatives to be equal to zero. The second
Refer to Example 7.6 and let us use the Hessian to check for the second order
conditiofi rrrder differential 02y will be the quadratic form. lts coefficients properly arranged will be the
Z : x2 * xy * y2 - 6x * 5, The above stated conditiofl
Given to find the extremum of ..ymmetric Hessian matrix.
implythatfora maximum, l7rl= A'Z < 0, lflrl = 022 > 0fora minimum, where lHt ft, fr,
I frnl
is the determinant of the first principal minor. This is consistent with (i) above. ln d'[ r,, .
examPle' we have IHt= l/?'
l-' -i " fnnl
-'|
41
Ih, f,z
ff = r, o I lhe principal minors for a maximum must all alternate in sign starting with negative. That
, +, -, +........, where /r, = lrrl,lk'
f;'l =V rt,
is

^O
 Mathcnratical Optin)isrtion lnd Programming Techniclues lirr freunorrric Analysis | ,U,
L62 STATIC OPTIMISATION: UNCONSTRAINED OPTIMISATION

I , ,r llte left hand side

I f, , [ ,, ]'t,,1

rr,"r = lt, ',,, : ,,,,1 n,*Tyy= ( 4)( B)

lr;. h, i^^l llr,,rcfore, rxxnyy ) (ttr)2

=32
and profits are indeed maximised at x = 40 and y : ltl.
For a minimum in a n-variable case, just like the two variable case, all the principal minors i\l llrat point, profit 7r - 1650
be positive.
7LL
7.11 Economic Applications
ln monopolistic competition, producers must determine the price that will maximise their
Optimisation the study of economics in many ways. The principal aim of
is useful is profit. Assume that a producer offers two different brands of a product, for which the
to optimise. of optimisation serve a vital role in economics or decision m
As such, tools demand functions are:
Though many areas of application can be cited, this section will deal with profit opti
See the examples below.
Qt=L4-0.25P1
Qz:24 - 0'5Pz
Example 7.10
The producer has a joint cost function
A firm producing two goods; x and y has the profit function given by
n = 64x - 2x2 + 4xy - 4yz + 32y - 14
The profit maximisation level of output, the price that should be charged for each brand,
What quantities of the two outputs should the firm produce in order to maximise and the profits are determined as follows:
profit?
First, establish the profit function 1r in terms of Q1 and Qr.folal revenue will be the sum
Partially differentiate the function with respect to the two choice variables. of revenues from the two brands. Since profit is total revenue less total costs, then
o):a+-4x1 4Y n: PtQt+ P2Q2-TC
n,:4x-8y+32 : hQt * PrQ, _ (Q? + sQQ2 + Qz)
The two demand functions given above can be converted to inverse demand functions so
Equate the partial derlvatives to zero and solve them simultaneously for x* and y*
that they are substituted into the profit function.
64-4x+4Y=g
Pt: 56 - 4Qt
n:-an+32=0
Pz:48-ZQz
After solving the two equations above simultaneously, the optimal levels of
The profit equation continues as
emerge to be,
x- :40 n = (56 - e)et+ (48 - zQ)ez - Q? - SQre, - ei
!. :24 : 56C] - sQ? + 48Q, - 3Q, - sQ.Q,
Next, take the second order direct partial derivatives and check whether both The profit function above is the final equation. This is the function that must be
negative, as required for a relative maximum The second order partial derivatives maximised. The process of maximisation must be more familiar now. Equate the partial
T*, = -4 derivatives to zero and solve the resulting equations simultaneously.
ftvv = -B 0n
Both the direct partials are negative but this is just the first condition' ln the second,
_:56_t0Q1_5Q2
oQt
must check whether n*rn"u > (tt*y)'. The cross partial derivative is required in 0n
to the two partial already at hand. aqr:
+e - 6Qz - SQt

tt*n= ttrr=4 Therefore

(1t*y)2 :75 56-1001-SQz=0
 I
MiLthernalical Optimisation and Progrrrnming l'echnitlucs for Ecitnortric Arralvsis 155
t64 srnrc oprtM tsATloN : UNCoNSTRAI N ED oPTl MlsATloN
I

48 - 6Q2 5Qr = t)
Chapter B
Take the secon(
Which, when solved simultaneously, give Qt:275 and Qz:5.7.
derivative to be sure profit is maximised:
02n It S'!-ATIC OPTXMISATION: CONSTRAINED OPTIII,tISATION WITH
dQi
--= -10 IJQUALITY CONSTRAINTS
-.
d2n
_=_6 lr I lntroduction
d()1
r ,rr lrained optimisation is a mathematical concept mostly essential to the subiect of
, ,,rrrmics. Economics is essentially a framework for understanding the world in rr,rhich
,,irvrduals, firms and goveTnments make therr seemingly best Cecislons given the intrinsic
rrrl,rlions confronting their behaviour. The most common examples of constratnts are
rrrl,rlions on tinle, money and other resources available to the ilse of man For the most part,
tl, r,constraints are likely to be binding, meaning that man tends to use up all the avallable
,. ,.r{ or in makirrg his best dccisions
\4,i,,()lten have problerns of constrained optirnisation, where a function has to be maximised or
rrrrrrntised !ubject to certain constraints or -siale relotrcns For instance, if we have a joint cost
lrrrrrltonC =.f(x,y),wetnayhavetochoosethevaluesofxandvwhichwill minimisethecost
rlrlt,cl to the condiiion that x + ! = Z, i.e the output of bo'.n goods conrbineci must be equal
ri, r (L.rtain value, this equal sign in the constraint gives it the name, equality constraint Or, if
u, Ir,rve a sales functron S - I ?,y') where x and y are the kwachas spent on two adveriising
we may want to choose the values of r and y which will maximise sales suLrject to the
rr, rlr,r,
(2Js)2 + -F r\z'
s(2'7s)(s'7) + (s'D'z)
= 4s(2.75)+- Jb'o(5'/J
36'6(s'7) - (Z"/)']- ,,rrr',lraintof agivenadvertisingbudgetZ whichmustbeusedfully,tex+y -Z.Constrained
:273'94 I ,lrtrrnisation thus implies optirnisingthe value
ofthe r:bjective function subjectto certain side
,,rrlrlions When these sirie conditions are absent, you have an unconstrained optimum
lr ,, rssed in the previous chapter

',,,rrrt,times, the constraints may be rn the forrn of inequalities ln the constrained cost
rrrrrrrrnis.rtion problem above, the constraint- may bex ty -.2 Z, ie, the quantities produced of
llr lwo goods must be at least equal to the value oi'Z llolvever, in this chapter, we focus on
corrstraints In the context of crptimisat;on, the function to be optimised is called tlre
, rlrr,rlity

"lttrtive function The variables whose values have to be chosen are called the instruments
rrrrl (if the optimisation is constrained) the set of values of the ;nstruments which satisfy all the
,,'lr',traints is called the opportunity sel.

t tlrviously, a constrainecj optimum can never givL. a better value for the objective function than
llr unconstrained optimum lt will generally give a less favourable value tor the objective
lrrrrrtion and at best may leave the value of the objective fLrnction given by an ulrconstrained
 WITH EQUALITY
Mathematical Optimisation and Programming Techniques for Economic Analysis | ,U,
166 ISTNTIC OPTIMISATION: CONSTRAINED OPTIMISATION
I corusrnnturs tc <nu
optimumunchangedTolllustratetheidea,considertheprofitcurveinthefigurebelow.ln As stated earlier, this chapter will mostly be devoted to the main techniques that deal with
produced and the maximum profit wi
absenceof constralnts, the output level oM will be optimisation with equality constraints. These are constraints that hold exactly. This is

ttu aquivalent to using up all the available resources or capacity. ln reality however, this may not
be the case. The case of second best is one such example. lfthe optimum is not permissible, the
which does not permit output beyond a cer
Suppose now a limit is imposed on resources rccond best or constrained optimum may require significantly deviating from the first best or
leveltobeproduced.ThislsaconstraintSinceitconstrainstheuseofresourcesorproduct unconstrained optimum. Secondly, the indivisibility of commodities will make it impossible to
Thisisbecause
lfthelimitissetabovetheoptimal level'thelattermustremainunchanged due to consume certain amounts. ln particular, it restricts quantities to integers. lt rules out the
limit will be ineffective' However'
within the permitted region' ln this case' the consumption of fractions of commodities as may be dictated by the available resources. Dealing
the limit' output will increase'
constraints, which imposes production at
nature of equality wlth such a situation involves the technique of lnteger programming which is not covered in
the limit but this will cause profits to dwindle' thls book.

Figure 8.1. Constraint in one choice case 8,2 Optimisation of a function in two variables subject to a single constraint
Two methods have been developed for solving constrained optimisation problems. The first,
and perhaps the simplest, is the substitution method. ln this method, an equality constraint is
used to substitute one variable for the other in the objective function. This reduces the problem
to a single variable case. The second is the Lagrangean method. This method introduces a new
variable called the Lagrangean multiplier and combines the objective function and constraint(s)
lnto a single function. This method is more robust and can be used to deal with multiple
constraint cases. The method is discussed in section 8.3.

The standard constrained optimisation problem can be stated as below:

oDtlmLse
' .r
{x' \xt
x'1 I xz)

s.t. g(xyxr) -- m

We can see that we have three variables in the above problem, namely x1, x, and f (xr,x2).
Our task is to choose the independent variables (xr,x2) so that the objective function f(xr,x2)
then it precludes production at the optl
lf however the limit is below the optimal level' can be made as large as possible, as long as when we plug these values into the constraint
level. Output must adjust downwards so
that it is in conformity with the new limit' For instat
g(xtx), we obtain exactly m. m is a parameter which is taken as a constant in the
this point is permissible' The constrai
if the rimit is set at oM,, then no output beyond optimisation problem. ln essence, the constraint imposes restrictions on the domaln of the
output level will now fallto the new level OM' function. Hence, the solution to a constrained optimisation problem is the optimum value that
best among all possible levels includir the function takes on, over the restricted part of the domain that is consistent with the
Since the inltial level, unconstrained optimum, was the
in any way make profits better' ln fact' profi constraint. The graph form is presented in Figure 8.2 below.
the new level, deviating from the former cannot
coming as a result of a constraint' ln general'
will decline in line with a reduction in output
that an unconstrained optimum ln the case
constrained optimum is never better
maximisation, this can be shown as
 li

Mathematical Optirrisatior] und Programnring ['echniques lir Econonric Analysis L69 l

wlrH EQUALITY
168 I ,rot,a oprlMtsATtoN: C.NSTRAINED oPTlMlsArloN
I corusrnnturs lrrrrlrlem The problem is evaluated with respect to the single choice variable x.. Once the
l

,'trlrmal value of .r, is known, then the other variable can be evaluated using the equality
Figure 8.2: Constrained Optimisation , r rr rstraint.
l

llrt' first order condition of this problem is given by the following expression:

df dh
ir= f"+ f"' an

wlrere /r, is a partial derivative with respect to the independent variable xr, frris a partial
rlr,rivative with respect to .rz' and $
dxt
ir tf," effect of xron h. Therefore, the solution to the
lrroblernisgivenbythevalueofrtthatmakesthefirstorderconditionexactlyequalto0lfthe
v,rlue of the parameterm is unknown, the maximising value of xr will be a function of m. To
lndicate that the maximising value of x is a number, we write it as rr.. Therefore, the
rrraximising value of 11 is written as x1.(m) and it solves the following equation:

dh
f*'+ f" 0

^=
lo obtain the maximising value of 12, use xr* in the implicit function obtained above:

x2.(m) = h( xr.(m),m)
I hc two equations above determine the optimum values ofx, and xr.
ln the above figure, point A is the maximum'
it is the optimal point' But because of the
point is no longer
of the independent variables' this I xample 8.1
constraint, which exclude some values to
constraint' must be found This corresponds
available. Another point, which satisfies the Optimise the followinB function

point B in the figure Z: xy

Thefirstthingtodowhenusingthesubstitutionmethodistosolvetheconstraintforoneof Subject to the constraint

theindependentvariablesasafunctionoftheothervariableandtheparameterm.lnthe
is' r' and m' That
x+y =m
'x2 as a function of
particular case given above, express
We want to find the maximum value for f (x,y) over the domain of x, y that satisfy x *
xz = h(xr'm) ! = m,lf we solve for h(x,m), we have the following:

of our given problem in order to obtain the

following
The next is to substitute x, out Y:m-x=h(x,m)
optimisation Problem; Substituting for y, we obtain the following equivalent maximisation problem:

""\7i1f (*''n{"'*)) ^f$x(m - x)

x2)
Theparameter?nisaconStant.Assuch,theaboveoptimisationisnowinonechoicevariable
possible to translate a problem into a
makes it single variable -^ffi(xm-
x.. The substitution method Taking the first order conditions of this problem, we have the following:
problem'Themethodisquitesimpleasitessentiallyreducestheproblemtoanunconstrained
I7o I
I
sTnTIc oPTIMISATION: CONSTRAINED Mathematical Optimisation and Programing Techniques for Economic Analysis t71
I CoNSTRATNTS
6r--5=0
, =e 5

lhe second order derivative is 6, signifying that the extremum point is a minimum.

Solving for y, we have,

The Second order condition is -2, sig

To obtain the optimal value of y, we u =t-x.

5
1__
-
6
7

Example 8.2
llr,\ubstitution method discussed above seems quite easy, but it gets rather complicated with
Optimise the following:
tlrl increase of variables and also their powers, therefore, replacing the constraint into the
mLn
{x,v} rrlrlcctive function by use of the substitution method becomes difficult and complicated. An
,rllr,rnative and easier way to go about such problems is the use of the Lagrange multiplier
s.
rrrcl hod discussed in the next section.

Hence,
ll..l The Lagrangean Multiplier Method
llrc Logrongeon multiplier method inlroduces one more variable,2, into the problem. This
f (x,y)
v,rriable is known as the Lagrangean multiplier and has an important economic interpretation
s@,y)
which will be explained later. The method of Lagrange relies on maximising an associated
lunction, called the Lagrangean function. We form the Lagrangean function by adding,tr times
ln this case, we want to find the min llte constraint to the objective function and maximising over the independent variables which
satisfiesr+Y=1. rrow include the Lagrange multiplier.

lf we solve for h(x,m), we have the Suppose we wish to optimise (maximise or minimise) a function y: f (xbx) subject to a

v r:onstraintg(.x1, x) = 0. We form a Logrongeon expression:

Substituting for y, we obtain the follo 7= f (xr,x) _tg(xr,x), i +0

ffi(xz +s(x2 where i is called the Logrongeon Multiplier.
+ ^(S(3x2 - 5x - 77) i
It can be seen that I is a function of three variables xr,x2 and 2. We differentiate t partially
This reduces to unconstrained optimisation in a single variable x rather than constrained with respect to x:-x2 and tr separately and set each of the partial derivatives to zero to solve
optimisation in two variables x andy, Hence we can repeat the same procedure as in, for xr,x2 and 2. We will get
Example 8.1.
dL Af 0o
To obtain the first order condition, equate the partial derivatives to zero whereby we
dxr dxr dxt
have
-=l-')'=L=A
172 Mathematical Optimisation and Programming Techniques for Economic Analysis
I

STATTC OPTTMISATON: CONSTRAINED OPTIMISATION WITH EQUALITY 773

|
I coNSTRATNTS

AL df , 0a" _0
0x, 0x, 0x,
dL
7Qt'x) = 0
-= I I ,\tu, vlr, since thereis a constraint, we have now to form a bordered Hession determinant by

Solving the above equations simultaneously, we get the critical values of x, and x2.I'he
1,,,r ,1,,r ng the Hessian with the partial derivatives of the constraint function as follows:
values have to be examined to check whether the function has a maximum or a minimum dg 09
those values. This is done by applying the same second-order conditions for unconstrainsd 0
0x, 0x,
optimisation in two variables. 0g 02L 02L
lHl= 0x, Ax? 0x10x2
Take an instance of a producer using two inputs, labour and capital. Both factors mutt
0g a2L a2L
ordinarily exhibit diminishing marginal productivity. Forthis kind of production, the productlon
will have two choice variables, labour and capital. The optimising firm has to determine tha 0x, 0xr)x, A13

optimal levels of capital and labour. The objective of the firm is to maximise output but thlt llr,,rccond order conditions then is:
must be accomplished within the available resources. The amount of labour and capital used
cannot be increased indefinitely becausethese have to be bought usingfinite resources. rn"*;-r- ,n, ,
I r rr ,r O

Take the following illustration. lor ,r minimum lHl ( 0

Optimise y = f (xr,x) subject to g(xr, xz) = 0 I x,rmple 8.3

The Lagrangean expression is

Find the optimal values of

y= z : xy subjectto x +'y -
f(xr,x)-hg(xr,xr), 1+0
L=xy-7(x+y-6)
The first step in optimisation is to equate the first order partial derivatives to zero which is
equivalent to equating the first order total derivative to zero. Thus,
r4L \

AL AL AL l*='-^=ol
dt=0{-=x-)=01
drr- Arr- A1-
+dL=0
v
l'ri
[r='+Y=o)
I

Once the optimum point(s)s has (have) been established using the first order condition, the Solving,x =3,y=3,2=9
step that follows is to determine whether such a point is maximum or minimum. ln the Second Order Conditions
unconstrained optimisation presented in Chapter 7, it was easy to check the second order
a2L a2L azl a2L 0o 0a
condition and verify whether the point is maximum or minimum. The Hessian determinant was
0x2 "' 0x0y dydx dy' 0x '' 0x
used to simplify this. The Hessian determinant is denoted by llll.
r0 11r
ln-l:lr o 1l:2>o
lrlol
 I
Mathematical Optimisation and Programming Techniques for Economic Analysis
ISTATIC OPTIMISATION: CONSTRAINED OPTIMISATION WITH
L74 EQUALITY 175

I CONSTRATNTS
(,
^o9dg
The function has a maximum -0x,
og azL
dx,
a2L
llll = 0x, A"l lxrdx2
Example 8.4 0g a2L AzL
0x, 0x20xy Ar3
Evaluate the critical points for the following problem:
I rurn the objective function, the second order partial derivatives are
optimise 3xl - xrx2 t 4xj
subjectto 2xr+xr:21 02L 02L azL
13=6
ax? d*l oxrox,
We form the Lagrangean function -:8
Arrd the constraint will give the following first order
-=-1
partial derivatives needed for the
L: 3xl - xrx2-l 4xZ - 1(2\l x2-21), llorder:
Next, we take partial derivatives with respect to (w.r.t.) xr, x2 and 7 0a do
=":2,
dxr ." =7
dxz
AL
-i-:6x1 -x2-2) =0 (1)
oxt Ihe border Hessian matrix lHl will be:
AL
(2) 1027
Un= -*'*8x2-1=0 11 =12 6 _L
AL
lr -1 B

xr l- 2l = o (3) = -42
a|= -2*, -
The bordered Hessian Determinant (lHl - -42) is negative. This implies that the second
Next, solve equations (1) and (2) simultaneously, you could do it easily bythe elimination
order total differential is positive. Therefore, the function has a minimum at the critical
method, that is, eliminate one variable. We choose to eliminate the variable ,t so that w0
point (8.s,4).
can remain with two variables, x, and x, which are also found in equation (3). To
eliminate .tr, multiply equation (2) by two, this does not change the equation in any way, lxample 8.5
Next, subtract (2) from (1). This process eliminates 2 to give the following equation:
OptimiseZ: x*! subjectto x2+y2:7
Bxr-1,7x, = 0 (4)
Setting up the Lagrangean, we geu
Next solve (3) and (a) simultaneously
L= xty-7(x2+y2-l)
Zxr+xr=21
8x1-17xr=Q Following the steps in the previous example, we leave it up to the reader to solve the
steps,
we obtain AL
^ 1-ZAx=0
/r : 8'5 dx
AL
xz:4 - 'l -Dl:0
dy
Thus, the critical point is (8.5,4)
We getx = Y
To check whether the critical point is a maximum or minimum, we use the bordered t2x2 =1-
Hessian determinant introduced earlier lt is given by
L75 STATIC OPTIMISATION: CONSTRATNED OpTtMtSATtON WITH EeUALtTy Mrthematical Optimisation zrnd Progranrming Techniques for Economic Anrlysis |
I

177
i CONSTRAINTS

af.
=t
I clm

'jz b, ll m were to increase by one unit, how much higher would /- go? ln utility
lon, m represents the available resources and /- is the highest possible level of
criticar points
",.e (f ,.E) ."0 (-f f) Economics variabres generaly . This question is therefore equivalent to knowing the effect of additional income on

optlmal utility. lf an additional unit of resource (to spend on buying consumables) is made
admit negative values. As such, we only take the point falling in the
what would be the resulting change in optimal utility?
quadrant. The second order partial derivatives are as follows:

azL AzL A2L

tnswer to all these questions lies in knowing the value of the multiplier.tr* in the optimal
axz
= -2^' al,'= -zl'' axay =
o . Ivlore formally, m is a parameter fixed over the optimisation. All the optimal values of
x.,y. and.7* are dependent on m. They can be written as
And the first order partial derivatives from the constraints given by:
nt),y"(m) and 7r(m) to emphasise their dependence on the parameter m. Therefore, the
lL : z* = ,l-2, 9: ru = A rr,rur)rum value of the objective function is obtained by plugging the maximisers into it:
dx, dx,
f " (m) = f (". (m),y. (m))
The resultant bordered Hessian at the critical point is
I rl rrrli the derivative of the objective function with respect to m, we obtain the following:
lo ^lZ Al
lHl=lJZ -z) o l=si df' dx' t l) dv'
la o -ztl d,^ - J' d^- l" d^
Given the positive restriction on the lambda 2, we know even without getting the
rrr,r .rll that the first order conditions ofthe Lagrangean function are
value that the product is positive and that is all that matters. The function is maxi
the critical point. f)-1.si=o
Note that the Lagrangean multiplier method can be generalised to the case of optimi fi-t.s;,=o
function in n variables subject to m constraints, where m < n.This is the major weaknesi
the substitution method which cannot easily solve such problems. ',rl)stituting into the former equation, the derivative with respect to the constraint, turns to:

df' .,t .dx' , . dy'\

8.4 Economic Interpretation of the Lagrangean multiplier n
d^= \5, d.+ Sy dm)
Another reason for using the Lagrangean method is that is provides additional insights into
nature of economic problem. Also, it enables us to carry out sensitivity analysis; that is, it ',rrrce the constraint mu t hold with equality, that is

how the solution varies when we change the parameters of the problem. one can work out
s @"(m),Y. (m)) = m
comparative statics properties.
We can take the derivative ofthe constraint with respect to m and obtain the following:
The insights that the Lagrangean multiplier 2 provides is in terms of the interpretation of
Lagrangean multiplier 2. ln general, it represents the shadow values of the constraints. dx' dv'
example, in consumer theory, it is called the marginal utility of income. lt shows the amount
il*+9i*=t
which the maximum value of the objective function changes when the constraint changes
one unit:
ln the equation
fi, tn" t"r, in brackets must sum to a unit This is proved by the preceding

equation which differentiates the constraint This leads to a proof that:

1'I

L78 I

|
STATIC OPTIMISATION: CONSTRATNED opTtM|SAT|ON W|TH EeUALtTy
Mrthematir:al Optinrisation and Proqrarnnring Techniques f clr Economic Analysis
I coNSTRA|NTS

, r,l,'r rOnditions
!r::
dm ^. Ju og dg 0g
0 ___:1
''
Thus ,tr shows how the objectlve function changes with the constraint parameters, 0x, dr, dr= 0x,
utility maximising individual, the multiplier shows how much additional utility the t)pl A2L azL dzL a2L
gain if the income was increased by a unit. Though this multiplier comes out as a 0x, A*tr 0xrdx2 dxr1x3 " o xrox,,
actually varies. lf all other factors are held constant and only income ls allowed to 0g a2L a2L a2L, a2L
lHl = dr, 0x2dx, A"3 0x2dx3 " dx2ax,
marginal utility of income, measured by the multiplier 2, will vary. Usually it will increase
but start to decline so that the consumer reaches a point of satiation. At the point of 011 A2 L o2L a2L a2L
0*. dx3dxl 0x30x2 i)x! " 0x30x,,
2 is zero since any additional income does not result in increased satisfaction or en

ln production, the Lagrange multiplier 2 shows how output changes in response to 0g d2L a2l, a2L a2L
,.,
the amount of resources spent on production. Since the input ratio remains u 0xr 0xrrdx, dxn)x2 0xrr)x3 dx?,

increasing the available resources will scale up all factors of production by the same
l maximum, we require the last n - 1 principal minors to alternate in sign starting with a
l-abour, land and capital will all increase by the same proportion. This is referred to as
. That is, of the last n - 1 principal minors, the first must be positive, followed by a
up/down of production. The resulting increase in output is called return to scole. ln this
rvc and so on.
the Lagrange multiplier measures the return to scale. When it is equal to one, then
constant return to scale. When 2 is greater than one, the production function ,r rrrrnimum, we require the last n - 1 principal minors to be all negative.
increasing return to scale and decreasing returns when it is lessthan one.
l,r',tn lprincipal minorsare lHzl, lH:1, ,lH"l =lHl Thusforinstance,
We shall have more discussion on in later chapters
I
,tr

0
og og
8.5 Optimisation of a function in n-variables subject to a single constraint 0x, 0x.
dg azL dr;.
With n-choice variables and a single constraint, the process of finding the optimal lHzl : dx, 0*? 0 x.Ox,
remains the same.
0g a2L A2I,

Y : f (xr,x2,...,xn) dx, 0 xr) x, A13

s.t. g(xr,x2, ...,xr) = 0

The Lagrange equation is formed as

a)g dg dg
0
L = f (x1, x2,..., xn) - lg(x1, xr,..., xn) d*, or, 0r,
0g azL a'2L a2L
First order conditions
0x, 0*l dxrlx, 0xr0x=
lH:l = 0g a2L a2L a2L
dL:0
+--nAL oxz 0xr0x, Ar3 0 xr) x.
dxi 0g a2L a2L a2L
AL oxs dxr)x, 0x.0x, a*!
=-=O
il,
And so on until lHnl : lHl
180 |
STATIC OPTIMISATION: CONSTRAINED OPTIMISATION WITH EQUALITY Malhernatical Optinrisation nnd Programning'J'cchniques for Econorric Analysis 181

I coNSTRAtNTS
ogt og, dg,
8.6 Optimisation ofa function in n variables subiectto m constraints (rn < n) a\ Ar, oxn
dg, dgz 0 gr.
The most general case is of optimisation of a function in n variables subject to m constra 00 Ar, Or, oxn
with m < n. The formulation of this general case is as follows: :

Optimisey - f (xr,xr,...,xn) 000 a;^ a;^ a;-

0x, lxz oxn
Subject to
H: 09, 09, 0g^ a2L azL a2L

gt(xr,xz,...,rr) = 0
0x, 0x, dx, a7, a-rar, 0xrdxn
g
09,,09, 0g^ a2L azl a2L
!2(xyx2,...,xn): dx, 0x, 0*, Arfrr, A4 0x20xn
p*(xtxz,"',x) = o
09, 09, 0g^ a2L a2L a2L
The solution procedure is as follows: .)xn 1xn lrn A-nr" Arrt." dxl
Form the Lagrangean function Nrrlr, that the above matrix is obtained by bordering the Hessian matrix by the Jacobian matrix
ilI t lrc constraint functions.

L = f (x1 x2 ,^1 -
i7,s,1*, ,, ,xn) lor rmaximum,thelastn-mprincipal minorsoflHl mustalternateinsign,thesignof lH-*rl
l,,,ing (-1)-+1. For the minimum, the sufficient condition requires that the bordered principal
Note that this is a problem in n variables and m lambdas (,tr). There are thus m * n variables to rrrirrors all take the same sign, namely, that of (-1)m. Note that it makes an important
be solved for. rlrlfcrence whether we have an odd or even number of constraints. This is so because raising
( l) to an even power will yield the opposite sign as when raised to an odd power.
The first order conditions are:
tl.7 EconomicApplications
dL=0
tl7.l Utility Maximisation and Consumer Demand
AL
:-=0, i:L,...,tt suppose the utility function is given as U(x,y). The available resources to buy the two
oxt
r:onsumables is M so that the resulting budget constraint is Prr + Pyy -- M. What is the
o' i : r' ""'m
AL optimum level of utility or satisfaction?
a\=
The standard consumer problem is that he must maximise utility such that he spends all his
For the second order conditions, we consider the bordered Hessian determinant corresponding income M on purchasing two goods x,y, and the prices of both goods are market determined
to the matrix: and hence exogenous. lt is also assumed that the marginal utility functions are continuous and
positive, that is, UfuU, ) 0.

We form a Lagrangean function and get the first order condition, thus;

t = U(x,y)+ l(M - P,x- P"t)

The first order conditions are:
 Mathematical Optimisation and Programming Techniques for Economic Analysis
L82 183
I

|
STATTC OPTTMTSATTON: CONSTRAINED OPTIMISATION WITH EQTJALITY

I coNSTRAtNTS

L,:U, )D
=0
Cha pter 9
= U., - 7Pn
1., :0
Ls.:M _D _ PY=0 9 STATIC OPTIMISATION: CONSTRAINED OPTIMISATION WITH
INEQUALITY CONSTRAINTS
Solving for 2, we obtain

9,1 InequalityConstraints
lu:!r-^
Px Py ln Chapter 8, we considered optimisation with equality constraints. That is to say. The
Constraints were of the (=) type. ln this chapter, we consider optimisation with inequality
ln this case, we can interpret the Lagrangean multiplier as the marginal r,rtility of income
Constraints. lnequalities can be strong or weok. Strong inequalities are of the (>) or (<) type.
utility is maximlsed, thus:
Weak inequalities are of the (>) or (<) type. ln other words, a weak inequality is one that
dU" prrmits the equality case.
du'-n
ln economics, we mostly deal with optimisation problems that involve weak inequalities. ln this
To verify whether the utility function is maximised, we check the second order chapter, therefore, we look at solutionsto problems of this type. Also, the objective function
forming the bordered hessian. The bordered Hessian is set here below lnd the constraints are assumed to be concave and convex respectively and hence the
D
problems are known as concove-programming problems. ln Chapter 13, we discuss linear
,y
programming which is a special case of concave programming.
urn
un,
9,2 Binding and Non-binding (Slack) Constraints
The largest principal minor is 2xZ Therefore, llll = lHrl. We only have to solve for lH, ln chapter 2, we briefly explained the concept of a constraint as binding or slack. Here, we
arrive at a conclusion. amplify on this concept a little more. Formally, we make the following propositions:

lT2l = ZPnP,t) *, - Pr' Ur, - Pl U r* l. lf an inequality constraint holds with equolity at the optimal point, the constraint is
binding;
It is not clear yet whether lErl 0. However, assuming the law of diminishing marginal
ll. lf an inequality constraint holds as a stflct inequality at the optimal point (that is, does
one can infer that Uyy, Uxx <=0. All the prices are positive numbers but the sign of not hold with equality), the constraint is non-binding.
remains unclear, lf positive, then lH2l is unambiguously positive. lf negative, the sign of lll. lf a constraint is non-binding, the solution for the optimisation problem would be the
will depend on the sign of which term outweighs the other, the negative 2P.P*U*, or same as in the absence of that constraint. To put it in another way, if a binding
positive (-P,2 U, ,- Pj U *) constraint is changed, the optimal solution will also change. But if a non-binding
constraint is changed, the optimal solution is unaffected.
Take a simple illustration. Suppose a manufacturing firm produces steel products that use two
lnputs, labour and steel. There is a limited supply of labour and steel available during a given
production period. The firm's objective is to maximise profits. Suppose now that when the
optimal (profit maximising) outputs are produced, all the labour is used up, but there is some
unused quantity of steel, say two (2) tonnes. This unused quantity of the resource is called s/ock
volue.
 \
I

184 STATIC OPTIMISATION: coNSTRAINED oPTIMISATION WITH INEQUALITY Mrlhertlatical Optinrislrtion tnd Prograrnnting lcchniques lirr ltcorrolric Artrl_vsis 185
I
I CoNSTRATNTS
tlrl same as minimising the negative of the objective function That is,
ln the above example, labour is a binding constraint. lf an additional unit of Iabour was McLx f(x) subject to g(x) < a.
available, the optimum solution would change and the profit would increase. On the I | ,,,]me aS
hand, steel is a non-binding constraint. The availability of an additional unit of steel will
affect the optimal solution or raise profits. ln economic terminology, labour has a Min f (x) subject to 9(x) < a
shadow cost, where as steel has zero shadow cost. ln general, binding constraints have
') I l(arush-Kuhn-TuckerConditions
shadow cost and non-binding constraints have zero shadow cost.
, rl rder the following problem:
Let us use diagrams to illustrate the possibilities for a firm that wishes to produce
output subject to two input constraints. The problem is: Mux f (xr,x2)
subjcctto g(xr,xr) < b
Maximiseg = f(xr,x2) x.,x2 2 0
subject to x7 < a llrlfirst-ordernecessaryconditionsforsolvingtheaboveproblemsareknown aslheKorush-
x"<b 1\ttltu Tucker Conditions, more frequently referred to as the l(uhn-Tucker conditions This is
*r,,, a O 1,, (,ruse they were initially named after American and Canadian mathematicians Harold W.

I'rrlrn and Albert w rucker respectively, who published the conditions in 1951 lt was later
The nature ofthe constraints is shown in the diagrams below.
,lr',tovered that another American mathematician William Karush had stated the conditions in
Figure 9.1. Binding and Slack constraints lrtr master's thesis in 1939 Here we shall refer to them as KKT conditions The KKT conditions in
l,rrt constitute one of the most important results in non-linear programming

I.r the above stated problern, the KKT conditions are given after forming the Lagrangean
x2
lrnction:
x2
L: f(x.,x) - ),lg(xr,xr) - bl
I he Kl(T conditions are :

x', x".'
"'5u, olt-lso, d)1>tt
1 oYr
lnput x'r, x'2 is binding Both inputs are binding lnpul x'r,x', is slack 2 xr}0, x.r>0,7>0
lnput zi', xi is slack lnput x'r' , x'r' is binding I x,lf' Jxt
-- ,.!!_ i'\7 = trrL
,l)
u

Conditions 3 are called cornplementary slackness conditions What do they imply? Now
ln the solution method that we shall explain in the next section, we assume we are dealing
!: S(rr,rr) - b lf this constraint is binding, then its value is zero in which case 2 > 0 But if
concave-programming problems; that is, we assume the objective function is concave and the the constraint is not binding, then 2 - 0 The parameter 2 is thus the shadow cost we spoke
constraint functions are convex. The reason for this is that the necessary conditions for an about A non binding constraint entails zero shadow cost, since there is some slack. A binding
optimal solution also become sufficient conditions. constraint implies no slack and hence can have positive shadow cost

Most problems in economics involving constrained optimisation wlth inequality constraints are dL Jt - d,t
Again,=--;-)i-U*h"r"/andgandobjectiveandconstraintfunctionsrespectively
maximisation problems, and hence we shall consider maximisation problems. However, it is not
as though the solution method for a minimisation is different Maximising an objective function lf we assume ,1 = 0, the above flrst order condition reduces to !J. = !t - 0 This lmolies that
,'lrr drL
 Mathematical Optimisation and Programming Techniques for Economic Analysis
186 Lao-,a oPTlMlSArloN: coNSTRAINED 78I

I cousrnatNrs
l,sing equation (3), we get

the objective function is optimised freely, 3(rr-10)=6

r, as well. )x2=t0
The overall upshot of the KKT conditions is Ihus the solution is xr : 6, x, = lQ, ): f

t. An inequality constraint gi7) 3 b1 is binding at some feasible point I0 if it holds wlth The value of the objective function is Z = 4(0) + 3(10) : 30
equalityls/.,)=bi]andnotbindingifitholdsWithstrictinequality[rsi@)<bi].
value' ,l Sufficient Conditions
ll. Only binding constraints matter since only they impact on the optimal 'l
lll. lf it is known from the beginning which of the constraints are binding, it is permissible to llr| KKT conditions constitute only the necessary conditions. They will also be sufficient
ignore the non-binding constraints from the problems The KKT
problem will then r rrrrditions if:
reduce to a Lagrangean multiplier problem with only equality constraints'
one important distinction between the KKT conditions and the Lagrangean condition
for I The objective function is differentiable and concave in the non-negative orthant
lv.
constraints is that > 0 for the KKT conditions whlla (quadrant for 2-dimentional), that is, the region where each x; > 0;
optimisation with only equality 'tr
i + 0 forthe Lagrangeanconditions'
ll constraint function is differentiable and convex in the non-negative orthant;
Each

between binding constraints and non-binding constraints is of relevance only ln lll A point x6 satisfies the KKT conditions.
The distinction
problems with equality lrr Lxample 9.1 above, both the objective function and the constraint are linear. Since all Iinear
cases of optimisation with inequality constraints. ln optimisation
lrltctions are both concave and convex, we can assume that conditions (l) and (ll) are fulfilled.
constraints, all constraints are binding.
ln addition, the point (0, 10) satisfies the KKT conditions. Hence we can be sure this is a point of
Example 9.1 il r,rxtmum

Consider the following Problem (r.l; Optimisation with Mixed Constraints

maxZ=4xr*3x2
Not all optimisation problems involve only equality constraints or only inequality constraints
subject to 2x1 * x2 31'O
Ihere can be problems where some constraints are in the form of equations and some in form
xyx2 >- 0
of inequalities. These are problems with mixed constroints
The Lagrangean for the problem is L=4xr*3xr-)'(2xr*x2-10)' The KKT
lhe general formulation of a constrained optimisation problem with mixed constraints would
conditions for the maximum are:
lle:
AL
xr7O, xJ4-27)=O (1)
dx.
-=4-21<0, max /(r) : f (xuxz,... xp) subject to
AL
1<o' xz>o' x2(3-1)=0 (2) gtQ) 3 bt gz(x) < bz, g^(x) 3 b^
axr=3-
t*="tix,-7o>o'
AL (3)
h1(x) = 6r, h2(x) = sr, hn(x): sn
l(Zx'*xz-10)=0
1>o'
That is xr: 0 or )' - This is a problem where an objective function in k-variables hasto be maximised subject to m-
Now xr(4 - 2A) :0 implies that either of the two factors is zero'
3 1 3 0' Clearly' this is inequality constraints and n-equality constraints. ln the optimal solution for such a problem,
2. We then test this value of 2 from the second equation, that is -
either 12 = 0 or namelyr*, one could have the first m0 of the inequality constraints binding and the remaining
false. Hence, we must have xr:0' Then x2(3 -1): o implies that
lm - mol not binding.
3:,tr.Wealreadyconcludedfromconditions(1)thatxl:0andtoassumethatx'=Q
would imply that 10,tr : 0. But this is not consistent with tr 2 2 from equations
(1).
To solve this kind of a problem, the Lagrangean function to formulate would be:
Hence,rz>0 and,tr=3.
188
I

| STATIC OPTIMISATION: CONSTRAINED OPTIMISATION WITH INEQI.JALITY Mlthenraticll Optintisation and Progranrming Tcchniques for Econonric AnaJysis
I corusrRnrrurs
trr>0
L(xr, ,xt,,7t,
(1
(x) L),l.q,lxl + 1">0
(6)
(7)
,7,,,,1.r1, ,p,,) = f bi) -
Lli@j(x) - cj)
i=7 )=1 x,>o (8)
The optimal solution (x' ,1' , tt') would be one that satisfies the conditions. x">0 (e)

0y
, (x'.)'.1.r')=9, Vx,
t)x From (1), we have

)ilgiG') - bil: 0, i = 1,2, ,m l*71=2P1,

Based on (4), either.tr1 or
,1 must be zero. But given the condition from (6) that r.l > 0, it
h1Q') ci = 0, i = t,z' ,n means ,1 cannot be zero. Equation (1) would never balance. Therefore, we have
,;> o, v i:1,2, ,m h:0, tt > 0 andr, > 0
From (2), we have
giQ.) < bi, i: L,2, ,m
2x2(l+tt):tr2
Example 9 2 Since 1 * pr ) x, and 22, are either both zero or
0 , there are two possibilities. The two,
Consider the problem: both positive. But from (5) they cannot be concurrently positive. Hence x2 = ).2 = 0.

maxZ-xt-x3 Looking at (3), we now obtain x1 : 2. Then from (1), we get I :1. ln conclusion, the
subject to xl + xl : +, solution thus is
x, >0.
,r=o xt=2, xz=0, ,:;, tr=0, tz=o
The first step is to form the Lagrangean equation

L: xt-xl- fixl +xj 4)r 7.x, t Lrx,

The Lagrange equation has five variables The five corresponding first order conditions
are:

AL
, 1-2pr, +rr -0
oxt
AL
-2x, - Zux, * ), 0
ix2
dT
(xi txi-a) o (3)
-dll
AL
1r .. )txt 0 (4)
oAt
AL
),:'d)- ),x,0
 Mathcrrirtical Op(iirisation lnd Progrtlnnting Tcchniques lirr Econontic Arralysis 191

Chapter 10
l0 IiCONOMIC DYNAMICS IN CONTINUOUS TIME: DIFFERENTIAT
IiQUATIONS

ll, I lntroduction
llr llrowth rate of GDP has become a common statistic used in many spheres. This also applies
l,' llre growth in population and many other variables Though these growth statistics are
, ,rilltnonly mentioned, the actual values of GDP, population and many other economic variables
,rr I reldom mentioned. This chapter (and its successor) is devoted to use techniques of calculus
lrr rleduce, from the known growth rate, the behaviour of the variables with time. The key
rlrr0stions are: Given a differential equation, what is the time path of the variable? When the
lr,,l.rntaneous rate of change is known, can we know how the actual variable behaves over
Itnro? The answer to the questions lies in finding the time path of the variable of interest. This
l rnre path is therefore a function expressing the variable as a function of time and time only.

llr(,chapter proceeds by first looking at the differential equation. An attempt is made to

,tnswer the question,'What is a differential equation?'in a manner that can be easily
rrnderstood by an ordinary reader. The chapter will then proceed to consider differential
r'(luations of different orders and their application in economics, particularly in mathematical
r,conom tcs.

I0.2 What is a differential equation?

lo attempt to answer this question, it is always helpful to reconsider what has been learnt so
l,tr or in the earlier chapters. Chapter 5 of this book dealt with Differential Calculus and the
roncept of differentiation must be clear at this stage. With this assumption at hand, the simple
rlefinition of a differential equation is that; 'this is an equation in which a derivative appears'. A
point of emphasis though is that since the main concern is the behaviour of a variable with
time, the derivative ought to be with respect to time. lt must be the change in a particular
variable for a one unit change in time. The unit of measurement of time must not arise, for it
matters not. A differential equation is therefore an equation in which a derivative,9, appears.

The power on the derivative determines the order of the differential equation. lf the derivative
appears only in the first degree, the equation is said to be of first order. When the derivative
appears with a second degree, the differential equation is of second order. With a higher
degree on the derivative, higher order differential equations set in with increasing complexity.
L92 I ECONOMIC DYNAMICS lN CONTINUOUS TIME: DIFFERENTIAL EQUATIONS Mathematical Optimisation and Programming Techniques for Economic Analysis 193

The chapter will however concentrate on the first order case from which inference will be two functions are categorised into three categories; zero, constant and varying with time. This
for the latter cases. hssens the complexity of working with differential equation and allows them to be tackled in a
tradual manner. lt enables us to start with a much simpler form of the differential equation,
10.3 First-order linear differential equations
from which the solution can be derived with ease. As the function assumes more complicated
The title of this section brings out two points. The first is that this is a first order lorms, we often have to rely on its simpler forms to understand the solution.
equation. This comes out quite clearly in the heading and implies that the derivative appe
10.3.1 Constant Coefficient and Constant Term
the first degree. Second, there is mention of linearity. The word linearity must be u
its ordinary meaning, which refers to /ine function. This is a function represented by a We start the exposition with an assumption that the two functions z(t) and a(t) are
line. To bring this to context, there is need to look at what is behind a linear function. constants. The differential equation is written as.
general discussion on functions, refer to Chapter 3.
!+or=u
dt
The function f (x) = x'is called quadratic function while a function of the form /(x) =
saldto be linear. So where does the difference lie? One may argue that the unit degree This is a first order linear differential equation. The derivative is of first degree and the
parameters are time invariant. The particular solution to such an equation will depend on the
independent variable defines a linear function. We don't challenge such an answer but
actual values of a and b. As usual, we set out with a simpler case.
worry is that it does not offer any help to solving the problem at hand, linear
equation. The second and perhaps more candid definition of a linear function is that it 10.3.L.1- The Homogeneous Case
function with a constant gradient, or precisely whose gradient is independent of A homogeneous equation is defined in various ways. One is the Euler's theorem which states
explanatory variable. The gradients of the two function given above and!* that for homogeneous functions of degree n, the sum of the products of each independent
"re9:2x
respectively. Clearly, the first derivative of the former is dependent on the explanatory varia variable and its partial derivative is equal to n times the value of the functions. lt however
ln the latter case however, the first derivative is constant and hence independent of suffices to just note that a homogeneous function is one which remains valid even after
explanatory variable. multiplying allthe variables by a constant factor. ln the case at hand, the variables arefiandy.
The differential equation can only be homogeneous if the constant b is zero so that the
ln short, this is referring to a differential equation whereffoccurs only in the first degree.
resulting equation is
equation has no product of the form !ff. Broadty, a first order linear differential equ
takes the form !+ou=o
dt
dv The particular solution for a homogeneous case is pretty simple. The rearrangement of the
fi+ u(t'1Y : ,111 equation, by dividing by y and multiplying by dt throughout, gives
wnere
fr occurs only once. The equation shows that the rate of growth of the variable y is I
only a function of time (t) but also its own
level. The functions u(t) and a(t) are -d.y = (-a)dt
v
functions of time and determine the role of time on the derivative. However, they need not Then applying integration on both sides of the equation will give
time dependent always since even a constant can be a function of some variable. For i ft f
l-dy:l(-a)dt
JY
a function may be given as f(x) = a even though the right side of the equation has no x. ln J
the same way, u(t) and ar(t) may actually be constants or in extremes zero but still be referred the left side of the equation is integrated with respect to y while the right side is integrated
to as functions of time. with respect to t. Techniques of integration are covered in chapter 6. This chapter assumes
familiarity with the technique. Therefore, it must proceed as follows.
The actual forms of the two functions are of critical importance here as they determine the
form of the differential equations To make the exposition simple, the possible values of the
Ii*: Ir-ov'
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| ECoNoMIC DYNAMICS lN CONTINUOUS TtME: DTFFERENTTAL EQUAT|ONS Mathematical Optimisation and Programming Techniques for Economic Analysis 195

lnY=-01 16zs A simpler route to the solution is to consider the complete equation as consisting of two parts;
y : e-at+c ll)e complementary function and the particular function. With y as the variable of interest, the
Y = gC t-at lwo parts are denoted by y, and /p respectively. The complementary function is the solution of
!(t) = Ae-"t lhe homogeneous version, referred to as the reduced equation. lts general solution is only
where / :
ec. Since C is an arbitrary constant, so is,4, its product. This is the general solution
rrnaltered to depict its new status, as a component of the complete equation by writing:
to the homogeneous differential equation The generality is based on the arbitrary constant
at
contained in the solution. lf any particular value is substituted for the arbitrary A, then the lc=Ae
solution is called a particular. Where / is not known, it can be definitised using the initial the particular on the other hand is defined as any particular solution ofthe complete equation.
condition. The initial condition requires that yt=o = y(0) where y(0) denotes the initial value Ihis definition is loose ended as it allows us to make assumption on the form of the complete
of y. This yields y(0) = ,4 which gives the definite solution as cquation. Of courseil is unwise to assume a complicated and indigestible complete equation,
Ihe choice of the form will not alter the solution since any details are taken care by the
y(t) = y(O)e-"t complementary part of the equation. To proceed, we assume the simplest form of the
The solution is now definite because it has ridden the arbitrary constant. The initial condition cquation, that y is constant. This selection has a bearing on the derivative and enables us to
which replaces the arbitrary constant is itself definite. Once a specific time point t is known, a find a non trivial solution. For a constant function y, its derivative with time I = 0.
definite value of y is calculated.
Using this condition, we find the solution to the particular equation by substituting
Example 10.1
fr = 0 into
the complete equation and solving the equation for y which we denote with a subscript p for a
particular equation. Thus
fr I 4y = y(O) = t, find the time path of /.
Given g. The time path can be found using two
routes The first is to derive like the general solution was derived. The second, if permissible, is b
to use the general solution by substitutin8 the specific values. We use the latter here. For this lpa
kind of differential equation, the general solution is given by
This is valid for a * 0. The solution to the complete equatlon involves summing the two
y(t) : y(O)e-at
components or parts. The sum is

By substituting the respective values, we obtain the time path as

Y(t) = e-at ! = !" * !, = 1s-at *b

o
10.3.1.2 The Non Homogeneous Cose
This is the solution to the complete equation. lt is however not definite but general because of
Homogeneity in the differential equatlon was defined based on the value of the constant.
the presence of an arbitrary constant /. The last step will hence involve definitising the solution
When the constant is zero, the equation is said to be homogeneous. Non homogeneous case is
using the initial condition. lt must be stressed that the presence of an arbitrary is not
the opposite and an equation is said to be non homogenous ifthe constant, b + 0
coincidental but a matter of procedure. Definitising must be the last step and no attempts
d!+ou=o should be made to eliminate the arbitrary before the full solution is derived. Once we arrive at
dt the final general solution, eliminating the arbitrary is pretty simple. Simply use the condition
The equation is non homogeneous as explained above. ln particular, the equation is referred to yt=o: y(0). This will yield a particular solution A = y(O)-9 which when substituted in the
asacompleteequotion withtheformerreferredtoasthe reducedequotion.ltcontainsall the general produces a given
solution definite solution by
components of a differential equation. Thought the interest lies in getting the solution for the tb1 b
latter, the former will prove useful. y(t): [y(0) -;1,-"'*A
this solution, like the particular equation solution above, requires that o + 0. The solution has
r5
With integration on both sides, the two emerginB arbitrary constants will sum to a single arbitrary constant
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| ECONOMIC DYNAMICS lN CONTINUOUS TIME: DIFFERENTIAL EQUATIONS 797

three parameters; y(0), a and b and varies with time t. Once a specific time point t is kn I llomooeneous Cose
definite value of the variable y(t) can be calculated. llnt order linear differential equation is homogeneous when the term a-r(t) equals zero. With
zero term, the differential equation simplifies to
Example 10.2 dy
*+u(t)Y=0
For the differential equation given by find the time path ,
#*rr:6;y(O):10 r,, llre equation is linear, the technique of separation of variables discussed in section l-0.5.1
Again, there is no need for now to go for derivation since a derived solution is at ,l,lrlrcable. Divide through by y and multiply by dt. This results in an equation with two
disposal. All that remains critical is to identify the two parameters as a:2 and b t rrrr',, both on one side. Take one term across the equal sign to form an equation
The initial condition is also given. Using the general solution given 1
rbtb -dv = - u(t)dt
v
.vtt) = [i,to) -';]" "' *'; I lr', l)rocedure is the same as that used in section 10.3.1 under a homogeneous equation case.
the specific solution or the time path for the problem at hand is given llr next step is to integrate both sides of the equation with respect to respective variables.
f1 f
rit) =
[ro -1]." *2 Jror=J-u(t)dt
=7e-zt +3 /\rl,)il1 refer to Chapter 6 on exactly how to integrate the above equation. The specific form of
rl/) is not given, hence its integral cannot be determined. For the left hand side, it is the
We have considered two scenarios based on the value of the constant in the complete rr,rlrrTal log technique
r
The homogeneous is based on zero constant while non homogeneous is based on the con lnY'-6 - | u(t),t
J
ln addition, both assumed a non zero coefficient o. ln the third scenario is a zero
taking antilog
scenario. With a zero coefficient, the differential equation reduces to

d! yt = Ae-l u(t)dt, where A = e-c

dt =u This gives an indefinite time path of y. Once the initial condition y(0) is known, the arbitrary
This can be solved without recourse to the complicated methods discussed earlier. lt is coefficient,4 is determined hence a definite time path.
ordinary derivative requiring an ordinary method.
Example 10.3
1.0.3.2 Variable Coefficient and Variable Term
eivenff + 3t2y : g,find the time path of y.
This section relaxes the assumption imposed in the preceding section that the coefficient a

term in the differential equation are constants. Of course that made the differential eq Multiply the equation by dt and then divide by y. Then take on term across the equal sign
simple, both to look at and solve. With that assumption gone, the differential equation reve to get the equation of the form
to the general form ! dv = _Et2 at
v
o\t) integrate the left and right side of the equation with respect to / and t respectively.
ff+ue)y: |!o,:-|zt'at
where the coefficient and term vary with time f. Like the preceding section, this section
ly' J
provides for a homogeneous and non-homogeneous differential equation. The only difference lny--t3+C
here is that the coefficient and term are function of time t. This however does not preclude Then apply antilog
them from being zero. We first deal with the homogeneous case before moving to a more
complex non-homogeneous case. lt=Ae-t', whereA:ec
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I
198 | ECONoMIC DYNAMICS lN CONTINUOUS TIME: DIFFERENTIAL EQUATIONS Mathematical Optimisation and Programming Techniques lbr Economic Analysis 199

0.3.2.2 N on-homogeneous Cose

= Br-" +l2
1

The linear differential equation is non-homogeneous when the term ar(f) is non-zero, t
when o(t) + 0. This is a more general form of a first order linear differential equation ll
llre constants A, K and C are arbitrary and so will B = (As-x + C). with this time path,
wc leave it to the reader to prove that
no restrictions on both the coefficient and term in the equation. The derivation of the
however is complex. A complete expose is given by Chiang and Wainwright (2005) dv
develop the solution using the method of exoct differentiol equdtions. ln this book, the *+ztl=t
will be provided in retrospective. We state the solution, without derivation, and provr
I ll.,l liconomic applications
correctness by proving its reverse process of differentiation. For a general first order I

IttllIrcntial equations play an important role in the understanding of how some economic
differential equation given by ff + u(t)y = o-r(f), the indefinite solution is given
v,rr r,rbles behave over time. A look at this kind of behaviour has up to now been overshadowed

lt = e- I u(odt (A + [ a@el "{ttat 4r) lry,r,,sumptions of equilibrium in the market. This assumption means the market is always in
Irlrrrlibrium and so does not put any stress on the price to change. ln reality however,
the arbitrary constant,4 makes the solution indefinite With information on the initial cond
,.rlrrrlibrium is more synonymous with the long run. ln the short run, some disequilibrium is
available, the constant can be definitised so that the solution is definite. Of course
differentiating the above equation is not easy To keep the exposition within reoch, we Irr,vitable There might be a period of low price followed by a period of high price, normally
trrlilicred by some shocks on the market. A shock is simply an unanticipated change (huge) in
,.rll)cr the demand or supply (or both) of a commodity. But how exactly does the price behave

"#:r.':"YoranexamPre L lil lhe short run? This question can be best answered by resorting to differential equations.

=, A,,\ume a simple market model with demand and supply given

Find the time path for the differential equation
ff + Zry by
t Qd=a-BP, (a,g>o)
To solve such a question, deriving the solution must be out of question now. The first r[
Qs = -y + 6P, (f,6 > 0)
is to identify the two functions u(r) and w(t) which after necessary integration !
liivcn this system of equation, the equilibrium price P can be solved which will lead to
substituted into the general solution formula. This is by far the easiest route. ln fl lquilibrium quantities traded. We leave it to the reader to show that equilibrium price is given
problem at hil l,y
o(t) -- t
u(t) =
'D-tr+Y
21
ff B+s
lu(t)dL=l2rdt=12+K
JJ Ior as long as P(t) = P the market clears and no pressure is exerted on the price. lf price
the solution then is given by it would be naive to think that it will simply
deviates from its long run equilibrium however,
jump back to equilibrium. There should be some adjustment that must take place over time
lt: until equilibrium is reached, if at all it can be reached.

lt: We know from theory that changes in prices are caused by the inequality of quantity demanded
and supplied at the initial price. ln particular, excess demand drives the price up. This allows
expressing the change in price (with respect to time) as a function of excess demand. This
should not limit the analysis to cases of excess demand only since an excess supply can be
regarded as a negative excess demand. The specific function can be expressed as follows.

dP
at-=i(Qo-Q'), i>0
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200 I ECONOMIC DYNAMICS lN CONTINUOUS TIME: DIFFERENTIAL EQUATIONS Mathematical Optimisation and Programming Techniques for Economic Analysis 207

whereTistheadjustmentcoefficient ltmeasurestheresponsivenessofpricetoadeficil,ll
clear from the above specification that price will only remain constant if and only if the
clears, qd = qs

Suppose now the price is disturbed by a shock such that P(0) + P, how will it behave
time? This requires solving the differential equation
dP P(r); case where P(0) > P
dt: i(Q" - Q')
= jl(a - !P)-v + 6P)
: jf(a + y) - (p + 6)Pl
:j(a+y)-j(p+6)P
when rearranged will t).;O a differential equation of the
P(t); case where P(0) < P
+ j(P + 6)P -- j(a + y)
dt
with P as the variable of interest, the differential equation corresponds to the general form

#* "n: b whose definite solution is given by l(t) : [l(O) -o;)"-"'+! . For the speclllc
problem at hand, identify the two parameters and substitute accordingly The time path h
given by
10.5 Non-linear differential equations of the first order and first degree
P(o =
[P(o) -;#le \P+6)t .# lhe previous section concentrated on linear differential equations. This subsection introduces
= lP(O) - Pfe-(P+6\t + P of the first order and first degree. These are equations for
rron-linear differential equation
The time path enables the examination of one important characteristic of time paths. Does tha which the variabley appears with an exponent other than unit. They are non-linear equations,
time path converge to its long run equilibrium P after a disturbance? Since both P(0) and P aru y appears with a
though still of first order because the differential is of first order, the variable
e (6+6)t g.in* the law o,
constants, the convergence (or divergence) will depend only on l){)wer. lt may be a quadratic power like y2. Alternatively, the variable may manifest in a
indices and limit theorem, it must be clear that lryperbolic form, written in a rather special form 1fl. Al these forms are non-linear and this
lim e (P+6)t = o, p,6 > o ,,r'ction provides the way out. Generally, non-linear differential equations of first order and first
which means the price always converges to the long run equilibrium. ln economics, such an rlcgree are of the form
equilibrium is said to be dynomicolly stoble since the variable has a tendency to revert back to
the equilibrium after any disturbance. lt requires the asymptotic vanishing of thr f (y,t) dy + s@,t) dt = 0 or
complementary function over time (t + 0) leaving only the particular integral. dv

The particular integral in this context is P, which is constantl5. With a constant remnant P in thc
o'= n1'tl
time path, the price is said to have a stotionory equilibrium, otherwise it would have been a llasically two methods are available for solving this kind of differential equations. Depending on
moving equilibrium. the exact form of the equation at hand, they may be separable variables or if not, conversion to
linear form is used to convert the non-linear equation to a linear differential equation and use
methods discussed under linear differential equations. We start with the former method.

36
With the same demand and supply schedules defined, there can only be one equilibrium price
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I ECONOMTC DYNAMTCS tN CONTTNUOUS TtME: DTFFERENTTAL EeUATtONS Mathematical Optimisation and Programing Techniques fbr Economic Analysis I
203

L0.5.1 Separable Variables final step is to make y the subject of formula. The
',nly one arbitrary constant C.37 The

The differential equation given above is of general form For some equations I lnrc path is
howevor,
function f (y,t) may only be a function of y Similarly, g(y,t) may also be
a function of I r 1
lf this holds simultaneousry, the differentiar equation appears in a much simprer It=rza6
I
f(y) dy + eQ) dt = o ll the initial condition was known, it could be used to make definite the arbitrary constant
The two variables are separable since each term only has one variable. I Without the initial condition or any information on known /t=a, the time path can only
lt is possible to tak6
on the other side of the equation so that the left hand side only has lx, solved up to this far.
one variable and
for the right hand side. separating the above equation transforms it to the
llr ', .i li(luations Reducible to Linear Form: Bernoulli Equations
f (y) dy = -s(t) dt
The variables have been separated and a simpre integration techniques
l,'r '.r)rne non-linear differential equation, it may not be possible or practical to separate the
is sufficient to
vrrl,rlrk,s This ls common with Bernoulli Equations. Consider a non-linear differential equation
the desired results. proceed by integrating on both sides of the equation, the
reft hand sr(.rF
integrated with respect to y whire the right is integrated with respect to t.
The actuar
9q pv:7v^
dt
of integration and the necessary steps to get the time path wiil depend on the
exact formt functions of t and m is between zero and one, that is, 0 < m < 1. Such
wlrr,r r, /? and T are both
thefunctions/(y)andg(t). Obviously,simple/(y) andg(t)will requiresimplestepsto l ron linear equation is called a Bernoulli Equation, named after its discoverer, the Swiss
at the time path while complex one will demand more complex procedures.
M,rtlr('matician, Astronomer and Theologian Jacob Bernoulli. He was one of the prominent
M,rtlromaticians of the Bernoulli family.
Example 10 5
l,'r llris family of equation, separation of variables is not tenable. Opportunely, they can be
! + yrt, = 0, find the time path
Given the differential equation of y. rlrlrrted to linear differential equations. The procedure is as follows. Divide both throughout by
with this kind of non-linear differentiar equation, the first step is to separate the 1,"' to get the equation
Ldv
so that the equation is expressed in the form
f(y)dy = _g(t)dr. This involves RYL-* -
simple algebra. Multiply the equation by dt and then divide by y2. After t^;* '
taking one term
across the equal sign, the equation is transformed to
llrlrr define a new variable Z as Z = y1-- Using the product rule of differentiation, the new
v.rrr,rble can be differentiated with respect to t as
dZ t rtZ dyt .^dy
1o" = (r
y"' arl= nil= - m)v-"'
dt
lntegrate both sides ofthe equation orr interest however is a transformed derivative equation
7 dZ_7dy
f1 !-md.t y^dt
JTon =
which can be substituted into the differential equation. Replace y with Z in the differential
The techniques of integration are discussed
the results without showing the requisit r,rlrration to get. |rff = nZ = f Multiply through by (l - m) to get
dZ
_i= at+G-m)RZ=(1 -m)T
v
both integrations will produce an arbitrary c
tlris is a linear first order differential equation in which Z has replaced /. lt has a coefficient
,r - (1 - m)R and a constant b -- (1 - m)I Techniques applicable to linear differential

Note that it does not matter the sign that an arbitrary C takes Here it is assigned a negative sign strategically
l', r .ruse all the other terms in the equation have negative stgns
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Marhemaricar optimisation and
programming Techniques for Economic Analysis | ,ot
I ECONoMIC DYNAMICS lN CONTINUOUS TIME: DIFFERENTIAL EQUATIONS

equation, discussed in section 10.3, can be used to derive the time path Z. After solving for This is the time path of Z and is not the final solution since the question is asking for the

time path ol Z , a reverse substitution replaces Z with / to give the time path of y. timepathofy.Togettoy,theequationthatdefinedthevariableZisused'inareverse
substitution. The reversed equation is y =; and the time path of y is'
Example 10.6 L
Yr= Z,
Find the time path given by the differential equation
fl + U = 3ty2 1

For this kind of differential equation, separation of variables is not feasible lts BeZ't3
however provides for reduction to a linear difference equation using the method t 1-, \' 1

discussed. We note that for the sake of conforming to the formula, m = 2. Divide = (8e2, + 3)
by y2. The new equation will be
10.6 Economicapplications
ldv I
7fr*'i=t' I orrsider the solow growth model and its implications. The model is described below.
then define a new variabte Z =! and#= -;sinceg*=dfid] which inoties
llrt, Solow growth model is an advancement of the Domar growth model. The Domar
model

,r,,sumes that capital and labour are used in the same proportion. That is to say the capital-

l,rlrour ratio is constant. As such, output can be determined exclusively by one

factor say
r,rpital Labour may be said to be complementing capital in the same proportion' Solow
grow at
ltowever relaxes this assumption. The latter recognises that capital and labour may
rlilferent rates which alter the capital labour ratio. Then output cannot be exclusively

r[,termined by capital alone.

n\sume a linearly homogeneous production function given by

Q: f (K,L)
(luantity Q is a function of capital I( and labour L where Q is net output after ad.lusting for
(lcpreciation. While labour is assumed to grow at a fixed rate equivalent to population growth

r,tte ! n, the change in capital is dependent on the presumably fixed savings rate. lf
a fixed
=
:
t)roportion of output is saved (and invested) then I? sQ
Since the production function is

linearly homogeneous, by scaling all factors by

ll7, tntt changes the above equation to
Q = Lf (:,1.) = Lf (k)
the
output is now expressed as a function of the capital labour ratio k. Given this expression,
directly determines how
tnterest is to know how k, the capital labour ratio grows, since this
zr" a")
I output grows. Since k is a quotient of capital and labour, it can be differentiated,
with respect

z"-lt'** + c) to time, using the quotient rrlet'. k :

f
t-2
e2'+3
whereB=(A+C)e-K "' Refer to chapter 5
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206 I ECoNoMtc DYNAMtcs tN coNTtNUous TlME: DTFFERENTTAL EeuATtoNS Mathematical Optimisation and Programing Techniques for Economic Analysis 207

. / r-lk' KI- Kl.

*(=z) t:
The new differential equation is linear. Since both the population growth rate and savings rates
ere constant, then we have a constant coefficient a = (l - a)n and a constant term
KL x- b = (1 - a)s. We proceed to get the time path of Z using techniques applicable to linear
--LL
dlfferential equations. We will state the solution without deriving it. For the full procedure of
sl.f (k)
t. - nk, Ll (k)
since Q deriving the general solution, we refer the reader to section 10.3.1.

dk Glven the two parameters a= (l- a)n and b: (l- a)s, the general solution ofZ is given
*--st'&)-nk by
s s
At this stage, the specific form of the function /(k) is unknown since the parent prod Zft) = lzrc\- l r-{'-')" *n
t" nJ
function f(K,L) was now specified. lf it is now assumed that the parent function takes
with the time path of Z known and how it relates to the variable of interest k, the time path of
Cobb-Douglas production function of the form
k is found by Z : k1-d e tc: Z*
Q = KdLl-d
1

Since the function is linearly homogeneous, it can be expressed as follows Q = Lka, t : ([t1o;'-" _s_,]e-rt-":"t *;)--
comparison with the earlier function leads to the conclusion that the specific
f(k) = k". Then the differentlal equation above is rewritten, taking the negative term to Since both (1 - a) and the population growth ?? are positive, the exponential expression will
hand side of the equation, as tend to zero as f J -. This means the capital labour ratio will ultimately converge to a stead
state
dk
**nk=sk" 1

r.: e'\i*
\n/
This is a Bernoulli equation discussed in the preceding section with R f : s. lt is
= n and n
linear but reducible to a linear form. This enables us determine the time path of k and It will increase with an increase in the savings rate and fall with an increase in the growth of
the time path of per copito income y (: i) = f fO. population. These conclusions follow quite naturally. With an increased savings rate, there
would be an increase in the build up of capital which, ceteris poribus, leads to an increase in the
To reduce the differential equation to a linear form, we follow the capital labour ratio. Population increase causes capital spread which reduces the concentration
discussed in the preceding section. divide the equation by ka to get. of capital per labour. So an increase in the population growth rate will cause a fall in the stead
state capital labour ratio.
t dk
W at+
nkr '- -s

10.7 Higher-order differential equations

Define a new variable Z as Z : kl d. Differentiating with time t, we get
The first order differential equations may not always be robust enough to explain the behaviour
_
dz tdk of many economic variables. Such variables require higher order differential equation, which
- I t _ "'k"dL
^\ _
dt-" become broader the higher the order. By definition, higher order differential equations are
I dZ 1, clk
differential equations with higher differentials. For instance, a second order differential
(1,
- tt) dt kd (lt equation has a second order differential. The third order will have a third order differential and
Then substitute into the non linear differential equation in k The new equation in Z is this goes on to the nth order differential equation.

dZ d2v
o)nZ=(1-a)s --1 = kv
dt-+(l d.t2
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I ECONOMIC DYNAMICS lN CONTINUOUS TIME: DIFFERENTIAL EQUATIONS Miithematical Optimisation ancl Programming Techniques lbr Economic Analysis 209

Specifically, the order of a differential equation is based on the highest order of differentlal I .,rrnple 10.7

the equation. This needs to be clear because often reader will come across
yi' + yt'
For the differential equation - 2y = 10, find the particular solution yp.
equations with two or more differentials of different order. Therefore, the highest order ruL',,
The coefficient on y is non-zero This allows assuming y as a simple constant function. All
It is also permissible to have a differential equation constituted by a polynomial of differentlah, its derivatives will be zero. The equation then reduces to
For instance, an nth order differential equation will have a chain of lower order differentials. A
simple variety of a linear differential equation of order n is: -2Y = -to
A simple division yields the particular solution
d'tv dn tv d2v dv
+ o, +...... + an-z t an-16 = b
77 dt,l dp _10
this is a general form of an nth order linear differential equation. The exact form need not havl /p
the entire decreasingly ordered differentials. Their absence may alternatively be simply _:'
indicated by a zero coefficient
llrrt what if ar: 97 Well, the answer is simple. The first assumption of a constant time path
Consider a differential equation given by illtlst be revisited.
dv
Yt" I arYt' * arY = b,where Yi = When o2 = y. Again it would not
0, there is no option but to assume a non-constant value for
,1, rkr any good to assume a more complicated function of y. The closest look is to think of y as a
This is a second order differential equation. For its solution, there might be need to review how
Itrtear function of time t This will meet the condition of non-constant since it changes with
the first order case was dealt with. Since this is a non homogeneous case, the solution must bo
tirne lf we take the simple form of
divided into two; the particular and the complementary solution denoted by y, and y,
respectively. The time path is the summation of the two. The former is of less controversy so !=kt
we start with the latter. where /r is an arbitrary constant, then y' = k andy" = 0 bearing in mind that this form is

r oming because a2: O, Ihe new equation is ark = b, t k = Cir"n this expression for /t,
The particular solution becomes easy when dealt with in phases or stages. The stages ara f
defined by the assumption made on the nature of the variable y. The first and simplest is to lhen y, = L1 represents the second possible particular solution for yr. Recall that the
assume that the variable is constant, that is, /: a With this assumption, both the first and particular solution represents the long run equilibrium of a time path Since this specific
second order derivatives will be zero. The differential of a constant is zero and any subsequent particular solution is non-constant, but changes with time, it represents a moving equilibrium
differentiation will yield zeros This simplifies the equation (forthe particular solution) to scenario. This is typical of most prices in the economy Prices fluctuate around some
az!:b cquilibrium but the equilibrium price itself may have a tendency to increase (or in rare
circu mstances, reduce).
so that by making y the subject of formula
Such a solution however is dependent on a, being non-zero. When the converse is true, that is
lp_ .rr = 0, the solution collapses. The zero left hand side of the equation is equated to a non-zero
a2
The subscript p indicates that this is a solution for the particular part of the equation. This right hand side The latter has b + 0 while the former will have zero because of zero
solution is strictly based on assuming that the variable is constant. This assumption however coefficients and a zero second order differential. This leads to the third and perhaps the last
comes with its own restrictions lt restricts a2 from attaining zero, otherwise the solution scenario, when both parameters a, and a, are zero. The simple linear function will not suffice t1l1iltl

collapses. to give a particular solution under these circumstances. A more complex function, specifically
with a non-zero second derivative, is needed.

llilil
27O
I
Mathematical Optimisation and Programming Techniques for Economic Analysis ztL
I ECONOMIC DYNAMICS lN CONTINUOUS TIME: DIFFERENTIAL EQUATIONS

It is easy to generate the function, in indefinite form, from the differential equation. ln the actual algebra however, only summation is prominent because the dividing is swallowed
and a2 are zero, the differential equation reduces to a second order ln the arbitrary constant(s). lc : lt I lz

!"=b Thls general result may need to be modified depending on the actual characteristics realised.
the process of integration, reverse differentiation, will enable finding y from its
ircall that there are three possible outcomes from a characteristic equation, depending on the
Though each function has an infinite number of integrals, we retain the right to
arbitrary constant is zero. This will produce a unique definite integral, giving the
Vtlue of the discriminant D = at2 - Aar. There may be two distinct roots, two repeated roots
lndtwo complex conjugotes, We explore the three in succession.
solution. Using the simple power rule of integration, the particular solution is

yp :
b" 10.7.1 Case 1: Two distinct roots
it" Thls case occurs when the discriminant D : arz - 4a2 is strictly positive. The characteristic
This is a quadratic function. lt is easy to prove that this is the correct integral by
aquation produces two distinct roots, t and 12. The two roots are used separately, to derive
the function twice and see if the outcome gets to the initial derivative. Like in the earlier
the two possible complementary solutions, y, andy, respectively which are summed to give
this again is a moving average case. ln particular, the long run equilibrium explodes with
given the nature of a quadratic function. the complementary solution. lt takes the form

The discussion of the particular solution might be lengthy but still falls short of all lc=lr*lz
scenarios. The three scenarios given are not exhaustive but meant to be a guide to = Aterrt * Arerzt
The solution still contain to two arbitrary constant 41 and Ar. No need to be bothered with
insurmountable scenarios not covered. They give a broader idea of dealing with various
flnding the definite values of the two at this stage. The two are only part of the complementary
of the functions.
solution. Getting the definite values of the arbitrary constants must always be the last step,
The complimentary solution, as earlier stated, does not need many suppositions. lts rfter putting together the complementary and particular solution, dealt with earlier, together.
more straightforward. lt relies on assuming a homogeneous differential equation, one
b = 0. lt is the general solution ofthe reduced equation yt" * atyt' * ary : g. Example L0.8

Given the differential equation yt" * yt' - 2y = -t0, find the time path of y.
that in the first order differential equation, the complementary solution was ofthe
Recall
! = Aert. lf this is adopted as a trial solution, then the first and second order derivatives This equation is one and the same as Example 10.7 which solved for the particular
!' =rAe't and y" - r2Aert respeclively. When substituted into the reduced form, solution. lt still remains the same and we simply adopt it here as
differential equation becomes
lp=5
rzAe't + alrAe't I a2Aert = 0
Even though the example asks for the general solution, the emphasis is on the
complementary solution. The way to get the particular solution is explained in the said
Ae't(rz+arr+az)=O example. The first step is to write the reduced form of the equation, one with a zero term.

ln the equation, there are two possible explanations. Either.4 : 0 or rz * arr * a, = g. yt"*yt'-2y=o
first case is not plausible since A = /r=0. The arbitrary constant,4 is the initial condition wh From the first order differential equation, the general solution is of the lorm y = Lsrt
cannot be assumed to be zero. This leads us to be certain that the second case holds. That is: which we adopt here. Then the first and second derivatives are

12+orr*or=g !' = rAe't

This is a characteristic equation ofthe reduced equation. lt is a quadratic equation in r and Y" = 72n"rt
solutions are called characteristic roots. As usual, such an equation will produce two root3, substitute into the reduced equation and factorise to generate
and 12 which will give rise to two complementary solutions lt = AGrLt and y2: A2e
rzAe't+rAe't-ZAe't=o
respectively. To get to a unique complementary solutlon, simply combine the two by aver
Ae't(rz+r-2)=0
 I
Mathematical Optimisation and Programming Techniques tbr Econtnric Analysis 273
272 I ECONoMIC DYNAMICS lN CONTINUOUS TIME: DIFFERENTIAL EQUATIONS

The first factor on the left hand side cannot be zero. Since the arbitrary We do not need to go into how to solve this pair of equations. We instead assume the
associated with the initial condition and the exponential has no zero in the ranqe. So
,ruthor is already conversant with simultaneous equation. So we just state the answers as

second must be. This gives the anticipated characteristic equation Ar=4
a
Az=3
-2t- -i so that the definite solution to the differential equation is
We do not have to labour to explain how to get the two roots. We can safely statc
and refer the reader to other texts on quadratic equations. The roots are 11 : 7, ti t lt:4et+3e-2t+5
and the two solutions are lll'/ 2 Case 2: Repeated Real Roots
,' Wlron the discriminant in the characteristic equation equals zero, the equation produces two
-Aot
lz: Aze-2t rr,;rr,ated real roots. ln other words, the two roots 11 and r2are equal. This means there is only
The complementary solution is a sum of the two rrrr| root / without a numerical subscript since this is meant to identify two different numbers.
( rlvcn the two roots are equal, there is no need to identify them.
Yc-rlt12
= Aft I Aze-2t Wilh a single root, the general form ofthe solution is slightly different from one presented in
This is the complementary solution of the differential equation Since the pa lltr two distinct root scenario. lf we were to continue with this form, the solution would
solution is available, the general solution of the complete differential equation is fourrrl r ollapse to a single constant. That is, in the expression
summing the two.
lr: At€tt * Are't : Ase't, 43 = A, * 42
lt=!,*lp with only one arbitrary constant. This is not sufficient for the second order differential
= Aft * Are-zt + 5 r,quation. As will be notice even in the next chapter on difference equation, the general
only after putting the two components together should definitising come. Emphasis m rolution in the case of two repeated roots is given by
be put that definitising is the last step in solving differential equations. ln this part
case, no initial information is provided Moreover, the solution has two unknowns lc:AGrt+Aztert
would require that two time points are known lhis does not collapse to a single constant. The rest ofthe procedure remains the same as in the
previous case. When the root r is known and two initial conditions, then a definite time path is
ln the interest of having a complete solutio
calculated by solving the resulting simultaneous equations.
known. These are Yo = 72; Yo' = -2. The i
its first derivative. Using this information, Example 10.9
simultaneously. The first equation is substi
Solve yl' + 6y't + 9!t = 27 for the time path.
equating to 12. The second equation is to s

indefinite solution and equating to -2. For the particular solution, since the coefficient on yt is non zero, it is permissible to
assume a simple constant for the function. this will have all the derivatives equal to zero.
lt = Ate
The equation will reduce to
!t=o = At
12=At 9!t:27
Ar+Ar:7 This gives a particular solution

!t' : AP' 27
_A
tt=O - ttl
Yr:i =3
-2: At
A7 - 2A2
274 | ,ao*or,a DYNAMtCS tN coNTtNUOUS TtME: DTFFERENTTAL EQUATTONS M{thematical Optirnisation and Programming Techniques for Economic Analysis | ,rt

The complementary solution is found by taking a reduced

form of the y't = -3Ate-3t - 3A2te-3t * A2e'3t
equation, the homogeneous one' This is given by Y't' + 6YI ]-9!t :0. Then Yl=a: -3A1 + A2 I

y : Ae't .The first and second order derivates are -s=-3(2)+Az

Az: I
!' = rAe't
Now the constants are definite and the definite general solution is
Y" =rzAe't
substitute the three into the reduced equation to
get an equation in r and then
!t:2e-3t+te-3t+3
the common,Ae't. l0 / :l Case 3: 'fwo complex conjugates
r2Ae't +6rAe't +9Ae't =o Wlrt'n the discriminant in the characteristic equation O is negative, the characteristic
Ae't(r2+6r+9)=0 /(r):
zero' of course it
r,r1il,rtion hasno real valued roots. The solution will require getting the square root of a
Again, we must determine which factor(s) makes the equation rr,li.rtive number, which is known to have no solution in the real number system- This case is
be the arbitrary constant,4 because this must be allowed to be arbitrary' The e)
,lr',r ussed after a discussion of Complex numbers.
alsoneverbecomeszero.Therangeofanexponentialfunctionisstrictlypositive.
root'
eliminotion method leaves one factor, called the characteristic ll).ll ComplexNumbers
12+6r+9=o Wr, have stated that when ar2 - 4a, 10 in the characteristic equation, the solution will lie
Since quadratic equations are no longer an issue at
this stage' it must suffice to state orrtside the real number system The problem is that the solution requires us to evaluate the
to show all the necessary steps' The reason we dodge the wo
root(s) without having ',(tuare root of a negative. To do this, a special number is defined. Let i = r/J. This is called an
is already wellvested on
is not that it is irrelevant but is on assuming that the reader tntoginory number. This number allows the evaluation of a square root of any negative number
This characteristic equation has the root ,,rrrce the negative can be separated using the law of indices. For example, ,,1= = A,,l= =

'l
li, ,14 : ,14J= = 3i. A complex number is a number that can be written as a sum of a
*J'** ",'"menta ution r,,rl part and an imaginary part.

:'ilH:::jJffJli"''
su bstitute' ry sor
"',"' "1
lrr general, a complex number will take the form (h + ui) where ft and u are two real numbers.
lr=AfrtlA2tett llrey can be negative or positive or even zero. This means any real number is also a complex
I

lc: Ate-3t * A2te-3t rrrrmber with (u = Ol. ln the same way, all imaginary numbers can be written as complex
step is to put the two together by
Since we already found the particular solution, the next nrrmbers with (h = 0). Thus sets of all real numbers and all imaginary numbers are both subsets
summation. Recall that the general solution is a sum of the two separate solutions' ol the set of all complex numbers. Examples of complex numbers include 5 +2i,8- i and so
(,n
lt=!"1lp
= Af-3t + A2te-3t + 3 ( omplex numbers can also be represented by an Argond diogrom or a complex plone. lt is
Thisisanindefinitegeneralsolution.ltisindefinitebecauseitcontainsindefinlt. r,rlled Argand after a nineteenth century French mathematician Jean-Robert Argand. This is
This information allowl
constants. suppose it is known that yr=o = 5 and yl-o = -5? ,,imilar to the Euclideon p/one used for functions. The Argand diagram is a plane measuring the
be made definite'
getting a definite solution since the two indefinite constants can now real part of a complex number in the horizontal axis and the imaginary part in the vertical axis.
generating two equation simultaneous equations using the two
The procedure involves
initial conditions.
!t=o= Atl3
5=Atl-3
Ar=2
 Mathematical Optimisrtion and Programming Techniques for Economic Analysis | ,r-,
EQUATIoNS
zt6 | ,ao*or'a DYNAMIcs lN coNTlNUous rlME: DIFFERENTIAL
llr' ',olution to the complementary function remains lc = Afrlt I Arerzt substituting the
Figure 10.2. Argand diagram lwil rlistinct roots

yc = A1e(h+vi)t * Are(h-vt\t

c(h,v) : eht(AG'it * A-re-'it)

llrr. solution has imaginary exponential functions. But how do we interpret imaginary
oxqronential functions? To aid understanding, we take a look at two key concepts in
M,rllrcmatics. The first is the Maclaurin series and the second is the Trigonometric functions.
llr| Maclaurin series is similar to the Taylor series. The two differ only on the point around
wlr( h the function is expanded. While the Taylor series expands the function around any point
r,, c in the domain, the Maclaurin series locks the point of expansion to the origin. That is
r,, 0. According to the Maclaurin series, a differentiable function can be expanded around
that R2 =h2+v2' where R is the modulus ol
Using Pythagoras theorem, we know rr,ro as follows:

complex number (h + 7i)'

Recall that the characteristic equation was

given by /(t)= /(0)
or
+f'(0)x f"(0)x2 f"'(o)r' f""{olro
1r - z - 3! + +r +
T2+arr*ar:g /(0) f'(0)x f2(0)x2 f3(0)x3 fo(o)xo
I\r)= ol - 1! - 2r - 3! - 4!
Its solution is

-o,t,[i71+6 ll then f(v) : svt

2 llrcn/(0): l,
This part can be rewritten I
When a1z - 4az .-0, the equation has no real valued roots'
the part in parentheses
-L(4ar- a:) <O-Though the left hand side still remains negative' f'(v) = ie"', f'(o) = i
square root'
now positive and we can proceed to find its
f"(u): i2e"i, f '(o) -- i'
-'o'+ '[=T@i'-i'\
2
f"'(u) = i'""', f'(o) -- i3

l-rJ(4a2- of) f""(v): i+""i, f'(o) = ia

-a +
f""'(u) = iseui, f'(o) = is
-o !i,[@a=o sincei =r/J,then i2:-1, i3:-i, ia =1, is =i, i6 =-7
50 that the series for the function can now be written as

_-o, *,,[G"=O
'2
--9: xn6 u @,gr.,
f (v) : eui =
+,.'+. +.
+.";+ S. #.
= _ ! +r:_y:_4*!1*r:l_t*
Now ret h =
r= hl ui "u, 0!|2l3!4!5!6l

I
 Mathematical Optimisation and Programrning Techniques for Economic Analysis 219
DYNAM.cs lN coNTlNUous
rlME: DIFFERENTIAL EQUATIoNS
278 | ,ao*o*'a
(cos 0) (sin 0)z (cos 0)u2 (sin 0)u3 (cos 0)ua
functions; the sine and cosine' + + 4l +.-'
Next we expand in the same
way the two trigonometric Ilu) = cosv = 0t - 1t - 2t 3!
plotted in Figure 1O'3 below
graphs of the two functions are
v2
functions =1-9-
0! \t
*o
2l 3t 4l -o5!-u"
*Y4
6l
*..,
Figure 10.3: Sine and Cosine

.or, : 10!- o -u'2! + o +{4t - o -ui6l + ...

Similarly, the sine u function can be expanded using the same procedure.

(sine 0) (cos 0)u (sine0)u2 (cos 0)u3 (sine 0) ua

ine ,i\ii=sinr=-OI- 1l - Zl -- 3! - 4!
-5
-cos :#.; .*-";-*.*.*-
-0.5

sinz=o*i-o ***o+$+
-t Horizonta axis measu'ed in
(^=3 142)

characteristics of the two

functions First' the functions As we delve into this algebra, it is important to restrain ourselves from being carried away. The
The graphs show two important That is' each function h objective should not be forgotten. lt was to find a way of dealing with the imaginary number. So
th"rn'"tues for every 2n radians'
periodicol.Their values '"pttt of fluctuation' far, this hasn't made any appearance in the trigonometric functions To introduce it, multiply on
have constant omplitude
period of 22. Second, the functions : fi.
both sides of the expanded sine z function by r
g by I the two will coincide'
The two will differ in
phases' lf the cos is shifted rightward '
iv iu3 ivs
isinu = - * *...
(t *\) 1! 3! St
cosg = sin
Now adding the new equation to the expanded cosu. Note that the cosu is zero for all odd
Given f(0) = sin0 the
Since the two are smooth
functions' they are differentiable' number powers of u while sine u is zero only for even number power of u. We can therefore
follows' rearrange the sum so that there is ascending power of u.
order derivatives3s are as

=-sinl' f"'=-cosg' f"tt - sine

f'(0)=cos|, J" cos / + , ,in, = I0!|2t3!4t5!6l
+'Y. -!: -'!: *t *'! -'! * .

0'the derivatives go as follows:

Similarly, given the function 9(g) = cos

g"'(0) = sinl g""(0) = cos9 ' The emerging expression is identical to the expression obtained by expanding evt. We obtain
g'(0) = -sin| , g"(0) =- cos0 ' ' the following Euler Relotion also called Euler's formula.
_ _sin0
5,,,,,16y
is equal to the function itself
and the process e'' = coslt + isin/
ln both functions, the forth derivative of the cosine function
We can then write the expansion e-vl : cosv - isinv
every after four derivatives'
cos(0) = 1 and sin(0) = 0
u : O. from Figure 10'3 above'

I\h" only if 0 is measured in radian

derivative results are valid
 Mathematical Optimisation and Programrning Techniques ftrr Economic Analysis | ,r,
2zo I rcouovttc DYNAMIcs lN coNTlNUous TIME: DIFFERENTIAL EQUATIoNS

elvi: cos.u * rsinu

10 8 1 Alternative Representation of Complex Numbers

Complexnumberscanalsoberepresentedusingpolarcoordinates.LetusUse the C

diagram presented in Figure 10.2 The figure is repeated with some details added.
h*vi:R(cos0*isinO)
=n("xoi), o<0<Ztt
Flgure 10 4. A detailed Argand diagram
I r,rrnple 10.10

Find the Cartesian form of the complex number 4e?'

The modulus R:4 and the angle I = 4. rhe Argand diagram will take the following
sh a pe.

The cartesian coordinates of the complex number are still given by (h,u) which defines
point

The modulus or absolute length is given by R = JET;V This is derived using the now fa
pythor;oros theorem dealing with a right angled triangle. Given the angle between the
number representation and the horizontal axis, it is possible to form some equations
trigonometry ln particular, /r and u can be expressed as functions of the angle and llto
Though the Argand diagram may not be needed, it is important to sketch it in order to
have insights on the expected values ln the above case, the given angle means the point
- .1" ;"rr,", R manifests as thehypotenuse, h is the odiocentwhile u is tt opporite. I
" fall in the second quadron, ,: a, < n). As such, we already know that h must be
generate the following equationt. t negative while v will be positive. We proceed as follows:

sina = u= sin0
f,-,
R
h = Rcos e :
3' - 4eo.s) -2
+coszl :
h
cos0--, h:Rcos0
2n a\7)
/\,5\ _
u : Rsino = 4sin ztt
by .5 3-
Thus, the complex number urill be given

ht ar =R cosd * i fi sind il The Cartesian form is h + ui : -2 + 2\Ei

: R(cos0 +,sin8, I0 8 2 de Moivre's Theorem
I The theorem is named after a French mathematician Abraham de Moivre. lt states that:
Where R and I are called polar coordinates of the complex number as opposed to h andL
which are Cartesian coordinates. Through Euler relations, the cornplex number can be furtfl (cos0 * isin d)n : cosn0 * i sinnd
translated into the exponential frorn. The Euler relations state that t By extension, the complex numLrer h * zi raised to power n will give:
222
I

| ECONOMIC DYNAMTCS tN CONTTNUOUS TtME: DTFFERENTTAL EeUATtONS Mathematical Optimisation and Programming Techniques for Economic Analysis 273

(h + vi)" : [R(cos d + i sin0)]u For the particular solution, simply assume the function is constant. Given this assumption,

: * rll the derivatives will be zero. The remnant of the differential equation is:
R" (cos d r sin 0)"
4Y =z
= R"(cosn9 * r sin nd) Thls leads to a particular solution

That is, to raise a complex number to the nth power, raise R to the nth power and multiply t
lpZ
angle d by n.
FOr the complementary solution, we must assume a homogeneous differential equation.
Now we return to the analysis of the complex root case that was suspended earlier
Further, assume the time path is of the form ! = Ae't' Given this form, the first and
question posed earlier regarding how to interpret the imaginary exponential function can
second order derivatives are y' = rAert and y" = 12Ae't respectively Then substitute
loosely answered The way out is to replace it with trigonometric functions in which tlu
lnto the assumed homogeneous equation.
imaginary number is stand-alone we now proceed solving for the complementary function
rzAe't +rAe't +4Ae't =o
yr'= eht(Are"ir + A2e-'it) This translates to the characteristic equation

y, = eht(Al(cos u * i sinu)t * A2(cosv - i sinu)r) 7'2+r+4=o

: ert(Ir(cos r:t + j sjn vt) + Ar(cosvt - j sin ut))+o This equation has the roots

= eht((Art A2)cosvt + l(,41 - Ar)sinut) -r+,[t2-4(4)

: eht (As cos ut + A, si n ut)
This is the complementary function. As shown in Figure 10.3, the sine and cosine functaonr
have a constant amplitude and period. we expect that the time path will have a smooth
oscillation around the around (constant or moving depending on the particular solution). Tho
number of complete cycles in a unit of time (also known as the frequency) is dependent on thrr
u. ln particular, the frequency f is given by / = 1. h: -1
2 u= ,TS
2
The complementary solution is
Since the term in brackets resutts in an oscillation of fixed amplitude (non converging and non
diverging) with a constant period, the question of convergence is only dependent on the ,"
exponential eht ln particular, h determines whether the time path converges or not. But h
derives from h = :91 Thus the coefficient of the first derivative in the differential equation
=""::il::",4, .,- +,r
]" 2l
determines the convergence of the time path. That is, when a, ) 0 so that h < 0, the time To get the general solution, add the two components.
path converges. lt diverges when the opposite is true.
lt=lc*lp
Example L0 11 -t,/ fis JIE\
yt=e7'(Ascos t+A4sin t)+,
1

Find the time path of the differential equation yr,, * yt, + 4y = 2 , Z

This is the general solution. lt is not definite because of the two arbitrary constants.
Suppose it is now known that y(q = 5: and y'(0) = 5, then what the definite time
a0Givenc'it:(cosu*isinu)r path? Using the two known initial conditions, we create two equations that are
lfweleta=vt,rheneai=cosa+isina=cosut+isinztwhichimplies
thate'i': cosut+ asinut ln thesameway,e ,it = cosut - isinyt simultaneously solved to get definite values of ,43 and Aa.
 I

224 | ,ao*or,a D'NAMrcs rN coNTrNUous.,ME: DTFFERENTTAL EeuATroNS Mathematical Optimisation and Programming Techniques for Economic Analysis I

the time path converges since the power on the exponential is negative This does not
y(o) = eo(Azcos(o) +.4o r,n1o;) *
] : I > 0 implying
come at any surprise. The coefficient ar that ft < 0. Since convergence
I
:At+Z=5.s hinges on h ( 0, we could conclude right away that the time path will be convergent.

',rrrrrrtimes, given an nth order differential equation,it may not be easy to find the time path.
"'Az=5 llr rharacteristic equation will be an n'n degree polynomial. This gives many possibilities on
_ _l _r.(,4, _1. llrr,outcome. There may be a mix of repeated roots, distinct roots as well as complex
y't = cosgt + /4sin At) + e 2'(-A30sindt +,4nQc<-rsQt), rrrrriugates. The problem of the higher degree characteristic equation is not an easy one.
-ez'
Nrrrrtrtheless, it is still possible to ascertain the convergence or divergence of a time path
y'(o) = +.4n sin(0)) * (-r,f ,'"(o) + e-f cos(o)) wrlhout necessarily finding the roots. This is done on the basis of the Routh theorem. lt was
](,+,.o,(o)
rr,rrned after an English mathematician Edward John Routh. lt states that for a normalised
-1 / fis\
=r(A)+lA+
(rr11 = 0) characteristic polynomial
, )=s i

a1r4-1 + "'+ an-7r * a. : g

agrn *
-(s)+(/4JG)=10 I lro real parts of all the roots are negative if and only if the first n of the following sequence of

15 term nant are all pos t ve

.. A,
'-:
v1s
:/15 a1 a3 a1 a3 a5 Z: :: Z:' Z"
ra,r, at a3
? i:, Z: o0aoa2a4
", ", d5

Therefore. the definite time path ts

lhe theorem makes it possible to determine the convergence of a time path without actually
finding the roots. Once the above condition is satisfied, we conclude that all the real parts (h) of
y, = e?'i (t."' (+,) *,ts,,,(f ,)) . ; the roots are negative. This is a sufficient test for convergence.

The graph is shown in Figure 10.5 below. Example 10.12

Figure 10.5: An oscillatory time path plot Determine whether the time path of the following differential equation will diverge or
converge to long run equilibrium. y't"' + 6yl" + l4yl' + l6y't + 8y :24
7
The characteristic root for a forth order differential equation is ofthe form
6
5 aor+ + arr3 + arrz + orr I an = O
4
for the particular example, the characteristic equation is
l
I r4+613+1412+16r*8:o
1 Without actually finding the roots, we straight away formulate the sequence of
0
determinants based on Routh theorem.
-7

lall=16l=6,
-l
l:: i;I:|" iil:*,
 Mathcrnatical Optimisation and Programming Techniques lbr Iiconorric Analysis 227
225 I ECONOMIC DYNAMICS lN CONTINUOUS TIME: DIFFERENTIAL EQUATIONS

tlr,,r0 are restrictions on 7 and d based on traditional supply theory, no restrictions are
ar Q4 asl 6 16 0l
an e2 a+l= | t4 Sl=l 800 , ,r r I he new parameters d and 2

oararlo616l
the given demand and supply schedules, the equilibrium equation is expressed as
ata3asatl 161,600
o6a2Q4oulhUB0 c-pP=-y+6P+eP'+7P"
0 a, o. asl-lO 6 1,6 0 = 6400
o ao a2 onl lo I 1,4 8 a+y=(6+B)P+eP'+LP"
All the principal determinants are positive The time path is convergent.
0 (6+ll) aty
P"+7P'.-;r- )
10.9 Economic applications
Higher order differential equations are very useful in understanding the behaviour of
ll, r,,,rsecondorderdifferentialequation Theaimhereistojustshowhowhigher(second)
markets. Take for instance a market governed by expectations Either the demand or the
, , l, r rlifferential equations can be used in economics The solution, which we do not attempt
of a commodity will not only depend on the current market price or what price is t, , rlvr:, will depend on the actual values of the parameters in the equation
be at a particular time. The long term behaviour of the price becomes critical. Players
Notlce that the order was restricted by the assumption that the producer only cares to know up
market are interested in knowing whether the price is on an increase or now 1fr lo the second differential. With complicated price functions however, third, fourth up to the nth
Secondly, players also take into consideration whether price is changing (increas Order differential equations emerge. The Taylor's formula is useful in understanding the link

decreasi ng) at an i ncreasing or decreasing rate (

bctween the order ofthe differential equation and the nature ofthe price function.
d rrt!\.

Take a specific example of the agriculture sector. The interplay of the demand and
10.10 Simultaneous differential equations
determine the ruling price. The quantity demanded will then be determined by the The word simultaneous is no longer new. lts prominence is in finding the point of intersection
price since each consumer will buy based on the price obtaining at that particular of two or more functions. lt is synonymous to the word concurrence which implies
Thus togetherness. ln this context, it is used when two or more differential equations have a
(r,P>o relationship or are interlinked. These may be rightly referred to as interacting patterns of
QD:a-BP, chonge.This is common with variable that are related so that a change in one variable is not
The supply side is however characterised by a long gestation period. There is a time lndependent of another. Their changes are interrelated.
between the time one decides to produce and the time the produce is actually offered for
on the market. At the time of making the decision, the price that will obtain on the For this kind of variables, their differential equations form a set of equations. Since each
unknown to the producer. Moreover, the producer will also be interested in future differential equation has more than one variable, it is then not possible to find its solution,
because the decisions to invest in certain fixed assets take into account the life time value independent of other equations. The number of unknowns exceeds the number of equations.
such assets. For instance, growing tobacco requires investing in tobocco kilns and the far Recall that in order for a system of equations to be solved simultaneously, the number of

may not profit from such in one period only. The asset is available for a much longer period
equations must at least be as many as the number of unknowns or variables This is what is
referred lo as simultoneous difJerentiol equdtions.
the farmer must consider the price of tobacco for the life-time of the kilns.

To make a rational decision, the farmer decides on the quantity to supply (based on how
The main area of focus is the analysis of a system of simultaneous dynamic equations or
to invest) based on the knowledge of the behaviour of the price. For a simple model, assu interacting patterns of change. Take for instance a multi-sector model where each sector is
the farmer only cares to know how the price changes and the nature of that change. Her described by a differential (dynamic) equation which impinges on at least one of the other
Qs :-y+6P +eP' +LP", y,6>0 sectors. The specific example may be sectors like the industry whose dynamic equation cannot
be independent of the education sector, let alone the health The health sector also relies on
228
I

| ECONOMIC DYNAMICS lN CONTINUOUS TIME: DIFFERENTIAL EQUATIONS Mathematical Optimisation and Programming Techniques for Economic Analysis I ZZS

the education sector for the personnel. Growth in agriculture cannot be independent of tht I ur lhe complementary solution, assume a reduced equation. Ihis is achieved by simply getting

performance of the industrial sector and vice-verso. The former depends on the latter lol rtrl o[ non-zero constants or terms in the complete equations. Then take trial solutions
equipment and the output from the Agriculture is raw material for the industry. With this klnd Xt:liert, lt=nett
of model, a single dynamic equation will contain more than one variable, in dynamic form llrr, two variables are of the same pattern but only differ in the coefficient. One rises or falls
because the different sectors are interlinked or interacting. To find the time path of each l.r',lcr than another depending on how the two arbitrary coefficients compare. The respective
sector's output, all the interacting equations must be solved simultaneously. r lr,r ivatives are

Consider a model with two sectors. The sectors are represented by x and y and the respectlvo x't = rme't, !'t = rne't
differential equations are given as
lil matrix form, the two variables and the respective derivatives can be summarised as

+ Zxt + lt -- 14
2x[ + y't ,, lTlU', *Mu= 0willgive
5Y[ + xs )- 3Y, = 12
":lnrertSubstitutingtheseintothereducedequation/z
This set of equation can be cast in matrix notation of the forrn tlTl,u'+u[ife"=o
lulMu=g er+M)lTlu,:o
where,i and M are coefficient matrices, u and u are vectors of variables and g is a vector of
constants. Taking this form, the various matrices are defined as
(tr + M)
[Tl =.
\ince the exponential is unambiguously non-zero, its multiplicative inverse can be multiplied on

t =13 ll,":ll,il,, :li, !1," =li',],'=lil)

both sides to eliminate it from the equation. The remnant is similar to the problem
cncountered under independent differential equations. There are two factors whose product is
rero. ln rnatrix however, a zero product does not imply that one of the factors is zero. Here it
Just as in independent differential equation, the first step to dealing with interacting differential
implies that the coefficient matrix, now Qr + M), is singular. ln algebra, singularity of a matrix
equations is to find the particular solutions. Since there are two equations for presumably two
is indicated by a zero determinant. Therefore, the resulting equation is
sectors, there must be found two particular solutions x, and y, To find these, we try with the
solution that the two variables are constant, that is rr : x and yt = y. ln the single variable lJr + Ml:0
case, this would equate to the term b. With more than one variable however, the term in each Sincethetwomatrices/andMareknown,wecanproceedtofindthevalueof r Thisbecomes

vG 1). (i 1)t:.
equation is a combination of the two variables. With an assumption that the variables are
constant, the first derivatives equal to zero. That is, x[: y', = g

The equation is now reduced to l('J ;,). G 1)l : .

t2r+2 r+11
Mv=g I r s.+:l:u
Using the formula for the determinanu we generate a non matrix equation
ti 116:1 = tiil (2r + 2)(sr + 3) - 7(r + 1) : 0
1,or2+1,5r+5=0
This is a set of simultaneous equation. Various methods are available on how to solve the
equation The actual details are presented in Chapter 4 of this book. Here we simply state the 2r2+3r+1:0
results as This is a characteristic equation, similar to what was encountered earlier. Again we state the
roots without showing the working as t = -1, and rz = -0.5
t;;)= ril With these characteristic roots, we solve the equation Qr + M)
[T] = , for rTt
corresponding to both roots. The solutions wrll not give specific values of the vector
 I

230 I ECONOMIC DYNAMICS lN CONTINUOUS TIME: DIFFERENTIAL EQUATIONS Mathematical Optimisation and Programming Techniques lirr Econornic Analysis 237

components. This because the coefficient matrix, with the given value of r, is singlllar 247-42+6=2
solution will merely be a relationship between the two variables. For L = -1, Aj+242+2=5
0llmrl
Wl, llirve it up to the reader to show or prove that At = -t and A, = 2. The definite solution
[0
l1 lrr,r r is
-zll"l=tt
I

m' = 2n' lxtl l-ze-t - ze-o.sL + 6l

lf n, = 4r, thenmr=24, [y,l= [ -"-r a4r-ost ,-2]
For r., : -g 5, Wlrr,n time paths have expressionse'r, the convergence or divergence will depend on the value
ill r time path converges when r is negative and diverges when positive. ln the particular
The

ri 3:lrtl =0 ,,x,rrnple at hand, r is throughout negative in both x, and yr. Therefore, both time paths
m2 = -0.5n2 r i,nverge to their respective equilibriums.
If n, :2A2, thenm, = -4.
l0 11 Economic applications
ln general, given 4, tni and ni, the complementary solutions are
differential equations are applied in many aspects of mathematical economics
"rnultaneous
llrcy provide useful tools for working with interrelated variables in a dynamic way. For
x- = I
l-/
m,e','
nstance, the inflation-unemployment model in which changes in either inflation orthe level of
rrnemployment cannot be isolated from the other variable. This is dealt with using the
v,:fn,"'r lramework presented above, the simultaneous differential equations.
t
in the particular case with r = -0.5, -1 the respective complementary solutions for thc
ln general, the framework can also be used with more than two interlinked variables. This is
variables are
ideal for an economy wide dynamic analysis where the fundamentals are assumed to change

xc:fllterlt +mzet't continuously. For the purpose of an example, take Leontief input-output model discussed in
: ZAte't - Are-o st Chapter 4. The model is static in nature and assumes that there is clearance in all the sectors. lf
we relax the two assumptions and assume production is done continuously so that the model
lc:fltet't !nrer't conforms to continuous time. Since markets do not clear, there must be a continuous
: Af-t * zA2e-ost adjustment of output. This adjustment will be a function of the deficit, that is, if there was a
Now both the particular solution and the complementary solution are known, with arbitrary shortage, then there must be some increase in output andvice-verso.
constants for the latter. The general solution is the summation of the two. To keep with tha
Take a two sector economy. The output for sector one is used as an input in all the sectors as
matrix format, the summation is done as
well as to meet the final demand for that particular output.

l;',1= l;:).;fr1 xt:d7rxl+dnx2+d1

where the left hand denotes the supply and the right hand denotes the demand. lf output
[ZAret-Are-osll . t61
= lor"-' zere-ot']+ l)l changes by exactly the difference, then the differential is written as.
+
L-A2e-ost+61
_lZAre d.x,
lAre -t + zAre-ost + 2J x', = a1x1 + a72x2 * d, - xv where x't = Oi
This is the general solution. lt is not definite since it still has arbitrary constants. The constants the equation is already a differential equation. When the change in output is not exactly the
can be made definite if the initial conditions are known. For instance, assume that information difference, a coefficient can be attached to the difference to take care of the less thon or more
is now available that lr=0 : 2, ft=o = 5. The resulting linear equations are thon deficit change. Rearrange by taking all variable terms to the left side of the equation. This
gives the equation of a more familiar form.
232 l"o*o*','D,NAMcs -'"1:;-::,.-"-,:;'*'"-' Mathemiitical Optimisation and Programming'I'cchniqucs lbr Economic Analysis 233

il!
Similarly, the differential equation involving sector 2, that is, ,2 can be Sotten. The tWO
Cha pter 1,1,

form simultaneous differential equations whose time paths can be gotten using the framcfl
+

I I ECONOMIC DYNAMICS IN DISCRETE TIME: DIFFERENCE EQUATIONS

illustrated earlier. I
I Ll lntroduction
hr lhe preceding chapter, we dealt with the behaviour or a variable over t,r,1e taking time as a
r orrtinuous variable. The change in the variable was assumed t. occur continuously. ln reality
Irowever, not all economic variables can be assumed to change (rntinuously. For instance,
wlrile the exchange rate, as a price of one currency in terms of i ,other, can be ctranging
r r)ntinuously when disturbed, the price of commodities like labour, seasonal ci?ps only change

trr discrete time. lnstead of taking time as a continuous variable, there is need to modify re
nrodel so that it takes into account cases where changes in the variable are not continuous.
',rrch dynamics are in discrete time and make use of difJerence equ ions as opposed to
rlilferentiol equotions studied earlier.

While in continuous time each time value represented a specific time point, in discrete time,
I ime is taken as an interval or a period. lf the price changes annually, then there would be year

I, 2, 3 and so on which represent periods of time. lnstead of looking at the instantaneous

rlrange in the variable, this type of dynamics is more concerned with how a variable changes
lrom one period to another. lf the price of say tobacco was higher than the long run
lquilibrium, how will it adjust to equilibrium, if at all it reverts to equilibrium? This type of
rluestions is best handled by difference equation.

11.2 What is a difference equation?

ll is much easier to define the difference equation by first looking at their use. Difference
oquations are used for dynamic analysis where time is treated as a discrete (categorical)
variable rather than a continuous variable. lt is a tool for deriving the solution to equations
where the current value of some variable depends upon the value of the same variable in
carlier time periods. fhus a difference equotion is one where a variable is a function of its past
values. More precisely, it is an equation of the form

y6:a*by5-1
where the current value of the variable depends on the previous value denoted by a lagged
lime indicator.

The highest time lag between the explained and explanatory variable defines the order of the
equation. For instance, in the above equation, the current value depends on the immediate
past value giving a unit time lag lt is therefore of first order. Second, the equation is also linear
234 1 ECONOMIC DYNAMICS lN DISCRETE TIME: DIFFERENCE EQUATIONS
Mathematical Optimisation and Programing Techniques for Economic Analysis 235

in the variable. Combining the two traits gives it the name of first order linear differenc!
lt = u * ab + abz + ab3 +... --. + abt-r I b'yr-,
eq u ation.
rrbviously !t-t = !o. Like in the differential equation case, the above equation can also be
The second order linear difference equation will have an added two-time-log variable y,-2 rlivided into two parts. The first, composing of constants only, is the particular sum. The term
ln
the equation. Recall that the order is determined by the highest time lag and will thus remaln rum is used here because in discrete time, summation is used in place of integration. Let's
definite even with multiple lagged variables on the right hand side of the equation. ln genenl,
(lenote this by y, which is given hy yp=o+ob+ab2+ob3 +......+cbt-1. This is a
an nth order linear difference equation is of the form (leometric progression with the initial term a and the common ratiob lts finite sum is given by
a(l - ht)
lt = d I brlt-r ! bzlt-z * "' "'l bnlt n
!p=St=i, b+t
the highest time lag between the regressed and the regressands is n, giving the order of tha
eq uation.
the second part is one with the lagged variable in it ln the continuous time case, it was
referred to as a complementing sum, mutotis mutondls. We can denote it with y" = btyo and
11.3 First-order linear difference equations requires no further modifications.

Given the above definition of linear difference equations, the first order linear differencc To derive the general solution of the first order difference equation, we sum the two
equation is then one given by components of the general sum. This process can be summarised as below

y7=a*by5-1 lt--lPlY"
It portrays a case where a current value of a variable depends only on the immediate past
a(l - bt) I hL i,
values of the variable. The knowledge of last year's level of output tells us how much output -
L-Ot
I u /o

will be produced this year. Or given this year's maize price, we can anticipate the price next o aht
year to be at a certain level. These expressions hinge on the first order linear differenca =r-b 1-o*brYo
-
equations, which provide a link between the current level and the immediate past level of a
tt=(to-h)* **, b+7
variable.
The equation has three components which deserve attention. These are (1"-ff), U
The solution of this order of linear difference equation can be found using a simple lagging
""a
-@ .
tb
11 ;5 easier to understand the three if we start with the third The third component gives
process. lt is known that the current value is a function of the immediate past. ln the same way,
the long run equilibrium of the variable y. When the long run equilibrium is attained, the
moving one-time back, the immediate past value is a function of value preceding it. Using thls
variable stabilised so that it does not change Given the primary difference equation !t = a *
link, the difference equation can be expanded up to the initial value y(0).
byr-r, at long run equilibrium will reduce to y = a * by. The time subscript disappears
The difference equation for this period is given by lt : a I by,-.. Using the same equation, because the variable no longer changes with time. The solution of the equation it y :
*
we can derive a chain of first order linear difference equations corresponding to all the past which gives the long run equilibrium level
periods. The equations are
The first component can now be defined using the definition ofthe third. lt is the deviation of
y5=a*bys-1 the initial level from the long run equilibrium level. ln other words, it shows how farthe starting
!r-r=o*byr-z point is from the long run equilibrium. Since this is a one-observable point, it is not expected to
!t-z=o*byt-z change, the reason it is not connected to time. lt can be looked at as a measure of the
...:o)-b...
magnitude of the disturbance on the variable. lf for instance price is disturbed from its long run
yr=a*bys level, the displacement of the price from its long run level gives an idea of how big the
This chain of equations is such that every equation can be substituted into the equation
disturbance was. Depending on whether the value falls or rises, this component which appears
immediately on top Substituting in this pattern will give rise to an equation of the form
in the equation as a coefficient of the second component, can take on both negative and
236 | ,.oror,a DyNAMtcs tN DtscRETE TIME: DTFFERENCE EeuATtoNS I
Mathematical Optimisation and Programming Techniques for Economic Analysis

positive value. When it is zero, the variable is at its long run equilibrium and the difftfl Second, the base is also less than unit. As time f becomes larger (with the passage of
equation collapses to the long run equilibrium level. rI time) (-0.5)r will tend to zero. Ultimately, the time path will just equal the constant.
Thus there is convergence to constant long run equilibrium. See Figure 11.1 below.
The second component shows how the variable is linked to time. The exampleU"nrriour o[
variable with time can be told depending on the value of the base b since time t is lnI igure 11.1: An oscillatory time path41
exponent. When the absolute value of the base is less than unit, the component approail

*'** jl: I;ff:r: ;I rn:::I :T ;: I

j:'*il| Tinre Path

*ffi :::'::
Base b Convergence
:'*
Oscillatory Behaviour
b<-1. Divergent Osci I latory Oscillating divergence
-1 <b<0 Convergent Dampened oscillatory path
0<b<1 Convergent Non oscillatory Smooth convergence
b>1 Divergent Smooth divergence

I.$
;-? ?;'T', 810

Example 11.1
Solveyl=3-05!t t, !o-4
This is a first order linear difference equation. lts solution comprises of two parts: thr,
particular solution and the complementary solution The general solution is given by
Example 11.2
y,=(yn
\-- J,t
l-b/
6, a-o
l-h
, b + t
Solvey, = 16*3!t-t, /o = 5 and determinethe nature of thetime path.
ln the current example, o = 3 and b = 0 5. Substituting these parameters into th('
The previous example did not derive the solution but merely used the provided formula.
general solution yields
ln this example however, we provide the derivation. The process is as follows.
r3r3
.Yr = (.Yo
I _Fosr)(-os)r+ I _(_os) lt = 16 * 3yt-r: 16 + 3(16 * 3yt-z)
= (4 - 2)(-0.s)t + 2 = 16 + 3(16) * 32yt-z = 16 + 3(16) + 32 (1,6 * 3yr-r)
= 2(-0.5)t + z = + 16(3) + 16(3'z) * 3'yr_s
16
All the parameters ofthe solution are definite. Therefore, the solution is also definite
= 16 + 16(3) + 16(32) + ...+ 16(3t-1) + 3tlo
We now comment on the above solution in line with Table 11 1 above. The solution is a Using the geometric series, the first t terms can be summarised as the geometric series
sum of two components. The first varies with time and the second is constant Our with a r = 3.
= 16 and the common ratio
interest lies in the first part which determines the behaviour of the time path Firstly the
r3t - 1)
base is negatlve The implication is that this part of the solution will be positive for even yt=t6#+3tyo
J-l
numbered time periods and negative tor odd numbered time periods. That means there
will be oscillations around the constant.
= 8(3t - t) .t 3tys

a1
The bars show deviations from the long run equilibrium as opposed to actual level
 I
Mathematical Optirnisation and Programming Techniques for Economic Analysis | ,r,
238 | eCoruovtc DYNAMICS lN DISCRETE TIME: DIFFERENCE EQUATIONS

=(yo+B)3t-B Q', = -y + 6EPt, (y,6 > 0)

=(s+B)3t-8 EP, = Pr-,
: 13(3)r _ 8
At cquilibrium, Qo, = Qt, a market clearing condition. The constant long run equilibrium price
The base in the time path is positive and greater than unit. Therefore, the time path lr t., tiiven by.
nonoscillotory and explosive in nature. It never reverts to any particular value as timo
d+v
grows
'D- -- B+6
_______L

11.4 Economic apPlications lhis example is similar to the one given in continuous time case. The difference though is that
ilow time has become important since there is time variation in the model. This was not the
The application of difference equations in economic problems is enormous. They can be applied
r,rse in the previous case. Solving for price in the above system of equation will produce
to any area of economics where a variable has an autoregressive process. This is where tho
d-\Pt:-y*6P5
variable is a function of its lagged values. Though these may not manifest to a form that is easlly
1

identifiable as difference equation, some modification and transformation may make them look , -olY -6 o
so. we put the question,'can the given equation be expressed in a form that shows a
B 8,,_,
Itris equation corresponds to the general form given above. Using the solution of the general
difference equation?'
lrrrm,'pB
we substitute 31 = a and -! = b to arrive at the oarticular solution
Take an example ofthe agriculture sector where a particular crop like cotton or tobacco can bo
taken. Their production is seasonal with a complete cycle corresponding to a year. This makes
time discrete. At any particular time (year), there is equilibrium because the price can adjust to
equate demand and supply. The demand for the product is not dlfferent from the ordinary
r,=(ro -;#)(il','ffi
market where it adjusts with the price. The supply on the other hand does not conform to
lhis is a particular solution for the linear difference equation looking at the price of an
traditional supply behaviour. Because of gestation lags between decisions to plant and
,rgricultural product. Given this particular solution, there is need to determine the behaviour of
subsequent harvest, the supply decision are not altered by current price but depend on price
price based on the dictates on Table 11.1 above. lt must be clear from the solution that the
expected when the crop is harvested.
rluestion of convergence or non convergence and whether there is oscillation or not are
This brings in the role of expectations. Depending on how expectations ofthe harvest price are dependent on the slopes ofthe demand and supply functions. The slopes are denoted by f and
formed, the decision to supply will be made Given the two types of expectations; rotional and rt respectively. These parameters are defined positive for downward sloped demand and
odoptive, and the complications of the former, it is acceptable to assume that farmers use the upward sloped supply. lt means therefore that for normal demand and supply function, the
latter. This is based on assuming the low level of technology and capacity among the farming ultimate b in the solution is negative. This means the price, when disturbed, will oscillate
communities in most developing countries to make use of the former. Besides, this assumption around its long run equilibrium level.
is outside the realm of this book and it must suffice to take it as an assumption.
l'he issue of convergence remains trivial. The information at hand is not sufficient to ascertain
With adaptive expectations at play, the price is expected to be the same as in the previous whether there will be convergence or not. ln the above solution, convergence will require the
period. The shorter form is to say the supply of a commodity will be a function of the previous absolutevalueofthebasetobelessthanunit.Thiswill holdifandonlyifjcf-d<Pthis
'tt
price. The long run equilibrium is reached only when expected price at time t is equal to actual
simplifies to having the demand schedule being steeper than that of the supply schedule. When
price at time t the opposite hold, there will be divergence and the market will be dynamically unstable.

Assume demand and supply of say tobacco is given by

The following graphs that show variations of the cobweb model illustrate the above situations
Qdr=a-FPr, (a'P>o) of convergence and non-convergence.
240 | ,aoror,a D'NAMrcs rN DT'.RETE TtME: DTFFERENCE EeuATtoNS Mathematical Optimisation and Programming Techniques for Economic Analysis 247

Figure 11.2. Cobweb model: Case of convergence I igure 11.4. Cobweb model: Case of divergence

Figure 11.3. Cobweb model: Non convergence and non divergence 11.5 Second-order difference equations
The second order linear difference equations are difference equations where the current level
is a function of two preceding values; the immediate past and the one preceding the immediate
past. lt is a situation where the current price for example is a function of the price two years
ago. Such equations are common in situations where the effects of a disturbance take time to
disappear. ln particular, second order linear difference equations apply where the recovery
takes two periods. The time lag on the regressands extends to two. ln notation, they are
written in the form

!t = a I Ft!t-ti Fz!>z
the parameters in the model are arbitrary. Though this may sound ordinary, it has a strong
implication on how the parameters can be treated in algebra. This is precisely what is about to
take place. As arbitrary numbers, changing their signs does not affect the model! We assume
the reader is familiar with this and we proceed to change the signs on the coefficients of the
lagged variables. The motivation is to change the form of the model. After changing the signs,
bring all the variables on one side so that their sum equates to a constant

This yields the model of the form.

lt*BJt r*Fzyt-z=a
The model remains unchanged except for the presentation. lt must be mentioned here that this
presentation does not invalidate the former; rather it facilitates a quicker and easy way to
derive the general solution. The equation is quite complex and deriving the solution needs a
242 I ECONOMIc DYNAMICS lN DISCRETE TtME: DTFFERENCE EeUATtoNS Matherratical Optimisation irnd [irogramming Techniques tbr Economic Analysis 243

well thought strategy. By strategy, it implies that what we apply is not the sole method 11.5.L Distinct real roots case
but
rather what we consider candid.
Example 11.3
Like in the continuous time case, this equation can be looked at in two forms; the !t - lt-r - 2!t-z : 0
Find the time path of the following difference equation:
homogeneous and non-homogeneous. The homogeneous part is found by lettingthe constant,
We start the solution by trying the solution of the form lt = Amt which reduces the
on the right hand side ofthe equation, equal to zero. This results in a homogeneous equation
of difference equation above to a quadratic equation of the form m2 -m - 2 = 0 The
the form
quadratic equation has two distinct real roots m = -L qnd 2. Given the two values of m,

lt*FJt-ttqzlt_z=0 substitute them into the general solution equation f t = AJnrt I Armrt. This gives the
The equation does not have a unique route to the solution but can be solved by trying different time path;
possible forms of the solution. This is often referred to as a triol ond error method.
we now try
lt= Amt as the solution to the equation. Then substitute this assumed solution into the
lt: Ar(-1)t + A2(2)t
equation, lagging appropriately for the lagged variables. The equation that emerges
Since there are now two arbitrary constants to be made definite, an initial condition will
is
not be sufficient This follows the general rule of equations that there must be at least as
Amt + PlAmt-r + prAmt-2 = o many equations as there are unknowns. To make definite the two constants, two
This equation can be simplified since Amt-2 is common in all the terms. After this conditions must be known, which can be used to generate two simultaneous equations.
simplification, the equation turns to These conditions need not includethe initial condition butcan be atanytime period.

m2+PlmlBz:0 For instance, suppose it ls known that y, = !7 and y2 = 13. The arbitrary constants can
This is an ordinary quadratic equation. since the problem is ordinary, the solution need not be be made definite using the derived equations
special. The solution of a quadratic equation is given by the formulae
lt=1-1'=-At*2Az
lz=13=Ar-l 4Az
Solving the two simultaneously gives At= -3 and Ar= 4. The definite solution ortime
m=
path
and the general solution ofthe homogeneous part ofthe difference equation is given by + 4(2)t
Yt: -3(-r)t
this definite solution because allthe parameters are definite. Once time
is a is known, the
lt=Atmtt*A2mrt
The twobases m. and m2 are the two roots of the quadratic equation above. ,4, and A2 are level of the variable at hand can be known as well
two arbitrary constants. once definitised, the solution will be definite. The homogeneous
11.5.2 Repeated real roots case
equation does not have the particular solution so that the solution of the complementary
function is the general solution. when we move to non homogeneous functions however, the Example 11.4
solution above will not be final, it will be a solution of the complementary function. At that Given the difference equation ys*t.2yr-r*0.36yr-r:0 and that yo:8, h:7.2,
stage, it will become necessary to add a superscript identifier to the solution. with a find the time path yr.
homogeneous equation, this is not critical.
The solution follows the same path as in Example 11.3 above. First use the solution form
But the solution to the quadratic equation come in three forms depending on the value of the
!t: Amt, simplify the equation to remain with a quadratic equation which is solved for
discriminant D = 0t2 - 4pr. These arc: distinct reol roots mL + m2; the repeoted reol roots the two values m, and m2. The first three steps will proceed as follows.
frt = rnz and the complex conjugotes where the roots are a pair of nonreol numbers. we use Amt + !.zAmt-' + O.36Amt-2 = O
examples to illustrate the three cases in succession. which after dividing by the highest common multiple of all the terms 36^4mc 2, the
equation reduces to a quadratic form.
 I

244 I ECONoMIC DYNAMICS lN DISCRETE TIME: DTFFERENCE EeUAT|ONS Mathematical Optimisation and Programing Techniques for Economic Analysis 245

m2+1.2m+0.36:0 Figure 11.5: Oscillatory time path

Using the formula for a quadratic equation, the roots turn out to be m, :mz= -0,6,
With repeated roots at hand, does y, : Armrt + Arm2t still apply?
Time path
Well, let's see. When modified tor m, = frz: ttl, it will still apply. The equation lr I5
factorised for the common root as lt = (At* A2)mt. Since the two parameters ar!
arbitrary, their sum will equally be arbitrary and can be denoted by.4 so that the solution
equation to use now is

Before going any further, this solution implies that

lt = Amt
lt = m!t_t, a first
homogeneous difference equation. The solution differs from the original equation, which
order
ilflIt'r-- ll 12 13

is of second order. Thus the form lt = Armtt I Armrt does not generally apply to
1

repeated root case.

-I5

When there are repeated roots, the solution of the function takes the form
lt: Atmt * tA2mt
And substituting the root, 11.5 3 The complex conjugate case

When the discriminant of the characteristic equation is negative, the characteristic

y t = A{-0.6)t + tAzeo.6)t
equation does not have real roots. Roots can nonetheless still be found in another
Using two known condition yo and yl, the arbitrary constants are made definite using domain. This is the complex number domain. The characteristic equation will produce
lo=At complex conjugates. This is a pair of complex numbers with an equal real number part
lt: -0'6At - 0'642 and an imaginary part of equal magnitude but of opposite signs.
The first equation is a complete definitising equation for,41. The reader should be able to
show that the remaining constant is We are looking for the two roots given by

.o_=-!t-0.6Yo -h+trP,'-402
0.6 m:2
so that the definite solution is expressed as

, /-!r - 0.6yn\ This time around howeve r, Br2 - 40:- < 0 so that it is not possible to get a square roots.
lt=!om,+r(__Z_j,Jmr we rewrite this inequality as (_l)(+pr- 0r') < 0 the second factor is now positive.
Placing this into the quadratic formula, we get
with yo : 8, yt = 7.2 given, the particular solution for the problem at hand is
I_
/ -7 .2 - 0.6(8\\
r(ff,) -p, 1 J(-1)(4 B, - p,'.)
Ir = 8(0.5)t + io,o)' m:2
lt = B(-0.6)t - 2ot(-0.6)t
= aQ - st)(-0.6)t _-8, ----l+or-Pr'
2- 2
246
I

I ECONOMIc DYNAMICS lN DISCRETE TIME: DIFFERENCE EQUATIONS Mathemal.ical Optimisation and Prograrnming Techniques for Ecororric Analysis 247

I tlirrre 11 5: Argand diagram (Difference equations)

-Br.- 48, - Br'
2-t'
- 2 2
= ntvl
o
The two roots can then be substituted into the general solution fr c
Eo
o
lr=Atrntt*A2m2t
E
Homogeneous difference equations do not have the particular solution. As such,
complementary solution is in fact the general solution.
e
lt = At(h + vi)t + Az(h - vi)t <
0
>
eal
Define R, the magnitude of the complex number as R = r,lF + v,. The above equ
can then be written, in view of the defined magnitude of the roots, as

th , \t rh ,, r'
lt = AtRL (E * E',) + /rR' (R - R
r,)
Given the angle d and the SOH-CAH-TOA4'z mnemonic, it is possible to link the
The effect of the R is neutrallt has the same power in the numerator as well as in thl trigonometric function with the parameters of the complex number. The length of the
denominator. This would ultimately cancel out but we have an interest in this format. vector OP was defined 65 p = al11z 1 yz using the pythagoras theorem. Trigonometry
states that
The solution requires raising a complex number to some power t. A similar scenario wrt
encountered in Chapter 10 when second order differential equations produce complcx h
cos0 =
-R
conjugates as roots to the characteristic equation. An available option is to usa
v
trigonometric functions. Since a complex number is a number in two dimensions, it can be sin6=R
presented in Euclidean plane as shown below.
Substitute the two equations into the solution

rh u ,' rh ,, .l
lt=ArR( (n*n',J +A,Rtlc- Ri)
lt = A$t(cos0 * isind)t * ArRt(cos0 - isind)t
lt = ArRt(cos0t * i sin 0t) * A2Rr(cos 0t - isin0t)
It = RtL(At * Ar) cos 0t + i(Ar - Ar) sin 0t)
yt = Rt(A3 cos 0t +.4n sin 0t)

Example L1.5

Find the time path for the difference equation !r-!t-t+]lr-r:0, lo=2 and
h=7

o'Sine
equals Opposite over Hypotenuse, Cosine equals Adiacent over Hypotenuse, TanBent equals Opposite over
Adjacent (SOH-CAH TOA)
748
I

| ECONOMIC DYNAMICS lN DISCRETE TIME: DIFFERENCE EQUATIONS I\4rtherlatical Optirnisation and Programming Techniques t'or Econorric Analysis 249

We make use of the general solution formyr: Amt.fhe difference equation tra
to
Ao sin(1.249) :L - Z cas(1,249)
,12.5
(
- Amt | +'=Z Atn'-2 = o
Am'
L- Zcos(1.249)
the resulting characteristic equation
,2 ism2 - m * I = 0 since the discriminant is A+= !2.5 =4
sin(1.249)
the roots will be complex numbers

b .,14a= 6z -t 12cos(l.249)t I 4sintt249)tl

lr =,12.5
m.= --2-+ i- 2
t , ,[+12s1-1-112 it lime
r.r.rL path
rsLtt

i::
I

2- 2
13 ?o I

= ztzi 60ffi ffi[**

To get the solution, two parameters are needed. The first is the magnitude of
complex number R and the second is angle 0 By definition,
tffi
::
ioffi ffi
ffi@

R:Jh,+v' 20
l
ro ffiffiffii
ilffim*
-*ffiHffim

= tlz.S
1*;;--ffiffiYYryl
o
-wffi-
i.:i.-",-..W.1
Given the two dimensions of the complex number, we can proceed to get the parametcf
0 using the fan function. This function, unlike the Cosine and Sine, does not require tha
The above figure shows the time path for the particular solution
magnitude of the number lt only relies on the imaginary and real dimension!, yt = !2.5 12cos(1.249)t + 4sin(1.249)tl. The time path oscillates around some long
.1 run averages and is divergent in nature. Though oscillatory in nature, the deviations from
ua
Tan? =: = i the long run average increase with time.
NI
2
suppose the assumption of homogeneity is relaxed, what happens? The assumption is relaxed
+0=tan-43:1.249 when the constant in the difference equation is allowed to assume values other than zero. This
is called the non-homogeneous second order linear difference equation. Having found solution
for the former, the solution for the latter is pretty simple. Recall that the solution to the former
The time path then is
is but a solution to the complementary part of the latter. when the non-homogeneous
equation is divided into two parts, the complementary and particular sum, the solution of the
y, = 25' ll r+ A a sin(1, 249) t)
"1 =cos(L.249) complementary is already known through the homogeneous case. The particular component of
Using the two initial conditions, we can find the particular solution as follows the equation is found by assuming the variable is constant.

Yo=lArl=2 For the difference equation

y, = nlTSlz cos(1,.249) + A+ sin(1..249)l = 7

lt I BJt-t, * \z!t-z: a
 Mathematicar optimisarion and program,ing Tcchniques firr Economic Anarysis
250 | ,ao*or,a DyNAMtcs tN DT'.RETE T.ME: DTFFERENCE EeuATtoNS
I
: ,4r(-1)t + Az(z)t + 2
the particular component requires equating all the variables, the present and lagged ,.rfaf This is an indefinite solution because it still has arbitrary
constants. lf there was
parent equation can be sorved forthe sorution to the natlcfl
information for some two periods, the constants could be made definite so that
JJ:,i'::lfilj;1,;';;i;T" solution is also definite. To use the information available to Example 11.3, two options
the
are
,rl available. The first one is to maintain
!t : 1,1 and y, = 13 which will result in a new parr
which holds for as long as 7
r,r:
-l p1+ Pz + 0
,*fi 6 of constants. The second option is to realise that changing the constant from zero
is equivalent
to four
to having all observations go up by four. New available information will then
t be y, = 15 andy., :17. This will leave the constants unchanged.
The solution to the homogeneous case, which is also the solution to the complementl
equation, will still depend on the kind of roots that come out of the characteristic equation, ll The second option is used, as it gives an opportunity to prove the resurts. The two
now denoted as yrc where lhe superscript as earlier mentioned, indicate that it is a solutlon[ simultaneous equation, when simplified become
the complementary equation. The procedure however remains the same as in I
homogeneous case alone. The general solution is found by summing the two components !t[ ll=-At*ZAz
L3=AtI4Az
This is the same set of equations used in Example 11.3 and so the results
should not be
different, At = -3 and A, = 4. The definite time path is
;:" .,:" -:,--::rj:r:,",,. us,ra,e,n" n.n.n.,,.,"n.ol
case. The examples are modified by adding a non-zero constant to the difference equation. Tl| yr: -3(-1)t +4(2)t +Z
solutions to the complementary parts of the equations will not change and so will be t.t en Example 11 7
{
already found. For the particular part of the equation, the solution only varies with the select{ Given the difference equation y1 I l.Zyr_, + 0.36yt z: IZ.B and that lo = g, lt:
constant a. 7 .2, tind the time path yr
I .

Example 11.6 t This is coming from Example 11.4, except the constant has been changed to
make it non-
Find the time path of the following difference equation: y6 - !t-t * Zly-z = 4 ,lt
homogeneous. Again, the sorution to its comprementary part is arready found
in Exampre
11.4 and there is absolutely no need to start the calculations again. Simply take
This example is coming from Example 11.3, where the complementary solution was foun{ it as given
in its indefinite form
as

!t' = Ar(-0.6)t + tA2(-o 6)t

!t':AJ-l)t+A2(2)t
Recall that the solutions of the two components must never be in definite form. The particular component is not provided. rt has to be found but the procedure
must be
Definitising is the last procedure, after combining the two components With a = 4, the familiar by now.
solution to the particular part is
ytp + L.2ytp + 0.36yro = 12 g
YtP-YtP*2Yrt=4
ytp(l*1.2+0.36):12.8
ltp(1 -L+2)-4 _ 72.8
li=7n=s
u.P =42 =2
The complete solution is
Then combining the two gives the indefinite generar sorution
Yt:ytc +ytP
Yt=Yt'+YLP = A{-0.6)t + tAzeo.6)t + s
252 | ,.oror,. DyNAMrcs rN DT'.RETE,ME: DTFFERENCE EeuATroNs 'I Mathematical Optimisation md Progrming Techniques for Economic Analysis 253

ln addition, the information lo B, : lt:7.2isavailable. ln the previous exeml G=C

given conditions were modified to keep the constants unchanged. tn this exrnfi rr,,w rr:lax a few assumptions. The consumption. lf time is measured in months, then it is
first is

keep the initial condition unchanged so as to observe what happens to the cdl Irr, ,rnceivable that this month's consumption depended on income this month. One would be

usins the two pieces of availab," ,"t;r];:"" the two resultins equations are lr,,l,comfortable to assume it aS a fUnction of last month's income. We get a salary at the end
,,1 tlrl month for consumption in the coming month. lf this lag is sustained on an annual basis,
I tlr,.rr the consumption function can be best described by the function

which give /r=3 and or:A';"'#;*'X3"lrrar sorution i, *,"n g*ll Ct=o*c(Ytt-T)

yr:3(-0.6)t-ft-0..1,*t
r- ;*"r* i
I
hrv|stment has always been aSsUmed to be exogenous. But this may not be close to reality.
Mrrr h as it is independent of current level of income, it is undeniable that it is a function of past
t1r ()mes. Particularly, it can be arguably assumed that it is a function of the increase in the

i"I n
rh'pcnds on how the income
.,rrrr:e
of investment (saving) in December
prr,vious period's income over its predecessor. The level
for November month end exceed that of October month end,
the December income is not yet received This is described by the function
10
It:j(!t-r-Yt-z)
B)vernment expenditure, as well as taxation can still be assumed to be lump-sum and
s

s:..- r.xogenous to the model. With this information, the economy is now best described by the
lr rllowing set of equations

:il
q.
1l 12 13 l4 15 t6
Yr=C1 *1, lG1
Ct:o*c(Yr-r-T)
It=j(lrr-Yez)
-2 the problem is to find the time path of income, and using this, the time path of consumption
,rnd investment.
Compare the solutions in Example 11.4 and Example 11.7. Maintaining the initial condition aftBr
t,utting the last two equations into the first, with some simplification leads to a second order
adding non-zero constants forces the constants to adjust.
hnear difference equation given by
11.6 Econornic applications
Yt - (c + i)Yt r* iY, z = a + (1 - c)G
The application of linear difference equation to economic problems is enormous. Some with p, : -(c + D and pr- 7. Once the autonomous part of consumption a, the marginal
economic variables may be too complex for the first order linear difference equations. Second propensity to consume c and the responsiveness of investment to change in income are known,
order equations become the best-fit expressions which can be used to work with such a particulartime path for income (as well as consumption and investment) can be found using
va riables. the second order linear difference equation method. Such time path is crucial for planning
because it enables a more accurate forecasting into fur future periods.
ln macroeconomics, the static Keynesian model is used to describe an economy. For an
economy closed to international trade, the model is 11.7 A note on higher-order difference equations
Y:C+l+G The principles discussed in this chapter on second order difference equation do apply, mutotis
C:o*c(-f) mutondis, to difference equations of higher orders. For the particular solution, very little
1:t changes. The solution proceeds in the same way as though one was dealing with a second order
254 I ECoNOMIC DYNAMICS lN DISCRETE TIME: DIFFERENCE EQUATIONS Mxthcnlillicrl Optinlisrtiotr rnrl ['rograrrnting T-echniqtLcs firr Ecouornrc Arralysis 255

difference equation. The idea is to assume that the variable is constant. The implicatiorr 1,,
the variable as well as its lagged values will be equal. Of course with higher order Chapter L2
equations, there will be more lagged variables.

With the complementary solution, the strategy of solving a characteristic equation still 12 DYNAMIC OPTIMISATION: AN INTRODUCTTON TO OPTIMAT CONTR0I
It is worth stating that the degree of the characteristic equation is the same as the ordet ol lltr THEORY
difference equation As such, higher order difference equation will require solving polyrrorrr
equations of higher degrees. Just as the second order difference equation results into a s,', rllrl
12.1 Introduction
degree polynomial equation commonly called quadratic equation, an nth order differr,rrr o Mining is a very significant activity in many countries in Africa. ln Zambia, copper mining has
equation will result into an nth degree polynomial equation. been the main stay of the economy even before the country attained political independence in
1964.
An nth degree polynomial equation must essentially produce n-roots. However, the actutl
number of roots may be less than the degree of the equation. This is because some roots mey 5ince the liberalisation of the economy began in 1991, a number of foreign companies have
be reaping (repeated roots). There is also a possibility of having complex conjugates among tht swarmed into the mining sector Harking back to our discussion in Chapter 2, these companies
n-roots. Once roots are established, their treatment follows that used in the second ordor are likely typically to be stickers, rather than snatchers. They come with their huge capital
difference equation. investment not to mal(e some quick profits and leave but to maximise returns from their
operations over a longer period of time. ln other words, their production decisions would be
Since a higher degree polynomial can produce roots in the three possible categories (distinct
made so as to choose a time path of investment that would maximise their profit over time. ln
roots; repeated roots and complex conjugates) at once, the general solution will also comblnt
short, the objective is not one of static optimisation 6ut dynomic optimisotion. Optimol control
the three methods for the three different kind of roots For instance, a fifth degreo
Theoryis a technique that is used to solve such dynamic optimisation problems
characteristic equation emanating from a fifth order difference equation can have the following
possibility. 12.2 An lllustrative Example

Distinct Roots Repeated roots (pair) Complex conjugates(pair) Let us start with a simple static production function
5 0 0
1 0
a = f(K)
where Q is mineral output and r( is capital that includes a slew of mining equipment. Let p be
3 0 1
the price per unit of output and c the unit cost of capital. Then the profit (n) will be given by:
1 1 1,

ft(K) - p. l(R) - cK
We already know that the first order condition for maximum profit is

n'(K)-p.f'(K)-c-0
But now, the firm may not be interested in simply maximising its current profit but the sum of
discounted profits over a per;od of time between now (f = 0) and a stipulated time horizon, T.
It would then want to maximise the function

T
r
sl/{(r)l = e-'LnlK(t)ldt
J
0

where r is the rate of discount and e rt is the discounting factor.

256 DYNAMIC OPTIMISATION: AN INTRODUCTION TO OPTIMAL CONTROL THEORY Mathematical Optimisation and Programming Techniques for Economic Analysis 257

S[K(r)] is in fact not a function as we understand it. lt is more appropriately called a 5ubjectto k = t(t) - d1((t) and K(0) = Ko where /(s isthe initialcapital stock
The distinction between a function and a functional is as follows.
Ihus, it will be clear that in order to maximise s given K6, the problem is to choose a time path
A function maps a single value for a variable like capital l( into a single value such as cu of investment, /(t). For once the path of /(t) is determined, the path of K(r) will also be
profit n A functional maps a function like /((t) into a single number like the discounted srrrrr determined given K(0) = Ko and the optimal value of s will be solved. The technique of
profits. To put it in other works, one has to choose a function of time, K(t) (a time path rrf l( choosingthe path of/(f) is optimal control theory.
values) to maximise S and not just choose a single value l( to maximise profit z.
12.3 Concepts Relating to Optimat Control Theory
But at this stage, the important thing to understand is that the maximisation of the sunt (,1
Let us begin with a general formulation of a dynamic optimisation problem. The problem is:
discounted profits does not necessarily mean dynamic optimisation! True, one has to choora
T
not just a single output but a stream of outputs that would maximise profits over time. Bul ll I
max.9 | ffx(t),y(L),t)dt
current output affects only current profits, then in choosing current output, one is concernad
I
only with its effect on current profit The solution to the optimisation problem in such a cnr!
would be nothing more than a sequence of solutions to a sequence of static optimisatlon subject to i = glx(t),y(t),t1, x(0) = 1o ;, g
problems.
The solution to the above problem is guided by a set of necessary conditions emanating from a

lf however, current output affects not only current profits but future profits, then in chooslrrg principal known as Pontryogin's Moximum principte, after the twentieth century soviet
current output, one has to be concerned with its effect on current and future profits. 'l'he mathematician Lev Semenovich Pontryagin. These necessary conditions are stated in terms of a
problem now becomes a dynamic one. Recall once again our quote in Chapter 2 from Silberborg Homiltonion function But before we come to this function, it is necessary to define the terms in
and Suen (2001) on what makes the problem dynamic the above optimisation problem.

Now, why and how would current output impact on future profit? The answer is this: currcnl o S is the value of the function to be maximised;
profit depends on current output. Current output depends on the amount of current capltal . x(t) is called the stote vorioble;
used. Although capital is durable, it has a limited lifetime Capital stock depreciates over timrl, . y(t) is called the control vorioble;
The more it is used to produce current output, the faster will be its physical wear and tear and ' x(T), the final value ofthe state variable is called the endpoint. lfthis final value is fixed,
shorter its effective lifetime. lt will have to be replaced through new purchase of capital it is known as a t'ixed endpoint. lf the value is unrestricted and free to be chosen
(investment) which will push up cost and cut into profits. optimally, it is known as a t'ree endpoint.
ln the example we have been dealing with, the amount of mining equipment (capital) the
Let l(t) be investment or amount of capital bought in time t and 6 be the rate of depreciation ln
company has, is the state variable The amount of investment the company makes is the
capital stock k is the change in the capital stock K(t) available is time t. Then, control variable. The problem has a free endpoint since no limit is placed on the amount of
k=/(t)-aK(t) capital stock. lf such a limit had been placed, it would have been a fixed endpoint problem.

The above equation means that at any given point in time, the firm's capital stock increases by The solution to the problem of maximising S is tantamount to finding the optlmal solution path
the amount of investment and decreases by the amount of depreciation. for the control variable, y(t).
Suppose cu(f)] is the cost of investment 1(t), then the profit at time t is given by the
functlonal 15.1 The Hamiltonian function and Necessary Condition for Optimisation
T
I The Hamiltonian function H, forthe problem specified in the preceding subsection is
SU(t)l - J| e-rLnlK(t) Ilt)ldt
0 H [x (r), y (t), 1(t), t] = f [x (t), y(r), r] + t(t) s lx (t), y (t), tl
258
I

| DYNAMIC OPTIMISATION: AN INTRODUCTION TO OPTIMAL CONTROL THEORY Mathematical Oprimisatbn and Prograrrming Techniclues fbr Economic Analysis 259

ln this equation, ,1(t) is called the co-stote vorioble or the shodow pricing function for x Nowi = -9!=
Ax -t
The optimal solution path for /(t) is one that satisfies the following necessary conditions: We now have the two linear differential equations

i L=o t: -1
..
dy
d) ; dH
x :27
at Ax
The boundary conditions are.x(0) = 2, 1(3) -- O
... Ax
iii -= i:;: AH slx(t),y(t),tl
The solution to the first differential equation is 2(t) : C, - t, where C, is an arbitrary
iv, x(0) = ao constant of integration Since the boundary condition 2(1) = 0, has to be satisfied, we
v. r(T) :0 will have Cr = 1. Hence
The first three equations together constitute the maximum principle The last two are called
boundary conditions. Note that one can readily discern a similarity between the t(t1=1-'
condition of the maximum principle and the Lagrangean function that was introduced Substituting this value in the second differential equation *. : 2), we get
Chapter 8 to solve constrained optimisation problems. The boundary condition in equation
x=Z(t-t)
assumes that x(7) is a fixed endpoint. lf it is a free endpoint, the equation would
substituted by 1(I) 0. This is called the tronsversolity condition.
=2-Zt
= this gives the solution x(t) = C2 + 2t - t2 which using the boundary condition on x will
12.4 Sufficient Conditions give C, = 2. Thus, the time path for x will be

The maximum principle provides only the necessary conditions for optimisation. lt can x(t)=2+2t-t2
shown that these conditions will also be sufficient if the following are satisfied: Given the time path of 2, we also substitute to get the time path for y. We get

i. f (x,y , t) is differentiable and jointly concave in x and y Y(t) : 1(t)

ii. Any one of the following holds; =l-t
as the solution path to the control variable. At f : o, y(t) = t. lt then declines over time
. g(x,y, t) is linear in (x,y);
and finishes at y(1) : 0.
o g(x,y,t)isconcavein (r,y)andl(t) > 0, t e (0,7);
. g(x,y,t)isconvexin(x,y)and,t(t) < 0, t € (0,I) 12.5 Economic Application: A Mining Problem
An investor has bought a mine and has been given rights to extract the ore between date 0 and
Example 12.1
date T. At time 0, there is t0 ore in the ground. The stock of ore remaining at any time t, r(t)
Solve maxlol(r - y2) dt subject to i : 2y and x(o) :2
will depend on the rate of extraction of the ore. The cost of extraction i, C = '(,t),'. The market
x(t)
We form the Hamiltonian function which takes the form price of the ore is p
H lx (t), y (t), t) : f lx (t), y (t), tl + t(t) s lx (t), y (t), tl
^(t), The decision to be made by the owner is the optimal rate of extraction in time t, y(t) that will
ThusH= x-y2+7.2y maximise profits over the period of ownership. For simplicity, let us assume there is no time-
The necessary conditions are discounting. The problem now is

AH
,#lr,
Thisgivesy=1(6)
ay:o=-zY+21 maxz:
il, ,u,
subject to x(t) : -y(t)
260 DYNAMIC OPTIMISATION: AN INTRODUCTION TO OPTIMAL CONTRoL THEORY

We form the Hamiltonian function LI - p ' y(t) - +1(r) - 2(r)y(rl

The necessary conditions (rearranged) are:

dH v(r)
- 'r 2:-
x(t) -
-- 0
dy '4(l)

. .tH
/(i)_ * _ /y(l)\
[r.,1l
x(0) - ro' 't(T) : 0

The two solutions to the differential equations are

. lP-l(r)l'
^\t)- +

Yf.t -'t.t)4 2

I
r
 Mrthcmatical Ol)tirrisrtion and Pr ogr antrning Techniclues tirr Econontic Analysis 263

Chapter 13
13 TINEAR PROGRAMMING

l:1.1 Introduction

lhe constrained optimisation problems studied in Chapter 8, involved an equality constraint. ln

,rddition, the objective and constraint functions were not simultaneously linear. That is, if the
objective function is linear, then the constraint needed to be nonlinear. This condition was very
r;ardinal, for a simple reason. The method of finding the optimal value relied on classical
rnethods of optimisation, based principally on the calculus. lt involved finding a point of
Langency between the two functions.

ln practice however, a firm is not obliged to use up all the available resources. A consumer is at
liberty to consume less than is actually at their disposal. These statements imply that the
constraint should not always be equality. Moreover, the indivisibility of certain commodities
rnake is impossible to spend all the money. For instance, consider a consumer with K5 income
to buy apples costing 1(2 each. An apple is indivisible, that is, it cannot be cut into half or
anything less than unit. The consumer then can only buy utmost two which leaves some
resources unutilised.
Further, the methods of Chapter 8 fail to deal with cases where the objective and constraint
functions are simultaneously linear. With linear objective and constraint function, the
respective slopes are constant Therefore, they are either equal throughout or never equal at
all This inhibits the use of tangency rule of optimisation. A more appropriate approach is
necessary to deal with such practical cases.

This chapter therefore brings out a non-classical method of optimisation known as linear
programming. As the name may suggest, this method deals the cases where both the objective
and constraint (inequality) functions are linear. The constraints are of the lorm g(x,y) < c
ratherthan g(x,y): c.

13.2 What Is Linear Programming?

Define the process by which an input is transformed into an output as an activity Thus an
activity is a process. lt links inputs to output. For instance, production is an activity whereby
inputs such as land, labour, raw materials and other factors of production are converted into
outputs of final goods and services. Consumption is an activity where final goods and services
are transformed into utility or satisfaction to the consumer.

Suppose an activity.A transforms inputs / into output O. The activity remains the same but it is
possible to alter the level of inputs. ln particular, consider
A
Ir )Ot
A
Iz -Oz
 I

266 I LINEAR PROGRAMMING Mathcrrtrlicrl OpLin)isrtioD antl Pro.trrrnrrinc'l'ccltniqucs tirr lJcononric Allrll,ris 267

K2.00 Themonthlydemandforthenextsixmonthsisforecastat800, 1000, 1400, 1500, 110(l F1 F E F^ F.

and 1550 units respectively. What ls the production schedule for the firm?
L 1,2 t4 15 10 4
13 4 1 7 Formulation of Problem 1 10 5 10 6
The optimum production schedule would be one that meets the monthly demands at minimurl L 9 8 0 5 9
cost The total cost would include cost of regular production, cost of overtime production artrl
L 7 6 2 3 l
storage cost. But since regular production costs have to be incurred anyway, one needs lrr
consider only storage costs and additional cost of overtime production to minimise total cost,
LeI X,, Xr, X=, Xa, X5, X5 be regular production in the six months The supply at each of the five factories is 1500, 4000, 3000, 2500 and 3500 respectively. The
demand at each ofthe places is 6000,3000, 1500 and 2500 respectively. The firm must meet all
Y1,Y2,Y3,Y4, Y5,l/6be overtime production in the six months.
these demands at minimum transport cost. What transportation strategy will minimise the
21,22,23,24,2., Zobe number of units held over f rom each month to the immediately following cost?
month
13.4.2.1 Formulation of Problem 2
p= unit penalty cost for unsold items
Let X,r= 3 6orn1 transported from factory F; to the place of demand L1. f hen,
Then the problem is:
: Minimise C= lZXt + I\Xn + gx1s + 7X7+
Minimise C 7 t?=ryi + ZL?l Z 1 -t pZ6
+].4x21+ 5x22 + gxn + 6X24
Subject to:
< +l-5x31 +'1Lx32 + 10x33 + 12X34
Xi 1000,i - 1,... b)
y, < 500,i = 1,.. .. 6J Productionlimitations +10x47 + 10x42 + 5x 4 + 3x 44
+4X1+6Xs2+9X*+7Xs4
X1+Yt-Z =800t Subject to:
X?+Y) b=l0O0l xlt+xn+xn+x,o<15001
X, + Y, Z" - :aOOI x2t + x22 + xB + x24 < 4000 |
X4 + y; Z't : l5O0l MonthLl' sales storcLge baLottce eqllatlon X3r + X32 + Xr + X34 < 3000 | Suppty conLraints
x., + Y\ - z5 :.:,ool x4t + x42 + x$ + x44 < 25ool
X,-+Y,.-2,.- 15501 xsL+ xsz+xs3 + xs4 < 3500./
x,>0. vtr X11 + Xzt+X3r + X4t + Xsl > 6000'l
f, , O, Vi f ,Vu ncgotiutly r(,slri( lir)ns. xp+ x22+ x32+ x42+ xs? > 3000 |
z,>0. vi) xa3 + xn + xT + xB+ xr, > tsool
D"*o'd requirements
x14 + x24 + x34 + x44 + xs4 > 2500)
1342 Problem2 Xt;20. i:1....5:)'l
' non - negatiuity conslraints
A blg firm has five factories which are manufacturing a product which is to be sold at four
i = j,...4
places The cost of transporting one unit of the product from any factory F; to any place 17 is This is developed into a transportation model. The model is discussed in Chapter 14.
shown below

1343 Problem3
A company owns a small paint factory that produces both interior and exterior house paints for
wholesale distribution. Two basic raw materials A and B are used to manufacture the paints.
268 I LTNEAR PRoGRAMMTNG I Mathematical Optimisation and Programming Techniques [br Econorric Analysis

The maximum availability of A is 6 tons a day; that of B is 8 tons a day. The a.ity r.qrlr"ml Let X be the amount of cereals consumed and Y represent the accompanying Meat consumed. X
of the raw materials per ton of interior and exterior paints are shown below. }I and Y are food activities which makes a program. The two commodities must be bought on the
market. Assume the price of cereal and meat is given by P* and Py respectively.
xterior nterior
The objective is to minimise the cost C : PxX + Pyy. With no lower limit on the quantities,
law materialA 1 2
one would decide to starve, by not buying anything at all The cost will be at its lowest, at zero.
law material B 2 I This, however, is not tenable because the continuation of life requires that certain amounts of
nutrition are consumed. Suppose now to maintain a healthy life, b1 of each food requirement
must be consumed. lf any nutrient is not necessary, a value of zero is assigned to it.
A market survey has established that the daily demand for interior paint cannot exceed that
of Therefore, the cost C = PxX + Pyf must be minimised subject to minimum nutrition
exterior paint by more than 1 ton. The survey also shows that the maximum demand for
interior paint is limited to 2 tons daily. The wholesale price per ton is K12 million for exterlor requirements, that is;
paint and K8 million for lnterior paint. How much interior and exterior paints should
tht arrX I ar2Y >- b,
company produce to maximise gross income? arrX I arrY )- b,
a=.rX I o32Y >- b,
13.4.3 1 Formulation of Problem 3 aorX I oorY >- bo
Let XE= tons of exterior paint produced daily and Xr= tons of interior paint produced daily. a5rX I a52Y >- bq
aurX I au2Y >- bu
The objective is to maximise total revenue R = TZXe * gX, subject to: The quantities of the two commodities consumed cannot fall below zero. That is, X > 0, y > 0
xE+2xr<6)
zk, + X', < g] 'o* material uuoilabilty constaints Optimal program: This is one that minimises the cost function. A vector which fulfils all the
constraints is called a feasible vector. There can be many feasible vectors but only the one with
-xE+x1<7',1 demand resliction\ a minimum cost is optimal. Thus the equation is solved using two things; feasible program of
X, <'2 J feasible vectors and select the optimal feasible program.
x->0)
i,; oJ non - negatiDity constrainls Example 13.1

1.5 L 1. 1 Problem 4 (Diet Problem) A firm produces three types of furniture. These are Bookshelves, TV cabinets and Dining
tables Let X, be the number of bookshelves, X, the number of TV cabinets and X3 the
suppose a meal consists of two goods; a cereal and meat, to get the requirements of proteins,
number of Dining tables. The market prices for the three products are Pr, P2 and P3
fats, calcium, iron, vitamin A and B. The table below shows the nutrition value (quantity per
respectively. the objective of the firm is to maximise revenue
unit of good) of the two ingredients of a meal
R = PtXt+ P2X2+ hX3
Cerea I Meat
Two inputs are required, in varying quantities to produce the three furniture, wood and
b, Protein a otz labour Though the firm can get different units of inputs are a constant prices, the supply
b, Fat of the two inputs is limited. Suppose there is only b, of wood and b, of labour available to
the firm The input-output matrix is given in the table below.
bl Calcium a" Oa)

bn lron a a a.c
Wood Labour

b( vit A Aq Bookshelf a a

b6 vit B 4.1 TV cabinet att att

Dining table ata atz

2]O LINEAR PROCRAMMINC Mathematical Optimisation and Programming Techniques for Economic Analysis 271

The objective is to maximise revenue R - P.X, + P2X2+ P3X3 subject to avail,rlrtltly c) The matrix of coefflcients of the constraints in the problem is transposed to obtain the
inputs The constraints are based on input availability matrix of coefficients of the constraints in the dual problem.

o11X1+anx2+o,.X.,{b, d) The sign of the inequalities in the constraints of the primal are reversed in the dual.
lrom (b) and (c) it can be understood that ifthere are m program variables and n constraints in
u,,X, I Lt",X" I Lt ,,X" < h, lhe primal, there will be n program variables and m constraints in the dual.
The feasible production must fulfil all constraints lil.5.1 Theorems on Duality
13.5 Duality The following theorems state the relationships between the primal and dual programs,
rspecially in respect oftheir solutions.
Consider the general form of the Linear programming problem:
i) The dual of the dual is the primal;
Minimise C'X
ii) lfeithertheprimal orthedual hasafiniteoptimal solution,sodoestheother;
subjecttoAX>B,X>O iii) If feasible solutions to both the primal and dual systems exist, then both have optimal
solutions;
iv) lf either problem has an unbounded optimal solutions then the other has no feasible
solution;
v) Both the primal and dual may be feasible;
vi) lf (ii) holds, then the objective functions of both problems have the same optimum
value

13.6 Graphical Solution of Linear Programmes

Unear programs can be solved using two methods. The first is the graphical method which
!mploys graphs to find the optimal solution. The second, and more robust, is the Simplex or
llgebraic method This section deals with the former and the latter is discussed in the
tucceeding section. The graphical method involves plotting all the constraints on one graph
which defines a feasible solution. Owing to the difficulties of drawing graphs of more than two
dlmensions, the discussion of this method is restricted to problems having two choice variables
only.

Example 13.2

Consider the following linear programming problem in mathematical form:

Maximise 6Xr + BXz subject to
2X1+X2<lO
x1+x2<6
x7+3x2<14
xt> 0, Xz> 0

Draw all the three constraints or feasibility conditions on the same graph. lf for each
constraint the unfeasible side is shaded, then the feasible region which satisfies all the
b) The right hand side constants of the primal problem become the coefficients of thn
constraints would emerge. lt is the five sided polygon lying on the left of all the
objective function of the dual problem and likewise the coefficients of the objectivt
constraints. lt is the region bound by the origin and the points A, B, C and D.
function functions of the primal problem become the right hand side constants of tho
dual problem.
 I

272 | LTNEAR PROGRAMMTNG Mathematicar optimisation and programming Techniques

for Economic Anarysis | ,-r..
Figure 13.1. lllustration of feasible solution though an extreme point is obviousry triviar and
need never be considered) need to be
examined.

Extreme point Value = 6Xr+BXz

2
A: (0,a5) 6(0)+B(-?=rI
B: (2,a) 6(2)+8(4)=44
A (0,4.7 C: (4,2.) 6(4)+B(2):40
D:(5,0) 6(s)+B(0)=30

Thus the objective function is maximised

with X1 = 2 and. X, = 4. The maximum value of
the objective function is 44.
ll r program had been a minimisation probrem,
Any point falling in the region, satisfies all the three constraints. Since the two rr,r our search for an
urdbetantamounttofindingthepointonthefeasibreregionwhich
cannot be negative, the vertical and horizontal axes are in fact boundaries li'", ne. The preference directioi wourd have been
respective sides. A feasible vector for this linear program is any point (Xr,Xz) towards ihe origin,
wlrr nimisation. This can be seen in the graph corresponding
satisfies all the three constraints simultaneously ln other words, such a point l,rlL
to the
the region, which may be called the feosible region, which at once satisfies all the
inequalities. I x.unple 13.3

ln the figure above, the feasible region is a closed, convex set. lt contains infinity of Minimise C : 20X * 40}/ subject to
or feasible solutions or program *oooff'], from which the optimal solution is 2X+Y>B
found. Since the objective function is one of maximisation, we select the point 4X+3Y>72
much away from the origin and upwards and to the right as possible. That is, x+3Y>9
preference direction is towards the north-east. lt is clear that the optimal point X>0, Y>O
somewhere on the boundary of the convex set. To find this, consider the iso-profit Again, draw the three constraints on one graph.
From the graph, identify the feasibre
solution. This is the region which satisfies all ihe
The objective is to maximise 6X1 + BX2. The unit value of X, is 6 and that of X, is 8. constraints.
is, their unit values are in the ratio of 3:4. This means that 4 units of X, would give
same profit as 3 units of X2 Hence the straight line joining points (4,0) and (0,3)
the other points or combinations of X1 and X, which give the same profit To
profit (objective function) subject to the constraints, push the iso-profit line
outwards and see which point is in feasible region that falls on the highest iso-profit
ln the diagram above, the optimal solution is at point B The coordinates of B, (2, 4)
the optimal values ofX, and Xr.
We have already stated that the feasible region is a closed, convex set lt can be
(though we shall not prove it) that an optimal feasible solution is one of the
points of this convex set. Thus, although there is an infinity number of feasible
to locate an optimal solution only the points A, B, C and D, in the diagram (the origin
274
I

| LTNEAR PROGRAMMTNG
Mathematical Optimisation and Programing Techniques for Economic Analysis 275

Figure 13.2. lllustration of irrelevant constraint

l.l7 2 Case 2: Multiple optimal solution
llris situation occurs ifthe slope ofthe iso-profit line (or iso-cost line) equalsthe slope ofthe
lroundary edges of the convex feasible region which is in fact the slope of the line
r orresponding to one of the constraints. This is depicted in the following two diagrams.

I igure 13.3. Case of multiple optimal solutions

One feature to be noted in the above graph is that the second constraint 4X + 3
not play any part in determining the feasible region. lt entirely lies outside the
region. Hence, it is irrelevant for solutions of the linear program.
We can examine the three possible points ,4, B and C and look for one giving the
cost because the problem is that of cost minimisation. Point.4 gives 320, point
ln each ofthe above diagrams, points a and b and all other points on the boundary edge or line
140 and point C gives 180. Thus point B is the optimal point. lt has the lowest cost.
segment connecting a and b represent the optimal feasible solutions.
Also recollect what we had said about closed and open sets in Chapter 3. ln the
13.7.3 Case 3: Unbounded optimal solutions
diagram, the feasible region is physically open but is nonetheless a closed convex set.
These are the kind of solutions that are not bound, in other words not exact as in the above
13.7 Nature of Linear Programming Solutions scenarios demonstrated. The necessary but not sufficient condition for this situation to
occur is that the feasible region be unbounded.
It was already stated in an earlier section that the optimal solution to a linear program
be finite and unique There are other possibilities. These are illustrated graphically.

13.7.1 Case 1: No feasible solution

Example 13.4
Suppose you have the following problem: Minimise crXr l crX, I caX3
Minimise Z : 10X + 15y subject to Subject to orrXl l or2X2 I agX, >- b,
x+2Y>8 arrX, + o22X2 * arrX3 )- b2
2X+Y>t0
4X+3Y<1,2 xr x2, x3>0
x>0, y>0 Maximise 8rY
There is no single point that will simultaneously satisfy the second and third constraint. Subject to.4rY < C
therefore, no point which can at once satisfy all the three constraints. ln other words,
y>0
feasible region is an empty set
276
I

| LTNEAR PROGRAMMTNG Mathematical Optimisation and Programming Techniques tbr Economic Analysis I Ztt

Example 13.5 rr..rr:hing the optimal solution. lt arrives at the optimal solution through an iterative process or
,rlrlorithm. The idea is to start with an initial feasible solution and go on improving the solution
Maximise brYt + b2Yz
,rl r'.rch iteration until no further improvement is possible and the optimal solution is arrived at.
Subject to arrYl * a21Y2 I C, li,lore illustrating this method, there is need to introduce a few prerequisite ideas and
r r Itcepts.
arrYr+orrY, 1C..
I I ll.1 Prerequisite ideas and concepts for the Simplex Method.
ar=Y1*anY2<C3
Y!, Y2>o i) Basicvariables are variables in a linear programming problem which are solved from
the constraints of the problem in terms of other variables which are called non-bosic
The difference ls that the program variables are different between the primal arrrl variables. Since we require as many equatlons as there are unknowns to be solved
problems The coefficients of the ob.iective function and right hand side constraints havt,
for, we can obviously have, at any time, only as many basic variables as there are
interchanged. The signs are different. There is also interchange of rows and column. ln
constraints. The collection of basic variables is termed the bosrs and the remaining
if you have n program variables and m constraints if ,,r < m, it is easier to solve than whcrr rr group of non-basic variables, the non-bosis.
m. Fewer program variables make a problem easier to solve. The program variables ol ll ii) Bosic solutions are solutions of the basic variables that result from equating the
primal and dual are different.
values of the non-basic variables to Zero. Basic solutions which satisfy all the
Maximise P,Xt + P2X2 + P3X3 Primal constraints are known as bosic feosible solutions (b.f.s.). These solutions correspond
to the extreme points of the convex feasible region.
Subject to
To find the optimal solution, only basic feasible solutions are considered. This is
arrXy*ar2X2*arsX3!b1 based on the Eoslc Theorem in linear programming (whose proof is not provided
a.rrX1* arrX, + a4X3 3 b2 here) which states that if a feasible solution exists, then a basic feasible solution
Xt> 0, Xr> 0, xz> o exists and that if finite optimal solutions exist, then at least one of them is a basic
This is a resource allocation problem amongthe produclsXr,Xr,Xr. feasible solution (b.f.s.). That is to say, if a unique optimum solution exists, then lt is
a bosic feosible solution. lf multiple optimum solutions exist, at least one of them
Minimise brYl + bzYz ......dua1
must be a b.f.s
Subject
There could be non-basic optimal solutions. Reverting to our diagram of case 2
arrYrlarrYr) P,
examined earlier, points'a'and'b'represent b f.s. because they are extreme points.
a12Y1*o22Y2) P2
Points lying on the line segment connecting 'a' and 'b'though optlmal solutions are
a1sY1*. a6Y3 ) P,
actually non-basic. They are non-basic because they are not extreme points. They lie
Yr>_0, Yz>_0
on a straight line and not a corner.
b1 and br= are units of resources
iii) Slack ond surplus voriobles:
Ir= price of one ton of e.g. labour Consider the inequality
YrandY2 are the input prices. arX, I a2X2 + ......+ anXn 1 b

This is an input pricing problem. arrY, is the price of producing one unit of Yr. ln actual sensa, The left hand side (LHS) is less than the right hand side (RHS). The actual difference
the constraints may encompass the equal sign due to the fact that if firms make super normel is not specific and can be treated as an arbitrary number. lf a non-negative arbitrary
profits, assuming perfect competition, the profits will be normal again. number assumed to be equal to the difference is added, then the inequality turns
into equality. The two sides of sides will now be equal since the lower side has been
13.8 The Simplex Method (Algebraic) raised by an arbitrary number equal to the initial difference. This will make up for
the difference, that is, how less the LHS was to the RHS
The algebraic analogue of the graphical method of solving a linear programme is known as tho
simplex method. lt was formulated by George Dantzig, the late American mathematician. ln th! arXr* a.rXr+'.....+ anXn*S: b, S > 0
former method, locating an optimal point required examining all the extreme points. These aro Similarly, the inequality below has the RHS less than the LHS
vertexes of the feasible solution shape. The simplex procedure is even more efficient in that lt
avoids examining all the extreme points. lt reduces on the points to be examined beforo alxt+ a2xz + """ + anXr) b
 Mathematical optimisation and Programming Techniques tbr Economic Analysis | ,r,
278 L N EAR' -:;,
coefficients in the objective function, thus increasing them will improve the objective

;;;I: ,*: ;r::H ff iiff i ril:ff i*:'ff "r;rtt :: .*:

I function. However, the simplex procedure allows only one variable to enter the basis at a
time. Further, to make room for the variable entering the basis, one of the existing basic
variables must leave the basis and become non-basic. The number of variables in the
constant from
""j:I:H;l: ::::l *" :l'1",'"'",
=
o il solution is limited by the number of equations will directly depends on the number of
constraints.
ln the resulting equations, s is called a slock variable and t, a surPlus varlab[
variable' But the model does not state which variable will enter and which one leaves. The
surplus variable may be regarded as a negative slack
,,f guidance nonetheless exists. Since the objective function is one of maximisation, then the
13.8.2 lllustration ofthe Simplexprocedure:
I

f, variable that add most is considered first. By adding most, we mean the value of the
To illustrate this method, it is much easier to make use of examples. Since
the granhical andf, coefficient since it measures how the objective function changes with a change in the
will as a strrtl concerned variable. Between the two variables, r, has a coefficient largerthan that ofxr.
simplex methods must ideally arrive at the same conclusion, this subsection
the former method. This will provide a check on the solutions at A unit of x2 will add 8 units to the value of Z while a unit of x, will add only 6 units.
use examples solved with I
as provide a detailed comparison of the two methods' I But which variable must leave the model to create room for xr? The strategy is as
follows The existing variables should be such that the solution corresponding to the new
Example 13.6 basis should be feasible. Suppose x, is made to leave the basis. Then our current basis
Revisit Example 13.2. Solve this example using the Simplex method' would consist ofx2,S, and S, and non-basis ofxr,Sr.

Maximise Z = 6xt*8xz From equation (1) we will gel

x2: 10. But if this value is substituted in equations (2)
Zx1lx2<lO and (3) it will give Sz= -4 S=: -1 respectively. These values violate the non-
and
negativity restrictions on xrandxo and hence is not feasible. Thus S, cannot be the
xr*.x236 i
exiting variable.
xa+3xz<14
Xt, xz )- 0 lf 52 is taken out of the basis, from equation (2), we will get the value oI x2 as 6. lf this
restate the problem by converting the constraint inequalities into value is inserted in equation (5), we get S: : -4. Thus 52 too cannot be made to leave
The first step is to
equations by adding slacks. The problem now is thebasis lf S, leavesthebasis, *r=t=+2=.Usingthisvalue of x2wegetSl:S]trom

Maximise Z = 6xt I Bx:- * 0x3 + 0r4 + 0x5 Subject to equation (1) and 52: fl. fnis is a feasible solution. our new basis is, therefore,
Zxrlxr*S, =10 lxr,S,,Sr) and non basis (xr,Sr) .the value of Z = fol (Sj) + (0) (1:) + fsl (+j) :
x1+xz * Sz =6 37!
x1*3xrf S: =14 3

We can improve the value of Z further by including .x1 now into the basis. Which variable
rr,12,Sr,52,53 2 0
will it replace? The choice is between S, and Sr. Supposei S, is chosen to leave the basis
We must start with an initial basic feasible solution. Usually, as a convenient starting (xr,x2,Sr) will constitute the new basis. We solve for x1,x2,52 in terms of S, and S=.
point, the slack variables are taken as the initial basis. This basis is, of course, trivial in the From equations (1) and (3) we get,
problem then, we
sense that the value of the objective function would be zero. ln our
Zxr+xr=19
beginwithSr,52and53asthebasis.Equatingnon-basicvariablesxlandx'tozero'we
get x1*3xr=lQ
1) 51= !0-Zxr-xz=10 Solvingthesetwoequationswegetrl =3l,xr=:l,itthesevaluesaresubstitutedin
2)Sr:6-*r- xz =6 equation(2), we will get Sz=6-ri-ri=-:.0. this is not a feasible solution.
3)Sr:14-xr-3xz=!4 Hence S. cannot leave the basis. lf 52 becomes part of non-basis, we shall now solve
equations (2) and (3) simultaneously. We get
z = (0)(10) + (0)(6) + (0)(14) = 0
The question now is: by including any non-basic variable into the basis can
the value of z xr*x2:6
improve?Theanswerisofcourseintheaffirmative.Bothxlandx,havepositive
 Mathematical Optimisation and Progranming Techniques fbr Econornic Analysis 287
280 | .-,rro* PR.GRAMMING

x1*3xr=\Q ',rrr,,lant in each constraint, or the Right Hand Side (RHS) of the constraint. ln the case of the
lrr',t (onstraint, this is br. ln the CBV column is the Coefficient of the Basis Variable in the
These equations yield 11 :2 and x2-- 4' And with these values, from equatlon ,'lrlr,rtive function. ln the initial basic solution, it is the slack variables that form the basis
set 51 : 2.The value 7 : (6)(2) + (8X4) + (0X2) = 44' Since the current
"1 ',hrt ron. Since these are not part of the objective function, the CBV will have zeros only. These
consists of the variables s2 and s3 and since both these have zero coefficit'tllr ltt wrll r hange as some variables enter and others leave the basis.
objective function, the value z cannol be improved further and hence is optimal,
llr, last row of the tableau is the Marginal Net Gain (G) for each variable including slack
The simplex method can also be worked in a more summarised way using the simplcx I v.rrr,rbles lt is calculated as follows
ltt
Assume the following linear programming problems with slack variables already addt'tl
inequality constraints into equality constraints. Gj = rj - lorlCBV, * o4CBV, * arlCBV, + oolCBVnl
maxZ=caxr+czxz+cax3 l or each constraint coefficlent in the column, multiply by the CBV in that row. Then subtract a
subject to of these from the Objective function coefficient. The marginal net gain shows how much
',rrrrr
tlrr,objective function would gain ifthe particular variable enters the basis.
a1rX1 I ar2X2 I apX3l S1 = b,
+ or=X3 * S, = b2 llr(' next step involves identifying the column with the highest Net Gain. Amongst variables not
arrX, * a22X2
rrr lhe basis, determine one with the highest positive value. The chosen column becomes the
a3rX31 * or2X2 I aTX3 + 53 = b3 ;rrvot column. The variable in that column will enter the basis. lf all the Net gains are non-
aa1X1 I oa2X2 * oa3X3 * Sa = ba t,o\itive, it means no change of variables will improve the objective function. This entails the
olrtimal solution has been reached. The searching ends
xa,x2,x3 > o
all tlre orrce the pivot column has been identified, we turn to the Ratio column (last column ln the
Since we have to start from some feasible solution, we start from the origin, where
',rrnplex tableau). This column is blank in Table 13.1. For each row, divide the RHS by the
are zero and form a special matrix called simplex tobleou as follows
r oefficient in the pivot column. This will give the ratio for each row. Then select the row with

llre lowest (positive) ratio. This is the pivot row, which identifies that variable that will leave the
Table 13.1. The SimPlex Tableau
I r, r sis.
C1 C2 Ca 0 o 0 0 Obj fn Coef
X X3 s s2 ss s+ RHS CBV Ratio once the pivot column and row are identified, a few operations are necessary ln the pivot
X1
ott dtz otz 1000 ba 0 row, divide (multiply) all the elements in the row by an appropriate number so that the
dzt dzz ozs 0100 b2 0 roefficient in the pivot column becomes one (1). Then make all the other elements in the pivot
azt azz dsz 0010 b3 0 rolumnzerobysubtractingfromthatrowmultiplesofthepivotrow Rememberthatwhatever
O+t A+Z O+S 0001 b" 0 rrultiplication or division in a row must be done for all the elements in the row. Once this is
c1 c C3 Marsinal Net Gain rlone, one of the slack variables leaves the basis as one choice variable enters. Virtually all the
llements will change in the tableau.
ln Table 13.1, the first seven columns pertain to the seven variables (choice variables plul
slack variables) indicated in the second row for each column. The first row of the table
lr lhis process must be repeated until all the Net gains are either zero or negative. When all the
representation of the objective function. lt gives, for each variable, the coefficient in net gains are non-positive, the optimal solution is found and no further searching is needed.
objective function. Since slack variables are not in the objective function, the coefficients lhe optimal values of the variable will be read using the coefficient one (1) in each column
2ero. that is, to read the optimal value of the variable in thefth column, check the row in which the
coefficient one (1) in that column is. The RHS element in that row is the optimalvalue.
After the first twO rows, there are as many rows as are constraints, each row representing
constraint. in the particular case used above, there are four constraints, therefore four roWt, Let us now take an example. Since the graphical and the simplex methods must ideally arrive at
The element c17 is the coefficient of the lth variable in the ith constraint. The first line lot the same conclusion, this subsection will as a strategy use an example solved with the former
instance pertains to the first constraint method This will provide a check on the solutions as well as provide a detailed comparison of
the two methods.
orrXl * anx2 + apX3 I 51 : b1

Only one slack variable appears in this constraint with a coefficient of one. The other thrra Example 13.7
therefore appear with coefficients of zero in the simplex tableau. The column headed RHS is tha Revisit Example 13.2. Solve this example using the Simplex tableau
2A2
1

LINEAR PROGRAMMING Mathematical Optimisation and Programming Techniques for Economic Analysis 283

MaximiseZ:6xr*\xz The second row is the pivot row. Therefore, we have to turn2frto one (1) in the pivot
zxt+x2<10 column and all the other elements to zero by subtraction. Multiply every element in the
xr*x2<6 pivot row ty3fr.then subtract 5/3,i."r row 2 from row 1 and a third of row 2 from
xr*3xr<14 row 3. This operation will ensure that the rest of elements in the pivot column are zero.
Xt, xz) O The new tableau is:
ThefirststepistorestatetheproblembyconvertinStheconstraintinequalltlot B 0 0 0 Obi fn Coef
equations by adding slacks. The problem now is
\I x s .s s RH5 CBV Ratio
Maximise Z : 6xt + 8x2 + 0S1 + 0S2 * 0S, Subject to t) 0 1
-t/, -, /, 2 0

2x7+x2+Sr = 10
xr+xz+ 52 =6 00
10
'1, -'lz ) 6

x, l3x2+ .93 = 14 0
-rt- 7/_ 4 8
0 0 0 -5 -1 Marginal Net Gain
x1,x2,51,5253) 0
The Net gains are all zero or below. No improvements are possible. Therefore, the
As explained already, we start with the origin as the basic solution The simplex
optimal values 3r€ )fr = 2, x2 = 4 and S, : 2. The first slack being non-zero means the
will be
first constraint is not binding lt has a slack at the optimal solution. The value of
6 8 0 0 0 Obj fn Coef z=6(2)+8(a)=aa.
x1 A2 s s2 s3 RH5 CBV Ratio

Z 1100 10 0 10 l:1.9 Comparison hetween the Simplex and Graphical Methods

1 1010 6 0 6
lr)the preceding section, we solved using the simplex method the same problem that we had
1 3001 t4 0 4.1
Marginal Net Gain
',olved by the graphical method in section 13.6. By dint of the linear programming theorem
6 8
which tells us that if an optimal feasible solution exists, it will be located in one of the extreme
coefficiottl
Since all the cBVs are zeto, the Net gains will equal the ob.iective function lroints of the closed convex set constituted by the feasible region. We had to evaluate only the
eachvariableThehighestNetGainisincolumn2,thatis,variablex2'Nowdividelllr' lroints A, B, C and D in the graph in Example 13.2. But all the four points had to be evaluated in
by the corresponding element in the pivot column, column 2 to identify
the lowesl I
order to identify the optimal feasible solution. The Simplex method however has two
The last row is identifled. The row and column are identified and we
now look to ch,rt .rdvantages over the graphical method.
the 3 in the pivot row and column to 1. . Starting from an arbitrary feasible solution which is the origin (which will obviously yield
in row 1
Divide every element in the pivot row by 3' Since the pivot column elements
6
a zero value for the objective function), it leads us step by step (that is, by changing one
2arealreadyl,wesUbtractthepivotrowwithoutnecessarilygettingitsmultiplesSlm variable at a time) to improved solutions until the optimal solution is reached. The
row 1 less the new row 3 and the same for row 2' The new tableau will be' procedure gives us a clear indication when the optimal solution is reached so that we do
6 B 0 0 0 Obi fn Coef not have to evaluate any remaining extreme points.
X1 xz s 1
s2 s3 RH5 CBV Ratio o
There is one further advantage of the simplex method. ln the example provided, one
5/^o1o-r/,
t:, 'J
t6t
t3 0 3.2 could have moved to the optimal point B by moving from O to A and then to B or by
moving from O to D, D to C and then C to B. The direction of movement in the first case
,/zoor-r/t a/z 0 2 would have involved two steps whereas the direction of movement in the second case
1/, 1oo13- 14t
t? 8 t4 would have involved three steps. The first case therefore provides a shorter route from
'l0t o Marginal Net Gain the origin to the optimal feasible solution in comparison to the second case. ln here, the
/2 simplex procedure tells us to move in the direction suggested by the first case and not
ln the new tableau, the first column has the highest gain and therefore
the pivot coltllllll the second case.
Wethengeneratetheratiotodeterminetheslackvariablethatleavesthebasis' the above two advantages show that the simplex (algebraic) method is more efficient than the
llraphical method in solving a linear programme involving only two programme variables.
284
I

| uuenn PRoGRAMMING
Mathematical Optimisation and Programing Techniques for Economic Analysis | ,,,
Needless to say in linear programmes in
no option but to resort to the algebraic
Chapter L4
13,10 Recent developments in Solvi
I4 SOME EXTENSIONS OF LINEAR PROGRAMMING
13.10.1 Algorithmic develoPmen
proceeding frottt l,l.l lntroduction
The simplex method explained in the preceding section basically involves
vertex of the feasible region until the optimal solution is attained lrl
ld
vertex to another llrr, linear programming model discussed in chapter 13 can be extended to analyze more
scale problems this could be very time-consuming
(sometimes weeks in optimisation prolllH ,,nrplex problems. lt is itself a tool for an easy way to reduce complex problems to something
involvingcommUnicationnetworks)andhenceinefficientAttemptshavebeenm.lrlo rrrrrch easy to grasp and understand. ln this chapter, we apply the notion of linear programming
develop more efficient algorithms One such algorithm was developed
in 1984 by an lrrrl trr understand three models. The Transportation model is presented in section 142 while its
,,1rr:cial case, the Assignment model is presented in section 14.4. The Transhipment model,
mathematician Narendra Karmarkar.
wlrich is an extension of the Transportation model is presented in section 14.5.
1.5.1.2 Karmarkar's algorithm
an interior point me
Karmarkar's alSorithm solves linear programming problems by using I 4.2 Transportation model
von Neumann. lt cuts through the requirement of proceeding from
first invented by John ()ne area of application of the linear programming techniques is in the decision making process
the solt
vertex to another by traversing the interior of the feasible region. consequently,
to days. This in turn enables faster p lry firms. Suppose we are given one hypothetical firm with multi production plants and several
time gets drastically reduced. weeks could be reduced
ilrrrkets scattered across the country or across the world for a multinational corporation. The
decisions.
lilm has m production sites and n market points. The production sites and markets are looked
Karmarkar,s algorithm has stimulated the development of several
other efficient methodr ,rt as points because the main interest is the transportation problem. lt is pretty easy to define a
solving linear programming problems. lrrnsportation problem with specific points of origin and destinations. From what is provided,
book' ll\ere are m sources of the product to satisfy the demand in n markets. There is need at this
Discussion of all these algorithms is beyond the scope of this
.,tage to assume the product from all the production points are identical so that there is no
rostriction on sources and destinations. Output from any source can be transported to any
(lestin ation

I he different plants have varying production capacities. We denote output from the ith plant as
\, Total output for the firm will be a summation of various outputs from all the plants. This
( onstitutes the supply ofthe commodity and can be denoted as

s=f,,
2',
Ihe demand side is also made up of individual demands from various markets. Let us denote
the demand from the jth market as d;. The total demand by the n markets is given by

o=ia,
this is summarised in Figure 1.4.1 below
?-'
 I Mathematical Optimisation and Programming Techniques for Economic Analysis 287
286 I soME EXTENSIONS OF LINEAR PROGRAMMING

The objective of the model is to find the combinations of output to be transported that will
Figure 14.1. Source and Destination nodes results in the lowest transportation cost. Formally, the model is stated as

-,nii,,,,,,
subject to

i,,,
L' =,,
j'-7

L,,=r,
e"
That is, output from all sources is transported and all markets are supplied needed quantities
respectively.
When there is a dummy source or destination in the model, there is possibility that the dummy
distorts the optimal outcome. ln the optimum case, the dummy source or destination may be
assigned a quantity different from the quantity it meant to account for. For instance, suppose
there is a deficit of 100 units so that a dummy source is added which is supposedly producing
the 100 units. To ensure the dummy source is assigned L00 units in the optimum outcome, the
Part (a) of the figure shows all the sources supplying to destination (2). The sum
unit transportation cost from the dummy source to any destination note is assigned at zero.
shipments will equal the demand from destination (2). Part (b) shows it the other way
has one source (3) supplying to all the destinations. This means the total output at sou Different transportation routes have different per unit costs. Finding the optimal transportation
(3) must equal the sum of shipments from this node to all the destination nodes. strategy involves two stages. The first is to find what is referred to as feosible solution. Since
these methods do not always result in the optimal strategy, there is need to test whether the
We know that the firm cannot overproduce nor can it under produce, at least in the
feasible solution found in the first step is the optimal solution. This becomes the second step.
This means the quantity supplied equals the quantity demanded and the model is said
ba la nced. To find the feasible solution, three alternative methods are used. The first is the Northlvest
corner rule. The second is the /owest cost entry method. As the name suggests, this method
The model is not balanced when total supply does not equal total demand. There is
involves searching for lowest individual transportation routes. The third is known as Vogel's
deficit (when the demand exceeds the supply) or a surplus (when supply exceeds the
The linear programming model cannot be used in an unbalanced model. This is
Approximotion Method (VAM). fhis method usually produces an optimal or near optimal
starting solution.
linear programming assumes a balanced model.
L4.2.1 Northwest corner rule
As an option for the above problem of unbalanced model, a dummy source or desti
assumed in the model. When there is a deficit, a dummy supply is brought into the mod The objective of this method is to find the feasible solution rather than an optimal solution. lt is
make up forthe difference. The same applies in the case of a surplus where a dummy concerned with ensuring that all supply is transported without paying attention to the resulting
assumed. cost. ln the transportatlon tableau with sources indicated in rows and destinations in columns,
The next variable we define is the cost of transporting the output from the production this method requires starting with the northwest corner, that is, the first column first row.
the destination markets. lf each and every plant supplied to all the markets, there Exhaust that cell by transporting everything from the first source to the first destination. lf the
m x n transportation routes. ln reality however, production plants will not necessarily source is exhausted before the destination, move to the next row in the same column. lf it is
supply all the markets and markets do not have to receive from all the plants. The output
the destination satisfied before exhausting the source, then move to the next column in the
all the plants is identical so the specific source is of no interest. The cost of transporting
same row. Once a row or column is satisfied, move to the next until ending in the southeast
unit of output from source I to destination I is ci;. The total cost will depend on how
output is transported. corner.
288 | ,or, ,r.r*r,oNs oF LTNEAR PRoGRAMMTNG Mathematical Optimisation and Programing Techniques for Economic Analysis 289

This method is equivalent to the first-come first-served rule. lt deals with 14.2.2 Lowest Cost Entry method
destinations as they come without paying attention to the unit costs. When any llnce the objective is to minimise the total transportation cost, then we must use least cost
columns are interchanged, the method will identify a different feasible solution. Let toutes whenever possible. Given different available routes, with different unit costs, the
the following example. decision must be using one that is least cost. Only when such a route is no longer available,
because the source or destination in that particular route has been exhausted should the next
Example 14.1 hast cost be used. ln summary, given the cost matrix, exhaust the routes in the order of
The Ministry of Education has received funding to buy lockers for distribution tscending cost. This is easier to illustrate with specific examples as given below.

boarding schools in the following districts: Monze (50 lockers), Kabwe (65
Example 14.2
(25 lockers) and Chongwe (40 lockers). The tenders have been awarded to three
Zambeef is a leading producer of beef in the country with many abattoirs and retail
for each to make 50. One bidder is in Lusaka, another in Livingstone and another
outlets across the country. To make the illustration simple, assume there are only three
The unit cost of transporting each locker from each source and destination is abattoirs located in Sinazongwe, Chisamba and Chipata with respective outputs of 350,
table below. 400, 27O units of beef. The three market destinations are Lusaka, Livingstone and the
Destination
Copperbelt. The demands from the market are 380, 200 and 44O respectively. The unit
Monze Kabwe Kaoma Chongwe cost of transporting from each abattoir to a market is as given in the table below.
(s0) (6s) (2s) (40)
o -usaka -ivingstone )opperbe t
f Lusaka (60) 2 1.5 4 0.5 t80 100 t40
o
Livingstone (60) 4 6 7 5
iinazonswe 350 75 5.0 10 5
Ndola (60) 5.5 8 4
-hisamba 400 1.0 1.0 ;.0
The method requires starting with the first row first column. Transport the lhipata 270 7.5 4.5 3.5
possible from Lusaka to Monze. Lusaka has 60 lockers but Monze only needs
Determine the optimal transporting strategy.
transport the 50 to Monze and the whole Monze column is exhausted. Then move
There is a total of 9 transportation routes available with different unit costs. Of the 9, the
next destination (column). Transport the remaining 10 from Lusaka to Kabwe. Ka
Chisamba-Lusaka is the cheapest. Chisamba has a total of 400 units of the commodity but
remains with a deficit of 55 lockers and everything from Lusaka is taken. Lusaka only takes 380. So, 380 units is transported from Chisamba to Lusaka. Lusaka now
Next transport 55 to Kabwe from Livingstone. Kabwe is satisfied but Livingstone has all it needs but Chisamba still has the remainder of 20 units. Since the Lusaka
destination is exhausted, then the whole column referring to Lusaka must be removed as
remaining. Take these to Kaoma. Next take 20 from Ndola to Kaoma and the
it refers to an exhausted destination.
from Ndola go to Chongwe. The feasible solution will be given by.
Now 6 possible routes remain. The least cost route is now Sinazongwe-Livingstone. Using
Destination the same procedure as above, Sinazongwe has 350 and Livingstone only needs 200 units,
Monze Kabwe Kaoma Chongw€
Total so 200 units is transported which exhausts the Livingstone destination. Sinazongwe still
o (s0) (6s) (2s) (40) remains with 150 units.
-o Lusaka (60) 5t 10 60
ivinestone (60) 55 5 60
With two destinations exhausted, there is only one destination remaining. All the
\dola (50) 20 40 60 remaining output should then be destined to the Copperbelt. Extra care must be taken
fotal 50 65 25 40 180 with numbers here especially the remainders. Just to remind ourselves, Chisamba
remained with 20 units, Sinazongwe with 150 units and Chipata still has all the 270 units.
We leave it to the reader to verify that the total cost will be (800. The feasible solu lf there was need for prioritisation, we would start with the 20 units from Chisamba since
will change when the order in which sources and destinations are presented and it has the least cost to the Copperbelt. But there is no need to set any order since
solution from this method will only be optimal by chance. everything must be transported to the same destination. The table below is a summary of
the optimal transportation strategy.
 I

2so | ,or, ,rrrr'oNs oF LTNEAR PR.GRAMMTNG Mathematical Optimisation and Programming Techniques for Economic Analysis I
297

-usa ka Livi n sstone lopperbelt otal Using the same procedure, the next is Kabwe-Katete where all the remaining units in
)lnaZOngWe 0 200 50 150 Kabwe of 200 is transported. Katete still remains with unmet demand of the remainder.
lhisamba 380 0 0 400 Then Kitwe supplied 2800 to Zambezi which exhausts the source.
lhipata 0 l 70 270
Three sources are exhausted and there is only one remaining to satisfy the two
total 380 200 440 1020
destinations. Simply apportion according to destination needs. Zambezi will take 1700
To get the total cost, slmply multiply the quantity transported with the unit Loll while Katete takes 900. The summary table is given below
transportation for each used route. This gives the cost incurred on each route. The Zambezi Mpika Katete Total
cost will be a sum of all the individual route costs. The calculation is provided in the Kabwe 2700 200 2900
below
Kitwe 2800 2800
Route QuantitV Unit Cost Total Cos Lusa ka 2500 2500
Sinazongwe-Livingstone 200 5.0 1,000 Monze L700 900 2600
Sinazongwe-Copperbelt 150 10.5 t,575 Total 4500 2700 3500
Chisamba-Lusaka 380 30 !,740
Ch isam ba-Coppe rbelt 20 6.0 720
Ch i pata-Copperbelt 270 13.5 3,645 This allocation satisfies all the sources and destinations. lt is therefore a feasible solution.
Total 1020 7,44O An optimality test is needed to ascertain the optimality of this allocation. The total cost
will be given by
Route Quantity Unit Cost Total Cos
Example 14.3 Kabwe-Mpika 2700 50 135,000
Kabwe-Katete 200 53 72,600
that is, Zamberl,
Suppose there is a critical food shortage in three districts of Zambia,
Kitwe-Zambezi 2800 66 184,800
Mpika and Katete. The actual deficits are estimated to be 4500, 2700, and 3600
Lusaka-Katete 2500 49 722,500
respectively. Fortunately, government has stocks of maize available in four of the six graln
Monze-Zambezi 1700 95 163,200
storage silos across the country. The available quantities are Kabwe 2900; Kitwe 2800;
Lusaka 2500; and Monze with 2600. How should these quantities be transported to tho
Monze-Katete 900 58 6L,200

needy districts so that minimum transportation cost is incurred? Assume the followlnS Total 10800 679,300
cost matrix based on the distance chart.
Zambezi Mpika Katete 14.2.3 Vogel's Approximation Method (VAMJ
4500 2700 3500
Vogel's method is unique and a little complex compared to the former. lt is a realisation that
Kabwe 2900 87 50 63
the lowest cost may not always be optimal. lnstead of looking at one route at a time, Vogel
Kitwe 2800 66 58 85
insists that some routes, though cheapest, may prevent us from using other cheap routes. The
Lusaka 2500 77 64 49
absolute cost is not enough to determine the optimality of a route but when considered in
Monze 2600 95 83 58
relation to others. Take a simple case of two source two destination model with the cost matrix
Using the lowest cost entry method, search for the smallest number in the matrix. The given in Table 14.1. Assume that the sum of output at the two sources equal total demand from
cheapest route is the Lusaka-Katete so allocate the maximum possible of 2500 which the two destinations so that the model is balanced.
exhausts the supply at Lusaka. So delete the Lusaka source (row) and reduce the quantity
for Katete by 2500 since it has received this from Lusaka. Table 14.1.Simple Transportation problem

With the Lusaka source gone, the next cheapest is the Kabwe-Mpika route. A maximum of D1 D"
2700 is transported which satisfies the Mpika needs. This destination is eliminated as well .s, 3 4
and Kabwe now has 200 units (2900 - 2700) remaining. s, 4 7
2s2 | ,or, ,*rrr'oNs oF LTNEAR PR,GRAMMTNG Mathematical Optimisation and Programming Techniques for Economic Analysis | ,,,

with the above cost matrix, the lowest cost method will begin with s1D1 because it lr,rr Lusaka Copperbelt
lowest cost of 3. The next potential number is 4, because it is less than 7. However, thr' 380 440
routes which cost 4 are no longer available because all the output at 51 has been exhau',1|rl Sinazongwe 150 7.5 10.5 5

satisfy the demand at D1 lt now becomes mandatory to use the most expensive rout('lll Chisamba 400 3.0 5.0 3
model because what looked 'more attractive' has precluded the neighbourhoods. The tot,rl Chipata 270 7.5 13.5 6
wlll be higher than if the absolutely low cost was avoided. 4.5 4.5

To find the feasible solution, the VAM applies five steps. They are as follows: The third row has the highest difference. Allocate the maximum possible cargo for the
Chipata-Lusaka route. The route takes all the supply from Chipata but the Lusaka market
Step 1: Determine for each row and column the difference between the two lowc\t llllll
still has a deficit of 110. We delete the Chipata source and make the necessary
costs or cell elements in the cost matrix, including dummies.
adjustments to the difference row and column.
Step 2: identify the row or column with the highest difference. When there is a tie, takl rttt:
Lusaka Copperbelt
from the highest arbitrarilY. f 110 440
Step 3: Allocate as much as possible of the good to the lowest cell(s) in the row/colf Sinazongwe l-50 75 10.5 3
identified in step 2 above. I Chisamba 400 3.0 6.0 3
Step 4: Eliminate the exhausted row or column and repeat the first three steps. Stoe !il 45 45
all the rows and columns are satisfied. ,[ There is a tie again between Lusaka and Copperbelt destinations. Arbitrarily, take all the
output from Chisamba to the Copperbelt. This only leaves one source. So the remaining

:ilJ:d.::,:.:"-:,",::]':""",*",,
When working with the VAM, the cost matrix is very important and so is the qufif,
table. We take the matrix table and replicate it below. A row and a column are addfl
t output in Sinazongwe is apportioned to Lusaka and Copperbelt. This completes the
transportation strategy which we present below.

Sinazongwe
Chisamba
Lusa ka
110
Livi ngstone
200
Copperbelt
40
400
Total
350
400
which the difference between the two lowest cells is presented. L Chipata 270 270
Total 380 200 440 1020

This strategy has the following cost

? Route Quantity Unit Cost Total Cost
1 Si n a zo ngwe- Livi n gsto ne 200 5.0 7 000
6
Si n a zon gwe-Co p pe rbe lt 40 10.s 420
Sinazongwe-Lusa ka 110 75 425
Therearetwohighestvalues,thesecondcolumnandthirdrow Wearbitrarilypickonlhe Ch isa m ba-Copperbe lt 400 6.0 2,400
former which has the lowest cost on the Sinazongwe-Livingstone route. The maximttttt Ch i pata-Lusaka 270 7.5 2,O25
quantity on this route is 200 units corresponding to the total demand in Livingstone 5n Total to20 6.670
the Livingstone market is satisfied and hence the second column is eliminated.

After adjusting for the remaining quantity from Sinazongwe, because it has alre,l(lV
The strategy has not utilised the least cost Chisamba-Lusaka route but the total cost falls far
supplied Livingstone, and calculating the new differences, we have the following table. below the total arrived using the previous method. This is a 'litmus-test' which proves that the
earlier method did not yield an optimal outcome. However, it must not imply that the VAM has
itself produced an optimal strategy.
 I

294 I soue rxrrlrsroNS oF LTNEAR pRoGRAMMtNG Mathematical Optimisation and Programming Techniques for Economic Analysis 295

Example 14.5 Zambezi Katete

Redo Example 14.3 using the VAM method. Which method produces the lower cost? L700 1100
Kabwe 200 6t 63 24
The example in question has the following cost matrix Monze 2500 96 68 78
Tambezi Mpika Katete 9 5
4500 2too 3600 From Monze transport 1100 to Katete The Katete demand is satisfied so eliminate the
Kabwe 2900 8l 50 63 corresponding column.
Kitwe 2800 66 58 85
Lusaka 2500
Tamhczi
64 49
Kabwe 200 )4
Monze 2600 96 83 68
Monze 1500 g5 )R
For each row and column, calculate the difference between the lowest and second I 9
cell element. These are inserted as follows
There is only one destination remaining so no need to calculate the cost difference. The
Zambezi Mpika Katete output is transported accordingly. From Kabwe to Zambezi 200 units and Monze to
4500 2700 3600 Zambezi 1500. The final table will appear as follows.
Kabwe 2900 aa 50 63 13
Zambezi Mpika Katete Total
Kitwe 2800 65 58 85 8
Kabwe 200 2700 2900
Lusaka 2500 77 64 49 15
Kitwe 2800 2800
Monze 2600 96 83 68 15
Lusaka 2500 2500
LL 8 L4
Monze 1s00 1100 2600
There is a tie between the last two rows or sources. So we plck the Lusaka source whi( lt l'. Total 4500 2700 3500 10800
all transported to Katete. The Lusaka source is no more so we eliminate tlrp
The total cost is given in the table below.
corresponding row. The resulting table with new differences calculated is
Route Quantitv Unit Cost Total Cosl
Zambezi Mpika Katete
Kabwe-Mpika 2700 50 135000
4500 2700 1100
Kabwe-Zambezi 200 87 L7400
Kabwe 2900 87 50 63 73
Kitwe-Zambezi 2800 66 184800
Kitwe 2800 65 58 85 8
Lusa ka-Katete 2500 49 1,22500
Monze 2600 96 83 68 15
Monze-Zambezi 1500 96 L44000
7L 8 5
Monze-Katete 1100 68 74800
Now we have the first column. The lowest cost is from Kitwe. Transport all output frorrr Total 10800 678,500
Kitwe to Zambezi and the Kitwe source is closed.

Zambezi Mpika Katete

7700 2700 L100 14.3 Test for Optimality
Kabwe 2900 6t 50 63 13
As stated earlier, the solutions that come from the above three methods are not always
Monze 2600 96 83 68 15
optimal. The solution may be optimal or may not at all be optimal. This calls for the need to test
9 33 5
for the optimality of the solution. Because the optimality of the solution is not certain yet, the
Then transport 2700 units from Kabwe to Mpika This closes the Mpika destination trrrt solution cannot be called optimal. An alternative name, bosic feosible solution is used to
the source still has some remaining output The resulting table is: distinguish it from a tested optimal solution.
To test for optimality, we must start with a non-degenerote basic feasible solution. The solution
is non degenerate if it possesses the following two properties
296 | ,or, ,rrrr,oNS oF LTNEAR PR.GRAMMTNG Mathematical Optimisation and Programming Techniques for Economic Analysis 297

The solution must contain exactly m * n - 1 number of individual allocations Zambezi Mpika Katete Total
variables m and n denote the number of sources and destinations respectively Kabwe rFt- +q0-1 2900
Kitwe )0 2800
The allocations must be such that it is impossible to form any closed loop by d
Lusa ka 00 2500
vertical and horizontal lines through these allocations.
Monze L7 €00+1 2500
Two methods are used to test for the optimality of the solution. These are: Total 4500 2700 3500
. fhe Stepping stone method. The change in total cost is87 - 63
- 96 + 68 = -4
o lhe u-v method (also called lhe Modified distribution method or MODI\.
The change in total cost is negative. That means this reallocation will lead to a reduction in cost
14.3.1 The stepping stone method
implying that the basic solution is not optimal yet.
ln the stepping stone method, the objective is to evaluate the effect (on cost) of using
Reallocate the whole 200 units from the Kabwe-Katete to Kabwe-zambezi and then take
more of the unused routes. The method determines whether there is a cell with no
another 200 from Monze-Zambezi to Monze-Katete. we obtain the following matrix
that would reduce the cost if used. We ask ourselves whether it is possible to reduce
allocating one unit to an unused route. This means trying all the empty or u Zambezi Mpika Katete Total
transportation routes. To illustrate this method, let us use the basic solution obtained by Kabwe 200 2700 2900
lowest cost entry method in Example 14.3. The basic solution is Kitwe 2800 2800
Lusa ka 2s00 2500
Zambezi Mpika Katete Total Monze 1500 1100 2600
Kabwe 200 2900
Total 4500 2700 3500
Kitwe 2800 2800 There is no guarantee yet that this allocation is optimal. We still need to test it for optimality by
Lusa ka 00 2500
repeating the method on the new allocation.
Monze 170e- -ed0 2600
Total 4500 2700 3600 There are 5 empty slots in the matrix. Nonetheless, the Kabwe-Katete route was already ruled
The solution has 6 : 4 x 3 - (4 + 3 - 1) as required by the non degenerate condition. out. So we start with the Kitwe-Mpika route. We form the following loop.
are Kabwe-Zambezi route, Kltwe-Mpika, Kitwe-Katete, Lusaka-Zambezi, Lusaka-Mpika a
Monze-Mpika routes. Let us start with the first route. Allocate one unit to the Kabwe-Zam Tamhez Mpika Katete Total
route. To ensure the basic solution is not changed, subtract one unit from the Kabwe Kabwe 20ffiO0-1 2900
route, add a unity to the Monze-Katete rout and subtract one from the Monze-Zambezi Kitwe 2800-1- _____J 1 2800
Lusa ka 2500 2500
This will ensure that the row and column total which represent the demands and supplies
Monze 1500 1100 2600
not altered. This must be done for all the empty or unused cells. lf all reallocations
Total 4500 2700 3500
positive change in total cost, the basic solution is optimal since no reallocation The change in cost resulting from allocating one unit to this route is gr
successfully reduce cost. lf however one cell shows a negative change in cost, then the
- 66 - 50 + 5g : 29.
This reallocation would result in increase in cost and therefore inferior to the current allocation.
solution is not optimal yet. The total cost can be reduced by utilising the 'discovered' route. So
allocate as much as possible to that route and recheck the optimality. We try the Kitwe-Katete route. We form the following loop.

We show the changes in the table below. Zambezi Mpika Katete Total
Kabwe 200 2700 2900
Kitwe 280S4- ------+1 2800
Lusa ka 25oo 2500
Monze 1sdffi -1100-1 2600
Total 4500 2700 3600
298 SOME EXTENSIONS OF LINEAR PROGRAMMING Mathematical Optimisation and Programing Techniques for Economic Alalysis 299

The change in cost resulting from allocating one unit to this route is 85 68 + 96 : entails the existence of another optimal allocation. This is a case of multiples optimal
- 66 -
Again, this reallocation will lead to increased cost and therefore not ideal. a llocations.

Now we try allocating one unit to the Lusaka-Zambezi route. The following loop is used. 14.3.2 The u-v method

The method has this name because it uses u and u in the procedure. The following steps
Zambezi Mpika Katete Total summarise this procedure:
Kabwe 200 2700 2900
Kitwe 2800 2800 Step 1: Determine a set of m+.nnumbers ui(i = t,2,. m) and v1Q = l'2'...n) in such as

Lusa ka #t- -2sq0-1 2500 way that for each occupied cell, the unit cost
Monze 1sd+ -++O0+1 2600 cij=ui+uj
Total 4500 2700 3600 The method uses the occupied cells or utilised routes to determine a new set of
The change to cost resulting from this reallocation is77 - 49 - 96 + 68 = 0. This variable ui and vj. However, there is only a total of m+n - 1 occupied cells
has a zero additional cost. Whether it is used or not, the cost would stay the same. equivalent tom+n - 1- independent equations to determine m+n unknowns. fhis
defies a mathematical condition on simultaneous equations, that the number of
We now move on to consider the Lusaka-Mpika route. The loop will be slightly different equations must at least be as many as are unknowns. To counteract this problem, one
the rest because it affects many routes. unknown must be arbitrarily determined.

Zambezi Mpika Katete Total

Step2: usingtheu;(i=1,2,...m)andvj(i =1,2,...n)fromstepl,calculateforeachempty
Kabwe 2O@++- --2790-7 2900
cell (r,j) lhe cell evaluotion. This is a unit cost difference di; given by di1 - ci1 -
Kitwe 2900 2800 (ui + u)
Lusa ka +!t- -25q0-1 2500 Step 3: examine the matrix of cell evaluation and conclude as follows:
Monze 1s0+ -++eb+t 2600
l. lf all drj ) O, that is, all cell evaluations are strictly positive, basic feasible solution is not
Total 4500 2700 3500
only optimal but also unique. lts total cost is the lowest. Any alternative solution will
The resulting change in cost is 87 - 50 + 64 - 49 + 68 - 96 = 24. The reallocation will car.rse
have higher total transportation cost.
total cost to go up. ll. lf all dij > O, that is, all cell evaluations are non negative, the basic feasible solution is
optimal. However, an alternate solution also exists with the same level of total cost.
Lastly, we try the Monze-Mpika route. Reallocate one unit to this route and adjust the others
lll. lf at least one d"ij < 0, the solution is not optimal. This outcome means repeating the
accordingly to ensure the model remains feasible. We form the following loop.
above process.
Zambezi Mpika Katete Total Example 14.5
Kabwe 200++- --il80-7 2900 t4'2 on pa9e 289.
Test the optimality of the basic feasible solution from Example
Kitwe 2800 2800
Lusa ka 2500 2s00 The basic feasible solution in the stated example is given in the table below. The table
Monze 1s0F -----+1 1100 2500 shows the unit cost of transportation for the utilised routes.
Total 4500 2700 3600 usaka Livingstone Copperbelt
The change in total cost is 87 - 50 - 96 + 83 = 24. fhis reallocation Will lead to increased Sinazongwe 5 10 .5 4
cost. Therefore, it is not favoured. Chisamba 3 6 -n .5
Chipata 13 .5 1
We have shown that all the reallocations to the empty cells or unused routes have positive u 3 5 L 6 .5
changes in costs except in one case, that is, the Lusaka-Zambezi route. This means there is no
Ithas3*3-1:5utilisedrouteswhichareinindependentpositions.Thereforethe
feasible reallocation that will succeed in reducing cost. Therefore, the obtained allocation is
basic feasible solution is non-degenerate. We can proceed with the u-v method.
optimal. The existence of a zero additlonal cost in utilising the Lusaka-Zambezi route only
With the us and us now determined, the cell evaluations table is given below.
300 | ,or, ,rrr*r,oNs oF LTNEAR PR.GRAMMTNG Mathematical Optimisation and Programming Techniques for Economic Analysis 301

,usaka Livingstone Coooerbelt Step 3: Conclusion On examination, all the cost differences dij : cij - (r, + u;) of the
Sinazongwe 0 4 unused routes are non-negative We conclude that the basic feasible solution is optimal.
Chisamba 10.5 -0.5 There nonetheless exists another optimal solution with the same level of cost based on
Chipa ta -3 6.5 7 the zero element in the cell evaluation.
u 3.5 L 6.5
14.4 Assignment Model
The cell evaluation matrix has one negative element. There is no need here to talk about
the other zero since the negative element overrides the zero elements. The conclusion is ln the Transportation model, we dealt with many sources and many destinations. Each source
that this solution is not optimal. lt is possible to change the transportation arrangement and destination is producing or demanding more than a unit of the commodity The problem
was to minimise the total cost of transporting all output from all the sources to satisfy all the
and reduce the costs.
demand by all the destinations. The question we pose here is what happens when each source
Let us then test the allocation obtained by the VAM method in Example 14.4 which uses is only producing one unit of output and each destination takes only one unit of output?
the same data. The following feasible allocation was obtained where the number in
parenthesis is the unit cost. The question still under consideration is how much output from which source goes to which
destination. This is a value-laden question but is precisely what the transportation model deals
with. When each node (source and destination) only takes one unit of output, the question now
is which source supplies which destination. The problem reduces from considering the
quantities to only looking at the pairing of source and destination nodes. Each source must
have a corresponding destination. The pairing could be done haphazardly since all nodes either
produce one unit or demand one unit of output. However, source nodes will not supply to
destination nodes at the same cost. With differing costs for different pairs, pairs must be
There are three steps to follow.
strategically made so that the cost (transportation) is minimised. This is called the ossignment
Step 1: Determining the u. and /,. We use the following equations model.
u11.v1 =f.$ It is named assignment model because instead of deciding on quantities, its central problem
u2*vr:5 involves assigning each source to a particular destination in a way that minimises
us -l u, - 1-0.5 transportation cost, As a special case of the transportation model, it is best described by the
u3lu2=$ assignment of workers to jobs or chores ln the model, there are n workers and n jobs so that
u1*v3=f.J the model is balanced. Each worker can take any job but because of differing skills and job
Let u, = 3, we have the following: requirements, the performance on the job will vary depending on the job assigned to.

-usa ka -ivi ngstone -opperbelt v This can be an output maximisation problem. With the duality theorem at hand, the problem
]80 100 [40 can be turned into cost minimisation. The cost is measured as the cost of producing a unit of
iinazongwe 350 7.5 .0 r-0.5 3 output. Alternatively, it means a more productive worker will be less costly. We denote cij as
lhisamba 400 5.0 -1.5 the cost of assigning the ith worker to the;th job. This will vary depending on how productive a
lhipata 270 7.5 3 worker is at that particular job, which depends on skill-match. That is, how one's skills match
u 1.5 7 .5 the job requirements. The objective of the model is to determine the optimal (least cost)
assignment of workers to jobs.
Step2: Cell e\aluation of emptycells, we have
Formally, the assignment model is defined by thenoptimisation problem of
Lusaka -ivingstone v
380 100 440 .in)),,;.,;
inazonswe 350 3 i=r j=r
lhisamba 400 0 05 15 subject to
Chipata 270 9.5 3 3 trfr
u 1.5 7 r5
L*',:
j=t
L*', = '
i=1
 I

302 I soME EXTENSIONS OF LINEAR PROGRAMMING Mathematical Optimisation and Programming Techniques for Economic Analysis 303

where x1; defines whether a particular worker has been assigned to a particular job. lt is

defined as Bar Laundry Restau ra nt

-. l1', if worker i assigned iob i
^ij - 10, otherwise Ba nda 72 15 11

When there is a mismatch in the number of workers and available jobs, the model is said to be Mwiya 9 7 6
unbalanced. lt is unbalanced in the sense that the number of workers and jobs do not equal.
There may be more workers than jobs in which case some workers will not be assigned any job. Phiri 72 13 t7
The opposite is where the number of jobs exceeds the number of workers so that some jobs
will not be assigned any workers. Since linear programming only workers with a balanced How should he assign the three workers?
model, we must make-up for the lesser jobs or workers so that the model is balanced. This ls Let us follow the five step Hungarian method outlined earlier.
achieved by adding Jictitious workers or jobs as the case may be. When there are fewer jobs
than workers, add as many fictitious jobs as is the short-fall. The same applies when the Step 1. Subtract 11from row L,7lrcm row 2 and 12 from row 3 to obtain.
number of workers exceeds the number of available jobs.
Bar Laundry Restau ra nt
t4.4,1 The Hungarian method
Ba nda 1 4 0
The optimal solution is found using the Hungarian method. The method was developed by an
American mathematician Harold william Kuhn based on the work of two Hungorion Mwiya 0 1
mathematicians; Denes Konig and Jeno Egervary from which the method derived its name. The
method has five steps involving the cost matrix. Given the cost matrix Phiri 0 1 5
Step 1:Subtract the smallest entry in each row from all elements of its row.
step 2: subtract 0 from column 1, 0 from column 2 and 0 from column 3 and we get (the
Step 2:Subtract the smallest entry in each column from allthe entries of its column. same matrix in this particular example).
Step 3: Draw lines through appropriate rows and columns so that all the zero entries of the cost
matrix are covered and the minimum number of such lines used. Bar La u ndry Resta u rant

Step 4:Test for optimality Banda 1 4 0

. lf the minimum number of covering lines is equal to n, (assuming the cost matrix is
Mwiya 2 0 1
nxnl, an optimal assignment of zeros is possible and we have reached the end of
the process. Phr 0 1 5
. lf the minimum number of covering lines is less than n, optimality is not yet achieved
and we go for step 5. step 3: cover all zeros with the minimum number of horizontal or vertical lines. we get

Step 5: Determine the smallest entry not covered by any line. Subtract this entry from each Bar Laundry Restau ra nt
uncovered row and then add it to each covered column Then return to step 3.
This has the lowest cost of assigning the n tasks to ?2 workers Any re-assignment will result in Banda ----+- -------€-_
higher total cost of performing the tasks. _______€_ --------+-
Mwiya
--2-
Example 14.7
Phiri ----€-- --+--- -------5--
A lodge supervisor has three workers to perform three tasks. He must assign one to work
in the Bar, one in the laundry and another in the restaurant. The cost, measured in man- Step4: Since the minimum number of lines is 3 (the number of assignments to be
hours per unit of output, is given form each worker in the table below made), we have an optimal assignment as follows:
304 | ,or, ,rrrr'oNS oF LTNEAR PR.GRAMMTNG Mathematical Optimisalion and Programing Techniques for Economic Analysis 305

Bar Laundry Resta u ra nt

Step 1: subtract 75 from row 1, 35 from row 2, 90 from row 3 and 45 from row 4. We
obtain the following matrix.
Banda 7 4
!_l Mining Area

tI Caterpillar

r
Mwiya 2 t A B c D
1 15 0 0 5
Phiri L 5 2 0 50 20 30
3 35 5 0 15
The optimal assignment is shown by the zeros n independent positions. 4 0 65 50 70
corresponding cost will be:
step 2: subtract 0 from column 1, 0 from column 2,0 from column 3 and 5 from column
Bar Laundry Resta u rant 4 to obtain

Ba nda 72 L4
!I Mining Area

Mwiya 9
E 1
Caterpillar
1 15
A B

0
c
0
D
0

Phr E 13 L7
2
3
4
35
0

0
50
5
65
20
0
50
25
10
65

So Banda should be assigned to the restaurant, Mwiya to the laundry while Phiri to Step3: Cover all the zeros of the matrix with the minimum number of horizontal or
Bar. The total cost will be 11 + 7 + 1,2 :30 manhours vertical lines. We get:

Example 14.8 Mining Area

Cateroillar A B c D
Keren Motors Ltd provides a wide range of equipment for hire to mining companies -1
Zambia. These include bulldozers, excavators and other earth moving machines.
1
2 50 o
--€-
25
3 5 ) 10
Suppose the company has four large caterpillar machines at four different sites.
4 0 65 50 65
caterpillars have to be moved to four different mining areas. The distances in kilometres
between the caterpillar locations and the mining areas are given below.
Step 4: Since the minimum number of lines is less than 4, we have to proceed to Step 5.

Mining Area
Step 5: The smallest entry not covered by any line is 5 So subtract it from each
CaterD ar A B c D
uncovered row. We get:
7 90 75 75 80
Mining Area
2 35 85 65 Caterpillar A B c D
3 125 95 90 105 7 15 0 0 0
4 45 110 95 11s 2 -5 45 15 20
3 30 o -5 5
4 -5 60 45 60
The problem is: Which caterpiilar should be assigned to which mining area in order to
Now add 5 to each covered column. We get:
minimise the total 'mobilisation'cost measured by distance travelled?

To find the optimal solution, we have to follow the five steps ofthe Hungarian method.
 Mathematical Oprinrisation and Programming Techniques for Economic Analysis 30/
I

306 I sovE rxtrrustoNs oF LINEAR PRocRAMMING

Step 4: Now the minimal number of covering lines is 4. So an optimal assignment

Mining Area possible.
Catero I ar A B c D

t 20 0 5 0 Mining Area
0 45 20 20 Cateroillar A B c D
2
0 0 5 L 40 o s lr,
3 35
4 0 60 50 60 2 0 25 fi 0
3 55 ln 0 5
Cover all the zeros with the minimum number of horizontal or
vertical lines'
Step 3: 4 td 40 30 40
Mining Area The optimal assignment is shown by the zeros in independent positions. The
Caterpillar A B c D

1 _*- ----e-
corresponding cost will be:

z 45 20 20 Mining Area
3 -----*- Caterpillar A B c D
4 0 60 50 60 1 90 75 75
2 l5 85 lssl 65
Step4:AnoptimalaSsignmentisstillnotpossiblesincetheminimalnumberofcovering 3 725 locl go 105
get to Step 5 again'
lines is less than 4, the number of assignments needed' So we 4 145l 11.0 95 115

Step5:Thesmallestelementnotcoveredbyanylineis20.subtract20fromeach So caterplllar l should be sent to mining area D. Caterpillar 2 sent to mining area c,
uncovered row. We get caterpillar 3 to mining area B and caterpillar 4to mining area A. The total distance (a

Mining Area measure of mobilisation cost) is B0 + 45 + 95 + 45 = 265km

Caterpillar A B ( D
14.4.2 Ki)nig's Theo rem
1 70 0 5 0
z 25 0 0 Underlying the assignment algorithm are the following theorems:
3 35 0 0 5

4 -20 40 30 40 lf a number is added to or subtracted from all of the entries of any one row or column of
a cost matrix, then an optimal assignment for the resulting cost matrix is also an optimal
Then add 20 to each covered column and obtain:
assignment for the original cost matrix
Mining Area
2. The maximum number of independent zero positions in a matrix is equal to the
CaterDillar A B c D
40 0 5 0 minimum number of lines (known as covering indexl required to cover all the zeros in
1
2 0 25 0 0 the matrix
3 55 0 0 5

4 0 40 30 40 14.5 Transhipment model

Then return to SteP 3. Transhipment, as the name suggests is the shipment of goods to an intermediate destination
of horizontal or and later on to the final destination. There are many reasons for transhipment, legal and illegal.
Step 3: Cover all the zeros of the matrix with the minimum number Here we just dwell on the legal and economical reasons. One reason is to change the means of
vertical lines transport during the journey, for instance, from road to rail or vice versa. This is known as
Minins Area transload i ng.
Caterpillar A B c D
The other is to combine small shipments into large shipments and vice versa. Quantities of
1 --4€|l- ----€- commodities to be transported from satellite depots may not be large enough to warrant the
2 use of large trucks which have low per unit costs. So smaller vehicles, with high per unit cost are
3 -*-
--o- used to ferry goods to centres. These centres are not the final destinations so the large trucks
4 ---30- ---40-
 Mathematical Optimisation and Progranming Techniques tbr Economic Analysis 309
308 I ,o*, ,*rr*ttoNs oF LTNEAR PR.GRAMMING

to an intermediate destination k. Then it will cost Cki to transport a unit of output from an
thentransporttofinaldestinations.Thisisknownasconsolidationandtheoppositelr intermediate destination k to the final destination I. This allows categorising costs at three
d econsol idation levels. The first is the cost of direct routes, from a source to a final destination. The second will
be costs incurred to transport from sources to intermediate destinations and the third category
Transhipmentbringsintheconceptofintermeciatedestination,adestinationwhichisnotfinal
The goods are meant for anothor is the shipment cost from these intermediate destinations to final destinations.
and itself having no demand {or the goods transported'
destinationcalledthefinaldestination.ltisnonethelessstillpossibleforsomegoodstobo Formally, the transhipment model is stated as follows.
transporteddirectlyfromthesourcetothefinaldestinationswithouthavingtopassthrouBh mnmlLn
some transient destinations'
s or source may actually be used as transient destinations'
.,"I I,ipi1 +\Z,,rr,n +\\,u1*r1
provincial i=1 j=1 i=1 k=1 k=7 j=t
cational or health supplies is normally through This is the total cost that will be incurred in the shipment. There are conditions however that
the provincial centre is also a district with a
stricts. But must be met regarding the total quantities from the sources and destinations. There are
case of a final destination functioning as a
transient point'
basically three conditions, in addition to having all quantities rs as non negative.
reversed, some sources become transient destinations'
The first is that the output from every source must be transported
nI
ThediagramshowingthedifferenttransportationroutesisportrayedinFigure14.2.lthas
threeprimarysources,twointermediatedestinationsandfourfinaldestinationsThemodel
model by allowing for transloadinB 2,,, *2,,n: o,, vi
must be treated as an extension of the transportation j=r k=1
and/orconsolidation.therefore,thedirectroutesthatwereavailableunderthetransportation
or to different points' Second, all final destinations must be satisfied
model are still available, ,ar" *ith a possibility of combining cargo from mI

Figwe 14.2'. A transhiPment model 2,,, *2,u,: 0,, v i

i=1 k=l
Third, everything arriving at a transient point must depart to final destinations. Every such point
must dispatch as much as it receives.

t.^*=f'r''
i=7 j=t
vk

For illustration purpose, consider the following example

Example 14 9
Consider the Zambeef case used in Example 14.2. ln that example, we looked at a case
where beef emanates from abattoirs and has to be transported to various markets. ln this
example, we now take into account the source of animals for the abattoirs. The full model
will now have three points: The farms will be the primary sources which will supply
animals to abattoirs. ln this case, abattoirs will be destinations, but only intermediate
destinations. After the slaughtering and all the required processes, the beef now has to
be transported from abattoirs to final destinations, the markets. ln the second stage,
These do not need to be used all together'
The model provides for many shipments routes' abattoirs are acting as sources (intermediate sources) while the markets are final
Economicrationalewilldictatewhichroutesareused.Moreover,usageofcertainroutesrules destinations.
out the possibility of using certain other routes'
Assume that Zambeef has m farms producing beef cattle at different scales, I abattoirs
Therationaleagainistodeterminetheroutesthatminimisetotaltransportationcosts.We and n destination markets (of different sizes).These could be town or district centres. The
continuetotakeciT.rtn".o'tott'ansportingaunitofoutputfromsourceitodestinationi' i
of transporting a unit of output from source
This is a direct route. Then define c;1 as the cost
 Mathenratical Optimisation ancl Prograrrnring Techniqucs trr Eeonomic Analysis 31,L
310 | ,or, ,rtrtttoNs oF LTNEAR PR'GRAMMING

this example' The first is that transportation

reader should realise two notions from
destination is not permissible since the product has to
unt Chapter 15
the primary source to final
points or abattoirs'
some processing at what we call transhipment
unit cost will also include the cost of
processing' 15 INTRODUCTION TO GAME THEORY
Second, because of this processing'
since the mother company will n
this cost is more likely to be equal for
all abattoirs
into the 15.1 Introduction
the cost is simply factored
likely build equal capacity in them' Nonetheless'
supply nor fixed demand' ln reality howt ln the optimisation problems studied so far, a decision by the firm on price is what solely
cost. The abattoirs neither have a fixed determines the quantity and consequently, its profits. For instance, a farmer's profit will be
of capacity where an upper limit exists'
there may be issues solely determined by his own decision regarding quantity and price which will depend on the
a transhipment mechanism that will
result in lov rnarl<et lf the firm is operating in a perfectly competitive market, its profit will solely depend on
The problem for Zambeef is to devise
in different outlets This cost wlll the quantity it decides to produce. lf operating in a monopolistic market, it can either decide on
cost of turning cattle from different
farms into beef
to the abattoir or using an aba' price or quantity (never on both) and the profit will be determined accordingly.
be minimised sinrply by minimising the
distance
ln this chapter however, we introduce a different scenario. A scenario where a firm's profit or
nearest to the market That ls
gain does not only depend on its own decisions. They will depend on what other players in the
mLLn
rnarket decide. Think of a tender process A tender will specify the size and/or quality of
-inI/_)/_/I c11x11 tFH F.u.,,u,
H expected output in what is known as bill of quontities. 1-he bidding firm will then maximise
i=1 k=1 k=t J=L
profits by working on the mark-up With a higher mark-up, the firm will make a higher profit.
that the totar quantities dispatched But there is a limit to this Many firms will participate in the tender and the tender is offered to
This minimisation is subject to the requirement
at final destinations and the non-neg' the lowest bidder. That is, increasing the mark up will increase the profits but only so long as
the primary sources equal the total arriving
other bidders have put a much higher mark up. When the mark-up exceeds that of the next
constraints on the quantities' lowest bidder, the firm looses the bid.
The bidders each makes a decision on the mark-up with a view to maximise own profits. Each
bidder does not l<now what the others have decided until the opening of bids. When bids have
been opened, only one firm gets the job. The acquisition of the job here is not only dependent
on the firm's bid price but also on what all the other firms have put Once the job is secured,
the actual profit willthen be a function of its own price.
lhus each firm will have to play it'right' if it has to be given the job lt must strategise on the
price to set ln so doing, it has to take into consideration what the others are likely to decide.
This is not different from how a game like poker is played. ln the same manner, firms play
games when they decide on strategies to maximise profits They must decide on whether to
;rdvertise or not, to lower prices or not and many other decisions which will only impact on
profits depending on the steps taken by other players in the market

This chapter is devoted to introducing the theory behind such decision making, the gome
theory. As an introduction, the chapter will not provide a 'mouth-full' of a discussion. lt will
rnerely introduce the theory by simply mentioning the various aspects of game theory.

15.2 What is a game?

ln answering this question, there is no need to restrict oneself to economics. Games are played
in virtually all spheres of life and take different forms. Think of a game like soccer, poker and
 Mathenraticrl Optirrisalion ancl Programnring Te chniqucs lirr Economic AnaJysis 313

312 | ,*r*orua,'o* To GAME THE.RY

llut the other one does not confess
because of the possibility of discharge He confesses
is more lrIcausehe/shetooisworriedofwhattheotherisworriedof.Thetwoareinadilemma Atthe
teller. ln each of these games' there
which normally are i ,'nd, they both confess. This is a decision not in their best interest They were going to be better
io or,'r't" own desires
oI f not confessing.

llre above illustration is a traditional narrative of the prisoner's dilemma ln economics, we can
Ixplain it with an economic illustration from international trade involving two countries.
irrppose that two countries Zambia and Zimbabwe are contemplating on liberalising trade.
I rom a theoretical view, trade liberalisation would benefit the two countries on condition that
L;oth liberalise. lf one liberalises and the other doesn't, the one liberalising will lose out. Let us
,rssume the measured losses and gains are as given in the payoff matrix below

Iable 15.2 Game of Trade liberalisation

Zimbabwe
GAME
Liberalise Control
Li bera lise 10,10 -L5,20
Zambia
Control 20, -15 00
The two countries are making decision in the 'dark'. That is, each does not know what the other
will decide upon. Collectively, the two countries make the highest gains if they both liberalise.
With only one liberalising, the total net gain is only 5 to the non-liberalising country.

Though each country knows the benefit from liberalisation, they are both worried of the losses
if the other does not follow suit. As such, the optimal decision for each will be to retain controls
confinement, a bargain'
on trade Each country will after liberalising because it believes the other may not liberalise.
rhe matrix shows llrn
Eachsuspectistoldthatifhe/sheconfessesa.ndtestifiesagainstheother,hegetsdischar|l,r.rl Thus both will remain in the'no trade liberalisation'where they forgo the positive benefits
with the following porsi;itity mdtrix.
Each suspect is therefore faced combination of pleas The first nutrtlrt't from trade liberalisation.
to for each
number of years that "..h ,;";;i;;.ed
r."t"rr," Chanda and the second to Mweemba The subject of Game Theory specifically deals with behaviours in a game. lt seeks to explain
lllustration how players will behave depending on the nature of games. By nature of games, we mean the
Table 15.1.Prisoners' Dilemma way a particular game relates the players. Does the game provide the scope for conflict or
cooperation? Two extreme levels of relationship are pure conflict and whole heorted
cooperotion The latter subsist in situations where the objective functions of two or more
players are positively related. The move by one to increase one's profits automatically makes
the other player better-off.
Don't Confess The latter on the other hand exists where objective functions are negatively related. When
l-
firms are fighting for buyers, they can never cooperate since the behaviour of one will always
be to the detriment of the other. As a consequence, each player will take actions to avoid being
made worse-off.

decision is to confess also

31.4 | TNTRODUCTTON TO GAME THEORY
Mathematical Optimisation and programming Techniques for Economic
Analvsrs 315

15.3 The terminology of game theory

Table 15.3. Decision to Advertise or Not
Two key terminologies in game theory deserve attention. The first is'players'and the second lc
'strategies'. In game theory, the term "players" still retains its natural connotation. lt refers to Player B
GAIV]E
the entlties that are taking part In the game. These are the people or firms or institutionr Y"r N"
depending on the nature of the game that are in the competition. ln the case of a bid, the firmr
participating are the players. Player A
Yes L0,1.2-J 22 2

No 2,23 76,17
A strategy on the other hand is the steps or moves that players make. These are choices ot
decisions made by players. These choices are not just made randomly but are carefully
calculated to improve the welfare of the player. They are strategically made. When deciding on
whether to reduce the prace or not, the producer has to consider the likely reaction from the
rivals as the effect on profitability will depend not only on the choices of the player but also on
the rival's reaction.
For instance, advertising would definitely lead to increased sales as long as rival firms don't
decide to advertise also A firm sends unchallenged messages that its product has improved
qualities and therefore worth trying This leads to increased sales as more consumers buy the
commodity following the advertising. lf however, other competing firms decide to advertise
also, the customers will learn of the good attributes for all the goods and are unlikely to be
swayed by a single firm This is the reason why they are called 'strategies'
Some strategies may however dominate others ln a multi strategy game, one strategy is said to
dominate another if it is preferred over the other regardless of the rival's strategy. Given a
payoff matrix, if the payoff from one strategy is always better than another for all the possible
choices of the rival, then the strategy is said to dominate another lt is better regardless of what
the rival will play, lf in the extreme, is single strategy pays better than all other regardless of
what the rival plays, then such is strategy is referred to as a dominont strategy A rational player
will always go for the dominant strategy (if it exists) because doing otherwise reduces the
payoff. Example 15.1
Payoff Function: in every game, there is an objective function. This ls precisely what each player suP minerar
wants to achieve. ln the case of profit maximising firms, the profit is the objective function. wat her they
That is the payoff a firm gets by playing the game. We have stated already that the payoff to a se ll
particular player does not only depend on his/her choices. lt also depends on the choices of all ch must
cho es).
the other piayers That is to say, the payoff is dependent on the strategies played by all the
players. This completes the meaning of the name 'payoff function'. lt is a payoff because it At the quantity demanded equal 50,000 while at K1.00, the quantity
shows what players are paid or benefit after playing the game. lt is a function because it is
deman lf both companies charge the same price, they split the sales
determined by the different strategies played by all the players.
evenly Otherwise the company with the lower price selis the whole
amoun off as the profit, formulate the problem as a two person game and
Let's tal(e a look at Table l-5 3 below. lt shows a game with two players, A and B. Each player is obtain
making a decision on whether to advertise or not. The strategy 'Yes' means 'Advertise' while
To answer the question, we must start by carcurating the profit each
'No' means 'Not Advertising'. ln the payoff matrix, each cell shows two numbers. The first is the wiI make for every
price combination. Note that the companies have no variable
payoff or profit for'Player A' and the second pertains to 'Player B' cost, only a fixed cost. The
profit or payoff table is given below.
316 | ,*r*orr.r,o,u ro .AME THE.RY Mathematical Optimisation and Programing Techniques for Econornic Analysis 31,7

Because the benefits for the two players are equal, save for the negation, the payoff function
Aquamina (B)
GAME can be sufficiently shown by only showing the payoff for one player. This is typical of
unproductive games such as gambles. This kind of game has no room for cooperation lt is a
K50,000, -K50,000 pure conflict game.
Aquarite
(A) K50,000, K50,000 0,0 The second type is ca lled non-zero sum gome. The su m of payoff for the players is not zero. lt
|
depends on the strategies played. One player can lose while the other gains or they can all lose
ln the table, each cell has two numbers which really are profits. The first related to player orgaintogether Thiskindof gameprovidesroomforbothconflictandcooperation.
A, in this case Aquarite. The second is for Aquamina. The table shows that the firm that The third type is called constont dit'ference gomes As the name may suggest, the difference
charges a higher price will not sell anything but will still incur the fixed cost. This is the between the payofffor one player and the other is constant whatever the strategy. This entails
explanation for the negative value which is indicative of a loss. that the players lose or gain together. As a consequence, the players will always cooperate. This
The strategy for each player is to consider the outcome for each choice. The player will is because the gain made by one player always equal the gain made by the other. Each player
get the worst case scenario for each choice and will choose the strategy with the highest benefits from the optimal strategy of the other. That is, if Player 1 chooses a strategy which
worst case scenario. This is synonymous to maximising the minimum. lt is often called the maximlses his welfare, Player 2's welfare is also maximised by the same strategy.
maxmln strategy. Clearly, both companies will charge the lower price. 15.4.4 A Two Person Zero-sum game
15.4 Types of Games Assume a two person zero sum game. Player A has m strategies while player B has n strategies.
The payoff matrix is given in the table below.
Not all games are the same. They will differ on a number of aspects. This subsection therefore
discusses the different differentiating aspect of a game. Particularly, it discusses the factors that Table 15.4.4 general payoff matrix
define the type of a game. Basically, three factors define the type of a game. These are:
Player B
15.4.1 Number of Players
L 2 n
You will know by reasoning that a game cannot have one player only. lt needs a minimum of
at, bt ap, o,n'b,,
--II
two players Since two is only a minimum, it is possible to have three or even more players. L b12
Thus, games would be differentiated based on the number of players involved. The example
given in Example 15.1 is an example of a two-player game.
Player A 2 oz., bzt azz, bzz
'*b'"
15 4.2 Number of Strategies
m a-1, b^1 amz, bmz
The number of strategies refers to the options that each player can choosefrom ln the case of | "^'b-
deciding to do something, there are only two strategies; to do or not to do. ln the case of Since it is a zero sum game, the table above can be simplified by looking at the payoff for player
changing price, every firm has three options; to increase, to maintain or to reduce. This is a A only The reader must know by now that the payoff for player B is simply negative of the
three strategy game. ln general, the number of strategies can be increased indefinitely. payoff for player A. The new table is shown below

15.4.3 Type of Payoffs Player B

The payoff functions come in different forms. We basically distinguish three types of pay-off 1 2 n
functions.Thefirstisknown asazerosumgome.Thisisagameinwhichthesumof payoffsfor
att atz ot
the players is always zero The profits one player makes exactly equals the loss of the other. For
instance, take a two player game with players A and B. When player,4 plays strategy i and B Player A 2 azt azz ozn
plays strategyT, the two players A and B will each get ai1 and bi1 respectively. As a zero sum
game, then the following equation must hold.
m om1 omz a^n
ai1 lbil=O
ai j = -bij
To optimise the objective function, Player A (whose gain is shown in the table) will choose a
strategy yielding the highest possible payoff. But A does not know yet the strategy that B will
318 INTRODUCTION TO GAME THEORY Mathematical Optimisation and Programming Techniques for Economic Analysis 319

play. As such, A will have to prepare for whatever step the rival plays. To do this, player A Player B

have to resort to the mqxmin strotegy. That is, for each strategy, mark the worst case scena
Player 1
or the minimum gain. Given these minimums, player A chooses a strategy with the hig
A -2 0 5
minimum This is a strategy with the maximum of the minimums.
For player B, the gain is the opposite or negative of A's As such, B will optimise the obje The two players have opposite objectives on the same payoff matrix. Player A wants the
function in pretty the same manner as A. The only difference is that B is maximising own best while B is looking for the waste. To get to the optimal strategies, we have to create a
indirectly by minimising A's gain. The steps will be to get the maximum (for A) for each s column of minimums for each of A's strategies and a row of maximums for each of B's
and then find the lowest among these maximums. This will be a minmox strotegy. lt d strategies. This is shown in the table below.
from A's only in the order in which maximisation and minimisation are applied.
Row
Formally, plaver A will maxmin strategy I with Player B
mtn
t1,)dXInln il;7
3 1 5 1
This represents the lowest that player A is expecting from the game, Player B on the other hand Player A
-2 0 6 -2
will minmax strategyJ with
Col max 1 6
o-Itllt11?X tt;i
This will represent the highest that player B is willlng to concede. The two players are Clearly, player A will go for the first strategy. Of the minimums, it gives the maximum of 1.
strategising on the same variable, except in different directions One maximises and the other This implies that this is the lowest A will accept as a gain from the game. Player B on the
minimises. lf they both arrive the same answer, that is, other hand has (among the maximums) the lowest of 1. This is what B is willing to accept
maxmlll all - lnlll[lax (Lt j = (Lij as loss Of course, a loss to B is a gain to A. Since the two are equal, the game has a saddle
point.
then the point ai, point of the game lt is a saddle point in the ordinary sense
is called a saddle
of optimisation Recall that a saddle point is a maximum when viewed from one angle and a Example 15.3
minimum when viewed from another ln game theory, it is maximum for one player and a
minimum for another lt is the lowest that the maximising player is willing to accept from the Consider another game similar to one in Example 15.2 but with a different payoff matrix.
game and at the same time, the highest that the minimising player is willing to concede from Does the game have a saddle point?

the game Player B

The saddle point is also referred to as a Nosh Equilibrium, named after an Anrerican P ayer 8 3
mathematician John F Nash, Jr ln simpler definition, a Nash Equilibrium is where each player in A -5 5 4
a game feels they have done the best There is no incentive for any play to play otherwise Since
there is no incentive to change, a Nash Equilibrium is generally stable.
To get to the answer, we follow the same strategy. Player A will have to look for the
But not all games will have a saddle point! lf a two person zero sum game has a sacidle point, maximum of all minimums. For each strategy, a minimum outcome must be identified
then it is called a strictly determined game lt has only one point where the two players agree and then go for the highest among minimums. Player B will look at the payoff matrix in
Since there is only one acceptable outcome, the game is deterministic lf such a point does not the opposite. The player knows too well that the payoff pertain to Player A and since this
exist hcrwever, the game ls l<nown as non-strictly determined game The game fails to precisely is a zero-sum game, he must minimise the function. To achieve this, Player B will look for
point on the strategies that the player will take the maximum for each of the strategy and then pick the minimum of all. Notice that
indirectly, Player B is also getting the maxmin of own payoff.
Example 15 2
Given the above strategies, we construct the payoff matrix showing the column
Consider the following two person zero sum game and determine whether it has a saddle minimums and row maximums for the two players. The new table is presented below
point or not The payoff shown pertain to player A
320 | ,*-*oora,o* To GAME THE,RY Mathematical Optimisation and Programming Techniques for Economic Analysis 321,

Player B mtn The probability vector for player A is in row format because A is a row player. His/her strategies
are listed in rows.
Player 6 -3 3 -3
Corollary, for player B (column player) the mixed strategy is given by the probability vector
A
// o.\
-5 4 -5
5
"\
q,'l
max 8 5 4
o =l
Player A will go for the first strategy. Given this
\'/
\q"/
It is clear from the above table that
strategy, the lowest A is expecting to gain is -3. This is actually a loss of 3 But player B,
whlch must also meet the same conditions.
the rival, will play strategy 3 because it gives the lowest of all minimums. ln this strategy,
B is willing to concede or lose at the most 4. a1>o,viand
+ =1
/q1
j
ln this game, Player B is willing to lose or concede an amount which is seven (7) more =t'
Given the mixed strategies for the two players, the problem now is to find the optimal values of
than what Player A is expecting to gain. lt is clear the two players are not maximising on
the vectors P and Q. This is done by calculating for p ( and e) which will give the highest
the game. Player A, by adopting the maxmin strategy hopes to gain no less than -3 (or loss possible return to player A regardless of what the opponent plays.
of not more than 3) but the opponent is prepared to concede 4. This means Player A is
not maximising his gains. There is still room to increase gains. Player B on the other hand Example 15.4
by adopting the minmax strategy hopes not to lose more than 4 but the opponent does
A two player zero sum game between players A and B has the following payoff matrix M.
not expect him to lose, let alone that amount. This mean Player B is also not maximising
his position.
The payoffs relate to player A.

ln such games, the solution for each player is to behave unpredictably in the selection of M=(s 3\
strategies in order to keep the rival guessing. The probability of selecting each strategy should
\2 4/
Find the optimal values of the probability vectors P and Q.
provide minimum knowledge if any regarding the strategy he will choose. ln this way, each
player can increase own payoff over time. This leads to the discussion of Mixed Strotegies. Let us start with player A whose probability vector is p = (h pz): (h l-p). tt
player B goes for the first strategy, then the value G for player A will be
15.5 Mixed strategies Gt: \pr + 2(l - pr)
ln the succeeding subsection, we looked at pure games. These are games where there is only :3h+2
one optimal strategy. Each player will know, based on the payoffs, what the rival will go for. This results from multiplying the first column elements by the probability vector which
This was the case because the games have what was referred to as a saddle point. For Sames shows precisely how they will be played. lf player B will go for the second strategy, then
that have no saddle point however, there is scope for each player to increase payoff by the value of the game for Player A will change. The new value denoted by G2 will be
selecting strategy other than the maxmin or minmax. The opponent must not know which based on the second column values.
strategy a player will go for. This is known as Mixed Stotegy game. ,, p,)
More precisely, Downing and Clark (1988) have defined a mixed strategy as a decision to play
_3ir:Ia(l-
P7
various choices or strategies at various probabilities, and is described by a probability vector. ln Player A can hope to gain G : min(GuG). Since A is looking for the lower value of the
short, a player does not just go for a single choice or strategy as in pure strategy. Under mixed two, he is better off ifthe two are equalised. He knows too well that increasing one over
strategy, a player can play any of more than one choices. Each strategy is played by a fixed the other' will make him worse off. This piece of information will allow the creation of two
probability. equations in lwo unknowns. That is

For a two player game, a mixed strategy for Player A is a probability vector G=3ht2
P = (Pt Pz P*) G=4_pt
pi i. As a probability vector, each element in the Solving the two simultaneously, we arrive at
where is the probability of playing strategy
vector must be non-negative and the summation of all elements must equal unit. That is
/" -= /.5\
(.;/, r
G -
= 3.5
sa
> 0,Vi and
p1
/Oi =
1

i=7
I

I INTRODUCTTON TO GAME THEORY Mathematical Optimisation and Programming Techniques for Economic Analysis 323

At the optimal, player A will play the two strategies half the time each. ln this way, A wlll lowest player 1 is expecting from the game is the value of the game itself. For player 2, since
be assured of gaining on average G = 3.5 the payoff is the converse of what playerl gets, the objective is to make the gain as low as
possible. Similarly to player 1, the maximum player 2 is expecting to concede is the value of the
Similarly, Player B will have to consider his take given the different strategies that A can
game. This illustration gives two linear programmes which are duals of each other.
play. rhe mixed stratesy for B is given bt a = (X) : G! tf player A goes for the
rr) ln the first, player 1 wants to maximise the objective function. Using a mixed strategy, player 1
first strategy, the value for B is
has the following expectations which depend on what player 2 plays.
Gt=5h+3(1 -qr) /m m m \
=zqt+3 G =min( Io,,p, Ir,,p, Fo,,r,)
lf A plays the second strategy, the value for B will be \fiHA/
There are n possible outcomes for the mixed strategy of player 1. Each related to a particular
Gz=2h+4(]--qr)
:4_Zet strategy that player 2 can go for. These represent different gains for player 1. ln the worst case,
the objective of player B is to minimise the gains for player A so as to indirectly optimise player2 isnotexpectinganythinglessthantheminimumofthenpossibleoutcomes.Thusthe
his. He will thus choose the minimum of the two game values. At the optimum, he minimum in the vector provides the lower limit on the gains for player 1. Anything above is
chooses a mixed strategy that will give equal value regardless of what the opponent plays. desirable for the player.
This will result in a system of simultaneous equations The linear programming formulation of the above problem is

G=2qtt3
Pf!,G' i=t'z'"''m
G:4_Zqt
By solving the two equations simultaneously, we conclude that Player two will play the subject to
first strategy a quarter of the times. This means the second strategy will be played three- m
quarters of times.The value of the game for player B is 6= 3.5.
\-a

Lo,iP,' G' j:1'2""'n

i--t
s=(12), G=35 Y,,=r
With mixed strategies, both players are maximising gains from the game. ln the above example, L"
Player A is hoping to gain 3.5 from the game. This is exactly equal to what player B (the rival Pi20, i=t,2,...,m
player) is willing to concede. Player B is also only willing to concede an amount equal to what A
The problem above has many unknowns than are equations. lt would therefore be difficult to
is willing to accept a gain from the game.
arrive at a desirable point. Moreover, it does not yet look like the linear programming
The solution for a two player zero sum game can also be evaluated using the theory of Linear encountered in the earlier chapter. One more step is needed to transform the problem into the
Programming. This is discussed in the section that follows. familiar linear programming problem. Let us introduce a new system of variables which will
serve a very critical purpose, that is, to swallow one unknown.
15.6 Game theory as a linear programme
Let r, :';,, = r,2, ,m. We use this new system of variables to transform the problem.
A two player zero sum game can be formulated as a Linear Programming problem. Consider a
Though the order is to start with the objective function and then the constraints, it is inevitable
game involvingtwo players; player 1 and player 2. The corresponding payoff matrix (for player)
that the order is altered here. First consider how the constraints will change. The new
is given by the matrix
constraints are now
t: (",), i = t, ...,m and j = !, ...,n
j:t'z'
The payoff matrix is of dimensions m x n. This means player t has m strategies while player 2
has n. Let us now assume that the game has a value of G = G. The mixed strategies for the two
io,,*,, t'
i=1
'n

players are as given in the previous section. They pr : (Pt Pz P*) and, :
s":r
L^' G
"r" (?_)
xr) 0, i=t2 m
ln the given payoff matrix, the objective of player 1 is to get the highest possible value. The
324
I

| TNTRODUCTTON TO GAME THEORY Mathematical Optimisation and Programing Techniques for Economic Analysis 325

ln the first constraint, the value of the game, G, was common on both sides of the inequality Example 15.5
and so was factored out. The requirement now is that the left hand side does not fall below a
A two person zero sum game for player 1 and player 2 has the following payoff matrix
known value of 1. ln matrix notation, this is expressed as X'A > 1 which replaces the earlier /5 9\
one with P. ln the second constraint, the action of bringing the new variable is equivalent to which pertaintoplayerf. isA: (9 6 )
dividing by a common value G on both sides of the equation. since by supposition x, =p;, *e \o sl
left hand side has the new variable. a. Formulate the linear programming problem.
The third constraint needs some closer attention. Each probability value is divided by a constant b. Find the optimal strategies for the two players.
G. The resulting z;'s would only remain non-negative if the divisor is positive. lf this is not met,
To get to the linear programming, there is need to look at the nature of player 1's mixed
then the third constraint would not hold This has the potential to put the solution in jeopardy.
strategies. The player has three cholces and the mixed strategy must reflect this. But let
To prevent this from occurring, it is made a condition that the value of the game is always
us look at the gain to player 1 for each of the three strategies. lt should suffice to state
positive. But the value of the game is part of what we are looking for. lt is not known yet. So,
that the benefits of the first strategy are greater that the third, irrespective of what the
how can we ensure it is positive? Well, this is achieved by ensuring that every entry in the
opponent plays. The same hold between the second and third strategy. ln game theory,
payoff matrix is positive. lf this is not the case, an arbitrary number large enough to make all
we say that the first two strategies dominate the third. As such, the player will never
entries positive is added to every element in the matrix. This is equivalent to player 2
consider the third strategy. His choice will now be between the two. This information
transferring a given sum to player 1 before the start ofthe game. This can be reversed once the
helps is simplifying the payoff matrix. The effective matrix is thus A : (l :)
actual value is known. \9 6)
Let us now turn to the objective function. lt remains to maximise the value of the game G. But Therefore, p' : (h Pz). the P vector is transformed to the X vector by dividing by the
G no longer appears in any ofthe constraints. lts equivalent must be used instead. To do this, yet to be known value of the game G. Therefore X = (xt 12)' The linear programming
we examine the second constraint. lt is problem is
m
\-"'=1 ,nirr$r,
a'G
from the binding equation above, the objective of maximising G is equivalent to minimising the subject to
a'
left side of the equation. This is because the G being maximised is in the denominator.
Maximising it actually minimises the quotient. Thus, the new objective is now that of (x, .i(Z ). t
minimising lllr t, subject to the constraints listed above. xi> 0, i= 7,2,3.
Formally, the resulting linear programming formulation is summarised as Or in a more familiar way, the linear programme is presented as
minxl I x2

-inf ,, subject to

subject to
2 5x, t 9x2> I
9xr+ 6x2> I
m
\- xyx2 >- 0

/)) oiix 21 t-1,2 n

Two methods can be used to find the optimal values of x1 and x2. Once these are known,
xr>0, the value of the game C and the mixed strategy P can be gotten. We leave is to the
t=12..,m
reader to verify the following solution.

once the optimal values of xis are know, the value of the game G can be calculated. Then the
individual p;'s can be calculated based on the x1's and the value of the game G.
.: (l,i',)
326 | ,rr*orra,o, ro cAME THE.RY Mathematical Optimisalion and Programming Techniques fbr Economic Analysis 327

/3r \ Recall thatfor player 2, the objective is not to maximise the value of the game. player 2 seeks
c =st/t =(ill, to minimise it. ln so doing, player2 will be indirectly maximising own gain. But as stated in the
dual case, minimising G is equivalent to maximising)T=J1.ln summary, player 2 will be faced
\ 6',l
The mixed strategy P must show that there is one option that player 1 will dare not
with the following problem

consider. A zero rn the third element shows that this option is available but because of it
being dominated by other options, the player will never cons der it. max
+
rlli
i=t
ln the second problem, we look at the game from player 2's perspective. Given the payoff subject to
matrix, player 2's objective is to minimise the value of the outcome. Given player 2's mixed
strategy, the following are the possible outcome depending on Player 1's unknown mixed
strateg,.
/^ il n \
fo,,r,= 1. i:t2 m
n
o,,q, o,iqi
\FFF/\ Ir-,r,)
c' = sal
^or(\ LY'= c
The outcome vector has a total of m elements. Like in the earlier case, each element
corresponds to the possible strategy played by the rival ln the worst case for of the game, !120' j=72 n
player 2 is only expecting to lose the maximum of the m possible outcomes. once the values of y have been established, the value of the game and consequently the mixed
strategy would be known.
The linear programming formulation for player 2 is to

i=1'2""'n Example 15.6

olri,\rG'
su bject to Redo Example 15.5 from Player 2's perspective.
I The summary of the problem in Example 15.4 is as follows (from player 2's perspective)
sa
Lo,tq'. C, i: 't,2, . m
subject to
miny1 + y2
n
sf syt+gyz<L
) a.=l
,L'J 9yr*6y2<L
Qt20, l:t'2...'n !t,lz 31
Using either the graphical or simplex method, the optimal values of y are
Like in the previous case where we looked at the mixed strategy for player 1, the above
formulation is not yet the familiar linear programming lt must be transformed further, to a r7t \
stage where the G will not appear in the inequality To attain this, a new variable must be
introduced. ln the dual case, the variable that was introduced was x This tlme around, y is
,:li,"
\ /5tl
l

used. From this, the value of the game and the mixed strategy are given as

Let ,jy, = I. Witf, the new variable, we can straight away go into changing the constraints The /3 1\
c
method that was applied to player 1 also applies, mutotis mutdndrs The new constraints are
c=sr/t, ,=(^:ll
n
Ytt/
Player 2 will play the first strategy thrice every seven times and play the second strategy

/J
) liiYi
'''' S 1, i:1,2, .,m four times every seven times.

I,, 1
U

!120'
 I

30 I AN TNTRODUCTTON TO DYNAMTC PROGRAMMTNG Mathematical Optimisation and programming Techniques for Economic Analysis 33L

16.2 Steps in dynamic programming problem Figure 16.1: Cost of moving from one state to another

The following are the steps involved in any dynamic programmlng problem:

1. Divide the problem intostoges with a policy decision required at each stage; BCD EFG HI
2. Each stage has a number ofstotes associated with it;
FT. frl B
PT4I6-] ' FTI
3. The effect of a policy decision is to transform the current state into a state associated
4
with the next stage;
Given the current state, an optimal policy for the remaining stages is independent of the
policy adopted in the previous stages; (technically this is known as lhe Morkovion
,ffi
c E-T-r-l-4] F E-ftl
"1,_]_,_l
property);
5. The solution procedure begins by finding the optimal policy for each state in the last
stage;
6. A recursive relotionship is available which identifies the optimal policy for each state Figure 16 2: Route Network with associated costs available to the traveler
with n stages remaining, given the optimal policy for each state, with (n'll stages
remarnrng;
7. Using the recursive relationship, the solution procedure moves backward stage by stage,
each time finding the optimal policy for each state at that stage, until it finds the
optimal policy when starting at the initial stage.

16.3 The classic stagecoach problem

The above steps have been frequently illustrated in many books on dynamic programming with
what is considered as a classic problem, namely, the stagecoach problem. A mythical fortune
seeker in Missouri decided to go west to join the gold rush in California in the mid-nineteenth
century. The main transport available was the stagecoach which would pass through unsettled
country with the danger of being attacked by bandits. Although the starting point and
destination were fixed, the traveller had a choice of the possible routes he could take. The
possible routes are shown in Figure 15.1 where each state is shown by a circled letter. The
direction of the travel was from left to right starting from Missouri (A) to California (J), and To give the problem a more contemporary and local flavour, let us assume that a man working
required four stages to be passed through by using stagecoaches. in Mfuwe in Zambia's Eastern Province has just heard that one of his long-lost uncles is actually
living in Dundumwezi in Southern Province and so he wants to visit him. The roads are not easy
ln view of the danger present in the travel, the stagecoaches offered life insurance policies to
to pass and the distance is long (over 1000 km). There could be many mishaps along the way.
the traveller and the cost of the policy would vary directly with the extent of danger perceived.
But there are many alternative routes to choose from with varying costs. He wants to choose
Hence, in order to maximise his security, the traveller decided to choose the route with the
the route with minimum cost
minimumtotal costoftheinsurancepolicy. Thecostofthepolicyforgoingfromstateitostate
7, C;; is shown below: The data on the route network between Mfuwe and Dundumwezi could still be similar to what
is shown in Figure 15.1 for the stagecoach problem. so let us see how this can be solved as a
problem in dynamic programming.
The insightful point that we shall gain at the end of this exercise is that the overoll optimol
decision is not necessorilv the sum of optimol decisions ot eoch stoge. For example, the
cheapest route by each successive stage ts:

A+B+f+l+J=lJ
But the following route yields a lower cost:
 I

turnooucrtoN To DYNAMIc PRoGRAMMING Mathematical Optimisation and Progrming Techniques for Economic Analysis 333
32 I nru

A+D+F)l-J:L!
With the above point in mind, we shall now see the analytics of the solution'

16.4 Solving a dynamic programming problem s B c D Fo'(s) x4

Let X, be the immediate destination when there are ?1 more staSes to go. Then the route to be A 13 LT 1-1- 71, CorD
chosen is:
A+X+)X3+Xr; )(r, wlrcreXr=J'
Thus the optimal routes are:
Let Fr(S, Xr) be the cost of transport for the last n stages, given that the traveller is in state
s
and selects X, as the immediate destination. A +C + E ) H ) J;

Givensandr-,letx;denotethevalueofx.whichminimisesFn(S,X).Thatistosay, A+D+E+H+J;and
ff(s) : F"(s,x;) A +D + F; I + J;
We have to solve for {(1)' We do so by successively finding F1-(S), Fr-(S), F3-(S) and Fa-(l)' The All the above routes yield a minimal total cost of 11.
sequence of backward computations would then be as follows: The recursive relation is given by:
With onlv one more stage to go
4i(s) : min{csy*+ F;_{x)}
5 F,'(S) x1
16.5 Bellman's Principle of Optimality
H 10 The above recursive relation is known asthe Bellmon equation afterthe mathematician Richard
Bellman who formulated the Principle of Optimolity. The Bellman Principle of Optimality states
4 10
I
that whatever the initial state and initial decision(s), the remaining decisions must constitute an
optimal policy with regard to the state resulting from the initial decision.
Another writer Aris (1964) summarised this principle as: lf you don't do the best with what you
with two more stages to go
happen to have got, you'll never do the best you might have done with what you should have
I F,'(5) X.' had.
5 H

This is the essence of the Markovian property stated in section 16.2. The optimal decision in the
4 8 4 H
E
current state depends only on the current state and not on how you got there. Given the
7 7 current state, an optimal policy for the remaining stages is independent of the policy decisions
F 9 I

in the previous stages. This is because the knowledge of the current state of the system conveys
G 6 7 6 H all the information about its previous behaviour necessary for determining the optimal policy
henceforth. Any problem lacking this Markovian property cannot be formulated as a dynamic
programming problem. (see Hillier et al,2OI2l.
With three more stases to go

5 E F G F.'(S) x,

B 11 LL L2 11 EorF

c 1 9 10 7 E

D 8 8 17 8 EorF
II
 Mathematical Optimisation and Programming Techniques for Economic Analysis 331

SOME KEY MESSAGES

The Main Takeaway

lf there is one takeaway that we hope readers of this book will carry, it is this: the relation
between problems and techniques for their solution in not a one-to-one relation. A given
problem can often be solved using more than one technique. And a given technique can often
be used to solve more than one type of problem. lt is this f/exiblity of the problems that allow
plural techn iques for their solution and the versotility of the techniques that have the capabil ity
to solve a variety of problems that one must learn to appreciate.

Flexibility in Problem Solving

As an illustration of problem flexibility in terms of its solution, consider the problem below
maxZ = 2x, I 5x2 subject to

2x7+xr< 430
Zx2 a 460
x'' x2 2 0

this can be solved as a linear programming problem. We transform the inequality constraints
into equality constraints by adding slack variables. We then have
2xr*xr*Sr=439
2x2+Sz=460
We start with an initial arbitrary basis consisting of S, and 52 and using the simplex procedure,
we finally attain the optimal basic feaslble solution.

But the above problem can also be solved as a constrained optimisation problem involving
inequality constraints And slnce the objective function and the constraints are all linear, it can
be viewed as an exercise in concave (convex) programming problem for which the fulfilment of
the necessary conditions stipulated by the Karush-Kuhn-Tucker conditions will also ensure
fulfilment of the sufficient conditions.

The Karush-Kuhn-Tucker conditions for the above problem will be attained as follows: from the
Lagrangean function

L :2xt * 5x, - 7r(2x, t x, - 43o) - )r(Zx., - 460)

the necessary conditions for a maximum are:

AL AL
oxt
< 0, x. )-0, x1 7=0
oxt
-AL AL
. (0,
o xz
xz)0, x27=0
oxz
 Mathematical Optimisation and Programing Techniques for Economic Analysis | ,r,
SOME KEY MESSAGES

Blood group O individuals can receive blood only from individuals of group O but can donate
AI, AL
d),
2,>u, i,-'-0
(tA\
blood to individuals with type A, B, AB or O (they are universal donors).

_>o
OL AL, The above characteristics are summarised in the following table
> 0. ),-: 0
dAz ^,' 'dlt
Donor
Recipient
B AB o

Using the above equations, one can obtain the optimal solution.
B

AB
The problem can also be solved as a problem in dynamic programming! There are
o
resources and hence there are two states. Let (ry,n) be the states are stage j,0 : 1,2)
Now consider a hospital which has patients belonging to different blood groups who are in
Then the recursive equation is given by:
need of blood. Let us indicate them as PA, PB, PAB and Pp, respectively. the hospital has has a
At stage 2 blood bank which stores supplies of different types of blood. Let us indicate them as SA, SB, SAB
and So, respectively.
l2(m2,n2) - ,=lll,Q;, 15rrJ
o<2r !r17 There is a cost associated with administering each unit of blood to each patient. Let us denote

At stage 1
this as C1i where i and j represent the blood group and the patent group respectively that is,
C1; is the cost of administering a unit of blood in the ith group to a patient in the 7th group. The
f.(m. nr) - n.tlil,,,[2r,
-t fr(m' - 2x.,n.)f problem is to use the quantities of blood available in the blood bank to meet the patients' need
for blood at minimum total cost.
Solving the recurslve equations will yield the optimal solution. We leave it to the reader to
verify that whatever technique is applied, the optimal solution will be l{ = 100, xj = 230 Here, the different types of blood stored in the blood bank are the "origins" and the patients
belonging to different blood groups are the "destinations". The quantities of blood available in
As another example of problem flexibility, we had shown earlier in Chapter 15 how a gamc the blood bank are the supplies of blood and the patients' requirements are the demands.
theoretic problem can also be expressed as a linear programming problem
The problem then is: How should the supplies of blood be "transported" to meet the demand at
Versatility of Techniques the destinations so as to minimise "total transportation cost"?
And now we give an example to illustrate the versatility of techniques
The problem can thus be viewed as a "transportation" problem and be solved by the
Most reader would know that there are four blood groups: A, B, AB and O And individuals techniques described in chapter 14, even though there is no transport involved. No doubt, in a
belonging to these different blood groups have different capabilities in terms of donating and very elementary sense, one can think of blood being physically transported from the blood
receiving blood: bank to the different wards where the patients are admitted. But then, the cost we a trying to
minimise is not this physical transport cost.
Blood group A individuals can receive blood only from individuals of groups A or O and can
donate to individuals with type A or AB; Efficiency

Blood group B individuals can receive blood only from individuals of groups B or O and can While problem flexibility and technique versatility may be advantageous, it is important to bear
donate blood to individuals with type B or AB; in mind another critical factor- efficiency in problem solving. Efficlency is determined both by
the amount of time taken and by the complexity involved in problem solving. More often than
Blood group AB individuals can receive blood from any group (they are universal recipients) but not, time and complexity are directly related. The more complex the solution process, the more
can donate blood only to individuals with type AB.
I sovr rev MESSAGFs
Mathematica[ Optimisation and Progranrming Techniques fbr Economic Analysis 341

time it will take to complete. we had some simple illustrations

of this in the chapter on Linear
Progra m ming.
SELECTED REFERENCES
we noted that the comprexity of sorution of a rinear programming depends more on Allen, R G D (1938): Mathemoticol Anolysis for Economists, Macmillan and Company Limited,
the
number of programme variables than on the number of constraints. London.
We also noted that the
number of programme variables and constraints are interchanged
between a primal and a dual
problem. Hence if a primal problem has more programme Anthony, M. & Biggs, N. (1996): Mothematics for Economics ond Finonce: Methods ond
variables than constraints, then it is
more efficient to sorve the duar probrem since it wiil yierd the same Model ling, Cambridge University Press, Cambridge.
optimar sorution

Again, we noted Aris, R. (1954): Discrete Dynomic Progromming, Blaisdell Publishing Company, New York.
that large scale programmes can take an enormous amount of time for their
solution if they are sought to be sorved by the standard argorithmic procedure Baldani, G., Bradfield, J. & Turner, R. (1996): Mothemotical Economics, Harcourt Brace and
of moving from
one extreme point of a feasible region to another extreme point.
Hence, efforts are being made Company, Fort Worth.
to develop new algorithms (such as Karmarkar's argorithm) that wiil significantry
cut down on
the solution time. Chiang, A. C. & Wainwright, K. (2013): Fundomentol Methods of Mothemoticol Economics,
McGraw-Hill, New York, Fourth Edition.
with the advent of computer technorogy, one discipre that is fast deveroping
in economics is
computotionol economics computational economics uses computer Bailey, D. l2OO4l: Mothemotics in Economics, McGraw-Hill, New York.
based economic modelling
for solution of analytically and statistically formulated economic problems.
Dixit, A. (1990): Optimisotion in EconomicTheory, Oxford University Press, Oxford.
The quest for time-efficient solutions for computationally hard optimisation problems
will Dowling, E. T. (2006): lntroduction to Mothemoticol Economics, Schaum's Outline Series,
continue, unbounded in time
McGraw Hill, New York.

Downing, D. and Ciark, J , (1988) Quantitative Methods Barron's Education Series, lnc, New
York;

Glaister, S. (1984): Mothemoticol Methods for Economists, Blackwell, Cambridge, Third Edition

Hillier, F. S, Lieberman, G J., Nag, B. & Basu, P (201,21: lntroduction to Operotions Reseorch,
Tata McGraw-Hlll, New Delhi, Ninth Edition

Hoy, M., Livernois, J., Rees, R. & Stengos, T. (2011): Mathemotics for Economics, MIT Press,
Cambridge.

lntrilligator, M. D. (2002): Mothemoticol Optimisotion ond Economic Theory, Society for

lndustrial and Applied Mathematics, Philadelphia.

Ostaszewski, A. (1993): Mothemotics in Economics: Models ond Methods, Blackwell, Oxford.

Rowcroft, J. (1994): Mothemoticol Economics: An lntegroted Approoch, Prentice Hall, Ontario.

SOME KEY MESSAGES Mathematical Optirnisrtion and Progranrr.ning Technic;ues lor Economic Analysrs 341,

time it will take to complete. We had some simple illustrations of this in the chapter on Linear
Program ming. SELECTED REFERENCES

we noted that the complexity of solution of a linear programming depends more on the Allen, R. G. D. (1938): Mothemoticol Analysis t'or Economists, Macmillan and Company Limited,
number of programme variables than on the number of constraints. We also noted that the London.
number of programme variables and constraints are interchanged between a primal and a dual
Anthony, M. & Biggs, N. (1996): Mothemotics for Economics ond Finance: Methods ond
problem. Hence if a primal problem has more programme variables than constraints, then it is
Modelling, Cambridge University Press, Cambridge.
more efficient to solve the dual problem since it will yield the same optimal solution.
Aris, R. (1964): Discrete Dynomic Progromming, Blaisdell Publishing Company, New York.
Again, we noted that large scale programmes can take an enormous amount of time for their
solution if they are sought to be solved by the standard algorithmic procedure of moving from Baldani, G., Bradfield, J. & Turner, R. (1996): Mathemoticol Economics, Harcourt Brace and
one extreme point of a feasible region to another extreme point. Hence, efforts are being made Company, Fort Worth.
to develop new algorithms (such as Karmarkar's algorithm) that will significantly cut down on
the solution time.
Chiang, A. C. & Wainwright, K. (2013): Fundomentol Methods of Mothemoticol Economics,
McGraw-Hill, New York, Fourth Edition.
With the advent of computer technology, one disciple that is fast developing in economics is
Bailey, D. (2004): Mothemotics in Economict McGraw-Hill, New York.
computcttionol economics. Computational economics uses computer based economic modelling
for solution of analytically and statistically formulated economic problems. Dixit, A. (1990): Optimisotion in EconomicTheory, Oxford University Press, Oxford.

The quest for time-efficient solutions for computationally hard optimisation problems will Dowling, E. T. (2006): lntroduction to Mathemoticol Economics, Schaum's outline Series,
continue, unbounded in time. McGraw Hill, New York.

Downing, D. and Clark, J., (1988) Quantitative Methods. Barron's Education Series, lnc, New
York;

Glaister,S.(1984): Mothemoticol MethodsforEconomists, Blackwell,Cambridge,ThirdEdition.

Hillier, F. S., Lieberman, G. J., Nag, B. & Basu, P. (2Ot7l: lntroduction to Operotions Reseorch,
Tata McGraw-Hill, New Delhi, Ninth Edition.

Hoy, M, Livernois, J., Rees, R& Stengos, T. (2011): Mothematics for Economics, MIT Press,

Cambridge.

lntrilligator, M. D (2002): Mothematicol Optimisotion and Economic Theory, SocietY fot

lndustrial and Applied Mathematics, Philadelphia.

Ostaszewski, A. (1993): Mothemotics in Economics: Models and Methods, Blackwell, Oxford.

Rowcroft, J. (199a): Mothemoticol Economics: An lntegroted Approoch, Prentice Hall, Ontario.

342
I

I SELECTED REFERENCES Mathematical Optimisation and Programming Techniques f'or Economic Analysis 343

Silberberg, E. & Suen, W (2001): The Structure of Economics: A Mothemoticol Anolysis, lrwin
McGraw-Hill, Boston. Appendix I

Spivey, A. & Thrall, R M. (1970): Lineor Optimisation, Holt, Rinehart and Winston, New york.
CONCEPTS AND THEOREMS NAMED AFTER MATHEMATICIANS
Srinivasan, G. (2012): Operotions Reseorch, PHI Learning Private Limited, New Delhi, Second
Edition.
Argand Diagram: Named after the French mathematician Jean-Robert Argand (L768 - 1822l.

Bellman's Principle: Named after the American mathematician Richard E. Bellman (tgZO -
Taha, H. A (2010): Operotions Research: An lntroduction: lnternotionol Edition, Prentice Hall,
1984) celebrated as the inventor of Dynamic Programming in 1953.
New York.

Taylor, R. & Hawkins, S. (2008): Mothemotics for Economics ond Business, McGraw-Hill, London.
Bernoulli equation: named after the Swiss mathematician Daniel Bernoulli (L7OO - t782\.

Georg Cantor: German mathematician (1845 - 1918), known as the inventor of set theory.
Yamane, T. (1968): Mothemotics for Economists: An Elementory Suryey, Prentice Hall of lndia,
New Delhi. Cayley-Hamilton Theorem: Named after the British mathematician Arthur Cayley (1821 - 1895)
who founded the modern British school of pure mathematics; and the lrish
mathematician William Rowan Hamilton (1805 - 1865) noted for his contributions to
classical mechanics, optics and algebra

Cartesian product: Named after the French mathematician Ren6 Descartes (1596 - 1650)
credited as the father of analytical geometry.

Cobb-Douglas function: Named after the American mathematician and economist Charles Cobb
(1875 - 1949) and the American economist Paul Howard Douglas (1892 - 1976) who
tested the functional form of production functions.

Cramer's Rule: Named after the Swiss mathematician Gabriel Cramer (17O4 - !7521.

De Moivre's Theorem: Named after the French mathematician Abraham de Moivre (1667 -
1754) well-known for his formula that links complex numbers and trigonometry and
his work on normal distribution and probability theory.

Euler relation or Euler's formula: Named after the Swiss mathematician and physicist Leonhard
Euler (1707 - 1783) who, among his many contributions, introduced the concept of
a mathematical function.

Euler's Theorem: also named after Leonhard Euler

Hamiltonian function: Named after the lrish mathematician and astronomer William Rowan
Hamilton (1805 - 1865).

Hessian matrix, determinant: Named after the German mathematician Ludwig Otto Hesse
(1811 - 1874) who developed the matrix and the determinant.
 I

I coNCEPTS AND THEOREMS NAMED AFTER MATHEMATICIANS Mathematical Optimisation and Programming Techniques for Economic Analysis 345
344

Hungarian method: Named after two Hungarian mathematicians D6nes Konig (1884 - 1944) Schur/s Theorem: Named after the mathematician lssai Schur (1875 - 1.94Il. He was born in
and Jen6 Egervdry (1891 - 1958) Belarus (then Russian Empire) and died in Tel Aviv, (then Palestine, now lsrael) but
worked in Germany most of his life.
Jacobian matrix, determinant: Named after the German mathematician Carl Gustav Jacob
Jacobi (1804 - 1851) who made fundamental contributions to areas such as Venn Diagram: Named after the British lo8ician and philosopher John Venn (1834 - 1923).

differential equations and number theory.

Vogel's Approximation Method: Named after William R. Vogel.
Karmarkar's algorithm: Named after lndian mathematician Narendra Karmarkar (b 1957).
Young's Theorem: Named after the British mathematician William Henry Young (1863 - 1942).

Kiinig's Theorem: Named after the Hungarian mathematician D6nes K6nig (1884 - 1944) who He made significant contributions to the study of functions of several complex
wrote the first textbook in the field of Graph Theory' variables. He was the husband of an equally renowned mathematician Grace
Chisholm Young who initially published her papers under her husband's name! Their
Karush-Kuhn-Tucker conditions: Named after American mathematicians William Karush (1917 son Lawrence Chisholm Young was also a brilliant mathematician who contributed
- tg97l, Harold W. Kuhn (1925 - ) and the Canadian mathematician Albert W. significantly to measure theory, calculus of variations and optimal control theory.
Tucker (1905 - 1995).

Lagrangean multiplier: Named after the ltalian mathematician and astronomer Joseph-Louis
Lagrange (1735 - 1813).

Laplace expansion: Named after a French mathematician and astronomer Pierr-Simon Laplace
(L749 - L827l

Leontief lnput Output Matrix: Named after a Russian-American economist Wassily Leontief
(1906-1999). Though Leontief was not a mathematician, his matrix, the Leontief
input Output matrix is actually a mathematical concept'

['H6pital's rule: Named after the French mathematician Guillaume de l'H6pital (1661- 1704).

Markovian property: Named after the Russian mathematician Andrey Markov (1856 - 1922).

Nash Equilibrium: Named after the American mathematician John Forbes Nash, Jr. (b. 1928)

Perron.Frobenius Theorem: Named after the German mathematicians Oscar Perron (1880 -
1975) and Ferdinand Georg Frobenius (1849 - 79171.

Pontryagin's Maxlmum Principle: Named after the Soviet mathematician Lev Pontryagin (1908
- 1s88).

Riemann lntegral: Named after the German mathematician Bernhard Riemann (1826 - 1866)

Routh theorem: Named after the British mathematician Edward John Routh (1831- 1907).
 I
| ,*or,*or,o* rYPE euEST..NS Mathematical Optimisation and Programing Techniques for Economic Analysis 347

An individual is HIV positive. Doctors have prescribed two drugs for him and his health status
Appendix ll depends on the consumption levels of the two drugs. lf the health status of the individual is
measured on a scale of 0 to 10 (0= extremely poor health, 1O=perfectly normal health) and if his
health function is given as:
EXAMINATION TYPE QUESTIONS
Y = -(Xt- l), - (X2- 2)2 + l0
PAPER ONE QUESTIONS Where I: health status, X1 = dosage ofDrugl, Xr: dosage of Drug 2.
question One a) What is the optimal dosage of the two drugs

a. Given the following matrix b) What is the best health status he can attain with the optimal consumption of the two
d rugs?
t2 1 -11
lo r rl c) Suppose we are also told that the individual can tolerate only one drug dose per day.
lz o -2ll What will be the best health status he can now attain?
i. Construct the matrix of Eigen vectors for the following matrix: Question Four
ii. Using the Eigen values, determine the sign definiteness of the above matrix?
a. Check the sign definiteness of the following quadratic form:
b. The coordinator of a research project wants to recruit researchers and research
assistants A researcher is paid K500 per day and a research asslstant is pald K300 per day. Q =2xt2 +5xr2 +73xrz +6xlxz+l0x1x3*l4x2x3
The project is estimated to cost: b Given the following market model:
Z:600O*6x3-36xy*3y2 QP : r80 -o'7sPt
where x is the number of researchers, and y is the number of research assistants
Qt:-30+o.3Pt-1
i. How many researchers and research assistants should be recruited for the
project to minimise cost? Po = 22O

ii. What will be the actual staffing cost of the project. i. Find the price at all times t in the market

iii. Use the second order condition to show that the answer above is indeed
ii. Comment on the dynamics of the price path.
mrnrmum question Five
Question Two
a. The demand function for a good X is given as:
a. Diagonalise the following matrix: Q = 3000 - 4Px + Sln (Py)
t4 lr Where 0 is quantity demanded of good X, Px is the price of the good X and Py is the
lg zl price of another good, good Y.

b. LetPyP2andP3 denotetheprofitswhichamulti-productcorporationearnsfromthe i. What is the cross price elasticity of demand when P : 5 and P' : 10?
production and sale oftea, coffee and cocoa respectively. The corporation's economics
ii. On the basis of your answer, what can you say about the relationship between
department believes that he profits are linked as follows: the two goods?
zPa- * P.:5
Pz
b.
Pl-3P2+2h=2
For the utility function U = U(x,y) to be maximised subject to the budget constraint

2P1+P2*4Pr:-3 xP, * yP": B,

x Set out the first and second order conditions that must be fulfilled.
Using a 3 3 marix, work out the profits of each product

Question Three
I

EXAMTNATTON TYPE QUESTIONS

Mathematical Optimisation and Programing Techniques for Economic Analysis | 349
I

ii. Does the optimal decision look inferior? lf yes, why is it still optimal?
PAPERTWO QUESTIONS
b. Consider the following equation:
question One Yt:16*3Yt-r
Analyze the time path of y with the first value of
I being given as 5.
a. Given the following differential equatlon:
Question Four
dy
*+
aY = 12
a. State the order and degree of the following differential equation:
i. Obtain the general solution showing clearly the complementary function and d.zy tdyl3
the particular solution. dp-*\i) tx2=o
ii. Test for the dynamic stability of the equilibrium. b. A firms profit function is given by:
n : l2Xt - Xr' + 24x2- l.sx22
The firm is faced with a resource constraint given by:zxt + X2 =
27
Question Two i. Calculate the profit-maximising outputs and the maximum profit

a. A manufacturing company estimates that the marginal costs for its business activities
ii. Compare the results in (a) with the results you would have obtained in the
absence of the resource constraint
follows the following function:

MC=4Q2+T+ 2Q+"12
a
the fixed cost is 150,000
i. Find the total cost function and average cost functions

ii. Calculate the total cost if 50 units are made

b. Due to large-scale mining activity on the Copperbelt, there are opportunities for small
companies to do business by way of providing a variety of ancillary services to large-
scale mining companies. At any time, there are 6000 small business companies that are
actually engaged in business or are seeking to do business. From past data, we know
that a company that is doing business this year has a 30 percent chance of losing its
business next year, and for a company without business this year, there is 60 percent
chance that it will have business next year. Find the equilibrium number of small
companies that will be engaged in business with the large companies.

Question Three

a. Suppose there are two firms in the market selling a product at some fixed price-
Advertising does not affect the total market demand but each firm's share of the
market will depend on the relative advertising levels chosen by it Each firm chooses
between two advertising levels: 'high' H and 'low', L. The total gross profit for both
firms is K1000. The cost of H for each firm is K400 and of L it is K200. When both
firms advertise at the same level, they split the market and hence the net profit, fifty-
fifty else the highest advertising firm gets K800 while the other gets K200 For
example, if both firms advertise at a high level, each firm has a gross profit of I(500
and a net profit (gross profit minus advertising cost) of 100.
i. What is the optimal decision for firm 1 and firm 2?
 Mathematical Optimisation and Programming Techniques for Economic Analysis 351
| ,ror,ro,o,u rYPE euESroNS

Given the consumption function

PAPERTHRE
c=o.oryz+0.4y+1oo
Question One calculate the marginal propensity to save when
^ ,. 1E
a. Given d, -
Y_LJ
b Y= 10

Question Five

i. The following set of equations describes the behaviour in the market of a particular commodity:
ii. at the equ br um
QP =t2o-o.SPt
b. Deter s.
0f=-30+0.3Pt
Pt = Pt-t + a(Q?-, - Qf-r)
i.

where QD is quantity demanded, Qs is quantity supplied, P is the price and a is a positive

ii. parameter. The market does not always clear but the price adjusts depending of the deficit
from the previous period.
question Two
a. Solve for the long run equilibrium price P-
a. Find the time path for the following differential equation. b. Solve the first order difference equation and find the particular solution if Pr=0 = 200.
d2v c. d
7V
+ o.zst = +, Y(0) = 16,Y'(0) = 2
For what values of will there be damped oscillations in the price?

b. Given the demand and supply function for a market below:

. dP dzP
' =42-4P-4-*-dt
Od
dtz
Qs : -6 + 8P, P(0) = 6, P'(0) = 4
i. Find the time path P1
ii. Comment on the stability of the market.

Question Three

a. Given the production function Q = K3 + 3I2, what is the marginal rate of technical
substitution between capital and labour?
b. An enterprise invested K20,000 in the development of a new product. They can
manufacture it for K2, per unit. They then hire a marketing consultant, Conda
Marketing Agency Limited in Lusaka, whose conclusions were: lf the enterprise spends
X kwacha on advertising and sell the product at price P per unit, the quantity sold will
be:
Their profit function will then be:
zo,ooo,+ rx - 2op

F (p,x) : (zo,ooo + 4lV - zop)(p - 2) - x - z0,ooo

i. Obtain the optimal level of advertising X* and price P*
ii. Using the Hessian determinant, obtain the second-order conditions for profit
maximisation.
question Four

\
 I

I EXAMINATION TYPE QUESTIONS Mathematicar optimisation and programming Techniques fbr Econornic
'2
Analysis | ,r,
PAPER FOUR QUESTIONS question four

Question One a One major problem in the developed countries is the mental disorder schizophrenia.
Three views on the cause ofthe development ofschizophrenia are: a) environmental
a. Suppose Zambia had an initial stock of 32,000 bags of maize in the year 2000. Each
year half of the existing stock was consumed and another 8000 bags of maize were conditions b) heredity interaction between environment and c) heredity.
prod uced.
At a convention of 80 psychologists,50 psychorogists felt that schizophrenia was due
i. What is the equilibrium quantity of maize?
ii. What will happen to the actual quantity of maize in Zambia over time?
to the interaction between environmental conditions and heredity. Another i-0
psychologists felt that schizophrenia was due to heredity arone. Determine:
b. A consumer is known to have a Cobb-Douglas utility function ofthe form
u(x,Y) : xaYl-d
where the parameter a is unknown. However, it is known that when faced with the
i How many psychologists believed that schizophrenia was due to
following utility maximisation problem: environmental conditions alone?
max.royl-o subiectto, + y = S
ii. How many fert that heredity had something to do with the deveropment of
the disease?
The consumer chooses.r = l,y = 2,
i. find the value of a b. Determine if the forrowing equations will give rise to a convergent time path:
ii. How much utility can be obtained given the choices of x and y. y"'(t) + t1.y" (t) + 3ay'Q) + 24y = 5

Question two

a. The amount of money deposited in a bank is proportional to the interest rate the bank
pays on this money. Furthermore, the bank can reinvest the money aITo/o.
i. Find the interest rate the bank should pay to maximise its profit.
b. An economy's output at time t : 0 was 100 and the rate of change of output is given
by,
dv
i= o.r,
i. Find the time-path of output ofthe economy.
ii. Comment on the time path pattern.
question three

Given the following diffr rential equation:

d2v dv : 2' : +,
dt; +z:dt+ Y Y(0) /'(0) = -1
i. Find the time path
ii. Test for convergence
b. Find the second-order partial derivatives of the following function:
7=
"xz+3xY+Yz
Solve the following problem using the Lagrange multiplier method:
maxZ = log(x) + Iog(y) subject to x * y = 7n
I
MNrthemalical Optimisatiori antl Programnring Techniques lbr Econonric Analysis 355
EXAMINATION TYPE QUESTIONS

Petroda
PAPER FIVE QUESTIONS

Question one 1 2 3 4 5 6

a. Diagonalise if you can the following matrices. L 50 10 20 30 40 50

/-L -6\
i. /=(s, ;) z 90 50 30 40 50 60

,. /5 -3\
.,:\3 _l) 6
E 3 80 70 50 50 60 10
a
C

4 70 60 50 50 70 80
b. A consumer,s utility function is given as: IJ(X1,X2,Xs) and he wants to maximise this
function subject to the constraint that Pl Xt + P2X2 * \X3 = | 5 bU 50 40 30 50 90
i. Write out the first-order condition for utility maximisatlon
ii. Use the Hessian to show the s cond order conditions. 6 50 40 30 20 10 50
c. Evaluate:
r /? f \ i. Using concepts of dominating and dominated strategies, reduce the above
I J6I ysinxdxdy
Jn matrix to an effective payoff matrix
ii. Does the resulting matrix in (i) have a saddle point?
Question two
Question Four
a. Given the following market model
Although South Sudan seceded from Sudan in July 2011, the relationship between the two
Qdt -- ].B0 - 0.75Pt countries remains tense and is often on the brink of war. ln May 2012, despite the United
Q,t=-30+0'3Pt-1 Nations Security Council Resolution 2046 calling for immediate cessation of hostilities,
Po = 220
President Bashir of Sudan stated: "lf they [South Sudan] want to change the regime in
i. Find he time Path Khartoum, we will work to change the regime in Juba. lf they want to attrite us, we will attrite
ii. com ent on the dynamics of the time path of price P6 them. And if they want to support our rebels, we will support theirs".
b. consider the differential equation:
ln the context of the above sltuation, explain the following concepts in Game theory and
lZYtdY + 4Y2dt = o
illustrate them with examples of defence strategies that may be adopted by the two countries
i. rs it exact? and their consequences:
ii. lf not, make it exact and obtain its solution.
a. Nash equilibrium
Question three
b. Prisoner's dilemma
a. suppose two oil marketing compani( s Petroda and Puma have to decide where to
Question Five
locate their service statlons along Great East Road between Kafue roundabout and
Manda hill. Their market share will depend on their choice of six possible locations on Find the extreme point of:
this route. lt is assumed that if both choose the same location, the market will be
divided equally between them The table below shows the market shares for Petroda.
Y = Sln X1 1- ].0lnX2 * 15lnX3
Puma's share is Petroda's share subtracted from 1,OOo/. Subject to

X1 + X2* Xt = 5

ls the extreme point a maximum or a minimum?

I
Mathematical Optimisation and Programming Techniques for Economic Analysis 351
I EXAMTNATTON TYPE QUESTIONS

Typist Rate per hour Typing speed Job Size (pages)

PAPER SIX QUESTIONS
A 5 72 P 199
Question one
B 6 14 a 1,75
a. Solve the problem:
maxZ = Xr2 + Xr2 subject to x1 * X, = 1 c 3 R 1-45

b. ln each of the following cases, comment on the continuity/discontinuity of the D 4 10 S 198

fu nction.
4 1T T 778
i. A welfare program offers unemployed individuals K600 per month. Once an E

individual earns income, the payment is stopped. Suppose an individual can question four
earn K10 per hour. Then the income, y, of the person is a function of the
hours worked, h. That is: a. A firm uses one input, Labour (I) to produce output (Q). The marginal production
function is MP(L) = tO - tOL2/z.Assume that 0 = 0 if L = 0. Find the production
,(') =
{13;' i-:0 function Q(L).
ii. The salary of a salesperson in a company has 3 components: (i) a basic salary b. t = 0 with a capital stock K(0) : K500,000, and in addition to
A firm begins at time
of l(800; (ii) a commission of 2 percent of one's sales, and (iii) a bonus of replacing any depreciated capital, is planning to invest in new capital at the rate
K500 if the sales person's sales for the month reach or exceed K20,000 per 1(t) = 66662 over the next 10 years. Compute the planned level of capital stock 10
month. years from now.

Question two Question five

a. For the matrix A given below. An international organisation intends to initiate a large research project. lt is willing to pay
o _14 2l principal researchers K250 a day and research assistants K50 a day. The man power cost ofthe
^-12 tl project is estimated to be:
Find:
i. The Eigen values and their respective eigenvectors c = 6000 + 6x3 -36xY +3Y2,
ii. Diagonalise the matrix Where X is the number of principal researchers and f is the number of research assistants.
b. Consider the following specific Cobb-Douglas production function:
Y = 50(tK)os a. How many principal researchers and research assistants should be assigned to the
i. Find the second-order partial derivatives and determine the signs; project to minimise cost?
ii. What is the economic interpretation of the signs of these derivatives? b. Calculate the total cost given these values.

Question three

a. Find the solution y(t) of the following differential equation, given that y(0) = -2
dy tY'
dt ,lTT7
b. A publishing company employs typist on an hourly basis' There are five typists for
service and their charges and speeds are different. According to an earlier
understanding, only one job is given to one typist and the typist is paid for full hour
even if he/she works for a fraction of it. Given the data in the following tables, find the
optimal assignment of typist to jobs.
)d EXAMINATION TYPE QUESTIONS Mathematical Optimisation and Programing Techniques for Economic Analysis

equilibrium is reached only when expectations are realised and the expected price at time t
PAPER SEVEN QUESTIONS
equal to the price which is actually on served at time t
Question one i. Formulate the model corresponding to the above situation

a. Check the sign definiteness of the following matrix using eigen value and ii. Obtain the particular solution for the model
determinantal tests: iii. Under what conditions will convergence on the long-run equilibrium price occur?
r-3 0t
a=lq -1 -3 o I Question Five
Lo o _zl
A taxi company has to assign each taxi to each passenger as fast as possible. The following
b. A Publisher agrees to pay the author of book royalty of 1-5%.rhe demand for the
a a
matrix shows the time to reach the passenger in minutes.
book is
x=200-5p Taxi 1 Taxi 2 Taxi 3 Taxi 4 Taxi 5
costis b
c:10*2x*xz Passenger 1 1-2 8 11 18 11
i. Find the optimal sales from both the author's and the publisher's perspective
ii. Comment on your result Passenger 2 t4 22 8 L2 t4
Question two Passenger 3 74 t4 16 L4 15

Given the demand function: Passenger 4 19 L1 14 L7 15

Qt=60-3P1 +2P2+0.25Y Passenger 5 9 77

L3 20 L1,
Where P. = 5, Pz: 10 and Y = 80, Calculate:
Find the optimal assignment of the taxies.
i. Own price elasticity of demand
ii. Cross price elasticity of demand

iii lncome elasticity of demand

iv. What is the relationship between good 1 and Eood 2?

v. ls good 1 an inferior or a superior good?

Question three

a. Given the production function Q : K3 + 3L2, what is the marginal rate of technical
substitution between capital and labour?
b. solve9+
dx
ex+r =o

Question four

A market for some agricultural product, because of gestation lags between the decision to plant
and the subsequent harvest, supply decisions are conditioned by the price which is expected to
rule at the time the crop is harvested; so the supply at time t depends upon the expected price
for the crop at time t which is assumed to equal to the actual price at timet - 1. The demand
for the crop at time t depends upon the actual price at time t. the actual price moves to clear
the market in each period, so the market is always in short-run equilibrium. But long-run
I

I EXAMTNATTON TYPE QUESTTONS Mathematical Optimisation and Programming Techniques for Economic Analysis 351

Question four
PAPER EIGHT QUESTIONS
Question one
a. Solve the following system of equations using Cramer's rule:
3X1+ 5X3 = 69
Find the second-order partial derivatives of the following function: 2X2+4h =46
z=xzeY+y2+x2+ln(X'z) Xt+ xz + 4X, : 39
b. The following is the input coefficient matrix A for a two-sector input-output model:
b. Using integration techniques find .
t0.10 0.031
4 = [o.os -,/
i. tzx,l7+tax o.2o]
i. Calculate the total input requirement tnatrix (- l)-' for producing a unit
ii I:,3@+2)2 dx output of each sector.
question two ii. Comment on the sign definiteness of the above matrix

Question five
a. Consider the first order linear difference equation:
Yt: The following table provides data on unit transportation cost from each of the three production
bt(Yo - Y') + Y' factories A, B and C to each of the three distributors of the output X, Y and Z. for each
Draw pictures showing the stability or instability of the system when destination, a second indicates an initial basic feasible solution obtained by the Matrix
i. 0<b<1, Minimum Method,
ii. -t< b<0; Factory x z Supply
iii. b > 7,
A 2 10 10
iv. b : -]^; 1-

v. b:t B 7 13 3 t2 4 25

b. Two companies A and B are promoting two competing products. Each product
C 6 z 5 18 20
currently controls 50% rif the market. Because of recent modifications in the two
products, the two companies are now preparing to launch a new advertisement
Requirement 15 22 18 55
campaign. lf no advertisement is made by either of the two companies, the present
status of the market shares will remain unchanged. However, if either company
i. Check if the cost of the initial feasible solution is optimal
launches a stronger campaign, the other company will certainly lose a proportlonal
percentage of its customers. A survey ofthe market indicated that 50% ofthe potential ii. lf not, what is the optimal cost?
customers can be reached through television, 30% through newspapers, and the
remaining 20% through radio. The objective of each company is to select the
appropriate adveft ising media.
i. Formulate the problem as a two-person zero-sum game
ii. Does the problem have a saddle point?

question three

A refinery must transport a finished good to some storage tanks. There are two pipelines A and
B to do the transporting. The cost of transporting .r units on A isax2; and the cost of
transporting/ units on Bisby2, where a > 0 and b > 0 are given.
a. What will be the minimum cost of transporting Q units?
b. What happens to the cost if Q increases by r0lo?
I
Mathematical Optimisation and Programming Techniques for Economic Analysis 363
I EXAMTNATTON TYPE QUESTIONS

lf price is initially 20, deduce an equation for price, P, at time f.how does P
PAPER NINE QUESTIONS
approach P-.
Question one Question four
a. civen the profit function r=L60x-3x2-2xy-2y2+I20y-1.8 for a firm a. A firm's production technology can be specified by the following Cobb-Douglas
producing two goods x and y production function:
i. Find the profit maximising level of outputs
Q : F(/(, N) = l0Ko sruo s
ii. What is the profit in (i) above What are the cost-minimising quantities of its two inputs, capital services, K, and
iii. Test the seco d order conditions labour services, N if the firm wishes to produce an output, Q of 500 units; given that
b. A monopolistic firm has the following demand functions for each of its products x and the wage rate is 8 and the price of a unit of capital services is 2?
v. b. Given the following differential equation,
x :72 - 0.5Px y"'(t) + 6y" (t) + y'(t) + 8y = I
Y :120 - Pv i. State the Routh Theorem;
The combined cost function is ii. Using the Routh theorem, check whether the differential equation has a
c=x2+xy+y2+35 convergent time path
and the maximum joint production is 40. Thus, the constraint is x + y = 40. Find the
Question five
i. Output
a. The payofftable for a two-person zero-sum for a mixed strategy game is presented in
ii. Price
iii. Profit the table below. The numbers relate to player X's gain:

question two Y1 Y2
xa 2 10
Find the following integrals
x2 6 7

i. I(4x3 *9x2)(xa +3x3 + 6)3dx Fi nd the best combination of strategies for each player and the value of the game.
..
tt. eS 3xJ2 @+1y b. Test the dynamic stability of the following system:
_--"01.
lll. I cLx
5x 05Y+12
question three-ClX E=
dv
:=6.1 _y.g
a. A profit maximising monopolist has the following demand function: p = 100 - Q2 lhe dr
marginal cost facing the monopolist is given by: MC : 2 + 3Q x(0) = 3, Y(0) = 6
i. Find the profit maximising level of price and quantity
ii. Find the consumers'surplus at the profit maximising price and quantity.
b. Mrs. Patricia Banda is the owner-manager of a beauty salon specialising in
contemporary hairstyles. The annual demand and supply for a standard treatment are
given by:
QD = P2 - t75P +7500
Qs = P2 +25P - l25o
Over the relevant ranges ofP and Q. Price adjusts according to excess demand as
follows:ff = 0.01(QD - Qs)
i. Find the equilibrium price, P-
I

EXAMTNATTON TYPE QUESTTONS Mathematical Optimisation and Programming Techniques for Economic Analysis 36s
I

PAPER TEN QUESTIONS (K aKZ l2)dt

1",
Question One
Subject to
Suppose a firm producing chemical products can buy a chemical for l(10 per ounce. There are
only 17.25 ounces available. The firm can transform this chemical into two products: A and B. K=l-6K
Transforming to A costs K3 per ounce, while transforming to B costs K5 per ounce. lf x1 ounces K(0) = 6o (given)
of A are produced, the price the firm will get for A is P, : 30 - x; if .r2 ounces of B are Where K is the capital stock and I is investment.
produced, the price the firm will get for B is P2 = 50 - x2. The question is: how much chemical
should the firm buy and what should it transform it to? Let x3 be the amount of chemical the a. Form the Hamiltonian for the above problem;

firm purchases. The model to be maximised is:

b. State the necessary conditions based on Pontryagin's maximum Principle;
c. State the boundary conditions;
Maximise d. Form the system of differential equations and obtain the solution.

x1(30- x) + 12(50 - x2)- 3xr- Sxr- 70xt question Four

Subject to Believe it or not, the following poem whose author is not known is a linear programming
problem! Read the poem and
xr lxz -x" < 0
x"< 77.25 a. Formulate the problem mathematically;
b, Solve the problem using the Simplex algorithmic procedure.
a. State the Karush-Kuhn-Tucker conditions for the above problem;
b. Obtain the solution for the problem,
SEREN DI PITY
The three princes of Serendip
question Two
Went on a little trip.
The demand for some agricultural product produced at time t is given by: They could not carry too much weight.
More than 300 pounds made them hesitate.
Q? :720- aPt They planned to the ounce. When they returned to Ceylon
Where QD is quantity demanded and P is the price. They found their supplies were just about gone
When, what to their joy Prince William found
The supply of the produce at time t is given by: A pile of coconuts on the ground.
"Each will bring 60 rupees", said Prince Richard with a grin
Ql=-20+3EPt When he almost tripped over a lion skin.
Where Qs is quantity supplied and EP is expected price. Assume that EP1 = Pr-, and that the "Look out!" cried Prince Robert with glee
market clears period after period. As he spied more lion skins under a tree.
"These are worth even more - 300 rupees each
a. lf Po : 25, obtain thetime path of price and commenton its nature;
lf we can carry them just down to the beach".
b. Assume the coefficient on the expected price in the supply function rises from 3 to 5.
Each skin weighed fifteen pounds and each coconut five.
Does the market converge to the long-run equilibrium price?
But they carried them all and made it alive.
The boat back to the island was very small,
Question Three Fifteen cubic feet capacity - that was all.
Each lion skin took up one cubic foot
Consider the following problem: While each coconut the same space took.
Maximise When everythlng was stowed, they headed to the sea
EXAMINATION TYPE QUESTIONS Mathematical Optimisation and Programing Techniques tbr Economic Analysis 367

And on the way calculated what their new wealth might be.
INDEX
"Eureka" cried Prince Robert, "our wealth is so great
That there is no other way we could return in this state. odditivity property of complete equation, 189 Derivative
And any other skins or nut which we might have brought integrotion, L28 Completely decomposable first order, 144
Might have left us poorer. And now I know what- olien cofactors, T3 matrix, 86, 87 second order, ttz, t43,
I will write my friend Horace in England, for surely, Argond diogrom, 2LO, 21.L, Complex Numbers, 210 !45, 1.67 ,204
Only he can appreciate our serendipity". 275,216,334 com plex pl one. See Argand Derivative, Concept of, 97
Assignment model, 279, diagram derivatives, Meaning of
295 concave downwards, 139 the signs, 138
Question Five covering index, 301 concave function, 151, determinant, of a matrix,
unbalanced,296 180, 183 68
A person wants to go from City 1 to City 10 by the shortest distance. The journey lnvolves four
Associative property of concave upwards, 139 determinantal test for Sign
legs and the distances (in kms) for each leg are shown below. conformability, 64, 78 Definiteness, 94
Sets, 29
Bellmon equotion,326 Constrained optimisation, Determinants, Properties
Leg2 Leg 3 Leg 4
Bellman's Principle, 325, 15, 153, 166, 183 of,71,

cn 334 equality constraints, deterministic

cs c6 c7 C. cs
Bernoulli equation, 201, r5L,27 4 Deterministic Situations,
334 lnequality Constraints, t3,312
c2 40 70 80 cs 60 80 c8 80
binding constraint, 15, 779 diogonol motrix, 6t, 57,
179, 180, 181 many constraints, 176 74,79,80,87,92
c3 80 100 50 c6 90 70 ce 90
Block triangular matrix single constraint, 174 Diagonalisation, 79
block triangular, 87 constroint, 15, 16, L37, Difference equations, 17,
c4 40 50 70 c7 50 70
bordered Hessian, 159, t6t, t62, t63, t64, L65, 228,229,233,236, 24L,
a. ln the above problem, identify the stage, state and policy decision variable. 770, 171, t72,176,778 767, 769, t1t,172, t73, 247,248
b. Write out the recursive relationship. boundary values, 122 774, 777, t79,1.80, 787, complex conjugates,
c. Assuming Markovian property and using Bellman's Principle of Optimality, solve the Cartesian products, 21, 32, 182, 183,318 240
problem through backward iterations. 33 continuity,49,50,97 Distinct roots, 238
Cayley-Hamilton Theorem, continuous function, 48, First-order linear, 229
82,334 49,50,5L,97,775 higher-order, 248
chain rule, 103, 7L9,12o continuous lime, 135, 226, homogeneous, 237
choracteristic roots, la, 228,230,234,237,745 homogeneous equation,
80,88, 89, 96,205,224 control variable , 252,254 237
Characteristic roots convergence, 49, 795, 2\7 , non homogeneous, 237
Repeated Real Roots, 220,226,23t,232,234 Repeated roots,238
208 convex function, 180, 183 Second-order, 236
Two distinct roots, 205 Convex set, 25 What is, 228
choice. Optimisotion cosine function, 213 Differential calculus, 97
Cofactor expansion. see Cramer's rule,75,76,77 Differential equations, 17,
Laplace expansion cross portial derivotives, L86, La7,193,794, 796,
cofactors, 69 1-09 798, 202,207,227, 222,
Commutative law,28 de Moivre's Theorem, 216, 223,224,226,227, 228,
comporotive stotics, 16, 334 247,254,255,335
772 decomposable matrix, 85 Bernoulli Equations, 198
 \

I |NDEX Mathematical Optimisation and Programing Techniques for Economic Analysis | ,.'
First order, 187 eigen vectors, 78, 79, 80, saddle point, 312 inflexion. See point of Linear programming, 257, symmetric,95
Higher order, 202 81, 95 strategy, 308 inflexion 258, 264, 265, 267 , 277, Maximax, 13
homogeneous, 189, Eigen vectors, 78 Types of, 310 lnitial conditions, 122 274,277 , 278 Maximin, 13
192,205,2L8 elosticity, T06,7LL zero sum game, 310 lnput-output models, 82 Linear Programming, 26, maximisation, t2, 14, 75,
Homogeneous, L88 equality constraints, 15, Gome Theory, 8 lnteger progromm i ng, 163 316 76, 39, 40, L37, 738,
non homogeneous, 189, 75, 76t, L62, L63, 719, general solution, 189, 190, lntegra I basic feasible solutions, Lsg, L62,165,766,772,
191, 193, 203 182, 183 191, 193, 202, 205, 206, definite, ]-ls, 122, 124, 27r 180, 2s1, 2s9, 266, 273.
Non-linear, 196 Euler diagram, 23 207 , 208, 209, 21O,2t8, 128,205 basis,277, 272 295,372
Simultaneous, 222,225 Euler relation, 2L5, 334 225,230,237,236,237, indefinite, 714, I75, Graphical solutions, 265 maximise, L2, 74, 75, t6,
variable coefficient, 191 Euler's formula. See Euler 238, 24t, 243, 245, 246, 722,723,724,73s Multiple optimal 137, 738, L45, 746, 7s8,
variable term, 191 relation 249 INTEGRAL CALCULUS, 114 solutions, 259 159, 161, L67, 768, 777,
What it is, 186 exponsion by columns, 68 geometric progression, integral sign, 115 No feasible solution, r79, 250, 257, 252, 254,
Differentiation, 101 exponential function, 43, 230 integrand,115 268 262, 263,264,266,30s,
discontinuous function, 44, LL7,209,2L7 globol moximum,L42 lntegration non-basis,27t,273 317,378,32t,323
50,51, 100, 128 Exponential Rule of global minimum, 142 definite, 124 Unbounded optimal Minimax, 13
discrete time, 17, 135, lntegration, 117 Hamiltonian function, 252, indefinite, l34 solutions,269 minimisation, 12, 16, 738,
228,230,233 feosible region, 25, 265, 253,255,334 lntegration by parts, 120 Linear Programming 161, 180, 259, 267 , 268,
discriminant, 94, 206, 2O8, 266,267,26a,269, 271, Necessary condition, interior point method, 278 Problems, 259 295,3t2
270,231 , 240 277 ,278 252 lnverse differentiation, Linearity, LT minimise, 72, 74, 738, 767,
Disjoint sets,24 flexibility, 328 Hessian determinant, 148, LL4 logarithmic function, 44, 767, 260,267,263,283,
divergence, 195, 220, 226, free endpoint, 252,253 168,169,770,776 inverse of a matrix, 73 118 295, 302, 373, 376, 320,
231,234 Frobenius root, 88, 89 Hessian matrix, 148, L49, Jacobian matrix, 148, L-l'|, Logarithmic Rule of 32L
domoin,34 full ronk. See Rank t52, ts6, L57, l7t, L77 , 335 integration, 118 Minors,69
Domar groMh model, 200 function, 37 334 Karmarkar's algorithm, Maclaurin series, 212 mixed portiol derivotives.
dominant root, 88 Function of o function. See Higher Order Derivatives, 278 mapping, 35, 35 See cross partial
double integrols,729 Chain rule 707 Konig, Denes, 296,335 Morkovion property, 323, derivatives
Duality,264,255 functional,25l homogeneous equation, Konig's theorem, 301, 335 325, 33s Mixed strategies, 314
dynamic optimisation, 16 fundomentol theorem of t88, 1.92,237 ,244 l'H6pital's rule, 131, 132, Mathematical economics, Modified distribution
Dynamic optimisation, 250 colculus,124 hostile 335 4 method, 290
Dynamic programming, Fundamental theorem of hostile game, 14 La Place, 13 mathematical statistics, 5 Multiple integrals, 129
322 calculus, 115 Hungarian method, 296 Lagrangean multiplier matrices, 53, 58, 60, 61, Nosh equilibrium, 18
Milk pouring problem, gome theory,3O5 Hurwicz Economic 62, 63, 54, 65, 66, 67 , non-negative matrix, 82,
322 Game theory, 8, 14, 305, Hurwicz Rule, 13 lnterpretation, 172 68, 69, 77,12,73,75, 89
Steps in, 323 307, 308, 372,3L9 identity motrix, 61, 67, 73, Logrongeon multiplier 76, 77, 7a,19, 81., aZ, normalization, 79, 80, 8L
Econometrics, 4, 7 Definition, 305 78,82 method,167,172 85, 87, 88, 90,92,223, null vector, 55
efficiency, 5, 7, 2L, 33O dominant strategy, 308 lmplicit differentiation, Loploce Expa nsion, 58, 7O 224 optimal, l, 73, L6, 78,90,
Egervary, Jeno, 296, 335 linear programme, 316 10s Leontief lnput-Output. See Matrices !45, 757, L58, L62, L64,
Eigen value test for Sign maxmin strategy, 312 lmproper integrals, 130 lnput-Output model doubly stochastic,9l 1.6s, 166, t68, t69,172,
Definiteness, 94 Mixed Strategy, 314 indecomposoble motrix, Leontief inverse, 84 stochastic matrices, 90 t74, L79,180, 182, 183,
eigen volues, 78, 79, 80, Nash Equilibrium, 312 85, 87, 88,89 limits of a function, 45, 48, stochastic matrix, 90, t84,252,753,254, 287,
8r., 96 players, 308 inequality constraints, 15, 49, 50, 97, L24, L28, 97,92 283, 285,286,287,289,
Eigen Values, 78 possibility matrix, 305 t6, 779, L80, 782,183 729,730,131, 134, 13s transition matrix, 91 293,294,295,296, 377,
pure conflict, 307 linear aclivity,258 matrix, 59 3L3,3t4,315,316,318,
r0
I Mathematical Optimisation and Programming Techniques for Economic Analysis 37.1
I rNDEX

379,321,323,324, 326, Perron-Froben ius return to scale, L74

stagecoach problem, 323 Transportation model, U nconstra ined
state variable, 252, 253 279,29s Optimisation, 75,
336 theorem, 88 Riemann integrol, L26, L27
static optimisation, 16 balanced model, 280 138, 150, 76L, 1,62,
Optimal Control Theory, point of inflexion, 146 Riemonn sum, L25,726
Statistical economics, 4 Lowest Cost Entry, 283 763,164,156,168
250 polar coordinates, 215 risk, 13
Stepping stone method, Northwest corner rule, Unconstrained
optimal decision making, political economy, 2 Routh theorem, 22O,335
13, t4 Polynomial Functions, 40 saddle point, 752, 372,
290 28t optimisation
stochqstic Test for Optimality, 289 n choice variables, 157
optimisation, 12, 13, 14, Pontryogin's Maximum 3L3,3L4
Stochasitic situations, Transpose,66 Single choice variable,
15, t6, L7, 79, 2t, 7L3, Principle,252,335 scarce, 2, t2, Scorcity
13, 90 tronsposing, 54 74L
L37, L38,74L, t44,748, positive motrix,81, 82, 88 Scarcity, l2
Subset,23 tronsversolity condition, Two choice variables,
150,151, 156,157, 158, principal minors, 89, 94, semi positive matrix, 88
Substitution method, 119 253 150
76L,763,764, t66,768, 95, 757 ,758, Lls, 777 se mi-positive matrix, 82
symmetricol matrices, 61 triangulor motrix,67 unit vector,55
172, L76,179,180,182, Prisoner's dile mmo, 306 Separable variables, 197
L83,250,25L,252,253, Product of two functions. sets, 14, 2L, 22, 23, 24, 25,
Taylor series, 212 Trigonometric functions, u-v method, 29O,293
Total differentials, 111 212 vector, 53
257, 258, 278, 295, 305, See Product rule 26, 27, 28, 29, 30, 32,
3t2 product rule, 7O2, L2l, 33,34,35,53, 56,57,
Tranportation model turning point, 739, 74t, Venn diagram, 23,24,32
unbalanced model, 280 153 versatility,32S
What is, 137 198 L26,146,zLO
Transhipment model, 279, uncertainty, 13 Vogel's Approximotion
Optimisation, 79, L37, 755, progrommtng side relotion. See
301
t58, 764, L74, L76,252 unconstrained Method,287,285,336
Mathematical Constra int
optimise, 78, 758, 767, Programming, 13, 17, side relotions, LGI
transloading. See U nconstra ined Young's theorem, Lt2, 752
Transhipment optimisation, 161
L70,258,306,309,311, t63, L79, 180, 181, Sign Definiteness, 93
3L2,3L6 279, 280, 296, 3L7, Simplex method, 271, 272,
ordered pairs, 2L, 32, 33, 318, 319, 320, 322, 277
34,35,36,37 323,324,325,326 Basic solutions, 271
Poreto optimality, 18 proportionolity, non-basic variables,
portial derivatives, 108, chorocteristic of,258 27L,272
109, 110, LtL,129, t48, Pythagoros theorem, 21,1, simplex tableau, 274
749, tso, L5t, 1.52, 753, 275,242 Slack variables, 271
757, L58, Lsg, 760, 156, quadratic form, 92, 93, 94, surplus variables, 271
767, L68, L69,170, t77, 9s,96, ts7 ,238 Simplex Method, 270
L72 Quadratic forms, 92 Simplex procedure,
Partial Derivatives, 107 Quotient of two functions. lllustration,2T2
partial integration, 129 See Quotient rule sine function, 213
Particular solution, 188, Quotient rule, LO3 singulqr matrix,73,74
790, 203, 204, 20s, 206, ronge, 34 skew symmetricol motrix,
207 , 208, 209, 277 , 278, rcnk,77 62
225,237,234,237, 239, rational function, 42, 43, slock
247,243,244, 248 48 Slack Constraint, 15,
pavoff, 74,308, 309, 310, rectangular hyperbola, 43 779, L81.
317, 3L2, 3t3, 314, 3ts, reduced equation, 189 Smoothness, 17
316,318,319,320 relation, 33 spectral decomposition.
Payoff Function, 308 relotive moximum, 1,42, See Diagonalisation
perfect competition, 13 151, 158 squore motrix, 60, 6L, 62,
relotive minimum, L42 68,77,78,82, 85, 90
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