Ees 300

KENYATTA UNIVERSITY
INSTITUTE OF OPEN DISTANCE & e-LEARNING
IN COLLABORATION WITH
SCHOOL OF ECONOMICS
DEPARTMENT: ECONOMETRICS AND STATISTICS
UNIT CODE: EES 300
MATHEMATICS FOR ECONOMISTS III
WRITTEN BY: DR.AFLONIA MBUTHIA EDITED BY: DR. DIANAH MUCHAI

Copyright © Kenyatta University, 2009
All Rights Reserved
Published By:
KENYATTA UNIVERSITY PRESS

This course is an advanced course in mathematics for economics. A student
should have learnt mathematics for economics two for her/him to be able to understand this
course. The course introduces the student to more optimization techniques, both for
constrained and unconstrained optimization problems.
The learner should be able to:
1. Formulate linear programming problems in a standard way.
2. Solve linear programming problems using the graphical and simplex
method.
3. Formulate the dual of a linear programming problem.
4. Test linear dependence using the Jacobian Determinant
5. Solve simultaneous equations using Gaussian and Gauss Jordan
elimination methods.
6. Test the convexity and concavity of a non-linear function using Hessian
matrices.
7. State the Kuhn Tucker conditions for a maximization and minimization
problem.
8. Solve a non-linear Programming problem.
9. Integrate logarithmic and non-algebraic functions.
10. Integrate functions using Integration by parts
11. Get solutions to First order differential equations used in Economics
12. Get solutions to First order difference equations used in Economics
13. Determine stability of economic variables.
INTRODUCTION
OBJECTIVES

Lecture One: Introduction to linear programming.
 Formulation of LPP
Lecture Two: Graphical method of solving LP problems
Lecture Three: Simplex method for maximization problems, Duality in LPP
Lecture Four: Tests of linear dependence using the Jacobian Determinant.
 Solution of simultaneous equations using Gaussian and Gauss
Jordan elimination methods.
Lecture Five: Tests of convexity and concavity using Hessian matrices and
determinants
Lecture six: Non-linear Programming
 Kuhn Tucker problem for maximisation
 Kuhn Tucker problem for minimization
 Economic applications of non-linear programming
Lecture Seven: Integration of logarithmic and non-algebraic
functions,Integration by parts
 Economic applications of integration of non-algebraic functions.
Lecture Eight: First order differential equations
 FODE with constant coefficient and term.
 FODE with variable coefficient and term
 Exact differential equations
 Economic applications of FODE
Lecture Nine: First order difference equations
 Homogenous and non-homogenous difference equations.
 Solution to FO difference equations
Lecture Ten: Stability of an equilibrium.
 Economic applications in cobweb model and market
model.
TABLE OF CONTENTS

By the end of this lecture the learner should be:
1. Able to define key terms used in a linear programming problem.
2. Able to formulate a linear programming problem in the standard way.
This lecture introduces the student to the method of formulating economic problems with
linear objective functions and also linear constraints/limitations. Standard steps of
formulating an LPP will be introduced.
LECTURE ONE: LINEAR PROGRAMMING
1.1 INTRODUCTION
1.2 LECTURE OBJECTIVES

1.3 Introduction to linear programming
• In most business organizations, allocation of limited resources to unlimited needs is a
major problem. Resources have to be allocated to the alternative that makes an economic
agent most well off.
• The management of a firm makes many allocations decisions within any given time.
• Linear programming is a mathematical technique concerned with allocation of scarce
resources. It involves optimization of an objective function subject to one or several
constraints faced by the economic agent.
• Optimization could either be maximization or minimization problem.
For a problem to be solved using linear programming technique, a number of conditions must
hold.
 Problem can be stated in numeric terms.
 All factors involved must have linear relationships
 Problem must permit a choice/s between alternative courses of action
 There must be one/more constraints involved
Definition of key terms
Programming – process of getting optimum values.
Linearity- In all relevant relationship, there are linear relationships between the
variables. For example, in production kind of problems, we have constant input prices,
output prices and we have constant returns to scale, constant profit per unit of output sold

and factors of production are used in fixed ratios. Although linearity may not always
apply, the relationship can be linear within a given range
1.4 Expressing Linear programming problems in a standard manner
We need to express all linear programming problems in a standardized manner. This
helps in calculations required in getting solutions. In addition, it ensures that no important
element is overlooked.
Steps involved
 Decide on the objective of firm
 Decide on decision variables
 Formulate the objective function
 Formulate the constraints
 Set up the non-negative constraints
 Solve by graphical or simplex method
i) Objective
What results are required/what do we want to achieve. e.g. maximise profit,
minimize cost/time or other. Once the objective is decided upon, the elements in
achieving it are stated mathematically.
ii) Decision variables
In order to achieve the objective, the decision maker has to decide on something.
For example, number of units of a product to produce.
iii) Objective function – This is the function to be maximized or minimized.

Example1
A firm produces 2 goods, x and y. The profit per unit of x = 100/= and y is 300/=.
The firm wants to maximize the profits.
 Objective: maximize profits
 Decision variables
x = number of units x to produce
y = number of units y to produce
 Objective function: Maximize Profits = 100x + 300y
iv) Constraints/limitations
These are the factors that could hinder the decision maker from achieving the
objective set. These are conditions that need to be met when achieving the
objective.For example, as a student, the objective is to maximise the marks
obtained in a given course. But this objective is limited by the time available for
reading, resources available to buy relevant books among others.
The limitations in any problem must be clearly identified, quantified and
expressed mathematically.
If a problem is concerned with scarce resources, constraints take the form of:
Resources used≤Resources available
In a profit maximization problem, the firm would want to maximize the profits by
producing as much as possible. However, the amount of raw materials and
machine hours available in the firm could act as limitations. A consumer would
want to maximize the utility derived from consuming two products, but the
income/budget constraints him/her.

v) Non-negative constraints
Most of the decision variables in LP problems cannot take negative values. To
ensure that decision varibles are non-negative, we have the non-negativity
constraints.
For example, the number of units of a product manufactured should be greater or
equal to zero but cannot be negative. i.e. X≥0
vi) Solve using either graphical or simplex method.
1.5 Examples of Formulating linear programming problems in a standard way:
We will look at several examples that will assist us to formulate the linear
programming problems in a standard way.
PROBLEM 1.1
A firm makes two types of furniture: chairs and tables. The contribution for each product
as calculated by the accounting department is Kshs. 2000 per chair and Kshs. 3000 per
table. Both products are processed on three machines M1, M2 and M3. The time
required by each product and total time available per week on each machine is as
follows:
Machine Chair Table Available Hours
M1 3 3 36
M2 5 2 50
M3 2 6 60
Formulate the linear programming problem in the standard way given that the
manufacturer wants to maximize contribution.

SOLUTION
Let x1 = Number of chairs to be produced
X2 = Number of tables to be produced
Since the objective is to maximize the profit, the objective function is given by:
Maximise Z= 20x1+ 30x2
Subject to constraints:
3x1 + 3x2 ≤ 36 ( Total time of machine M1)
X1, X2≥ 0 [ Non-Negativity constraint)
PROBLEM 1.2
The ABC manufacturing company can make two products P1 and P2. Each of the
products requires time on a cutting machine and a finishing machine. Relevant data are:
Products
P1 p2
Cutting Hours ( per unit) 2 1
Finishing Hours ( per Unit) 3 3
Profit ( $ per week) 6 4
Maximum Sales ( Unit per week) 200
The number of cutting hours available per week is 390 and the number of finishing hours
available per week is 810. Formulate the linear programming problem in the standard
way given that the manufacturer wants to maximize profits.

SOLUTION
Let X1 =Number of product p1 to be produced
X2 = Number of product p2 to be produced
Since the objective is to maximize profit, the objective function is given by:
Maximise Z = 6x1 +4x2
Subject to constraints
2x1 +x2 ≤ 390 ( Availability of cutting hours)
3x1 +3x1 ≤ 810 (Availability of finishing hours)
X2 ≤ 200 ( Maximum sales)
X1, X2≥ 0 ( Non-Negativity constraint)
• PROBLEM 1.3
• An animal feed company must produce 100 kg of a mixture consisting of ingredients X1,
and X2 daily. X1 cost Kshs. 3 per kg and X2 Kshs. 8 per kg. Not more than 80 kg of X1
can be used, and at least 60kg of X2 must be used. Formulate the problem in a standard
way to enable the company minimize cost.
Solution
Let X1 = Number of kg of ingredient X1 to be used
X2 = Number of kg of ingredient X2 to be used
Since the objective is to minimize the cost, the objective function is given as:
minimize Z = 3x1 +8x2
X1 +X2 = 100 (Total mixture to be produced)
X1 ≤ 80 ( Maximum use of X1)

In this lecture we have learnt how to formulate minimization and
maximization problems in a standard way. In our next lecture we will look at how
we solve the linear programming problems using the graphical method.
Activity
As a distant learning student, try to think of the limitations you encounter when
trying to maximize your study time.
Linear programming problems can only be used if all relationships have been
expressed as linear functions
X2≥ 60 (Minimum use of x2)
X1 ≥ 0 (Non-negativity constraint
Note: There is no need for a non-negativity constraint for X2 since it is already supposed
to be greater or equal to 60.
1.6 SUMMARY
NOTE
1.7 ACTIVITIES

References
1. Gupta S.K. Cozzolino J.M. (1975) Fundamentals of Operations Research
for Management
2. Lucey T. Quantitative techniques
3. Monga G.S: Mathematics and statistics for Economists, second revised
edition
4. Tulsian and Pandey V. Quantitative Techniques, Theory and Problems.
1.8 FURTHER READING

• The Tusome Printing Company is facing a tight financial squeeze and is attempting to
cut costs wherever possible. At present it has only one printing contract and, luckily,
the book is selling well in both the hardcover and paperback editions. It has just
received a request to print more copies of this book in either the hardcover or
paperback form. Printing cost for hardcover books is Kshs. 700 per 100 while
printing cost for paperback is only Kshs. 600 per 100. Although the company is
attempting to economise, it does not wish to lay off any employees. Therefore, it feels
obliged to run its two printing presses at least 90 and 70 hours per week, respectively.
Press I can produce 100 hardcover books in 2 hours or 100 paperback books in 1 hour,
while Press I can produce 100 hardcover books in 1 hours or 100 paperback books in
2 hours. Formulate the problem in a standard way, in order to minimize cost.
1.9 SELF-TEST QUESTIONS 1

By the end of this lecture the learner should be able to solve maximization and minimization
linear programming problem using the graphical method
This lecture discusses the method of solving linear programming problems (LPP) using the
graphical method.
LECTURE TWO: GRAPHICAL SOLUTION TO LINEAR
PROGRAMMING
2.1 INTRODUCTION

2.3 Graphical solution to linear programming.
In our lecture one, we were able to see how a LPP can be formulated in a standard way. In this
section we will see how a LPP can be solved using the graphical method.
Graphical methods of solving Linear programming can only be used for problems with TWO
unknowns or decision variables. Problems with THREE OR MORE unknowns must be solved
by techniques such as the simplex method. Graphical methods are the simplest to use and should
be used wherever possible. Graphical method can be used to solve both maximization and
minimization problems. It can deal with limitations of the form ‘greater than or equal to’ type
( , the ‘less than or equal to’ type. ( , and even constraints of ‘equal to’( =) type.
The following is a step by step procedure for solving LP maximisation problems graphically.
Example 1
A manufacturer produces two products, toothpaste and detergents. Toothpaste has a contribution
of Kshs. 3000 per carton and Detergent Kshs. 4000 per carton. The manufacturer wishes to
establish the weekly production plan which maximizes contribution.
The machine hours, labor hours and raw materials (in kgs) required to produce a carton of
toothpaste and detergent respectively are as follows:
PER UNIT REQUIREMENTS
Toothpaste
Detergent
Machine (Hours)
4
2
Labour (Hours)
4
6
Material (kgs)
1
1
Total Available
Per week
100 180 40

Because of a trade agreement, sales of Toothpaste are limited to a weekly maximum of 20 units
and to honor an agreement with an old established customer at least 10 units of Detergent must
be sold per week.
Solution
STEP 1. Formulate the linear programming model in the standardized manner.
• Decision variables
Let X1 be the number of units of toothpaste to be made and:
X2 be no. of cartons of detergents to be made
Objective function
Maximise: Contribution = 3000x1 + 4000x2
Constraint
• A: 4X1+2X2≤100 machine hours constraint
• B: 4X1+6X2≤180 Labour hours constraint
• C: X1+X2≤ 40 Materials constraint
• D: X1≤ 20 maximum toothpaste sales
• E: X2≥10 minimum detergent sales
• X1 ≥ 0 non-negativity constraint

• (No need for having a non-negativity constraint for X2 since it is already supposed to be
greater than 10 which is non-negative).
• The source and sales constraints include both types of restrictions (i.e. ≥ and ≤).
• The problem has only 2 decision variables ( X1 and X2) and can it can be solved
graphically.
• The number of limitations does not exclude a graphical solution.
• STEP 2. Draw the axes of the graph to represent the X1 and X2
You should determine the scale once you know your limitations. Each axis must start at
zero and the scale must be constant (i.e. linear) along the axis but it is not necessary for
the scales on both axes to be the same.
STEP 3. Draw each limitation as a separate line on the graph.
(For illustration purposes each of these lines is drawn separately but they should all be
combined and drawn on the same graph).
0
X2
X1

• Since the constraints are all linear, they will be represented by straight lines.
• In order to draw a straight line, we only need two points on a graph.
• Each constraint will be drawn as follows:
• 4X1+2X2≤100 machine hours constraint
• We assume there is no inequality to enable us get the two points, to aid us in drawing the
graph.
• i.e. Assume 4X1+2X2=100
• When X1 = 0, X2=100/2=50
• When X2=0, X1=100/4=25
• We have two points : (X1,X2)=(0,50)and (25,0)
• Drawing and shading of constraint 1

• Shade off the region that is not required. In our case it is area to the right of the line.
• Constraint two and three
• 4X1+6X2≤180 Labor hours constraint
• Assume 4X1+6X2=180
• When X1=0, X2=30 and when X2=0, X1= 45
• We can plot the constraint using the coordinates (X1,X2)=(0,30)and (45,0)
• X1+X2≤ 40 Materials constraint
• Assume X1+X2= 40
• When X1=0, X2=40, and when X2=0, X1=40
• We can plot the constraint using the coordinates (X1,X2)=(0,40)and (40,0)
X1
X2
0
Equation of the line:4X1+2X2≤
50
25

• Drawing and shading of constraint 2
• Shade off the region that is not required. In our case it is the area to the right of the line.
•
Constraints four and five
• Sales limitations : this only affects one of the products at a time and in our case, the sales
restrictions were
X1 ≤ 20 and X2 ≥ 10
If we assume X1= 20 and X2=10, we can represent each of them by one line.
The vertical line represents X1 = 20 and the shaded area to the right of the line represents the area
containing the values greater than 20. The horizontal line represents X2 ≥ 10 and the shaded line
below the line represents values of X2 which are less than 10.
X1
X2
0
Equation of the line: 4X1+6X2≤ 8
30
45

Graphing and shading of the non-negativity constraint
X1
X2
0
X1
X2
0 20
10
X2
X10

Step 4: Identify the feasible region(the region satisfying all the constraints)
• All the constraints are supposed to be drawn on the same graph to help in identifying the
area that satisfies all the constraints.
0
X2
X1
50
40
30
20
10
10 20 30 40 50
R
The region labeled R represents the feasible region i.e. the area that satisfies all the 5 constraints.
Note : The constraint X1+X2≤ 40 does not touch the feasible region, it is overshadowed
by the others. It is called a redundant constraint in that it is not affecting the decision
made by the producer. It is non-binding.

Step 5: Identify the point within the feasible region that ensures the contribution is maximum.
• This can be identified by marking the corner points within the feasible region.
•
A=(0,30), B=(15,20), C=(0,10), D=(20,10)
0
X2
X1
50
40
30
20
10
10 20 30 40 50
R
A
B
C D
Once the corner points are identified, the one that maximizes the contribution is identified
by evaluating the value of the objective function at each of the points . This has been
done in the next slide.
• Contribution = 3000x1 + 4000x2
• At point A, contribution = 3000(0) + 4000(30)
• = kshs.120,000

• At point B, contribution = 3000(15) + 4000(20)
• = kshs.125,000
• At point C, contribution = 3000(0) + 4000(10)
• = kshs.40,000
• At point D, contribution = 3000(20) + 4000(10)
• = kshs.100,000
• Since it is a maximization problem, we choose the point giving us the maximum
value. Point B= Kshs.125,000 and it is the maximum contribution
Example Two: Minimisation problem
• A chemical manufacturer uses 2 chemicals X and Y in different proportions to produce 3
products A, B, and C. The manufacturer wants to produce at least 150 units of A, 200 units of
B and 60 units of C. Each ton of X yields 3 units of A, 5 of B and 3 of C. Each ton of Y
yields 5 of A, 5 of B and 1 of C. If chemical X costs $2000 per ton and Y costs $2500 per
ton. Question: Formulate the linear programming problem and solve it using the graphical
method.
• Solution
• Decide on the decision variables:
• Let X=number of tons of chemical X to be used
• Y=number of tons of chemical Y to be used

• Objective:
• To minimize the total cost of purchasing the chemicals
• Cost=2000X+2500Y
1. At least 150 units of A should be produced
3X+5Y≥150
2. At least 200 of B should be produced
5X+5Y≥200
3. At least 60 of B should be produced
3X + Y ≥60
4. Non-negativity constraints
X ≥0, Y ≥0
Graph and shade the constraints on the same graph:
3X+5Y=150
when X=0, Y=30,when Y=0, X=50
(X,Y) = (0,30)(50,0)
5X+5Y=200
When X=0, Y=40, whenY=0, X=40

(X,Y)= (0,40), (40,0)
3X + Y =60
When X=0, Y=60, when Y=0, X=20
•
10 20 30 40 50
60
50
40
30
20
10
0
Y
X
A
B
C
D
A=0,60, cost=150,000
• B=10,30 cost=20,000+75000=95,000
• C=20,25 cost=102,500
Point B gives the lowest cost, Kshs. 95,000. We should therefore buy 20 units of chemical X and
25 units of chemical Y.

In this lecture we have learnt how to solve LPP using
graphical method. It can only deal with problems involving 2 decision
variables. In our next lecture we will look at how to solve LPP using
simplex method.
Ensure you do more practice questions from your reference
texts.
When using the graphical method, ensure your
graph is drawn to scale.
2.4 SUMMARY
NOTE
2.5 ACTIVITIES

References
1. Gupta S.K. Cozzolino J.M. (1975) Fundamentals of
Operations Research for Management.
3. Monga G.S: Mathematics and statistics for Economists,
second revised edition.
4. Tulsian and Pandey V. Quantitative Techniques, Theory
and Problems.
2.6 FURTHER READING

A manufacturer produces two products, toothpaste and detergents. The
contribution per unit is $3 and $4 for toothpaste and detergents respectively. The manufacturer
wishes to establish the weekly production plan which maximizes contribution. Both products are
processed on three machines M1, M2 and M3. The time required by each product and total time
available per week on each machine is as follows:
Machine chairs tables Available hours
M1 3 3 36
M2 5 2 50
M3 2 6 60
Use the graphical method to determine how the manufacturer should schedule his production in
order to maximize contribution.

By the end of this lecture the learner should be:
1. Able to solve a linear programming problem using the simplex method.
2. Able to formulate the dual of any linear programming problem.
We have already looked at how we solve an LPP using the graphical method.
We noted that the graphical method can not be used if we have more than 2 decision
variables. In this lecture we will discuss the method of solving (LPP) using the simplex
method.In addition, we will look at how a maximization/minimisation problem can be turned
into a dual.
LECTURE 3: SOLUTION OF LPP USING SIMPLEX METHOD
3.1 INTRODUCTION

3.2 Solution of maximisation LPP using the simplex method.
In our lecture two, we were able to see how a LPP can be solved using the graphical method. In
this lecture we will learn how to a LPP can be solved using the simplex method.
Steps
• Formulate the problem in the standard way.
• Convert the inequalities into equations through the addition of a slack variable if the
inequality is of a less than type (≤).
• A slack variable represents unused capacity in the constraint.
E.g. 4X1+6X2≤180 Labour hours constraint
4X1+6X2 + S1 =180
S1= unused capacity
• It will have the coefficient of 1 if both X1 and X2=0 or can be 0 if the labour hours are
fully utilised.
• Each constraint has its own slack variable.
• Once the slack variable is incorporated, the simplex method assigns it a value during the
step by step procedure.
A company can produce 2 products A and B. The profit per unit of A produced is 6 dollars while
the profits per unit of B produced is 8 dollars. To produce a unit of A the company requires 30
labour hours and 20 labour hours. The machine hours required are 5 hours and 10 hours for A

and B respectively. The company has a maximum of 300 and 110 labour and machine hours at its
disposal. Solve the problem using the simplex method.
Step 1
Formulate the problem in the standard way.
• Let X=no. of units of A produced and
Y= no. of units of B produced.
Objective function
Profits = 6X+8Y
Constraints:
6X+8Y≤300 labour hours constraint
5X+10Y≤110 Machine hours constraint
X≥0, Y≥0, non-negativity constraints
Step 2:
Change the constraints into equations by introducing slack variables
Profits = 6X+8Y+0S1+0S2
Constraints:
6X+8Y+S1=300 labour hours constraint
5X+10Y+S2=110 Machine hours constraint
X, Y, S1, S2 ≥0, non-negativity constraints

Step 3: Prepare the initial simplex table
Solution
variable
Products Slack variables Solution
QuantityX Y S1 S2
S1 30 20 1 0 300
S2 5 10 0 1 110
Z=Profit 6 8 0 0 0
Note:
• The values from the table are the values from the objective function and the constraints in
step 2.
• Z is used to represent total profits.
• The table shows that S1=300 and S2=110 and profits (Z)=0.
• Nothing is produced hence the slacks are at their maximum values.
Step Four
Profits can be improved by producing the products A and B in varying quantities.
• Improve the profits by producing as much as possible of the product with highest profit
per unit.
• i.e. highest value in Z row.
• The number of the units to be produced will be limited by one or more constraints
becoming operative.
• E.g. in our case, product B has the highest profit.
• It becomes the incoming variable.

• Determine the row it should be come into by dividing the solution quantity column by the
corresponding values in the Y column.
• 300/20=15
• 110/10=11
• Select the row with the lowest quotient. (Why: it is the one that will limit the maximum
amount of product B that can be produced.
• The selected row is known as the key row. It gives us the outgoing variable.
• The column with the highest profit per unit is called the key column and gives us the
incoming variable.
• The value at the intersection of the key row and the key element is known as the pivot
element/key element.
This is shown in the table below:
Solution
variable
Products Products Slack variables Solution
Quantity
X Y S1 S2
S1 30 20 1 0 300
S2 5 0 1 110
Z=Profit 6 8 0 0 0
Divide all elements in the pivot row by the pivot element.
• Old row
5 10 0 1 110
New row
0.5 1 0 0.1 11
Enter this new row in a new simplex table

Second simplex table
Row no. Solution
variable
Quantity
X Y S1 S2
1 S1 30 20 1 0 300
2 Y 0.5 0 0.1 11
3 Z=Profit 6 8 0 0 0
• The entire table is similar to the initial simplex table except that row 2 is replaced with
new row 2.
• This implies that we will now produce 11 units of product B.
• As a result of producing product B, the labour and machine hours available have reduced.
In addition, the profits have also been improved. We therefore need to adjust the other
two rows to reflect this change in resources and also the profits.
• This is done by an iterative procedure that ensures that all the other values in the pivot
column become 0. In order to ensure the equality of the row, we ensure that the entire
row is altered.
• New Row 1=Row 1-20row 2
• Old row 1
• 20 1 0 300
• New row 1
• 20 0 1 -2 80
• Old row 3
• 8 0 0 0
• New row 3=row 3-8row 1

• 2 0 0 -0.8 88
Third Simplex table
Row
no.
Solution
variable
Quantity
X Y S1 S2
1 S1 20 0 1 -2 80
2 Y 0.5 0 0.1 11
3 Z=Profit 2 0 0 -0.8 -88
• If in the Z row, in the products columns we still have a positive number, the profits can
be improved further. So we need to introduce the product whose profits can still be
positive.
The above steps are repeated as follows:
• If we introduce a unit of X, the profit will increase by 2 per unit.
• X becomes the incoming variable.
• X column is the key column.
• To identify the key row and pivot element, we divide solution quantity with
corresponding values in the X column.
80/20=4
11/0.5=22
We choose the row with smallest quotient.
Row 1 is chosen, 20 becomes the pivot element.
• Divide the key row values by 20.
New row 1:
1 0 1/20 -0.1 4

Fourth Simplex table
Row
no.
Solution
variable
Quantity
X Y S1 S2
1 S1 1 0 1/20 -0.1 4
2 Y 0.5 0 0.1 11
3 Z=Profit 2 0 0 -0.8 -88
• Adjust the other rows in the table to ensure that all the other elements in key column are
zero.
• New Row 2=old row 2 - 0.5New Row1
• Old row 2: 0.5 1 0 1/10 11
• New row 1: 1 0 1/20 -2/20 4
• New row 2: 0 1 -1⁄40 3/20 9
• New row 3=old row 3-2newrow 1
• Old row 3: 2 0 0 -8/10 -88
• New row 1: 1 0 1/20 -2/20 4
New row 3: 0 0 -2/20 -12/20 -96
Fifth Simplex table
Row
no.
Solution
variable
Quantity
X Y S1 S2
1 S1 1 0 1/20 -0.1 4
2 Y 0 1 -1/40 3/20 9
3 Z=Profit 0 0 -2/20 -12/20 -96
• The optimal solution is producing 4 units of A and 9 units of product B.

• The total profits will be: Z=6X+8Y=6*4+8*9=96
The values in the Z row and in the columns of the slack variables S1 and S2 are the shadow prices
of the labour hours and machine hours respectively. If we have an extra unit of labour hours, we
will be able to increase the profits by 1/10 dollars while an extra unit of machine hours will
increase the profits by 6/10 dollars.
3.4 DUALITY IN LPP
Every maximization or minimization problem in a linear programming has a corresponding
minimization or maximization problem.
The original problem is referred to as a PRIMAL while the corresponding one is the DUAL.
The relationship between the two can best be expressed through the use of parameters that they
share in common.
Consider a primal maximization problem generally expressed as.
= + + +. . . +
.
+ + +
+ + +
+ + +
, …
Rewriting the above in matrix form

Max = [ … ] ⌈ ⌉
.
⌊ ⌋ ⌈ ⌉ ⌈ ⌉
, …
RULES OF TRANSFORMATION
The direction of optimization changes i.e maximization becomes minimization and vice versa.
e.g. min Q
 The row vector of coefficient of the objective function in the primal problem is
transposed into a column vector of constants in the dual.
 The column vector of constants in the constants in the primal problem is transposed
into a row vector of coefficients in the objective function of the dual.
 The matrix of coefficients in the constraint equation of the primal problem is
transposed into a matrix of coefficients in the matrix of the dual problem.
 The inequality sign in the primal problem is reversed, i.e. in a primal maximization
problem becomes in the dual problem.
Note. 1. The non-negativity constraints remain unchanged.
2. The decision variables will change say from Xi to Zj.

The dual problem therefore becomes, in matrix form:
= [ … ] ⌈ ⌉
.
⌊ ⌋ ⌈ ⌉ ⌈ ⌉
, …
= + +
.
+ +
+ +
+ +
, …
Example
Let the primal problem be
� = +
.
+
+

+
Rewriting into matrix
� = [ ]
.
[ ] [ ] [ ]
The dual will be,
= [ ] [ ]
.
[ ] [ ] [ ]
In the linear form,
min = + +
.
+ +

This lecture has looked at how you can solve LPP using
the simplex method. It can be used for problems involving more than
2 decision variables. .Formulation of dual from the primal has also
been learnt.
Solve the LPP problems in lecture 2 using the simplex
method and check whether the answers obtained are similar.
Shadow prices are obtained only for the completely utilized
resources.
Duality in LPP is important especially if you want to use
maximization instead of minimization problem.
+ +
3.5 SUMMARY
NOTE
3.6 ACTIVITIES

Use the simplex method to solve the following:
• Z=6X+8Y
• Subject to:
• 2X+3Y≤16
• 4X+2Y≤16
X, Y≥0
References
1. Gupta S.K. Cozzolino J.M. (1975) Fundamentals of
Operations Research for Management.
4. Tulsian and Pandey V. Quantitative Techniques, Theory
and Problems.
3.7 FURTHER READING

By the end of this lecture, the learner should be able to:
1. Determine if functions of two and three independent variables
are functionally dependent using the Jacobian determinant.
2. Solve three simultaneous equations using the gauss Jordan
elimination method.
In this lecture you will learn how to test for functional
dependence in non-linear functions. In addition, the solution to
simultaneous equations using the Gaussian and Gauss Jordan
eliminitain methods will be learnt.
LECTURE FOUR: FUNCTIONAL DEPENDENCE AND SOLUTION TO
SIMULATNEOUS EQUATIONS
4.1 INTRODUCTION

4.3 Jacobian Determinant
The Jacobian matrix can help to test whether there exists functional linear dependence in a set of
functions in variables. It is a matrix of the first order partial derivatives. If the functions are
linearly dependent, the determinant of the Jacobian matrix is 0.
E.g. = +
= + +
�
�
=
�
�⁄ =
�
�
= +
�
�
= +
=
[
�
�
�
�
�
� ]
= |
+ +
| =
The functions are linearly dependent.
Example 2
= +
= +

= [ ]
| | = | |=− unless X2
Is zero, the determinant is not zero, hence the 2 functions are functionally independent.
If we have several functions,
= | | = |
′ ′
… ′
|
=
=
A Jacobian | |is identically zero for all values of , … , if the n functions are functionally
dependant.
4.4 Solution of simultaneous equations using the Gaussian method
Gaussian method involves solving for simultaneous equations using matrix algebra by getting the
lower echelon matrix for the matrix of coefficients.
If we have a matrix, given as:
+ + =
+ + =
+ + =
We can get the values of X, Y and Z by rewriting the functions in matrix form such that:
[ ] [ ] = [ ]
The lower echelon matrix for matrix of coefficients is given as:
[ ]
This is done through a step by step iteration procedure.
Let us use a practical example.

+ − =
+ + =
− − = −
If we rewrite the equations in matrix form, we get:
−
− −
[ ] = [
−
]
As we do the iterations, we need to maintain the row equality. This will be done by some row
operations, similar to those we used when we were looking at the simplex method in LPP. We
will only affect the matrix of the coefficients and the constants matrix. We can join them
together into one augmented matrix as we undertake the step by step procedure.
−
− −
|
−
We want to make the first element in the first row to be 1. This can be accomplished by dividing
the entire row 1 by 2. This gives us:
−
− −
|
−
The elements 2 and 3 in column 1 Row 2 and 3 respectively will be turned into 0 using the
following row procedure:
= −
= −
This gives us the following matrix:
−
−
−
| −
−
We then change the − in the second row , second column into by dividing the entire row by
− . This gives us the following matrix.
−
−
−
|
−

We change the value -21 in the second column into a 0 by the following procedure:
= − −
It will give us:
−
−
−
|
−
The next step involves changing the element in 3rd
row, 3rd
column into a 1. This can be done by
dividing the entire row 3 by − giving us the following matrix
−
− |
If we bring back our variables, X, Y and Z we have the following matrices:
[
−
− ] [ ] = [ ⁄ ]
+ − =
− =
=
If we replace Z=3 in the second equation, it will give us the following
− = − = → = + = =
= =
By replacing the values of Y and Z in the first equation, we get:
+ − =
= − =

= = =
These will be the values that will satisfy the three equations.
4.5 Gauss Jordan Elimination Method
This method is just an extension of the Gaussian method. It involves ensuring that the matrix of
coefficients is turned into an identity matrix.
Using our previous example, the matrix [
−
− ] will be changed to an identity matrix
through the same row procedures.
We will start by changing the elements − and − / in the 3rd
column, 1st
and 2nd
row
respectively into 0 using the 3rd
row. This will be done by the following process:
= − −
= − −
[ ] [ ] = [ ]
Next step involves changing the 6 in row 1 column 2 into a 0. This will be done by using row 2.
= −
It will give us:
[ ] [ ] = [ ]
We can then rewrite our matrix into equations and we will have:
=
=
=

Example 2
After using the Gaussian method to solve for three simultaneous equations, the following three
equations were obtained:
Determine the values of , , using the Gauss Jordan elimination method.
− . − . =
− . =
=
[
− . − .
− . ] [ ] = [ ]
Work from the last column to eliminate the coefficients of in the other rows
i.e.
= + .
= + .
− .
|
Finally change − . into a 0 by using the second row. This will be done through the following
operation:
= + .
|
= = =

In this lecture, you have learnt about the Jacobian matrix which is a
matrix of second order partial derivatives. It is very helpful when
testing for functional dependence. You have also learnt an additional
method for solving three simultaneous equations.
Determine if the following functions are functionally
dependent
In order to use the gauss Jordan elimination method, you need to
understand how the Gaussian elimination method is done.
4.6 SUMMARY
NOTE
4.7 ACTIVITIES

Solve the following three simultaneous equations using
the gauss Jordan elimination method.
− =
− =
+ =
References
1. Main Text: Chiang A.C. and Wainwright K:
Fundamental methods of mathematical Economics.
3. A cook-book of Mathematics, Viatcheslav
VINOGRADOV, June 1999
4.8 FURTHER READING

By the end of this lecture the learner should be able to:
1. Determine the convexity of two and three independent
variables.
2. Get the determinants of a hessian matrix.
In this lecture we will learn how to test for convexity and concavity
of a function using the hessian matrix. In order to understand better,
you may need to remind yourself about the rules of partial
differentiation.
LECTURE FIVE: TESTS OF CONVEXITY AND CONCAVITY USING HESSIAN
DETERMINANTS
5.1 INTRODUCTION

5.3 CONCAVE AND CONVEX FUNCTIONS
A concave function has an absolute maximum when plotted on a graph, while a convex function
gives an absolute minimum when plotted on a graph.
It is easy to check for concavity and convexity in functions of single independent variable by just
checking the second order derivative.
If the second order derivative is negative, the function is concave and if positive, the function is
said to be convex.
In functions of two independent variables, there is a corresponding test which also uses the
second order derivatives.
If we have a function Z given as = ,
=
� ,
�
+
� ,
�
It can be written as:
= +
The second order partial derivative will be given as:
= + +
a) The function is convex if: , . −
b) The function is concave if: , . −
c) The function is strictly convex if: > , . − >
d) The function is strictly concave if: < , . − >
Example one
= − − + − Is the function convex or concave
= − + = − = − + − = − =
= − < , = ( ) =
ℎ
− ( ) =

The function Z is concave but not strictly so.
Determinantal test for sign definiteness
The matrix of second order partial derivatives: H= [ ] is called a hessian matrix.
A function Z is positive definite if the first and second principal leading minors of matrix H are
positive.
i.e. |H | = |fxx| >
|H |= | | >0 If we get the determinant, it will be given as: − >0
A function Z is negative definite if the first leading minor is negative while the second principal
leading minors of matrix H are positive.
i.e. |H | = |fxx| <
|H |= | | >0
Example Two
Determine if the following function is positive definite or negative definite using the Hessian
determinants.
= − +
= [ ]
|H | = >
|H |= | | = − = >
Since the determinants of both H1 and H2 are greater than 0, we say the function is positive
definite and by extension the function is convex.
The determinantal test can be extended to the case of more than 2 independent variables.
= , ,
= [ ]

The function Z is positive definite (convex) iff
|H | = |fxx| >
|H |= | | >0
| | = | | >0
It is negative definite (concave iff
|H | = |fxx|
|H |= | |
| | = | |
If Z is a function of n independent variables such that:
= , … … . ,
The function is negative definite (convex) iff:
|H | < , |H | > … … … … … − n|Hn| >
Its positive definite iff:
|H | > , |H | > … … … … … |Hn| >
Example three
Determine if function Z is convex or concave
= − − − − − + +
The matrix of second order derivatives will be given as:
= [
− − − − −
− − − − −
− − −
]
|H | = |− − | <
|H |= |
− − − −
− − − −
| = + >

We have learned how to check for concavity and convexity using the
hessian determinant.
Convex functions have a minimum as an extreme point while
concave functions have a maximum as the extreme point.
| | = [
− − − − −
− − − − −
− − −
] = − <0
The function is therefore strictly concave.
5.4 SUMMARY
NOTE

Determine if a cobb-douglas production function given
as = � is concave or convex given that β+α≤1
• Practice Question Five
• Determine if the following function is concave or convex
= − + − −
References
Fundamental methods of mathematical Economics.
5.5 ACTIVITIES
5.6 FURTHER READING
5.7 SELF-TEST QUESTIONS FIVE

By the end of this lecture the learner should be able to:
1. Formulate non-linear programming problems.
2. State the Kuhn tucker necessary conditions for a minimum and for a
maximum.
3. Solve for optimal values for both maximization and minimization
problems.
This lecture is an advancement of constrained optimization involving
inequality and non-negative constraints.
LECTURE SIX: NON-LINEAR PROGRAMMING
6.1 INTRODUCTION

6.3 Non Linear programming
This is a constrained optimization technique that makes it possible to handle problems that
involve non-linear inequalities and non-negativity constraints.
In normal optimization problems, first order necessary conditions for a relative extremum require
that the first order partial derivatives of the function with respect to each choice variable be zero.
In non-linear programming the first order necessary conditions are called Kuhn Tucker
conditions. In normal optimization first order conditions are necessary but not sufficient for an
extreme. The Kuhn Tucker conditions are necessary and sufficient conditions as well.
Effect of Non-negativity constraints
1. Maximisation case
Let’s assume a firm wants to maximize the profits subject to ensuring that the quantity X is non-
negative. This can be written as:
=
.
We can have three cases:
In case A, the profits are maximized (the maximum point on curve) when X is greater than 0. It
is an interior solution. At the point where profits are maximum, X>0 and the first order
derivative of the profit function with respect to X is 0 i.e. > , ′
= .
A
B C

In case B, the profits are maximized when X is equal to 0. It is a boundary solution. At the point
where profits are maximum, X=0 and the first order derivative of the profit function with respect
to X is 0 i.e. = , ′
= .
In case C, the profits are maximized when X is less than 0. Since X is not allowed to take
negative values, the firm will not produce. The firm will therefore operate at the point where
X=0, and first order derivative of the profit function with respect to X is negative i.e. =
, ′
< .
These conditions can be summarized as:
1. ′
=
>
2. ′
=
=
3. ′
<
=
If the three conditions are combined, maximization will occur at the point where:
′
and . ′
= This is known as the Complementary slackness condition between
and ′
.
The three conditions are the Kuhn Tucker first order necessary conditions for a maximum.
If a problem has got choice variables,
i.e.
= , ,
= ,
Then
′ ′ = = ,
6.4 Effect of non equality constraints
= , ,
, ,

, ,
, ,
Can be written as:
= , ,
, , + =
+ =
, , , ,
Using classical approach:
= , , + − + + – , +
In normal constrained optimization, first order conditions are:
�
�
=
�
�
=
�
�
=
�
�
=
�
�
=
�
�
=
�
�
=
But now, due to the presence of non-negativity constraints, we modify them to become:
1.
�
�
.
��
�
=
2.
�
��
.
��
=
3.
�
��
=
�
=
− . =
: − ℎ
If we ignore the non-negativity constraints and inequality signs in constraints we can write it as:
= , , + − , , + − , ,

a.
�
�
.
�
�
=
�
��
.
�
��
=
How?
, ,
, ,
�
�
= −
Example
=
. +
,
= + − − + −
K.T.C.
�
�
= − − − − − − − =
�
�
= − − − − − − =
�
��
= − − − − − − − =
�
��
= − − − − − =
Use trial and error: but ensure all inequalities hold. If they do not, try another solution
We can try: = = → Not possible for = ignore this and assume that both
are non zero.
− − > ℎ ℎ ℎ

− − = → to hold.
= + − = =
→ = + → − =
= ℎ =
− − =
ℎ
− =
=
=
ℎ ℎ ℎ ℎ ≠
Other alternative
> ℎ :
= ∗
= = =
− −
= +
= + ∗
=
6.5 Kuhn Tucker conditions for a minimization case.
Let’s assume a firm wants to minimize the costs subject to ensuring that the quantity X is non-
negative. This can be written as:
=
.
We can have three cases:

Y
In case A, the costs are minimized (the minimum point on curve) when X is greater than 0. It is
an interior solution. At the point where costs are minimum, X>0 and the first order derivative of
the cost function with respect to X is 0 i.e. > , ′
= .
In case B, the costs are minimized when X is equal to 0. It is a boundary solution. At the point
where costs are minimized, X=0 and the first order derivative of the cost function with respect to
X is 0 i.e. = , ′
= .
In case C, the costs are minimized when X is less than 0. Since X is not allowed to take negative
values, the firm will not produce. The firm will therefore operate at the point where X=0, and
first order derivative of the cost function with respect to X is positive i.e. = , ′
> .
These conditions can be summarized as:
a. ′
=
>
′
=
=
′
>
=
In each of the cases,
′ ′
=
Combined, these are the Kuhn tucker necessary conditions for a minimum
′ ′
=
Example two
:
= − + −
.
A B
C
f(x)

+
− −
,
= − + − + − − + − + +
�
= − − +
�
�
− − − +
�
�
= − −
�
�
= + +
� > , >
+ =
+ =
= ⁄
= − ⁄
This violates the non-negativity condition that
>
Assume both X1 and X2 are greater than 0
> >
�
�
= =
�
�
=
− − + =
− − + =
− − + =
− = +

Kuhn tucker first order conditions are important when
dealing with inequality constraints and non-negativity constraints in
non-linear programming.
Complementary slackness conditions are very
important when determining the solutions that satisfy all the
constraints in non-linear programming.
− + =
=
= ⁄ = −
� ≠
�
�
= + =
= = ⁄
= ⁄
>
= = ⁄
→ all choice variables are non-negative and all constraints are satisfied.
6.6 SUMMARY
NOTE

What are the Kuhn-tucker necessary conditions for a
minimum and a maximum?
References
Fundamental methods of mathematical Economics
6.7 ACTIVITIES
6.8 FURTHER READING

Max REVENUE =R = X1(10 -X1) + X2(20 - X2)
subject to 5X1 + 3X2 ≤40
X1 ≤ 5
X2 ≤10
X1 ≥ 0; X2 ≥0:
6.9 SELF-TEST QUESTIONS SIX

By the end of the lecture, the learner should be able to:
1. Integrate non-algebraic functions
2. Integrate economic functions by parts.
This lecture is an advanced topic in integration that was learnt in
mathematics for economics two. It will help the learner to be able to
integrate exponential and logarithmic functions. It also introduces the
concept of integration by parts especially when one is dealing with
complex functions.
LECTURE SEVEN: INTEGRATION
7.1 INTRODUCTION

7.3 Integration of non-algebraic functions.
Recall from the differentiation of non-algebraic functions:
If =
=
And
if = then
=
7.1.1 Exponential Rule
∫ . = +
∫ ′ �
= �
+
7.1.2 Logarithmic rule
∫ = +
∫
�′
�
= + >
| | + [ = ]
Example 1
∫
Using the exponential rule,
∫ = +
Example 2
∫ +

∫
+
. = ∫
( )
. = +
= + +
Example 3
∫( − −
+ ⁄ )
= ∫ . − ∫ −
. + ∫ ⁄ .
= ∫ . = + . + −
+ +
7.4 Integration by substitution
Used if appropriate
∫ . . = ∫ . = +
Comes from chain rule
= . = . =
∫ [ . ] = +
e.g.
∫ . + = = +
=

=
∫ = ∫ =
= +
=
+
+ = + + [ + ]
+ +
Definite integral after substitution
∫
+
.
= +
= =
∫
∫ = [ + ]
= .
= + = + =
=
= + =

[Lnv + ] = | | − Ln| |
*Ensure you change the limits.
Definite Integral as on area under the curve
=
=
=
0
∫ . = [ + ]
∫ . = − ∫ .
∫ =
∫ = ∫ .
∫ − = − ∫
∫ + = ∫ + .

= + =
=
= ∫ .
.
+ = +
= +
∫ +
.dx = +
= =
= = ∫ .
+
+
+c
This rule applies whenever it is possible to express the integral as .
∫ =
∫ + � +
Substitution = � +
= +
= +
= ∫ .

8
+ . =
( + )
8
+
7.5 Integration by Parts
The rule applied when integrating by parts is given as:
∫ . = − ∫ .
Comes from the product rule:
i.e.
= . +
∫
∫ . + ∫
= ∫ . + ∫ .
− ∫ = ∫ .
Example 1
∫ + ⁄
. → Substitution cannot be applied.
= + ⁄
=
+ +⁄
⁄
+ ⁄
=
= + ⁄
.
∫ = + ⁄

∫ + ⁄
= =
= + ⁄
. − ∫ ⁄ + ⁄
.
= − + ⁄
+
Find
∫ + + ⁄
= +
= =
= + ⁄
=
+ ⁄
∫ . = − ∫ .
=
+ ⁄
+ − + ⁄
+
Example 2
∫ .
= − ∫ .
− ∫ . .
= − +
= − +

In this lecture, we have learnt how to integrate non-algebraic
functions and also integrating by parts. The knowledge of integrating
the non-algebraic functions will be very helpful when looking at the
next topic of differential functions.
Integrate the following functions:
1. ∫ . 2. ∫ −
. + .
When integrating logarithmic functions by parts you should always
make the log function the part that will be differentiated.
7.6 SUMMARY
NOTE
7.7 ACTIVITIES

Problem question seven
Integrate the following function by parts
∫ + + ⁄
References
7.8 FURTHER READING
7.9 SELF-TEST QUESTIONS SEVEN

By the end of this lecture, the learner should be able to:
1. Formulate a differential equation.
2. Solve homogenous and non-homogenous first differential
equations.
3. Solve exact differential equations.
Differential Equations are very important in Economics. In order to
be able to solve differential equations, the knowledge derived from
integral calculus will be very important.
LECTURE EIGHT: FIRST ORDER DIFFERENTIAL EQUATIONS
8.1 INTRODUCTION

8.3 Definition of Differential Equations
A differential equation helps in expressing the relationship between a variable with its past
value/s if it changes continuously with time. A differential equation is an equation involving
derivatives. It is an implicit functional relationship between a variable and its differentials.
For example
= −
- The order of a Differential Equation is the highest order of any of the derivatives
contained in it. It’s said to be of order if the ℎ
derivative is the highest order
derivative in it.
= + = This is a of order 1
- Highest power to which the derivative of the highest order occurs is the degree of a
differential equation. For example:
− + → Order 2, degree 3
A differential equation is linear if the dependent variable is of 1st
degree e.g.
� + �
−
−
+ � =
-Linear . of order
Assuming � , � and � are constants and = the equation is said to be a linear
homogenous differential equation of order .
If ⁄ appears only in the 1st
degree and so does the dependent variable and we do not
have any product of the form:

( ⁄ ) the equation is linear.
8.4 First order linear differential equation (FOLDE)
If
+ =
If : = constant ie. Coefficient of is constant and is a constant additive term then we
have a FOLDE with constant coefficient and constant term.
Given a differential equation of the form:
+ � =
Since = , we say it is a homogenous differential equation.
Solution to a differential equation
Involves getting all solutions and if it does not have one, to show it doesn’t have a solution.
A differential equation of order has a general solution containing independent
constants.
A particular solution is obtained from the general solution, by assigning specific values to the
arbitrary constants which are obtained from the information given as initial conditions.
Example
1. If: =
=
In order to get back Y, we need to integrate the function
∫ = ∫ . = ∫ . +
= + .
2. . =0

∫ = ∫
= +
= +�
= . � �
=
=
3. =
∫ = ∫ .
= − + →
= ℎ =
= − +
= − . + =
= − +
8.5 Non Homogenous differential functions
+ � = −
The Solution is composed of 2 parts:
1. Complementary function and
2. Particular integral ( )
If = , then,
=
Then
=
= ⁄
= ⁄ ≠
= + = � −
+ ⁄

To solve, assume = . Whats ?
The solution is the particular integral that ensures that A has a definite value,
When = � −
+ ⁄
= � + ⁄ � = − ⁄
= [ − ⁄ ] −
+ ⁄
8.6 Separable differential equations
These are equations that can be specifically separated into two parts, one a derivative of one
variable and the other a derivative of another variable. For example:
+ =
. + =
∫ . + ∫ =
Example
. = −�
∫ = ∫
= − +
= − +
= −
� =
Then
Y = Αe−at
− General solution
At t =
= Αe−
= Α
= −
−

Definite solution of a differential equation
– Is the only one that make solution satisfy initial condition.
The solution is not numerical rather it is a function of .
is free of any derivative or differential expressions
e.g.
=
= ∫ = ∫
= + =
= � +�
=
If = ℎ =
=
Then
= +
= . =
+ − =
. + . − =
+ + = +
+ = ⁄ +
√ +
+
= √ = √ + +
= +
= .

∫ = ∫ .
= ∫ −
= ∫ .
− −
= +
+ −
=
Example
=
−
=
− ∫ = ∫
− = +
+ =
=
8.7 Exact Differential Equations
If , + , . = We say the function is exact if =
= is a necessary and sufficient condition for the equation to be exact.
Solution of E.DE
, = ∫ . +
. + =
ℎ
, = ∫ +
. . Takes care of any terms containing that dropped out in the process of
differentiating

. + . =
=
∫ . = =
Steps
1. , = + −
2.
� ,
= = + ′
=
3. ′
=
′
= ∫ ′
= ∫ . =
4. , = + =
=
=
= ⁄
= − ⁄
+ =
=
, = ∫ . +
= ∫ . +
= +

, = +
= = + ′
=
=
, = + =
=
= − +
= − ⁄
+ = − +
=
∫ . = +
= +
= + =
+ =
− = =
=
=
= + =
=
=
= ( )
⁄

Homogenous differential equation (HDE)
If , + , = is homogenous first differential equation of degree .degree
A . . is solved by reducing it to a separable equation after making necessary
substitution. If
= ℎ = . + .
=
=
=
∫ = ∫
=
= +
= +�
= +�
−
=
−
+
=
−
= +
− −
= −
=
−
=
=
= .

= +
= ⁄
= ⁄
=
[ =] .
= +
= . . = → .
e.g.
= =
= → − . .
→ � ℎ .
8.8 Non linear D.E of 1st
order 1st
degree
appears in a higher power or lower power not equal to e.g.
+ =
Separable function e.g.
− =
+ . =
+ =
+ − =
−
+
−
−
=
∫
−
+ ∫ =
∫ = − ∫
+ = − +
= − +
= − +
⁄

In this lecture we have learnt how to get the general solution to first
order homogenous and non-homogenous differential equations.
Differentiate between a homogenous and non-
homogenous differential equation.?
A differential equation has a general and a particular solution.
8.9 SUMMARY
NOTE
8.10 ACTIVITIES

1. Solve for Y in the following differential equation
2. + =
=
References
8.11 FURTHER READING
8.12 SELF-TEST QUESTIONS EIGHT

At this end of this lecture, the student should be able to:
1. Form difference equations from economic relationships.
2. Get general and particular solutions to first order homogenous
difference equations.
3. Get general and particular solutions to first order non-
homogenous difference equations
This lecture introduces the student to first order difference equations.
These are equations used to model; the relationship between a
variable over time, if the variable changes over discrete time
intervals. We will look at the applications of these equations in
economic relationships.
LECTURE NINE: FIRST ORDER DIFFERENCE EQUATIONS
9.1 INTRODUCTION

9.3 Definition of a Difference Equation
A difference equation describes the relationship between a variable overtime. It describes the
relationship between dependent variables (for example investments) and the independent
variable time.
When we denote as ‘ ’ it means the is a discrete function. i.e. changes at fixed points
in time only.
On the other hand when we denote as , it implies is a continuous function which
changes continuously overtime.
If the following functions:
= + →
+ = + +
= + →
A difference equation represents the relationship between a variable with its value in the
previous period e.g.
= −
+ =
A difference equation is used to model dynamic systems in which changes occur at fixed
intervals, which are equally spaced.
For example:
1. If salary increments are 20% per annum, then in any given year, the salary will be 120%
of the salary in the previous year.
e.g. + = = −
− − =
− . − =

2. If a machine depreciates at the rate of 10% per annum, then the value of the machine in
year will be 90% of its value in year − i.e.
= −
− − =
− . − =
Order of difference equations
This is the number of time intervals spanned by the difference equation.
e.g . = . − + . − + . − −
= . − + . − = −
= . − = −
Homogeneity of a difference equation
A difference equation said to be homogenous if the =
I.e. − . − =
− . − − . − =
On the other hand, if the is not equal to zero, it is classified as non-homogenous.
e.g. + − + − =
9.4 Solution to difference equations
The solution can be obtained using the iteration method: i.e.
If + − . =
Get the values when = , , , , by assuming = , in year 1.
+ − . =
+ = . . =
+ = . = . ,

= ,
= .
= . , = ,
= . , = ,
= . , = ,
9.5 General solution for a Homogenous first order difference equation
+ − . =
If we get the income for years , , , , in terms of we will have:
+ = . =
= .
= . [ ] = . [ . ] .
= . [ ] = . [ . ]
= .
In general
= . −
Assume = = ,
= . −
= . ,
= . ,
= , .
General and particular solutions of difference equations
We have shown that:
− . − = can be generalized as:
= . −
…………………………1

This can be rewritten as:
= . . −
……………........2
= . −
. . . ………………………3
Since is a constant, . −
is also a constant, we can combine them into one constant to be
represented as � so that equation 3 becomes:
= � . = �. . ……………………4
. is also a constant. If we represent it with a small ′�′ we can have equation 4 written as:
= Α. � …………………………………5
This is called a general solution where ′�′ will be determined from the difference equation.
When we evaluate equation 5 at the given values of , and specific values of ′Α′we get a
particular solution.
Example:
Assume + = .
a) Find the general solution.
b) If = , find the particular solution.
c) Evaluate , , .
Solution
a) = Α �
+ = Α +
if + − . = then we can substitute this into equation of & + to
get:
Α � +
− . Α α t
=
Α � � − . Α � =
Α αt
α − . Ααt
=
Α at [α − . ] =

Either Α � = or � − . =
� = .
If Α = then
=
If on the other hand � = then:
= Α = - This is called a trivial solution.
Therefore:
� = .
= Α. . General solution
b) If = , find the particular solution
= Α . t
= = Α. .
→ Α =
.
=
.
=
The particular solution is therefore: = .
c) Evaluate , ,
When = = . =
= = . =
= = . = .
9.6 Solution to Non-homogenous difference equations
If
+ − . = , the equation is called a non-homogenous difference equation.
The solution will consist of 2 parts:
1. Complementary function (CF)
2. Particular Integral (P.I)
i.e. = + .

Complementary Function
Solution to the homogenous part of the difference equation e.g.
If
+ − . =
then the complementary function is the solution to:
+ − . =
Particular Integral
It is a function that satisfies the full difference equation. If the is fully a constant then, the
particular integral will be given as: = (another constant).
If is constant , then will be given as
, =
To show how we get the particular integral lets use the following example:
a) Find the General solution of:
+ − . =
b) Find the particular solution if = ,
Complementary Function
+ − . =
= Α �
+ = Α � +
Α � +
− . Α� =
Α � . � − . Α� =
Α� � − . = � = .
,� = Α . → This is the complementary function.

In this lecture we have learnt about first order difference
equations which are used to model variables which change discretely
with time. Solution to difference equations has also been learnt.
Particular Integral
, =
+ , =
Constant does not change irrespective of time
+ − . =
− . =
. = = ,
Therefore
,� = ,
General solution is
= � + � = Α . + ,
If:
= , , then:
= ,� + = Α . + , = ,
Α . = , − =
Α =
.
=
The particular solution is therefore:
= . + ,
9.7 SUMMARY

1. Solve for by iteration method:
a) + − . − = Given = find
2. Find the general and particular solutions for”
+ = + Given =
3. Ι − . Ι − =
Get the general solution
Given that Ι = , get the particular solution
Most economic variables can be modeled using difference
equations. Difference equations are important when forecasting
variables in the future.
NOTE
9.8 ACTIVITIES

Practice question nine
Determine the general and particular solution of the following
difference equation
+ + = . =
References
9.9 FURTHER READING
9.10 SELF-TEST QUESTIONS

By the end of the lecture the learner should be able to:
1. Determine the conditions necessary for a variable to be stable
2. Solve difference equations used in economics.
The lecture discusses how to determine whether a difference
equation is stable.
LECTURE TEN: STABILITY AND APPLICATIONS IN ECONOMICS
10.1 INTRODUCTION

10.3 Stability and the time path to stability
The general solution of a difference equation
= Α �
 The expression can be used to forecast the dependent variable Y. To determine if Y will
stabilize i.e. approach some fixed value with time, we can trace how income changes each
year until stability is reached (trace the time path to stability).
The stability of the solution as ′ ′ increases are deduced from the exponential term � .
� as increases i.e. > .
If:
−∞ < � < − then −� → ±∞,
Solution is unstable e.g. − = − − = −
Time path alternates − = − =
If:
− < � <
−� → e.g. − . = − .
− . = .
− . − .
Value decreases as → ∞
If
< � < e.g. � = .
= .
� → . = .
Solution is stable . = .
Time path tends to zero . = .

< � < ∞ e.g. � =
Solution unstable =
Time path tends to infinity =
=
Examples
Find the general and particular solutions of the difference equations given as:
+ − = =
= Α �
+ = Α� +
Α � +
− Α at
=
Α � . − Α at
=
Α � − = � − = � =
General solution
= Α Since � = , the solution is unstable.
Particular solution
= Α t
=
= Α Α = =
=
= t
- Particular solution
Example 2:
+ + . = …. (1) =
= Α
+ = Α � +
Substituting this in equation 1:
Α � +
+ . Α αt
=

Α � . � + . Α � =
Α � � + . = � + . =
� = − .
Since
− < � <
Solution is stable and time path alternates
= Α − . → General solution
=
= Α − . = Α = − .
= −
= − − . - Particular solution
Applications of Difference equations:
Cobweb Model:
This is a microeconomic application of difference equations.
Consider the following demand and supply functions over time:
, = � − - Quality demanded is a function of the price at time .
, = + − - Quality supplied at time is a function of the price prevailing in past
period i.e. − .
At equilibrium:
= ,
� − = + −
− − − = − � This is a difference equation
Example:
= −
= − + . −
At equilibrium
− = − + . −

+ = − +
→ . − + =
Since this is a non-homogenous difference equation, the solution will have two parts.
= . + Ι
= ,� + ,�
Find the Complementary functions:
+ . − =
= Α �
− = Α � −
Α � + . Α � −
=
Α � + . Α � . �−
=
Α � [ − . �− ] =
+
.
�
=
= −
.
�
� = −
.
= − .
Therefore:
,� = Α − .
Particular Integral
Since the right hand side is a constant, the
,� =
− ,� =
Hence: in + . − =
+ . =
. =
= =
, =

In this lecture we have learnt how to check for stability in
a difference equation, especially in the cobweb models.
The stability of a difference equation is determined
by the value of ′ ′ in the following expression:
�
General solution
= ,� + ,�
= Α − . +
Particular solution
If =
Then
= Α − . +
= Α + =
Α =
= − . + → Particular solution
Is the price stable?
� = − . Since � is between − and , the price is stable.
1.2 SUMMARY
NOTE

Practice question
Find:
i. The general solution
ii. Determine if solution is stable
iii. Find the particular solution
2. Find the general and particular solution given that
Dt=10-2Pt St= -5+4Pt-1
P(0) = 10
. + + = =
References
1.3 ACTIVITIES
1.7 FURTHER READING

Practice question 10
The supply and demand function of cabbage is given as:
= − + − = −
Required:
i) Determine the equilibrium price
ii) Is it stable?
1.8 SELF-TEST QUESTIONS

Answers to self-test questions
Solution to self-test one
Let x1 = Number of hardcover books ( per 100) to be produced
X2 = Number of paperback books ( per 100) to be produced
Since the objective is to minimize the cost, the objective function is given by-
Minimise Z = 700x1 + 600x2
2x1 + x2 90 ( Minimum running of press I)
x1 +2x2 70 ( Minimum running of press II)
x1 ,x2 ≥ 0 ( Non-negativity constraint)
Solution to self test two
• Let x1 = Number of chairs to be produced
• X2 = Number of tables to be produced
• Z=contribution
• Since the objective is to maximize the contribution, the objective function is given by
Maximise Z= 20x1+ 30x2
• Subject to constraints:
• 3x1 + 3x2 ≤ 36 ( Total time of machine M1)

• X1, x2≥ 0 [ Non-Negativity constraint)
• X1= X2 =
Solution to self test three
• Y=4
• X=2
• Z==44
Solution to self test four
 X=6 Y=4 Z=2
Solution to self test five
• = [
−
−
−
]
• |H | <
• |H | >
• |H | <
• Z=Concave
Solution to self test six
• X1 =2; X2 = 10:
•
Solution to self test seven
• = +
• = =
• = + ⁄

• =
+ ⁄
• ∫ . = − ∫ .
• =
+ ⁄
+ − + ⁄
+
Solution to self test eight
1. ∫ . = ∫ .
+ = +
− =
− =
= +
= √ +
±
= � −
+
Solution to self test nine
Complementary function
,� = Α (− ⁄ )
Particular Integral
Since the right hand side = . , the general form of
, = .
+ ,� = . +
= . .
Substituting this in the original difference equation it will be:
+ + = .
. . +
+ . = .
. . . + . = .
. . [ . + ] = .
Divide all through by . , we get

[ . + ] =
. =
=
.
=
Therefore:
,� = . .
General solution:
= ,� + ,�
= Α (− ⁄ ) + .
Particular solution
=
= Α (− ⁄ ) + . =
Α + = Α =
Yt = (− ⁄ )
t
+ .
Solution to self-test ten
 Pt = A − t
+
 Solution is stable

REFERENCES
1. Main Text: Chiang A.C. and Wainwright K: Fundamental
methods of mathematical Economics
2. Monga G.S: Mathematics and statistics for Economists, second
revised edition.
3. A cook-book of Mathematics, Viatcheslav VINOGRADOV, June
1999.
4. Tulsian and Pandey V. Quantitative Techniques, Theory and
Problems.
5. Gupta S.K. Cozzolino J.M. (1975) Fundamentals of Operations
Research for Management
6. Kothari C.R. (2004) Quantitative Techniques, 3rd
revised edition
7. Kothari C.R. An introduction to Operations Research, Vikas
Publishing Press.
8. Lucey T. (2004) Quantitative Techniques, 4th
edition

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