Business Mathematics Final

FINANCIAL MANAGEMENT IN VALUE CHAIN

MODULE CODE: ABVM 322

TOTAL ECTS: 20

August 2012

Financial Management in Value Chain: Business Mathematics and its Application

FINANCIAL MANAGEMENT IN VALUE CHAIN

(ABVM 322)

(20 ECTS)

LIST OF LEARNING TASKSLT1: Business Mathematics and its Application (5 ECTS)LT2: Accounting Principles (5 ECTS)LT3: Financial Management (5 ECTS)LT4: Cost and management Accounting (5 ECTS)

AUGUST 2012

Jimma, Haramaya, Ambo, Hawassa, Adama, Bahirdar, W.Sodo,Semara Universities

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1. General information

1.1. Module introduction

Hello! Dear student welcome to the module/educational unit Financial Management in Value Chain.

This educational unit consists of four learning tasks :( LT1) Business Mathematics and its Application (LT2) Accounting Principles (LT3) Financial Management and (LT4) Cost and management Accounting. The learning tasks under this educational unit contain examples and learning activities. Some also contains self-test questions at the end of each learning tasks and or sections under learning tasks. The learning activities and self test questions are designed to assist you analyze and grasp the ideas and issues discussed in the learning task/sections. You are strongly advised to attempt these activities and questions by yourself to check your level of understanding before moving ahead to the next topics or section within the learning tasks.

1.2. Relation with the curriculum

This educational unit is designed to introduce and equip Agribusiness and Value chain Management (ABVM) graduates with technical capacity and knowledge of basic accounting principles, mathematics of finance, financial management, and cost and management accounting. It enables graduates to provide information used to make reasoned choice among alternative uses of scarce resources in the business and economic activities.

It also gives students an opportunity to evaluate how earning an income or having access to credit can serve to bring benefits for women. Students need to ask critical questions about how increased access to resources can be translated into changes in the strategic choices that women are able to make – at the level of the household and community, as well as at work. Students will learn to be conscious of continuous financing mechanism, dynamism of financial rules, regulations and principles, and financing risks to contribute to the agribusiness and value chain management on continual basis.


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2. Module objectives

The aim of this educational unit is to introduce and equip ABVM graduates with technical capacity and knowledge of basics for managing finance in value chains. After completing this educational unit, the learner will:

o Explain the concept of long-term financing,

o Use the concept of time value of money,

o Differentiate the various techniques of mathematics that can be employed in solving

business problems,o Apply basic concepts and principles of accounting and cost and management

accounting for business decisions,o Prepare and Analyze financial Statements of agribusinesses,

o Analyze costing system in the value chain development,

o Evaluate the gender and financial empowerment procedures, and

o Make financial planning and forecast,


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FINANCIAL MANAGEMENT IN VALUE CHAIN (ABVM 322)

LT 1: BUSINESS MATHEMATICS AND ITS APPLICATION

(5 ECTS)

Prepared by:Adare Assefa (MBA)Ebisa Deribie (MA)

AUGUST 2012


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Table of Contents3.1. Business mathematics and its application.......................................................................1

3.1.1. Introduction.............................................................................................................1

3.1.2. Objectives................................................................................................................1

3.1.3. Sections...................................................................................................................1

3.1.3.1. Section I: Linear equations and their interpretative applications.....................1

3.1.3.2. Section II: Matrix algebra and its applications..............................................18

3.1.3.3. Section III: Linear programming...................................................................34

3.1.3.4. Section IV: Mathematics of Finance..............................................................53

3.1.3.5. Section V: Introduction to Calculus...............................................................73

3.1.4. Proof of Ability.....................................................................................................86

Major References.........................................................................................................87


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3.1. Business mathematics and its application

3.1.1. Introduction

Dear student, welcome to the learning task Business Mathematics and its application. This learning task is designed to expose you to the basic concepts and area of managerial application of mathematics. It is divided into five sections: The first section deals with the linear equations and its applications; the second section is about the matrix algebra and its applications; the third section deals with linear programming, the fourth section is dedicated to mathematics of finance and the fifth section is about elements and application of calculus. You will find learning activities in each sections. To successfully accomplish this learning task 140 study hours is allotted.This learning task is executed both in class room and through students self learning, students are expected to attend lectures in class rooms, engaged in group and individual assignment/works and various kinds of assessments/PoA as stated under the assessment plan.

3.1.2. Objectives

After working through this learning task, you will be able to:

o Differentiate the various techniques of mathematics that can be employed in

solving Agribusiness problems,

o Appreciate the importance of mathematics in solving real world business

problems, and

o Use different mathematical techniques for supporting decisions


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3.1.3. Sections

3.1.3.1. Section I: Linear equations and their interpretative applications

Dear student! What is linear equation? In what way you think is useful in agribusiness management? -----------------------------------------------------------------------------------------------

Introduction

Today the business environments are changing. In such environment, organizations encounter diverse set of problems as well as opportunities. Consequently, managers are expected to make appropriate decisions to tackle the challenges and to feed the breast of the opportunity. .In practices the concept and interpretative applications of linear equations have a considerable importance. This is because, it is common to face so many cases demanding the application of mathematics of linear algebra and geometry in making a viable decision that enhance the achievement of organizational objectives. For instance sales volume and advertisement expense, output level and number of employees engaged on some activity and cost of production, demand for and supply of a given product can be well approximated and explained by a linear equation.

Cognizant to the above fact, we need to be well acquainted with the fundamentals of linear equations algebra and geometry as related to its agribusiness application. This section, therefore, is dedicated to our study of linear equations.

Linear equations, functions and graphs:

Basic Concepts of Linear Equations and Functions: An equation is a statement of equality, which shows two mathematical expressions are equal. Equations always involve one or more unknown quantities that need to be solved. Among the different types of equations, linear equation is the one that we are going to deal with in some detail.

Linear equations: are equations whose terms1 are a constant times a variable to the first power. Accordingly, equations that can be transposed to the form,

a1 x1+ a2 x2+ …+ an xn = care said to be linear equations. Where,

a1, a2, a3, … an and c are constantsx1, x2, x3, …xn are variables (unknown quantities)

1 Terms of a linear equation represent the parts of equation that are separated by plus, minus, and equal signs.Jimma, Haramaya, Ambo, Hawassa, Adama, Bahirdar, W.Sodo,Semara Universities

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a1 x1, a2x2, … an xn and c are the terms of the equation (terms of a linear equation represent the parts separated by plus, minus, and equal signs)

As it occurs in many business application cases, a linear equation may involve two variables, x and y, and constants a, b, and c in which case the equation relating x and y takes the form,a x + b y = c

The following are all examples of linear equations. 2x + 3y = 9, 3x – 9y + z = 23, 4y + 7.5x – 11 = 14

On the other hand, 4xy + 7x = 8 is not a linear equation because the tem 4x y is a product of a constant and two variables. Likewise 5x2 + 3y = 25 is not linear because of the term 5x2 which is a constant times one variable to the second – power.

Example: Assume that Ethiopian Electric Power Corporation charges Birr 0.55 per kilowatt-hour consumed and a fixed monthly charge of Birr 7 for rent of electric meter. If y is the total monthly charge and x is the amount of kilowatt-hours consumed in a given month, write the equation for y in terms of x.

Solution: The total monthly charge will be, 0.55 times the number of monthly KWh consumption plus Birr 7 for meter rent. Thus, using the symbols given, y = 0.55x + 7

The equation of this example is linear with two variable x and y. In such linear equations, we need to note that the constants can be positive or negative, and can be fractions when graphs of these equations is plotted it will be a straight line. This is the reason for the term equation.

Linear Functions: functional relationship refers to the case where there is one and only one corresponding value of the dependent variable for each value of the independent variable. The relationship between x and y as expressed by

Y = 0.55x + 7

is called a functional relationship since for each value of x (independent variable), there is a single corresponding value for y (dependent). Thus if we write y as expression involving x and constants x is called the independent variable, then the value of y depends upon what value we may assign to x and as a result it is called the dependent variable. Therefore, a linear function refers to a linear equation, which does have one corresponding value of dependent variable for each value of the independent variable.Learning activity 1.1


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Suppose that a car rent company charges Birr 65 per hour a car is rented. In addition Birr 150 for insurance premium. Write the equation for the total amount charged by the company in terms of the hours the car is rented.

Graph of a Linear Equation: Linear equations in two variables can be plotted on a coordinate plane with two dimensions. Such equations have graphs that are straight lines. This means that the graph of the relationship between the variables takes the form of a straight line. Any straight-line graph can be sketched by plotting just two points which satisfy the linear equation and then joining them with a straight line. Now let us further develop this approach by considering the following example.

Example: Sketch the graph of the equation 2y - 3x = 3.

Solution: To plot the graph, you may arbitrarily select two values for x and obtain the corresponding values for y. Therefore, let’s set x = 0. Then the equation becomes 2y – 3(0) = 3. That is, 2y = 3

y = 3/2

This means that when x = 0, the value of y is 3/2. So, the point with coordinates (0, 3/2) lies on the line of 2y – 3x = 3.

In the same way, let y = 0. Then the equation becomes -3x = 3.That is, x = 3/-3 = -1.

This means, when y = 0, the value of x is -1. So, the point with coordinates (-1, 0) lies on the line of equation 2y – 3x = 3. These two points are plotted on the coordinate plane with horizontal “x – axis” and vertical “y-axis” as follows.

Learning Activity 1.2


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0 1 2 3 4 x

(-1, 0)

-2 -1

y

5

4

3

2

1

(0, 3/2)

Fig 1. Linear Equation Graph


Find two coordinate points that satisfy the equation 3x + 4y = 24. Then using the two coordinate points plot the graph of the given function.

The distance between two points: The distance between two points is the length of a straight-line segment that joins the points. To determine the length of a given segment in coordinate geometry, algebraic procedures are applied to the x – and y coordinates of the end points of the segment. Distance on horizontal and vertical line segments are used in the computation of the distance. Distance on a vertical segment (also called vertical separation) is found by computing the positive difference of the y- coordinates of the end points of the segment. Distance on the horizontal segment (also called horizontal separation) is found by computing the positive difference of the x-coordinate of the end points of the segment. Thus, given two points (x1, y1) and (x2, y2), the quantity / x2 – x1 /, is called the horizontal separation of the two points. Further, the quantity / y2 – y1 / is the vertical separation of the two points.

Example: Given the points A (-5, 7), B (-3, -9), C (-5, 15), D (12, 6), find the horizontal and vertical distance of the segment,

a. AB b. AD c. BD

Solution:a. The horizontal distance (Separation) of the points A (-5, 7) and B (-3, -9) is given by

Horizontal distance = / x2 - x1 / = /-3-(-5)/

= /-3+5/ = 2Vertical distance = / y2 - y1 /

= / -9 - 7 /= /-16/ = 16

b. Horizontal distance = / x2 - x1/ = / 12 - (-5)/ = 12+ 5/ = 17 Vertical distance AD = / y2 - y1 / = / 6 – 7 /

= / -1 / = 1c. Horizontal distance BD = / x2 - x1/ = /12- (-3)/

= / 12+3/ = 15Vertical distance BD = / y2 - y1/ = / 6- (-9)/

= / 6 + 9/ =15

Learning activity 1.3


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Find the vertical and horizontal separation of the following points. a. (5, 7) and (-3, 2)b. ( 5, - 3) and (-11, -7)c. (6, 2) and (6, -4)d. (3, 4) and (9, 4)

Dear student, as you recall all lines in a coordinate plane are not vertical and/or horizontal. Hence, in case the segment is slant to any direction the actual distance between (x1, y1) and (x2, y2) may be calculated from Pythagoras’ Theorem, using their horizontal and vertical separations.

In the above diagram, AB2 = AC2 + BC2. That is, the distance d between point A and B is given by:

d2 = (horizontal separation) 2 + (vertical separation) 2

d2 = (x2 – x1)2 + (y2 – y1)2

d =

Example: Calculate the distance d between the points (5, -3) and (-11, -7).


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B (x2, y2)

X

C (x2, y2)A (x1, y1)

Y

Fig1. 2. Places

of coordinates


Solution :

d =

d =

That is, d = = d = 16.5


Find the distance between the points given below. a. (5, 10) and (11, 8) b. (0, 0) and (9, 12)c. (-2, -5) and (3, -4) d. (4, 7) and (6, -5)

Developing Equation of a Line: We have three alternative forms of developing the equation of a straight line. These are, slope-intercept form, slope-point form, and two-point form. Let us consider these approaches further. The Slope – Intercept Form: Dear student, before considering slope intercept form of developing equation of a line let’s have a brief look at the concept of slope or gradient. Slope is a measure of steepness or inclination of a line and it is represented by the letter m. the slope of a non-vertical line is defined in several ways. It is the rise over the run. It is the change in y over the change in x. Thus, given coordinates of two points (x1, y1) and (x2, y2)

Slope = m = Run

eRis =

12

12

xx

yy

, Where, x1 ≠ x2

If the value of the slope is positive, the line raises form left to right. If the slope is negative, the line falls from left to right. If the slope is zero, the line is horizontal. If the slope is undefined then the line is vertical. Dear student, please consider equally ranged graph and try to find slope of any coordinates on the graph before reading the next example.

Example (a): Obtain the slope of the straight-line segment joining the two points (8, -13) and (-2, 5).Solution:

Therefore, the line that passes through these two points falls downwards from left to right. On the other hand, if the equation of a line is given, then the slope can be determined more


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simply. Thus, if a liner equation is written in the form y = m x + b, “m” is the slope and “b” is often referred as the intercept term and it is the value at which the straight line intercepts the Y-axis.

Example (b): An agent rents cars for one day and charges Birr 22 plus 20 cents per mile driven.

a. Write the equation for one day’s rental (y) in terms of the number of miles driven (x). b. Interpret the slope and the y – intercept. c. What is the renter’s total cost if a car is driven 100 miles? What is the renter’s cost

per mile?

Solution:

Given fixed (constant) cost of Birr 22 = b Slope = m = 20 cents = Birr 0.2

y = Total cost for one day rental x = Number of miles driven

a. The equation y = m x + by = 0.2x + 22

b. Interpretation The slope, m = 20 cents (Birr 0.2) means that each additional mile driven adds 20 cents to total cost. b = Birr 22 is the fixed cost (the amount to be paid irrespective of the mile driven). Hence, it will be the total cost when no mile is driven.

c. Total cost of driving 100 miles (x = 100) = 0.2 (x) + 22

Total cost of the renter = 0.2 (100) + 22 = 20+22 = Birr 42

Cost per mile when x = 100 miles is given by total cost of driving 100 miles divided by 100 miles. Putting it in equation form,

= 42 ÷ 100 = Birr 0.42


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Learners form a small group and attempt and discuss on the question that follows

It costs Birr 2500 to set up the presses and machinery needed to print and bind a paperback book. After setup, it costs Birr 2 per book printed and bound.

a. Write the equation for the total cost of making a number of booksb. State the slope of the line and interpret it.c. State the y-intercept of the line and interpret it.

The Slope – Point Form: In this form, we will be provided with the slope and points on the line say (x1, y1). Then, we determine the intercept from the slope and the given point and develop the equation. Accordingly, the expression we need further is the equation that is true not only for the point (x1, y1) but also for all other points say (x, y) on the line. Therefore, we have points (x1, y1) and (x, y) with slope m. The slope of the line is y - y1 / x - x1 and this is equal for all pair of points on the line. Thus, we have the following formula for slope-point

form:

Alternatively,

y – y1 = m (x – x 1)Example: Find the equation of a line that has a slope of 3 and that passes through the point (3, 4).

Solution: Given m = 3 and (x1, y1) = (3, 4). By substituting these values in the formula y – y 1 = m (x – x 1) We will obtain y – 4 = 3 (x – 3). Then, y – 4 = 3 x – 9

y = 3 x – 9 + 4 y = 3 x – 5.

In another approach, y = m x + b where (x, y) = (3, 4) y = 3x + b 4 = 3(3) + b 4 = 9 + b4 – 9 = b = - 5 is the y-intercept.


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Thus, y = 3x – 5 is the equation of the line.


If the relationship between total cost (y) and the number of units made (x) is linear and if cost increases by Birr 3 for each additional unit made and if the total cost of making 10 units is Birr 40 find the equation of the relationship between cost and number of units made.

The Two – Point Form: In this case, two points that are on the line are given and completely used to determine equation of a straight line. In doing so, we first compute the slope and then use this value with either points to generate the equation. Taking two points designated by (x1 – y1) and (x2 – y2) and another point (x, y), we can develop the expression for the equation of the line as follows.

Therefore, (y – y 1) (x 2 – x 1) = (y 2 – y 1) (x – x 1) is the expression for the two-point form of generating equation of a straight line.

Example: A publisher asks a printer for quotations on the cost of printing 1000 and 2000 copies of a book. The printer quotes Birr 4500 for 1000 copies and Birr 7500 birr for 2000 copies. Assume that cost (y) is linearly related to the number of books printed (x).

a. Write the coordinates of the given points b. Write the equation of the line

Solution:

Given the values x 1 = 1000 Books y1= Birr 4500x 2 = 2000 books y2= Birr 7500

a. Coordinates of the points are:(x1, y1) and (x2, y2)

Thus, (1000, 4500) and (2000, 7500)

b. To develop the equation of the line, first let’s compute the slope. m = y2 – y1

x2 – x1

= 7500 – 4500 2000 – 1000 = 3000 ÷ 1000 = 3


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Then, consider the formula of two-point form of developing equation of a line as given by,

y – y1 = y2 – y1

x – x1 x2 – x1

We have obtained the value for the slope m = 3 as it’s expressed by

.

Subsequently, by substitution this value in the above formula will result in;

In continuation, substitute the value (1000, 4500) in place of x1 and y2 in the equation y – y1 = 3 (x – x1). As a result, you will obtain,

y – 4500 = 3 (x – 1000)y – 4500 = 3x – 3000y = 3x – 3000 + 4500y = 3x + 1500 ……………………… is the equation of the line.

Learning activity 1.7 Dear student form a small group and attempt the following question:

As the number of units manufactured increases from 100 to 200 manufacturing cost increases form 350 birr to 650 birr. Assume that the given data establishes relationship between cost (y) and the number of units made (x) and assume that the relationship is linear. Find the equation of this relationship.

Applications of linear equations: As of the very beginning, we aimed at developing our understanding on the interpretative application of linear equations in business. Consequently, our interest and purpose in this section is to learn how we can approximate and relate the mathematical terminology and technique of linear equations in addressing real world business issues. In dealing, we are going to consider three application areas of linear equations. These are the linear cost – output relations analysis, break – even analysis, and market equilibrium analysis.


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Linear Cost – Output Relations Analysis: In order to grasp the concept of linear cost output relations, let us consider the relationship among different types of cost on the following a coordinate plane.

Fig1.3 Classification of costs

Definitions: Cost is resource sacrificed to produce a given good or render service. Different classification of costs based on different basis for classification is possible but for our purpose here let’s define fixed cost, Variable cost and the sum of the two totals cost as hereunder.

Fixed cost is a cost component that does not change with the number of units produced whereas variable cost is a cost component that varies with change in number of units produced. Then at each level of production, total cost is the summation of fixed cost and variable cost. Marginal cost is the additional cost incurred in producing one more unit of output.

Illustration: Assume that total manufacturing cost and the number of units produced are linearly related. The total cost originates from the fixed cost line because of zero level of production the total cost will be equal to the fixed cost (see the above figure (Fig 1.2.1)). Accordingly,

- Fixed costs (FC) = AD = BE = CF- The segment BG is the Total Cost (TC) of producing AB units of outputs.- The segment CI is the TC of producing AC units of outputs.- The segment AD is the TC of producing zero units of outputs. - The ratio


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Total cost (TC) H

F

I

Variable

cost (VC)

Fixed cost (FC)

Total cost line

FC

Total

cost

FC

E

A

VC

B C Number of units (Q)

D

D

G


- Marginal cost is given by change in TC divided by change in Quantity (Q). Thus,

Marginal cost (MC) = = VC per unit.

- Therefore, marginal cost and VC/unit are the considered as the slope of TC line and they are constant as long as total cost and quantity produced are linear.

- When TC ÷ Q = Average cost per unit (AC).- Unlike MC and VC per units, AC per unit is not constant although cost and quantity,

produced are linearly related.

Break – Even Analysis: Definition: Breakeven point is the level of sales at which profit is zero. According to this definition, at breakeven point sales are equal to fixed cost plus variable cost. This concept is further explained by the following equation:

[Break even sales (BS) = fixed cost (FC) + variable cost (VC)]Breakeven sales= Selling price (SP)*Quantity (Q)

VC= VC per unit*Quantity (Q)SP*Q=FC+ VC/unit*QSP*Q-VC/unit*Q=FC

Q(SP/unit-VC/unit)=FC Q=FC/Sp/unit-Vc/unit- is quantity to be produced or sold at

breakeven pointTo further our understanding of break-even analysis, let us consider the following break-even chart.


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Profit (R > C)

R= PQe and

C = VQe + FC

Loss

(R < C)

Revenue

Variable cost

Fixed cost (FC)

Number of units (Q) Q e 0

Rev

enue

/ cos

t

FC

BEP

http://www.accounting4management.com/variable_cost_definition.htm

http://www.accounting4management.com/fixed_cost_definition.htm


Dear student, what did you observe from the above figure?--------------------------------------------------------------------------------------------------------------------------------------------------------The following are important points to note from the above breakeven chart.

i. As such, the total revenue line passes through the origin and hence has a y-intercept of zero while the total cost line has a y – intercept which is equal to the amount of the fixed cost.

ii. The fixed cost line which is parallel to the quantity axis (x – axis) is constant at all levels of output.

iii. To the left of the break – even point the revenue line is found below the cost line and hence any vertical separation indicates a loss while to the right the opposite is true.

iv. The total variable cost, which is the gap between the total cost and the fixed cost line increases as more units are produced.

v. Important linear cost – output expressions (equations): Total cost (TC) =VC+ FC Revenue (R) = SPQ Average Revenue (AR) = R ÷ Q = PQ ÷ Q = SP Average Variable Cost (AVC) = VQ ÷ Q = VC = Slope (m) Average Fixed Cost (AFC) = FC ÷ Q Average Cost = C ÷ Q = AVC + AFC Profit ( ) = R – TC

Example: A book company produces children’s books. One time fixed costs for Little Home are $12,838 that includes fees to the author, the printer, and for the building. Variable costs amount to $14.50 per book the books are then sold to bookstores around the country at $39.00 each. How many books must be printed and sold to break-even?

Solution:

Given, V = $14.50 FC = $12,838 Sp = $39Let Q = the number of books printed and sold Thus, C = VQ + FC TC = 14.5Q + 12,838 is the cost equation. The revenue (R) is also given by, R = SP Q = 39Q


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Then to obtain the quantity of books to be printed and sold to break-even, you need to equate the R and C equations. 39Q = 14.5Q + 12,838 39Q – 14.5Q = 12838 24.5Q = 12838 Q = 12838/24.5 Q = 524 books must be printed and sold to break – even. Learning activity-1.8

Dear learners form a small group and attempt the following question

A manufacture has a fixed cost of Birr 60,000 and a variable cost of Birr 2 per unit made and sold at selling price of Birr 5 per unit. Required:

a. Write the revenue and cost equations b. Computer the profit, if 25,000 units are made and soldc. Compute the profit, if 10,000 units are made and soldd. Find the breakeven quantity e. Find the break-even birr volume of sales f. Construct the break-even chart

Market Equilibrium Analysis: Market equilibrium analysis is concerned with the supply and demand of a product in a case they are linearly related.

Demand of a product: is the amount of a product consumers are willing and able to buy at a given price per unit. The linear demand function has a negative slope (falls downward from left to right see the figure below) since demand for a product decreases as price increases.

Supply of a product: is the amount Q, of a product the producer is willing and able to supply (make available for sell) at a given price per unit P. A linear supply curve or function has a positive slope (rises upward from left to right t see the figure below) and the price and the amount of product supplied are directly related. This is because of the fact that suppliers are more interested to supply their product when the selling price increases.

Market equilibrium: shows a market price that will equate the quantity consumers are willing and able to buy with the quantity suppliers are willing and able to supply. Thus, at the equilibrium,


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Demand (DD) = Supply (SS).

Graphically,

Example: Suppose the supply and demand equation for a given product on a given day reveal the following.

Demand (DD): P = 300 – 15QSupply (SS): P = 500 + Q

a. Find the market equilibrium price and quantity. b. Plot the demand and supply equation on a graph.

Solution:a. First, let us compute the equilibrium quantity for the given supply and demand

functions. Hence, at equilibrium: DD = SS3000 – 15Q = 500 + 5Q-15Q – 5Q = 500 – 3000-20Q = -2500Q = -2500 ÷ -20Q = 125 units -The market equilibrium quantity is 125 units.

Now, we progress to find the equilibrium price for the supply and demand function of the given product. In the same manner with the above one, we can obtain the market equilibrium price by simply substituting the market equilibrium quantity in either of the supply or demand equations. Thus, let us take the supply function of P = 500 + 5 Q.

Then substitute market equilibrium quantity of 125 units in place of Q. P = 500 + 5(125) P = 500 + 625

P = Birr 1125


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Market equilibrium point

Quantity (q)

Pri

ce (

p)

DD

SS


You will obtain the same result (i.e. Birr 1125), if you take the demand function of P = 3000 – 15 Q and substitute Q = 125.

a. Graph of demand and supply function: In plotting the graph, first we need to get the x and y intercept for both the supply and demand equations. The Y – intercept for demand equation is obtained by setting Q = 0 in the equation P = 3000 – 15Q. Thus, P = 3000 – 15(0) = 3000Therefore, (0, 3000) is the y-intercept. Likewise, the X intercept is obtained by setting P = 0 in the equation P = 3000 – 15Q. Consequently,

0 = 3000 – 15Q15Q = 3000 = 3000 ÷ 15

Q = 200The point (200, 0) is the x-intercept of the demand function. The same procedure is to be followed in computing the Y and y intercept for the supply function of P = 500 + 5Q. The Y – intercept is the value of P when Q = 0.

Therefore, P = 500 + 5(0) P = 500The y – intercept is the point with coordinate of (0, 500). In the same manner, the X – intercept is the value of q at P = 0. Thus, 0 = 500 + 5Q -5Q = 500 = 500 ÷ - 5

Q = -100Hence, the X – intercept is given by the point (-100, 0). Off course, the graph (the straight line) of the supply function is also passes through the equilibrium point of (125 units, 1125 birr). Thus, we do not need to extend the line to the negative direction.


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Quantity (q)

3000

2000

1125

500

Price (P)

125 200 300 400

DD: P = 8000 – 15q

SS: P = 500 + 5q



Kalifa Plc. is a national distributor of Dell Computers. The selling price and quantity of computers distributed are linearly related. Further, the company’s market analyst found out the following demand and supply functions for a particular year. Demand (DD): P = 3500 – 2q Supply (SS): - q = 950 – p

a. Find the excess demand for computers at a price of Birr 1400.b. Find the excess supply of computers at a price of Birr 2100.c. Find the market equilibrium quantity.d. Find the market equilibrium price. e. Sketch the demand and supply functions.

Continuous Assessment

The assessment methods for this criterion include tests, quiz, exam, assignments, case analysis and group works

Summary

Dear student, with confidence, you have already acquired knowledge about the concepts and the interpretative applications of linear equations, functions, and graphs in business. In this unit, we have considered the managerial applications of linear algebra and geometry so far. In so, we have considered that linear equations are mathematical expressions written in the form of The graph of such equation on coordinate plane is a straight line. As a result, the slope of the line is constant for any given points on the line. The slope of a straight-line m, given two points on the line with coordinates of (x1, y1) and (x2, y2) is expressed by the equation

Further, we have considered how to compute the distance between two points on a coordinate plane. Subsequently, approaches of developing equation of a line are discussed in the present unit. Above all, we have seen the interpretive applications of linear equations: analysis of linear cost-output relations, break-even analysis, and market supply and demand equilibrium analysis. In the next section, we will advance with the study of the matrix algebra and its application in solving business problems and backing management decisions that further organizational interests.


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3.1.3.2. Section II: Matrix algebra and its applications

Dear student! What do you know about matrix algebra? Why we learn matrix?______________________________________________________________________________________________________________________________________________________Introduction

It is evident that managerial problems are amenable to quantification thereby calling up for the application of mathematical models. Of the various quantitative techniques, this section tries to introduce students of business stream about major topics in matrix algebra. The section deals with basic concepts of matrix algebra, dimension and types of matrices, matrix operations and techniques, inverse of a matrix and major applications including solving system of linear equations. In total, this part of the learning task introduces students of business stream about matrix algebra principles and ways of applying them in handling real life business problems at individual or organizational level scientifically.

Matrix concepts:

Why we learn matrix? There are three major reasons for learning matrix: 1. Matrices are used to handle large linear systems 2. Matrices are used to solve complex linear equations3. Matrices are an effective means for summarizing voluminous business data.

Definition of a Matrix: A matrix is a rectangular array of numbers, parameters, or variables each of which has a carefully ordered place within the matrix. The numbers (parameters or variables) are referred to as elements of the matrix. The numbers in the horizontal like are called rows; the numbers in a vertical line are called columns. It is customary to enclose the elements of a matrix in parentheses, brackets, or braces to signify that they must be considered as a whole and not individually. A matrix is often denoted by a single letter in bold face type. The first subscript in a matrix refers to the row and the second subscript refers to the column.

A general matrix of order m x n is written as:

X = x11 x12 x1n

x 21 x22 x2n

Xm1 xm2 xmn (mxn)


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Matrix X above has m rows and n columns or it is said to be a matrix of order (size) m x n (read as m by n).

Example:A = a11 a12 a13 a 21 a22 a 23

a 31 a 32 a33

Here A is a general matrix composed of 3x3 = 9 elements, arranged in three rows and three columns. The elements all have double subscripts which give the address or placement of the element in the matrix; the first subscript identifies the row in which the element appears and the second identifies the column. For instance, a23 is the element which appears in the second row and the third column and a32 is the element which appears in the third row and the second column.

Dimensions and Types of Matrices: Dimension of a matrix is defined as the number of rows and columns. Based on their dimension (order), matrices are classified in to the following types: A. A row matrix: is a matrix that has only one row and can have many columns. E .g. A = 2 5 7 is a row matrix of order 1x3.

B.A column matrix: is a matrix with one column and can have many rows.E.g. B = 1 2 6

is a column matrix of dimensions 3x1.

C.A square matrix: is a matrix with equal number of rows and columns. 1 4 3 E.g. C = 6 ; D = 2 6 E = 2 2 5 3 8 ; 8 6 9


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3x3


D. A diagonal matrix: is a square matrix where its all non- diagonal elements are zero.

E.g. x = 2 0 0 0 6 0 is a diagonal matrix of order 3x3. 0 0 11

E. A scalar matrix: a square matrix is called a scalar matrix if all its non- diagonal elements are zero and all diagonal elements are equal.

6 0 0E.g. Y = 2 0 Z = 0 6 0 0 2 0 0 6

F. A unit matrix (Identity matrix): is a type of diagonal matrix where its main diagonal elements are equal to one.

1 0 0E.g. B = 0 1 0 0 0 1

G. A null matrix (zero matrix): a matrix is called a null matrix if all its elements are zero. 0 0 0E.g. A = 0 0 0 0 0 0

H. A symmetric matrix: a matrix is said to be symmetric if A = At. E.g. A = 8 2 1 2 3 4 1 4 5

I. Idempotent matrix: this is a matrix having the property that A2 =A.


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E.g. If A = ; then AA= A2 =


Dear student, as we have seen above there are various dimensions and types of matrices. In line with this, what do you conclude about the relationship of scalar matrix and diagonal matrix? And about unit matrix and scalar matrix?

Remark: It is seen above that every scalar matrix is a diagonal matrix; whereas a diagonal matrix need not be a scalar matrix. Every unit matrix is a scalar matrix; whereas a scalar matrix need not be a unit matrix.

Matrix Operations and Properties:

1. Matrix equality: two matrices are said to be equal if and only if they have the same dimension and corresponding elements of each matrix are equal.

3 0 3 -4 3 0E.g. A = B = C = 1 -4 1 0 1 -4

A ≠ B; A = C; B ≠ C.

2. Transpose of a matrix: If the rows and columns of a matrix are interchanged the new matrix is known as the transpose of the original matrix. If the original matrix is denoted by

A, the transpose is denoted by or At. Transposition means interchanging the rows or columns of a given matrix. That is, the rows become columns and the columns become rows.

E.g B = 3 5 6 9 0 11 13 8 6 8 3 4


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The transpose of matrix B, denoted by or Bt is given as:

3 0 6 Bt = 5 11 8

6 13 3 9 8 4

The dimension of B is changed from 3x4 to 4x3.

A = 1 3 = 1 0 2 (2 X 3) 0 4 (3x2) 3 4 8 2 8

The following properties are held for the transpose of a matrix: Property 1: (At)t =A Property 2: (aA)t = aAt, where (a) is a scalar (at = a) Property 3: (A+B)t = At + Bt

Property 4: (AB)t = BtAt

3. Addition and subtraction of matrices: Two matrices A and B can be added or subtracted if and only if they have the same order, which is the same number of rows and columns.Example: A= 2 0 B = 3 6 -5 6 4 1 Then;

2+3 0+6 5 6 A+B = -5+4 6+1 = -1 7

1 5 10 2 If A = 6 7 B= 8 6 8 9

A+B is not defined, since orders of A and B are not the same.


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2 3 4 3 2-4 3-3 -2 0 A - B = 1 0 - 2 1 = 1-2 0-1 = -1 -1

4. Matrix Multiplication

Two matrices A and B can be multiplied together to get AB if the number of columns in A is equal to the number of rows in B.

E.g. 1 2 2 1 4 A= 3 4 B= 3 0 5 0 1 (3x2) (2x3)

Then, A x B = (1x2) + (2x3) (1x1) + (2x0) (1x4) + (2x5) (3x2) + (4x3) (3x1) + (4x0) (3x4) + (4x5) (0x2) + (1x3) (0x1) + (1x0) (0x4) + (1x5)

8 1 14 = 15 3 32 3 0 5 (3x3)

Solved problems: Finfine Furniture Factory (3F) produces three types of executive chairs namely A, B and C. The following matrix shows the sale of executive chairs in two different cities.

Executive chairs A B C C1 400 300 200 Cities C2 300 200 100 (2x3)

If the cost of each chair (A, B and C) is Birr 1000, 2000 and 3000 respectively, and the selling price is Birr 2500, 3000 and 4000 respectively;

a) Find the total cost of the factory for the total sale made.b) Find the total profit of the factory.

Solution:Jimma, Haramaya, Ambo, Hawassa, Adama, Bahirdar, W.Sodo,Semara Universities

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Given: Let the quantity matrix be q Let the price matrix be p Let the unit cost matrix be v

400 300 200 q = p = 1500 V = 1000 300 200 100 3000 2000 4000 3000

Total cost = (unit cost) (Quantity)

= 400 300 200 * 1000 300 200 100 2000

3000 = 1,600,000

1,000,000

Total cost = Birr 1,600,000 + Birr 1,000,000 = Birr 2,600,000 Total profit = Total Revenue - Total Cost Total Revenue = (price) (quantity) 400 300 200 * 1500 = 300 200 100 3000

4000 2,300,000 = 1,450,000

Total Revenue = Birr 2,300,000 + Birr 1,450,000 = Birr 3,750,000Profit = Birr 3,750,000 – Birr 2, 600,000 = Birr 1,150,000



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Dear student, having seen the properties of matrix, we will now turn our face to some activity. Interest at the rates of 0.06, 0.07 and 0.08 is earned on respective investments of $3000, $2000 and $4000.a) Express the total amount of interest earned as the product of a row vector by a column vector.b) Compute the total interest by matrix multiplication.

Determinant of a Matrix:Let A = a11 a12

a21 a22 ( 2x2)

= a11 a12

a21 a22 is known as a determinant of order two

and its value is given as: = a11a22 - a12a21.

E.g. A= 6 4 = 6 4 = 6(9)-7(4)=26

7 9 ; 7 9

3. Let A = a11 a12 a13

a21 a22 a23

a31 a32 a33

= a11 a12 a13

a21 a22 a23 is called a third order determinant a31 a32 a33

a22 a23 a21 a23 a21 a22

= + a11 a32 a33 - a12 a31 a33 +a13 a31 a32

= a11 (a22 a33 - a32 a23) – a12 (a21 a33-a31a23) + a13 (a21a32-a31a22)

E.g. Let A = 1 2 4


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0 -1 0

-2 0 3 ; Find .

= 1 2 4 -1 0 0 0 0 -1

0 -1 0 = +1 0 3 -2 -2 3 + 4 -2 0 -2 0 3 = 1 (-1x3 – 0x0) -2 (0x3- (-2x0)) + 4 (0x0 – (-2x-1)) = -3 -0 -8 = -11

Inverse of a Matrix: In scalar algebra, the inverse of a number is that number which, when multiplied by the original number, gives a product of 1. Hence, the inverse of x is simply 1/x; or in slightly different notation, x-1. In matrix algebra, the inverse of a matrix is that which, when multiplied by the original matrix, gives an identity matrix. The inverse of a matrix is denoted by the superscript “-1”. Hence, AA-1 = A-1A = I.

Note that: A matrix must be square to have an inverse, but not all square matrices have an inverse. The necessary and sufficient condition for a square matrix to possess its inverse is that /A/ ≠ 0. Finding the inverse of a matrix requires the concept of row operations to be performed. The row operations are the following:

A. Multiply or divide a row by a non- zero constant;

If A = 2 3 6 9

multiply row one (R1) by -2 to get matrix B.

Then, B = -4 -6 6 9

Divide row two (R2) by 3 to get matrix C. Then, matrix C = 2 3

2 3

B. Add a multiple of one row to another row;Jimma, Haramaya, Ambo, Hawassa, Adama, Bahirdar, W.Sodo,Semara Universities

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If A = 1 2 multiply R1 by 2 and add to R2 to get matrix x. 3 4

Matrix X = 1 2 5 8

C. Interchanging of rows;

If A = 1 0 Interchange R1 and R2 ( R1 ↔ R2 ) ; to 2 4 get matrix D.

D = 2 4

1 0

Note: The first row elements in the original matrix become second row elements in the new matrix and vice versa. The most important methods to find inverse of a given matrix is Gauss- Jordan Inversion method. This method was developed by a mathematician called Gauss and it was named so by the founder. Example: Find the inverse of the following matrix using the Gauss- Jordan method.

A = 3 2 1 1

Solution

Steps:First: write the given matrix at the left and the corresponding identity matrix at the right;

3 2 1 0

A/I = 1 1 0 1

Second : Interchange R1 and R2;

3 2 1 0 1 1 0 1 1 1 0 1 3 2 1 0

Third: Multiply R1 by -3 and add the result to R2; Jimma, Haramaya, Ambo, Hawassa, Adama, Bahirdar, W.Sodo,Semara Universities

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-3 R1 = -3 -3 0 -3 + R2 = 3 2 1 0 0 -1 1 -3 The resulting matrix is given by: 1 1 0 1 0 -1 1 -3

Fourth: Simply add R2 entries to R1 entries; R2 = 0 -1 1 -3 R1 = 1 1 0 1 1 0 1 -2

The resulting matrix is given by:

1 0 1 -2 0 -1 1 -3

Fifth: Multiply R2 by -1; (-1) (R2) = 0 1 - 1 3

The resulting matrix is given by;

1 0 1 -2 0 1 -1 3

Thus; the inverse matrix A, denoted by A-1 is given as:A-1 = 1 -2 -1 3

Check! A.A-1 = A-1. A = I = 3 2 1 -2 1 0


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1 1 * -1 3 = 0 1

Matrix Applications: A system of linear equations can be solved by the following method using matrix algebra: Cramer’s rule (the determinant method): works according to this formula:

Xi = /Ai/ /A/ Where, xi = indicates the variables we want to solve for. /Ai/ = is the determinant obtained by putting the right-hand side of the system in place

of the column of coefficients of the variable whose solution is needed; and /A/ = is the determinant of the system.

Given a system of equations:i) a11x+a12y = b1 → algebraic form

a21x+a22y = b2

The above system of equations can also be rewritten in expanded matrix form as follows: a11 a12 x b1

a21 a22 * y = b2

Matrix of coefficients column vector of column vector of denoted by A variables (X) constants (B)

Using Cramer’s rule, the solution is given by: b1 a12 a11 b1

X= b2 a22 ; and Y = a21 b2

Example; x- y = 1 = 1 -1 x 1 → matrix expression x+ y = 2 1 1 * y = 2

Then, x= 1 -1 2 1

= 1 -1


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2 1 =3/2 1 -1

1 1

y = 1 1 1 1 1 2 = 1 2 =1/2

1 -1

1 1

ii) Given a system of equations: a11x+ a12y+ a13z = b1

a21x+ a22y+ a23z = b2

a31 x+a32y+ a33z = b3

Expanded form:a11 a12 a13 x b1

a21 a22 a23 . y = b2

a31 a32 a33 z b3

A X B

Then; the value of x is given by: b1 a12 a13 a11 b1 a13

b2 a22 a23 a21 b2 a23

b3 a32 a33 ; y = a31 b3 a33 a11 a12 a13 a11 a12 a13

a21 a22 a23 a21 a22 a23

a31 a32 a33 a31 a32 a33

z = a11 a12 b1

a21 a22 b2

a31 a32 b3


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a11 a22 a13

a21 a22 a23

a31 a32 a33

Example; Solve using Cramer’s rule: 2x + y – z = 0 x + y + z = 0 y – z = 1

Expanded form of the above system is: 2 1 -1 x 0 1 1 1 * y = 0 0 1 -1 z 1 A X B

= +2 1 1 -1 1 1 + (-1) 1 1

1 -1 0 -1 0 1 = -4

Thus; x = 0 1 -1 0 1 1 = 2/-4 = -1/2 1 1 -1

Y = 2 0 -1 1 0 1 = -3/-4 = 3/4 0 1 -1

Z = 2 1 0 1 1 0 0 1 1 = -1/4


The assessment methods for this criterion include tests, quiz, exam, assignments, case analysis and group works


31


Summary

Dear learner! We have seen about types of matrices, matrix operations, inverse of a matrix, ways of finding an inverse and applications of matrix algebra. The following gives the summary of major points. The equality of matrices is assured by equality of corresponding elements of the same dimension. Matrix addition and subtraction is defined for matrices of the same dimension but matrix multiplication is defined by considering the equality of inner dimensions. Inverse of a matrix is defined only for square matrices. Inverse of a matrix is unique. If matrix B is the inverse of matrix A, then matrix A is the inverse of matrix B. Every square matrix may not have an inverse. If a matrix has no inverse, then it is said to be singular and if a matrix has an inverse, it is said to be invertible or non- singular. Matrix algebra is applied in solving system of linear equations.

Exercises

1. Find the inverse of the following matrices using the Gauss – Jordan method. A = 3 3

2 2

2. Write the expanded matrix form of a 2 by 3 general matrix.

3. If matrix p is 1 by 2 and we have py = q, where q is a 1 by 1 matrix:a. What are the dimensions of matrix y?b. Write the expanded vector form of the equation.c. Write the usual algebraic form.

4. Find the inverse of A = 3 4 2 1

5. If A = 1 -1 2 and B = -2 0 1 2 3 4 2 -1 4 0 1 3 1 1 2 Find A-B and B-A.

7. Given the matrices:

1 -2


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A= 0 3 1 3 0 0 4 B = 2 0 -1 ; Determine where possible:

a) ABb) BA c) 2A

8. Verify whether AB = BA for matrices:

A = 2 1 0 1 2 -1 1 -1 2 and B = 1 -1 2 0 1 3 1 1 2

3.1.3.3. Section III: Linear programming

Dear learner! Have you ever experienced the problem of allocating scarce resource to various activities to attain the predetermined goal? If so, in what way you approached the allocation?....................................................................................................................................


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......................................................................................................................................................

.Introduction

A large number of decision problems faced by a business manger involve allocation of resources to various activities, with the objective of increasing profits or decreasing costs, or both. When resources are in excess, no difficulty is experienced. Nevertheless, such cases are very rare. Practically in all situations, the managements are confronted with the problem of scarce resources. Normally, there are several activities to perform but limitations of either the resources or their use prevent each activity from being performed to the level demanded. Thus, the manger has to take a decision as to how best the resources be allocated among the various activates. These decision problems can be formulated and solved as mathematical programming problems and hence knowhow of linear programming is imperative.

Definition: Linear Programming (LP):o Is a mathematical process that has been developed to help management in decision-

making involving the efficient allocation of scares resources to achieve a certain objective/s.

o Is a method for choosing the best alternative from a set of feasible alternatives?

Requirements to Apply Linear Programming: To apply LP, the following conditions must be satisfied.

a. There should be an objective that should be clearly identified and measurable in quantitative terms. Example, maximization of sales, profit, and minimization of costs etc.

b. The activities to be included should be distinctly identifiable and measurable in quantitative terms.

c. The resources of the system which are to be allocated for the attainment of the goal should also be identifiable and measurable quantitatively. They must be in limited supply. These resources should be allocated in a manner that would trade off returns on investment of the resources for the attainment of the objective.

d. The relationship representing the objective and the resource limitation considerations represented by the objective function and the constraint equations or inequalities, respectively, must be linear in nature.

e. There should be a series of feasible alternative courses of actions available to the decision-maker that is determined by the resource constraints.


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When these stated conditions are satisfied in a given solution, the problem can be expressed in algebraic form called linear programming problem (LPP), and then solved for optimal decision.Diagrammatically Linear Programming (LP) model can be presented as shown below.

Fig 3.1. Linear Programming Model

Dear learner! Let’s first illustrate the formulation of linear programming problems and then consider the method of their solution.

Formulation of linear programming model: Problem Formulation or Modelling is the process of translating the verbal statement of a problem in to mathematical statements. Formulating model is an art that can only be measured with practice and experience. Even though every problem has certain unique features, most problems have common features. Therefore, some general guidelines for model formulation are helpful. Thus, to understand the formulation let’s see cases on maximization and minimization problems.

The Maximization Case: A firm is engaged in producing two products, A and B. Each unit of product A requires two kgs of raw material and four labor hours for processing whereas each unit of product B requires three kg of raw material and three hours of labor of the same type. Every week the firm has an availability of 60 kgs of raw material and 96 labor hours. One unit of product A sold yields Birr 40 and one unit of product B sold yields Birr 35 as profit. Formulate this problem as a linear programming problem to determine as to how many units of each of the products should be produced per week so that the firm can


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Scares Resource

To be allocated

to:

Objectives

Constraints

Resource constraints

Non-negativityConstraints

Optimization

Maximization

Minimization


maximum the profit. Assume that there is no marketing constraint so that all that is produced can be sold.

The objective function: the first major requirement of linear programming problem (LPP) is that we should be able to identify the goal in terms of the objective function. This function relates mathematically the variables with which we are dealing in the problem. For our problem, the goal is the maximization of profit, which would be obtained by producing (and selling) the products A and B. If we let x1 and x2 represent the number of units produced per week of the products A and B respectively, the total profit, Z, would be equal to 40 x 1 +35x2

is then, the objective function, relating the profit and the output level of each of the two items. Notice that the function is a linear one. Further, since the problem calls for a decision about the optimal values of x1 and x2, these are known as the decision variables.

The constraints: As has been laid, another requirement of linear programming is that the resources must be in limited supply. The mathematical relationship which is used to explain this limitation is inequality. The limitation itself is known as a constraint. Each unit of product A requires 2 kg of raw material while each unit of product B needs 3 kg. The total consumption would be 2x2 and 3x2, which cannot be the total the availability of 60 kg every week. We can express this constraint as 2x1 and 3x2 < 60. Similarly, it is given that a unit of A requires 4 labor hours for its production and one unit of B requires 3 hours. With an availability of 96 hours a week, we have 4 x1 and 3x2 < 96 as the labor hour's constraint. It is important to note that for each of the constraint, inequality rather than equation has been used. This is because the profit maximizing output might not use all the resources to the full leaving some unused, hence the < sign. However, it may be noticed that all the constraints are also linear in nature.

Non-negativity Condition: Quite obviously, x1 and x2, being the number of units produced, cannot have negative values. Thus both of them can assume values only greater than or equal to zero. This is the non-negativity condition, expressed symbolically as x1 > 0 and x2, > 0.Now we can write the problem in complete form as follows.

Maximize Z = 40 X1 + 35 X2, Profit Subject to

2X1 + 3X2 < 60 Raw material constraint4X1 + 3X2 < 96 Labor hours constraint X1, X2, > 0 Non-negativity restriction

The Minimization Case: The agricultural Research institute has suggested to a farmer to spread out at least 4800 kg of a special phosphate fertilizer and no less than 7200 kg of a


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special nitrogen fertilizer to raise productivity of crops in his fields. There are two resources for obtaining these mixtures A and B. Both of these are available in bags weighing 100 kg each and they cost Birr 40 and Birr 24 respectively. Mixture A contains phosphate and nitrogen equivalent of 20 kg and 80 respectively, while mixture B contains these ingredients equivalent of 50 kg each. Write this as a linear programming problem and determine how many bags of each type the farmer should buy in order to obtain the required fertilizer at minimum cost.

The Objective Function: In the given problem, such a combination of mixtures A and B is sought to be purchased as would minimize the total cost. If x1 and x2 are taken to represent the number of bags of mixtures A and B respectively, the objective function can be expressed as follows:

Minimize Z = 40x1 + 24 x2 Cost

The constraints: In this problem, there are two constrains, namely, a minimum of 4800 kg of phosphate and 7200 kg of nitrogen ingredients are required. It is known that each bag of mixture A contains 20 kg and each bag of mixture B contains 50 kg of phosphate. The phosphate requirement can be expressed as 20x1 + 50x2 > 4800. Similarly, with the given information on the contents, the nitrogen requirement would be written as 80 x1 + 50x2 > 7200. Non-negativity condition: As before, it lays that the decision variables, representing the number of bags of mixtures A and B, would be non-negative. Thus x1 > 0 and x2 > 0. The linear programming problem can now be expressed as follows:

Minimize Z = 40x1 + 24 x2 Cost Subject to

20x1 + 50x2 > 4800 Phosphate requirement 80x1 + 50x2 > 7200 Nitrogen requirement x1, x2 > 0 Non negativity restriction

In general, linear programming problem can be written as

Maximize Z = c1x1 + c2x2 ………+ cn xn Objective Function

Subject to a11x11 + a22 x2 +………………………..+ ain x n < b1a21x1 + a22 x2 + ………………………..+ ain x n < b2am1 x1 + am2 x2 + ............................+ amn < xn bm

x1, x2, ……………………….., xn > 0


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Where cj, aij, bi (i = 1, 2, ……..…., m; j = 1,2,……n) are known as constants and x’s are decision variablesc’s are termed as the profit coefficients aij’s the technological coefficients b’s the resource values

Note:

Although, generally, the constraints in the maximization problems are of the < type, and the constraints in the minimization problems are of > type , a given problem might contain a mix of the constraints, involving the signs <, >, and /or =\Assumptions Underlying Linear Programming: A linear programming model is based on the assumptions detailed hereunder.

1. Proportionality: A basic assumption of linear programming is that proportionality exists in the objective function and the constraint inequalities. For example, if one unit of a product is assumed to contribute Birr 10 toward profit, then the total contribution would be equal to 10x1 where x1 is the number of units of the product. For 4 units, it would equal Birr 40 and for 8 units it would be Birr 80, thus if the output (and sales) is doubled, the profit would also be doubled. Similarly, if one unit takes 2 hours of labor of a certain type, 10 units would require 20 hours, 20 units would require 40 hours….and so on. In effect, then, proportionality means that there are constant returns to scale and there are no economies of scale.

2. Additively: Another assumption underlying the linear programming model is that in the objective function and constraint inequalities both, the total of all the activities is given by the sum total of each activity conducted separately. Thus, the total profit in the objective function is determined by the sum of the profit contributed by each of the products separately. Similarly, the total amount of a resource used is equal to the sum of the resource values used by various activities.

3. Continuity: It is also an assumption of a linear programming model that the decision variables are continuous. Therefore, combinations of output with fractional values, in the context of production problems, are possible and obtained frequently. For example, the best solution to a problem might be to produces 5¾ units of product B per week. Although in many situations we can have only integer values, but we can deal with the fractional values, when they appear.


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4. Certainty: A further assumption underlying a linear programming model is that the various parameters, namely, the objective function coefficients, the coefficients of the inequality/equality constraints and the constraint (resource) values are known with certainty. Thus, the profit per unit of the product, requirements of materials and labor per unit, availability of materials, labor etc. are given and known in a problem involving these. The linear programming is obviously deterministic in nature.

5. Finite Choices: A linear programming model also assumes that a limited number of choices are available to a decision maker and the decision variables do not assume negative values. Thus, only non-negative levels of activity are considered feasible. This assumption is indeed a realistic one. For instance, in the production problems, the output cannot obviously be negative, because a negative production implies that we should be above to reverse the production process and convert the finished output back in to the raw materials!

Solution approaches to LPPS: Now we shall consider the solution to the linear programming problems. They can be solved by using graphic method or by applying algebraic method, called the Simplex Method. The graphic method is restricted in application – it can only be used when two variables are involved. Nevertheless, it provides an intuitive grasp of the concepts that are used in the simplex technique. The Simplex method, is the mathematical technique of solving linear programming problems with two or more variables. Graphial Solution to Linear Programming Problems: To use the graphic method, the following steps are needed:

i. Identify the problem – determine the decision variables, the objective function, and the constraints.

ii. Draw a graph including all the constraints and identify the feasible region.iii. Obtain a point on the feasible region that optimizes the objective function –

optimal solution.iv. Interpret the results.

Note: Graphical LP is a two-dimensional model.

a) Maximization Problem: This is the case of Maximize Z with inequalities of constraints in < form. Consider two models of color TV sets; Model A and B, are produced by a company to maximize profit. The profit realized is $300 from A and $250 from set B. The limitations are

a. Availability of only 40 hrs of labor each day in the production department,b. A daily availability of only 45 hrs on machine time, and


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c. Ability to sale 12 set of model A.

How many sets of each model will be produced each day so that the total profit will be as large as possible?

ConstraintsResources used per unit

Maximum AvailableHoursModel A

(x1)Model B

(x2)

Labor Hours 2 1 40

Machine Hours 1 3 45

Marketing Hours 1 0 12

Profit $300 $250

Solution:

1. Formulation of mathematical model of LPPMax Z=300X1 +250X2

St:2X1 +X2< 40X1 +3X2< 45X1 < 12X1, X2 > 0

2. Convert constraints inequalities into equalities2X1 + X2 = 40 X1 + 3X2 = 45 X1 = 12

3. Draw the graph by finding out the x– and y–intercepts2X1 +X2 = 40 ==> (0, 40) and (20, 0)X1 +3X2 = 45 ==> (0, 15) and (45, 0)X1 = 12 ==> (12, 0)X1 , X2 = 0


40

LP Model

2X

1

+X

2

=

40

C

B15

40

12 20 45

X1

X2

AD

X1 +

X2

= 4

5

(12, 11)

X1=12

X1=

0

X2=0Feasibl

e

Region


4. Identify the feasible area of the solution which satisfies all constrains. The shaded region in the above graph satisfies all the constraints and it is called Feasible Region.

5. Identify the corner points in the feasible region. Referring to the above graph, the corner points are in this case are:

A (0, 0), B (0, 15), C (12, 11) and D (12, 0)

6. Identify the optimal point.

Corners Coordinates Max Z = 300 X1 +250X2

A (0, 0) $0

B (0, 15) $3750

C (12, 11) $ 6350 (Optimal)

D (12, 0) $3600

7. Interpret the result. Accordingly, the highlighted result in the table above implies that 12

units of Model A and 11 units of Model B TV sets should be produced so that the total profit will be $6350.

b) Minimization Problem: In this case, we deal with Minimize Z with inequalities of constraints in > form. Suppose that a machine shop has two different types of machines; Machine 1 and Machine 2, which can be used to make a single product. These machines vary in the amount of product produced per hr., in the amount of labor used and in the cost of operation. Assume that at least a certain amount of product must be produced and that we would like to utilize at least the regular labor force. How much should we utilize on each machine in order to utilize total costs and still meets the requirement?

Resource Used Minimum Required Hours


41


Items Machine 1 (X1) Machine 2 (X2)

Product produced/hr 20 15 100

Labor/hr 2 3 15

Operation Cost $25 $30

Solution:

1. The LP Model:

Dear student, can you graph the above constraints? See step 2 below.

2. The Graph of Constraint Equations:20X1 +15X2=100 ==> (0, 20/3) and (5, 0)2X1 + 3X2=15 ==> (0, 5) and (7.5, 0) X1 , X2 = 0

3. The Corners and Feasible Solution:

Corners Coordinates Min Z= 25 X1 + 30X2

A (0, 20/3) 200


42

LP Model

B (2.5, 3.33)

A (0, 20/3)

C (7.5, 0)

X1

X2

5

X2

=0

X1

=0

Feasible

Region


B (2.5, 3.33) 162.5 (Optimal)

C (7.5, 0) 187.5

Since our objective is to minimize cost, the minimum amount (162.5) will be selected.X1 = 2.5X2 = 3.33 and Min Z= 162.5

Note:- In maximization problems, our point of interest is looking the furthest point from the

origin (Maximum value of Z).- In minimization problems, our point of interest is looking the point nearest to the origin

(Minimum value of Z).


A company owns two flourmills (A and B) which have different production capacities for HIGH, MEDIUM and LOW grade flour. This company has entered contract supply flour to a firm every week with 12, 8, and 24 quintals of HIGH, MEDIUM and LOW grade respectively. It costs the Co. $1000 and $800 per day to run mill A and mill B respectively. On a day, mill A produces 6, 2, and 4 quintals of HIGH, MEDIUM and LOW grade flour respectively. Mill B produces 2, 2 and 12 quintals of HIGH, MEDIUM and LOW grade flour respectively. How many days per week should each mill be operated in order to meet the contract order most economically. Solve the problem graphically.

Algebraic Simplex Method: The graphical method to solving LPPs provides fundamental concepts for fully understanding the LP process. However, the graphical method can handle problems involving only two decision variables (say X1 and X2). In 19940’s George B.Dantzig developed an algebraic approach called the Simplex Method, which is an efficient approach to solve applied problems containing numerous constraints and involving many variables that cannot be solved by the graphical method. The simplex method is an ITERATIVE or “step by step” method or repetitive algebraic approach that moves automatically from one basic feasible solution to another basic feasible solution improving the situation each time until the optimal solution is reached at. The simplex method starts with a corner that is in the solution space or feasible region and moves to another corner improving the value of the objective function each time until optimal solution is reached at the optimal corner.

a) Maximization ProblemsMaximize Z with inequalities of constraints in ‘<’ formSolve the problem using the simplex approach


43


Max. Z = 300x1 +250x2

Subject to: 2x1 + x2 < 40 (Labor)

x1 + 3x2 < 45 (Machine) x1 < 12 (Marketing) x1, x2 > 0

Solution:

Step 1 Formulate LP Model: It is already given in the form of linear programming model. Step 2 Standardize the problem: Convert constraint inequality into equality form by introducing a variable called Slack variable. A slack variable(s) is added to the left hand side of a < constraint to covert the constraint inequality in to equality. The value of the slack variable shows unused resource. A slack variable emerges when the LPP is a maximization problem. Slack variables represent unused resource or idle capacity. Thus, they do not produce any profit and their contribution to profit is zero. Slack variables are added to the objective function with zero coefficients. Let that s1, s2, and s3 are unused labor, machine, and marketing hrs respectively.

Max.Z=300X1 +250X2 + 0 S1 +0 S2+ 0 S3

St:2 X1+X2 +0S1 = 40 X1+3X2 +0S2 = 45 X1 + +0S3= 12 X1 , X2, S1, S2, S3 > 0

Step 3 Obtain the initial simplex tableau: To represent the data, the simplex method uses a table called the simplex table or the simplex matrix. In constructing the initial simplex tableau, the search of the optimal solution begins at the origin indicating that nothing can be produced. Thus, first assumption, No production implies that x1 =0 and x2=0

==>2 x1+x2 + s1 +0 s2+ 0 s3= 40 ==> x1+3x2 +0 s1 + s2+ 0 s3= 45 2(0) +0 + s1 +0 s2+ 0 s3= 40 0 +3(0) + 0s1 + s2+ 0 s3= 45 s1= 40 – Unused labor hrs. s2= 45 – Unused machine hrs.

==> x1+0s1 +0s2+ s3= 12 0 +0s1 +0 s2+ s3= 12 s3= 12 – Unused Marketing hrs.

Therefore, Max. Z=300x1 +250x2 + 0 s1 +0 s2+ 0 s3

=300(0) +250(0) + 0(40) +0(45) + 0(12)= 0


44

Standard form


Note:

In general, whenever there are n variables and m constraints (excluding the non-negativity), where m is less than n (m < n), n – m variables must be set equal to zero before the solution can be solved algebraically.

a. Basic variables are variables with non-zero solution values. Or: Basic variables are variables that are in the basic solution. Basic variables have 0

values in the Cj – Zj row .

b. Non-basic variables are variables with zero solution values. Or: Non-basic variables are variables that are out of the solution.

==> n = 5 variables (x1, x2, s1, s2, and s3) and m = 3 constraints (Labor, Machine and Marketing constraints), excluding non-negativity.

Therefore, n – m = 5 – 3 = 2 variables (x1 and x2) are set equal to zero in the 1st simplex tableau. These are non-basic variables. Three Variables (s1, s2, and s3) are basic variables (in the 1st simplex tableau) because they have non-zero solution values.

Step 4 Construct the initial simplex tableau: To set up the tableau, we first list horizontally all the variables contained in the problem. Here, there are five variables: x1, x2, s1, s2 and s3. Next, the coefficients in the constraint equations are written listing vertically the coefficients under their respective variables. It may be noted that each of the slack variables appear only in one equation. Therefore, the coefficient of each of the slack viable is taken to be zero in all the equations except the one in which it appears. After putting the coefficients, the constraint values are mentioned on the right hand side against the rows. Finally, the row titled indicates the coefficients of the various variables in the objective function, mentioned respectively in the various columns representing the variables.

Initial simplex tableau


45

Sla

ck

vari

ab

les

colu

mn

s

Solu

tion

qu

an

tity

colu

mn

Pro

fit

per

un

it c

olu

mn

Basic

or

Solu

tion

vari

ab

le c

olu

mn

Real

or

decis

ion

vari

ab

les

colu

mn

Profit per unit

row

Constraint

equation

rows

Gross Profit

rowNet Profit

row

/Indicator

row/


Cj 300 250 0 0 0

SV X1 X2 S1 S2 S3 Q

0 S1 2 1 1 0 0 40

0S2 1 3 0 1 0

45

0 S3 1 0 0 0 1 12 Zj 0 0 0 0 0 0

Cj - Zj 300 250 0 0 0

Entering Column (Pivot Column)

Step 5 Choose the “incoming” or “entering” variables:: The entering variable is the variable that has the highest positive value in the Cj - Zj row also called as indicator row. Alternatively, the entering variable is the variable that has the highest contribution to profit per unit.

a. X1 in our case is the entering variable because it has the highest positive c-z value.b. The column associated with the entering variable is called key or pivot column ( X1

column in our case )

Step 6 Choose the “leaving “or “outgoing” variable: In this step, we determine the variable that will leave the solution for X1 (or entering variable)

Note:

The row with the minimum or lowest positive (non-negative) replacement ratio shows the variable to leave the solution.

Note: RR > 0

- The variable leaving the solution is called leaving variable or outgoing variable.- The row associated with the leaving variable is called key or pivot row (s3 column in our

case)- The element that lies at the intersection of the pivot column and pivot row is called pivot

element(No 1 in our case)


46

Replacement Ratio (RR) = Solution Quantity (Q)

Corresponding values in pivot column

R1=20

R2=

45

R3=12 Leaving Row


Step 7 Derive the revised tableau: For improved solution using the information obtained earlier, another tableau is derived where in the various elements are obtained as given here. a. Divide each element of the pivot row (including bi) by the pivot element to get the corresponding values in the new tableau. The row of values so derived is called the replacement raw.

For each row other than the pivot row,- Divide each element of the pivot row by the pivot element to find new values in the

key or pivot row.- Perform row operations to make all other entries for the pivot column equal to

zero.

2nd simplex table

Cj 300 250 0 0 0

SV X1X2 S1 S2 S3 Q

0 S1 0 1 1 0 -2 16

0S2 0 3 0 1 -1

33

300 X1 1 0 0 0 1 12Zj 300 0 0 0 300 3600Cj - Zj 0 250 0 0 -300

3rd simplex table

Elementary Row Operations


47

Cj 300 250 0 0 0

SV X1 X2 S1 S2 S3 Q

0 S1 0 0 1 -1/3 -5/3 5

250 X2 0 1 0 1/3 -1/3 11

300 X1 1 0 0 0 1 12Zj 300 250 0 250/3 650/3 6350Cj - Zj 0 0 0 -250/3 -650/3

R’’1=R’1-R’2

R’’2=R2/3

R’’3=R’3

R’1=16

R’2=11

R’3=Undefin

ed


Since all the Cj - Zj < 0, optimal solution is reached at and this tableau is the final one.Therefore, X1=12, X2=11, S1=5 and Max Z=6350

b) The Minimization Problem

The solution procedure for the linear programming problems that have the objective function of the minimization type is similar to the one for the maximization problems, except for some differences. To illustrate, let us again consider the examples.

Minimize Z= 40x1 + 24x2 Total cost Subject to:

20x1+ 50 x2 > 4800 Phosphate Requirement 80 x1+ 50 x2 > 7200 Nitrogen Requirement

X1, x2 > 0

Following the approach already discussed, we first introduce some new variables to convert inequalities of the system in to equations. The variable required for converting a greater than type of inequality in to an equation is called surplus variable and it represents the excess of what is generated (given by the LHS of the inequality) over the requirement (shown by the RHS value bi). With surplus variables, S1 and S2 respectively for the first and the second constraints, the augmented problem shall be:

Minimize Z= 40x1+24x2 +0S1 +0S2

Subject to: 20x1+ 50 x2- S1= 4800

80 x1+ 50 x2- S2 =7200 X1, X2, S1 ,S2 > 0

Now, as soon as we proceed to the next step we experience a problem, which is like this. We know that the simplex method needs an initial solution to get the process started. In this case, it is easy to visualize that an initial solution does not exist because, if we let x1 and x2 each equal to zero, we get S1 = -4800 and S2 = -7200, which is not feasible as it violates the non negativity restriction. In terms of the simplex tableau, when we write all the information, we observe that we do not get identity because unlike in case of slack variables, the co-efficient values of surplus variables S1 and S2 appear as minus one (-1).


48


To provide an initial solution, we add artificial variables in to the model. Unlike slack or surplus variables, artificial variables have no tangible relationship with the decision problem. Their sole purpose is to provide an initial solution to the given problem. When artificial variables are introduced in our example, it appears as follows,

20x1+ +50-S1+A1= 480080 x1+50x2 –S2 + A2 =7200

Before we set up the initial tableau, a few words on the artificial variables follow. The artificial variables are introduced for the limited purpose of obtaining an initial solution and are required for the constraints with > type, or the constraints “=” sign. It is not relevant whether the objective function is of the minimization or the maximization type. Obviously, since artificial variables do not represent any quantity relating to the decision problem, they must be driven out of the systems and must not show in the final solution (and if at all they do, it represents a situation of infeasibility). This can be ensured by assigning an extremely high cost to them. Generally, a value M is assigned to each artificial variable, where M represents a number higher than any finite number. For this reason, the method of solving the problems where artificial variable are involved is termed as the Big M Method. When the problem is of the minimization nature, we assign in the objective function a coefficient of + M to each of the artificial variables. On the other hand, for the problems with the objective function of maximization type, each of the artificial variables introduced has a coefficient –M. For our present example, the objective function would appear as

Minimize Z=40x1+24x2 + 0S1+0S2 + MA1+MA2

It is significant to note that the initial solution obtained using the artificial variable is not a feasible solution to the given problem. It only gives the starting point and the artificial variables are driven out in the normal course of applying the simplex algorithm. A solution to the problem which does not include an artificial variable in the basis represents a feasible solution to the problem. The initial simplex tableau giving the initial solution to our problem is given in the following table.

Simplex Table 1: Non optimal solution

Cj 40 24 0 0 M MBasis X1 X2 S1 S2 A 1 A 2 Qty RR

A1 M 20 50 -1 0 1 0 4800 96A2 M 80 50 0 -1 0 1 7200 144


49


Zj 40 24 0 0 M MCj - Zj M M 0 0

For the minimization problem, the optimal solution is indicated when the values in the Zj-Cj row are zero or positive. The presence of the negative Zj- Cj value and the column headed by this variable is called, as before, the pivot column. The selection of the pivot row (and the outgoing variable) is done exactly the same way as for the maximization problems- the row that has the least (non-negative) quotient is our row of interest. Finally, the revised simplex tableau is derived in the same way as discussed earlier. We proceed in this manner until optimal solution is obtained. In respect of our problem, the initial solution is not optimal. Here the incoming variable is x2 while A1 is the outgoing variable. The revised tableau is given in the following table. In a similar way we proceed until the optimal solution is found. The optimal tableau for minimization problem is a tableau that consists in its C-Z row all

zeros and positive values. (i.e. C-Z values> 0). Conduct elementary row operations (ERO) to arrive at optimal solution before going to the next solution part.

Cj 40 24 0 0 M MBasis X1 X2 S1 S2 A 1 A 2 Q RRX2 24 2/5 1 -1/50 0 1/50 0 96 240

A2 M 60* 0 1 -1 -1 1 2400 40Zj 40 24 0 0 M M 2400

Cj –Zj 152/5-60M

0 1225- M

- M 122M- 25

0

Simplex Table 3 : Non optimal Solution

Cj 40 24 0 0 M MBasis X1 X2 S1 S2 A 1 A 2 Q RRX2 24 0 1 -2/75 1/150 2/75 -1/150 80 -3000x2 40 1 0 1/60* -1/60 -1/60 1/60 40 2400Zj 40 24 0 0 M M 3520Cj – Zj 0 0 -2

753875

M+ 38 75

M+ 38 75


50


Simplex Table 4 : Optimal Solution

Cj 40 24 0 0 M MBasis X1 X2 S1 S2 A 1 A 2 QX2 24 8/5 1 0 -1/150 0 1/50 144x2 0 60 0 1 -1 -1 1 2400Zj 40 24 0 0 M M 3456Cj – ZJ 8/5 0 0 12/25 M M-12/5

According to the optimal solution, the objective function value is

Z=40x0+24x144+0x2400+0x0xM+0xM=Birr 3456

The value of S1 = 2400 indicates the surplus phosphate ingredient obtained by buying the least cost mix.


Find the optimal solution using simplex method.1. Min Z=10x1 +5x2

Subject to: 2x1 + 5x2 > 150

3x1+ x2 > 120 x1, x2 > 0


The assessment methods for this criterion include tests, quiz, exam, assignments, case analysis and group works Summary

The standard form of LP problem should have the characteristics of(1)All the constraints should be expressed as equations by slack or surplus and/or artificial variables (2)The right hand side of each constraint should be made non-negative; if it is not, this should be done by multiplying both sides of the resulting constraint by -1 and (3) Three types of additional variables, namely (1) Slack Variable(S)(2) Surplus variable (-S), and (3) Artificial variables (A) are added in the given LP problem to convert it into standard form for two reasons: the extra variables needed to add in the given LP problem to convert it into standard form is given below:


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Types ofconstraint

Extra variables to be added

Coefficient of extra variablesin the objective function

MaxZ MinZ

Presence of variables in the initial solution

mix

< Add only slack variable 0 0

Yes

>Subtract surplus variable and 0

0No

Add artificial variable -M +M

Yes

= Add artificial variable -M +M

Yes

2. Test of optimality- If all Cj - Zj < 0, then the basic feasible solution is optimal (Maximization

case).- If all Cj - Zj > 0, then the basic feasible solution is optimal (Minimization

case).

3. Variable to enter the basis- A variable that has the most positive value in the Cj - Zj row (Maximization case)- A variable that has the highest negative value in the Cj - Zj row (Minimization case)

N:B-‘Highest negative’ values in this case refers to the negative value which is far from zero on the number line!

4. Variable to leave the basis- The row with the non-negative and minimum replacement ratio (For both

maximization and minimization cases i.e: RR > 0

3.1.3.4. Section IV: Mathematics of Finance

Dear student! What do you know about mathematics of finance? Why we need to learn mathematics of finance?______________________________________________________________________________________________________________________________________________________

Introduction

Mathematics of finance is concerned with the analysis of time-value of money. The fundamental premise behind such analysis is the concept that entails the value of money


52


changes overtime. Putting it in simple terms, the value of one birr today is not the same after a year. Mathematics of finance has an important implication in organizations as transactions and business dealings are mostly pecuniary. Such matters as lending and borrowing money for various purposes, leasing materials, accumulating funds for future use, sell of bonds are some of the cases that involve the concept of time value of money. Likewise, finance mathematics is equally important in our personal affairs. For example, we might be interested in owning a house, in financing our educational fees, having a car, having enough retirement funds etc. All these cases and others involve financial matter. Cognizant of this fact, we proceed to the study of mathematics of finance in this section.

Simple interest and discounts:

Simple Interest: Interest that is paid solely on the amount of the principal P is called simple interest. Simple interest formula:

I = p i n Where, I = Simple interest (in dollars or birr)

P = Principal (in dollar, or birr) and it is the amount i = Rate of interest per period (the annual simple interest rate) n = Number of years or fraction of one year

In computing simple interest, any stated time period such as months, weeks or days should be expressed in terms of years. Accordingly, if the time period is given in terms of,

i. Months, then n= Number of months

12ii. Weeks, then

n= Number of Weeks 52

iii. Days, then a. Exact interest

n= Number of days 365

b. Ordinary simple interest n= Number of days 360

Maturity value (future value) represents the accumulated amount or value at the end of the time periods given. Thus,

Future value (F) = Principal (P) + Interest (I)


53


Example: A credit union has issued a 3 year loan of Birr 5000. Simple interest is charged at a rate of 10% per year. The principal plus interest is to be repaid at the end of the third year.

a. Compute the interest for the 3-year period. b. What amount will be repaid at the end of the third year?

Solution: Given values in the problem

3 – Years loan = Principal = Birr 5000Interest rate = i = 10% = 0.1Number of years (n) = 3 years

a. I = p i n I = 5000 x 0.1 x 3 I = Birr 1500

b. The amount to be repaid at the end of the third year is the maturity (future) value of the specified money (Birr 5000). Accordingly, F = P + I

F = 5000 + 1500 F = Birr 6500

Or, using alternative approach, F = P + I

Then, substitute I = P i n in the expression to obtain F = P + PinF = P (1 + in)

Consequently, using this formula we can obtain F = 5000 (1+ (0.1x3)F = 5000 x 1.3F = Birr 1500

Ordinary and Exact Interest:

In computing simple interest, the number of years or time, n, can be measured in days. In such case, there are two ways of computing the interest.

i. The Exact Method: if a year is considered as 365 days, the interest is called exact simple interest. If the exact method is used to calculate interest, then the time is


54


n = number of days / 365

ii. The Ordinary Method (Banker’s Rule): if a year is considered as 360 days, the interest is called ordinary simple interest. The time n, is calculated as

n = number of days/ 360

Example: Find the interest on Birr 1460 for 72 days at 10% interest using, a. The exact method b. The ordinary method

Solution:

Given a) P=1460 b) P=1460 P = Birr 1460 n=72/365 n= 72/360

n = 72 days i= 0.1 i=0.1 i = 10% = 0.1 I=Pin =1460*72/365*0.1=28.8 I=Pin= 1460*72/365*0.1=29.2

Simple Discount: Present Value The principal that must be invested at a given rate for a given time in order to produce a definite amount or accumulated value is called present value. The present value is analogous to a principal P. It involves discounting the maturity or future value of a sum of money to a present time. Hence, the simple present value formula is derived from the future value (F) formula as follows.

Future Value = Principal + Interest F = P + I but I = Pin

Thus, F = P + Pin F = P (1+ in)

Then from this, solve for P.

If P is found by the above formula, we say that F has been discounted. The difference between F and P is called the simple discount and is the same as the simple interest on P.

Example: 90 days after borrowing money a person repaid exactly Birr 870.19. How much money was borrowed if the payment includes principal and arch nary simple interest at 9 ½ %?

Solution:


55

inFP 1


1. Given values in the problem, n in ordinary method = Number of days / 360

= 90 /360 n = 0.25

F = the amount repaid = Birr 870.19 i = 9 ½% = 9.5% = 0.095

Required:

The amount borrowed which is the same as simple present value, P.

= 870.19 / (1+ (0.095 x 0.25)) P = 870.19 ÷ 1.024 P = Birr 849.795

Promissory Notes and Bank Discount:

Definitions: A promissory note is a promise to pay a certain sum of money on a specified date. It is also considered as a written contract containing an unconditional promise by the debtor called the maker of the note to pay a certain sum of money to the creditor called the payee of the note, under terms clearly specified in the contract. Promissory note is unconditional in a sense that it gives the maker of the note an exclusive right either to sell, borrow, or discount it against the value of the note.

A bank discount is the amount of money received or collected after discounting a note before its due date. It is not unusual when borrowing money from a bank that one is required to pay a charge based on the total amount that is to be repaid (maturity value), instead of the principal used. If the maturity value is used in determining the charge for use of money, we say that the promissory note (or simply the note) is discounted. Consequently, a charge of loan computed in this manner is called ‘Bank Discount’ and it is always computed based on the maturity value. Bank discount is the amount that is charged on maturity value. Hence, the amount of money payable to the debtor or the amount that the borrower receives is called ‘Proceed.’ The amount that the borrower is going to pay to the creditor (lender) is called ‘maturity value.’ To further our understanding of this concept, let’s develop mathematical expressions (formula) for computation of the variables at stake.

Proceed = Maturity Value – Bank Discount Symbolically, P = F – D , and D = Fdt


56


Where, P = Proceed F = Maturity value D = Bank discount d = Rate of discount t = Time of discount

Now we can further elaborate the above formula for proceed. To begin with, P = F – D , but D = Fdt

Therefore, P = F – Fdt = F (1 – d t)

In sum, proceeds can be calculated by P = F (1 – d t)

For example, if Birr 1000 is borrowed at 12% for 6 months, the borrower receives the proceeds, P, and pays back F = Birr 1000. The proceeds will be Birr 1000 minus the interest on Birr 1000. This will be:

P = 1000 – (1000 x 0.12 x 6/12) = Birr 940 Or, P = 1000 (1 – (0.12 x 6/12) P = Birr 940

i. Proceeds are an amount received now for payment in the future. Therefore, they are analogous to present value. Yet, proceeds are not equal to present value because the proceeds from a futures obligation to pay are always less than the present value of that obligation if, of course, the same rate of interest is used in both adulations.

ii. Proceeds should be completed when the interest rate is stated by the qualifier word as discount rate or a bank discount or interest deducted – in – advance, and present value should be computed where the interest is given without such qualifiers, discount.

iii. The computation of simple interest and bank discount is the same except in the former case principal and in the later case the maturity values are used for between trimmings the amount discount.

Having the idea of promissory notes and bank discounts, we may now progress to consider some illustrative problems.

Example: Find the bank discount and proceeds on a note whose maturity value is Birr 480 which is discounted at 4% ninety days before it is due.

Solution: Given values in the problem F = Birr 480


57


d = 4% = 0.04 t = 90 days or 3 months = 3/12 = 90/360 = 0.25 D = ? and P = ? To find the value of the bank discount, we use the formula D = Fdt. Accordingly, D = 480 x 0.04 x 3/12 D = Birr 4.8 is the amount of bank discount. In the same manner, the proceed can be obtained as follows.

P = F – D or P = F (1 – d t)P = 480 – 4.8 or P = 480 (1 – (0.04 x 0.25)P = Birr 475.2 or P = 480(0.99) = Birr 475.2


Dear learner, we have seen above the concepts of promissory notes and bank discount with examples now, would you try to do this question. A borrower signed a note promising to pay a bank Birr 5000 ten months from now.

a. How much will the borrower receive if the discount rate is 6%?b. How much would the borrower have to repay in order to receive Birr 5000

now?

The Compound Interest:

If an amount of money, P, earns interest compounded at a rate of I percent per period it will grow after n periods to the compound amount F, and it is computed by the formula:

Compound amount formula: Fn = P (1 + i) n

Where, P = Principal i = Interest rate per compounding periods

n = Number of compounding periods (number of periods in which the principal earn interest)

F = Compound amount

A period, for this purpose, can be any unit of time. If interest is compounded annually, a year is the appropriate compounding or conversion or interest period. If it is compounded monthly, a month is the appropriate period. It is important to know that the number of compounding period/s within a year is/are used in order to find the interest rate per


58


compounding periods and it is denoted by i in the above formula. Consequently, when the interest rate is stated as annual interest rate and is compounded more than once a year, the interest rate per compounding period is computed by the formula:

i = j / m, where j is annual quoted or nominal interest rate m number of conversation periods per year or the compounding periods per year

n = m x t, where t is the number of years

Example: Assume that we have deposited Birr 6000 at commercial Bank of Ethiopia which pays interest of 6% per year compounded yearly. Assume that we want to determine the amount of money we will have on deposit (our account) at the end of 2 years (the first and second year) if all interest is left in the savings account.

Solution: Give values in the problem, P = Birr 6000, j = 6% = 0.06, t = 2 years m = compounded annually = i.e. only once n = m x t = 1 x 2 = 2 i = j / m = 0.06 / 1 = 0.06

Then, the required value is the maturity or future value F = P(1 + i )n = 6000 (1 + 0.006)2 = Birr 6000 (1.06)2 = Birr 6741.6

Example: An individual accumulated Birr 30,000 ten years before his retirement in order to buy a house after he is retried. If the person invests this money at 12% compounded monthly, how much will be the balance immediately after his retirement? Solution:

Given values, P = Birr 30,000 , t = 10 years , i = 12% = 0.12m = compounded monthly = 12 i = j / m = 0.12 / 12 = 0.01n = m x t = 12 x 10 = 120 and what is required is the Future Value F.

Then, F = P (1 + i) n

= 30,000 (1.01)120

= 30,000 (1.01)120

F = Birr 99,011.61


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Having the understanding of how compound interest works and computation of future value, in subsequent example, we will consider how to determine the number of periods it will take for P birr deposited now at i percent to grow to an amount of F birr.

Present Value of a Compound Amount:

Future (maturity) value is the value of the present sum of money at some future date (time). Conversely, present value (or simply principal) is the current birr or dollar value equivalent of the future amount. It is the sum of money that is invested initially and that is expected to grow to some amount in the future at a specified rate. If we put the present and future (maturity) values on a continuum as shown below, we can observe that they are inverse to one another. And, future value is always greater than the present value or the principal since it adds/earns interest over specified time-period.

0 1 2 3 . . . nPresent Value (P) Future Value (Compound Amount)

Future value is obtained by compounding technique and the expression (1 + i) n is called compound factor. On the other hand, present value is obtained by discounting techniques and the expression (1 + i) -n is referred to as the compound discount factor. The formula for present value of compound amount is simply derived from compound amount formula by solving for P.

Examples:

1. What is the present value of

a. Birr 5000 in 3 years at 12% compounded annually? b. Birr 8000 in 10 years at 10% compounded quarterly?

2. Suppose that a person can invest money in a saving account at a rate of 6% per year compounded quarterly. Assume that the person wishes to deposit a lump sum at the beginning of the year and have that some grow to Birr 20,000 over the next 10 years. How much money should be deposited at the beginning of the year?

3. A young man has recently received an inheritance of birr 200,000. He wants to make a portion of his inheritance and invest it for his late years. His goal is to accumulate Birr 300,000 in 15 years. How much of the inheritance should be invested if the money will


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earn 8% per year compounded semi-annually? How much interest will be earned over the 15 years?

Solution: 1. (a) Given the values, Fn = F3 = Birr 5000, t = 3 years m = 1 (compounded

annually)n = t x m = 3 x 1 = 3j = 12 % = 0.12i = j / m = 0.12/1 = 0.12 and we are required to find Present Value P.

Thus, P = Fn (1 + i) -n = 5000 (1 + 0.12)-3

= 5000 (1.12-3) = 5000 (0.7118) P = Birr 3559

(b) Fn = F40= Birr 8000, t = 10 years , m = quarterly = 4 n = t x m = 10 x 4 = 40 , i = 10% = 0.1 , i = j / m = 0.1 / 4 = 0.025 p = ? but P = Fn (1 + i)-n

= 8000 (1 + 0.025)-40 = 8000 (1.025) -40

P = Birr 2979.5

2. Given the values, i = 6% = 0.06 , m = quarter = 4 times a year i = j ÷ m = 0.06 ÷ 4 = 0.015F = Birr 20,000 shall be accumulated t = 10 yearsn = m x t = 10 x 4 = 40 interest periods P = how much should be deposited now?

P = Fn(1 + i)-n

= 20,000 (1+0.015)-40 = 20,000(1.015-40)P = Birr 11,025.25

3. Inheritance = Birr 200,000 Fn = Birr 300,000 (the person's goal of deposit) , t = 15 years , j = 8% = 0.08m = semi-annual = 2 times a yeari = j ÷ m = 0.08 ÷ 2 = 0.04n = t x m = 15 x 2 = 30 interest periods/semi-annuals P = how much of the inheritance should be invested now? P = Fn(1 + i)-n

I = Amount of interest?


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= 300,000 (1+0.04)-30 = 300,000(1.04)-30 = 300,000(0.3083) = Birr 92,490

The present value of Birr 300,000 after 15 years at 4% semi-annual interest rate is equal to Birr 92,490. Therefore, from the total inheritances received Birr 92,490 needs to be deposited now.

Amount of compound interest = Future Value – Preset Value = 300,000 – 92,490Amount of compound interest = Birr 207,510

Annuities:

Annuity refers to a sequence or series of equal periodic payments, deposits, withdrawals, or receipts made at equal intervals for a specified number of periods. For instance, regular deposits to a saving account, monthly expenditures for car rent, insurance, house rent expenses, and periodic payments to a person from a retirement plan fund are some of particular examples of annuity. Payments of any type are considered as annuities if all of the following conditions are present:

i. The periodic payments are equal in amountii. The time between payments is constant such as a year, half a year, a quarter of a

year, a month etc. iii. The interest rate per period remains constant. iv. The interest is compounded at the end of every time.

Annuities are classified according to the time the payment is made. Accordingly, we have two basic types of annuities.

i. Ordinary annuity: is a series of equal periodic payment is made at the end of each interval or period. In this case, the last payment does not earn interest.

ii. Annuity due: is a type of annuity for which a payment is made of the beginning of each interval or period.

It is only for ordinary annuity that we have a formula to compute the present as well as future values. Yet, for annuity due case, we may drive it from the ordinary annuity formula. To proceed, let us first consider some important terminologies that we are going to use in dealing with annuities.

i. Payment interval or period: it is the time between successive payments of an annuity.


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ii. Term of annuity: it is the period or time interval between the beginning of the first payment period and the end of the last one.

iii. Conversion or interest period: it is the interval between consecutive conversions of interest.

iv. Periodic payment/rent: it is the amount paid at the end or the beginning or each payment period.

v. Simple annuity: is the one in which the payment period and the conversion periods coincides each other.

Sum of Ordinary Annuity: Maturity Value

Maturity value of ordinary annuity is the sum of all payments made and all the interest earned there from. It is an accumulated value of a series of equal payments at some point of time in the future. Suppose you started to deposit Birr 1000 in to a saving account at the end of every year for four years. How much will be in the account immediately after the last deposit if interest is 10% compounded annually?

In attempting this problem, we should understand that the phrase at the end of every year implies an ordinary annuity case. Likewise, we are required to find out the accumulated money immediately after the last deposit which also indicates the type of annuity. Further, the term of the annuity is four years with annual interest rate of 10%. Thus, we can show the pattern of deposits diagrammatically as follows.

The Present The Future 0 1st 2nd 3rd 4th 1000 1000 1000 1000

Birr 1000 Birr 1100 Birr 1210 Birr 1331 Total Future Value = Birr 4641The first payment earns interest for the remaining 3 periods. Therefore, the compound amount of it at the end of the term of annuity is given by,

F = P (1 + i) n = 1000 (1 + 0.1)3 = Birr 1331

In the same manner, the second payment earns interest for two periods (years). So, F= 1000 (1+0.1)2 = 1210


63


The 3rd payment earns interest for only one period. So, F=1000(1+0.1)1 = 1100

No interest for the fourth payment since it is made at the end of the term. Thus, its value is 1000 itself. In total, the maturity value amounts to Birr 4641. This approach of computing future value of ordinary annuity is complex and may be tiresome in case the term is somewhat longer. Thus, in simple approach we can use the following formula for sum of ordinary annuity (Future Value).

Where, n = the number of payment periods i = interest rate per period R = payment per period Fn = future value of the Annuity or sum of the annuity after n periods

Now, let us consider the above example. That is, R = Birr 1000i = 0.1 and n = 4

Future Value = Birr 4641

Example: A person plans to deposit 1000 birr in a savings account at the end of this year and an equal sum at the end of each following year. If interest is expected to be earned at the rate of 6% per year compound semi-annually, to what sum will the deposit (investment) grow at the time of the fourth deposit?

Solution: The known values in the problem are,

R = 1000, j = 6% = 0.06 , m = semi-annual = twice a year i = 0.06 ÷ 2 = 0.03n = 4


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F4 = ? F4 = R

= 1000

= 1000

= 1000 x 4. 183627F = Birr 4183.63

Learning Activities 4.2

A 12 year old student wants to begin saving for college. She plans to deposit Birr 50 in a saving account at the end of each quarter for the next 6 years. Interest is earned at a rate of 6% per year compounded quarterly. What should be her account balance 6 years from now? How much interest will she earn?

Ordinary Annuities: Sinking Fund Payments A sinking fund is a fund into which periodic payments or deposits are made at regular interval to accumulate a specified amount (sum) of money in the future to meet financial goals and/or obligations. The equal periodic payment to be made constitute an ordinary annuity and our interest is to determine the equal periodic payments that should be made to meet future obligations. Accordingly, we will be given the Future Amount, F, in n period and our interest is to determine the periodic payment, R. Then we can drive the formula for R as follows.

Multiplied both sides by That is

Then,

Where, R = Periodic payment amount of an

annuity


65

is the sinking found formula.


i = Interest per period which is given by j ÷ m

j = Annual nominal interest rate

m = Number of conversion periods per year

n = Number of annuity payment or deposits (also, the number of compounding

periods)

F = Future value of ordinary annuity

In general, a sinking fund can be established for expanding business, buying a new building, vehicles, settling mortgage payment, financing educational expenses etc.

Example:

A corporation wants to establish a sinking fund beginning at the end of this year. Annual deposits will be made at the end of this year and for the following 9 years. If deposits earn interest at the rate of 8% per year compounded annually, how much money must be deposited each year in order to have 12 million Birr at the time of the tenth deposit? How much interest will be earned? Solution:

1. Future level of deposit desired = Fn = Birr 12 million Term of the annuity = t = 10 yearsConversion periods = m = annual = 1

n = t x m = 10 x 1 = 10 annuals j = 0.08 i = j ÷ m = 0.08 ÷ 1= 0.08

R = the amount to be deposited each year to have 12 million at the end of the 10th year = ? Then to obtain the value of R, we shall use the formula for sinking fund.


66

1)1( nn i

iFR


R = Birr 828,353.86

On the other hand, the amount of interest, I, is obtained by computing the difference between the maturity value (Fn = 12,000,000) and the sum of all periodic payments made. Thus,

I = Fn - R (10) = 12,000,000 – 823,353.86 (10) = 12,000,000 – 8,283,538.6 = Birr 3,716,461.4

Present Value of Ordinary Annuity:

The present value of annuity is an amount of money today, which is equivalent to a series of equal payments in the future. It is the value at the beginning of the term of the annuity. The present value of annuity calculation arise when we wish to determine what lump sum must be deposited in an account now if this sum and the interest it earns will provide equal periodic payment over a defined period of time, with the last payment making the balance in account zero. Present value of ordinary annuity is given by the formula:

Where, R= Periodic amount of an annuity i = Interest per period which is given by j ÷ m j = Annual nominal interest rate m = Interest/ conversion periods per year n = Number of annuity payments / deposits (also, the number of compounding periods)P = Present value of ordinary annuity

Example: A person recently won a state lottery. The term of the lottery is that the winner will receive annual payments of birr 18,000 at the end of this year and each of the following 4 years. If the winner could invest money today at the rate of 6% per year compounded annually, what is the present value of the five payments?

Solution:

R = Annual payments of Birr 18,000 Term of the annuity = t = this year and the following 4 years = 5 years


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i = 6% = 0.06 (since the conversion period per year is annual)n = 5Present value of payments = P = ? P = Birr 75,822.55

Mortgage Payments and Amortization: Another main area of application of annuities in to real world business situations in general and financial management practices in particular is mortgage amortization or payment. Mortgage payment is an arrangement whereby regular payments are made in order to settle an initial sum of money borrowed from any source of finance. Such payments are made until the outstanding debt gets down to zero. An individual or a firm, for instance, may borrow a given sum of money from a bank to construct a building or undertake something else. Then the borrower (debtor) may repay the loan by effecting (making) a monthly payment to the lender (creditor) with the last payment settling the debt totally.

In mortgage payment, initial sum of money borrowed and regular payments made to settle the respective debt relate to the idea of present value of an ordinary annuity. Along this line, the expression for mortgage payment computation is derived from the present value of ordinary annuity formula. Our intention in this case is to determine the periodic payments to be made in order to settle the debt over a specified time – period. Hence, we know that

Now, we progress to isolate R on one side. It involves solving for R in the above present value of ordinary annuity formula. Hence, multiply both sides by the interest rate i to obtain:

P i = R [1 – (1 + i) –n]

Further, we divide both sides by [1 – (1 + i) –n] and the result will be the mathematical expression or formula for computing mortgage periodic payments as follows.


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i

iRP

n)1(1


Where, R = Periodic amount of an annuity i = Interest per conversion period which is given by j ÷ m j = Annual nominal interest rate m = Interest or conversion periods per year n = the number of annuity payments/deposits (number of compounding periods) P = Present value of an ordinary annuity

Example: Emmanuel purchased a house for Birr 115,000. He made a 20% down payment with the remaining balance amortized in 30 years mortgage at annual interest rate of 11% compounded monthly.

a. Find the monthly mortgage payment?b. Compute the total interest.

Solution: 1. Total cost of purchase = Birr 115,000

Amount paid at the beginning (Amount of down payment) = 20% of the total cost = 0.2 x 115,000 = Birr 23,000

Amount Unpaid or Mortgage or Outstanding Debt = 115,000 – 23,000 = Birr 92,000

t = 30 years j = 11% = 0.11 , m = 12 , i = 0.11 ÷ 12 = 0.00916n = t x m = 30 x 12 = 360 months

The periodic payment R = ?

= 92,000 (0.009523233) R = Birr 876.14

b. Total Interest = (R x n) – P

= 876.14 x 360 – 92,000 = Birr 223,409.49


69

ni

iPR

)1(1


Over the 30 years period Emmanuel is going to pay a total interest of Birr 223,409.49, which is well more than double of the initial amount of loan. Nonetheless, the high interest can be justified by the fact that value of a real estate is usually tend to increase overtime. Therefore, by the end of the term of the loan the value of the real estate (house) could be well higher than its purchase cost in addition to owning a house to live in for the 30 years and more.

Continuous Assessment The assessment methods for this criterion include tests, quiz, exam, assignments, case analysis and group works

Summary

The following are some of important formula that deal with mathematics of finance. Simple Interest

= P i n Future Value Of A Simple Interest

= P (1 + i n) Compound Amount

= P (1 + i) n

Present Value Of A Compound Amount = F (1 + i) –n

Maturity Value Of Ordinary Annuity (Fn)

Sinking Fund Payment Formula

Present Value Of Ordinary Annuity


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i

iRP

n)1(1

1)1( nn i

iFR

i

1 -i)(1 n

RFn


Mortgage Payments and Amortization

Exercises

1. Suppose, a small handicraft enterprise has requested a two year loan of Birr 6500 from the commercial Bank of Ethiopia. If the bank approves the loan at an annual interest rate of 7.5%,

a. What is the simple interest on the loan?b. What is the maturity value of the loan?

2. A person signs a note promising to pay a bank Birr 1500 eight months from now and receives Birr 1350. Find the discount rate.

3. Find the bank discount and proceeds on a 120-day note for Birr 720 bearing 5% interest if discounted at 4% 90 days before it is due.

4. If Birr 6500 is invested at 8 ½ % compound a. Annually b. Semi annuallyc. Quarterly d. Monthly,

What is the amount after 7 years? 5. At what interest rate compounded quarterly will a sum of money double in 4 years?

6. How long will it take for Birr 4750 to accumulate to Birr 7500 at 51/3% compounded semi-annually?

7. If, in 11 years, Birr 1200 accumulates to Birr 1482, what is the compound interest rate provided it is converted annually?

8. 8. If money is worth 14% compounded semi-annually, would it be better to discharge a debt by paying Birr 500 now or Birr 600 eighteen months from now?

9. Find the accumulated value of an ordinary annuity of Birr 65 per period for 23 periods if money is worth 4% per period.


71

ni

iPR

)1(1


10. If birr 500 is deposited each quarter in to in account paying 12% compounded quarterly,

a) How much is the value of deposit at the end of the 3 years?

3.1.3.5. Section V: Introduction to Calculus

Dear learner! What do calculus means? Why we study calculus?______________________________________________________________________________________________________________________________________________________

Introduction

It is a dried fact that the application of concepts of calculus in the business arena specially; in marginal analysis and optimization problems is paramount. In this part of the module, basic concepts in calculus to be seen include: concept of limit and continuity, derivatives, definite and indefinite integration, and their major application areas in business; typically, marginal analysis, optimization problems and area functions.

Differential Calculus:

It is one aspect of calculus that measures the rate of change in one variables as another variable changes. It broadens the idea of slope.

A function: if for every value of a variable x, there corresponds exactly one and only one value of the variable y, we call y is a function of x, written as:


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Y = f(x).Limits:

Definition : the limit of f(x) as x c is ℓ which is written as ; (x) = ℓ, if

and only if the functional value f(x) is close to the single real number ℓ, whenever x is close to but not equal to c ( on either side of c).Example-1:For the function f(x) =x2 + 2, find the limit of f(x) as x approaches 1.

Solution:

X 0.8 0.9 0.99 0.999 1 1.0001 1.001 1.01 1.1f(x)=x2+2 264 2.81 2.9801 2.998 3.0002 3.002 3.021 3.21

ℓ- = 3 ℓ+ = 3

f(x) = 3

Example-2:For the function f(x) = /x/, find; X

a. f(x) b. f(x)

Solution:

X 1.9 1.99 1.999 2 2.001 2.001 2.01

f(x)=/x/ x

1 1 1 1 1 1

ℓ- = 1 ℓ+ = 1

Thus; f(x) = 1

b.X -0.999 -0.99 -0.9 0 0.0001 0.001 0.01 0.1f(x)=/ / x

-1 -1 -1 1 1 1 1

ℓ- = -1 ℓ+ = 1

ℓ-≠ ℓ+ → So, f(x) doesn’t exist.


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Limit Theorems:

1. If k is any constant , k = k

E.g. 10 = 10

c = c

2. kf(x) = k f(x)

E.g. 3x2 = 3( x2) = 3(a2) = 3a2

3. (f(x) +g(x)] = f(x) + g(x)

E.g. (x2-2+3) = x2 – 2 x + 3

= a2-2a +3

4. [f(x).g(X) ] = ( f(x) ] ( g(x) ]

E.g. (x+3)(X-2) = [ (x+3) ] [ (x-2)

= [5] [0] = 0

5. [f(x)n] =[ f(x)n]

E.g. (x-1)5 = [ (x-1)]5 =25 = 32

6. If f(x) = L and g(x) = m, then;

a, If m0, then [f(x)/g(x) ] = L/m

b. If m = 0 and L 0, the [f(x)/g(x)] = doesn’t exist

c. If m = 0 and L = 0, then f(x) and g(X) have a common factor and the limit can be evaluated after employing the process of cancellation.

Continuity of a Function:

Definition: a function f is continuous at the point x = c if:

1. f(x) → exists


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2. f(c) is defined

3. f(x) = f(c)

Example:

Using the definition of continuity, discuss the continuity of the function f(x) = x 2 -4 x-2

at c =1 and c = 2

Solution:

At c =1, f(x)

X 0.9 0.99 0.999 1 1.0001 1.001 1.01 1.1f(x) = x 2 -4 x-2

2.9 2.99 2.999 3.0001 3.001 3.01 3.1

ℓ- = 3 ℓ+ = 3

f(x) =3 f(x) exists

f(1) =1 2 -4 = 3 → f(c) is defined 1-2

f(x) = 3 = f(1)

Therefore; f(x) = x 2 -4 is continues at c =1 x-2At c = 2

X 1.9 1.99 1.999 2 2.0001 2.001 2.01 2.1F(x)

3.9 3.99 3.999 4.0001 4.001 4.01 4.1

ℓ- = 4 ℓ+ = 4

f(x) = 4 It exists

f(2) =2 2 -4 = 4-4 = 0= ¢ 2-2 0 0

f(x) f(c) →4¢

Therefore; f(x) is not continuous at c = 2.

Derivatives:

Definition: for y = f(x) we define the derivative of f at x, denoted by f′′(x) to be; Jimma, Haramaya, Ambo, Hawassa, Adama, Bahirdar, W.Sodo,Semara Universities

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f′(x) = lim f(x) = lim f(x+ x)- f(x) x 0 x x 0 x

Example-1Find f′(x) for f(x) = 2x +4

f(x) = lim f(x+ x) –f(x) x 0 x1st → find f(x) = f(x) = f(x+x)-f(x) = [2(x+x) +4] – (2+4)] = 2+2x+4-2x-4 f(x) = 2x2nd find the limit: f′(x) = lim f(x) x 0 x = lim 2 x x 0 X f′(x) = 2Example-2

For the function f(x) = 4x-x2 find f′(x) f′(x)= lim f(x+ x) – f(x) x 0 x1st → find f(x) = 4(x+ x)-(x+ x) 2 - (4x-x 2 )

x x = 4x+4 x-x 2 – 2x x - x 2 -4x+x 2 x = x (4-2x- x) x = 4-2x-x2nd → find the limit of the resulting function f′(x) = lim f(x ) = lim 4-2x-x x 0 x x 0 f′(x) = 4-2x

Rule of differentiation:

1. A constant function rule If f(x) = c then f′(x) = 0E.g. If f(x) = 5 then f′(x) = 0


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2. The power rule The derivative of the power function is the power times the function raised the power minus one. If f(x) = axn, then f′(x) = anxn-1

E.g. If f(x) = x5, then f′(x) = 5x5-1 = 5x4

If f(x) =3x3, f′(x) =33x3-1 = 9x2

3. The sum and difference rule The derivative o f the sum or difference of two functions is the derivative of the first function plus or minus the derivative of the second function.

If f(x) = u(x) + v(x), then f′(x) =u′(x) + v′ (x)E.g. f(x) = 3x+8 , f′(x) =3+0= 3 f(x) = 4x-x2 ; f′(x) = 4-2x

4. The product rule The derivative of the product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first.If f(x) = u(x). (v(x)) then; f′(x) =u(x). v′(x) + v(x) .u′(x)E.g. f(x) = 3x2 (4x-1), the f′(x) = 3x2(4) + (4x-1) (6x) = 12x2 +24x2-6x = 36x2-6x

5. The quotient rule ; The derivative of the division of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator over the denominator square. If f(x) = u(x), then; V(x) f′(x) = v(x) .u ′ (x) – u(x). v ′ (x) [V(x)] 2

If f(x) = x2/2x-1f′(x) = (2x-1)(2x) –(x 2 )(2) (2x-1)2

= 4x 2 -2x-2x 2 4x2-4x+1 = 2x 2 -2x (2x-1)2


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Application of Differential Calculus to Marginal Analysis: The word marginal refers to rate o f change→ that is a derivative. Let x be the number of units of a product produced Total cost function → C(x)Total revenue function→ R(x)Total profit function→ P(x) = R(x) –C(x)Marginal cost function → C′(x)

Marginal cost is the rate of change in total cost per unit change in production at an output level of x-units.

Marginal revenue function→ R′(x) Marginal profit function→ P′(x) = R′(x) - C′(x) Average cost→ č(x) = C(x) x Marginal average cost→ č′ (x) Average revenue → Ř(x) = R(x)

x Marginal average revenue→ Ř′(x Average profit → Pˉ (x) =P(x) x Marginal average profit → Pˉ′(x)

Example-1

A company manufactures and sells x transistor radios per week. Its weekly cost and demand equations are:

C(x) = 5000 +2xP = 10 – x find 1000a) Production level that maximizes revenue and the maximum revenue.b) The production level that maximizes profit and the maximum profit.c) The MR and MC at the profit maximizing output level. d) The average cost per unit if 1000 radios are produced. e) The marginal average cost at a production level of 1000 radios and interpret the result.

Solution:

a) R(x) = P.X = (10-x) (x)


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1000 R(x) 10x- x 2 1000R(x) = 10 –x = 0

5000 10= x → x = 5000 units 500R′(x) = -1/5000; R′(x) <0 x= 5000 units is the revenue maximizing output level. Maximum revenue is at x = 5000 units R(x) = 10x – x 2 1000R(5000) =10(5000)- (5000) 2 1000 = Birr 25000 b) P(x) = R(x) – C(x) [10x- x 2 ]- [5000+2x] 1000= 10x –x 2 - 5000-2x 1000P(x) = 8x-x 2 - 5000 1000P′(x) 8-2x = 8-x

1000 5008-x= 0 5008= x 500x = 4000 units

P″(x) = -1 P″(x)<0 x = 4000 units is the profit maximizing output level 500At x = 4000 units P(x) = 8x-x 2 - 5000 1000 = 8(4000) - (4000)2-5000 1000 = Birr 11,000

c) C′(x) = 2 production cost increases by birr 2 at each level of out put R′(x) = 10- 2x = 10-x 1000 500


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At x = 4000 units R′(x) = 10 -4000 = 10-8 =2 birr 500

At each level of output TR increases by birr 2 At the profit maximization; MR = MC, i.e. 2 Birr

d) x =1000 radios; Average cost č(x) = C(x) x

= 5000 +2x x At x = 1000 radios; č(x) = 5000 +2(1000) 1000 = 7 Birr

e) Marginal average cost = č′(x) MAC = - 5000 x2

č′(x) = - 5000 = -0.005 Birr (1000)2

Interpretation: At a production level of 1000 units a unit increase in production will decrease average cost by approximately 0.5 cents or by 0.005 Birr.


A certain manufacturing company has the following information: Average total cost is given by the equation:č(q) = 0.5q -500+ 5000 and, qThe demand function is: P = 2500-0.5qA. Find the firm’s: i. Total profit function ii. Marginal cost function iii. Marginal average cost function B. Find the quantity level that: i. maximizes total revenue ii. Maximizes total profit iii. Minimizes total cost C. Find the firm’s: i. Maximum revenue ii. Maximum profit D. Find the price level that leads to maximum: i. Revenue ii. Profit

Integral Calculus:


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Indefinite Integral: Given F(x) which is the anti-derivative of f(x), the indefinite integral of f(x) is defined to be:

f(x) dx = F(x) +C,

Where; = the integral symbol f(x) = the integrand (the function to be integrated)F(x) = the integral (the outcome of integration)C = the constant of integration dx = indicates the variable to be integrated

The rules of integration:

1. A constant function rule;If f(x) = kf(x)dx = (k)dx = kx 0+1 + C = kx+C 0+1E.g. If f(x) = 5f(x) dx = (5)dx = 5x 0+1 + C = 5x+C 0+1

2. The power rule If f(x) = xn f(x) dx = (xn)dx = x n+1 + C n+1E.g. If f(x) = x5;f(x)dx = (x5)dx = x 5+1 + C = x 6 + C 5+1 6

3. A constant times a function rule;If f(x) = axn

(x)dx = (axn)dx = a(xn)dx = a(x n+1 ) + C n+1

E.g. If f(x) = 3x3

f(x)dx = (3x3)dx = 3(x3)dx = 3(x 3+1 )+ C 3+1


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= 3/4x4+C

4. The sum and difference ruleIf f(x) =g(x) ± h(x)f(x)dx = [g(x) ± h(x)]=g(x)dx ± (h(x)dx

E.g. If f(x) = 5x+9 f(x)dx = (5x+9)dx = (5x)dx+ (9)dx

= 5x 1+1 + 9x 0+1 1+1 0+1 = 5/2 x2 + 9x + C

5. The product ruleIf f(x) = (ax+b)n

f(x)dx = (ax+b)n = (ax+b) n+1 + C a(n+1)E.g. If f(x) =(x+2)2= f(x)dx = (x+2)2dx

= (x+2) 2+1 + C = (x+2) 3 + C (2+1) 3

6. The quotient rule; If f(x) = g(x) + h(x) k(x)f(x)dx = (g(x)dx + (h(x))dx

k(x) k(x)E.g. If f(x) = 8+x 3 x2

f(x)dx =(8 )dx + (x 3 )dx

x2 x2

(8x-2)dx + (x)dx

= 8x -2+1 = x 1+1 + C -2+1 1+1= 8x -1 +x 2 + C -1 2= -8x-1 +1/2x2 + C= -8/x+x2 + C 2Indefinite integral for finding total functions:


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Example: The function describing the marginal cost of producing a product is given by f(x) = x+100, where x is the number of units produced, determine the total cost function if the total cost of producing 100 units is birr 40 000.

Solution: C(x) = f(x)dx= f(x) = x+100 (x+100)dx = x 1+1 +100x 0+1 + C 1+1 0+1 x 2 + 100x+ C 2C(x) =1/2x2 + 100 x +c (fixed cost)40,000 = ½(100)2+100(100) + C40,000 = 5000+ 10,000+ C 40,000 = 15000+ C40,000 -15000 = CC = 25000 BirrTherefore; C(x) = ½x2+100x+25,000


Dear student, attempt this question. The marginal revenue function for a company’s product is given by f(x) = 50,000-x, where x is the number of units produced. Develop the total revenue function if revenue is zero when no units are produced and sold.

Definite integral:

Definition: If f(x) is a continuous function on the interval [a, b], the definite integral of f(x)

is defined as f (x)dx = F(b) –F(a)

Where

F(x) = the anti-derivative for f(x) F(b) = the upper limit F(a) = the lower limit F′(x) = f(x)

A definite integral has a single numerical value associated with it and can be obtained through the indefinite integral by using the following steps.


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Step-1: get the indefinite integral of the function Step-2: substitute the value x = a in the indefinite integral Step-3: substitute x = b in the indefinite integral Step-4: subtract the numerical value obtained in Step 2 from step 3 and the result gives the definite integral value of the function between the limits x = a x = b.

ExampleIf marginal revenue is given by: F(q) = 200-6q , what extra total revenue is obtained by increasing sales (q) from 15 to 20?

Solution:

Extra revenue = f(q)dq = (200-6q)dq

= 200q – 6q 1+1 + C 1+1= [200(20) -3(20)2+ C] - [200(15) -3(15)2+C]= 2800+C- 2325 – C= Birr475


The assessment methods for this section include tests, quiz, exam, assignments, case analysis and group works.

Summary

The major formulas for application of concept of calculus in marginal analysis and optimization problems include the following.

- Marginal cost is the rate of change in total cost per unit change in production at an output level of x-units.

- Marginal revenue function→ R′(x)- Marginal profit function→ P′(x) = R′(x) - C′(x)- Average cost→ č(x) = C(x)

x- Marginal average cost→ č′ (x) - Average revenue → Ř(x) = R(x)


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x- Marginal average revenue→ Ř′(x - Average profit → Pˉ (x) =P(x)

x- Marginal average profit → Pˉ′(x)

f(x) dx = F(x) +C

f (x)dx = F(b) –F(a)

Exercises

1. Find the limit of the function f(x) = 2+1 as x approaches:a) 2 b) 1 c) 0

2. Find the limit of the function f(x) = 2x2+1 as x approaches:a) 0 b) 1 c) 3

3. Find the limit of the function f(x) = 1/(x2+1) as x approaches:a) 0 b) 2

3.1.4. Proof of Ability

At the end of this learning task the student will be evaluated through summative exam, attachment, simulation (50 or 40%). It incorporates all the sections in the learning task.

Product: The student has; Criteria and Methods of assessmentApplied systematic approach of the linear equation, algebra and geometry in solving real world situations..

Developed linear equations, functions, graphic representation of linear equations, Compute and formulate slope and equation of a line and demonstrated correct computations on the application of the liner equation and algebra on supposed real world representing cases (Written exam, case study)

Applied matrix operations in real world agribusiness problems

Matrices and its dimensions, and operation elaborated, and the application of matrix algebra and Markov Chain analysis using hypothetical real world allied cases(Written exam, case study)

Recommended decision using the systematic approach of the linear programming to solve practical agribusiness problems

Formulated linear programming problems in both cases of minimization and maximization, optimal solution determined using systematic techniques/graphical and Simplex methods) on real world related(Written exam, case study)


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Compile reports that analyze the time value of money applications for financing decision in agribusiness value chain

Time value of money concepts and its application demonstrated(Written exam, case study)

Applied the concepts of calculus in the agribusiness arena specially; in marginal analysis and optimization problems

Rules of limits and continuity, rules of differentiation and integration as well as their interpretative applications in the agribusiness stream are well understood and applied(Written exam, case study)

Major References

Ann, J. Hughes 1983. Applied Mathematics: For Business, Economics, and the social Science.

Barnett, A. Raymond and Ziegler, R. Michael. Essentials of College Mathematics for Business and Economics, Life Science and Social Science. 3rd ed., 1989.

Bowen, K. Earl, Prichett, D. Gordon, and Saber, C. John 1987. Mathematics with Applications in Management and Economics. 6 th ed., Richard Irwin Inc., USA.

Dexter, J. Booth and John, K. Turner 1996. Business mathematics with Statistics.

Orema, M and Agarwal, K. 1988. Quantitative Techniques. Kings Books, Delhi.


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Business Mathematics Final

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