Business Mathematics

Financial Management in Value Chain: Business Mathematics and its Application

FINANCIAL MANAGEMENT IN VALUE CHAIN (ABVM 322) LT 1: BUSINESS MATHEMATICS AND ITS APPLICATION (5 ECTS)

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

Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

i

Financial Management in Value Chain: Business Mathematics and its Application AUGUST 2012

Table of Contents

3.1. Business mathematics and its application............... .............................. .............................. .............................. ......................... ................ ......... ...1 a. Introduction.............................................................................................................................1 b. Objectives... Objectives............. .................... .................... .................... .................... .................... .................... .................... .................... .................... ..................................1 ........................1 c. Sections...................................................................................................................................1 i. Section I: Linear equations and their interpretative applications.......................................1 ii. Section II: Matrix algebra and its applications................................................................17 iii. Section III: Linear programming....................................................................................33 iv. Section IV: Mathematics of Finance..............................................................................52 v. Section V: Introduction to Calculus.................................................................................72 d. Proof of Ability.....................................................................................................................85 Major References.............................................................................................................85


ii



iii


3.1.

Business ma mathematics an and it its ap application

a. Int Introdu roducction tion

Dear ear stude student, nt, welcom elcome to the the learn learning ing task task Busin usines esss Mathem athematic aticss and and its appl applic ica ation tion . This This learning task is designed to expose you to the basic concepts and area of manager managerial ial applicat application ion of mathem mathematic atics. s. It is divide divided d into five section sections: s: The The first section deals with the linear equations and its applications; the second section is about about the matr matrix ix alge algebr braa and and its its appl applic icat atio ions ns;; the the thir thirdd sect sectio ionn deal dealss with with line linear ar

programming, programming, the fourth section section is dedicated to mathematics mathematics of finance and the fifth section section is about elements and application of calculus. You will find learning activities in each sections. sections. To successfu successfully lly accom accomplish this t his learning learning task 140 study study hours 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. b. Ob Objjecti ective vess

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

Differentiate the various techniques of mathematics that can be employed in solving Agribusiness problems,

o

Appreci Appreciate ate the import importance ance of mathema mathematic ticss in solving solving real real world world bus busines inesss problems, problems, and o

o

Use different mathematical techniques for supporting decisions


1


c. Sections i.

Sect Sectio ion n I: Line Linear ar equa equati tion onss and and thei theirr inte interp rpre reta tati tive ve app appli lica cati tion onss

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

Today the business business environments environments are changing. changing. In such environment, environment, organizations 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 practices the concept and interpretative interpretative applications applications of linear equations equations have a considerable considerable importance. importance. This is because, because, it is common common to face so many cases demanding the application application of mathematics of linear algebra and geometry in making a viable decision that enhance the achieve achievement ment of organiz organizati ational onal objective objectives. s. For instance instance sales sales volume volume and advertise advertisement ment expens exp ense, e, output output level level and number number of emplo employee yeess enga engaged ged on some some acti activit vityy and and cost cost of production, production, demand demand for and supply of a given given product can can be well approxima approximated ted and explained explained by a linear linear equation. equation. Cognizant to the above fact, we need to be well acquainted with the fundamentals of linear equation equationss algebra algebra and geo geometr metryy as related related to its agribus agribusines inesss applicat application. ion. This sectio section, n, therefore, is dedicated to our study of linear equations. Linear equations, functions and graphs:

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. Basic Concepts of Linear Equations and Functions:

are equations whose terms1 are a constant times a variable to the first power. According Accordingly, ly, equations equations that can be transpose transposedd to the form, form, a1 x1+ a2 x2+ …+ an xn = c are said to be linear equations. Where, a1, a2, a3, … an and c are constants x1, x2, x3, …xn are variables (unknown quantities) Linear equations:

1

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

1

Financial Management in Value Chain: Business Mathematics and its Application a1 x1, a2x2, … an xn and c are the terms of 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, ax+by=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. Assume that Ethiopian Electric Power Corporation charges Birr 0.55 per kilowatthour 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.

Example:

The The tota totall month monthly ly charg chargee will will be, be, 0.55 0.55 time timess the numbe numberr of mont monthly hly KW KWhh consumption plus Birr 7 for meter rent. Thus, using the symbols given, y = 0.55x + 7

Solution:

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

Y = 0.55x + 7 is called called a functional functional relations relationship hip since since for each each value of x (independent (independent variable), variable), there is is a single corresponding value for y (dependent). Thus if we write y as expression expression involving x and constants constants x is called called the independent independent variable, variable, then then the value of of y depends upon upon what value value we may assign assign to x and as a result result it is called called the dependent dependent variable variable.. Theref Therefore ore,, 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 Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

2

Financial Management in Value Chain: Business Mathematics and its Application 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. 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.

Graph Graph of a Linear Linear Eq Equat uation ion:: Linea Linearr

Example:

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

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

Solution:

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. y 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. 5 4 Fig 1. Linear Equation Equation Graph

3 (-1, 0)

-2

(0, 3/2)

2 1 0

-1

1

2

3

4

Learning Activity 1.2

Find two coordinate points that satisfy the equation 3x + 4y = 24. Then using the two Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

3

x

Financial Management in Value Chain: Business Mathematics and its Application coordinate points plot the graph of the given function. The distance between two points is the length of a straightline 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 – x – x1 /, is called the horizontal separation of the two points. Further, the quantity / y2 – y1 / is the vertical separation of the two points. The distance between two points:

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

Example:

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/ = 2 Vertical distance = / y2 - y1 / = / -9 - 7 / = /-16/ = 16 b. Horizontal Horizontal distance distance = / x2 - x1/ = / 12 - (-5)/ = 12+ 5/ = 17 Vertical distance AD = / y2 - y1 / = / 6 – 7 / = / -1 / = 1 c. Horizontal distance BD = / x2 - x1/ = /12- (-3)/ = / 12+3/ = 15 Vertical distance BD = / y2 - y1/ = / 6- (-9)/ = / 6 + 9/ =15

Learning activity 1.3


4

Financial Management in Value Chain: Business Mathematics and its Application Find the vertical and horizontal separation of the following points. a. (5, (5, 7) 7) and and (-3, 2) b. ( 5, - 3) and (-11, (-11, -7) c. (6, (6, 2) 2) and and (6, -4) -4) d. (3, (3, 4) 4) and 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 may be calcul calculate atedd from from Pythag Pythagor oras as’’ Theo Theore rem, m, using using their their horizo horizonta ntall and vert vertic ical al separations. Fig1. Fig1. 2. Places Places of coordinates

Y B (x2, y2)

X

C (x2, y2)

A (x1, y1)

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 = ( x2 − x1 ) 2 + ( y 2 − y1 ) 2 Example: Calculate

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


5

Financial Management in Value Chain: Business Mathematics and its Application Solution :

That is,

d =

( x2 − x1 ) 2 + ( y2 − y1 ) 2

d =

(-11 - 5) 2 + (-7 (-3)) 2

d = d =

256 + 16 = 272 16.5


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

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)

The Slope Slope – Interc Intercept ept Form: Form:

Slope = m =

Rise Run

=

y 2 − y1 x 2 − x1

,

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 straight-line segment joining the two points (8, -13) and

(-2, 5). Solution:

m=

y 2 - y1 5 − (−13) 18 = = x 2 - x1 −2 −8 − 10


6

Financial Management in Value Chain: Business Mathematics and its Application Therefore, 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 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 Write the equation equation for one day’s rental rental (y) in terms terms of the number of miles miles driven (x). (x). b. Interpret Interpret the slope slope and the the y – intercept. intercept. c. What is the renter’s renter’s total total cost if a car car is driven driven 100 miles? miles? What is the renter’s renter’s cost 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 equat uation y=mx+b y = 0.2x + 22 b. Interpretation 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 Total cost cost of of driving driving 100 mile miless (x = 100 100)) = 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, Cost per mile =

Total

Cost

Total Number Miles Driven

=

42 ÷ 100 = Birr 0.42


7



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 After setup, setup, it costs costs Birr Birr 2 per book book printed printed and bound. bound. a. Write the equation equation for the the total total cost of making making a number of books books b. State the slope of the line and interpret interpret it. it. c. State State the y-inter y-intercept cept of of the line line and and interpre interprett it. 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 The Slope – Point Form:

y − y1 x − x1

=m

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=3x–9 y=3x–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+b Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

8

Financial Management in Value Chain: Business Mathematics and its Application 4 – 9 = b = - 5 is the y-intercept. Thus, y = 3x – 5 is the equation of the line. Learning Activity 1.6

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. In this case, two points that are on the line are given and completely 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. The Two – Point Form:

y − y1 x − x1

=

y 2 − y1 x 2 − x1

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. 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 Write the equation equation of the the line

Example:

Solution:

Given the values x 1 = 1000 Books y1= Birr 4500 x 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 develop the equation equation of the line, line, first let’s let’s compute compute the slope. slope. m = y2 – y1 x2 – x1 = 7500 – 4500 2000 – 1000 Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

9

Financial Management in Value Chain: Business Mathematics and its Application =

3000 ÷ 1000 = 3

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

y 2 − y1 ÷ x2 − x1 . Subsequently, by substitution this value in the above formula will result in; y − y1 x − x1

=3

Then,

y − y1 = 3( x − x1 )

In continuation, substitute the value (1000, 4500) in 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 – 3000 y = 3x – 3000 + 4500 y = 3x + 1500 ……………………… is the 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 establishes relationship between between cost (y) and the number of units made (x) and assume that the relationship is linear. Find the equation of this relationship. 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 mathematical terminology terminology and technique of linear equations in addressing 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. Applications Applications of linear equation e quations: s:


10

Financial Management in Value Chain: Business Mathematics and its Application 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. Linear Cost – Output Relations Analysis:

Total

I

cost

G

Total cost l ine VC

Variable

H

Total cost (TC)

cost (VC)

D

F

E

D

FC

Fixed cost (FC)

FC

A B

C

Fig1.3 Classification of costs

Number of of units units (Q)

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

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. 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 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 Variable Cost FI ÷ DF = Unit ) = Variable Cost Per Unit (VC unit or VC per Unit Number of Units - Marginal cost is given by change in TC divided by change in Quantity (Q). Thus, Illustration:


11

Financial Management in Value Chain: Business Mathematics and its Application Marginal cost (MC) =

∆TC = VC per unit. ∆Q

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 produced are linearly linearly related. related. -

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*Q SP*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 breakeven point To further our understanding of break-even analysis, let us consider the following break-even chart. Break – Even Analysis:

Profit (R > C) t s o c / e u n e v e R

Revenue

R= PQe and

BEP

C = VQe + FC Variable cost FC

Dear student, what did you observe from the above Fixed cost (FC) Loss figure?---------------------------------------------------------------------------------------------------------(R < C) ---------------------------------------------0 The following are important points to note from the above breakeven chart. Qe

Number of units units (Q)


12

Financial Management in Value Chain: Business Mathematics and its Application i.

As such, such, the total total reven revenue ue lin linee pass passes es thro throug ughh the the ori origi ginn and and hence hence has a y-i y-inte nterce rcept pt of zero while the total cost line has a y – intercept which is equal to the amount of the fixed cost. The fixed fixed cost cost line line wh whic ichh is paral paralle lell to the the qua quanti ntity ty axi axiss (x – axis axis)) is const constant ant at all all levels of output. To the the lef leftt of the the brea breakk – even even point point the the reve revenu nuee line line is is foun foundd belo below w the cos costt line line and hence any vertical separation indicates a loss while to the right the opposite is true. The The total total vari variab able le cos cost, t, whi which ch is the the gap gap betwe between en the the tota totall cost cost and and the fix fixed ed cos costt line increases as more units are produced. Impo Import rtan antt lin linea earr cos costt – ou outp tput ut expr expres essi sion onss (eq (equa uati tion ons) s):: • 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 ( π ) = R – TC

ii. iii. iii.

iv. iv. v.

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?

Example:

Solution:

Given, V = $14.50 FC = $12,838 Sp = $39 Let 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 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 Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

13

Financial Management in Value Chain: Business Mathematics and its Application 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 Write the revenue revenue and cost cost equat equations ions b. Computer Computer the profit, profit, if 25,000 units units are made and sold c. Compute Compute the the profit profit,, if 10,000 10,000 units units are are made and sold sold d. Find Find the the bre breake akeve venn quant quantit ityy e. Find the break-e break-even ven birr birr volume volume of of sales sales f. Co Const nstru ruct ct the the bre breakak-ev even en char chartt Market equilibrium analysis is concerned with the supply and demand of a product in a case they are linearly related. Market Equilibrium Analysis:

•

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. Demand of a product:

• 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 directly related. 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, ) p ( e c i r Graphically, P

Demand (DD) = Supply (SS).

DD


14

SS Quantity (q)


Market equilibrium point

Suppose the supply and demand equation for a given product on a given day reveal the following. Demand (DD): P = 300 – 15Q Supply (SS): P = 500 + Q a. Find the mark market et equili equilibriu brium m price price and and quantit quantity. y. b. Plot the demand and and supply equation equation on a graph. graph. Example:

Solution:

a. Firs First, t, let let us comp comput utee the the equi equili libr briu ium m quan quanti tity ty for for the the give givenn supp supply ly and and dema demand nd functions. Hence, at equilibrium: DD = SS 3000 – 15Q = 500 + 5Q -15Q – 5Q = 500 – 3000 -20Q = -2500 Q = -2500 ÷ -20 Q = 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 demand equations. Thus, let us take the supply function of 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

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 Graph of demand demand and supply supply functi function: on: In plotti plotting ng the graph, graph, first first we need need to get the x and y intercept for both the supply and demand equations. The Y – intercept for Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

15

Financial Management in Value Chain: Business Mathematics and its Application demand equation equation is obtained obtained by setting Q = 0 in the equation equation P = 3000 – 15Q. Thus, P = 3000 – 15(0) = 3000 Therefore, (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 – 15Q 15Q = 3000 = 3000 ÷ 15 Q = 200 The point (200, 0) is the x-intercept of the demand function. The same procedure is to be followed followed in computing computing the Y and y intercept 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 = 500 The 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 = -100 Hence, 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. Price (P) 3000

SS: P = 500 + 5q

2000 1125 DD: P = 8000 15q 500

125

200

300

400 Quantity (q)



16

Financial Management in Value Chain: Business Mathematics and its Application 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 Find the excess excess demand demand for for compute computers rs at a price price of Birr Birr 1400. 1400. b. Find the excess supply supply of computers computers at a price price of Birr Birr 2100. 2100. c. Find Find the the marke markett equil equilibr ibrium ium quan quantit tity. y. d. Find Find the the marke markett equil equilibr ibrium ium price. price. e. Sketch Sketch the demand demand and sup supply ply functio functions. ns. 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 considered the managerial applications applications of linear algebra and geometry so far. In so, we have considered that linear equations are mathematical expressions written in the form of y = m x + b 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 m=

y 2 − y1 x 2 − x1

Further, we have considered how to compute the distance between two points on a coordinate plane. Subsequently, Subsequently, approaches approaches of developing developing equation equation of a line are discussed 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. ii. ii.

Sec Section tion II: II: Ma Matrix trix alg algebra ebra and and its its app ppli lica cati tion onss


17

Financial Management in Value Chain: Business Mathematics and its Application 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, algebra, dimension dimension and types of matrices, 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. Matrice Matricess are used used to handle handle larg largee linear linear system systemss 2. Matrice Matricess are used used to solve solve complex complex linear linear equat equations ions 3. Matrices Matrices are an effective effective means means for for summarizi summarizing ng voluminous voluminous business business data. data. 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. Definition of a Matrix:

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

x11 x 21

Xm1

x12 x22

x1n x2n

xm2

xmn (mxn)

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). Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

18

Financial Management in Value Chain: Business Mathematics and its Application Example: Example:

A=

a11 a 21 a 31

a12 a22 a 32

a13 a 23 a33

3x3

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. 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: Dimensions and Types of Matrices:

A. A row matrix:

E .g. A =

2

is a matrix that has only one row and can have many many columns.

5 7

B.A column matrix:

E.g. B =

is a row matrix of order 1x3.

is a matrix with one column and can have many rows. 1 2 6

is a column matrix of dimensions 3x1. is a matrix with equal number of rows and columns. 1 4 3 6 ; D= 2 6 E= 2 2 5 3 8 ; 8 6 9

C.A square matrix:

E.g. C =

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

E.g. x =

2

0

0


19

Financial Management in Value Chain: Business Mathematics and its Application 0 0

6 0

0 11

is a diagonal matrix of order 3x3.

a square matrix is called a scalar matrix if all its non- diagonal elements are zero and all diagonal elements are equal. 6 0 0 E.g. Y = 2 0 Z= 0 6 0 0 2 0 0 6

E. A scalar matrix:

F. A unit matrix (Identity matrix):

is a type of diagonal matrix where its main diagonal

elements are equal to one. 1 0 0 0 1 0 0 0 1

E.g. B =

G. A null matrix (zero matrix):

0 E.g.

A=

0 0

0

a matrix is called a null matrix if all its elements are zero.

0 0

0

0

0

H. A symmetric matrix: matrix: a matrix is said to be symmetric if A = A t.

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.

E.g. If

A=

2

1

3

3

; then AA= A2 =

2

1

3

3

2

1

2

1

3

3

3

3


20



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 matrix need not be a scalar scalar matrix. matrix. Every Every unit matrix matrix is a scalar scalar matrix; matrix; whereas whereas a scalar matrix need not be a unit matrix. Matrix Operations and Properties:

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

1.

Matrix equality:

3 E.g.

0

3

A=

-4

3

B= 1

-4

0

C= 1

0

1

-4

A ≠ B; A = C; B ≠ C. If the rows and columns of a matrix are interchanged the new matrix is known as the transpose transpose of the original matrix. matrix. If the original matrix matrix is denoted by A, the transpose is denoted by A′ or At. Transposition Transposition means interchanging interchanging the rows or columns of a given matrix. That is, the rows become columns and the columns become rows. 2. Transpose of a matrix:

E.g

B=

3 5 6 9 0 11 13 8 6 8 3 4

The transpose of matrix B, denoted by

t

B=

3 5

0 11

B '

or Bt is given as:

6 8


21

Financial Management in Value Chain: Business Mathematics and its Application 6 9

13 8

3 4

The dimension of B is changed from 3x4 to 4x3. A′ = A= 1 3 1 0 4 (3x2) 3 2 8

0 4

2 8

(2 X 3)

The following properties are held for the transpose of a matrix: t t  Property 1: (A ) =A t t t  Property 2: (aA) = aA , where (a) is a scalar (a = a) t t t  Property 3: (A+B) = A + B t t t  Property 4: (AB) = B A 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; 3. Addition and subtraction of matrices:

2+3 0+6 -5+4 6+1

A+B =

1 6

If A = 8

5 7

B=

5 =

10 8

-1

6 7

2 6

9

A+B is not defined, since orders of A and B are not the same. 2 A-B=

1

3 0

-

4 3 2-4 3-3 2 1 = 1-2 0-1 =

-1

-2 -1

0


22

Financial Management in Value Chain: Business Mathematics and its Application 4. Matr Matrix ix Mult Multip ipli lica cati tion on

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. A=

1 2 3 4 0 1

Then, A x B =

2 1 4 3 0 5

B= (3x2)

(1x2) + (2x3) (3x2) + (4x3) (0x2) + (1x3)

=

8 1 15 3 3 0

(2x3)

(1x1) + (2x0) (1x4) + (2x5) (3x1) + (4x0) (3x4) + (4x5) (0x1) + (1x0) (0x4) + (1x5)

14 32 5 (3x3)

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.

Solved problems:

Executive chairs

Cities

A C1 400 C2 300

B 300 200

C 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 total cost of the factory factory for for the total total sale made. b) Find the total total profit profit of the the factory. factory. Solution:

Given: Let the quantity matrix be q Let the price matrix be p Let the unit cost matrix be v Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

23

Financial Management in Value Chain: Business Mathematics and its Application 400 300 200 q=

p= 300 200 100

1500 3000 4000

V = 1000 2000 3000

Total cost = (unit cost) (Quantity) = 400 300 =

300 200 200 100

*

1000 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 * 300 200 100

=

2,300,000 1,450,000

1500 3000 4000

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



24

Financial Management in Value Chain: Business Mathematics and its Application Dear student, student, having seen the properties 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 Compute the total total interest interest by matrix matrix multipli multiplication. cation. Determinant of a Matrix:

Let A = a11 a12 a21 a22 A = a11 a21

( 2x2)

a12 a22

and its value is given as: E.g.

A= 6 4 7 9

3. Let

A

A

A=

=

A

= a11a22 - a12a21.

A

= 6 7

;

4 9

= 6(9)-7(4)=26

a11 a12 a13 a21 a22 a23 a31a32 a33

a11 a21 a31

= + a11

is known as a determinant of order two

a22 a23 a32 a33

a12 a13 a22 a23 a32 a33

- a12

is called a third order determinant

a21 a23 a31 a33 +a13

a21 a31

a22 a32

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

E.g. Let A =

1 2 4 0 -1 0


25

Financial Management in Value Chain: Business Mathematics and its Application -2 A

=

0

1 2 0 -1 -2 0

4 0 3

3

;

-1 = +1 0

Find 0 3

A

-2

. 0 0 -2 3

0 -1 + 4 -2 0

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

Inverse of a Matrix:

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 (R 1) by -2 to get matrix B. Then,

B =

-4 -6 -6 6 9

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

3 3

B. Add a multiple of one row to another row; If A = 1 2 multiply R 1 by 2 and add to R 2 to get matrix x. Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

26

Financial Management in Value Chain: Business Mathematics and its Application 3

4

Matrix X =

1 5

C.

2 8

Interchanging of rows;

If A = 1 2 D=

2 1

0 4

Interchange R 1 and R 2 ( R 1 ↔ R 2 ) ; to get matrix D.

4 0

Note:

The first row elements in the original matrix become second row elements in the new matrix and vice versa. The most important important methods to find inverse of a given matrix is GaussJordan Inversion Inversion method. This method was developed by a mathematician 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 1

2 1

Solution

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

A/I

=

3 1

2 1

1 0 0 1

Second : Interchange R 1 and R 2;2; 3 2 1 0 1 1 1 0 1 3

1 2

0 1 1 0

Third: Multiply R 1 by -3 and add the result to R 2; -3 R 1 = -3 -3 0 -3 Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

27

Financial Management in Value Chain: Business Mathematics and its Application + R 2

=

1 0

1 -1

3 2 1 0 0 -1 1 -3 The resulting matrix is given by: 0 1

1 -3

Fourth: Simply add R 2 entries to R 1 entries; R 2 = 0 -1 1 -3 R 1 = 1 1 0 1 1 0 1 -2 The resulting matrix is given by: 1 0

0 -1

1 1

-2 -3

Fifth: Multiply R 2 by -1; (-1) (R 2) = 0 1 - 1 3 The resulting matrix is given by;

1 0

0 1 1 -1

-2 3

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

A.A-1 = A-1. A = I 3 2 1 -2 1 1 * -1 3

Check!

=

=

1 0

0 1


28


A system of linear equations can be solved by the following method using using matrix matrix algebra algebra:: Cramer Cramer’s ’s rule rule (the determinan determinantt method) method):: works works accordi according ng to this this formula:

Matrix Applications: Applications :

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 denoted by A

column vector of variables (X)

column vector of constants (B)

Using Cramer’s rule, the solution is given by: b1 a12 a11 b1 X= b2 a22 ; and Y= a21 b2 A

A

Example; x- y = 1 = 1 -1 x x+ y = 2 1 1 * y

=

1 2

→ matrix expression

Then, x= 1 -1 2 1 A

=

1 2

-1 1 =3/2


29


1

-1 1

1

y= 1 1

1 2

=

A

1 1

1 2

1

-1

1

=1/2

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 a21 a22 a23 . a31 a32 a33 A

x y z X

=

b1 b2 b3 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 a21

a12 a22

b1 b2

a31 a11

a32 a22

b3 a13


30

Financial Management in Value Chain: Business Mathematics and its Application a21 a31

a22 a23 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 A

X

= +2 1 1

Thus; x =

0 0 1

1

B

-1

-1

1

1

+ (-1) 1

0 -1

0

1

1 = -4

1 -1 1 1 = 2/-4 = -1/2 1 -1 A

Y=

2 1 0

0 -1 0 1 = 1 -1

-3/-4

= 3/4

=

-1/4

A

Z=

2 1 0

1 0 1 0 1 1 A

Continuous Assessment

The assessment methods for this criterion include tests, quiz, exam, assignments, case analysis and group works Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

31

Financial Management in Value Chain: Business Mathematics and its Application 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 matrix A is the inverse of matrix 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 2

5. If A = 1 2 0

-1 3 1

2 4 3

4 1

and B =

-2 0 1 2 -1 4 1 1 2

Find A-B and B-A. 7. Given the matrices:

A=

1 0

-2 3

1

3

0


32

Financial Management in Value Chain: Business Mathematics and its Application 0

4

B=

2

0 -1 ; Determine where possible:

a) AB b) BA c) 2A 8. Verify whether AB = BA for matrices: A= 2 1 0

iii.

1 -1 -1 1

0 2 3

and B =

1 1 1

2 -1 -1 2 1 2

Section II III: Li Linear pr program ramming

Dear learner! Have you ever experienced the problem of allocating scarce resource to various acti activi viti ties es to atta attain in the the pred predet eter ermi mine nedd go goal al?? If so, so, in wh what at wa wayy you you appr approa oach ched ed the the allocation?............ allocation?................................ ....................................... ........................................ ........................................ ....................................... ....................................... ........................ ..... ...................................... .......................................................... ....................................... ........................................ ........................................ ..................................... ............................... ............. Introduction Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

33


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 difficulty is experienced. experienced. Nevertheless, Nevertheless, such cases are very rare. Practically in all situations, the managements are confronted with the problem of scarce resources. resources. Normally, Normally, there are several activities to perform perform but limitations 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 acti activa vate tes. s. Thes Thesee deci decisi sion on prob proble lems ms can can be form formul ulat ated ed and and solv solved ed as math mathem emat atic ical al programming programming problems problems and and hence knowhow knowhow of of linear programming programming is is imperative. imperative. Definition: Linear Programming (LP): o

o

Is a mathematical process that has been developed to help management in decisionmaking involving the efficient allocation of scares resources to achieve a certain objective/s. 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. satisfied. a. Ther Theree shou should ld be an ob obje ject ctiv ivee that that shou should ld be clea clearl rlyy iden identi tifi fied ed and and meas measur urab able le in quantitative terms. Example, maximization of sales, profit, and minimization of costs etc. b. The activities activities to to be included included should should be distinctly distinctly identifia identifiable ble and measurabl measurablee in quantitative quantitative terms. c. The resources resources of the system system which which are to to be allocated allocated for the attainment attainment of the goal goal should 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 objective. To of bethe allocated Scares d. The rela relati tions onshi hipp repre represe senti nting ng the the objec objecti tive ve and and the reso resour urce ce limi limita tati tion on cons conside iderat ration ionss to: Resourc represe sente ntedd by the the obje object ctive ive funct function ion and the the const constra raint int equ equat ation ionss or inequa inequali liti ties es,, e repre respectively, must be linear in nature. e. There should should be a series series of feasible feasible alternati alternative ve courses of of actions availabl availablee to the decisiondecisionResource maker that is determined by the resource constraints. constrain problemtscan be expressed

When these stated conditions are satisfied in Constrain a given solution, the in Objectiv alge algebra braic ic form form es calle calledd line linear ar progr program ammi ming ngtsprob proble lem m (LPP (LPP), ), and then then solve solvedd for for optim optimal al decision. Nonnegativity Diagrammatically Linear Programming (LP) model can be presented as shown below. Constraints Optimization Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

34

Maximizatio

Minimizatio


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. Problem Formulation or Modelling is the process of translating translating the verbal statement statement of a problem problem in to mathematical mathematical statements. 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. Formulat Formulation ion of linear linear programm programming ing model: model:

A firm is engaged in producing two products, A and B. Each unit of product A requires requires two kgs of raw material material and four labor hours for processing whereas 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 maximum the profit. Assume that there is no marketing constraint so that all that is produced can be sold. The Maximization Case:

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 The objective function:


35

Financial Management in Value Chain: Business Mathematics and its Application problem, problem, the goal is the maximization maximization of profit, profit, which would be obtained by producing producing (and selling) the products A and B. If we let x 1 and x2 represent the number of units produced per week of the products A and B respectively, respectively, the total profit, Z, would be equal to 40 x1 +35x2 is then, the objective function, relating the profit and the output level of each of the two items. Notice that that the function function is a linear linear one. Further, Further, since the the problem calls for a decision decision about the optimal values of x1 and x2, these are known as the decision variables. 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, constraint, inequality rather 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. The constraints:

Quite obviously, x1 and x2, being the number of units produced, 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 can write write the problem problem in complet completee form as follows. follows. Non-negativity Condition:

Maximize

Z = 40 X1 + 35 X2,

Profit

Subject to 2X1 + 3X2 < 60 4X1 + 3X2 < 96 X1, X2, > 0

Raw material constraint Labor hours constraint Non-negativity restriction

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 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 this as a linear programming programming problem and determine determine how many bags of each type the farmer should buy in order to obtain the required fertilizer at minimum cost. The Minimization Case:


36

Financial Management in Value Chain: Business Mathematics and its Application 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: The Objective Function:

Minimize Z = 40x1 + 24 x2

Cost

In this problem, there are two constrains, constrains, namely, a minimum minimum of 4800 kg of phosphate and 7200 kg of nitrogen nitrogen ingredients ingredients are required. 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 requirement can be expressed expressed as 20x1 + 50x2 > 4800. Similarly, with the given information on the contents, the nitrogen requirement would be written as 80 x1 + 50x2 > 7200. The constraints: constraints:

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:

Non-negativity Non-negativity condition:

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 < b1 a21x1 + a22 x2 + ………………………..+ ain x n < b2 am1 x1 + am2 x2 + ............................+ amn < xn bm x1, x2, ……………………….., ……………………….., xn > 0 Where

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

Note:


37

Financial Management in Value Chain: Business Mathematics and its Application 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. 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 assumed to contribute contribute Birr Birr 10 toward profit, profit, then the total total contribution 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, proportionalit proportionalityy means that there there are constant returns returns to scale scale and there are no economies economies of scale.

1. Prop Propor orti tion onal alit ity: y:

Another assumption underlying the linear programming model is that in the objective objective function and constraint constraint inequalities 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.

2. Addi Addittivel ively: y:

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.

3. Con Contin tinuit uity:

A further assumption underlying a linear programming model is that the various parameters, namely, the objective function coefficients, the coefficients of the inequal inequality/ ity/equa equalit lityy constra constraints ints and the constra constraint int (resour (resource) ce) values values are kno known wn with 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.

4. Certainty:


38

Financial Management in Value Chain: Business Mathematics and its Application 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!

5. Fin Finite ite Cho Choices ices::

Solution approaches to LPPS: Now we shall consider consider the solution solution to the linear programming programming

problems. problems. They can can be solved solved by using graphic method method or by applying applying algebraic algebraic method, method, called 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 Graphial Solution Solution to Linear Linear Progra Programmin mming g Problem Problems: s:

i. ii. iii. iv.

To use the graphic method, the

following steps are needed: Identify the the problem – de determine th the de decision va variables, the the objective fu function, an and the constraints. Draw a graph including all the constraints and identify the feasible region. Obta btain a po poin intt on the feasible region that hat optim ptimiizes the obje bjectiv ctivee func functtion – opti ptimal solution. Interpret the results. Note: Graphical LP is a two-dimensional model. a) Maximization 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 Availability of only 40 hrs of labor labor each day day in the the production production department department,, b. A daily availability availability of only only 45 hrs on machine machine time, time, and c. Ab Abil ilit ityy to sal salee 12 set set of of mode modell A. How many sets of each model will be produced each day so that the total profit will be as large as possible? Resources used per unit Constraints

Labor Hours

Model A (x1) 2

Model B (x2) 1

Maximum Available Hours 40


39

Financial Management in Value Chain: Business Mathematics and its Application Machine Hours

1

3

45

Marketing Hours

1

0

12

$300

$250

Profit

Solution:

1. Formulation of mathematical model of LPP Max Z=300 X 1 +250 X 2 St: 2 X 1 + X 2< 40 X 1 +3 X 2< 45 LP Model X 1 < 12 X 1 , X , X 2 > 0 2. Convert constraints inequalities into equalities 2 X 1 + X 2 = 40 X 1 + 3 X 2 = 45 X 1 = 12 3. Draw the graph by finding out the x– and y–intercepts 2 X 1 + X 2 = 40 ==> ==> (0, (0, 40) 40) and and (20, (20, 0) 0) X 1 +3 X 2 = 45 ==> ==> (0, (0, 15) 15) and and (45, (45, 0) 0) X 1 = 12 12 ==> ==> (12 (12,, 0) 0) X 1 , X 2 = 0

= 2

X

+ 1

X = X 0

X

1

2

2 4

0

40

5 4

X =12 1

= 2

X

+ 1

15

X

B

Feasibl

C(12, 11)

e

X =0 2

X D 1 Jimma, Harama Haramaya, ya, Ambo, Amb Hawassa, Adama, Adama,2Bahirdar, Bahirda r, W.Sodo,Sema W.Sodo,Semara Universities ies A 12 o, Hawassa, 0 45 ra Universit

Region

40

Financial Management in Value Chain: Business Mathematics and its Application 4. Identify Identify the feasible feasible area of the solution solution which satisfi satisfies es all constrains constrains.. The shaded shaded region in the above graph satisfies all the constraints and it is called Feasible Feasible Region. 5. Identif Identifyy the corner points points in the feasibl feasiblee region. region. Referring Referring to the above above graph, the corner corner points are are in this this case are: are: A (0, 0), B (0, 15), C (12, 11) and D (12, 0) 6. Ident Identif ifyy the the opti optima mall point 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. Interpre Interprett the result. result. Accordi Accordingly ngly,, the highlig highlighted hted result result in the table table above implies implies that 12 units of Model A and 11 units of Model B TV sets should be produced so that the total profit will will be $6350. $6350. b) Minimiza Minimization tion Problem Problem:: In this case, we deal with Minimiz Minimizee 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

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 Mod Modeel: Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

41


Min. Z =

25 X 1+30 X 2

St :

20 X 1+15 X 2≥ 100 100 2 X 1+3 X 2≥ 15 X 1 , X 2 ≥ 0

LP Model Model

Dear student, can you graph the above constraints? See step 2 below. 2. The Graph Graph of of Const Constrai raint nt Equat Equations ions:: 20X1 +15X2=100 ==> (0, 20/3) and (5, 0) 2X1 + 3X2=15 =15 ==> ==> (0, (0, 5) and and (7.5, (7.5, 0) X1 , X2 = 0 X

1

X

2

=0

A (0, 20/3)

Feasible Region X

2

B (2.5, 3.33)

=0 X

1

5

C (7.5, 0)

3. The Corners Corners and Feasibl Feasiblee Solut Solution: ion: Corners

Coordinates

Min Z= 25 X1 + 30X2

A

(0, 20/3)

B

(2.5, 3.33)

C

(7.5, 0)

200 162.5 (Optimal)

187.5

Since our objective is to minimize cost, the minimum amount (162.5) will be selected. X1 = 2.5 X2 = 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). Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

42

Financial Management in Value Chain: Business Mathematics and its Application -

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 MEDIUM and LOW grade flour. This company has entered contract supply flour to a firm firm ever everyy we week ek with with 12 12,, 8, and and 24 qu quin inta tals ls of HIGH, MEDIUM 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, HIGH, MEDIUM and LOW grade flour respectively. Mill B produces 2, 2 and 12 quintals of HIGH, HIGH, MEDIUM 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. The graphical method to solving LPPs provides fundamental concepts for fully understanding the LP process. However, the graphical method can handle problems problems involving involving only two two decision decision variables 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. Algebraic Simplex Method:

a) Maxim Maximiza izatio tion n Proble Problems ms

Maximize Z with inequalities of constraints in ‘<’ form Solve the problem using the simplex approach Max. Z = 300x1 +250x2 Subject to: 2x1 + x2 < 40 (Labor) x1 + 3x2 < 45 (Machine) x1 < 12 (Marketing) x1, x2 > 0 Solution: Step 1 Formulate Formulate LP Model: It is already given in the form of linear programming model. Step Ste p 2 Stand Standar ardiz dizee the the proble problem: m: Co Conve nvert rt con const stra raint int inequa inequali lity ty into into equal equality ity form form by introducing a variable called Slack variable. A slack variable(s) is added to the left hand side Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

43

Financial Management in Value Chain: Business Mathematics and its Application of a < constra constraint int to covert covert the constra constraint int inequalit inequalityy in to equalit equality. y. The value of the slack variable shows unused resource. A slack variable emerges when the LPP is a maximization problem. problem. Slack variables variables represent represent unused resource resource or idle capacity. 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

Standard Standard form

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, Thus, first first assumption, assumption, No No production production implies 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. s 2= 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 Note:

In general, whenever there are n variables and m constraints constraints (excluding (excluding the non-negativity) 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 Basic variabl variables es are variab variables les with with non-zer non-zeroo solution solution values values.. Or: Basic variables are variables that are in the basic solution. Basic variables have 0 values in the C j – Z j row . Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

44


b. Non-basic Non-basic variables variables are variables variables with with zero solutio solutionn 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. 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 coefficients in the constraint constraint equations equations are written listing vertically vertically the coefficients 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. Step 4 Construct the initial simplex tableau:

n Initial simplex tableau o i r e n p m u l t o c i f t i o r n P u

s e l b a i r a v s n k m c u a l l o S c

300

250

0

SV

X1

X2

S1

S1

2

1

C j

0

s n n i o c i t m e u l u d s l o o r e n S c o l b m r e l i a u o l a r l b e a c o a i i s r R v c a a B v

0 S2

1

0 S3

0

n m u l o c y n t o i i t t u n a l o u S q

Q

0

40

0

S2

1

3

0

1

0

45

0

S3

1 0

0 0

0 0

0 0

1 0

12 0

Z j

Profit per unit row

R1=20 R2= 45 R3=12

Constraint

Leaving Row

equation

Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies rows

45

Gross Net row

Profit Profit

row /Indicator /Indicator

Financial Management in Value Chain: Business Mathematics and its Application C j - Z j

300

250

0

0

0

Entering Column (Pivot Column)

Step 5 Choose Choose the “incomi “incoming” ng” or “entering “entering” ” variabl variables:: es:: The entering variable is the variable that has the highest positive value in the C j - Z j 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 associated with the entering entering variable is called key or pivot column ( X1 column in our case ) Step 6 Choose the “leaving “or “outgoing” variable: that will leave the solution for X1 (or entering variable)

In this step, we determine determine the variable

Note:

• The row with the minimum or lowest positive (non-negative ) replacement replacement ratio shows the

variable to leave the solution. Replacement Ratio (RR) = Solution

Quantity (Q)

Corr Corres es ondi ondin n

valu values es in

ivot ivot colu column mn

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 No 1 in our case) element( No

For improved solution using the information obtained earlier, another tableau is derived where in the various elements are obtained as given here. a. Divi Divide de each each elem elemen entt of the the pivo pivott row row (inc (inclludin udingg bi) bi) by the the pivo pivott elem elemen entt to get get the the corre correspo spondi nding ng valu values es in the new tableau tableau.. The row row of value valuess so deri derive vedd is calle calledd the the replacement raw. Step 7 Derive the revised tableau:

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 . zero . Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

46

Financial Management in Value Chain: Business Mathematics and its Application 2nd simplex table

300

250

SV

X 1

X 2

0

S 1

0

0

S 2

0

C j

0

0

0

S 1

S 2

S 3

1

1

0

-2

16

3

0

1

-1

33

1 300 0

0 0 250

0 0 0

0 0 0

1 300 -300

12 3600

300

250

0

0

SV

X 1

X 2

S 1

S 2

S 1

0

0

1

-1/3

-5/3

0 1 300 0

1 0 250 0

0 0 0 0

11//3 0 250/3 -250/3

-1/3 1 650/3 -650/3

300 X 1 Z j C j - Z j

Q

R’1=16

R’2=11 R’3=Undefined

3rd simplex table C j

0

250 X 2 300 X 1 Z j C j - Z j

0 S 3

Q

5 11 12 6350

Elementary Row Operations R’’ 1=R’ 1-R’ 2 R’’ 2=R2 /3 R’’ 3=R’ 3

Since all the C j - Z j < 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 minimization type is similar to the one for the maximization maximization problems, except for some differences. To illustrate, let us again consider the examples. Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

47


Minimize Z= 40x1 + 24x2 Subject to: 20x1+ 50 x2 > 4800 80 x1+ 50 x2 > 7200 X1, x2 > 0

Total cost Phosphate Requirement Nitrogen Requirement

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 experience a problem, 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 x 1 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). 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 artificial variables are introduced in our example, it appears as follows,

20x1+ +50-S1+A1= 4800 80 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 Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

48

Financial Management in Value Chain: Business Mathematics and its Application for the constra constraints ints with > type, type, or the constrai constraints nts “=” sign. sign. It is not releva relevant nt whe whether ther 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 systems and must not show in the final solution (and if at all they do, it represents 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 artifi artificia ciall varia variabl bles es.. On the other other han hand, d, for for the the proble problems ms with with the objec objecti tive ve funct functio ionn of maximization type, each of the artificial variables introduced has a coefficient –M. For our present example, example, the objective objective function function would would appear as 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 Basis

40 X1

24 X2

0 S1

0 S2

M A1

M A2

A1 M A2 M Zj Cj - Zj

20 80 40

50 50 24

40 − 100 M

24 − 100 M

-1 0 0 M

0 -1 0 M

1 0 M 0

0 1 M 0

Qty

RR

4800 7200

96 144

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 Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

49

Financial Management in Value Chain: Business Mathematics and its Application 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 Basis X2 24

40 X1 2/5

24 X2 1

0 S1 -1/50

0 S2 0

M A1 1/50

M A2 0

Q 96

RR 240

A2 M Zj

60* 40

0 24

1 0

-1 0

-1 M

1 M

2400 2400

40

Cj –Zj

152/560M

0

12 25- M

-M

12 2M- 25

0

0 S2 1/150 -1/60 0 38 75

M A1 2/75 -1/60 M M+ 38 75

M A2 -1/150 1/60 M M+ 38 75

Q 80 40 3520

RR -3000 2400

0 S2 -1/150 -1 0 12/25

M A1 0 -1 M M

M A2 1/50 1 M M-12/5

Simplex Table 3 : Non optimal Solution Cj Basis X2 24 x2 40 Zj Cj – Zj

40 X1 0 1 40 0

24 X2 1 0 24 0

0 S1 -2/75 1/60* 0 -2 75

Simplex Table 4 : Optimal Solution Cj Basis X2 24 x2 0 Zj Cj – ZJ

40 X1 8/5 60 40 8/5

24 X2 1 0 24 0

0 S1 0 1 0 0

Q 144 2400 3456

According to the optimal solution, the objective function value is Z=40x0+24x144+0x2400+0x0xM+0xM=Birr 3456


50

Financial Management in Value Chain: Business Mathematics and its Application The value of S1 = 2400 indicates the surplus phosphate ingredient obtained by buying the least cost mix. Learning Activity 3.2

Find the optimal solution using simplex method. 1. Min Min Z=1 Z=100 x1 +5 x2 Subject to: 2x1 + 5 x2 > 150 3x1+ x2 > 120 x1, x2 > 0 Continuous Assessment


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:

Types of constraint

Extra variables to be added

<

Add only slack variable Subtract surplus variable and

> Add artificial variable =

Add artificial variable

Coefficient of extra variables in the objective function MaxZ MinZ 0 0 0 0 -M +M -M +M

Presence of variables in the initial solution mix

2. Te Test st of of opti optim malit ality y Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

51

Yes No Yes Yes

Financial Management in Value Chain: Business Mathematics and its Application If all C j - Z j < 0, then the basic feasible solution is optimal (Maximization case). If all C j - Z j > 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 C j - Z j row (Maximization case) A variable that has the highest negative value in the C j - Z j row (Minimization case) N:B-‘Highest N:B-‘Highest negative’ negative’ values in in this case case refers refers to the the negative negative value which which is far far from from zero on the number line!

4. Variable to leave the basis

-

iv. iv.

The row with the non-negative and minimum replacement ratio (For both maximization and minimization cases i.e: RR > 0 Sec Sectio tion IV: IV: Mathem themat atic icss of of Fin Fina ance

Dear student! What do you know about mathematics of finance? Why we need to learn mathematics of finance? ____________________ _____________________________ ____________________ ______________________ ___________________ ___________________ _________________ ______ ____________________ _____________________________ ____________________ ______________________ ___________________ ___________________ _______________ ____ Introduction

Mathe Mathema mati tics cs of fina finance nce is conce concerne rnedd with with the the analy analysi siss of time time-va -value lue of money money.. The fundamental premise behind such analysis is the concept that entails the value of money 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 mostly pecuniary. pecuniary. Such matters matters as lending and borrowing 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 Cognizant of this fact, we proceed to the study of mathematics of finance in this section. Simple interest and discounts: discounts:

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

I = pin Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

52

Financial Management in Value Chain: Business Mathematics and its Application Where, I = Simple interest (in dollars or birr) 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 12 ii. 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. given. Thus, Thus, Future value (F) = Principal (P) + Interest (I) A credit union has issued a 3 year loan of Birr 5000. Simple interest 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 will be repaid repaid at the end of the the third year? year?

Example:

Solution:

Given values in the problem 3 – Years loan = Principal = Birr 5000 Interest rate = i = 10% = 0.1 Number Number of years years (n) = 3 years a. I = p i n I = 5000 x 0.1 x 3 I = Birr 1500 Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

53

Financial Management in Value Chain: Business Mathematics and its Application b. The amount to to be repaid at the end end of the third year year is the maturity maturity (future) (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 + Pin F = P (1 + in) Consequently, using this formula we can obtain F = 5000 (1+ (0.1x3) F = 5000 x 1.3 F = 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 The Exact Exact Method Method:: if a year year is consid consider ered ed as 365 days, days, the intere interest st is calle calledd exact exact simple interest. If the exact method is used to calculate interest, then the time is n = number of days / 365 ii. The Ordinary Ordinary Method Method (Banker’s (Banker’s Rule): Rule): if a year year is considered considered as 360 360 days, the interest 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 exac exactt met metho hodd b. The ordinary ordinary method method

Solution:

Given P = Birr 1460 n = 72 days i = 10% = 0.1

a) P=1460 n=72/365 i= 0.1 I=Pin =1460*72/365*0.1=28.8 =1460*72/365*0.1=28 .8

b) P=1460 n= 72/360 i=0.1 I=Pin= 1460*72/365*0.1=29.2

Simple Discount: Present Value Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

54


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 present value formula 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. P = F

1 + in 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. 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 %?

Example:

Solution:

1. Give Givenn valu values es in in the the probl problem em,, 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. P = F

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

Promissory Notes and Bank Discount:


55

Financial Management in Value Chain: Business Mathematics and its Application 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 exclusive right either to sell, borrow, or discount 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 determining the charge for use of money, we say that the promissory promissory note (or simply the note) is discounted. discounted. Consequently, 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 borrower receives is called ‘Proceed.’ The amount that the borrower is going to pay to the creditor 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 Where, P = Pr Proceed F = Maturity value D = Bank discount d = Rate of discount t = Time of discount Now we can can further further elaborate elaborate the above above formula formula for proceed. proceed. To begin with, 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, proceeds, P, and pays back F = Birr 1000. 1000. The proceeds proceeds will be Birr 1000 1000 minus the the interest interest on Birr 1000. This will be: P

=

10 1000 00 – (100 (10000 x 0.12 0.12 x 6/12 6/12)) = Birr Birr 940 940


56

Financial Management in Value Chain: Business Mathematics and its Application Or,

P = 1000 (1 – (0.12 x 6/12)

P = Birr 940

i. Proceed Proceedss are an amount amount receive receivedd now for payment payment in the the futur future. e. There Therefore fore,, they they are analogous to present value. Yet, proceeds are not equal to present value because the proceeds proceeds from a futures obligation 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. Proceed Proceedss sho should uld be compl complete etedd when when the the inte interes restt rate rate is is stat stated ed by by the the qualif qualifier ier word 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. iii. The compu computa tati tion on of simpl simplee inter interes estt and bank bank disc discou ount nt is the the same same except except in the 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. 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.

Example:

Solution: Given values in the problem

F = Birr 480 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 P = 480 – 4.8 or P P = Birr 475.2 or P

= F (1 – d t) = 480 (1 – (0.04 x 0.25) = 480(0.99) 480(0.9 9) = Birr 475.2



57

Financial Management in Value Chain: Business Mathematics and its Application 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 bank Birr 5000 5000 ten months months from from now. a. How much much will will the borro borrower wer receiv receivee if the discou discount nt rate rate is 6%? b. How much would the borrower borrower have to repay in order to receive 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 within a year is/are is/are used used in order to to find the interest interest rate per compounding compounding periods 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 computed by the formula 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 interest of 6% per year compounded compounded yearly. Assume Assume that we want to determine 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


58

Financial Management in Value Chain: Business Mathematics and its Application m = compounded annually = i.e. only once n=mxt =1x2 = 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 An individual accumulated Birr 30,000 ten years before his retirement in order to buy a house after he is retried. retried. If the person invests this money at 12% compounded compounded monthly, how much will be the balance immediately after his retirement? Example:

Solution:

Given values, P = Birr 30,000 , m = compounded monthly = 12 i = j / m = 0.12 / 12 = 0.01 n = m x t = 12 x 10 = 120 Then, F = = = F =

t = 10 years ,

i = 12% = 0.12

and what is required is the the Future Value F.

P (1 + i) n 30,000 (1.01)120 30,000 (1.01)120 Birr 99,011.61

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 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, Conversely, present value (or simply principal) 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 . . .

n


59

Financial Management in Value Chain: Business Mathematics and its Application Present 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. P =

F n

(1 + i )

n

= F n (1 + i) n

n F n = P (1 + i)

Examples:

1. What is the present value of a. Birr Birr 5000 5000 in 3 year yearss at 12% 12% compou compounded nded annu annuall ally? y? b. Birr 8000 8000 in 10 years years at 10% 10% compounded compounded quarterly? quarterly? 2. Suppos Supposee that a person person can inves investt money money in a savi saving ng account account at a rate of 6% per year year compounded quarterly. Assume that the person wishes to deposit a lump sum at the beginning of the year year and have that some grow to Birr 20,000 20,000 over the next 10 years. How much money should be deposited at the beginning of the year? 3. A young man man has recentl recentlyy received received an inheritan inheritance ce of birr 200,000 200,000.. He wants to make make a portion of his inheritance inheritance and invest it for his late years. His goal is to accumulate accumulate Birr 300,000 in 15 years. How much of the inheritance should be invested if the money will earn 8% per year compounded semi-annually? How much interest will be earned over the 15 years? Solution:

1. (a) (a) Give Givenn the the valu values es,, Fn = F3 = Birr 5000, 5000, t = 3 years m = 1 (compounded (compounded annually) n = t xm = 3x 1= 3 j = 12 % = 0.12 i = 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


60

Financial Management in Value Chain: Business Mathematics and its Application 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 Given the values, values, i = 6% = 0.06 0.06 , m = quarter quarter = 4 times times a year i = j ÷ m = 0.06 ÷ 4 = 0.015 F = Birr 20,000 shall be accumulated t = 10 years n = 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. Inher Inherit itanc ancee = Birr Birr 200 200,00 ,0000 Fn = Birr 300,000 (the person's goal of deposit) , t = 15 years , j = 8% = 0.08 m = semi-annual = 2 times a year i = j ÷ m = 0.08 ÷ 2 = 0.04 n = t x m = 15 x 2 = 30 interest periods/semi-annuals P = how much much of the inheritance should be invested now? P = Fn(1 + i)-n I = Amount of interest? = 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 – 92,490 Amount 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 The per perio iodi dicc pay payme ment ntss are are equa equall in in amo amoun untt


61

Financial Management in Value Chain: Business Mathematics and its Application ii. ii. iii. iii. iv. iv.

The tim timee betwe between en payme payments nts is is const constant ant suc suchh as a year, year, half half a year, year, a quart quarter er of a year, a month etc. The inter interes estt rate rate per period period rema remains ins const constan ant. t. The inter interes estt is compou compounde ndedd at the end of every every time time..

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

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

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 important terminologies terminologies that we are going to use in dealing with annuities. i. Paym Paymeent int inteerval rval or or peri period od:: it is the the tim timee betw betwee eenn succ succes essi sive ve pay payme ment ntss of an an annuity. ii. Term of ann annui uitty: it is is the the per period iod or tim timee int interval bet between the the beg beginni nning of 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. iv. Peri Period odic ic paym paymen ent/ t/re rent nt:: itit is is the the amou amount nt paid paid at the the end end or the the beg begin inni ning ng or each each payment period. v. Simple ple annu annuiity: is the the one one in in whi which the the pay payment pe period aand nd the the co conve nversion periods coincides coincides each other. other. Sum of Ordinary Annuity: Maturity Value

Maturity Maturity value of ordinary ordinary annuity is the sum of all payments 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 Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

62

Financial Management in Value Chain: Business Mathematics and its Application the annuity is four years with annual interest rate of 10%. Thus, we can show the pattern of deposits diagrammatically as follows. The Present 0

st

1 1000

nd

rd

2 1000

3 1000

The Future 4th 1000

Birr 1000 Birr 1100 Birr 1210 Birr 1331 Total Future Value = Birr 4641 The first payment earns interest for the remaining 3 periods. 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 The 3rd payment earns interest for only one period. So, F=1000(1+0.1)1 = 1100 No interest 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).  (1 + i) n - 1 Fn = R   i  

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 consider the above above example. example. That is, R = Birr 1000 i = 0.1 and n=4 Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

63


F 4

 (1 + 0.1) 4 - 1 = 1000   0.1  

Future Value = Birr 4641 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? Example:

Solution:

The known values in the problem are, R = 1000, j = 6% = 0.06 , i = 0.06 ÷ 2 = 0.03 n=4

m = semi-annual = twice a year

 (1 + i ) n − 1  F4 = R   i  

F4 = ?

 (1 + 0.03) 4 − 1  = 1000   0.03  

= 1000 (1.03) 4 − 1 / 0.03 = 1000 x 4. 183627 F = 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 Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

64

Financial Management in Value Chain: Business Mathematics and its Application 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.

(1 + i ) n −1  F n = R   i   Multiplied both sides by That is i n  (11+ i ) n − 1 (1i +i ) − i F n × = R  ×  n n (1 + i ) − 1 i   (1 + i ) − 1

Then, 

  n  (1 + i ) − 1

R = F n 

Where,

i

is the sinking found formula.

R = Periodic payment amount of an annuity i = Interest per period which 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) 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 interest at the rate of 8% per year compounded 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? Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

65

Financial Management in Value Chain: Business Mathematics and its Application Solution:

1. Future Future level level of depos deposit it desi desire redd = Fn = Birr 12 million Term of the annuity = t = 10 years Conversion 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 amount to be deposited deposited each year to to have 12 million million at the end of the the 10th year = ? Then to obtain the value of R, we shall use the formula for sinking fund. R

  i = F n   n (1 +i ) −1 



  10  (1 + 0.08) − 1 0.08

R = 12,000,000

 0.08  R = 12,000,000  10  (1.08) − 1

 0.08  1.158925 

R = 12,000,000 

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 calculation arise when we wish to determine what lump sum must be deposited deposited in an account now if this sum and the interest interest it earns will provide provide equal periodic payment over a defined period of time, time, with the last payment making the balance in account zero. Present value of ordinary annuity is given by the formula:


66


1 −(1 +i ) −  P = R   i   n

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) periods) P = Present value of ordinary annuity 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?

Example:

Solution:

R = Annual payments of Birr 18,000 Term of the annuity = t = this year and the following 4 years = 5 years i = 6% = 0.06 (since the conversion period per year is annual) n=5 Present value of payments = P = ? P = Birr 75,822.55 1 − (1 + i ) −n  P = R   i   1 −(1 +0.06 ) −5  1 −(1.06 ) −5  P =18 ,000   =18 ,000   0.06 0.06     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 an arrangement arrangement whereby whereby regular payments payments are made made in order to settle settle an initial 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 Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

67

Financial Management in Value Chain: Business Mathematics and its Application (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 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

1 − (1 + i ) −n  P = R   i   Now, we progress progress to isolate isolate R on one side. It involves involves solving for for R in the above present 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.

R

  i = P  −  1 −(1 +i )  n

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 number of annu annuity ity payment payments/de s/deposi posits ts (number (number of compoun compounding ding periods) 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 Find the the month monthly ly mort mortgag gagee payme payment? nt? b. Compute Compute the total total interest. interest. Solution:

1.

Total cost of purchase = Birr 115,000 Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

68

Financial Management in Value Chain: Business Mathematics and its Application 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 = 0.00916 0.00916 n = t x m = 30 x 12 = 360 months The periodic payment R = ?

R

  i = P  −  1 −(1 +i )  n

  0.00916 =92 ,000  −360  1 −(1 +0.00916 )  = 92,000 (0.009523233) R = Birr 876.14 R

b. Total Interest Interest

= (R x n) – P = 876.14 x 360 – 92,000 = Birr 223,409.49

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 following are some of important formula that deal with mathematics of finance. • Simple Interest = Pin Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

69

Financial Management in Value Chain: Business Mathematics and its Application • 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)

 (1 + i) n - 1  Fn = R   i   • Sinking Fund Payment Formula



  n ( 1 i ) 1 + −  

R = F n 

i

• Present Value Of Ordinary Annuity

1 −(1 +i ) −  P = R   i   n

• Mortgage Payments and Amortization R

  i = P  −  1 −(1 +i )  n

Exercises

1. Suppose, Suppose, a small handicraft handicraft enterprise enterprise has requested requested a two year year loan of Birr 6500 6500 from from the commercial Bank of Ethiopia. If the bank approves the loan at an annual interest rate of 7.5%, a. What What is the simple simple interest interest on the the loan? loan? b. What is the maturity maturity value value of the loan? loan? Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

70

Financial Management in Value Chain: Business Mathematics and its Application 2. A person person signs signs a note promisi promising ng to pay a bank Birr 1500 1500 eight months months from from now and receives Birr 1350. Find the discount rate. 3.

Find Find the bank disco discount unt and proce proceeds eds on a 120120-day day note note for Birr 720 bear bearing ing 5% interest if discounted at 4% 90 days before it is due.

4. If Birr Birr 650 65000 is investe investedd at 8 ½ % compoun compoundd a. Annually b. Semi annually c. Quarterly d. Monthly, What is the amount after 7 years? 5. At what interest interest rate compounde compoundedd quarterly quarterly will a sum of money double double in 4 years? 6. How long long will it take take for Birr Birr 4750 to accumul accumulate ate to Birr Birr 7500 at 51/3% 51/3% compounde compoundedd semi-annually? 7. If, in 11 11 years, Birr Birr 1200 accumul accumulates ates to Birr Birr 1482, 1482, what is the compound compound interest interest rate rate provided provided it is converted converted annually? annually? 8. 8. If money money is worth 14% 14% compounded compounded semi-annu semi-annually, ally, would would it be be better better to discharge discharge a debt by paying Birr 500 now or Birr 600 eighteen months from now? 9. Find the accumulated accumulated value value of an ordinary ordinary annuity annuity of Birr 65 65 per period period for 23 periods periods if money is worth 4% per period. 10. 10. If birr 500 is depos deposit ited ed each each quar quarte terr in to in accou account nt pay paying ing 12% comp compou ounde ndedd quarterly, a) How much is the value of deposit at the end of the 3 years?


71


v.

Sec Sectio tion V: V: In Intro trodu ducction tion to Calc Calcul ulus us

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. 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:

A function:

Y = f(x). Limits: Definition : the limit of f(x) as x

c is ℓ which is written written as ;

lim (x) = ℓ, if

x → c

and only if the functional value f ( x 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 f ( x x) as x approaches 1. Solution:

X f(x)=x f(x)=x2+2

0.8 264

0.9 2.81

0.99 0.999 2.9801 2.998 ℓ- = 3

1

1.0001 3.0002 ℓ+ = 3

1.001 3.002

1.01 3.021

lim f ( x x) = 3 x →1


72

1.1 3.21


Example-2:

For the function f ( x x) = /x/, find; X a. xlim f(x) →2

b. xlim f(x) →0

Solution:

X

1.9

1.99

1.999

f(x)=/x/ f(x)=/x/ x

1

1

1

2

ℓ- = 1

2.001

2.001

2.01

1

1

1

0.0001 1

0.001 1

0.01 1

ℓ+ = 1

Thus; lim f(x) = 1 x →1 b.

X f(x)=/ × × / x

-0.999 -1

-0.99 -1

-0.9 -1

ℓ- = -1

0 ℓ+ = 1

ℓ-≠ ℓ+ → So, xlim doesn’t exist. exist. → 0 f(x) doesn’t Limit Theorems:

1. If k is is any any cons consta tant nt , xlim k = k →a E.g. xlim → 5 10 = 10 lim c = c a →b

2.

lim

x → a

kf(x) = k xlim f(x) →a

E.g. xlim 3x2 = 3( xlim x2) = 3(a2) = 3a2 →a →a 3. xlim ( f(x) f(x) +g(x)] +g(x)] = xlim f(x) + xlim g(x) →a →a →a E.g. xlim (x2-2× +3) = xlim x2 – 2 xlim x + xlim 3 →a →a →a →a = a2-2a +3 4. xlim [f(x).g(X) ] = ( xlim f(x) ] ( xlim g(x) ] →a →a →a Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

73

0.1 1

Financial Management in Value Chain: Business Mathematics and its Application E.g. xlim (x+3)(X-2) = [ xlim (x+3) ] [ xlim (x-2) →2 →2 →2 = [5] [0] =0 5. xlim [f(x)n] =[ xlim f(x)n] →a →a E.g. xlim (x-1)5 = [ xlim (x-1)]5 =25 = 32 →3 →3 6. If xlim f(x) = L and xlim g(x) = m, then; →a →a a, If m≠ 0, then lim [f(x)/g(x) ] = L/m

b. If m = 0 and L ≠ 0, the lim [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:

a function f is continuous at the point x = c if: 1. lim f(x) → exists 2. f(c) is defined 3. lim f(x) = f(c)

Definition:

Example:

Using the definition of continuity, discuss the continuity of the function f(x) = x2-4 x-2 at c =1 and c = 2 Solution:

At c =1, lim f(x) x →1 X 2

f(x) = x -4

0.9 2.9

0.99 2.99

0.999 2.999

1

1.0001 3.0001

1.001 3.001

1.01 3.01

x-2 ℓ- = 3

ℓ+ = 3


74

1.1 3.1

Financial Management in Value Chain: Business Mathematics and its Application lim f(x) =3

lim f(x) exists x →1

x →1

f(1) =12-4 = 3 → f(c) is defined 1-2

lim f(x) = 3 = f(1) x →1

Therefore; f(x) = x2-4 is continues continu es at c =1 x-2 At c = 2 X 1.9 1.99 1.999 2 F(x 3.9 3.99 3.999

2.0001 4.0001

2.001 4.001

2.01 4.01

2.1 4.1

)

ℓ- = 4

ℓ+ = 4

lim f(x) = 4

It exists

x → 2

f(2) =22-4 = 4-4 = 0= ¢

2-2

0

0

lim f(x) ≠ f(c) →4≠ ¢

x → 2

Therefore; f(x) is not continuous at c = 2. Derivatives: Definition: for y y = f(x) f(x) we define the derivative of f f at x, denoted by f′′(x) to be;

f′(x) = lim ∆ x

∆ f(x)

0

= lim

∆ x

∆ x

f(x+∆x)- f(x) 0

∆ x

Example-1 Find f′(x) for f(x) = 2x +4 f(x) = lim lim ∆ x

f(x+ ∆x) –f(x) 0

∆ x

st

1 → find ∆ f(x) = ∆ f(x) = f(x+∆x)-f(x) = [2(x+∆x) +4] – (2× +4)] = 2× +2∆x+4-2x-4 ∆f(x) = 2∆x 2nd find the limit: f′(x) = lim ∆f(x) 0 ∆x ∆x = lim 2∆x ∆x 0 ∆X Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

75

Financial Management in Value Chain: Business Mathematics and its Application f′(x) = 2 Example-2 For the function f(x) = 4x-x2 find f ′(x) f ′(x)= lim f(x+∆x) – f(x) ∆x 0 ∆x st 1 → find ∆f(x) = 4(x+∆x)-(x+∆x)2- (4x-x2) ∆x ∆x = 4x+4∆x-x2 – 2x∆x -∆x2-4x+x2 ∆x = ∆x (4-2x-∆x) ∆x = 4-2x-∆x 2nd → find the limit of the resulting function f′(x) = lim ∆f(x) f(x) = lim 4-2x-∆x ∆x ∆x 0 ∆x 0 ′ f (x) = 4-2x Rule of differentiation: differentiation: 1. A constant function rule

If f(x) = c then f ′(x) = 0 E.g. If f(x) = 5 then f ′(x) = 0 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) =3× 3x3-1 = 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


76

Financial Management in Value Chain: Business Mathematics and its Application 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-1 f′(x) = (2x-1)(2x) –(x2)(2) (2x-1)2 = 4x2-2x-2x2 4x2-4x+1 = 2x2-2x (2x-1)2 Application of Differential Calculus to Marginal 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 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 Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

77

Financial Management in Value Chain: Business Mathematics and its Application  



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 +2x P = 10 – x find 1000 a) Production level that maximizes revenue and the maximum revenue. b) The producti production on level that that maximizes maximizes profit profit and the maximum maximum profit. 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) 1000 R(x) 10x- x2 1000 R(x) = 10 –x =0 5000 10= x → x = 5000 units 500 R′(x) = -1/5000; -1/5000; R′(x) <0 x= 5000 units units is the revenue maximizing output level. Maximum revenue is at x = 5000 units R(x) = 10x – x2 1000 R(5000) =10(5000)- (5000)2 1000 = Birr 25000 25000 b) P(x) = R(x) – C(x) C(x) Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

78

Financial Management in Value Chain: Business Mathematics and its Application [10x- x2 ]- [5000+2x] 1000 = 10x –x2 - 5000-2x 1000 P(x) = 8x-x2 - 5000 1000 P′(x) 8-2x = 8-x 1000 500 8-x= 0 500 8= x 500 x = 4000 units P″(x) = -1 P″(x)<0 x = 4000 units is the profit maximizing output level 500 At x = 4000 units P(x) = 8x-x2 - 5000 1000 = 8(4000) - (4000)2-5000 1000 = Birr 11,000 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 At x = 4000 units R′(x) = 10 -4000 = 10-8 =2 birr 500 each level of output TR increases by birr 2 • At each • 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) Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

79

Financial Management in Value Chain: Business Mathematics and its Application 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. Learning Activity 5.1

A certain manufacturing company has the following information: Average total cost is given by the equation: č(q) = 0.5q -500+ 5000 and, q The demand function is: P = 2500-0.5q A. Find the firm’s: firm’s: i. Total profit function function ii. Marginal 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: 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: Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

80

Financial Management in Value Chain: Business Mathematics and its Application 1. A constant function rule ; If f(x) = k ∫ f(x)dx = ∫ (k)dx = kx0+1 + C = kx+C 0+1 E.g. If f(x) = 5 ∫ f(x) dx = ∫ (5)dx = 5x0+1 + C = 5x+C 0+1 2. The power rule If f(x) = xn ∫ f(x) dx = ∫ (xn)dx = xn+1 + C n+1 E.g. If f(x) = x5; ∫ f(x)dx = ∫ (x5)dx = x5+1 + C = x6 + C 5+1 6 3. A constant times a function rule; If f(x) = ax n ∫ (x)dx = ∫ (axn)dx = a∫ (xn)dx = a(xn+1) + C n+1 E.g. If f(x) = 3x3 ∫ f(x)dx = ∫ (3x3)dx = 3∫ (x3)dx = 3(x3+1)+ C 3+1 = 3/4x4+C 4. The sum and difference rule If 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 = 5x1+1 + 9x0+1 1+1 0+1 = 5/2 x2 + 9x + C 5. The product rule If f(x) = (ax+b) n ∫ f(x)dx = ∫ (ax+b)n = (ax+b)n+1 + C Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

81

Financial Management in Value Chain: Business Mathematics and its Application 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) 3 E.g. If f(x) = 8+x x2 ∫ f(x)dx =∫ (8 )dx + ∫ (x3)dx x2 x2 ∫ (8x-2)dx + ∫ (x)dx = 8x-2+1 = x1+1 + C -2+1 1+1 -1 2 = 8x +x + C -1 2 = -8x-1 +1/2x2 + C = -8/x+x2 + C 2 Indefinite integral for finding total functions:

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 = x1+1 +100x0+1 + C 1+1 0+1 x2 + 100x+ C 2 2 C(x) =1/2x + 100 x +c (fixed cost) 2 40,000 = ½(100) +100(100) + C 40,000 = 5000+ 10,000+ C 40,000 = 15000+ C Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

82

Financial Management in Value Chain: Business Mathematics and its Application 40,000 -15000 = C C = 25000 Birr Therefore; C(x) = ½x2+100x+25,000 Learning Activity 5.2

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:

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

Definition:

defined as

b

∫ a

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. Step-1: g et et 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 between the limits x = a x = b. Example

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

Solution: Solution:

Extra revenue =

20

∫

15

f(q)dq =

20

∫

15

(200-6q)dq

= 200q – 6q1+1 + C Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

83

Financial Management in Value Chain: Business Mathematics and its Application 1+1 = [200(20) -3(20)2+ C] - [200(15) -3(15) 2+C] = 2800+C- 2325 – C = Birr475 Continuous Assessment

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

The major major form formul ulas as for for app appli licat cation ion of conce concept pt of calcu calculus lus in marg margina inall analy analysi siss and optimization problems include the following. - Marginal 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) • •

∫ f(x) dx = F(x) +C ∫ f (x)dx = F(b) –F(a) b

a

Exercises

1. Find Find the the lim limit it of of the the funct functio ionn f(x) = 2× +1 as x approaches: a) 2 b) 1

c) 0

2. Find Find the the lim limit it of of the the funct functio ionn f(x) = 2x2+1 as x approaches: Jimma, Harama Haramaya, ya, Ambo, Ambo, Hawassa, Hawassa, Adama, Adama, Bahirdar, Bahirdar, W.Sodo,Sema W.Sodo,Semara ra Universit Universities ies

84

Financial Management in Value Chain: Business Mathematics and its Application a) 0

b) 1

c) 3

3. Find Find the the lim limit it of of the the funct functio ionn f(x) = 1/(x2+1) as x approaches: a) 0 b) 2 d. Proo Prooff of Abilit Ability y

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. The student has; Appl Ap plie iedd syst system emat atic ic appro approac achh of the linear equation, algebra and geometry in solving real world situations. . Product:

Criteria and Methods of assessment

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 equation and algebra on supposed real world representing representing cases (Written exam, case study) Applied matrix operations in real world Matrices and its dimensions, and operation elaborated, and the agribusiness problems application of matrix algebra and Markov Chain analysis using hypothetical real world allied cases(Written exam, case study) Reco Recomm mmen ende dedd deci decisi sion on usin usingg the the Form Formul ulat ated ed line linear ar prog progra ramm mmin ingg prob proble lems ms in both both case casess of syst system emat atic ic appr approa oach ch of the the line linear ar minimization and maximization, optimal solution determined using programming programming to solve practical practical system systematic atic techniq techniques ues/gr /graphi aphical cal and Simplex Simplex methods methods)) on real real agribusiness problems world related(Written exam, case study) Compile Compile reports reports that analyze the time time Time value of money concepts and its application valu valuee of mone moneyy appl applic icat atio ions ns for for demonstrated(Written exam, case study) fina financ ncin ingg deci decisi sion on in agri agribu busi sine ness ss value chain Applied the concepts of calculus in the Ru Rule less of limi limits ts and and cont contin inui uity ty,, rule ruless of diff differ eren enti tiat atio ionn and and agr agribus ibusiiness ness aren arenaa speci pecial ally ly;; in integr integrat ation ion as we well ll as their their inter interpre pretat tativ ivee appl applica icati tions ons in the the marg margina inall analy analysi siss and optim optimiz izat atio ionn agribus agribusine iness ss strea stream m are well well underst understood ood and and applied applied(Wr (Writte ittenn problems problems 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.


85

Financial Management in Value Chain: Business Mathematics and its Application 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.


86

Business Mathematics

Recommend Documents