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School of Distance Education

Quantitative Methods for Economic Analysis – I 2

UNIVERSITY OF CALICUT

SCHOOL OF DISTANCE EDUCATION

STUDY MATERIAL

Core Course for

BA – ECONOMICS

III Semester

QUANTITATIVE METHODS FOR ECONOMIC ANALYSIS I

Prepared by: Module I Smt. Jameela. K,Assistant Professor, Department of Economics, Govt. College, Kodanchery.

Module II Sri. Shabeer. K.P, Assistant Professor, Department of Economics, Govt. College, Kodanchery.

Module III & V Dr. Chacko Jose P, Associate Professor, Department of Economics, Sacred Heart College, Chalakkudy, Thrissur.

Module IV Sri. Y.C. Ibrahim, Associate Professor, Department of Economics, Govt. College, Kodanchery.

Scrutinised by: Dr. C. Krishnan, Associate Professor , Department of Economics, Govt. College, Kodanchery .

Layout: Computer Section, SDE

© Reserved



CONTENTS Pages

MODULE – I ALGEBRA 5

MODULE II BASIC MATRIX ALGEBRA 33

MODULE III FUNCTIONS AND GRAPHS 52

MODULE IV LIMITS AND CONTINUTY OF A FUNCTION 90

MODULE V FINANCIAL MATHEMATICS 114



MODULE – I

ALGEBRA THEORY OF EXPONENTS Exponent

If we add the letter a, six times, we get a+ a+ a+ a+ a = 5a. i.e., 5 x a. If we multiply this, we get a a a a a = a , i.e. a is raised to the power 5. Here a is the factor. a is called the base. 5 is called the exponent (Power or Index)

1. Meaning of positive Integral power.

is defined only positive integral values. If a is a positive integer, is defined as the product of n factors. Each of which is a

a = a a a ……….. n times

Eg: 2 = 2 2 2 = 8

2. Meaning of zero exponent ( zero power)

If a≠0, a = 1, i.e., any number (other than zero) raised to zero = 1

Eg: 7 = 1, 2 3 = 1

3. Meaning of negative integral power (negative exponent) If n is a positive integer and a≠0, a is the reciprocal of a .

i.e., a = 1

Eg: 3 = 1 3 = 1 9

4. Root of a number

(a) Meaning of square root

If = b, then a is the square root of b and we write, a = √ or a = b½

Eg: 3 = 9, 3 = √9 or 3 = 9½

(b) Meaning of cube root

If = b, then a is the cube root of b and we write, a =√ , or a = b

Eg: 2 = 8, 2 = √8 or 2 = 8



(c) Meaning of nth root

If = b then a is the nth root of b and we write a=√ or a = b

Eg: 3 = 81, i.e 3 = √81, or 3 = 81

(d) Meaning of positive fractional power.

If m and n are positive integers, then ⁄ is defined as nth root of mth power of a.

i.e., ⁄ = n √

Eg: 16 = 4√16 = 2 = 4

(e) Meaning of negative fractional powers.

If m and n are positive integers, ⁄ is defined as 1 ⁄ or 1 √

Eg. 16 = 116

= √

= 1 4 = 1 64 Laws of Indices

1. Product rule:‐ When two powers of the same base are multiplied, indices or exponents are added. i.e., Eg. 2 2 2 = 25 = 32 31 34 = 3 = 35 = 243 3 35 = 3 = 33 = 27

2. Quotient rule:‐ When some power of a is divided by some other power of a, index of the denominator

is subtracted from that of the numerator. i.e, = 5 5 = 5 = 5

3. Power rule:‐ When some power of a is raised by some other power, the indices are multiplied. i.e., =



Eg. 3 = 3 = 36 = 729 4 = 4 = 4 = 41 = 4

4. =

2 3 = 2 3 = 4 9 = 36 3 4 = 27 64 = 1728

5.

43 = 4 3 = 16 9

Extension

1.

2. =

3. =

4. =

Examples

1. . . .

. .

= . .

2. 6 8 6 8 = 48

3. 6 4 24

4. 63 9

= · · = 7 . = 7

5. 15 5 3



= . .

= 15 35

= 45 5 = 9

6. =

= =

= =

= 1 1 = 1

7. Multiply (

) by

=

=

= , a = , b =

=

=

8. Multiply by

=

=

=

= = a + b



9. Prove that 1

By multiplying each term by , , respectively we get,

=

1

10. If bc, b ca and c ab, prove hat abc 1 b ca c ab

1 Since the bases are same

11. If y, z and x, prove that abc 1 x zc z yb



1 Since the bases are same

12. Simplify ·

2

13. Solve 2 4

2 2

2 2

x 7 2x 4

2x‐x 7‐4

x 3



LOGARITHMS Logarithm of a positive number to a given base is the power to which the base must be raised to get the number. For eg:‐ 42 16 logarithm of 16 to be the base 4 is 2. It can be written as log 16 2 Eg:‐ log 49 2

Eg: 10 10,000, log 10,000 4

LAWS OF LOGARITHMS

1. Product rule

The logarithm of a product is equal to the sum of the logarithms of its factors. i.e., Eg: log 2 3 log 2 log 3

2. Quotient rule

The logarithm of a Quotient is the logarithm of the numerator minus the logarithm of the denominator

⁄

Eg: log 52 log 5 log 2

3. Power rule The logarithm of a number raised to a power is equal to the product of the power and

the logarithm of the number.

Eg: log 2 3 log 2

4. The logarithm of unity to any base is zero

Eg: log 1 0

5. The logarithm of any number to the same base is unity

i.e., 1 Eg: log 2 1



6. Base changing rule

The logarithm of a number to a given base is equal to the logarithm of the number to a new base multiplied by the logarithm of the new base to the given base. .

7. The logarithm obtained by interchanging the number and the base of a logarithm is

the reciprocal of the original logarithm. Eg:‐ Find logarithm of 10,000 to the base 10 10 10,000

log 10,000 4 Eg:‐ Find logarithm of 125 to the base 5 5 125

log 125 3

Eg:‐ 7 343, log 343 3

Eg:‐ Find logarithm of

1 log 12 log 3 4 log 3 2

log 3 2 log 2

LOGARITHM TABLES The logarithm of a number consists of two parts, the integral parts called the characteristics and the decimal part called the mantissa Characteristic

The characteristic of the logarithm of any number greater than 1 is positive and is one less than the number of digits to the left of the decimal point in the given number. The characteristic of the logarithm of any number less then 1 is negative and it is numerically one more than the number of zeros to the right to the decimal point.



Number Character

75.3 1

One less than the number of digits

2400.0 3 144.0 2 3.2 0

.5 ‐1 One more than the number of zero immediately after the decimal point

.0902 ‐2

.0032 ‐3

.0007 ‐4

Antilogarithm

If the logarithm of a number ‘a’ is b, then the antilogarithm of ‘b’ is a For example if log 61720 4.7904, then antilog 4.7904 61720



EQUATIONS An equation is a statement of equality between two expressions. Eg:‐ 1 2x 3 2 3x 5y 9 are equations. An equation contains one or more unknowns. In the first equation, the unknown is x and the second equation the unknowns are x and y. The value or values of unknown for which the equation is true are called solution of equations. Eg:‐ In the equation 4x 2, the x is x 2/4 ½ Difference between an equation and an Identity An equation is true for only certain values of the unknown. But an identity is true for all real values of the unknown. Eg:‐ 2 3 0 is true for x ‐3 or x 1. So it is an equation. But x 1 2

2 1 is true for all real values of x. So it is an identity. Types of equations

1. Simple linear equations in one unknown

1. 4x 2, x 2/4 ½

2. x‐3 2, x 2 3 5

3. Find two numbers of which sum is 25 and the difference is 5 Let one number be x so, that the other is 25‐x. Since the difference is 5, 25‐x ‐x 5 25‐2x 5 ‐2 x ‐20 x ‐20/‐2 10 So, one number is 10, and other is 25‐10 15.

2. Simultaneous equations of first degree in two unknown 1. Solve 4x 3y 6 1

8x 4y 18 2

For eliminating y, make the co‐efficient of y in both the equations, same. For this multiply first equation by 4, and second equation by 3. 16x 12y 24 3 24x 12y 54 4



‐8x ‐30

x 30/8 15/4 Substituting x 15/4 in equation 1 4x 3y 6

4 3y 6

3y ‐9 Y ‐9/3 ‐3

The solution is and ‐3

2. 5x ‐ 2y 4

x ‐ 3y 6 by multiplying first equation by 1 and second equation by 5

5x ‐ 2y 4 1 5x – 15y 30 2 13y ‐26

y ‐26/13 ‐2 by substituting it in equation 2

x – 3 ‐2 6 x 6 6, x 0 So , the solution is ‐2, 0

3. Find the equilibrium price and the quantity exchanged at the equation price, if supply and dd functions are given by s 20 3p and D 160 – 2p, where p is the price charged.

Ans: s 20 3p D 160 – 2p

For Equation s D 20 3p 160 – 2p 3p 2p 160‐20 5p 140, P 140/5 28 Equation price Rs. 28 Quantity exchanged 20 3p 20 3 28 20 84 104



3. Simultaneous equations containing three unknowns.

From one pair of equations, eliminate one unknown. From another pair eliminate the same unknown. The two equations thus obtained contain only two unknowns. So they can be solved. The values of the two unknowns obtained may be substituted in one of the given equations. So as to get the value of the remaining unknown.

1. 9x 3y – 4z 35 1

x y – z 4 2 2x – 5y – 4z 48 0 3 1 is 9x 3y – 4z 35 2 9 9x 9y – 9z 36 ‐6y 5z ‐1 4 2 2 2x 2y – 2z 8 3 is 3x – 5y – 4z ‐48 7y 2z 56 5

7y 2z 56 5 ‐6y 5z ‐1 4 5 5 35y 10z 280 4 2 ‐12y 10z ‐2

47y 282

y 282/47 6

Substituting 6 in equation 4 ‐6 6 5z ‐1

‐36 5z ‐1, 5z 35, z 35/5 7

Substituting y 6, z 7 in equ. 2 x 6‐7 4 x – 1 4, x 4 1 5

2. Solve 7x – 4y – 20z 0 1 10x – 13y – 14z 0 2 3x 4y – 9z 11 3 7x – 4y – 20z 0 1 10x – 13y – 14z 0 2



1 13 91x – 52y ‐260z 0 2 4 40x – 52y – 56z 0 51x – 204z 0 4

7x – 4y – 20z 0 3x 4y – 9z 11 10x ‐29z 11 5

51x – 204z 0 10x – 29z 11 4 10 510x – 2040z 0 5 51 510x – 1479z 561 ‐561z 561 z ‐ 561/‐561 1

Substituting z 1 in equ. 5 10x 29 1 11 10x – 29 11 10x 11 29, 10x 40, x 40/10 4 Substituting x 4, z 1 in equ. 1 7x ‐ 4y – 20z 0 7 4 – 4y ‐20 1 0 28 – 4y – 20 0 8 – 4y 0, 8 4y, y 8/4 2 the solutions are x 4, y 2, z 1

QUADRATIC EQUATIONS The equation of the form 0 in which a, b, c are constants is called a quadratic equation in x (a≠0).

Solving the quadratic equation, we get the two values of x. These two values are known as the roots of the quadratic equation.

It may be pure or general Pure quadratic equation If in the equation 0, b is zero, then the equation becomes 0, this is called pure quadratic equation. General Quadratic Equation

0 is the general form of the quadratic equation. The general quadratic equation may be solved by one of the following methods.



1. By factorization 2. By completing the square 3. By quadratic formula

Eg:‐ Solve 7 12 0 7 12 0 (x – 4) (x – 2) Equation becomes (x – 4) (x – 2) = 0 x – 4 = 0 or x – 2 = 0 x = 4 or x = 2 the roots are 2 and 4.

2 Method of Completing squares or Shridhar’s method

Solve 9 24 16 0

Ans: 9 24 16 0

3 4 0 3x – 4 0, 3x 4, x 4/3

3 By Quadratic Formula

0

a Solve : 2 8 8 0 Here a 2, b 8, c 8

√

√ , √

, ‐8/4 ‐2

b Solve the equation 4 3 0

√

, 6/2 3, 2/2 1

the solutions are 3,1



PROGRESSIONS

ARITHMETIC PROGRESSION

A series is said to be in Arithmetic Progression, if any term of it is obtained by adding a constant number to its preceding term.

This constant number is called the common difference.

Eg: 2, 5, 8, 12 ………is in A.P. Since every term of it is obtained by adding a constant number i.e.,3 to its preceding term. When the common difference is positive, the terms are increasing and when the common difference is negative the terms are decreasing.

Eg:‐ 1 8, 10, 12, …… is an increasing A.P. 2 10, 8, 6, …….. is a decreasing A.P.

Common difference for an A.P. denoted by d is any term minus its preceding term.

The nth term of an A.P

nth term (tn) = a + (n1)d First term = a + (11)d = a Second term = a + (21)d = a + d Third term = a + (31)d = a + 2d Fourth term = a + (41)d = a + 3d

So series is a, a+d, a+2d, a+3d……… the nth term is also called general term of the A.P.

1. Find (1) 10th term of the series 1, 5, 9, 13…….. (2) 14th term of the series 8, 5, 2 …… (3) nth term of the series 3, 5, 7……. (4) 6th term of the series 8, 7½, 7, 6½ ……. (5) 9th term of the series 4, 2, 0, ‐2, ‐4 …….

(1) tn = a + (n1)d a = 1

= 1 + (10‐1)4 n = 10 = 1 + 9 4 1 36 37 d 4

(2) tn = a + (n1)d

= 8 + (14 – 1)‐3 a = 8 = 8 + 13 ‐3 n = 14 = 8 + ‐39 = ‐31 d = ‐3



(3) tn = a + (n1)d = 3 + (n – 1)2 a = 3 = 3 + 2n ‐ 2 n = n = 2n + 1 d = 2

(4) tn = a + (n1)d = 8 + (6 – 1) ‐½ a = 8 = 8 + 5 ‐½ n = 6

= 8 – 5/2 = = 11/2 d = ‐.5 (‐½)

(5) tn = a + (n1)d = 4 + (9 – 1)‐2 a = 4 = 4+ 8 ‐ 2 n = 9 = 4 – 16 = ‐12 d = ‐2

(6) Which term of the series 87, 84, 81 ….. is ‐3. tn = a + (n1)d ‐3 = 87 + (n‐1) ‐3 tn = ‐3 ‐3 = 87 + (‐3n + 3) a = 87 ‐3 =90 – 3n n = ?

3n = 90 + 3 = 93, n = = 31 d = ‐3

(7) Which term of the series 27 + 24 + 21 …… Is equal to (1) – 27 (2) – 80 ? tn = a + (n1)d ‐27 = 27 + (n – 1) ‐3 a = 27 ‐27 = 27 + (‐3n + 3) d ‐3 ‐27 = 30 – 3n n = ?

3n = 30 + 27, 3n = 57, n = = 19 tn = ‐27 ‐27 is the 19th term

tn = a + (n1)d ‐80 = 27 + (n – 1) ‐3 a = 27 ‐80 = 27 + (‐3n + 3) d ‐3 ‐80 = 30 – 3n n = ?

3n = 30 + 80, tn = ‐80



3n = 110, n = = 36.67

Which is impossible because n must be whole number. Hence there exists no tem in the series which is equal to ‐80.

(8) If the 9th term of an A.P. is 99 and 99th term is 9 find 108th term. Let a be the first term and ‘d’ be the common difference. So 9th term = t9 = a + (9‐1)d = 99 = a + 8d = 99 (1)

99th term t99 = a + (99‐1)d = 9 = a + 98d = 9 (2) a + 8d = 99 (1) a + 98d = 9 (2) ‐90d = 90 d = 90/‐90 = ‐1

Substituting d = ‐1 in equ. (1)

a + 8 ‐1 = 99

= a ‐8 = 99, a = 99 + 8, a = 107

t108 = a + (n1)d = 107 + (108 – 1) ‐1 = 107 + 107 ‐1 = 107 – 107 = 0

9 If the third term of an A.P is 7 and the 9th term is 19, find the 27th term. t3 = a + (3‐1)d = 7 = a + 2d = 7 (1)

t9 = a + (9‐1)d = 19 = a + 8d = 19 (2)

a + 2d = 7 (1) a + 8d = 19 (2) ‐6d = ‐12 d = ‐12/‐6 = 2

Substituting d = 2 in equ. (1)



a + 2d = 7 a + 2 2 = 7 a + 4 = 7, a = 7 – 4 = 3

t27 = a + (n1)d

= 3 + (27 – 1) 2 = 3 + 26 2 = 3 + 52 = 55

Sum of n terms of an A.P Let a be the first term and d be the common difference. Let Sn be the sum of n terms. Sn

Or Sn Ist term last term

1. Find the sum of 10 terms of the series 15, 13, 11 ……..

a 15, d ‐2, n 10

Sn =

= 10 2 2 15 10 1 2

= 5[30 + ‐18]

= 5[12] = 5 12 = 60

2. Find the sum of the following series. 1 +4+7+10……. to 20 terms.

S20 = 20 2 2 1 20 1 3

= 10[2 + 19 3 d 3

10 2 57 a 1

10 59 590 n = 20

1 Find the sum of first n natural numbers.

First ‘n’ natural numbers are 1,2,3,4 …… n This is an A.P with a = 1, d = 1, n =n Sum = 2 2 1

2 2 1 1 1

2 2 1



2 1

2 Find the sum of series 40 36 32 ……. 4

tn = a + (n1)d 4 = 40 + (n‐1)‐4 4 40 – 4n 4

4 44 – 4n 4n 44 – 4 40, n 40/4 10 Sum 2 I

st term last term

10 2 40 4

5 44 220

3. Find the sum of all natural numbers between 100 and 300 which are divisible by 8. The natural numbers between 100 and 300 which are divisible by 8 are 104, 112, …… 296.

tn = a + (n1)d 296 = 104 + (n‐1)8 a = 104

296 104 8n – 8 d 8 296 104 ‐ 8 8n n ? 296‐96 8n tn = 296 200 8n, n 200/8 25 Sum 2 I

st term last term

25 2 104 296

25 2 400 25 200 5000

Sum 5000 Arithmetic Mean AM

1. If a, b, c are A.P, then b is said to be the AM between a and c. 2. If a, a1, a2 ……. an, b are in A.P. Then a1, a2 ……. an are said to be arithmetic means

between a and b.

A.M. between a and c =

Insert 5 Arithmetic Means between 10 and 40



When 5 Arithmetic Means are inserted between 10 and 40, there are 7 terms. They form an A.P, First term of it is 10 and last term is 40. a 10, n 7, tn 40, d ?

tn = a + (n1)d

40 = 10 + (7‐1)d

40 10 7d‐d

40 10 – 6d

40‐10 6d,

30 6d, d 30/6 5 the terms are 10, 15, 20, 25, 30, 35, 40 the arithmetic means are 15, 20, 25, 30, 35

GEOMETRIC PROGRESSION A series is said to be in G.P, if every term of it is obtained by multiplying the previous term by a constant number. This constant number is called common ratio, denoted by r. Eg:‐ 4, 8, 16 …….. is in G.P. Here, common ratio 8/4 2 When the common ratio is positive and less than one the series is a decreasing G.P. When the common ratio is positive and greater than 1, the series is an increasing G.P. Eg:‐ 1 3, 6, 12, 24 …… is an increasing G.P. Since ‘r’ is greater than 1 r 2 2 4, 2, 1, ½ …….. is decreasing G.P. Since ‘r’ is less than 1 r ½ nth term of a G.P. Let ‘a’ be the first term and ‘r’ be the common ratio, The nth term is tn

1. Given the series 2, 6, 18, 54 ……….. find 1 12th term 2 nth term a 2, r 3 tn

t12 2 3 2 3 2 177147 554294



2 a 2, r 3 tn tn 2 3 Sum of n terms of a G.P Let a be the first term and r be the common ratio

Sn When r 1 numerically

or

Sn

When r 1 numerically

1 Find the sum of the following series.

1 1 3 9 27 ……… to 10 terms

2 1 1 2 1 4 1 8 ……… to 12 terms

1 Here a 1, r 3 and n 10

Sn

as r 1

29524

2 a 1, r 1 2 and n 12

Sn as r 1

Infinite G.P

Sum of the infinite G.P.

1 Find the sum of the infinite G.P, 4, 2, 1……



a 4, r 2 4 1 2

4 2 8

Note: ‐ When the product is known, three number in G.P. can be taken as ⁄ , a, ar.

1. The sum of three numbers in G.P. is 35 and their produt is 1000. Find the numbers.

Let the numbers be ⁄ , a, a r

⁄ a + a r = 35 (1)

⁄ a a r = 1000 (2)

= 1000

a = √1000 = 10

10 10 + 10 r = 35 (3)

Multiplying both sides by r

10 10r + 10 = 35r

10 + 10r + 10 = 35 r

10 +10r – 35r + 10 = 0

10 – 25r + 10 = 0 (4)

This is a quadratic equation

r =

= √

= √ = = 2

= = =

When r = 2, the numbers are ⁄ , a, ar

102, 10, 10 2



5, 10, 20 When r =

, the numbers are ⁄ , a, ar

10 ½, 10, 10 ½ 20, 10, 5 Since both are same, the required numbers are 5, 10, 20 Geometric Mean

1. If a, b, c are in G.P, then b is said to be the geometric mean between a and c 2. If a, G1, G2 ……. Gn, b are in G.P, then G1, G2 …….. Gn are said to be the geometric means

between a and b. Insert 5 geometric means between 2 and 1458. Let G1, G2, G3, G4, G5 be the geometric means, then G1, G2, G3, G4, G5, 1458 are in G.P. n = 7, a = 2

tn

1458 2

1458 2

1458 2 729

r √729 3 and ‐3

The geometric means are

6, 18, 54, 162, 486,

‐6, ‐18, ‐54, ‐162, ‐486

Economics Applications of Arithmetic and Geometric Progressions

Growth rate – Simple & Compound

Growth rate – Simple

The total simple interest at the end of nth year is 100 and the total amount at the end of nth year is P 100

1. The salary of an employee increases every year by 10% of his basic salary and his initial basic salary is Rs. 3,500/‐. Find his salary at the end of 8th year. Initial salary 3500, P 3500, n 8, r 10.



Salary at the end of 8th year P 100

3500

3500 2800 6300

2. A borrowed Rs. 800/‐ at simple interest from a bank at 12% per annum for 6 years. Find the amount paid by the firm to the bank to clear the accounts.

Amount borrowed 8000, p 8000, n 6, r 12 Amount at the end of the 6th year

P 100

8000

8000 5760 13760

Growth rate – Compound

Formula P

1. If the population of country rises by 2% every year and it was 30 crore in 1960. What was the population in 1970? Initial population 30

n 11, r 2

Population at the end of 1970 P 1 100

30 1 2100

30 1.02 37.3 crore.

2. Find the compound interest on 2000 at 7% for 5 years P 2000, n 5, r 7

Amount at the end of 5th year

P 1 100

2000 1 7100

2000 1.07 2000 1.40255 2805 P 2000, amount 2805 So, compound interest 2805 – 2000 805



SET THEORY Set:‐ A set is a well defined collection or a class of objects. i.e., there exists a rule with the help of which we will be able to say whether a particular object be longs to the set or doesn’t belong to the set. Eg:‐ Vowels in English alphabet, students in a class, flowers in a garden etc, are examples of sets. The objects of the set are elements or members of the set. Methods of describing a set. 1. Tabular or Roster or Enumeration Method. 2. Rule Selector or property builder Method.

Tabular Method In this method we define a set by actually listing its members.

Eg:‐ The set ‘A’ consisting of number 1, 2, 3, 4 is expressed as: A 1, 2, 3, 4 . The elements are separated by a comma and are enclosed with in bracket.

Set builder Method Rule Method Selector Method In this method we define a set by stating certain properties which its members must satisfy. For eg:‐ Consider the set of flower in the garden. It can be represented as A x:x is flower in the garden Here x is the general element and so it represents all the elements of the set. Examples of some sets and their representations.

1. The set of number 1, 3, 5, 7, 9 …… can be represented by a Tabular method: A 1, 3, 5, 7, 9 …… b Rule method: A x:x is an odd integer

Finite and infinite sets:‐

A set is said to be finite if it consists of a specific number of different elements. In other words, it means that if we count the different members of the set, the counting, process may come to an end. So, in an infinite set, the number of elements are not exhaustively countable. The set of students in a class, set of positive integers less than 10 etc, are examples of finite sets whereas the set of positive integers is an infinite set.



Equality of sets

The two sets a and B are equal i.e., A B as every element which be longs to set A also belongs to set B and vice versa.

Eg:‐ A 1, 3, 4, 7 and B 7, 3, 1, 4

Equivalent sets

In equivalent sets, the number of elements in the two sets are equal. Eg:‐ A a, b, c, d, e and B 1, 2, 3, 4, 5 then A & B are equivalent sets.

Null Set Empty Set A set which contains no element is called the null set or the empty set and is denoted by

Singleton A set which contains only one element is called Singleton or unit set. Thus 0 , 3 , a are all Singleton sets. Set or all positive integers less than 2 is a Singleton

Subset of a set

If every element of a set A is also an element of a set B then A is called subset of B and is written as A B, i.e., A is contained in B Eg:‐ Let A 1, 2, 3 and B 1, 2, 3, 5, 7 . Then A is a subset of B as each element 1, 2, 3 of set A also belongs to B.

1 The null set is a subset of every set. 2 Every set is a subset of itself.

Superset If A B, then B is called superset of A and is written as B A i.e., B contains A

Proper subset

A set ‘A’ is said to be a proper subset of B and is denoted as A B if A is a subset of B and A B. For proper subset of A to B, all the elements of A shall be long to B and there must exist at least one element of B which doesn’t belong to A.

Eg:‐ A 1, 2, 3, 4, 5 B 1, 2, 3, 4, 5, 6 So, a is a proper subset of B.



SET OPERATIONS

Basic set operations are

1. Union of two sets 2. Intersection of two sets 3. Difference of two sets.

1. Union of two sets.

It is a set of all those elements which belong to A or B or to both. Union of a and B is written as A B and is read as ‘A Union B’

If A 2, 3, 4, 5 and B 3,5,7,8 , then A B 2, 3, 4, 5, 7, 8

2. Intersection of two sets. It is a set of all those elements which belong to both A & B. It is written as A B. A B contains elements common to A & B Eg:‐ A 1, 2, 3 and B 1, 3, 5 then A B 1, 3

3. Difference of Two sets. It is a set of all those elements which belong to A, but do not belong to B. It is written as A – B and is read as A minus B.

Power Set If a be a given set, then the family of sets each of whose members is a subset of the given set A is called the power set of the set A and is denoted as P S If A a, b , then its subsets are , a , b , a, b P s , a , b , a, b and it has 22 4 elements

Universal set

If all the sets under consideration are subsets of a fixed set say , then this set is called Universal set.

Eg:‐ If A is the set of vowels in English alphabet, then its universe is the set of all letters of the alphabet

Disjoint sets

Two sets A and B are said to be disjoint sets if no element of A is in B and no element of B is in A.



i.e., A 1, 2, 3 , B 5, 7, 9 4. Complement of a set The complement of a set A is the set of all those elements belonging to the Universal set but not belonging to A.

Therefore complement of A, denoted by A or A is ‐A Domain and Range of a set Let A and B are any two sets. A relation from A to B is defined as a subset of A B and is denoted by R. Then R A B. The domain X of the relation R is defined as the set consisting of all the first elements of the ordered pairs belonging to R. The range Y of the relation is defined as the set consisting of all the second elements of the ordered pairs belonging to R. In the language of notation Domain of R X a A: a, y R Range of R Y b B: x, b R Evidently X A, Y B

Let R be the inverse of the relation of R, i.e., R is a relation of B in to A Then we define

Domain of R range of R

Range of R domain of R



MODULE II

BASIC MATRIX ALGEBRA MATRICES : DEFINITION AND TERMS

A matrix is defined as a rectangular array of numbers, parameters or variables. Each of which has a carefully ordered place within the matix. The members of the array are referred to as “elements” of the matrix and are usually enclosed in brackets, as shown below.

A=

The members in the horizondal line are called rows and members in the vertical line are called colums. The number of rows and the number of columns together define the dimension or order of the matrix. If a matrix contains ‘m’ rows and ‘n’ columns, it is said to be of dimension mxn (read as ‘). The row number precedes the column number. In that sence the above matrix is of dimension 3 x 3. Similarly

B = 3 5 12 7 4

C =7810

D = 10 2

E = 2 01 4

TYPES OF MATRICES

1. Square Matrix

A matrix with equal number of rows and colums is called a square matrix. Thus, it is a special case where m=n. For example

2 13 4 is a square matrix of order 2

2 1 34 0 69 7 5

is a square matrix of order 3

2. Row matrix or Row Vector

A matrix having only one row is called row vector of row matrix. The row vector will have a dimension of 1×0. For example



2 5 0 1

2 1

0 2 3

3. Column matrix or Column Vector

A matrix having only one column is called column vector or column matrix. The column vector will have a dimension of m 1 . For example

58

89214

025

4. Diagonal Matrix

In a matrix the elements lie on the diagonal from left top to the right bottom are called diagonal elements. For instance, in the matrix 2 5

4 6 the element 2 and 6 are diagonal elements. A square matrix in which all elements except those in diagonal are zero are called diagonal matrix. For example

2 00 6

2 2

4 0 00 9 00 0 2

5. Identity matrix or Unit Matrix

A diagonal matrix in which each of the diagonal element is unity is said to be unit matrix and denoted by I. The identity matrix is similar to the number one in algebra since multiplication of a matrix by an identity matrix leaves the original matrix unchanged. Than is, AI = I A =A

1 00 1 2 2

1 0 00 1 00 0 1

3 3

are examples of identity matrix

6. Null Matrix or Zero Matrix

A matrix in which every element is zero is called null matrix or zero matrix. It is not necessarily square. Addition or subtraction of the null matrix leaves the original matrix unchanged and multiplication by a null matrix produces a null matrix.

0 0 00 0 0 2 3

0 00 0

2 2

0 00 00 0

3 2 are examples of null matrix

7. Triangular Matrix



If every element above or below the leading diagonal is zero, the matrix is called a triangular matrix. Triangular matrix may be upper triangular or lower triangular. In the upper triangular matrix, all elements below the leading diagonal are zero, like

A = 1 9 20 3 70 0 4

In the lower triangular matrix, all elements above leading diagonal are zero like

B = 4 0 02 9 05 6 3

8. Idempolent Matrix

A square matrix A is said to be idenpoint if A = A2.

TRANSPOSE OF A MATRIX

A matrix obtained from any given matrix A by interchanging its rows and columns is called its transpose and is denoted by or A’. If A is matrix A’ will be dimension. For example

A = 6 7 92 8 4

2 3 =

6 27 89 4

B = 1 2 2 3 3 4

3 44 12 5

=

1 2 32 3 43 4 24 1 5

C =121925

= 12 19 25

D =

2 17 83 09 5

2 7 1 8

3 90 5

Symmetric and skew Symmetric Matrix

Any square matrix A is said to be symmetric if it is equal to its transpose. That is, A is symmetric if A = Consider the following examples

A = 1 55 3

1 55 3 A = , hence A is symmetric



B =5 2 62 3 96 9 7

=5 2 62 3 96 9 7

B = B is symmetric

At the same time, any square matrix A is said to be skew symmetric if it is equal to its negative transpose. That is A = ‐At, then A is skew symmetric consider the following examples

A = 0 44 0 = 0 4

4 0 = 0 44 0

A = A is skew symmetric

B = 0 3 53 0 25 2 0

= 0 3 53 0 25 2 0

= 0 3 53 0 25 2 0

OPERATION OF MATRIXES

1. Addition and subtraction of Matrices

Two matrixes can be added or subtracted if and only if they have the same dimension. That is, given two matrixes A and B, their addition or subtraction that is, A + B and A – B requires that A and B have the same diamension. When this dimensional requirement is met, the matrices are said to be “conformable for addition or subtraction”. Then, each element of one matrix is added to (or subtracted from) the corresponding element of the other matrix.

For example, if A = 4 92 1 and B = 6 3

7 0

A + B = 4 92 1 +

6 37 0 =

4 6 9 32 7 1 0 =

10 129 1

A ‐ B = 4 92 1 ‐

6 37 0 =

4 6 9 32 7 1 0 =

2 65 1

Example :2

If A = 8 9 73 6 24 5 10

B = 1 3 65 2 47 9 2

A+B = 9 12 138 8 611 14 12

Example : 3

A = 3 7 1112 9 2 B = 6 8 1

9 5 8 A – B = 3 1 103 4 6

Example : 4



A = 2 2 21 1 31 0 4

B = 3 3 33 0 56 9 1

C = 4 4 45 1 02 3 1

A + B – C = 1 1 11 2 25 6 2

Example 5

A = 12 16 2 7 8 B = 0 1 9 5 6 A + B = 12 17 11 12 14

2. Scalar Multiplication

In the matrix algebra, a simple number such as 1,2, ‐1, ‐2 ……. is called a scalar. Multiplication of matrix by a scalar or number involves multiplication of every element of the matrix by the number. The process is called scalar multiplication.

Let ‘A’ be any matrix and ‘k’ any scalar, then the matrix obtained by multiplying every element of A by K is said to be the scalar multiple of A by K, because it scales the matrix up or down according to the size of the scalar.

Example 1

If A = 3 10 5 and scalar k = 7 then KA = 7 3 1

0 5 = 21 70 35

Example 2

Determine KA if K = 4 and A = 3 29 56 7

KA = 12 836 2024 28

Example 3

K = ‐2 and A = 7 3 25 6 82 7 9

KA = 14 6 410 12 164 14 18

Example 4

If A = 2 3 10 1 5 B =

1 2 10 1 3 Find 2A ‐3B

2A = 4 6 20 2 10 3B =

3 6 30 3 9 2A – 3B = 1 0 5

0 1 1

3. Vector Multiplication

Multiplication of a row vector ‘A’ by a column vector ‘B’ requires that each vector has precisely the same number of elements. The product is found by multiplying the individual



elements of the row vector by their corresponding elements in the column vector and summing the product. For example

If A = [a b c] B = def

AB = [ ad + be + cf ]

Thus the product of row – column multiplication will be a single number or scalar. Row –column vector multiplication is very important because it serves the basis for all matrix multiplication.

Example 1

A = [ 4 7 2 9 ] B = AB = (4 12) + (7 1) + (2 5) + (9×0) = 119

Example 2

C = [ 3 6 8 ] D = 245 CD = 70

Example 3

A = [ 12 ‐5 6 11 ] B = AB = 44

Example 4

A = [ 9 6 2 0 ‐5 ] B AB = 101

4. Matrix Multiplication

The matrices A and B are conformable for multiplication if and only if the number of columns in the matrix A is equal to the number of rows in the matrix B. That is, to find the product AB, conformity condition for multiplication requires that the column dimension of A (the lead matrix in the expression AB) must be equal to the row dimension of B (the lag matrix)

In general, if A is of the order m × n then B should be of the order n × p and dimension of AB will be m × p. That is, if dimension of A is and 1 × 2 and dimension of B is 2 ×3, then AB will be of 1 × 3 dimension. For multiplication, the procedure is that take each row and multiply with all column. For example if



A = and B =

AB =

Similarly if A = 3 6 712 9 11 B =

6 125 1013 2

Since A is of 2 × 3 dimension and B is of 3 × 2 dimension the matrices are conformable for multiplication and the product AB will be of 2 × 2 dimension. Then

AB = 3 6 6 5 7 13 3 12 6 10 7 212 6 9 5 11 13 12 12 9 10 11 2

AB = 139 110260 256

Example 1

A = 3 54 6 B = 1 0

4 7 AB = 17 3520 42

Example 2

A = 1 32 84 0

B = 59 AB = 328220

Example 3

A = 7 112 910 6

B = 12 4 53 6 1 AB =

117 94 4651 62 19138 76 56

Example 4

A = B =[ 2 6 5 3 ] B = 2 6 5 3 AB =

6 18 15 92 6 5 3 8 24 20 12 10 30 25 15

Matrix Expression of a System of Linear Equations

Matrix algebra permits the concise expression of a system of linear equations. For example, the following system of linear equation

+ =



Can be expressed in matrix form as

A X =B where,

A = X= and B =

Here, A is the coefficient matrix, x is the solution vector an B is the vector of constant terms. X and B will always be column vector. Since A is 2x2 matrix and x is 2x1 vector, they we conformable for multiplication, and the product matrix will be 2 x 1.

Example 1

7 + 3 = 45

4 + 5 = 29

In matrix from AX=B

7 34 5 = 4529

Example 2

7 + 8 = 120

6 + 9 = 92

In matrix form AX=B

7 86 9 = 12092

Example 3

2x1 + 4x2 + 9x2 = 143 2x1 + 8x2 + 7x3 = 204 5x1 + 6x2 + 3x3 = ‐168

In matrix form AX = B

2 4 92 8 75 6 3

= 143204168

Example 4

8w + 12x ‐ 7y + 22 = 139

3w ‐ 13x + 4y + 92 = 242



In matrix from AX=B

8 123 13

7 2 4 9 = 139242

Concept of Determinants

The determinant is a single number or scalar associated with a square matrix. Determinants are defined only for square matrix. In other words, determinant denoted as | |, is a uniquely defined number or scalar associated with that matrix

If A= is a 1×1 matrix, then the determinant of A, ie | | is the number itself. If A is a 2 × 2 matrix then the determinant of such matrix, like

A =

called the second order determinant is derived by taking the product of two elements on the principal diagonal and subtracting from it the product of two elements off the principal diagonal. That is,

| | =

Thus | | is obtained by cross multiplication of the elements. If the determinant is equal to zero, the determinant is said to vanish and the matrix is termed as singular matrix. That is, a singular matrix is one in which there exists linear dependence between at least two rows or columns. If | | ≠ 0, matrix A is non‐singular and all its rows and columns are linearly independent.

Example 1

If A = 10 48 5

| | = (10 × 5) – (4 × 8) = 18

Example 2

B = 2 13 2

| | = 7

Example 3

C = 6 47 9

| |= 26

Example 4

D = 4 66 9

| |=0



RANK OF MATRIX

The rank (P) of a matrix is defined as the maximum number of linearly independent rows and columns in the matrix. For example, if

A = 2 33 6

| |= 3 and the matrix A is non singular and its rows and columns are linearly independent and the rank of the matrix A, ie, P(A) = 2. If

B = 4 28 4

= 0 and matrix B is singular and a Linear dependence exists between its rows and columns. Hence the rank of the matrix P(B) = 1

Third order Determinants

A determinant of order three is associated with a 3 × 3 matrix. Given.

A = 11 12 1321 22 2331 32 33

Then

| |= ‐ +

| |= ( ) – ( ‐ ) + ( ‐ )

| |= a scalar

| | is called a third order determinant and is the summation of three products to desire three products.

1. Take the first element of the first row, ie, a11 and mentally delete the row and column in which it appears. Then multiply a11 by the determinant of the remaining elements.

2. Take the second element of the first row, ie, a12 and mentally delete the row and column in which it appears. Then multiply a12 by ‐1 times the determinant of the remaining element.

3. Take the third element of the first row, ie, a13 and mentally delete the row and column in which it appears. Then multiply by the determinant of the remaining elements.

In the like manner, the determinant of a 4 × 4 matrix is the sum of four products. The determinant of a 5 × 5 matrix is the sum of five products and so on.



Example 1

A = 8 3 26 4 75 1 3

| | = 8 4 71 3 – 3

6 75 3 + 2

6 45 1

| | = (8 × 5) – (3 ×‐17) + 2(‐4)

| |= 63

Example 2

A = 3 6 22 1 87 9 1

| | = 3 1 89 1 ‐ 6

2 87 1 + 5

2 17 9

| |= 166

Example 3

B = 3 6 21 5 44 8 2

| |= ‐3 5 48 2 – 6

1 44 2 + 2

1 54 8

| |= 98

Example 4

C = 5 7 22 3 14 6 2

| |= 5 3 16 2 – 7

2 14 2 + 2

2 34 6

| |= 0

PROPERTIES OF A DETERMINANT

1. The value of the determinant does not change if the rows and columns of it are interchanged. That is, the determinant of a matrix equals the determinant of its transpose. That is .For Example

A= 4 35 6

| |=9 At = 4 53 6

| | 9

2. The interchange of any two rows or any two columns will alter the sign, but not the numerical value of the determinant. For example, if

A = 3 1 07 5 21 0 3

| | = 3 5 20 3 ‐1

7 21 3 + 0

7 51 0 = 26



Now if we interchange first and third column, 0 1 32 5 73 0 1

⇒0 5 70 1 ‐ 1

2 73 1 + 3

2 53 0 = ‐26 =

‐| |

3. If any two rows or columns of a matrix are indentical or proportional, ie linearly dependent, the determinant is zero. For Example

A = 2 3 14 1 02 3 1

| |=2 1 0

3 1 ‐ 3 4 02 1 + 1

4 12 3

| |=0, since first and third row are identical

4. The multiplication of any one row or one column by a scalar or constant ‘k’ will change the value of the determinant k . For example

If A = 3 5 72 1 44 2 3

| | = 35

Now forming a new matrix B by multiplying the first row of A by 2, then

B = 6 10 142 1 44 2 3

| |=70, ie, 2 × | |

Thus, multiplying a single row or column of a matrix by a scalar will cause the value of determinant to be multiplied by the scalar.

5. The determinant of triangular matrix is equal to the product of elements on the principal diagonal For example, for the following lower triangular matrix

A = 3 0 02 5 06 1 4

| |= 60, ie ‐3 × ‐5 × 4

6. If all the elements of any row or column are zero the determinant is zero. For example

A = 12 16 130 0 015 20 9

| |= 0 Since all elements of second row is zero

7. If every element in a row or column of a matrix is sum of two numbers, then the given determinant can be expressed as the sum of two determinant.

| |= 2 3 14 1 5 = 20



ie 2 14 5 +

3 11 5 = 6 + 14 = 20

8. Addition or subtraction of a non zero multiple of any one row or column from another row or column does not change the value of determinant. For example

A = 20 310 2

| |= 10

Now subtract tow times of second column from first column and for a new matrix.

B = 14 36 2 ⇒ | |=10

MINORS AND COFACTORS

Every element of a square matrix has a minor. It is the value of the determinant formed with the elements obtained when the row and the column in which the element lies are deleted. Thus, a minor, denoted as is the determinant of the submatrix formed by deleting the ith row and jth column of the matrix.

For example, if A =

Minor of Minor of Minor of =



A cofactor (cij) is a minor with a prescribed sign. Cofactor of an element is obtained by multiplying the minor of the element with where i is the number of row and j is the number of column.

That is | | = 1

In the above matrix A

Cofactor of

A cofactor matrix is a matrix in which every element of is replaced with its cofactor cij.

Example 1

A = 7 124 3 Matrix of cofactors Cij = 3 4

12 7



Example 2

B = 2 513 6 Matrix of cofactors Cij =

6 135 2

Example 3

C 2 3 14 1 25 3 4

= Matrix of Cofactors Cij = 2 6 79 3 95 0 10

Example 4

D = 6 2 75 4 93 3 1

Matrix of Cofactors Cij = 23 22 319 15 1210 19 14

ADJOINT MATRIX

An adjoint matrix is transpose of a cofactor matrix that is adjoint of a given square matrix is the transpose of the matrix formed by cofactors of the elements of a given square matrix taken in order.

Example 1

A = 13 1719 15

Matrix of Cofactors Cij = 15 1917 13

Adjoint of A = c = 15 1719 13

Example 2

A = 6 712 9 Cij=

9 127 6

Adj A = c = 9 712 6

Example 3

B = 0 1 21 2 33 1 1

= 1 8 51 6 31 2 1

Adj A = c = 1 1 18 6 21 2 1

Example 4



A=13 2 89 6 43 2 1

Cij=2 3 014 11 2040 20 60

A Cij A = Cijt = 2 14 403 11 200 20 60

INVERSE MATRIX

For a square matrix A, if there exists a square matrix B such that AB=BA=1, then B is called the inverse matrix of A and is denoted as A‐1 =

Example 1

A= 3 21 0

| |= ‐2

Adj A= 0 21 3

A‐1 = | | =

A‐1 = 0 1

Example 2

A = 7 96 12

| |= 30

A‐1 = | | =

A‐1 =

Example 3

A = 1 2 35 7 42 1 3



| |= ‐24

Adj A = 17 3 137 3 119 3 3

A‐1 = | | =

Example 4

A = 4 2 53 1 89 6 7

| |= ‐17

AdjA = 41 16 1151 17 179 6 2

A‐1 = | | = 3 1 1

CRAMERS RULE FOR MATRIX SOLUTIONS

Cramers rule provides a simplified method of solving a system of linear equations through the use of determinants. Cramer’s rule states

= | || |

where is the unknown variable, | |is the determinant of the coefficient matrix and | |is the determinant of special matrix formed from the original coefficient matrix by replacing the column of coefficient of with the column vector of constants

Example 1

To solve

6 + 5 = 49

3 + 4 = 32



In matrix from A×=B

6 53 4 = 4932

| |= 9

To solve for , replace the first column of A, that is coefficient of with vector of constants B, forming a new matrix ,

= 49 532 4

| 1|= 36

= | || |

= = 4

Similarly, to solve for , replace the second column of A, that is coefficient of x2, with vector of constants B, forming a new matrix A2,

A2 = 6 493 32

| | = 45

= | || | = = 5

the solution is = 4 and = 5

Example 2

2 6 = 22

5 = 22

Ax = B

2 61 5 = 2253

| | = 16

A1 = 22 653 5

| | =‐208

= | || |

= =‐13



= 2 221 53

=128

= | || | = = 8

the solution is = ‐13 and = 8

Example 3

7 ‐ ‐ =0

10 ‐ 2 + = 8

6 3 2 ‐ 2 = 7

AX = B 7 1 110 2 16 3 2

= 087

| | = ‐61

| |= 0 1 18 2 17 3 2

| | = ‐61

= | || |

= = 1

| |= 7 0 110 8 16 7 2

| |= ‐183

= | || |

= = 3

A3 = 7 1 010 2 86 3 7

| |= ‐244

= | || |

= = 4



the solution is x1 = 1 x2 = 3 x3 = 4 Example 4

5x1 – 2x2 +3x3 = 16

2x1 + 3x2 ‐ 5x3 =2

4x1 – 5x2 + 6x3 = 7

AX = B ⇒ A = 5 2 32 3 54 5 6

=16217

| |= ‐37

A1 = 16 2 32 3 57 5 6

| |= ‐111 = | || |

= = 3

A2 = 5 16 32 2 54 7 6

| |= ‐259 = | || |

= = 7

A3 = 5 2 162 3 24 5 7

| |= ‐185 = | || |

= = 5

solution is = 3, = 7 and = 5

References

1. Edward Dowling : Introduction to Mathematical Economics, Shaum series 2. Manmohan Gupta : Mathematics for Business and Economics 3. Alpha .C. Chiang : Fundamental Methods of Mathematical Economics 4. Barry Bressler : A unified Introduction to Mathematical Economics



MODULE III

FUNCTIONS AND GRAPHS

Part A

FUNCTIONS

Suppose you worked in a shop part time in the evening. You are paid on an hourly basis and you earn Rs. 100 for an hour. The more you work the more you are paid. That means if you work for one hour you get Rs. 100, if you work for two hours, you get Rs. 200 and so on. This implies that the amount of money you earn depends on the time you work. If this sentence is written in mathematical format, we can write the amount of money you earn is a function of the time you work. Here the amount of money you earn is a dependent on the time you work. But the time you work is independent. This is the crux of a functional relation. For example, if we represent the amount you earn by ‘y’ and ‘the time you work by x’ and write in mathematical form, it can be written as y = f (x).

Before seeing the formal definition of a function, let us first see understand the concept of a variable better.

Variable: A variable is a value that may change within the scope of a given problem or set of operations. Thus a variable is a symbol for a number we don't know yet. It is usually a letter like x or y. We call these letters ‘variables’ because the numbers they represent can vary‐ that is, we can substitute one or more numbers for the letters in the expression.

In the above example y = f (x), both y and x are variables. But there is one difference. The amount you earn (y) is depend on the time you work (x). But the length of time you work (x) is independent of the amount you earn(y). Here ‘y’ is called a dependent variable and ‘x’ is called an independent variable. Sometimes the independent variable is called the ‘input’ to the function and the dependent variable the ‘output’ of the function.

Thus dependent variable is a variable whose value depends on those of others; it represents a response, behavior, or outcome that the researcher wishes to predict or explain. Independent variable is any variable whose value determines that of others. Sometimes we also see a extraneous variable, which is a factor that is not itself under study but affects the measurement of the study variables or the examination of their relationships.

Constant: In contrast to a variable, a constant is a value that remains unchanged, though often unknown or undetermined. In Algebra, a constant is a number on its own, or sometimes a letter such as a, b or c to stand for a fixed number.

Coefficients : Coefficients are the number part of the terms with variables. In 3x2 + 2y + 7xy + 5, the coefficient of the first term is 3. The coefficient of the second term is 2, and the coefficient of the third term is 7. Note that if a term consists of only variables, its coefficient is 1. Expressions consisting of a real number or of a coefficient times one or more variables raised to the power of a positive integer are called monomials. Monomials can be added or subtracted to form polynomials.



Variable Expression: A variable expression is a combination of numbers (or constants), operations, and variables. For example in a variable expressions 5a + 3b, ‘a’ and ‘b’ are variables, 5 and 3 are constants and + is an operator.

Function: A mathematical function relates one variable to another. There are lots of other definitions for a function and most of them involve the terms rule, relation, or correspondence. While these are more technically accurate than the definition that we are going to use is

A function is an equation for which any x that can be plugged into the equation will yield exactly one y out of the equation. Let us explore further.

When one value is completely determined by another, or several others, then it is called a function of the other value or values. In this case the value of the function is a dependent variable and the other values are independent variables. The notation f(x) is used for the value of the function f with x representing the independent variable. Similarly, notation such as f(x, y, z) may be used when there are several independent variables.

Here we used ‘f’ to represent a function. We may also represent a function by using symbols like ‘g’ or ‘h’ or the Greek letter ϕ phi. Just as your name signifies all of the many things that make ‘you,’ a symbol like ‘f’’ serves as a shorthand for what may be a long or complicated rule expressing a particular relationship between variable quantities. Then a function y = f (x) shows that f is a rule which assigns a unique value of the variable quantity y to values of the variable quantity x.

Example : (1) y = 3x – 2 (2) h = 5x + 4y

In other words, a function can be defined as a set of mathematical operations performed on one or more inputs (variables) that results in an output. Consider a simple function y = x + 1. In this case, x in the input value (independent variable) and y is the output (dependent variable). By putting any number in for x, we calculate a corresponding output y by simply adding one. The set of possible input values is known as the domain, while the set of possible outputs is known as the range.

x 1 2 3 4 5 6 Domain y 2 3 4 5 6 7 Range

Note that the above table shows that as we give different values to the independent variable x, the function (y) assumes different values.

Now consider the function y = 3x – 2. Here the variable y represents the function of whatever inputs appear on the other side of the equation. In other words, y is a function of the variable x in y = 3x – 2. Because of that, we sometimes see the function written in this form f(x) = 3x – 2. Here f(x) means just the same as “y =” in front of an equation. Since there is really no significance to y, and it is just an arbitrary letter that represents the output of the function, sometimes it will be written as f(x) to indicate that the expression is a function of x. As we have discussed above, a function may also be written as g(x), h(x), and so forth, but f(x) is the most common because function starts with the letter f.



Thus the key point to remember is that all of the following are same, they are just different ways of expressing a function.

y = 3x – 2, y = f (3x – 2), f(x) = 3x – 2, h(x) = 3x – 2, g(x) = 3x – 2

Evaluate a function: To evaluate a function means to pick different values for the independent variable x (often named the input) in order to find the dependent variable y (often named output). In terms of evaluation, for every choice of x that we pick, only one corresponding value of y will be the end result. For example if you are asked to evaluate the function y = 3x – 2 at x = 5, substitute the value at the place of x. Then you get y = 3(5) – 2 = 13. Note that since a function is a unique mapping from the domain (the inputs) to the range (the outputs), there can only be one output for any input. However, there can, be many inputs which give the same output.

Thus, strictly speaking, a function is a rule that produces one and only one value of y for any given value of x. Some equations, such as y =√x , are not functions because they produce more than one value of y for a given value of x.

To plot a graph of a function, as a matter of convention, we denote the independent variable as x and plot it on the horizontal axis of a graph, and we denote the dependent variable as y and plot it on the vertical axis.

Types of functions: (Classification of Functions)

Functions take a variety of forms, but to begin with, we will concern ourselves with the three broad categories of functions mentioned earlier: linear, exponential, and power.

Linear functions take the form y = a + bx where a is called the y‐intercept and b is called the slope. The reasons for these terms will become clear in the course of this exercise.

Exponential functions take the form y = a + qx

Power functions take the form y = a + kxp

Note the difference between exponential functions and power functions. Exponential functions have a constant base (q) raised to a variable power (x); power functions have a variable base (x) raised to a constant power (p). The base is multiplied by a constant (k) after raising it to the power (p).

Here are some more functions with examples.

Linear function: Linear function is a first‐degree polynomial function of one variable. These functions are known as ‘linear’ because they are precisely the functions whose graph is a straight line. Example: When drawn on a common (x, y) graph it is usually expressed as f (x) = mx + b. This is a very common way to express a linear function is named the 'slope‐intercept form' of a linear function.

Equations whose graphs are not straight lines are called nonlinear functions. Some nonlinear functions have specific names. A quadratic function is nonlinear and has an equation in the form of y = ax2 + bx + c where a ≠ 0. Another nonlinear function is a cubic function. A cubic function has an equation in the form of y = ax3 + bx2 + cx + d, where a ≠ 0.



Or, in a formal function definition it can be written as f(x) = mx + b. Basically, this function describes a set, or locus, of (x, y) points, and these points all lie along a straight line. The variable m holds the slope of this line. The variable b holds the y‐coordinate for the spot where the line crosses the y‐axis. This point is called the 'y‐intercept'.

Quadratic function: A polynomial of the second degree, represented as f(x) = ax2 + bx + c, a ≠ 0. where a, b, c are constant and x is a variable. example, value of f(x) = 2x2 + x ‐ 1 at x = 2. Put x = 2, f(2) = 2.22 + 2 ‐ 1 = 9

Single valued function: If for every value of x, there correspond only one value of y, then y is called a single valued function. Example: f(x) = 2x + 3, when x=3, f(3) = 2(3)+3 = 6+3 = 9

Many valued function: If for each value of x, there corresponds more than one value of y, then y is called many or multi valued function.

Explicit function: If the functional relation between the two variables x and y is expressed in the form y = f(x), y is called an explicit function of x. Example: y = 4x ‐5

Implicit function: If the relation between two variables x and y is expressed in the form f(x,y)=0, where x cannot be expressed as a function of y, or y cannot be expressed as a function of x, is called an implicit function. Example: y ‐ 4x ‐ 5 = 0

Odd and Even Functions: When there is no change in the sign of f(x) when x is changed to –x, then that function is called an even function. (i.e) f(‐x) = f(x). The graph of an even function is such that the two ends of the graph will be directed towards the same side.(Figure 1)

Figure 1

When the sign of f(x) is changed when x is changed to –x, then it is called an odd function.

(i.e) f(‐x) = ‐ f(x). The following graph (Figure 2) shows the odd function, f(x)=x3 , and its reflection about the y‐axis, which is f(‐x) = −x3.



Figure 2

Inverse functions: From every function y = f(x), we may be able to deduce a function x=g(y). This means that the composition of the function and its inverse is an identity function. In an inverse function we may be able to express the independent variable in terms of the dependent variable. Function g(x) is inverse function of f(x) if, for all x, g(f(x)) = f(g(x)) = x. A function f(x) has an inverse function if and only if f(x) is one‐to‐one. For example, the inverse of f(x) = x + 1 is g(x) = x ‐ 1

Polynomial functions of degree n : A polynomial is an expression of finite length constructed from variables (also called indeterminates) and constants, using only the operations of addition, subtraction, multiplication, and non‐negative integer exponents. For example, x2 − x/4 + 7 is a polynomial, but x2 − 4/x + 7x3/2 is not, because its second term involves division by the variable x (4/x), and also because its third term contains an exponent that is not an integer (3/2).

Example f(x) = anxn + a n‐1 x n‐1 + … + a0 ( n = nonnegative integer, an ≠ 0)

Rational function: The term rational comes from ‘ratio’

Where g(x) and h(x) are both polynomials and h(x) ≠ 0

Algebraic Functions: A function which consists of finite number of terms involving powers and roots of independent variable x and the four fundamental operations of addition, subtraction, multiplication and division is called an algebraic function. Polynomials, rational functions and irrational functions are all the examples of algebraic functions.

Transcendental functions : Functions which are not algebraic are called transcendental functions. Trigonometric functions, Inverse trigonometric functions, exponential functions, logarithmic functions are all transcendental functions.



Figure 3

Example, f(x) = sinx, g(x) = log(x), h(x)= ex , k(x) = tan−1 (x)

Modulus functions: Modulus functions are defined as follows.

y = |x | = { x, if x ≥ 0

‐ x , if x < 0.

For example, | 3 | = 3, and | ‐4 |= ‐(‐4) = 4 , since ‐4 < 0.

The graph of the modulus function y = |x| is shown below. (Figure 4)

(Figure 4)

Onto functions: A function f(x) is one‐to‐one (or injective),if f be a function with domain D and range R. A function g with domain R and range D is an inverse function for f if, for all x in D, y = f(x) if and only if x = g(y).

One‐to‐one functions: A function f(x) is one‐to‐one (or injective) if, for every value of f, there is only one value of x that corresponds to that value of ‘f’. Eg. f(x) = x + 3 is is one‐to‐one, because, for every possible value of f(x), there is exactly one corresponding value of x.

Identity functions: A polynomial of the first degree, represented as f(x) = x, example, values of f(x) = x, at x = 1,2 are f(1) = 1 and f(2) = 2

Constant functions: It is a polynomial of the ‘zeroth’ degree where f(x) = cx0 = c(1) = c. It disregards the input and the result is always c. Its graph is a horizontal line. For example f(x) = 2, whatever the value of x result is always 2.



Linear functions: It is a polynomial of the first degree, the input should be multiplied by m and it adds to c. It is represented as f(x) = m x + c such as f(x) = 2x + 1 at x = 1.f(1) = 2 . 1 + 1 = 3 that is f(1) = 3

Trigonometrical functions: Trigonometric functions are often useful in modeling cyclical trends such as the seasonal variation of demand for certain items, or the cyclical nature of recession and prosperity. There are six trigonometric functions: sin(x), cos(x), tan(x), csc(x), sec(x) and cot(x)

Analytic functions: All polynomials and all power series in the interior of their circle of convergence are analytic functions. Arithmetic operations of analytic function are differentiated according to the elementary rules of the calculation, and, hence analytic function.

Differentiable functions: A differentiable function is a function whose derivative exists at each point in its domain. If x0 is a point in the domain of a function f, then f is said to be differentiable at x0 if the derivative f '(x0) is defined. The graph of differential functions are always smooth.

Smooth functions: A smooth function is a function that has continuous derivatives over some domain or we can say that, it is a function on a Cartesian space Rn with values in the real line R if its derivatives exist at all points

Even and odd functions: A function for every x in the domain of f is an even function if f(‐x) = f(x) and odd function if f(‐x) = ‐ f(x) for example, f(x) = x2 is an even function because f(‐x) = (‐x)2 = x2 = f(x) and f(x) = x is an even function because f(x) = (‐x) = ‐x = ‐ f(x)

Curves functions: A space curve C is the set of all ordered triples (f(t),g(t),h(t)) together with their definition parametric equations x= f(t) y = g(t) and z = h(t) where f, g, h are continuous functions of t on an interval I.

Composite functions: There is one particular way to combine functions. The value of a function f depends upon the value of another variable x and that variable could be equal to another function g, so its value depends on the value of a third variable. If this is the case, then the first variable which is a function h, is called as the composition of two functions( f and g). It denoted as f o g = (f o g) x = f(g(x)). For example, let f(x) = x+1 and g(x) = 2x then h(x) = f(g(x)) = f(2x) = 2x + 1.

Monotonic functions: A monotonic function is a function that preserves the given order. These are the functions that tend to move in only one direction as x increases. A monotonic increasing function always increases as x increases, that is f(a) > f(b) for all a>b. A monotonic decreasing function always decreases as x increases, that is . f(a) < f(b) for all a>b

Periodic functions: Functions which repeat after a same period are called as a periodic function, such as trigonometric functions like sine and cosine are periodic functions with the period 2∏.

Evaluation of Functions

To evaluate a function, substitute the value of the variable. If a is any value of x, the value of the function f(x) for x = a is denoted by f(a).



Examples

1. Given y = f(x) = x2 + 4x – 5 find f(2) and f(–4)

To find f(2), substitute x = 2

f(2) = 22 + 4(2) – 5 = 7

To find f(–4), substitute x = –4

f(–4) = –42 + 4(‐4) – 5 = 16 –16 – 5 = –5

2. Given y = f(x) = 3x3 – 4x2 + 4x – 10, find value of the function at f(2) and f(‐3)

f(2) = 3(2)3 – 4(2)2 + 4(2) – 10 = 6

f(–3) = 3(–3)3 – 4(–3)2 + 4(–3) – 10 = – 139

3. Given y = f(x) =√ 2 find value of the function when x = 0, x = 7, x= –2

f(0) =√0 2 = √2

f(17) =√7 2 = √9 = 3

f(‐2) =√ 2 2 = 0

4. Given f(x) = x2 + 4x – 11 find f(x+1)

This means we have to evaluate the function when x = x + 1

f(x + 1) = (x + 1)2 + 4(x + 1) – 11

Simplifying using (a+b)2

f(x + 1) = x2 + 2x + 1 + 4x + 4 – 11 = x2 + 6x – 6

5. f(x) = find f(2)

f(x) = = f(x) = = 1

6. Given y = x2 + 1, find f(0), f(–1) and f(3). Also find –

f(0) = 1, f(–1) = 2, f(3) = 10

–

7. Given a function f(x) = 2x + 7, find f(4), f(1) and f(–2)

f(4) = 2(4) + 7, so f(4) = 8 + 7 = 15

f(1) = 2(1) + 7 = 10

f(–2) = 2(–2) + 7 = –4 + 7 = 3

Quantitative M

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Midpoint Formula

The midpoint is an ordered pair formed by finding the average of the x‐values and the average of the y‐values of the given points. The point that bisects the line segment formed by two points, (x1, y1) and (x2, y2), is called the midpoint and is given by the following formula:

2 , 2

Example1: Find the midpoint of the line segment joining P(–3, 8)and Q(4, –4) Use the midpoint formula:

3 42

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12 ,42

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Example 2: Find the midpoint of the line segment joining P(‐5, ‐4)and Q(3, 7)

Use the midpoint formula:

5 32 ,

4 72

22 ,

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Graphs

A graph is a picture of one or more functions on some part of a coordinate plane. Our paper usually is of limited size. The coordinate plane is infinite. Therefore, we need to pick an area of the coordinate plane that would fit on the piece of paper. That means specifying minimum and maximum values of x and y coordinates. To draw the graph of a function y = f(x), the values of the independent variable, x, is marked on the horizontal axis. The values of the dependent variable, y, is placed on the vertical axis. To graph a function we should know the slope of the function. Slope: Slope is used to find how steep a particular line is, or it can be used to show how much something has changed over time. The slope of a line measures the change in y ∆ divided by change in x ∆ . We calculate slope by using the following definition. In Algebra, slope is defined as the rise over the run. This is written as a fraction like this:

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( ‐3/4 )x + 12( 0 ) = 9 ( ‐3/4 )x + 0 = 9 ( ‐3/4 )x = 9 x = 9/( ‐3/4 ) x = ‐12

To find the y‐intercept, set x = 0 and solve for y.

( ‐3/4 )( 0 ) + 12y = 9 0 + 12y = 9 12y = 9 y = 9/12 y = 3/4

Therefore, the x‐intercept is ( –12, 0 ) and the y‐intercept is ( 0, 3/4 ).

The graph of the line looks like this:

Example 4

Find the x and y intercepts of the equation ‐2x + ( 1/2 )y = ‐3.

To find the x‐intercept, set y = 0 and solve for x.

‐2x + ( 1/2 )( 0 ) = ‐3 ‐2x + 0 = ‐3 ‐2x = ‐3 x = ‐3/( ‐2 ) x = 3/2

To find the y‐intercept, set x = 0 and solve for y.

‐2( 0 ) + ( 1/2 )y = ‐3 0 + ( 1/2 )y = ‐3 ( 1/2 )y = ‐3 y = ‐3/( 1/2 ) y = ‐6

Therefore, the x‐intercept is ( 3/2, 0 ) and the y‐intercept is ( 0, ‐6 ).

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Example 2: y = –3x + 2

Y‐intercept = 2, slope = 3

Start by graphing the y‐intercept by going up 2 units on the y‐axis.

From this point go down (rise) 3 units, then 1 unit to the RIGHT (run), and put another point. This is the second point. Connect the points and it should look like this

Steps for Graphing a Line With a Given Slope

Plot a point on the y‐axis. (In the next lesson, Graphing with Slope Intercept Form, you will learn the exact point that needs to be plotted first. For right now, we are only focusing on slope!)

Look at the numerator of the slope. Count the rise from the point that you plotted. If the slope is positive, count up and if the slope is negative, count down. Look at the denominator of the slope. Count the run to the right. Plot your point. Repeat the above steps from your second point to plot a third point if you wish. Draw a straight line through your points. The trickiest part about graphing slope is knowing which way to rise and run if the slope is negative! If the slope is negative, then only one ‐ either the numerator or denominator is negative (NOT Both!) Remember your rules for dividing integers? If the signs are different then the answer is negative! If the slope is negative you can plot your next point by going down and right OR up and left. If the slope is positive you can plot your next point by going up and right OR down and left. Example 1 This example shows how to graph a line with a slope of 2/3.



Example 2

Graph the linear function f given by f (x) = 2x + 4 We need only two points to graph a linear function. These points may be chosen as the x and y intercepts of the graph for example. Determine the x intercept, set f(x) = 0 and solve for x. 2x + 4 = 0

x = ‐2 Determine the y intercept, set x = 0 to find f(0). f(0) = 4 The graph of the above function is a line passing through the points (‐2 , 0) and (0 , 4) as shown below.

Quantitative M

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makes the whole process simpler. Point‐slope form is also used to take a graph and find the equation of that particular line.

Point slope form gets its name because it uses a single point on the graph and the slope of the line.

The standard point‐slope equation looks like this:

Using this formula, If you know:

• one point on the line • and the slope of the line,

you can find other points on the line.

Example 1 :

You are given the point (4,3) and a slope of 2. Find the equation for this line in point slope form.

Just substitute the given values into your point‐slope formula above y – y1 = m(x–x1). Your point (4,3) is in the form of (x1,y1). That means where you see y1, use 3. Where you see x1, use 4. Slope is given, so where you see m, use 2. So we get Y – 3 = 2(x–4)

Example 2 : You are given the point (– 1,5)and a slope of 1/2. Find the equation for this line in point slope form.

Substituting in the formula y – y1 = m(x–x1) we get y – 5 = 0.5(x– ‐1). Ie. y – 5 = 0.5(x+1).

Point‐slope form is about having a single point and a direction (slope) and converting that between an algebraic equation and a graph. In the example above, we took a given set of point and slope and made an equation. Now let's take an equation and find out the point and slope so we can graph it.

Example:

Find the equation (in point‐slope form) for the line shown in this graph:



To write the equation, we need a point, and a slope. To find a point because we just need any point on the line. This means you can select any point as per your convenience. The point indicated in the figure is (–1,0). We have chosen this point for our convenience since it is the easiest one to find. We choose it also because it is useful to pick a point on the axis, because one of the values will be zero.

Now let us find slope. Just count the number of lines on the graph paper going in each direction of a triangle, like shown in the figure. Remember that slope is rise over run, or y/x. Therefore the slope of this line is 2. Putting it all together, our point is (–1,0) and our slope is 2, the point‐slope form is y – 0 = 2(x + 1)

Example

Write the equation of the line passing through the points (9,‐17) and (‐4,4).

Calculate the slope:

m = – 17 – 4 9 4

m = – 21 / 13

Use the point‐slope formula:

y – 4 = (– 21/13)(x + 4)

y – 4 = (– 21/13)x – 84/13

y = (– 21/13)x – 32/13

Quantitative M

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4. Find the equation of a line which passes through the point (3, 4) and makes intercepts on the axes equl in magnitude but opposite in sign.

Solution : Let the x‐intercept and y‐intercept be a and –a respectively

The equation of the line is

1

x – y = a … (i)

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3 – 4 = a or

a = –1

Thus, the required equation of the line is

x – y = – 1

or x – y + 1 = 0

4. Determine the equation of the line through the point (– 1,1) and parallel to x – axis Since the line is parallel to x‐axis its slope ia zero. Therefore from the point slope form of the equation, we get

y – 1 = 0 [ x – (– 1)]

y – 1 = 0

which is the required equation of the given line.

StraightLine Equations: Standard Form

In the Standard Form of the equation of a straight line, the equation is expressed as: ax + by = c where a and b are not both equal to zero. 1. 7x + 4y = 6 2. 2x – 2y = –2 3. –4x + 17y = –432

Exercise

1. Graph a linear function 3

To draw graph of this function we should first find two points (the x intercept and the y intercept) which satisfy the equation. Then we join these two points using a straight line.



This is because what we are given is the equation of linear function. For a linear function, all the points satisfying the equation must lie on the line.

To find the y intercept, let us use our knowledge of the definition of intercept, ie, the y intercept is the point where the line touches the y axis. This means at this point x = 0. So to find the y intercept, set x equal to zero.

14 0 3

3

So the y intercept is (x,y) = (0,3)

Similarly to find the x intercept, put y = 0.

0 14 3

14 3

12

So the x intercept is (x,y) = (10,0)

Now plot the two intercept points (0,3) and (12,0) and connect them with a straight line.

2. Graph the linear function 2y + 10x = 20

To find the y intercept, put x = 0

2y + 10(0) = 20, 2y = 20, y = 10


To find the x intercept, put y = 0

2(0) + 10x = 20, 10x = 20, x = 2


‐0.6

‐0.5

‐0.4

‐0.3

‐0.2

‐0.1

0

‐2 ‐1.5 ‐1 ‐0.5 0



Now plot the two intercept points (0,10) and (2,0) and connect them with a straight line.

Alternatively, you may first of all rewrite the equation 2y + 10x = 20 by solving for Q,

2y = 20 – 10x, , 10 5 , 5 10

Now the function is in y = mx + c from. From this form, we can easily find the slope (m)of the function which is – 5.

3. Graph the linear function 2y ‐10x = 20


2y + 10(0) = 20, 2y = 20, y = 10



2(0) – 10x = 20, –10x = 20, x = –2

So the x intercept is (x,y) = (–2,0) Now plot the two intercept points (0,10) and (–2,0) and connect them with a straight line.

‐0.6

‐0.5

‐0.4

‐0.3

‐0.2

‐0.1

0

‐1.6 ‐1.4 ‐1.2 ‐1 ‐0.8 ‐0.6 ‐0.4 ‐0.2 0



4. Graph the linear function 2y – 10x + 20 = 0


2y – 10(0) – 20 = 0, 2y = –20, y = –10

So the y intercept is (x,y) = (0, –10)


2(0) – 10x + 20 = 0, –10x = –20, x = 2


Now plot the two intercept points (0, –10) and (2,0) and connect them with a straight line.

5. Given a linear function 6y + 3x – 18 = 0. Also find its slope.

Rewrite in the y = mx + c form

6y + 3x – 18 = 0

‐0.6

‐0.5

‐0.4

‐0.3

‐0.2

‐0.1

0

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‐0.4

‐0.3

‐0.2

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6y = – 3x + 18

3

Slope =


0 3 , y = 3

So the y intercept is (x,y) = (0, 3)


0 3

3

3 2 3 6


Now plot the two intercept points (4, 12) and (8,2) and connect them with a straight line.

6. Find the slope m of a linear function passing through (0,10), (2,0)

Substituting in the formula

2 128 4

104

Graphing of Non Linear Functions

‐0.6

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‐0.3

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MODULE IV LIMITS AND CONTINUTY OF A FUNCTION

LIMITS The concept of limits is useful in developing some mathematical techniques and also in analyzing various economic problems in economics. For example, the rate of interest ‘i’ can never fall to zero even if the quantity of capital available in the economy is very large. There is a minimum below which ‘i’ cannot fall. Let this minimum be 2%, then if ‘K’ stands for capital and ‘C’ is a positive constant we can write the function as

i 2

If K is small ‘i’ is large

If K is large ‘i’ is small If K is very large infinite ‘i’ will never be less than 2. Therefore the limit ‘i’ as K increase is 2.

It can be written as LtK ∞ i 2

A function f x is said to approach the limit ‘l’ as ‘x’ approach to ‘a’ if the difference between f x and ‘l’ can be made as small as possible by taking ‘x’ sufficiently nearer to ‘a’. Example 1: Limit x2 as x approach to 4 is 16. This can be expressed as

Ltx 4 x 16

i.e., the difference between x and 16 can be made as small as possible by taking ‘x’ sufficiently nearer to 4.

Ex. 2: If f x 5x 10 find Ltx 0 f x

Substituting x 0 Ltx 0 f x 10

Ex. 3: Find Ltx 3 10x 5

Ans: Ltx 3 10x 5

Ltx 3 10 Lt

x 3xLt

x 35

9 30 5 44



Theorems on Limit of Function 1. If ‘a’ is a constant

Ltx K a = a

2. If a, b are constants

Ltx K ax b ak b

3. If Ltx K f x l, Lt

x K g x m then

i Ltx K f x g x l m

ii Ltx K f x . g x l m

iii Ltx K , m 0

iv Ltx K √

v Ltx K f x l , Lt

x K g x 0

Ltx K f x . g x 0

Ex. 3: Find Ltx 2

Ans: Ltx 2

Ex. 4: Find Ltx 3 2 5



Ans. Ltx 3 2 5 3 2 5

5 5 25 5 30

Indeterminate Form

If the value of the function becomes indeterminate like , on substituting the value of the variables, mere substitution method may not be feasible. In such case factorization of the function may remove the difficulties.

For example: Ltx 3

write as

x 3 6

Note a b a – b

Ex. 2: Ltx ∞

Ltx ∞

division of numerator and denominator by

Since

0 as x ∞

We go 1

Ex. 3: Ltx 3

Factorizing the function we get



Problems

1. Find Ltx 2

Ans: ‐1

2. Find Ltx 1 Ans: 2

3. Find Ltx 2

Ans:

4. Find Ltx 0

Ans: 8

5. Find Ltx ∞

Ans: 2

6. Find Ltx 6

Ans:

CONTINUITY

A function is said to be continuous in a particular region if it has determinate value or a finite quantity for every value of the variable in that region.

i.e., f x may or may not be the same as f a

If f x f a a finite quantity, then f x is said to be a continuous function of x at x a.

If a function is continuous over a range it is continuous at every point of the range and can be represented by a curve which has no gaps and no jumps in the range concerned. Therefore a function f x is said to be continuous at x a if f x tends to f a from either side.

f x f x f a

Where x refers to f a + h) and x refers to f(a – h) as h 0.



Condition for continuity

1. F a exists

2. f x f a , a finite quantity.

If the conditions are not satisfied for any value of x, f x is said to be discontinuous for that value of ‘x’.

Eg: If f x and f x

If x 2. f x does not have a finite value

f x is continuous for any value of ‘a’ except 2.

Eg. 1: Check for continuity of function at x 1

f x

Solution : 1 f x 1

1 x 4 5

But f 1 does not exist

i.e., substituting 1 for x in f x function

∞ indeterminate

Thus f x is not continuous at x 1

Eg. 2: Prove that the function 3 2 is continuous for x 2

Ans: f 2 2 3 2 2 8 is finite.

2 f x 0 2 3 2 2

2 3 2 2 8

2 f x 2 f x f 2

f x is continuous at x 2.



DIFFERENTIATION

Differentiation is the process of finding the rate at which a variable quantity is changing. It concerned with determination of the rate of change in the dependent variable for a given small change in the independent variable.

Let ‘x’ be an independent variable and ‘y’ be depending upon ‘x’, then ‘y’ is a function of ‘x’.

Let ∆x be small change in ‘x’ corresponding change in ‘y’ is ∆y. Then is called ‘incremental ratio’.

The value of incremental ration when ∆x is very small is called differential coefficient or derivative of y with respect to ‘x’ and is denoted by

i.e., ∆x 0

implies derivative of ‘y’ with respect to ‘x’.

Derivative or differential coefficient of f x is generally written as f x or f x or

or where y f x .

Derivation is the rate of change of y with respect to a change in x.

Rules of Differentiation

Rule No. 1: Power Function Rule or Polynomial Function Rule.

The derivative of a power function y f x where ‘n’ is any real number is

i.e., If y

Examples

1 y , 8 8

2 y , 10 10

3 y , 5 3 5 3

4 y ,



5 y √ , y , ½ ½

6 y , 1 1

Rule No. II: Derivative of a Constant

The derivative of a constant function is zero

If y k where k is constant

Then c 0

Eg:‐ If y 10 0

Rule No III: Derivative of the Product of a constant and a function.

The derivative of the product of a constant and a function is equal to the product of the constant and the derivative of the function.

If y where ‘a’ is constant and a≠0

a

Ex. 1 If y 5 find

5 5 7 35

Ex. 2 If y ‐30 find

‐30 ‐30 ‐5

150

Rule No. IV Linear Function Rule

If y mx c where ‘m’ and ‘c’ are constant then y

mx c m

Ex: 1 If y 5x 10 find

5



Ex:2 If y 23 5 find

23

Rule No. V: Addition Rule or Sum Rule

The derivative of a sum of two function is simply equal to the sum of the separate derivatives.

If y u f x and v g x are the two function then u v Derivative of

the first function Derivative of Second

Ex.1: If y 5 3 2

5 3 2

35 6

Ex.2: If y 10 3 4 10 find

60 12 8

Rule No. VI: Subtraction Rule

The derivative of a difference two function is equal to the difference of the separate derivatives.

If y u – v then

ie derivative of the first functions ‐ minus derivative of the second function

Ex. 1: If y 4 50 find

4 50

9 24

Ex. 2 If y 20 2 4 ‐ 50 find

20 2 4 50

60 8 – 4



Rule No. VII: Product Rule or Multiplication Rule

The derivative of the product of two functions is equal to the first function multiplied by the derivative of the second function plus the second functions multiplied by the derivative of the first function.

If u f x and v g x are the two given function then

u. v u

v

Example:

1. If y 3 5 5 7 find

Solution: consider u 3 5, v 5 7

6x,

15

3 5 15 5 7 6

45 75 30 42

75 75 42

2. Find if y

u v

2 1 4 3

2 2 4 3 4 3

6 10 4

Rule No. VIII: Quotient Rule or Division Rule

The derivative of the quotient of two functions is equal to the denominator multiplied by the derivative of the numerator minus the numerator multiplied by the derivative of the denominator, all divided by the square of the Denominator.

If y Where u f x and v g x then

.



Example

1. If y find

–

2. If y find

Rule No. IX : Chain Rule or Function of Function Rule

If ‘y’ is the function of ‘u’ where ‘u’ is the function of ‘x’, then the derivative of ‘y’ with respective to x i.e is equal to the product of the derivative of ‘y’ with respect of ‘u’ and the derivative of ‘u’ with respect of ‘x’.

i.e., If y f u , where u f x

then

Example:

1. If y 7u 3 where u 5x find

7, 10x



7 10x 70x

2. If y z 10 and z x find

3z , 7x

3z 7x Sub. x for z we get

3 x 7x 3x 7x 21x

3. Find if y x 3x we find by the chain rule

Let u x 3x

y u

2u 3x 3

2 x 3x 3x 3 since u 3

2x 6x 3x 3 6 6 18 18

6 24 18

Rule No. X: Parametric Function Rule

If both ‘x’ and ‘y’ are the differential function of ‘t’ then the derivative of ‘y’ with represent to ‘x’ is obtained by dividing the derivative of ‘y’ with respect to ‘t’ by the derivative of ‘x’ with respect to ‘t’ . i.e., if x f x and y g t then

÷

Examples:

Find if x and y 4

÷



4a, = 4

= =

Rule No. XI: Implicit Function Rule.

An equation y = 4 2 10 directly expresses ‘y’ in terms of ‘x’. Hence this function is called ‘Explicit Function’.

An equation 2 = 0 does not directly express ‘y’ in terms of ‘x’. This types of function is called ‘ Implicit Function’.

In order to find the derivative of Implicit Function every term in both sides of the function is to be differentiated with respect to ‘x’.

Example:

1. Find of the following function 3x – 2y = 4

3x ‐

2y =

(4)

3 ‐ 2 = 0, ‐ 2 = ‐3, = = 1.5

2. xy 6

x y + y

x = (6) (Product Rule)

+ y = 0

3. 2 = 10

x + 2y

(x) +2x (y)+ (y ) = 10

= 2x +2y+2x + 2y = 0

= 2x +2y = ‐(2x +2y)

= (2x+2y) = ‐(2x +2y)

‐1

Rule No. XII: Inverse Function Rule



The derivative of the inverse function i.e., is equal to the reciprocal of the

derivative of the original function i.e.,

i.e.,

Example 1. Find the of the function y

4

2. Find the derivative of the inverse function

y x 2 x 4

x 2 4x x 4 2x

4x 8x 2x 8x

6x 8x 8x

Rule No. XIII: Exponential Function Rule.

The Derivative of Exponential function is the exponential itself. If the base of the function is ‘e’ the natural base of an exponent.

Example

1. If y , then

2. If y , then

3. If y , then 2

4. Find if y ,

Let u 2

y



2x

2x since u 2

Note:‐ if y when u g x then

.

5. Find if y

Let u

2x

2x

6. Find if y

Let u 4 6 10

12 12

12 12

12 12

Rule No XIV: Logarithmic Function Rule

The derivative of a logarithm with natural base such as y log x, then

If y log u, when u g x , then .

.

.

Example



1. If y log find

Let u y log u

. 3x

since u

Rule No. XV: Differentiation of one function with respect to another function.

Let f x and g x be two functions of ‘x’, then, derivative of f x with respect to g x is

i.e., differentiate both the function and divide

Examples:‐

1. Differentiate 1 with respect to 1

2x

2x

1

2. Differentiate with respect to

Let u

Let v



÷ ÷

Rule No XVI: Derivative of a Derivative or Higher order Derivative or Successive Derivative

The first order derivative of ‘y’ with respect to ‘x’ is or , then second order derivative is obtained by differentiating the first order derivative with respect to ‘x’ and is written as

and is denoted as

Example:

1. If y 5 3 8 2 find , and

15x 6x 8

30x 6

30

2. If y x x x find ,

4x 2x

12x 2

24x

MAXIMA AND MINIMA

Maxima

A function f x is said to have attained its ‘Maximum value’ or ‘Maxima’ at x a, if f x increase up to x a, stops to increase and begins to decrease at x a.

Minima



A function f x is said to attained its ‘Minimum Value’ or Minima at x a, if the function decrease up to x b and begins to increase at x b

Condition for Maxima & Minima

1. A function is said to be maximum at x a if is zero and 0 or negative.

2. A function is said to be minimum at x b if 0 and 0 or positive

Y A B

O a b X

In the diagram, point A is the maximum point because y has a maximum value at this point when x O a. Therefore we say that at point A, Y is maximum i.e., the value of Y at A is higher than any value on either side of A. As value of X increase from ‘a’ to ‘b’, the slope of the curve decreasing or the slope curve changes from zero to negative. Therefore A is maximum when

0 and 0 or negative

Similarly point B is Minimum when 0 and 0 or positive

Point of Inflection



For a single valued function, the function may become concave from convex or convex from concave at a point, that point is called the point of inflection.

Criterion for point of Inflection

1. 0

2. 0

3. 0

Examples:

1. Find the maximum or minimum of the function y x 4x 10

y x 4x 10

2x – 4

At the maximum or minimum 0

Therefore 2x – 4 0 2x 4, x 4/2 2 The function may be maximum or minimum at x 2

2 0 positive

Therefore the function has minimum at x 2. 2. Find the maximum or minima of the function

y 2x 6x Solution

6x 6

At maxima or minima 0

6x 6 0

6x 6

x 1

x 1 At x ‐1 pr 1 the function is either maxima or minima.

12x

When x 1, 12 1 12 positive



The function is maxima.

When x ‐1, 12 ‐1 ‐12 negative

The function is maxima Problems

Ex. 1: The total revenue function of a firm is given by R 21x x and its total cost function is C 1 3 3 7 10 where x is the output. Find a the output at which the total revenue is maximum b the output at which the total cost is minimum. Ans: a By differentiating the total revenue function we can obtain the maximum value

R 21x x

21 – 2x

‐2 negative

Taking 0 we get

21 – 2x 0

21 2x, x 21 2 10.5

Since is negative. Revenue is maximum

When output x 10.5 b By differentiating the total cost function we can obtain the minimum cost.

C 1 3 3 7 10

6 7

2x – 6

Taking 0 we get

6 7 Solving this quadratic equation, we get, x 7, ‐1 Since output cannot be negative x 7

Substituting x 7 in we get

2 7‐6 8 positive So total cost is minimum when output is 7



Increasing & Decreasing Function

A function is an increasing function if the value of the function increases with an increase in the value of the independent variable and decrease with a decrease in the value of the independent variable. i.e., y f x is an increasing function if , for , domain of the function. Similarly the value of a decreasing function increasers with a fall in the value of independent variable and decrease with an increase in the value of independent variable. i.e., , for , domain of the function. The derivative of a function can be applied to check whether given function is an increasing function or decreasing function in a given interval. If the sign of the first derivative is positive, it means that the value of the function f x increase as the value of the independent variable increases and vice versa. i.e., a function y f x differentiable in the interval a, b is said to be an increasing function if and only if its derivatives on a, b is non negative.

i.e., 0 in the interval a, b

A function y f x differentiable in the interval a, b is said to be a decreasing function is and only if its derivative on a, b is non positive.

i.e., 0 in the interval a, b

The derivative of a curve at a point also measure the slope of the tangent to the curve at that point. If the derivative is positive, then it means that the tangent has a positive slope and the function curve increases as the value of independent variable increase. Similar interpretation is given to the decreasing function and negative slope of the tangent. Y

O x1 x2 X

Y

O x1 x2 X

Increasing Function Decreasing Function



Ex: Check whether the function 3x 3x x 10 is increasing or decreasing Solution

3x 3x x 10

9x 6x 1 3x 1

The function is always positive because 3x 1 is positive since square of a real number is always positive.

Therefore 3x 3x x 10 is increasing.

Convex and Concave Function

For a monotonic function we know the functions increases or decreases. But we do not know whether the function increases or decreases at an increasing rate or at a decreasing rate. The sign of second derivative shows it.

Convex Function If second derivative of a function f x is positive i.e., 0, then the function is said to be a convex function in the given interval.

Concave Function If the second derivative of a function f x is negative i.e., 0, then the function is said to be a concave of function in the given interval. Y O X

Y O X

Convex Function Concave Function



Ex: Show that the curve of y 2x – 3 1/x convex from below for all positive values of x. Solution y 2x – 3 1/x

2 1

2

Since 2 0 for x 0 the curve is convex from below for all positive value of x.

Ex:2 Show that the demand curve P ‐8 is downward sloping and convex from below.

We have,

P ‐8

Since slope is negative the demand curve is downward sloping.

is positive, the demand curve is convex from below.

PARTIAL AND TOTAL DIFFERENTIATION

A variable may depend up on more than one independent variables. For example quantity demand for a commodity may depend on price, Price of related commodity, income of the consumer, tastes and preferences etc. Thus we can write Q f Px, Y, T, …. where Q represent quantity demand of x, Px price of x, Y, income of consumer.

Partial Derivatives When a dependent variable depends on a number of independent variables, then differential coefficient of the dependent variable with respect to one of the independent variable, keeping the other independent variables as constant is called partial differential coefficient. Let Z f x, y

Then derivative of Z with respect to x when y is held constant is called the partial derivative of Z with respect to x and is denoted by . Similerly the derivative of Z with respect to y, when ‘x’ is held constant is called the partial derivative of Z with respect to ‘y’ and is denoted by .



Example

5xy 5y

5xy 5x

2x 0 2x

0 2x 2y

x .

Example

If Z 2 3 4 find and at x 5, y 4

4x ‐ 3y

‐3x 8y

at x 5 and y 4

4 5 – 3 4 20 – 12 8

at x 5 and y 4

‐3 5 8 4 ‐15 32 17 Second order partial derivatives

1. 2.

3. 4.



Example : Find the first and the second order partial derivatives for Z 3 2 28

Ans:

9 4 2

2 2 2 3 2 4 3

‐4x 4y

‐4x 4y

18x – 4y

4x 6y

TOTAL DIFFERENTIATION Let ‘x’ and ‘y’ be two independent variables and ‘z’ be depending on them, so that ‘z’ is a function of x and y. We can write z f x, y The change in ‘z’ corresponding to a very small change in ‘x’ when ‘y’ remain constant is .

Let ∆x be small change in x, then the change in z will be ∆x

Similarly when ∆y is a small change in y then ∆y is the corresponding change in z.

The total change in Z ∆z ∆x ∆y

when ∆x is very small.

In the limit case.

dz dx dy

This is called total differential of z with respect to ‘x’ and ‘y’

Ex: Find the total differential of z 3 2

Ans: dz dx dy

6x y dx x – 6y2 dy



MODULE V

FINANCIAL MATHEMATICS Part A Concepts Financial Mathematics [

“The special sphere of finance within economics is the study of the allocation and deployment of economic resources...in an uncertain environment.” From Robert Merton’s Nobel Prize lecture

Financial Mathematics is a collection of mathematical techniques that are used in the area of finance. The discipline has close relation to financial economics, which is concerned with much of the underlying theory. Thus financial mathematics is applied in nature. This has been included in your syllabus, because without the knowledge of the basic tools of financial mathematics, one can not apply the theories of macro economics or monetary economics in a real world situation. Examples are Asset pricing: derivative securities, Hedging and risk management, Portfolio optimization and Structured products.

Financial economics is the branch of economics studying the interrelation of financial variables, such as prices, interest rates and shares, as opposed to those concerning the real economy. Financial economics concentrates on influences of real economic variables on financial ones, in contrast to pure finance.

It studies the following: Valuation ‐ Determination of the fair value of an asset, How risky is the asset? (Identification of the asset appropriate discount rate), What cash flows will it produce? (Discounting of relevant cash flows), How does the market price compare to similar assets? (Relative valuation), Are the cash flows dependent on some other asset or event? (Derivatives, contingent claim valuation)

Financial Econometrics is the branch of Financial Economics that uses econometric techniques to parameterize the relationships.

The financial markets worldwide have seen a tremendous growth in the last four decades. This was driven largely by financial innovation riding on complex pricing and hedging formulas provided by pioneering developments in mathematical finance. Mathematical finance also contributed strongly by providing quantitative models for investment in portfolio of assets and in managing diverse and complex market, credit and operational risks. Financial derivatives were introduced in Indian markets in 2000. Since then they have grown tremendously in trade volume. However, India currently lacks a critical mass of researchers and practitioners adept in further developing and implementing sophisticated ideas in Financial Mathematics.

It is important to acknowledge that financial markets serve a fundamental role in



economic growth of nations by helping efficient allocation of investment of individuals to the most productive sectors of the economy. They also provide an avenue for corporates to raise capital for productive ventures. Financial sector has seen enormous growth over the past thirty years in the developed world. This growth has been led by the innovations in products referred to as financial derivatives that require great deal of mathematical sophistication and ingenuity in pricing and in creating an insurance or hedge against associated risks. History of Financial Mathematics: The origins of much of the mathematics in financial models traces to Louis Bachelier’s 1900 dissertation on the theory of speculation in the Paris markets. Completed at the Sorbonne in 1900, this work marks the twin births of both the continuous time mathematics of stochastic processes and the continuous time economics of option pricing. While analyzing option pricing, Bachelier provided two different derivations of the partial differential equation for the probability density for the Wiener process or Brownian motion. In one of the derivations, he works out what is now called the Chapman‐Kolmogorov convolution probability integral. Along the way, Bachelier derived the method of reflection to solve for the probability function of a diffusion process with an absorbing barrier. Not a bad performance for a thesis on which the first reader, Henri Poincaré, gave less than a top mark! After Bachelier, option pricing theory laid dormant in the economics literature for over half a century until economists and mathematicians renewed study of it in the late 1960s. Jarrow and Protter speculate that this may have been because the Paris mathematical elite scorned economics as an application of mathematics. The first influential work of mathematical finance is the theory of portfolio optimization by Harry Markowitz on using mean‐variance estimates of portfolios to judge investment strategies, causing a shift away from the concept of trying to identify the best individual stock for investment. Using a linear regression strategy to understand and quantify the risk (i.e. variance) and return (i.e. mean) of an entire portfolio of stocks and bonds, an optimization strategy was used to choose a portfolio with largest mean return subject to acceptable levels of variance in the return. Simultaneously, William Sharpe developed the mathematics of determining the correlation between each stock and the market. For their pioneering work, Markowitz and Sharpe, along with Merton Miller, shared the 1990 Nobel Memorial Prize in Economic Sciences, for the first time ever awarded for a work in finance. The portfolio‐selection work of Markowitz and Sharpe introduced mathematics to the study of investment management. With time, the mathematics has become more sophisticated. Thanks to Robert Merton and Paul Samuelson, one‐period models were replaced by continuous time, Brownian‐motion models, and the quadratic utility function implicit in meanâvariance optimization was replaced by more general increasing, concave utility functions. The next major revolution in mathematical finance came with the work of Fischer Black and Myron Scholes along with fundamental contributions by Robert C. Merton, by modeling financial markets with stochastic models. For this M. Scholes and R. Merton were awarded the 1997 Nobel Memorial Prize in Economic Sciences. Black was ineligible for the prize because of his death in 1995. Why as study of financial economics is needed: Finance may be defined as the study of



how people allocate scarce resources over time. The outcomes of financial decisions (costs and benefits) are spread over time. They are not generally known with certainty ahead of time, i.e. subject to an element of risk. Decision makers must therefore be able to compare the values of cash flows at different dates and take a probabilistic view.

Modern mathematical finance theory begins in the 1960s. In 1965 the economist Paul Samuelson published two papers that argue that stock prices fluctuate randomly . One explained the Samuelson and Fama efficient markets hypothesis that in a well‐functioning and informed capital market, asset‐price dynamics are described by a model in which the best estimate of an asset’s future price is the current price (possibly adjusted for a fair expected rate of return.) Under this hypothesis, attempts to use past price data or publicly available forecasts about economic fundamentals to predict security prices are doomed to failure. In the other paper with mathematician Henry McKean, Samuelson shows that a good model for stock price movements is geometric Brownian motion. Samuelson noted that Bachelier’s model failed to ensure that stock prices would always be positive, whereas geometric Brownian motion avoids this error .

The most important development in terms of practice was the 1973 Black‐Scholes model for option pricing. The two economists Fischer Black and Myron Scholes (and simultaneously, and somewhat independently, the economist Robert Merton) deduced an equation that provided the first strictly quantitative model for calculating the prices of options. The key variable is the volatility of the underlying asset. These equations standardized the pricing of derivatives in exclusively quantitative terms. The formal press release from the Royal Swedish Academy of Sciences announcing the 1997 Nobel Prize in Economics states that the honor was given “for a new method to determine the value of derivatives. Robert C. Merton and Myron S. Scholes have, in collaboration with the late Fischer Black developed a pioneering formula for the valuation of stock options. Their methodology has paved the way for economic valuations in many areas. It has also generated new types of financial instruments and facilitated more efficient risk management in society.”

Growth rate: Simple and Compound

Growth Rate is the amount by which a variable increases over a given period of time as a percentage of its previous value. For example, a 3% growth rate in GDP for a year means that the value of an economy is 103% of the value of the previous year.

Simple Growth Rate

Hope you remember the Aesop’s fable of the golden goose that lays golden eggs. The simple growth rate pays a fixed amount of return over time. Like the goose, every day it laid a single golden egg. It could not lay faster (more than one egg per day), and the eggs did not grow into golden geese of their own. A simple growth rate is linear. That is to say, the increase in each period is some fraction of the initial value.

A compound growth rate is exponential. It is the measure of how much something grew on average, per year, over a multiple‐year period, after considering the effects of compounding. Increase in each period is a specified fraction of the value in the prior period. The distinction between simple growth rate and compound growth rate is the same as the difference between simple interest and compound interest. In compound interest after the



fixed period of say 6 months or one year the interest accrued till then is added to the capital. Therefore for the next period of 6 months or 1 year the added interest also gets interest. i.e. the interest is compounded. Albert Einstein called compound interest “the greatest mathematical discovery of all time”.

The compounded annual growth rate (CAGR) is the rate at which something (e.g., revenue, savings, population) grows over a period of years, taking into account the effect of annual compounding. A compound is composed of two or more parts. In the case of compound growth, the two parts are principal and the amount of change in the principal over a certain time period, which is called ‘interest’ in some circumstances (as we saw above). This is sometimes called ‘growth on growth’ because it measures periodic growth of a value that is itself growing periodically. If we are calculating the annual compound growth rate, then each year the new basis is the previous basis plus the growth over the previous period. In looking at an investment, the CAGR is a measure that is commonly used to show how quickly the investment, or certain aspects of it, such as gross sales, have been growing. Investment analysts often look at five‐year periods to discern a trend. A specific company’s rate of growth is often then compared with that of competitors or with the industry as a whole.

For example consider an investment of Rs.100 at 10% annual interest. Let us see the difference between simple interest and compound interest. As per simple growth rate, Total Value at the end of year 1 will be Rs. 110, at the end of year 2 will be Rs. 120 and year 3 will be Rs. 130. As per compound growth rate, Total Value at the end of year 1 will be Rs. 110, at the end of year 2 will be Rs. 120 and year 3 will be Rs. 130. The Compound Total Value at the end of years 1,2, 3 will be Rs. 110, Rs. 121, Rs. 133.10 respectively.

Depreciation

Depreciation may be defined as a method of allocating the cost of a tangible asset over its useful life. Businesses depreciate long‐term assets for both tax and accounting purposes. Depreciation is also defined as a decrease in an asset’s value caused by unfavorable market conditions. For accounting purposes, depreciation indicates how much of an asset’s value has been used up. For tax purposes, businesses can deduct the cost of the tangible assets they purchase as business expenses; however, businesses must depreciate these assets in accordance with tax rules about how and when the deduction may be taken based on what the asset is and how long it will last.

Depreciation is a measure of the wearing out, consumption or other loss of value of a depreciable asset arising from use, effluxion of time or obsolescence through technology and market changes. Depreciation is allocated so as to charge a fair proportion of the depreciable amount in each accounting period during the expected useful life of the asset. Depreciation includes amortization of assets whose useful life is predetermined. Depreciation is used in accounting to try to match the expense of an asset to the income that the asset helps the company earn. For example, if a company buys a piece of equipment for Rs10 lakh and expects it to have a useful life of 10 years, it will be depreciated over 10 years. Every accounting year, the company will expense Rs.100,000 (assuming straight‐line depreciation), which will be matched with the money that the equipment helps to make each year. Time Value of Money



Time literally is money ‐ the value of the money we have now is not the same as it will be in the future and vice versa. So, it is important to know how to calculate the time value of money so that you can distinguish between the worth of investments that offer you returns at different times.

The time value of money serves as the foundation for all other notions in finance. Time Value of Money. The concept is based on the concept that a dollar that you have today is worth more than the promise or expectation that you will receive a dollar in the future. Money that you hold today is worth more because you can invest it and earn interest. After all, you should receive some compensation for foregoing spending. For instance, you can invest your one rupee for one year at a 6% annual interest rate and accumulate Rs. 1.06 at the end of the year. You can say that the future value of one rupee is Rs. 1.06 given a 6% interest rate and a one‐year period. It follows that the present value of the Rs. 1.06 you expect to receive in one year is only Rs. 1. But why is this? Rs. 100 has the same value one year from now, doesn't it? Actually, although the amount is the same, you can do much more with the money if you have it now because over time you can earn more interest on your money.

In the above table Option A shows what happens to Rs. 10000 after 3 years. The Rs. 10000 you invest today grows to Rs. 10000 + interest at the end of three years. Option B shows what is the value of Rs. 10000 you get after three years. That’s is, the future value for Option B, would only be Rs. 10,000. So how can you calculate exactly how much more Option A is worth, compared to Option B? Future Value If you choose Option A and invest the total amount at a simple annual rate of 5%, the future value of your investment at the end of the first year is Rs.10, 500, which of course is calculated by multiplying the principal amount of Rs.10,000 by the interest rate of 5% and then adding the interest gained to the principal amount: Future value of investment at end of first year: = (Rs. 10,000 × 0.05) + Rs.10,000 = Rs.10,500 This is a direct method of calculation and it is easy for calculation of one year. But if you want to calculate for more than one year, this calculation may turnout to be complex. So we use a formula to make the same calculation, which will make calculations for

Present Value

0 1 2 3

←Years

Option A

Rs. 10000 Rs. 10000 + interest

Option B

Rs. 10000 – interest Rs. 10000



more than one year easy.

Future value = original amount × (1 + interest rate per period)Number of periods

Using the above formula for calculation future value of investment at end of first year:

Future value = 10000 × (1 + 0.05)1 = 10000×1.05 = 10500 Future value of investment at end of 2 years:

Future value = 10000 × (1 + 0.05)1+1 = 10000×1.052 = 10000 × 1.1025 = 11025 Future value of investment at end of 3 years:

Future value = 10000 × (1 + 0.05)1+1+1 = 10000×1.053 = 10000 × 1.16 = 11600 Future value of investment at end of 10 years:

Future value = 10000 × (1 + 0.05)10 = 10000×1.0510 = 10000×1.63 = 16288.95 In general future value of investment at end of n years:

Future value = P(1+i)n

Present Value

Now consider option B in the above table. Here the question is to find the present value of a future sum. If you received Rs.10,000 today, the present value would nothing but Rs.10,000 because present value is what your investment gives you now if you were to spend it today. If Rs.10, 000 were to be received after one year, the present value of the amount would not be Rs.10,000 because you do not have it in your hand now, in the present. To find the present value of the future Rs.10, 000 we need to find out how much we would have to invest today in to receive that Rs.10,000 in the future, say after one year.

To calculate present value, or the amount that we would have to invest today, we must subtract the (hypothetical) accumulated interest from the Rs. 10,000. To achieve this, we should do discounting, that is to discount the future payment amount (Rs. 10,000) by the interest rate for the period. For this all we have to do is rearranging the future value equation above so that you may solve for P. The above future value equation can be rewritten by replacing the P variable with present value (PV) and manipulated as follows: Original equation: Future Value = original amount × (1 + interest rate per period)Number of periods Future Value = P(1+i)n Replacing the P variable with present value PV, and writing FV for future value, we get F V = PV(1+i)n Divide both sides by (1+i)n

=



= PV Bringing (1+i)n to the numerator, PV = FV(1+i)n

Consider option B. We walk backwards from the Rs.10,000 offered after three years. Note that, the Rs.10,000 is to be received after three years. It is another way of looking at the future value of an investment. So, here is how we can calculate today's present value of the Rs.10,000 expected from a three‐year investment earning 5%.

PV of three year investment = 10000 1 0.05 = 10000 0.86 = 8638.38

This means that if you invest Rs. 8638.38 today, @ 5 % rate of interest, it will grow to Rs. 10000 in three years. In other words the present value of a future payment of Rs10,000 is worth Rs 8638.38 today if interest rates are 5% per year. Similarly

PV of 2 year investment = 10000 1 0.05 = 10000 0.91 = 9070.29

This means that if you invest Rs. 9070.29 today, @ 5 % rate of interest, it will grow to Rs. 10000 in two years.

Similarly

PV of 1 year investment = 10000 1 0.05 = 10000 × 0.95 = 9523.81

This means that if you invest Rs. 9523.81 today, @ 5 % rate of interest, it will grow to Rs. 10000 in one year.

The in the equation for present value

PV = FV(1+i)n

All we are doing is discounting the future value of an investment, as we are going to see in the next section. For example, in the above case, the present value of a Rs. 10000 payment in one year would be Rs. 9523.81.

The formula PV = FV(1+i)n can be rewritten to find interest rate.

i = ‐1

So in the above example interest rate

i = .

‐1 = 1.05 – 1 = 0.5



Thus we get the interest rate as 5 %.

A related concept is NPV which is discussed below.

Net present value (NPV) The basic concept of a net present value procedure is that a rupee in hand today is worth more than a rupee to be received sometime in the future. A rupee is worth more today than tomorrow because today’s rupee can be invested and can generate earnings. In addition, the uncertainty of receiving a rupee in the future and inflation make a future rupee less valuable than if it were received today. The concept of net present value can be stated in another manner: The best way to evaluate a business case is to recognize the fact money received in the future is not as valuable as money received today. Most of us would prefer to receive Rs.100 today than get paid Rs. 100 five years from now because we can take that Rs. 100 today and invest it in an asset that provides us with a return. This concept is often referred to as the time value of money. This discounting procedure converts the cash flows that occur over a period of future years into a single current value so that alternative investments can be compared on the basis of that single value. This conversion of flows over time into a single figure via the discounting procedure takes into account the opportunity cost of having money tied up in the investment.

For example, assume that a manager can earn an 8% annual return on funds invested in his or her business. Based on this return, if the manager invests Rs. 681 today, then the investment will be worth Rs.1,000 in five years (Table 1).

Table 1 Year Value at Interest Annual Amount the rate Interest the end of beginning the year

1

of the year

681

×

8 %

=

54

735

2

735 × 8 % = 59

794

3

794 × 8 % = 64

858

4

858 × 8 % = 69

926

5

926 × 8 % = 74

1000

To adjust money for its future value, you use compounding (as in the above example) to obtain the future value of a current sum. Every year, you add an amount to the money you already have based on the rate of return you earn. You do the reverse to calculate the present value of money you could receive in the future.

To calculate a present value ‐ or discount future earnings to the present‐ assume,



for example, that a manager earning 8% on his or her capital will receive Rs.1,000 at the end of each year for the next five years. The discount factor for money received at the end of the first year, assuming an 8% rate, is 0.9259. Hence, the Rs.1,000 received at the end of the first year has a present value of only Rs.925.90 (Rs.1,000 x 0.9259). In similar fashion, you can calculate the present value of the Rs.1,000 received at the end of years two through five using the discount factors from the appendix for 8% and the appropriate years. You can then determine the present value of this flow of money as the sum of the annual present values. So the present value of an annual flow of Rs. 1,000 for each of five years (assuming an 8% discount rate) is only Rs. Rs.3,992.60. As a manager, you would be equally well off if you were to receive a current payment of Rs. 3,992.60 or the annual payment of Rs.1,000 per year for five years, assuming an 8% discount rate. In essence, discounting reverses the compounding process and converts a future sum of money to a current sum by discounting or penalizing it for the fact that you do not have it now, but have to wait to get it and consequently give up any earnings you could obtain if you had it today.

Calculating NPV

The NPV of a project is the difference between PV of benefits (cash inflows) and PV of costs (cash outflows i.e. investment). In other words the PV of net benefits of a series of a project is called NPV.

NPV = ∑ ‐ ∑ or

NPV = ∑ or

NPV = ∑ ‐ initial investment (C0) or NPV = total PV of cash inflows – total PV of cash outflows Where Rt = Benefits in year t.

Ct = cots in year t

Bt = Rt‐ Ct = Net benefit in year t

i = discount rate

For economic viability of a project, NPV≥0 (i.e. positive) Greater the NPV more profitable

is the project, so accept

If NPV is negative project is undesirable NPV = 0, It is a matter of indifference.

NPV = PV of benefits – PV of costs

E.g. Calculate the NPV for a project requiring an initial investment of Rs. 20000, & which provides a net cash inflow of Rs. 6000 each yr for 6 yrs. Assume the cost of capital to be 8% p.a. & that there is no scrap value?



NPV = TPV of cash inflows – TPV of cash outflows

= 6000 x PVF 8%,6 – initial investment

= 6000 x 4.623 – 20000

= 27738 – 20000 = 7738

or

NPV = ∑ ‐ Co

NPV = 6000/(1.08)1 + 6000/(1.08)2 + .... 6000/(1.08)6 – 20000

= 5555.6 + 5145.8 + 4763 + 4460.14 + 4081.63 + 3788.7‐20000

=27744.73 – 20000 ≈ 7744.73

What we discussed about PV can be re‐presented here as the principles of compounding and discounting.

Compounding and Discounting

Compounding finds the future value of a present value using a compound interest rate. Discounting finds the present value of some future value, using a discount rate. They are inverse relationships. This is perhaps best illustrated by demonstrating that a present value of some future sum is the amount which, if compounded using the same interest rate and time period, results in a future value of the very same amount. In economics these principles are used in the Cost Benefit Analysis – CBA.

Let us go back to our examples in the previous section. We saw that future value of investment of Rs. 10000 at 5% rate of interest at end of 10 years:

Future value = 10000 × (1 + 0.05)10 = 10000×1.0510 = 10500×1.63 = 17103.39

In other words the compounded value of Rs. 10000 in ten years at 5% is Rs. 16288.95. So what we did here is compounding.

Now let us put it in terms of discounting by looking in the other direction. What is the present value of Rs. 16288.95 received after ten years at 5% rate of interest. In other words how much money should we invest today if we ant to get Rs. 16288.95after ten years – what is its present value if the rate of interest is Rs. 5%. You might have noticed such advertisements by insurance companies – “Do you want to get Rs 10000 per month as pension when you retire? Start palling today”. So what they do is they do a discounting calculation.

So recalling our formula for finding NPV to find what is the present value of Rs. 16288.95 received after ten years at 5% rate of interest

PV of ten year investment = 16288.95 1 0.05 ‐10 = 16288.95 × 1.05‐10



= 16288.95 × 0.61 = 10000

Thus we see that compounding Rs. 10000 (at 5 % for 10 years) gives Rs. 16288.95 and discounting Rs. 16288.95 (at 5 % for 10 years) gives Rs. 10000. Thus it is evident that compounding and discounting are inverse relationships.

Discounting Rates

An important consideration when discounting future costs and benefits to present value is the discount rate applied. Discounting Rate is the interest rate used in discounted cash flow analysis to determine the present value of future cash flows. The discount rate takes into account the time value of money (the idea that money available now is worth more than the same amount of money available in the future because it could be earning interest) and the risk or uncertainty of the anticipated future cash flows (which might be less than expected). Discount rate is generated using the following equation:

r = ρ + µg

Where, r is the discount rate. ρ is pure time preference (discount future consumption over present consumption on the basis of no change in expected per capita consumption). µ is elasticity of marginal utility of consumption g is annual growth in per capita consumption

The ‘time preference’ comes from the principle that, generally, people prefer to receive goods and services now rather than later. Meanwhile, the latter part of the equation refers to the fact that growth means people are better off and extra consumption worth less.

Example

Let us say we expect to get Rs 1,000 in one year's time. To determine the present value of this Rs. 1,000 (what it is worth to you today) we would need to discount it by a particular rate of interest. Assuming a discount rate of 10%, the Rs. 1,000 in a year's time would be the equivalent of Rs. 909.09 to us today (1000/[1.00 + 0.10]).

Problems with Discounting

A key concern involved in the use of discounting is the value assigned to the discount rate. An incorrect value for the discount rate could easily result in an inaccurate estimate of the present value of a future cost or benefit, and equally could result in the Cost Benefit Analysis of an entire project providing an inaccurate net present value. Often government agencies provide a suggested discount rate to use when performing discounting.



A second limitation in the common method of discounting to present value is in the timing of costs and benefits. The standard method of discounting future costs and benefits to present values assumes that all costs and benefits occur at the end of each year. However, it may be that in some cases the timing of costs and benefits needs to be considered more minutely. If a cost occurs half way through a year its discounted value may be different.

The economic evaluation of investment proposals

The economic analysis specifies a decision rule for:

I) accepting or

II) rejecting investment projects

Consider a set of borrowers and lenders. The interaction of lenders with borrowers sets an equilibrium rate of interest. That is, if the rate of interest demanded by the lender is high, the borrower will not be willing to borrow. On the other hand, if the interest rate suggested by the borrower is low, the lender will not be willing to lend. Borrowing is only worthwhile if the return on the loan exceeds the cost of the borrowed funds. Lending is only worthwhile if the return is at least equal to that which can be obtained from alternative opportunities in the same risk class.

The interest rate received by the lender is made up of:

i) The time value of money: the receipt of money is preferred sooner rather than later. Money can be used to earn more money. The earlier the money is received, the greater the potential for increasing wealth. Thus, to forego the use of money, you must get some compensation.

ii) The risk of the capital sum not being repaid. This uncertainty requires a premium as a hedge against the risk, hence the return must be commensurate with the risk being undertaken.

iii) Inflation: money may lose its purchasing power over time. The lender must be compensated for the declining spending/purchasing power of money. If the lender receives no compensation, he/she will be worse off when the loan is repaid than at the time of lending the money.

In this regard the following values are important. (some of the concepts are already explained. So we skip explanation for them. Others are explained here)

a) Future values/compound interest b) Net present value (NPV) c) Annuities d) Perpetuities e) The internal rate of return (IRR)

Let us explain the two concepts viz Annuities, Perpetuities. The internal rate of return (IRR) is explained later.

Annuity

A set of cash flows that are equal in each and every period is called an annuity

An annuity is a fixed amount payable periodically @ equal intervals (may be 1 yr,



½ year, a month so on) it may take the form of interest, rent, pension cash inflow etc.

An annuity is a stream of constraint cash flow (payment or receipt) occurring @ regular intervals of time. When the cash flow occur the end of each period, the annuity is called regular annuity or deferred annuity. When cash flow occurs @ the beginning of each period, the annuity called annuity due.

The present value of an annuity

PVAnr = 1 1 where I = r/10

Here [1‐(1+i)‐n] is the PV of Rs. 1 for n years @ r%. This is the PV factor.

Total PVA = A x PVFnr

In the future value of an annuity (FVAn) for n years, 1st year’s annuity will earn interest for n‐1 years, 2nd year’s annuity for n‐2 yr & last yr’s annuity will have no interest.

FVavn = [(1+i)n‐1] where i=r/100

A perpetuity is an annuity of infinite duration. PV of a perpetuity = A/r

E.g.1Find the total PV of an annuity of Rs. 150 payable at the end of every year & for 10 years, rate of interest being 8% p.a?

PVAnr = 1 1 = A x P.V. factor (1038% ‘r’)

= 150 x 6.7101 = 1006.56

Perpetuities

Perpetuity is an annuity with an infinite life. It is an equal sum of money to be paid in each period forever.

PV of a perpetuity =

where:

C is the sum to be received per period r is the discount rate or interest rate

Example:

You are promised a perpetuity of Rs.700 per year at a rate of interest of 15% per annum. What price (PV) should you be willing to pay for this income?

PV of a perpetuity = . = Rs.4,666.67

A perpetuity with growth:



Suppose that the Rs.700 annual income most recently received is expected to grow by a rate G of 5% per year (compounded) forever. How much would this income be worth when discounted at 15%?

Solution:

Subtract the growth rate from the discount rate and treat the first period's cash flow as a perpetuity.

PV of a perpetuity =

= .

. . = .

.

= Rs. 735/0.10

= Rs. 3.350

Present Value and Investment Decisions

Firms purchase capital goods to increase their future output and income. Income earned in the future is often evaluated in terms of its present value. The present value of future income is the value of having this future income today. The firm's investment decision is to determine whether to purchase new capital. In determining whether to purchase new capital—for example, new equipment—the firm will take into account the price of the new equipment, the revenue that the new equipment will generate for the firm over time, and the scrap value of the new equipment. The firm will also take into account the interest rate, which represents the firm's opportunity cost of investing in the new equipment. It will use the interest rate to calculate the present value of the future net income that it expects to earn from its purchase of the new capital equipment. If the present value is positive, the firm will choose to purchase the new equipment. If the present value is negative, it is better off forgoing the investment in new equipment. The decision to invest in other types of capital goods can also be made on the basis of present value calculations. For example, the decision to invest in human capital by attending college is based on the present value of the future income that an individual can earn with a college degree. If the present value is positive, the individual will choose to attend college. If the present value is negative, the individual will not attend college and will perhaps take a job instead. Capital budgeting

Capital budgeting is vital in making investment decisions. Decisions on investment, which take time to mature, have to be based on the returns which that investment will make. Unless the project is for social reasons only, if the investment is unprofitable in the long run, it is unwise to invest in it now. [[[ Often, it would be good to know what the present value of the future investment is,



or how long it will take to mature (give returns). It could be much more profitable putting the planned investment money in the bank and earning interest, or investing in an alternative project. Typical investment decisions include the decision to build another grain silo, cotton gin or cold store or invest in a new distribution depot. At a lower level, marketers may wish to evaluate whether to spend more on advertising or increase the sales force, although it is difficult to measure the sales to advertising ratio. Capital Budgeting is the process by which the firm decides which long‐term investments to make. Capital Budgeting projects, i.e., potential long‐term investments, are expected to generate cash flows over several years. The decision to accept or reject a Capital Budgeting project depends on an analysis of the cash flows generated by the project and its cost.

Capital budgeting is a system of long term financial planning involving: 1. Identifying investment opportunities (long term projects)

2. Determining profitability of investment projects

3. Ranking projects in terms of profitability

4. Selecting projects

A Capital Budgeting decision should satisfy the following criteria: Must consider all of the project’s cash flows.

Must consider the Time Value of Money

Must always lead to the correct decision when choosing among Mutually Exclusive Projects. Capital Budgeting projects are classified as either Independent Projects or Mutually Exclusive Projects.

An Independent Project is a project whose cash flows are not affected by the accept/reject decision for other projects. Thus, all Independent Projects which meet the Capital Budgeting critierion should be accepted.

Mutually Exclusive Projects are a set of projects from which at most one will be accepted. For example, a set of projects which are to accomplish the same task. Thus, when choosing between ‘Mutually Exclusive Projects’ more than one project may satisfy the Capital Budgeting criterion. However, only one, i.e., the best project can be accepted.

The internal rate of return (IRR)

The Internal Rate of Return (IRR) of a Capital Budgeting project is the discount rate at which the Net Present Value (NPV) of a project equals zero. This rate means that the present value of the cash inflows for the project would equal the present value of its outflows.

Internal Rate of Return is the flip side of Net Present Value and is based on the same principles and the same math. NPV shows the value of a stream of future cash flows discounted back to the present by some percentage that represents the minimum desired rate of return, often your company’s cost of capital. IRR, on the other hand, computes a break‐even rate of return. It shows the discount rate below which an investment results in a positive NPV (and should be made) and above which an investment results in a negative



NPV (and should be avoided). It’s the break‐ even discount rate, the rate at which the value of cash outflows equals the value of cash inflows.

We use the following formula to find IRR.

NPV = 0 = = CF0 + + + …. +

Where CFt = the cash flow at time t and

The determination of the IRR for a project, generally, involves trial and error or a numerical technique. The decision rule is that a project should only be accepted if its IRR is not less than the target internal rate of return. When comparing two or more mutually exclusive projects, the project having highest value of IRR should be accepted.

The example below illustrates the determination of IRR. Consider Capital Budgeting projects A and B which yield the following cash flows over their five year lives. The cost of capital for both projects is 10%. Note that the cash flow in year zero is negative as the project is only in the planning stage.

Finding Internal Rate of Return Project A: Substituting in the formula. Remember that IRR of a Capital Budgeting project is the discount rate at which the Net Present Value (NPV) of a project equals zero. So we set NPV = 0 in the equation. That is why the equation starts with 0 =.

0= ‐1000 + + + + +

IRR = 16.82%

Project A Project B

Year Cash Flow(in Rs.)

Cash Flow(in Rs.)

0 ‐1000 ‐1000

1 500 100

2 400 200

3 200 200

4 200 400

5 100 700



Project B:

0= ‐1000 + + + + + IRR = 13.28%

Thus, if Projects A snd B are independent projects then both projects should be accepted since their IRRs are greater than the cost of capital. On the other hand, if they are mutually exclusive projects then Project A should be chosen since it has the higher IRR. The above example makes it clear that the IRR is presented as a break‐even discount rate‐i.e., it is defined as the discount rate that makes the NPV equal to zero. It is then argued that if the ‘appropriate’ discount rate falls below the IRR, the NPV must be positive—and the reverse must hold for discount rates above the IRR. However, IRR rankings have no relation with the relevant NPV rankings. Often the IRR can be very misleading, at best just somewhat misleading. Moreover, the informational requirements for computing the proper NPV and the IRR are the same except for the discount rate‐yet without thinking about the proper discount rate no proper CBA (discussed below) can be performed. There are no shortcuts. The IRR contains no useful information about the economic value of a project. What is the difference between net present value (NPV) method and internal rate of return (IRR) method? The net present value (NPV) method has several important advantages over the internal rate of return (IRR) method. First the net present value method is often simpler to use. The internal rate of return method may require hunting for the discount rate that results in a net present value of zero. This can be a very laborious trial‐and‐error process, although it can be automated to some degree using a computer spreadsheet. Second, a key assumption made by the internal rate of return (IRR) method is questionable. Both methods assume that cash flows generated by a project during its useful life are immediately reinvested elsewhere. However, the two methods make different assumptions concerning the rate of return that is earned on those cash flow. The net present value method assumes the rate of return is the discount rate, whereas the internal rate of return method assumes the rate of return is the internal rate of return on the project. Specifically, it the internal rate of return of the project is high, this assumption may not be realistic. It is generally more realistic to assume that cash inflows can be reinvested at a rate of return equal to the discount rate ‐ particularly if the discount rate is the company's cost of capital or an opportunity rate of return. Cost Benefit Analysis (CBA) As a student of Economics, you should have understanding of the fundamental concepts of CBA since it is here that most of the tools of capital budgeting / financial mathematics you studied in this chapter is being used. Here we make a brief discussion on CBA.



Cost‐benefit analysis (CBA or COBA) is a major tool employed to evaluate projects. It provides the researcher or the planner with a set of values that are useful to determine the feasibility of a project from and economic standpoint. Conceptually simple, its results are easy for decision makers to comprehend, and therefore enjoys a great deal of favour in project assessments. The end product of the procedure is a benefit/cost ratio that compares the total expected benefits to the total predicted costs. In practice CBA is quite complex, because it raises a number of assumptions about the scope of the assessment, the time‐frame, as well as technical issues involved in measuring the benefits and costs. Before any meaningful analysis can be pursued, it is essential that an appropriate framework be specified. An extremely important issue is to define the spatial scope of the assessment. Transport projects tend to have negative impacts over short distances from the site, and broader benefits over wider areas. Thus extending a runway may impact severely on local residents through noise generation, and if the evaluation is based on such a narrowly defined area, the costs could easily outweigh any benefits. On the other hand defining an area that is too broad could lead to spurious benefits. Costs associated with the project are usually easier to define and measure than benefits. They include both investment and operating costs. Investment costs include the planning costs incurred in the design and planning, the land and property costs in acquiring the site(s) for the project, and construction costs, including materials, labour, etc. Operating costs typically involve the annual maintenance costs of the project, but may include additional operating costs incurred, as for example the costs of operating a new light rail system. Benefits are much more difficult to measure, particularly for transport projects, since they are likely to be diffuse and extensive. Safety is a benefit that needs to be assessed, and while there are complex issues involved, many CBA studies use standard measures of property savings per accident avoided, financial implications for reductions in bodily injury or deaths for accidents involving people. Three separate measures are usually obtained from CBA to aid decision making: Net Present Value (NPV): This is obtained by subtracting the discounted costs and negative effects from the discounted benefits. A negative NPV suggests that the project should be rejected because society would be worse off. Benefit‐cost ratio: This is derived by dividing the discounted costs by the discounted benefits. A value greater than 1 would indicate a useful project. Internal rate of return (IRR): The average rate of return on investment costs over the life of the project.

NPV, IRR and Economists Though the tools of financial mathematics are presented as unrelated to the Economics theory, we can see the traces of the core idea in theories of economists like Keynes and Fisher etc. For your information, some examples are given below. The NPV concept is a derivative of Irving Fisher’s Rate of Return Over Cost. The



IRR is a derivative of John Maynard Keynes’ Marginal Efficiency of Capital. Fisher published his concept in 1930 and Keynes published in 1936. Irving Fisher and John Keynes were, without doubt, giants among economists. Fisher, in his book The Theory of Interest, introduced the concept of ‘the rate of return over cost’. This is the market‐determined rate of interest where the net present value (NPV) of two projects are identical. In Keynes’ book, The General Theory of Employment Interest and Money, he advanced the concept of the ‘marginal efficiency of capital’. This is the interest rate that sets the NPV of a project to zero. Today, it is known as the internal rate of return (IRR). There is a strong link between these two concepts. Fisher’s rate of return over cost is the internal rate of return of the marginal cash flows of the two projects. In essence, Fisher compared a new project with an existing project, using the example of a farmer contemplating forestry as an optional use of the land. Keynes’ IRR is the rate of return over cost where the project is compared with its opportunity cost derived from the market.

Annexure

(Here we see some information which is related to your syllabus and very useful) I Cash flow is an important concept to understand when evaluating a company's overall financial health. It's also useful when trying to understand the impact of a new investment or project on a company's finances. The reason so many investors are interested in cash flow is that it removes all of the accounting allocations, and delivers a clearer picture of the inflows and outflows of money. The easiest way to think about the concept of cash flow is this: it is the actual inflow and outflow of funds from a company. Cash flow removes all of the accrual accounting adjustments that appear on an income statement such as deferred income taxes and depreciation, and allows us to see a company's earning power and operating success in a slightly different way. Cash flow also takes into account some very practical considerations such as inflation. Perhaps the most important value that understanding cash flow provides to the investor is the ability to understand if a company has enough resources to meet its cash operating needs. Discounted Cash Flows When analyzing the after‐tax cash flow of our business case model, the convention is to discount those cash flows. The discount rate used for cash flows would typically be a company's after tax weighted average cost of capital, which is a proxy for the company's cost of money. Perhaps the most common measure that uses discounted cash flows, or DCF, is the calculation of net present value.



Net Present Value of Cash Flows The concept of net present value can be stated in this manner: The best way to evaluate a business case is to recognize the fact money received in the future is not as valuable as money received today. Most of us would prefer to receive Rs.100 today than get paid Rs.100 five years from now because we can take that Rs.100 today and invest it in an asset that provides us with a return. This concept is often referred to as the time value of money.

Internal Rates of Return

The internal rate of return, or IRR, is simply the discount rate (in terms of percentage) where the net present value of cash flow is equal to zero. The value provides executives / decision‐makers with a better feel for how "good" the investment is that's being modeled.

Some companies set hurdle rates for new investments, and these hurdle rates are usually expressed in terms of IRR. For example, a company might decide that a new computer system requires an IRR of 15% or more for the project to be approved.

Payback Period

Payback is a measure that allows us to see how quickly the initial investment is returned to us. To demonstrate this point, consider the following cash flow example:

Payback Cash Flow Example

Year

Y0 Y1 Y2 Y3 Cash Flow

−1000 500 500 500

In this example, the payback on this project would be 2.0 years. The Rs.1000 investment is returned after only two years. Sometimes companies wish to calculate payback on a discounted cash flow basis. So in this next example, our cash flows would be discounted by 8.95% and we'd have: Discounted Payback Example

Year

Y0 Y1 Y2 Y3

Cash Flow

−1,000 500 500 500

DCF

−1,000 459 421 387

In this example, the discounted payback on this project would be 2.3 years. Advantages of Payback Period • It is easy to understand and apply. The concept of recovery is familiar to every decision‐maker.



• Business enterprises facing uncertainty ‐ both of product and technology ‐ will benefit by the use of payback period method since the stress in this technique is on early recovery of investment. So enterprises facing technological obsolescence and product obsolescence ‐ as in electronics/computer industry ‐ prefer payback period method. • Liquidity requirement requires earlier cash flows. Hence, enterprises having high liquidity requirement prefer this tool since it involves minimal waiting time for recovery of cash outflows as the emphasis is on early recoupment of investment. Disadvantages of Payback Period • The time value of money is ignored. For example, in the case of project • A Rs.500 received at the end of 2nd and 3rd years are given same weightage. Broadly a rupee received in the first year and during any other year within the payback period is given same weight. But it is common knowledge that a rupee received today has higher value than a rupee to be received in future. • But this drawback can be set right by using the discounted payback period method. The discounted payback period method looks at recovery of initial investment after considering the time value of inflows. • Another important drawback of the payback period method is that it ignores the cash inflows received beyond the payback period. In its emphasis on early recovery, it often rejects projects offering higher total cash inflow. • Investment decision is essentially concerned with a comparison of rate of return promised by a project with the cost of acquiring funds required by that project. Payback period is essentially a time concept; it does not consider the rate of return. II Application in Economics

Economic profitability refers to an investment having a NPV greater than zero and an IRR greater than the opportunity cost of capital. This means the investment will earn a rate of return at least equal to this opportunity cost so there will be an economic profit. Cash flow needed to make principal and interest payments are not included in these analyses because it is assumed that the results are independent of how the investment is financed. Financial feasibility is concerned with the periodic net cash flows from the investment and becomes particularly important when capital must be borrowed to finance the investment. It is entirely possible for an investment to be economically profitable but result in negative net cash flows for several time periods. This often occurs when the loan must be paid off in a relatively short period of time and the investment produces little initial cash flow but larger flows in later time periods.

III Opportunity Cost of Capital

The difference in return between an investment one makes and another that one chose not to make. This may occur insecurities trading or in other decisions. For example, if a person has Rs.10,000 to invest and must choose between Stock A and Stock B, the opportunity cost is the difference in their returns. If that person invested Rs.10,000 in Stock A and received a 5% return while Stock B makes a 7% return, the opportunity cost is 2%. One way of conceptualizing opportunity cost is as the amount of money one could have



made by making a different investment decision. Importantly, opportunity cost is not a type of risk because there is not a chance of actual loss.

IV Social discount rate

Social discount rate (SDR) is a measure used to help guide choices about the value of diverting funds to social projects. It is defined as ‘the appropriate value of r to use in computing present discount value for social investments’. Determining this rate is not always easy and can be the subject of discrepancies in the true net benefit to certain projects, plans and policies. SDR refers to the rate used for converting future flows of costs and benefits from an investment project into the same units as current values (present values) from a social perspective. This rate is generally applied to public sector investment projects. In a without uncertainty world, when projects are independent and there are no externalities, using a given SDR, the net present value of benefit of each project can be calculated and the projects can be ranked on the basis of their net present values. The decision problem becomes choice of a project which yields the highest net present value of benefit. Part B Problems / Worked out Examples (Source: Financial Management, Prasanna Chandra, Tata McGrawHill) Growth Rate Simple

Let ‘P’ be the quantity / value in the beginning of the 1st year ‘r%’ be the rate of increase. Then increase in every year is P.r/100

[ The value of P grows at the end of nth year into –

Amount = P + = P 1 Note: Here P is the value at the beginning of the first year & P + is the value at the end of nth year.

( Total Growth in ‘n’ years = )

E.g. 1 ‐ The salary of an employee increases every year by 10% of his initial salary, which is Rs. 3500. Find salary at the end of 8th yr?

P = 3500, r = 10%, n = 8

Salary at the end of 8th year = P 1

= 3500 1

= 3500 = 3500 1.8 6300/‐



OR

Salary at the end of 8th year P +

= 3500 +

= 6300/‐

E.g. 2 – Find the amount at the end of 5th year for Rs. 5000 @ 10% p.a , simple interest

(SI). What is the total amount of growth?

P=5000, n=5, r‐10% Amount at the end of 5th year = P 1

= 5000 1 = 5000 1 = 5000 = 7500/

Total amount of growth = 7500 – 5000 = 2500/‐ E.g.3‐ Calculate the total interest on (1) Rs. 500 for 73 days (ii) Rs. 720 for 14 weeks & (iii) Rs.900 for 3 months all at 6% per annum? (i) P = 500. r = 6%., n = 73/365 SI = = 500 x 73/365 x 6/100 = 6/‐

(ii) P = 720, n = 14/52, n = 6%

SI = 120 x 14/52 x 6/100 = 11.63/‐

(iii) P = 900, n=3/12, n = 6%

SI = = 900 x 3/12 x 6/100 = 13.50/‐

Total interest = 6.0 + 11.63 + 13.50 = 31.13/‐

E.g.4‐ In what time will a sum of money be doubled itself at 10% p.a., SI?

Amount = P + = P 1 & Simple Growth =



P=P, A=2P, r =10%, n=?

A = P 1 = P 1

i.e. 2p = P 1 = P 1

2 = 1 = 1+

2‐1 = n/10

i.e. 1 = n/10, so n = 10 years

Growth rate – compound

When the growth raet is calculated every year on the value of the preceding year, instead the value at the starting year, the growth rate is compound. Let P be the value in 1st year, let r% be the growth rate over the preceding year. Then values at the end of every year are:

A = P 1 , P 1 , P 1

The value at the end of nth year A = P 1

Compound interest (CI) – A‐P

E.g.5 ‐Find the sum @ the end of r years of Rs. 10000 10% p.a. CI?

P= 10000, n = 4, r= 10%

A = P 1 = 1000 1 = 1000 x (1.1)4

Sum at the end of the 4 years = 10000 x 1.4641 = 14641/‐

E.g.6 ‐If the population of a country rises by 2% every year & it was 30 cr. in 1960, find population in 1970?

Initial population = P = 30, r = 2%, n = 1960 to 1970 i.e. 10 years

population at the end of 1970 = P 1

= 30 1 = 30(1.02)10 = (30 1.219) = 36.6 cr

(Note: Half yearly/semi annually = 2 times in a year i.e. N = given no. of years x 2)

Quarterly = 4 times in a year, n = no. of years x 4)

E.g.7‐ The population of a country increases every year by 2.4% of the population @ the beginning of that year. In what time will the population doubled itself?



Initial population =P,

population @ end = 2P, r = 2.4%, n=?

A = p 1

2P= p 1 . = = (1.024)n

Log 2 = n log 1.024 0.3010 = n x 0.0103 By 29 years population will not double, so rounding 29.22 to 30. Hence the required time = 30 yrs.

E.g.8 ‐Mr. A borrowed R.s 2000 from a money lender but he could not repay any amount in a period of a 4 years so the money lender demanded 26500 from him. What is the r/i (r) charged?

P = 20000, n=4, A = 26500, r =?

A = p 1 = 2000

p 1 = 26000

26500/20000 = 1 = 1.325

Taking log of both sides,

Log 1.325 = 4 log 1

0.1222/4 = log = 0.03055

1 = Antilog 0.03055 = 1.073

r/100 = 1.073 – 1 = .073

r = .073 100 = 7.3%

Depreciation

It is the decline in the value of fixed assets like buildings, plant, machinery etc. Due tot wear & tear in use. The residual value of the assets at the end of its life time us called scrap value.

Depreciated / scrap value D = p 1

P = original value, r = rate of depreciation, n = no. of yrs



E.g.9 ‐Vehicle costs 40000/‐. It depreciates at 10% per year & is sold after 5 yrs. Find its sales price. P = 40000, r = 10%, n = 5, D = ?

D = p 1 = 40000 1 = 40000(0.90)5

= 40000 .59049 = 23619.6/‐

E.g.10 ‐A machine depreciates in value each yr @ 10% of its previous value & at the end of 4th yr its value is Rs 131220. Find original value?

Given A = 131220, n = 4, r = 10%, p = ?

A = p 1

i.e. 131220 = p 1 = P (0.9)

Log 131220 = log P + 4 log 0.9 5.1179 = log P + 4(1.9542) Log P = 5.1179 – 4(1 .9542)

= 5.1179 – (.1832) Log p = 5.1179 + .1832 = 5.3011 P = antilog 5.3011 = Rs. 200000/‐

Simple interest (SI) is the interest computed on the original P for the time

during which the money is being used

I=Pnr/100, r = 100 I/Pn, n=100I/pr, P = 100I/nr

THE VALUE OF MONEY – PRESENT VALUE & FUTURE VALUE

Time value of money means, a rupee to be received in the future is not worth a rupee today

Present value

The present value (PV) of a sum of money is the principle which if placed on the interest will amount to that sum of money.

I. PV in the case of simple interest is,

PV =

here FV = Amount , So PV =

II. In the case of compound interest;



PV = = A(1+i)‐n, i =

(1+i)‐n is the PV factor i.e.,

(For various values of n & r, the PV factor is worked out & given at the statistical table (Refer logarithm book under title i.e, PV of Rs. 1 of nth yr = (1+i)‐n where i =

E.g. 11 Find the PV of a sum of Rs. 1000/‐ due after 3 years if the r/i charged is 10% p.a. (simple)

PV =

= = . =

. =769.23/‐

E.g.1 2 Find the PV of Rs. 400 due after 5 yrs compounded annually @ the rate of 8%?

A = 400, n=5, r = 8%

PV =

= A(1+r)‐n = A PV factor (PVnr)

= 400 0.681 = 272.40/‐

E.g. 13 Find the total PV of each cash inflows @ the end of each yr shown below?

Year 1 2 3 4 5 Cash inflow 2000 3000 3500 3000 4000

The rate of interest (discount rate) 8%

Year cash inflow PVF PV = A x PVF

1 2000 0.926 1852

2 3000 0.857 2571

3 3500 0.794 2779

4 3000 0.735 2205

5 4000 0.681 2724

Total PV 12131

Future Value/Compounded value



The process of investing money as well as reinvesting the interest earned theorem is called compounding. The future value (FV) compounded value or compounded value of an investment after n yrs when interest rate is r% is:

FVn = PV (1+r)n

If no interest is earned on interest, then investment earns only simple interest, then

FV = PV (1+nr)

E.g. 14 Find the amt due for a sum of Rs. 500 after 2 yrs @ 8% interest (simple). PV = 500, n = 2, r = 8%

FV = PV (1+nr) = 500 [1+ (2 x 8/100)] = 500 (1+0.16) = 580/‐

ANNUITY

E.g.15 Find the total PV of an annuity of Rs. 150 payable at the end of every year & for

10 years, rate of interest being 8% p.a?

PVAnr = 1 1 = A P.V. factor (1038%’r’)

= 150 x 6.7101 = 1006.56

i.e. PVFA r n =

= .

. . = .

. = 6.7076

PVA = A PVFA = 150 1006.56 Present value of a series

PV = + + ……. = ∑

Where PV n = PV of a cash flow stream

At = Cash flow occurring @ the end of yr t

PVFArn =

PVArn = A PVFA



r = discount rate

n = duration of cash flow stream E.g.16 Shows the calculation of the PV of an uneven cash flow stream using a discount rate of 12%.

Yr Cash flow PVIF 12%, n PV of Individual Cash flow

1 1000 0.893 893

2 2000 0.799 1594

3 2000 0.712 1424

4 3000 0.636 1908

5 3000 0.567 1701

6 4000 0.507 2028

7 4000 0.452 1808

8 5000 0.404 2020 ‐‐‐‐‐‐‐‐‐ TPV 13,376 ======

PV of a single Amount PV = FVn [1/(1+r)n]

PV = FVn [1/(1+r)n];

PV = FVn(1+r)‐n {The process of discounting used for calculating the PV, is simply the inverse of

compounding. The factor 1/(1+r)n is called the discounting factor the present value interest factor (PV1Fr n). }

E.g.17 What is the PV of Rs. 1000 receivable 6 years, (1) If the rate of discount is 10%

PV = 1000 x PVIF10%; 6 = 1000 (0.5645) = 564.5

(2) What is the PV of Rs. 1000 receivable 20 yrs hence if the discount rate is 8%?

Since we does not have the value of PVIF 8%, 20 we obtain the answer as follows. PV = 1000 (1/1.08)20

= 1000 (1/1.08)10 (1/1.08)10

= 1000 (PVIF 8% 10) (PVIF 8%, 10)

= 1000 (0.4632) (0.4632)



Investment Criteria

Discounting Non‐Discounting

NPV Benefit‐Cost Ratio IRR Payback Period Accounting Rate

= 214/‐ Net present value (NPV)

There are several methods for project evaluation /appraisal i.e. determining whether a project is worthwhile or not. This is also called capital budgeting.

An investor faces 2 problems in project evaluation. (a) Whether a project is to be undertaken or not (b) if there are many competing projects, which one is to be preferred first.

The investment criteria are classified into 2 categories:

E.g.18 A project requires investment of Rs. 100000 initially it is estimated to provide annual net cash flows of 4000 for a period of 8 yrs. The co.’s cost of capital is 10%. Find the NPV of the project (PV of 1 for 8 yrs @ 10% p.a. is 5.335] Annual estimated cash inflow = 40000 PV of estimated cash inflow = 4000 x PVIF 6, 8% = 40000 x 5.335 = 213400/‐

Initial investment (cash outflow) = 100000

NPV = 213400 – 100000 = 113400/‐

E.g.19 An investment of Rs. 1000 (having scrap value of Rs. 500) yields the following returns Years 1 2 3 4 5 Yields 4000 4000 3000 3000 2000



yr cash inflows PVF@10% PV (Rs)

1 4000 0.909 3636

2 4000 0.826 3304

3 3000 0.751 2253

4 3000 0.683 2049

5 2000 0.621 1552

Total PV of future cash inflows 12794 NPV = 12794 – 1000 (initial investment) = 2794 Since NPV is positive, the investment is desirable (Note: when cash flows are not uniform; multiply each cash flow by PV factor in table

(1+i)‐n and add all of them together, we get total PV of all cash flows) INTERNAL RATE OF RETURN (IRR) The IRR of a project is the discount rate which makes its NPV equal to zero. IRR is the discount rate (%) which equates the PV of future cash flows (i.e. net benefits) generated. By a project during its life span (i.e. t = 1 ton) with the initial investment cost. In economic theory it is also known as marginal efficiency of capital (MEC). It is the value of ‘r’ is the following equation.

Investment = ∑ where

Ct = cash flow @ the end of year t;

r = IRR, n = life of the project

The decision rule for IRR is as follows:

Accept: if the IRR is greater than the cost of capital

Reject: if the IRR is less than the cost of capital. IRR can be calculated by locating the factor in annuity table i.e. F = I/C where

F = factor to be located, I = Initial Investment C = cash inflow per year. E.g.20 equipment requires an initial investment of Rs. 6000. The annual cash inflow

is estimated @ 2000 for 5 yrs. IRR?

F = I/C = 6000/2000 = 3 Referring to the annuity table for 5 yrs @ 20%, PVF = 2.99 & @ 18%, PVF = 3.127

Since 3 is very close to 2.99, we consider IRR as 20%.



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