Download Q QUA ANTI ITAT

 QUA
Q
ANTIITAT
TIVE MET
THOD
DS
F
FOR
E ONOM
ECON
MIC ANA
A ALYSI
SIS I
III Se
emesterr CORE
E COURS
SE BA
A ECO
ONOM
MICS (2011 A
Admissio
on) UN
NIVER
RSITY
Y OF CALIICUT
T
SCHOOL
L OF DIST
TANCE E
EDUCATIO
ON Calicut Universiity P.O. Mallappuram, Kerala,
K
Indiia 673 635
2
263
School of Distance Education UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION STUDY MATERIAL Core Course for BA – ECONOMICS III Semester QUANTITATIVE METHODS FOR ECONOMIC ANALYSIS ­ I
Prepared by:­ Scrutinised by: Layout: 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. Dr. C. Krishnan, Associate Professor , Department of Economics, Govt. College, Kodanchery . Computer Section, SDE ©
Reserved
Quantitative Methods for Economic Analysis – I 2 School of Distance Education CONTENTS
Pages
MODULE – I ALGEBRA 5 MODULE II BASIC MATRIX ALGEBRA 33 MODULE III FUNCTIONS AND GRAPHS MODULE IV LIMITS AND CONTINUTY OF A FUNCTION 90 MODULE V FINANCIAL MATHEMATICS 52 114 Quantitative Methods for Economic Analysis – I 3 School of Distance Education Quantitative Methods for Economic Analysis – I 4 School of Distance Education 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, product of n factors. Each of which is a Eg: is defined as the a = a a a ……….. n times 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
i.e., a
= 1
Eg: is the reciprocal of a . 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 = b ½ 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
Quantitative Methods for Economic Analysis – I 5 School of Distance Education (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 i.e., Eg: 16
⁄
= n √
⁄
is defined as nth root of mth power of a. = 4√16 = 2 = 4 (e) Meaning of negative fractional powers. If m and n are positive integers, Eg. 16
= 1
16
⁄
is defined as 1
= √
⁄
or 1
√
= 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 35 = 3
= 33 = 27 3
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., = Quantitative Methods for Economic Analysis – I 6 School of Distance Education Eg. 4.
= 3
4
= 3
= 4
2
3 = 2
= 36 = 729 = 4
= 41 = 4 3
3 = 4 9 = 36 4 = 27 64 = 1728 5.
4
16
4
3 = 3 = 9 Extension 1.
2.
= 3.
= 4.
= Examples 1.
.
.
. = .
2. .
6
.
.
8
6
= 8
48
3. 6
4
24
4.
63
= 5. 15
9
·
·
= 7
5 3
.
= 7
Quantitative Methods for Economic Analysis – I 7 School of Distance Education = = 15
.
.
3
5
= 45 5 = 9
= 6. = = = = = 1 1
7. Multiply ( = = = = = = 1
) by , a = , b = 8. Multiply by = = = = = a + b Quantitative Methods for Economic Analysis – I 8 School of Distance Education 9. Prove that By multiplying each term by = , , 1 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 Quantitative Methods for Economic Analysis – I 9 School of Distance Education 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 Quantitative Methods for Economic Analysis – I 10 School of Distance Education 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 log 16 2 16 logarithm of 16 to be the base 4 is 2. It can be written as 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 5 2 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 Quantitative Methods for Economic Analysis – I 11 School of Distance Education 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
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
10,000 4 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. Quantitative Methods for Economic Analysis – I 12 School of Distance Education Number Character 75.3 1 2400.0 3 144.0 2 3.2 0 .5 ‐1 .0902 ‐2 .0032 ‐3 .0007 ‐4 One less than the number of digits
One more than the number of zero immediately after the decimal point
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 Quantitative Methods for Economic Analysis – I 13 School of Distance Education 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 Quantitative Methods for Economic Analysis – I 14 School of Distance Education ‐8x ‐30 Substituting x 15/4 in equation 1 4x 3y 6 4 3y ‐9 Y ‐9/3 ‐3 The solution is x 30/8 15/4 3y 6 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 Quantitative Methods for Economic Analysis – I 15 School of Distance Education 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 7y 2z 56 ‐6y 5z ‐1 2 2 3 is ‐6y 5z ‐1 4 5 5 4 2x 2y – 2z 8 3x – 5y – 4z ‐48 7y 2z 56 5 5 4 2 35y 10z 280 ‐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, 2. Solve 7x – 4y – 20z 0 10x – 13y – 14z 0 3x 4y – 9z 11 x 4 1 5 1 2 3 7x – 4y – 20z 0 10x – 13y – 14z 0 1 2 Quantitative Methods for Economic Analysis – I 16 School of Distance Education 1 13 2 4 91x – 52y ‐260z 0 40x – 52y – 56z 0 51x – 204z 7x – 4y – 20z 0 3x 4y – 9z 11 10x ‐29z 11 51x – 204z 0 10x – 29z 11 4 10 5 51 0 4 5 510x – 2040z 0 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 quadratic equation in x (a≠0). 0 in which a, b, c are constants is called a 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 this is called pure quadratic equation. General Quadratic Equation 0, b is zero, then the equation becomes 0 is the general form of the quadratic equation. The general quadratic equation may be solved by one of the following methods. Quantitative Methods for Economic Analysis – I 17 0, School of Distance Education 1. By factorization 2. By completing the square 3. By quadratic formula 7
12 0 Eg:‐ Solve 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
3
24
16
0 24
Ans: 9
3
3x – 4 0, 16
0 4 0 3x 4, x 4/3 By Quadratic Formula 0 a Solve : 2
8
8
0 Here a 2, b 8, c 8 √
b Solve the equation √
√
, ‐8/4 ‐2 4
, 3 0 √
, 6/2 3, 2/2 1 the solutions are 3,1 Quantitative Methods for Economic Analysis – I 18 School of Distance Education 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 (2) tn = 1 + (10‐1)4 = 1 + 9 4 1 36 37 a = 1 n = 10 d 4 = a + (n­1)d = 8 + (14 – 1)‐3 a = 8 = 8 + 13 ‐3 n = 14 = 8 + ‐39 = ‐31 d = ‐3 Quantitative Methods for Economic Analysis – I 19 School of Distance Education (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. = a + (n­1)d tn ‐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 ‐27 is the 19th term tn = ‐27 tn ‐80 ‐80 = a + (n­1)d = 27 + (n – 1) ‐3 = 27 + (‐3n + 3) a = 27 d ‐3 ‐80 = 30 – 3n n = ? 3n = 30 + 80, tn = ‐80 Quantitative Methods for Economic Analysis – I 20 School of Distance Education 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 a + 98d = 9 ‐90d = 90 (1) (2) d = 90/‐90 = ‐1 Substituting d = ‐1 in equ. (1) a + 8 ‐1 = 99 = a ‐8 = 99, a = 99 + 8, t108 = a + (n­1)d = 107 + (108 – 1) ‐1 = 107 + 107 ‐1 a = 107 9
= 107 – 107 = 0 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 a + 8d = 19 (1) (2) ‐6d = ‐12 d = ‐12/‐6 = 2 Substituting d = 2 in equ. (1) Quantitative Methods for Economic Analysis – I 21 School of Distance Education 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
= 10[2 + 19 3 d 3 10 2 57 a 1 10 59 590 n = 20 1
20
1
3 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 2 2 2 2 1 2 2 Sum = 1
1 1 1 Quantitative Methods for Economic Analysis – I 22 School of Distance Education 2
2 1
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 2 Ist 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. Sum 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 Ist term last term 25 2 104 296 25 2 400 25 200 5000 5000 Sum 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 Quantitative Methods for Economic Analysis – I 23 School of Distance Education 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 t n t12 2 3
2 3 2 177147 554294 Quantitative Methods for Economic Analysis – I 24 School of Distance Education 2 a 2, r 3 t n t n 2 3 Sum of n terms of a G.P Let a be the first term and r be the common ratio Sn Sn When r 1 numerically or 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, Sn 2
r 3 and n 10 as r 1 29524 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…… Quantitative Methods for Economic Analysis – I 25 School of Distance Education 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) a 10 10 + 10 r = 35 = 1000 = √1000 = 10 (3) Multiplying both sides by r 10
10 + 10r + 10 10 +10r – 35r + 10 = 0 10 – 25r + 10 = 0 10r + 10 = 35r = 35 r (4) This is a quadratic equation r = = = = √
√
= = 2 = = When r = 2, the numbers are ⁄ , a, ar 10 , 10, 10 2 2
Quantitative Methods for Economic Analysis – I 26 School of Distance Education 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 ¾ The total simple interest at the end of nth year is end of nth year is P 100 100 and the total amount at the 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. Quantitative Methods for Economic Analysis – I 27 School of Distance Education 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 8000 8000 5760 13760 100 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 1.02 37.3 crore. 30 1
100 2
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
2000 1
100 2000 1.07 2000 1.40255 2805 100 7
P 2000, amount 2805 So, compound interest 2805 – 2000 805 Quantitative Methods for Economic Analysis – I 28 School of Distance Education 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. Quantitative Methods for Economic Analysis – I 29 School of Distance Education 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. Quantitative Methods for Economic Analysis – I 30 School of Distance Education 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. Eg:‐ A 1, 2, 3 and B 1, 3, 5 then A B 1, 3 A B contains elements common to A & B 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. Quantitative Methods for Economic Analysis – I 31 School of Distance Education 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 Quantitative Methods for Economic Analysis – I 32 School of Distance Education 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
2
5 1
7 4
7
C = 8
10
D = 10 2
E = 2
1
0
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 1
is a square matrix of order 2 3 4
2
4
9
1 3
0 6 is a square matrix of order 3 7 5
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 Quantitative Methods for Economic Analysis – I 33 School of Distance Education 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 5
8
0
2
5
8
9
21
4
4. Diagonal Matrix In a matrix the elements lie on the diagonal from left top to the right bottom are called 2 5
diagonal elements. For instance, in the matrix the element 2 and 6 are diagonal 4 6
elements. A square matrix in which all elements except those in diagonal are zero are called diagonal matrix. For example 2 0
0 6 2 4 0 0
2 0 9 0
0 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
0
0
1 2 1 0 0
are examples of identity matrix 1 0 2 0 0 1 3 3
0
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 0 0 0
0 0
are examples of null matrix 0 0
0 0 0 2 3
0 0 2 2
3
2
0 0
7. Triangular Matrix Quantitative Methods for Economic Analysis – I 34 School of Distance Education 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 1
A = 0
0
9 2
3 7 0 4
In the lower triangular matrix, all elements above leading diagonal are zero like 4
B = 2
5
0 0
9 0 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
2 8
1
B = 2
3
12
C = 19
25
9 4 2 3
6
= 7
9
2 3 4
3 4 1
4 2 5
1 2 3
2 3 4
= 3 4 2
4 1 5
2 1
7 8
D = 3 0
9 5 = 12 19
2
1
2
8
4
25
7 3 9
8 0 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
5
5
3
1
5
5
A = , hence A is symmetric 3
Quantitative Methods for Economic Analysis – I 35 School of Distance Education 5
B = 2
6
2 6
3 9 9 7
5
=2
6
2 6
3 9 B =
9 7
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 0 4
0 4
= 4 0
4 0
A = 0
3
5
3
0
2
0 4
4 0
A is skew symmetric A =
B = = 5
2 0
0
= 3
5
3
0
2
5
2 0
= 0 3
3 0
5 2
5
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 = A + B = 4
2
4 9
2 1
and B = 6
9
+ 7
1
6 3
7 0
4
3
= 2
0
6
7
9
1
3
10 12
= 0
9
1
A ‐ B = 6 3
4
4 9
‐ = 7 0
2
2 1
6 9
7 1
3
= 0
2 6
5 1
Example :2 8 9 7
1
If A = 3 6 2 B = 5
4 5 10
7
3 6
9 12 13
A+B = 2 4
8
8
6 9 2
11 14 12
Example : 3 A =
3 7 11
6 8 1
3
B = A – B = 12 9 2
9 5 8
3
1
4
10
6
Example : 4 Quantitative Methods for Economic Analysis – I 36 School of Distance Education 2
A = 1
1
2
1
0
A + B – C = 2
3 3
3 B = 3 0
4
6 9
1
1
5
1
2
6
3
4
5 C = 5
1
2
4 4
1 0 3 1
1
2 2
Example 5 A = 12 16 2 7
8 B = 0 1 9
6 A + B = 12
5
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
0
1
3
and scalar k = 7 then KA = 7
5
0
1
21
= 5
0
7
35
Example 2 3 2
12 8
Determine KA if K = 4 and A = 9 5 KA = 36 20 6 7
24 28
Example 3 K = ‐2 and A = 7
5
2
3
6
7
2
14
8 KA = 10
9
4
6
12
14
4
16 18
Example 4 If A = 2
0
3 1
1
B = 1 5
0
2A = 4
0
6
2
2
3
3B = 10
0
2
1
1
Find 2A ‐3B 3
6
3
3
1
2A – 3B = 9
0
0 5
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 Quantitative Methods for Economic Analysis – I 37 School of Distance Education elements of the row vector by their corresponding elements in the column vector and summing the product. For example d
If A = [a b c] B = e f
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 2
C = [ 3 6 8 ] D = 4 CD = 70 5
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 Quantitative Methods for Economic Analysis – I 38 School of Distance Education A =
and B = AB = Similarly if A = 6 12
3 6 7
B = 5 10 12 9 11
13 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
12
6
6
6
9
AB =
139 110
260 256
5
5
7 13
3 12 6 10 7 2
11 13 12 12 9 10 11 2
Example 1 A = 3
4
5
6
1 0
4 7
B = 5
B = 9
11
9 6
12
B = 3
AB =
17 35
20 42
Example 2 1
A = 2
4
Example 3 7
A = 2
10
Example 4 3
8 0
A = B =[ 2 6 5 3 ] B = 2
4 5
6 1
32
AB = 82 20
117 94 46
AB = 51 62 19 138 76 56
6 18 15 9
2 6 5 3
6 5 3 AB = 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 + = Quantitative Methods for Economic Analysis – I 39 School of Distance Education 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
4
3
5
= 45
29
Example 2 7 + 8 = 120 6 + 9 = 92 In matrix form AX=B 7
6
8
9
= 120
92
Example 3 2x1 + 4x2 + 9x2 = 143 2x1 + 8x2 + 7x3 = 204 5x1 + 6x2 + 3x3 = ‐168 In matrix form AX = B 2 4 9
2 8 7 5 6 3
143
= 204 168
Example 4 8w + 12x ‐ 7y + 22 = 139 3w ‐ 13x + 4y + 92 = 242 Quantitative Methods for Economic Analysis – I 40 School of Distance Education In matrix from AX=B 8
3
12 7
13 4
2
9
= 139
242
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 A is a 2 × 2 matrix then the determinant of such matrix, like itself. If 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 4
| | = (10 × 5) – (4 × 8) = 18 8 5
Example 2 B = 2
3
1
| | = 7 2
Example 3 C = 6
7
4
| |= 26 9
Example 4 D = 4
6
6
| |=0 9
Quantitative Methods for Economic Analysis – I 41 School of Distance Education 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
3
3
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
8
2
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. 11
A = 21
31
12
22
32
13
23 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. Quantitative Methods for Economic Analysis – I 42 School of Distance Education Example 1 8
A = 6
5
3
4
1
| | = 8 4
1
2
7 3
7
6 7
6 4
– 3 + 2 3
5 3
5 1
| | = (8 × 5) – (3 ×‐17) + 2(‐4) | |= 63 Example 2 3 6 2
1 8
2
A = 2 1 8 | | = 3 ‐ 6 9 1
7
7 9 1
2 1
8
+ 5 1
7 9
| |= 166 Example 3 3
B = 1
4
6
5
8
2
4 2
| |= ‐3 5 4 – 6 1 4 + 2 1
4 2
8 2
4
5
8
| |= 98 Example 4 5
C = 2
4
| |= 5 3 1 – 7 2 1 + 2 2
6 2
4
4 2
7 2
3 1 6 2
3
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
5
3
4 5
| |=9 At = | |
6
3 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 3
A = 7
1
1
5
0
0
5
2 | | = 3 0
3
7 2
2
7 5
‐1 + 0 = 26 1 3
3
1 0
Quantitative Methods for Economic Analysis – I 43 School of Distance Education Now if we interchange first and third column, 0
2
3
1
5
0
3
5
7 ⇒0 0
1
2 7
7
2
‐ 1 + 3 3 1
1
3
5
= ‐26 = ‐| | 0
3. If any two rows or columns of a matrix are indentical or proportional, ie linearly dependent, the determinant is zero. For Example 2 3
A = 4 1
2 3
1
0 1
| |=2 4 1
1 0
4 0
+ 1 ‐ 3 2 3
3 1
2 1
| |=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 3 5 7
If A = 2 1 4 | | = 35 4 2 3
Now forming a new matrix B by multiplying the first row of A by 2, then 6 10 14
B = 2 1
4 | |=70, ie, 2 × | | 4 2
3
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 3
A = 2
6
0
5
1
0
0 | |= 60, ie ‐3 × ‐5 × 4 4
6. If all the elements of any row or column are zero the determinant is zero. For example A = 12 16
0
0
15 20
13
0 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
4
3
1
1
= 20 5
Quantitative Methods for Economic Analysis – I 44 School of Distance Education ie 2
4
1 3 1
+ = 6 + 14 = 20 5 1 5
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 3
| |= 10 10 2
Now subtract tow times of second column from first column and for a new matrix. B = 14 3
⇒ | |=10 6 2
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 = Minor of Minor of Minor of = 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
4
12
Matrix of cofactors Cij = 3
3
12
4
7
Quantitative Methods for Economic Analysis – I 45 School of Distance Education Example 2 6
2 5
Matrix of cofactors Cij = 5
13 6
B = 13
2
Example 3 2
C 4
5
3
1
3
1
2
2 = Matrix of Cofactors Cij = 9
4
5
6
3
0
7
9 10
Example 4 6 2 7
23
D = 5 4 9 Matrix of Cofactors Cij = 19
3 3 1
10
22
15
19
3
12 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 17
19 15
Matrix of Cofactors Cij = 15
17
= 15
Adjoint of A = c
19
13
19
17
13
Example 2 9
6 7
Cij= 7
12 9
A = Adj A = c
= 9
12
Example 3 0
B = 1
3
1 2
2 3 1 1
Adj A = c
12
6
7
6
1
= 1
1
1
= 8
1
1
6
2
8
6
2
5
3 1
1
2 1
Example 4 Quantitative Methods for Economic Analysis – I 46 School of Distance Education A=
13 2
9 6
3 2
8
2
4 Cij= 14
1
40
2
A Cij A = Cijt = 3
0
14
11
20
3
11
20
40
20 60
0
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 3 2
1 0
A=
| |= ‐2 Adj A=
A‐1 = A‐1 = 0
1
| |
0
2
3
=
1
Example 2 A = 7
6
9
12
| |= 30 A‐1 = | |
= A‐1 = Example 3 1
A = 5
2
2 3
7 4 1 3
Quantitative Methods for Economic Analysis – I 47 School of Distance Education |
|= ‐24 Adj A = 17
7
9
A‐1 = = | |
3
3
3
13
11 3
Example 4 4
A = 3
9
2 5
1 8 6 7
| |= ‐17 41
AdjA = 51
9
16
17
6
11
17 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 Quantitative Methods for Economic Analysis – I 48 School of Distance Education In matrix from A×=B 6 5
3 4
49
32
=
| |= 9 To solve for , replace the first column of A, that is coefficient of constants B, forming a new matrix , 49
32
= with vector of 5
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, 6 49
3 32
A2 = |
| = 45 |
= the solution is |
| |
= = 5 = 4 and = 5 Example 2 2
6 = 22 5 = 22 Ax = B 2
1
6
5
22
53
| | = 16 = 22
53
A1 = 6
5
| | =‐208 = |
|
| |
= =‐13 Quantitative Methods for Economic Analysis – I 49 School of Distance Education 2
1
=
22
53
=128 | |
= = | |
= 8 the solution is = ‐13 and = 8 Example 3 ‐ 7 ‐ =0 10 ‐ 2
+ = 8 6 3 2 ‐ 2
= 7 AX = B 7
10
6
1
2
3
1
1
2
0
= 8 7
| | = ‐61 0
|
1
2
3
|= 8
7
|
| = ‐61 = |
|
|
| |
7
0
6
7
= |= ‐183 |
= |
| |
= 7
A3 = 10
6
|
= 1 1
1 2
|= 10 8
|
1
1 2
= 3 1 0
2 8 3 7
|= ‐244 | |
= = | |
= 4 Quantitative Methods for Economic Analysis – I 50 School of Distance Education the solution is Example 4 x1 = 1 x2 = 3 x3 = 4 5x1 – 2x2 +3x3 = 16 2x1 + 3x2 ‐ 5x3 =2 4x1 – 5x2 + 6x3 = 7 5
AX = B ⇒ A = 2
4
2
3
5
16
= 2 17
3
5
6
| |= ‐37 16
A1 = 2
7
|
5
A2 = 2
4
| |
= = | |
2
3
5
= |
|
| |
= = 7 16
2 7
|= ‐185 solution is = 3 3
5 6
|= ‐259 5
A3 = 2
4
|
3
5 6
|= ‐111 16
2
7
|
2
3
5
= 3, = |
|
| |
= = 7 and = 5 = 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 Quantitative Methods for Economic Analysis – I 51 School of Distance Education 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. Quantitative Methods for Economic Analysis – I 52 School of Distance Education 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 y 1 2 2 3 3
4
4
5
5
6
6
7
Domain 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. Quantitative Methods for Economic Analysis – I 53 School of Distance Education 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. Quantitative Methods for Economic Analysis – I 54 School of Distance Education 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. Quantitative Methods for Economic Analysis – I 55 School of Distance Education 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. Quantitative Methods for Economic Analysis – I 56 School of Distance Education 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. Quantitative Methods for Economic Analysis – I 57 School of Distance Education 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). Quantitative Methods for Economic Analysis – I 58 School of Distance Education 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) =√
f(0) =√0
f(17) =√7
f(‐2) =√ 2
2 find value of the function when x = 0, x = 7, x= –2 2 = √2 2 = √9 = 3 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) = 6. Given y = x2 + 1, find f(0), f(–1) and f(3). Also find find f(2) f(x) = = f(x) = = 1 –
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 Methods for Economic Analysis – I 59 School of f Distance Educcation Part B RECTANGU
ULAR CO­O
ORDINATE S
SYSTEM AN
ND GRAPHS
S OF FUNCT
TIONS Equations can b
be graphed on a set of f coordinatee axes. The llocation of eevery pointt on a graph can b
be determin
ned by two coordinatess, written as an ordereed pair, (x,y)). These aree also known as Cartesian coordinates
c
s, after the French maathematiciaan Rene Deescartes, wh
ho is credited wiith their inv
vention. The rectangular coordinatte system, also called
d the Cartesian coordinate systeem or the x‐y coo
axis, a vertical axis, and ordinate sy
ystem consists of 4 qu
uadrants, a horizontal a
the origin. T
The horizon
ntal axis is u
usually calleed the x‐axis, and the vvertical axis is usually ccalled the y‐axis. T
The origin iis the point where the two axes crross. The co
oordinates o
of the origin
n are (0,0). This notation iss called an o
ordered paiir. The firstt coordinatee (or abscissa) is know
wn as the x‐coord
dinate, while the seco
ond coordin
nate (or ord
dinate) is th
he y‐coordiinate. Thesee tell how far and
d in what diirection we move from
m the origin.
The Rectan
ngular Coord
dinate Systeem he above figgure please note the folllowing feattures of the
e Quadrantts In th
The x‐axis an
nd y‐axis separate s
the coord
dinate plaane into four reggions called quad
drants. Thee upper righ
ht quadrant is quadrantt I, the uppeer left quadrant is quad
drant II, the loweer left quadrrant is quad
drant III, an
nd the lowerr right quad
drant is quaadrant IV. Notice that, as sho
own in Figurre 1, •
in quadrant I, x is always po
ositive and yy is always p
positive (+,+
+) •
in quadrant II, x is always neegative and
d y is alwayss positive (––,+) •
in quadrant III, xx is always n
negative and
d y is always negative ((–,–) •
in quadrant IV, xx is always p
positive and
d y is alwayss negative (+
+,–) Quantitative M
Methods for Econ
nomic Analysis –– I 60 School of f Distance Educcation Examples 1. Plot the o
ordered paiir (−3, 5) an
nd determine the quadrrant in whicch it lies. The coordinatees x=−3 and y=5 indicatte a point 3
3 units to th
he left of an
nd 5 units aabove the origin. The point iis plotted in
n quadrant II (QII) beccause the x‐‐coordinate is negativee and the y‐coord
dinate is possitive. Example 2
2 Plot ordereed pairs: (4, 0), (−6, 0),, (0, 3), (−2, 6), (−4, −6)) Ordered paairs with 0 as one of th
he coordinaates do not lie in a quaadrant; thesse points arre on one axis orr the otherr (or the po
oint is the o
origin if bo
oth coordinaates are 0). Also, the scale indicated o
on the x‐axiss may be different from
m the scale indicated on the y‐axiss. Choose a scale that is conv
venient for tthe given sittuation. Distance F
Formula Freq
quently you
u need to caalculate thee distance between b
tw
wo points in
n a plane. To T do this, form aa right trian
ngle using tthe two points as vertiices of the ttriangle and
d then apply
y the Pythagoreaan theorem.. The Distan
nce Formulaa is a variantt of the Pyth
hagorean Th
heorem. Quantitative M
Methods for Econ
nomic Analysis –– I 61 School of f Distance Educcation The distance fo
ormula can b
be obtained
d by creatingg a triangle and using tthe Pythago
orean Theorem to
o find the length l
of th
he hypotenu
use. The hy
ypotenuse of o the0 triangle will bee the distance beetween the ttwo points. x2 and
a y2 are th
he x,y coordinates for onee point.
x1an
nd y1 are thee x,y coordinnates for the second poin
nt.
d is the distance between thee two points.
For examplee, consider the
t two pointts A (1,4) annd B (4,0). Find
F
the distaance betweenn them.
so: x1 = 1, y1 = 4, x2 = 4, and y2 = 0.
Substitutingg into the disstance formuula we have:
Example 1:: Find the diistance betw
ween (−1, 2) and (3, 5).
Quantitative M
Methods for Econ
nomic Analysis –– I 62 School of Distance Education so: x1 = –1, y1 = 2, x2 = 3, and y2 = 5. Substituting into the distance formula we have: 3
1
5
4
2 3 √16
9 √25 5 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 4 8
4
,
2
2
1 4
, 2 2
1
,2 2
Example 2: Find the midpoint of the line segment joining P(‐5, ‐4)and Q(3, 7) Use the midpoint formula: 5 3 4 7
,
2
2
2 3
, 2 2
3
1, 2
Quantitative Methods for Economic Analysis – I 63 School of Distance Education 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: ∆
∆
Quantitative Methods for Economic Analysis – I 64 School of f Distance Educcation d direction of a line. The T greaterr the The value of slope indicaates the steeepness and
absolute vaalue of the sslope, the stteeper the lline. A posittively slopp
ped line moves up from
m left to right (fo
or example a supply cu
urve). A neggatively slop
pped line m
moves down
n (for example a demand cu
urve). The sllope of a ho
orizontal lin
ne (for exam
mple a perfeectly elastic demand cu
urve), for which is zeero. The slop
pe of a vertiical line (forr example aa perfectly in
nelastic dem
mand curve), for which , is undeffined. That is here slop
pe does not exist since dividing by zero is not possiible. Negative Slope Po
ositive slopee Quantitative M
Methods for Econ
nomic Analysis –– I 655 School of f Distance Educcation Rules for Calculating th
he Slope of a Line F
two po
oints on thee line. Everry straight line has a consistent slope. In other o
1. Find words, the slope of a line never changes. Th
his fundameental idea m
means that you can ch
hoose ANY two points on a line to find tthe slope. T
This should intuitively make sensee with your own understand
ding of a strraight line. A
After all, if tthe slope of a line could
d change, th
hen it would
d be a zigzag line and not a sttraight line. ount the rise (How man
ny units do you count u
up or down to get from
m one point tto 2. Co
the next?) R
Record this number as your numerator. 3. Co
ount the run
n (How man
ny units do you count left or right to get to thee point?) Record thiss number ass your denominator. 4. Siimplify yourr fraction if possible. Imp
portant notee: If you count up or rigght your num
mber is posiitive. If you count down
n or left your nu
umber is negative. Another ap
pproach to fi
finding slopee Find the diffference of b
both the y aand x coordiinates and p
place them iin a ratio. Piick two poin
nts on the line. Because yo
ou are usingg two sets off ordered paairs both haaving x and y values, a subscript m
must be used
d to distingu
uish betweeen the two v
values. Conssider the folllowing grap
ph. wo points. The two points we choose are (x1, y1) = (3,2). and (x2, y2) = Choose any tw
(‐1,‐1) Thiss is a simpler formulaa for findingg the slope of a line. where m is the variable used for the slope. Substitutin
ng Quantitative M
Methods for Econ
nomic Analysis –– I 66 School of f Distance Educcation Example 1: Find slope The slope of a lline through
h the pointss (1, 2) and (2, 5) is 3 b
because eveery time thaat the line moves up three (th
he change in y or the riise) the linee moves to the right (th
he run) by 1.. Example 2 : Find slopee The slope of a line is thee ratio of itts rise to itss run. The rise is the vertical ch
hange between tw
wo points on
n a line. Thee difference between th
he y‐coordin
nates createes the rise. In th
his example, the differen
nce between the y‐coordiinates of poiints A and B is 3 – (–3) = 6. The run iss the horizontal differrence betw
ween two points p
or th
he differencce between
n the x‐coordinattes. In this eexample thee difference is –3 – 1 = ––4. So h
here slope of the line = Quantitative M
Methods for Econ
nomic Analysis –– I 677 School of f Distance Educcation Example 3 Find Slope of the lines in the graph based on the informaation given
Line AE, co
oordinates ((‐3,2) = (x1, y1) (3,‐3) =
= (x2, y2) , sslope = Line DB, coordinates (‐4,‐1) = (x1
1, y1), (3,2) = (x2, y2) , slope = Line BC, co
oordinates ((3,2) = (x1, y
y1), (1,‐2) =
= (x2, y2) , slope = Intercept: We are going tto talk abou
ut x and y in
ntercepts. A
An x interceept is the po
oint where your line crossess the x‐axis.. The y interrcept is the point wheree your line ccrosses the y‐axis. Algebraicallly, An ‘xx‐intercept’’ is a point o
on the grap
ph where y is zero, thatt is, an x‐intercept is a p
point in the equaation where the y‐valuee is zero. Quantitative M
Methods for Econ
nomic Analysis –– I 688 School of f Distance Educcation A ‘y‐‐intercept’ iis a point on
n the graph where x is zzero, that iss. a ‘y‐interccept’ is a poiint in the equatio
on where the x‐value is zero. In th
he above illlustration, the t x‐interccept is the point p
(2, 0)) and the y‐‐intercept is the point (0, 3)). Example 1 d the x and y
y interceptss of the equaation 3x + 4
4y = 12. Find
To find the x‐interceptt, set y = 0 and solve forr x. 3x + 4( 0 ) = 12
2 3x + 0 = 12 3x = 12 x = 12/3 x = 4 To find the y‐interceptt, set x = 0 and solve forr y. 3( 0 ) + 4y = 12
2 0 +
+ 4y = 12 4y = 12 y = 12/4 y = 3 Therefore, the x‐inttercept is ( 4
4, 0 ) and th
he y‐intercep
pt is ( 0, 3 ).. The graph o
of the line lo
ooks like this: Quantitative M
Methods for Econ
nomic Analysis –– I 69 School of f Distance Educcation Example 2 d the x and y
y interceptss of the equaation 2x ‐ 3y
y = ‐6. Find
To ffind the x‐in
ntercept, sett y = 0 and solve for x. 2x ‐ 3( 0 )) = ‐6 2x ‐ 0 = ‐6
6 2x = ‐6 x = ‐6/2 x = ‐3 To ffind the y‐in
ntercept, sett x = 0 and solve for y. 2( 0 ) ‐ 3yy = ‐6 0 ‐ 3y = ‐6
6 ‐3y = ‐6 y = ‐6/(‐3
3) y = 2 Therefore, the xx‐intercept is ( –3, 0 ) aand the y‐inttercept is ( 0, 2 ). The graph o
of the line lo
ooks like this: Example 3 Find
d the x and y
y interceptss of the equaation ( ‐3/4 )x + 12y = 9
9. To ffind the x‐in
ntercept, sett y = 0 and solve for x. Quantitative M
Methods for Econ
nomic Analysis –– I 70 School of Distance Education ( ‐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 ). Quantitative Methods for Economic Analysis – I 71 School of f Distance Educcation The graph of th
he line lookss like this: Graphing o
of Function
ns A fu
unction is a relation (usually an eq
quation) in which no ttwo ordered
d pairs havee the same x‐coo
ordinate when graphed
d. Onee way to tell if a graph iss a function
n is the vertiical line testt, which say
ys if it is posssible for a verticcal line to meet m
a grap
ph more than once, th
he graph is not a functtion. The fiigure below is an
n example of a function. Functions are usually den
noted by letters such as f or g. Iff the first coordinate c
o an of ordered paair is represented by x, the second coordinate (the y coorrdinate) can
n be represeented by f(x). In tthe figure b
below, f(1) =
= ‐1 and f(3)) = 2. Quantitative M
Methods for Econ
nomic Analysis –– I 72 School of f Distance Educcation Wheen a function is an equaation, the do
omain is thee set of num
mbers that are replacem
ments for x that give a value
g
e for f(x) thaat is on thee graph. Som
metimes, ceertain replaacements do
o not work, such as 0 in the function: f(xx) = 4/x (beecause we cannot divid
de by 0). Straight­Liine Equatio
ons: Slope I
Intercept fo
form of a grraph The slope intercept forrm of a line is y = mx + b. Where m iss the slope aand b is the y‐interceptt. Everry straight line can be representeed by an equ
uation: y = mx + b. Thee coordinattes of every pointt on the linee will solve the equatio
on if you sub
bstitute them in the eq
quation for xx and y. The equation of any straight line, calleed a linear eequation, caan be writteen as: y = mxx + b, where m is the slope o
of the line an
nd b is the yy‐intercept.
axis The y‐interceptt of this line is the vallue of y at the t point where the w
lin
ne crosses tthe y Quantitative M
Methods for Econ
nomic Analysis –– I 73 School of f Distance Educcation Giveen the slopee intercept fform, = mxx + b, let's fiind the equ
uation for th
his line. Pick
k any two points,, in this diaggram, A = (1
1, 1) and B =
= (2, 3). We found that the slope m
m for this lin
ne is 2. By lo
ooking at th
he graph, w
we can see th
hat it intersects the y‐axis a
t
at the point (0, –1), so –1 is the vaalue of b, th
he y‐interceept. Substitu
uting these valuees into the eequation forrmula, we geet: y =
= 2x –1 The line showss the solutio
on to the eq
quation: that is, it show
ws all the vaalues that saatisfy the equatio
on. If we substitute thee x and y vaalues of a po
oint on the line into th
he equation
n, you will get a trrue statemeent. To ggraph the equation of aa line, we p
plot at least two pointss whose coo
ordinates saatisfy the equatio
on, and then
n connect th
he points wiith a line. W
We call these equations "linear" beccause the graph o
of these equ
uations is a sstraight linee. If th
he equation is given in tthe form y =
= mx + b, then the consstant term, w
which is b, iis the y intercept,, and the coefficient of x, which is m
m, is the slo
ope of the sttraight line. The easiest waay to graph such a linee, is to plott the y‐interrcept first. Then, writee the slope m in the form of o a fraction
n, like rise o
over run, and from the y‐intercep
pt, count up
p (or down) for the rise, ov
ver (right orr left) for th
he run, and
d put the neext point. Th
hen connecct the two points and this is y
your line. Example1: y = 3x + 2 Heree Y‐intercep
pt = 2, slopee = Starrt by graphin
ng the y‐inttercept by going up 2 units on the yy‐axis. From
m this point go UP (risse) anotherr 3 units, then 1 unit to
o the RIGHT
T (run), and
d put another point. This is tthe second p
point. Conn
nect the poin
nts and it sh
hould look liike this: Quantitative M
Methods for Econ
nomic Analysis –– I 74 School of Distance Education Example 2: y = –3x + 2 3 Y‐intercept = 2, slope = 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! left. 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 Example 1 This example shows how to graph a line with a slope of 2/3. Quantitative Methods for Economic Analysis – I 75 School of Distance Education 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 Methods for Economic Analysis – I 76 School of f Distance Educcation Example 6 by f (x) = ‐(1
1 / 3)x ‐ 1 / 2 Graph the llinear functiion f given b
Determine the x interccept, set f(x)) = 0 and solve for x. 1 / 2 = 0 ‐(1 / 3)x ‐ 1
x = – 3 / 2 =
= – 1.5 Determine the y interccept, set x = 0 to find f(0
0). f(0) = –1 / 2 = –0.5 of the abovee function iss a line passsing through
h the pointss (‐3 / 2 , 0) and (0 , ‐1 // 2) The graph o
as shown b
below. 0
‐2
‐1.5
‐1
‐0.5
‐0.1
0
‐0.2
‐0.3
Series1
‐0.4
‐0.5
‐0.6
Straight­Liine Equatio
ons: Point­S
Slope Form
m Poin
nt‐slope reffers to a meethod for ggraphing a linear equaation on an x‐y axis. When W
graphing a linear equaation, the w
whole idea iss to take paairs of x's an
nd y's and p
plot them on
n the graph. While you could
d plot severral points by
y just pluggging in valuees of x, the p
point‐slope form Quantitative M
Methods for Econ
nomic Analysis –– I 777 School of Distance Education 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: Quantitative Methods for Economic Analysis – I 78 School of Distance Education 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: – 17 – 4 m = 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 9 4 Quantitative Methods for Economic Analysis – I 79 School of f Distance Educcation Straight­Liine Equatio
ons: Two Point Form Two
o Point Slope Form is used to geneerate the Equ
uation of a sstraight linee passing through thee two given points. Thee two‐point form of a line in the Caartesian plan
ne passing through thee points (x1,y
, 1) and (x2,y
, 2) is given
n by or equivaleently, Or alternattively Two point fform diagraam is as follo
ows. Example 1 Deteermine the eequation of tthe line that passes thro
ough the poin
nts A = (1, 2) and B = (−2, 5). Example 2 Find
d the equation of the lin
ne passing tthrough (3, – 7) and (– 2,– 5). by Solu
ution : The eequation of a line passin
ng through two points (x1, y1) and
d (x2, y2) is ggiven Quantitative M
Methods for Econ
nomic Analysis –– I 80 School of f Distance Educcation Since x1 = 3
3, y1 = – 7 an
nd x2 = – 2, aand y2 = – 5, equation b
becomes, Or Or 2x + 5
5y + 29 = 0
Straight­Liine Equatio
ons: Interce
ept Form The intercept form of th
he line is the equation of the line segmen
nt based on
n the intercepts with both
h axes. Thee interceptt form of a line in th
he Cartesiaan plane with x w
intercept a and y interrcept b is giv
ven by a is the x‐in
ntercept. b is the y‐intercept. a and
d b must be nonzero. The values of a and b ccan be obtaiined from th
he general fform equation. Quantitative M
Methods for Econ
nomic Analysis –– I 81 School of f Distance Educcation If y = 0, x = a. If x = 0, y = b. A line does not have an
n intercept form equatiion in the fo
ollowing casses: 1.A line parrallel to the x‐axis, whicch has the eequation y =
= k. 2.A line parrallel to the x‐axis, whicch has the eequation x =
= k. 3.A line thaat passes thrrough the origin, which
h has equatiion y = mx.
Examples nd a y‐interccept of 3. Find its equattion. 1. A line has an x‐interrcept of 5 an
0 forms a triiangle with the axes. Determine th
he area of th
he triangle.
2. The line x − y + 4 = 0
n and its leggs are the axxes. The line forrms a right triangle witth the origin
If y = 0 xx = −4 = a. If x = 0 y
y = 2 = b. The interceept form is: The area is: 3. Find the equation off a line whicch cuts off in
ntercepts 5 and –3 on xx and y axess respectiveely. The intercepts are 5 and –3
3 on x and yy axes respeectively. i.e.,, a = 5, b = – 3 The required eq
quation of tthe line is 3 x – 5y –15 = 0 Quantitative M
Methods for Econ
nomic Analysis –– I 82 School of Distance Education 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) Since (i) passes through (3, 4) Thus, the required equation of the line is x – y = – 1 or x – y + 1 = 0 3 – 4 = a or a = –1 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. Quantitative Methods for Economic Analysis – I 83 School of Distance Education 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. 1
0
3 4
3 So the y intercept is (x,y) = (0,3) Similarly to find the x intercept, put y = 0. 0
1
4
1
4
3 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. 0
‐2
‐1.5
‐1
‐0.5
‐0.1 0
‐0.2
‐0.3
‐0.4
‐0.5
‐0.6
2. Graph the linear function 2y + 10x = 20 To find the y intercept, put x = 0 2y + 10(0) = 20, 2y = 20, y = 10 So the y intercept is (x,y) = (0,10) To find the x intercept, put y = 0 2(0) + 10x = 20, 10x = 20, x = 2 So the x intercept is (x,y) = (2,0) Quantitative Methods for Economic Analysis – I 84 School of Distance Education 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. 0
‐1.6
‐1.4
‐1.2
‐1
‐0.8
‐0.6
‐0.4
‐0.2
0
‐0.1
‐0.2
‐0.3
‐0.4
‐0.5
‐0.6
3. Graph the linear function 2y ‐10x = 20 To find the y intercept, put x = 0 2y + 10(0) = 20, 2y = 20, y = 10 So the y intercept is (x,y) = (0,10) To find the x intercept, put y = 0 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. Quantitative Methods for Economic Analysis – I 85 School of Distance Education 0
‐2
‐1.5
‐1
‐0.5
‐0.1
0
‐0.2
‐0.3
‐0.4
‐0.5
‐0.6
4. Graph the linear function 2y – 10x + 20 = 0 To find the y intercept, put x = 0 2y – 10(0) – 20 = 0, 2y = –20, y = –10 So the y intercept is (x,y) = (0, –10) To find the x intercept, put y = 0 2(0) – 10x + 20 = 0, –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
‐2
‐1.5
‐1
‐0.5
0
‐0.1
‐0.2
‐0.3
‐0.4
‐0.5
‐0.6
5. Given a linear function 6y + 3x – 18 = 0. Also find its slope. Rewrite in the y = mx + c form 6y + 3x – 18 = 0 Quantitative Methods for Economic Analysis – I 86 School of Distance Education 6y = – 3x + 18 3 Slope = To find the y intercept, put x = 0 0
3 , y = 3 So the y intercept is (x,y) = (0, 3) To find the x intercept, put y = 0 0
3 3 3 2 3 6 So the x intercept is (x,y) = (6,0) Now plot the two intercept points (4, 12) and (8,2) and connect them with a straight line. 0
‐2
‐1.5
‐1
‐0.5
0
‐0.1
‐0.2
‐0.3
‐0.4
‐0.5
‐0.6
6. Find the slope m of a linear function passing through (0,10), (2,0) Substituting in the formula 2 12
8 4
10
4
Graphing of Non Linear Functions Quantitative Methods for Economic Analysis – I 87 School of f Distance Educcation So far f we havee been seeing graphing of linear functions w
which givess a straight line. Now let us see the grap
phing of som
me non lineear functions. phs of Quad
dratic Functions Grap
Quaadratic Funcction The term quad
dratic comees from the word quad
drate mean
ning square or rectanggular. Similarly, one o of the definitions d
of o the term quadratic is a square. In an algeebraic sensee, the definition of o somethin
ng quadratic involves tthe square and no higgher power of an unkn
nown quantity; second degrree. So, for our o purposes, we will be workingg with quad
dratic equattions which meaan that the h
highest deggree we'll bee encounterring is a square. Norm
mally, we seee the standard quadratic eq
quation writtten as the sum of threee terms set equal to zzero. Simply
y, the three termss include on
ne that has aan x2, one haas an x, and
d one term iss "by itself" with no x2 or x. A qu
uadratic fun
nction in its normal form
m is written
n in the form
m wheere a, b, and
d c are consstants and aa is not equ
ual to zero. If a = 0, thee x2 term w
would disappear and we wou
a
uld have a linear equaation. Note that in a qu
uadratic fun
nction theree is a power of tw
wo on the in
ndependentt variable an
nd that is the highest po
ower. The graph of a quadratic ffunction is called a pa
arabola. It is basically a curved shape p or down. W
When we haave a quadraatic function
n in either fform, opening up
if a > 0, then the parab
bola opens u
up , and , if a < 0, then
n the parabola opens d
down . make thingss simple, lett us consideer a simple quadratic ffunction wh
here a = 1, b
b = 0 To m
and c = 0. SSo we get th
he normal quadratic eq
quation as y = 1x2 or y =
= x2. To grap
ph this funcction, let us try su
ubstituting v
values in for x and solvving for y as shown in th
he followingg table. x y = x2
y = x2 (x, y)) – 3 (‐3)2
9 (‐3, 9
9) – 2 (‐2)2
4 (‐2, 4
4) – 1 (‐1)2
1 (‐1, 1
1) 0 (0)2
0 (0, 0) 1 (1)2
1 (1, 1) 2 (2)2
4 (2, 4) 3 (3)2
9 (3, 9) owing graph
h. Plot the graaph on a graaph paper and we will gget the follo
Quantitative M
Methods for Econ
nomic Analysis –– I 888 School of f Distance Educcation As stated earlieer, If a>0, then the parabola has a m
minimum po
oint and it o
opens upwaards (U‐shaped)) For examp
ple see the fo
ollowing graph for the function y=
=x2+2x−3 Similarly, Iff a<0, then tthe parabola has a maxximum point and it opens downwaards (n‐shap
ped) For examplle see below
w the graph of the functtion y=−2x2
2+5x+3 Quantitative M
Methods for Econ
nomic Analysis –– I 89 School of Distance Education 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. Lt
i 2 K ∞
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’. It can be written as Example 1: Limit x2 as x approach to 4 is 16. This can be expressed as Lt
x 16 x 4
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 x
Lt
f x 0
Substituting x 0 x
Lt
0
f x 10 Ex. 3: Find x
Ans: Lt
3
10x
x
Lt
3
5 10x
5 Lt
Lt
10
x
x 3
x 3
9 30 5 44 x
Lt
3
5 Quantitative Methods for Economic Analysis – I 90 School of Distance Education Theorems on Limit of Function 1. If ‘a’ is a constant Lt
a = a x K
2. If a, b are constants 3. If x
Lt
Lt
x
f x l, K
x
Lt
i
x
K
Lt
ii
x
K
Lt
iii
x
K
Lt
iv
x
K
Lt
v
x
K
Lt
x
Ex. 3: Find Ans: Ex. 4: ax b ak b K
K
g x m then f x g x l m f x . g x l m , m 0 √ f x l , Lt
x
g x 0 K
f x . g x 0 K
Lt
x
2
Lt
x
2
Find Lt
x
Lt
3
2
5 Quantitative Methods for Economic Analysis – I 91 School of Distance Education Ans. Lt
x
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: Lt
x
3
write Ex. 2: Lt
x
∞
x
as x 3 6 Note ∞
Lt
a b a – b division of numerator and denominator by 0 as x ∞ Since We go 1 Ex. 3: x
Lt
3
Factorizing the function we get Quantitative Methods for Economic Analysis – I 92 School of Distance Education Ans: ‐1 Ans: 2 Ans:
Ans: 8 Problems 1. Find 2. Find 3. Find 4. Find 5. Find 6. Find Lt
x
2
Lt
x
1
Lt
x
x
2
Lt
0
x
∞
Lt
6
Lt
x
Ans: 2 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., If f x may or may not be the same as f a 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. Where x
f x f x f a refers to f a + h) and x
refers to f(a – h) as h 0. Quantitative Methods for Economic Analysis – I 93 School of Distance Education 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 But f 1 does not exist i.e., substituting 1 for x in f x function Thus f x is not continuous at x 1 1
1
f x 3
2 8 is finite. Ans: f 2 2
f x is continuous at x 2. f x 2
0
2
f x 2 is continuous for x 2 2
∞ indeterminate Eg. 2: Prove that the function 2 x 4 5 3
1
Solution : 2
3
2
3 2
2 2 2 8 f x f 2 Quantitative Methods for Economic Analysis – I 94 School of Distance Education 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. ‘x’. Let ‘x’ be an independent variable and ‘y’ be depending upon ‘x’, then ‘y’ is a function of Let ∆x be small change in ‘x’ corresponding change in ‘y’ is ∆y. Then ‘incremental ratio’. is called 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., Derivative or differential coefficient of f x is generally written as ∆x
0
implies derivative of ‘y’ with respect to ‘x’. 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
i.e., If y where ‘n’ is any real number is Examples , 8
1
y 2
y , 3
y , 5 3
4
y , 8
10
10
5 3
Quantitative Methods for Economic Analysis – I 95 School of Distance Education 5
y √ , y 6
y , , 1
½
½
1 Rule No. II: Derivative of a Constant The derivative of a constant function is zero Then If y k where k is constant Eg:‐ c 0 0 If y 10 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 Ex. 2 If y ‐30
5 7
find ‐30 35
‐30 ‐5
150 Rule No. IV Linear Function Rule Ex: 1 If y 5x 10 find y mx c m If y mx c where ‘m’ and ‘c’ are constant then 5 Quantitative Methods for Economic Analysis – I 96 School of Distance Education Ex:2 If y 2 3 5 find 2 3 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: 3
If y 5
2 5
35
Ex.2: If y 10
3
3
6 4
2 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 4
Ex. 1: If y 50 find 4
24
9
Ex. 2 If y 20 20
60
2
4 ‐ 50 find 50 2
8 4
50 – 4 Quantitative Methods for Economic Analysis – I 97 School of Distance Education 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 v 7 find u. v u Example: 1. If y 3
5 5
Solution: consider u 3
5, v 5
6x, 45
75
30
75
75
42 3
15 5
5 15
2. Find if y 7 7
42 6 u 2
6
10
v 2
1 2
4
4
4
3
3
4
3
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 .
Quantitative Methods for Economic Analysis – I 98 School of Distance Education 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 Quantitative Methods for Economic Analysis – I 99 School of Distance Education 2. If y z
7 10x 70x 10 and z x find 3z , 3z
3. Find 7x 7x Sub. x for z we get 7x 3x
3 x
7x 21x 3x we find by the chain rule if y x
3x Let u x
2 x
3x 3x
3 since u 3 2x
6x 3x
3 6
18
6
y u 2u 3x
24
3
6
18 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 ÷ Quantitative Methods for Economic Analysis – I 100 School of Distance Education 4a, = 4
= = Rule No. XI: Implicit Function Rule. 2 10 directly expresses ‘y’ in terms of ‘x’. Hence this An equation y = 4
function is called ‘Explicit Function’. 2 = 0 does not directly express ‘y’ in terms of ‘x’. This types An equation 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 3 ‐ 2
x y + y x = (6) (Product Rule) 3. 2
= 2x +2y+2x = 2x
= (2x+2y) = ‐(2x +2y) 3x ‐ 2y = (4) = 0, ‐ 2
= ‐3, = = 1.5 2. xy 6 + y = 0 = 10 x + 2y (x) +2x (y)+ (y ) = 10 + 2y = 0 +2y = ‐(2x +2y) ‐1 Rule No. XII: Inverse Function Rule Quantitative Methods for Economic Analysis – I 101 School of Distance Education The derivative of the inverse function i.e., derivative of the original function i.e., i.e., is equal to the reciprocal of the Example 1. Find the of the function y 4
2. Find the derivative of the inverse function 2 x
y x
4 x
4x
8x
2x
6x
8x
8x x
2 4x
4 2x 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
if y , 4. Find Let u 2 y Quantitative Methods for Economic Analysis – I 102 School of Distance Education 2x Note:‐ if y 2x since u 2 when u g x then 5. Find if y . Let u 2x 2x 6. Find if y Let u 4
12
6
10 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 Quantitative Methods for Economic Analysis – I 103 School of Distance Education 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 with respect to 1. Differentiate 2. Differentiate 2x Let u 1 Let v 2x with respect to 1 Quantitative Methods for Economic Analysis – I 104 School of Distance Education ÷ ÷ 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: 3
1. If y 5
2 find 8
30x 6 30 2. If y x
15x
x
x find 6x
, and 8 , 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 Quantitative Methods for Economic Analysis – I 105 School of Distance Education 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 2. A function is said to be minimum at x b if 0 and 0 or negative. 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 Quantitative Methods for Economic Analysis – I 106 School of Distance Education 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
10 4x
y x
2x – 4 4x
10 0 At the maximum or minimum Therefore 2x – 4 0 The function may be maximum or minimum at x 2 Therefore the function has minimum at x 2. 2x 4, x 4/2 2 2 0 positive 2. Find the maximum or minima of the function Solution y 2x
6x 6x
6 At maxima or minima 0 6x
6x x 1 6 0 6 x 1 At x ‐1 pr 1 the function is either maxima or minima. When x 1, 12x 12 1 12 positive Quantitative Methods for Economic Analysis – I 107 School of Distance Education 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 x R 21x
Taking 21 – 2x 0 21 2x, Since 21 – 2x ‐2 negative 0 we get x 21 2 10.5 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
6
10 7 2x – 6 0 we get Taking Solving this quadratic equation, we get, x 7, ‐1 Since output cannot be negative x 7 Substituting x 7 in So total cost is minimum when output is 7 6
7 we get 2 7‐6 8 positive Quantitative Methods for Economic Analysis – I 108 School of Distance Education 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. , for , i.e., y f x is an increasing function if the function. domain of 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. , for , i.e., 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 Y O x1 x2 X Increasing Function O
x1 x2 X Decreasing Function Quantitative Methods for Economic Analysis – I 109 School of Distance Education Ex: Check whether the function 3x
Solution 3x
x
10 is increasing or decreasing 3x
10 3x
x
9x
6x
1 3x
1 The function is always positive because 3x
number is always positive. Therefore 3x
3x
x
1 is positive since square of a real 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., to be a convex function in the given interval. 0, then the function is said Concave Function If the second derivative of a function f x is negative i.e., is said to be a concave of function in the given interval. 0, then the function Y O X O Y Convex Function X
Concave Function Quantitative Methods for Economic Analysis – I 110 School of Distance Education 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
Since 2 2
1
0 for x 0 the curve is convex from below for all positive value of x. Ex:2 Show that the demand curve P We have, Since P ‐8 is downward sloping and convex from below. ‐8 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 . Quantitative Methods for Economic Analysis – I 111 School of Distance Education Example 5xy 5y 5xy 5x 2x 0 2x 0 2x 2y x .
Example 3
4
find and at x 5, y 4 If Z 2
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. Quantitative Methods for Economic Analysis – I 112 School of Distance Education Example : Find the first and the second order partial derivatives for Z 3
8 2
2
Ans: 9
2
‐4x 4y ‐4x 4y 18x – 4y 4x 6y 4
2
2 2
3
2
4
3
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 Similarly when ∆y is a small change in y then The total change in Z ∆z ∆x ∆y when ∆x is very small. In the limit case. This is called total differential of z with respect to ‘x’ and ‘y’ ∆x ∆y is the corresponding change in z. dz dx dy Ex: Find the total differential of z 3
2
Ans: dz dx dy 6x y dx x – 6y2 dy Quantitative Methods for Economic Analysis – I 113 School of Distance Education 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 Quantitative Methods for Economic Analysis – I 114 School of Distance Education 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 Quantitative Methods for Economic Analysis – I 115 School of Distance Education 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 Quantitative Methods for Economic Analysis – I 116 School of Distance Education 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 Quantitative Methods for Economic Analysis – I 117 School of Distance Education 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. Present Value 0 1 2 3 ←Years Option A Option B Rs. 10000 Rs. 10000 + interest
Rs. 10000 – interest Rs. 10000 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 Quantitative Methods for Economic Analysis – I 118 School of Distance Education more than one year easy. Number of periods Future value = original amount × (1 + interest rate per period)
Using the above formula for calculation future value of investment at end of first year: 1 Future value = 10000 × (1 + 0.05) = 10000×1.05 = 10500 Future value of investment at end of 2 years: Future value = 10000 × (1 + 0.05)
1+1 2 = 10000×1.05 = 10000 × 1.1025 = 11025 Future value of investment at end of 3 years: Future value = 10000 × (1 + 0.05)
1+1+1 3 = 10000×1.05 = 10000 × 1.16 = 11600 Future value of investment at end of 10 years: Future value = 10000 × (1 + 0.05)
10 = 10000×1.05
10 = 10000×1.63 = 16288.95 In general future value of investment at end of n years: n
Future value = P(1+i) 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 = Quantitative Methods for Economic Analysis – I 119 School of Distance Education = 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 Quantitative Methods for Economic Analysis – I 120 School of Distance Education 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 1 2 3 4 5 Value at the beginning of the year 681 735 794 858 926 Interest rate
× 8 % Annual Interest = ×
8 %
=
×
8 %
=
×
8 %
=
×
8 %
=
54 59 64 69 74 Amount the end of
the year
735 794 858 926 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, Quantitative Methods for Economic Analysis – I 121 School of Distance Education 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 = ∑
‐ ∑
NPV = ∑
or NPV = ∑
‐ initial investment (C0) or 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? Quantitative Methods for Economic Analysis – I 122 School of Distance Education 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 Quantitative Methods for Economic Analysis – I 123 School of Distance Education = 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. Quantitative Methods for Economic Analysis – I 124 School of Distance Education 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, Quantitative Methods for Economic Analysis – I 125 School of Distance Education 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: Quantitative Methods for Economic Analysis – I 126 School of Distance Education 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, Quantitative Methods for Economic Analysis – I 127 School of Distance Education 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 Quantitative Methods for Economic Analysis – I 128 School of Distance Education 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. Project A Project B
Year Cash Flow Cash Flow
(in Rs.) (in Rs.) 0 ‐1000 ‐1000 1 500 100 2 400 200 3 200 200 4 200 400 5 100 700 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% Quantitative Methods for Economic Analysis – I 129 School of Distance Education 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. Quantitative Methods for Economic Analysis – I 130 School of Distance Education 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 Quantitative Methods for Economic Analysis – I 131 School of Distance Education 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. Quantitative Methods for Economic Analysis – I 132 School of Distance Education 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 Y1 Y2 Y3 Year Y0 Cash Flow −1,000 500 500 500 DCF 459 421 387 −1,000 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. Quantitative Methods for Economic Analysis – I 133 School of Distance Education • 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 Quantitative Methods for Economic Analysis – I 134 School of Distance Education 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 – = P 1
Amount = P + 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/‐ Quantitative Methods for Economic Analysis – I 135 School of Distance Education OR Salary at the end of 8th year P + = 3500 + = 6300/‐ Amount = P + = P 1
& Simple Growth = 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? Quantitative Methods for Economic Analysis – I 136 School of Distance Education P=P, A=2P, r =10%, n=? A = P 1
= P 1
i.e. 2p = P 1
2 = 1
2‐1 = n/10 i.e. 1 = n/10, so n = 10 years = P 1
= 1+ 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? Quantitative Methods for Economic Analysis – I 137 School of Distance Education Initial population =P, population @ end = 2P, r = 2.4%, n=? A = p 1
2P= p 1
Log 2 = n log 1.024 .
= = (1.024)n 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
p 1
26500/20000 = 1
= 2000 = 26000 = 1.325 Taking log of both sides, Log 1.325 = 4 log 1
0.1222/4 = log 1
= 0.03055 = 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 Quantitative Methods for Economic Analysis – I 138 School of Distance Education 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; Quantitative Methods for Economic Analysis – I 139 School of Distance Education = A(1+i)‐n, i = PV = (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 = = 400 0.681 = 272.40/‐ = A(1+r)‐n = A PV factor (PVnr) 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 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 PV = A x PVF 12131 Future Value/Compounded value Quantitative Methods for Economic Analysis – I 140 School of Distance Education 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 = i.e. PVFA r n = 1 = A P.V. factor (1038%’r’) 1
= 150 x 6.7101 = 1006.56 PVFArn = PVArn = A PVFA = .
.
.
.
= .
= 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 Quantitative Methods for Economic Analysis – I 141 School of Distance Education 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) Quantitative Methods for Economic Analysis – I 142 School of Distance Education = 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: Investment Criteria Non‐Discounting Discounting NPV Benefit‐Cost Ratio IRR Payback Period
Accounting Rate 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 Quantitative Methods for Economic Analysis – I 143 School of Distance Education 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%. Quantitative Methods for Economic Analysis – I 144 School of Distance Education *********
Quantitative Methods for Economic Analysis – I 145