Financial Mathematics [1 ed.] 9789350432600, 9788184885859

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FINANCIAL MATHEMATICS

A.. LENIN JOTHI Department of Management Kasturi Ram College of Higher Education GGSIP University, Delhi

ImI GJlinmlaya GFublishing GJlouse • MUMBAI • DELHI. BANGALORE • HYDERABAD • CHENNAI • ERNAKULAM • NAGPUR • PUNE • AHMEDABAD • LUCKNOW

ISBN : 978-81-84885-85-9

(ii)

©Author No part of this book shall be reproduced, reprin~ or translated for any purpose whatsoever without prior permission of the publisher in writing.

IFIRST EDITION :2009 Published By

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IBranch Offices

I

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Pune

No. 527, "Laksha" Apartment, First Floor, Mehunpura, Shaniwarpeth, (Near Prabhat Theatre), Pune-411030, MAHARASHTRA. Phones: 020-24496333/24496333/24496323 E-mail: [email protected]

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I Typeset at

Time~ Printographi~s

Printed at

~I

I

.Shri Krishna offsetPress Delhi-93

Page 1. Permutation and Combination

1

2. Logarithm

32

3. Simple Interest

49

4. Compound Interest

56

5. Nominal and Effective Rates of Interest

76

6. Equation of Value

88

7. Discount

99

8. Depreciation

114

9. Bills of Exchange

126

10. Immediate Annuity

142

11. Annuity Due

177

12. Deferred Annuity

198

. 13. Perpetuity and General Annuity

215

14. Amortisation of Loan

235

15. Sinking Fund

250

16. Leasing, Capital Expenditure and Bond

263

17. Theory of Probability

278

~8. C~nstruction

303

of Mortality Table

, 19. A 'Complete Mortality Table

315

20. Probabilities on Survival and Death

332

21. Well-known Mortality Tables

349

22. Rate-making in Insurance

370

23. Determination of Net Single Premium.

382

24. Determination of Net Level Premium

403

25. Determination of Premium. for Annuity Plans

424

26. Determination of Gross Premium.

447

27. Credibility Theory

458

APPENDICES Table I

Logarithm Tables

Table II

Amount of Re. 1 at Compound Interest

464 468

Table III

Present Value of Re. 1 at Compound Interest

477

Table IV

Amount of an Annuity of Re. 1

486

Table V

Present Value of an Arinuity of Re. 1

495

Table VI

Value of eX and e-«

504

Table VII

LIC (1970 - 73) Ultimate Table

508

Table VIII: Hm Table (Makeham Graduation)

510

Table IX

512

Area under the Standard Normal Curve

Permutation 'and Combination INTRODUCTION It is often necessary to calculate the number of different ways in which something can be done or can happen. In this chapter, we shall study some techniques of answering some counting problems without actually counting or listing all of them. These techniques are useful in determining the number of different ways of arranging or selecting different items.

FUNDAMENTAL PRINCIPLE OF COUNTING PRINCIPLE I : ADDITIVE PRINCIPLE

This principle states that: Let A and B be two events that cannot occur simultaneously. Then if A can occur in rn ways and B can occur in n ways, it follows that the number of ways in which either A or B can occur in m + n. This plinciple can be generated, for more than two events, as follows: "If one event can happen in ml ways, following which another event can happen in m2 ways, following which another event can happen in rna ways and so on that all events, in succession can happen in ml + m2 + rn3 + ...... different ways.

For example, in a class room there are 20 boys and 15 girls. If the teacher wants to select either a boy or a girl to represent the class, the number of ways in which the teacher can perform the selection is 20 + 15, i.e., 35 ways. PRINCIPLE II : MULTIPLICATION PRINCIPLE

This principle states that: If an event can happen in 'm' different ways following which another event can happen in 'n' different ways, then both event in succession can happen in exactly 'rnn' different ways.

2

FINANCIAL MATHEMAnCS

This principle can also be generalised, for even more than two operations, as follows: If an event can happen in 'ml' different ways, following whh:h another event can happen in 'm2' different ways, following which another event can happen in 'm3' different ways and so on, then all event in succession can happen in exactly ml x m2 x m3 x ...... different ways. For example, in a class room there are 20 boys and 15 girls. If the teacher wants to select one boy and one girl to represent the class, the number of ~ays in which the teacher can perform the selection is 20 x 15, i.e., 300 ways. Consider the following illustration: \ There are 3 stations X, Y and Z and 3 routes to go from X to Yand 2 routes from X to Z. This is shown in the following figure: Route 1

I

X

R

()

e:'·9 yCo~ jr-----'z Route 3

Route (b)

A person can go from station X to Y in three different ways and from Y to Z in 2 different ways. By the fundamental principle of counting, the number of ways for a person to go from X to Z is 3 x 2 =6, i.e., (1, a), (1, b), (2, a), (2, b), (3, a) and (3, b). E:x;ample 1 : There are eight doors in the college hall. In how many ways can a student enter the college hall through one door and come out through a different door?

Solution : The student can enter the hall in 8 ways and corresponding to each way of entering, there are only 7 ways of coming out, since he has to come out through a different door. Hence the required number of ways = 8 x 7 = 56. Example 2 : Given 7 flags of different colours, how many different signals can be generated if a..signal requires the use of two flags, one below the other ?

Solution : The upper flag can be anyone of the 7 flags and the lower flag can be anyone of the remaining 6 flags. :. Required no. of signals which can be generated = 7 x 6 = 42. Example 3 : A customer forgets a four-digit code for an Automatic Teller Machine (ATM) i1t a bank. However, he remembers that this code consists of digits 3, 5, 6 and 9. Find the largest possible number of trials necessary to obtain the correct code.

SQlution : The digits are 3, 5, 6 and 9. rio. of ways of choosing first digit == 4

3

PERMUTATION AND COMBINA TlON

No. of ways of choosing second digit = 3 No. of ways of choosing third digit = 2 No. of ways of choosing fourth digit = 1 . . Total number of ways of choosing four digits = 4 x 3 x 2 x 1 = 24 . . In all there could be 24 codes and customer would have to try 24 trials to obtain the correct code.

Example 4 : A coin is tossed three times and the outcomes are recorded. How many possible outcomes are there ? Solution : In a single toss of a coin, there are 2 possible outcomes. Either 'Head' of 'tail' can appear on the upper-most face. Therefore, when a coin is tossed three time, number of possible outcomes are 2 x 2 x 2 = 23 = 8. The possible outcome are HHH, HHT, HTH, HTT, THH, THT, TI'H, TTT

Example 5 : How many 3-digit numbers can be formed from the digits I, 2, 3, 4 and 5,if (a) repetition of digits is allowed? (b) repetition of digits is not allowed ? Solution: Total number of digits = 5 (a) When repetition of digits is allowed: Unit's place can be filled by anyone of the five digits . . No.. of ways to fill units place = 5 Since the repetition is allowed, the ten's place can also be filled by anyone of the five digits. " No. of ways to fill the 10's place = 5 'I'he 100's place can also be filled by anyone of the five digits. No. of ways to fill the 100's place = 5 This is shown as follows : 100's

5

10's 5

1's 5

. . The required number of 3-digit numbers = 5 x 5 x 5 = 125

(b) When repetition is not allowed: Unit's place can be filled by anyone of 5 digits :. No. of ways to fiU units place"= 5 Since repetition is not ellowed, there are only 4 choices :. No. of ways to fiU10's place = 4 Now only 3 digits are left :.

No. of ways to fiU100's place = 3

4

FINANCIAL MATHEMA TICS

This is shown as follows: 100's

1O's

l's

3

4

5

. . The required number of 3-digit numbers

=5 x 4 x 3 =60

Example 6 : How many 3-digit even numbers can be formed using the digits 1, 2, 3, 4, and 5, if (a) repetition of digits is allowed? (b) repetition of digits is not allowed?

Solution: Tota.l number of digits =5 (i.e. 1, 2, 3, 4, 5) =2 (i.e., 2, and 4) Total number of ev.en digits (a) When the repetition is allowed: 100's

10's

l's

5

5

2

No. of ways to fill unit's place No. of ways to fill10's place No. of ways to fill100's place

=2 =5 =5

,'. Required number of 3-digit even number =2 x 5 x 5 = 50

(b) When the repetition is not allowed : 100's

10's

l's

3

4

2

No. of ways to fill unit's place =2 No. of ways to fi1l10's place =4 No. of ways to filllOO's place =3 :. required number of 3-digit even numbers

=2 x 4 x 3 =24

Example 7 : How many 3-digit numbers can be formed from the digits 0, 1, 2, 3, and 4, if .a I repetition is allowed? (b) repetition is not allowed?

Solution: Total number of digits = 5 ~a)

When repetition is allowed: 100's

10's

l's

4

5

5

'0' cannot be placed in 100's place while forming 3-digit number .. No. of ways to fill100's place =4 No. of ways to fill10's place

=5

5

PERMUTA TlON AND COMBINA TlON

No. of was to filll's place =5 · . required number of 3-digit numbers =4 x 5 x 5 =100 (b) When repetition is not allowed: 100's

10's

1's

4

4

3

No. of ways to fi1l100's place =4 Now 4 digits are left : .. No. of ways to filll0's place = 4 Now 3 digits are left · . No. of ways to fill l's place = 3 · . required number of 3-digit. number =4 x 4 x 3 =48

Example 8 : How many different numbers below 1000 can be formed form the digit.'!! 3, 4, 6, 7 and 8, if: (a) no digit is repeated? (b) digits can be repeated ? Solution : The numbers less than 1000 can be : (i) One-digit numbers

(ii) Two-digit numbers

(iii) Three-digit numbers.

(a) No digit is repeated: (i) One digited numbers are 3, 4, 6, 7 and 8. These are 5 numbers. (ii) No. of ways of filling unit's place =5 No. of ways of filling ten's place =4 :. No. of 2 digited numbers =5 x 4 =20. (iii) No. No. No. ..

of ways of filling unit's place =5 of ways of filling ten's place =4 of ways of filling hundred's place = 3 No. of 3 digited numbers = 5 x 4 x 3 = 60 Total numbers =5 + 20 + 60 =85.

(b) Digits can be repeated: (i) One digited numbers = 5 (ii) No. of ways of filling unit's place = 5 No. of ways of filling ten's place = 5 . . No. of 2 digited numbers =5 x 5 =25 (iii)

No. of ways of filling unit's place =5 No. of ways of filling ten's Jllace =5 No. of ways of filling hundred's place =5 .. No. of 3 digited numbers =5 x 5 x 5 =125 Total numbers =5 + 25 + 125 =155.

(.: Repetition is allowed)

FINANCIAL IfATHEliA ncs

6

Example ~ : Ten people compete in a race. In now many ways the first three prices can be distributed ? -

Solution: First price goes to anyone of the 10 people Second price goes to anyone of the remaining 9 people Third price goes to any of the remaining 8 people :. No. of ways to distribute the prices =10 x 9 x 8 =720. Example 10 : How many automobile license plates can be made. if the inscription on each contains two different letters followed by three different digits ?

Solution: Out of 5 inscriptions, first 2-are of different letters (out of 26) and the other 3 are of different digits (out of 10). No. of ways of 1st inscription = 26 No. of ways of 2nd inscription No. of ways of 3rd inscription

=25 =10

No. of ways of 4th inscription No. of ways of 5 th inscrip.tion.

=9 =8

.. Total number of license plates =26 x 25 x 10 x 9 x 8 =4,68,000.

L A hall has 3 entrances and 4 exits. In how many ways can a man enter and-exit from

2. 3.

4.

5.

6.

the hall? [Alls. 12] In a railway compartment, 6 seats are vacant on a bench. In how many ways can 3 passengers sit on them? [ADs. 120] If there are 20 steamers playing between places A and B. in how many ways could the round trip from A be made if the return was made on (a) the same steamer, (b) a different steamer. [ADs. (a) 20, (b) 380] There are 5 routes between city X and city Y. In how many different ways can a man go from city X to city Y and return, if for return journey: (a) any of the routes is taken (b) the same routes is taken (c) the same route is not taken. [ADs. (a) 25, (b) 5, (c) 20] Four coins are tossed simultaneously. In how many ways can they fall? [Ans. 32] In how many ways can 4 students draw water from 4 taps, if no tap remain unused? [ADs. 24]

7. It has been decided that the flag of a newly formed forum will be in the form of three blocks one below the other, each coloured different. If there are five different colours on the whole to choose from, how many such designs are possible? [ADs. 60] 8. Given 6 flags of different colours, how many different signals can be generated, if a [ADs. 30] signal requires the use of two flags, one below the other?

PERMUTATION AND COMBINA TlON

7

9. For a set of 6 true. or false questions no student has written all correct answers and no two students have given the same sequence of aJ).swer. What is the maximum number of students in the class for this to be possible? [ADs. 63] 10. A team consists of 6 boys and 4 girls and the other has 5 boys and 3 girls. How many single matches can be arranged between the two terms, when a boy plays against a [ADs. 42] boys and a girls plays against a girl ? U A class consists of 20 girls and 15 boys. In how many ways can a president, vice president, treasurer and secretary be chosen if the treasurer must be a girls, the secretary must be a boy and a student may not hold more than one office? [Ans. 31680] 12. How many four digit numbers can be formed out of the digits 1, 2, 3, 4, 5, and 6, when no digit is repeated in the same number? [Ans. 360] 13. How many three digit numbers can be formed without using the digits, 1, 2, 3, 4 ? [Ans.180] 14. From the numbers 1, 2, 3, 4, 5, and 6, how many 3-digit odd numbers can be formed when (a) the repetition' of the digits is allowed, (b) the rep~tition of the digits is not allowed. [Ans. (a) 180, (b) 60] 15. How many words (with or without meaning) of four distinct letters of the English [Ans. 358800] alphabet are there? . 16.. Find the number of possible even numbers which have three digits? [Ans. 450] 17. How many numbers can be formed using the digits 1, 2, 3, and 9, if repetition of digits is not allowed? [Ans. 64] 18. How many 6-digit telephone number can be constructed if each number starts with [Ans. 1680] 32 and no digit appears more than once? 19. How many 2-digit even numbers can be formed from the digits 1,2,3,4 and 5, if (a) the repetition of digits is allowed (b) the repetition of digits is not allowed. . , [Ans. (a) 10, (b) 8] 20. How many 3-letter code words are possible using the first 10 letters of English alphabet, if (a) no letter can be repeated? (b) letters are repeated? [Ans. (a) 720, (b) 1000] 2L How many numbers are there between 100 and 1000 in which all the digits are distinct? [Ans. 648] 22. How many number are there between 100 and 1000 such that every digit is either 2 [Ans. 8] or 9 ? 23. How many numbers are there between 100 and 1000 which have exactly one of their [ADs. 225] digits as 7 ? 24. Twelve students compete'in a race. In how many was can the first three prices be [Ans.1320] distributed? 25. In how many ways can 4 different prizes be awarded among 6 contestants, so that a contestant may receive: (a) at most one prize, (b) any number of prizes. LAns. (a) 360, (b) 1296]

FINANCIAL MATHEMATICS

26. A number oflock on a suitcase has 3 wheels each labelled with ten digits from 0 to 9. If opening of the lock is a particular sequence of three digits with no repeats, how many such sequences will be possible. ? [ADs. 720] 27. The licence plates for vehicles registered in Delhi consists 'of 3 letters from English alphabet followed by 1, 2, 3 or 4 digits. The letter on the extreme left has be 'D'. For the I-digit number plates the number '0' is not allowed. For others, the digits and the letters of course can repeat ~ut the number should be significant. Determine the possible number of licence plates. [ADs. 6759324] 28. A code word is to consist of two distinct English alphabet followed by two distinct numbers from 1 to 9. For example GH 79 is.a code world. How many such code words are possible? [ADs. 46800]

FACTORIAL NOTATION There are some occasions when we wish to consider the product of fIrst n natural numbers. The continued product of first n natural numbers (beginning with 1 and ending with n) is called n-factorial or factorial n and is denoted by n ! or Ill.. Thus, n!

= 1. 2. 3 ... , (n -

1) . n

For example, 5 I = 1 x 2 x 3 x 4 x 5 =120· . ,3 !

= 1 x 2 x 3 =6

Remarks : (a) O! = 1

(c) n! = (n - 1) ! x n

(b) 1!;= 1

for example 7 ! =7 x 6 ! Example 11 ; Evaluate the following : 91 12! (a) 77' (b) 7! 51 '

Solution: 9!

(a)

(1))

7f = 9 x 87!x 7 ! -_ 9 x 8 --

61 (c) 2)(41

72

12 ! 12-x 11 x 10 x 8 x 9 x 7 ! _ 792 7! 5! :;: 7!x1x2x3x4x5 -

(c.) 2 6x !4!

=

6 x 5 x ~! 2)( 4 !

=15

Example 12 : Which of the following statements are correct (a) 2! + 3 ! =5 ! (b) (4!) (2 !) =81 (c)

5(41)=(5)(4)1

6! (d ') 21=

3

I

.

9

PERMUTA TlON AND COMBINATION

Solution: 2 ! + 3 ! = (2 x 1) + (3 x 2 x 1) = 2 + 6 = 8

(a)

5 ! = 5 x 4 x 3 x 2 x 1 = 120 :. 2! + 3!

06

5!

(4 !) (2!) = (4 x 3 x 2 x 1) x (2 x 1) = 24 x 2 = 48

(b)

8 ! = 8 x 7 x 6 x 5 x 4 x 3 x 1 = 40320 :. (4!) (2 !)

06

8

!

5 x (4!) = 5!

(c)

..

5 (4!)

¢

06

20 !

20!

~ _ 6 x 5 x 4 x 3 x 2 ! _ 360

(d)

2! -

2!

-

3! = 3x2x1=2

6! 063 I .. 2T .

PERMUTATION An arrangement in a definite order of a number of things taking some or all of them at !:l time is called a permutation. The total number of permutations of n distinct things taking r (1 s r s n) at a time is denoted by npro or by P(n, r).

The number of permutations ofn thlDgs taking r at a time is given by nPr = n (n - 1) (n - 2) ...... r factors, 1 s r s n. Alternatively

I nPr =(n

iIiil

,Is r s n.

Remark 1. The number of permutations of "n' different thinks taking at

R

time is equal

to n! npn=( :!)I=no;=n!

n

n.

.

Remark 2. It should be noted carefully that in nPn one count only those permutation in which repetition of things is not allowed. Remark 3. The number of all permutation of "n' different objects taking r at a time, when a particular object is always included in each arrangement is

r.n-1Pr _ 1

Remark 4. The number of all permutations of "n' different objects taking r at a time, when a particular object is always excluded in each arrangement is n-lPr Example 13 : Evaluate the following: (a)1P 3,

(b) 6P4

10

FINANCIAL MATHEMATICS

Solution: 7

Pa

(a) ,

71

71

7x6x5x4!

= (7 _' 3) ! =4t =

41

=21

0

Alternatively, 7Pa = 7 x 6 x 5 = 210 6n _ r 4 -

(b)

6 I _ ~ _ 6 )( 5 x 4 x 3 x 2 1 _ 360 (6 _ 4) I - 2 ! 2! -

Alternatively 6'1'4 = 6 )( 5 x 4 x 3 = 360 Example 14 : If 5P (5, r) =2P (6, r -I), find r.

Solution: Given that 6Pr = 2. 6Pr _ 1 -2

5!

(5 - r)! -

5!

6i . 6 - (r -1) !

_ 2

(5-r)l-

-2

51

(5 -r) I -

61 . (7-r)!

6x51 . (7 -r)(6-r)(5 -r) I

1 _

2x6

- (7-r)(6-r) =>

(7 - r)(6 - 1')

=> . (7 - r)(6 - r) =>

= 12

= 4)( 3

(7 - r)(6 - r) = (7 - 3)(6 - 3) ... r= 3

Example 16 :If 16"'P3 =13. ",+lPa. find n.

Solution : 16 . nPa = 13.

11,

+ IPa

=>

n! (n + 1) I 16 . (n _ 3) I = 13· (n + 1- 3) I

=>

n! (n + I)! 16 . (n _ 3) I = 13· (n - 2) !

=>

16'

n! _ 13. (n'+ 1) n I (n - 3)! (n - 2)(n - 3) 1

16 _ 13 (n + 1)

=>

-

(n-2)

16 (n - 2) = 13 (n + 1)

=> => .

...

3 n = 45 => n = 15 n = 15

11

PERMUTATION AND COIIBINA TlON

Example 18 : Prove thatnPr = n-IPr + r. n-1Pr _ l

Solution: RHS

= n-1Pr+r.n-1Pr_l (n - 1) ! - (n - 1 - r)

-

-

(n - 1)

!

!

- (n - r -1) ! 1)

_ (

1) , [

!

(n - 1) ! +r (n - r)!

!

[(n _ rl_ 1) + (n

~ r) !]

(n - r)

r]

. (n-r)(n-r-1)! +'(n-r)!

= (n -

=(n -

(n - 1) ! (n - 1- (r - 1»

!

= (n - n-

+r

1)

![(~~~\

1) ! (n

+ (n

1

~ r) !] =(n - 1) ! ~n-~ ~ ~ ]

n n (n-1)! _ r) ! = (n - r) !

n!

=(n _ r) ! =nPr =LHS. Hence proved. Example 17 : In how many ways can five children stand in a queue?

Solution: Number of ways in which 5 children

stan~ in queue

=Number of ways in which 5 children can be arranged among themselves =5P5 =5! =5 x 4 x 3 x 2 x 1 =120. The required number of ways is 120.

[.,' nPn=n!]

Example 18 : S-even persons are participating in a race. In how many can the first three prizes be own ?

Solution: Total number of participant n =7 No. of prizes r =3 .. The required number of ways = 7P3 =7 x 6 x 5 = 210 The required no. of ways =210. Example 19 : (a) It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible ? (b) If it is required to seat 5 men and 2 women in a row so that the women occupy the even places, how many such arrangements are possible ?

Solution: (a)

Total number of position

=9

12

FINANCIAL MATHEMATICS

No. of even positions =4 and No. of women =4 Given that the women occupy even places. :. No. of ways to arrange the women =4P4 =4! =4 x 3 x 2 x 1 = 24 No. of remaining position =5 No. of persons to be occupied =5 (men) :. No. of ways to occupy the remaining positions =5P 5 = 5 ! =5 x 4 x 3 x 2 x 1 = 120 :. No. of ways to occupy all 9 position =24 x 120 =2880 : .. Required number of arrangements = 2880 (b)

Total number of positions = 7 No. of even positions =3 (2nd , 4th , and 6th) No. of women =2 Given that women occupy the even places. :. No. of ways to arrange the women =3P2 = 3 x 2 = 6 No. of remaining position = 5 No. of remaining persons = 5 (men) .. No. of ways to occupy the remaining positions = 5P5 = 5 ! = 5 ! = 5 x 4 )( 3 x 2 x 1 = 120 . . The number of ways to occupy all 7 positions =4 x 120 =720 . . Required number of arrangement of 720

Example 20 : In how many ways can 6 exam papers can be arranged so that the best and the worst papers are never placed iogether ? Solution: Total number of papers = 6 No. of best papers =4 No. ofwors:, papers =2 First we arrange the best papers. No. of ways to arrange the best papers =4p 4 = 4 ! = 24 There are 5 places for the worst papers as shown in the following figure.

xBxBxBxBx No. of ways in which the worst papers can be arranged = 5P 2 = 5 x 4 No. of arrangements of all 6 papers =24 x 20 =480 The required number of arrangements = 480.

=20

Alternative Method: No. of ways in which ail 6 papers can be arranged with out any restriction = 6! Now, coniider the t.wo worst papers as one units. :. No. of units to be arranged =5 (i.e. 4 + 1) No. of ways to arrange this 5 units = 5 !

. 13

PERMUTATION AND COMBINA TlON

In each such arrangements, the worst papers can be arranged among themselves in 2 ! ways. The number of ways in which the worst papers are placed together is 5 !

x

2!

The number of arrangement in which the worst papers never place together

= Total no.

of possible arrangement - The no. of arrangement in which they are placed together. =6!-5!x2! =6x5!-5!x2 [.: 2! =2]

= 5! [6 -

2] = 5 ! x 4 = 120 x 4 = 480

The required number of arrangement =480 Example 21: A family of 4 brothers and:J sisters is to be arranged for a photograph in one row. In how many way~ can the.y be seated if (aJ all sisters sit together? (bJ no two sisters sit together?

Solution: Given, No. of brothers =4 No. of sisters (a)

=3

We take 3 sisters as one unit

= 4 + 1 =5 No. of ways to arrange all units =5 !

No. of units to be arranged

No. of ways to arrange all sisters among themselves

=3 !

The number of ways in which all sisters sit together

= 5 ! x 3 ! = 120 x 6 = 720 The required no. of arrangements = 720. (b)

First we arrange 4 brothers No. of ways to arrange the brothers =4!

xBxBxBxBx There are 5 places in which 3 sisters can be seated. ..

No. of ways to arrange the sisters = 5P3

..

No. of ways in which all can be seated

= 5 ! x 5P 3 = 60 x 24 = 1440 The required no. of arrangements = 1440. Example 22 : A team has 7 players. In how many ways they can be seated in a row, if the captain sits in the middle seat ?

Solution : Since the captain is fixed in the middle seat, the remaining six players can be seated in the remaining six seats in 6 ! was, i.e., 720 ways.

14

FINANCIAL MATHEMATICS

Example 23 : In how many ways can 5 English 4 Tamil and 2 Hindi books can be arranged if the books of each different language are kept together? Solution: We take 5 English books as one unit; 4 Tamil books as one unit; and 2 Hindi books as one unit. No.' of ways to arrange 3 units =sPs =3 ! Now we arrange each language books among themselves. No. of ways to arrange English books =5P5 =5 ! No. of ways to arrange Tamil books =4P4 =4! No. of ways to arrange Hindi books = 2P 2 ="2! .. No. of ways to arrange all books = 3 ! x 5! x 4! x 2 ! = 6 x 120 x 24 x 2 = 34560 :. The required no. of arrangements =34560. Example 24 : How many four digit numbers are there with distinct digits 1. Solution: There are 10 digits (0, 1, 2, ... , 9). The 1000's place cannot be filled with o. No. of ways to fill the 1000's place =9 The remaining 3 places can be filled with any three of the remaining 9 digits. No. of ways to fill the remaining places =9Pa No. offour digits numbers with distinct digits = 9 x 9Pa = 9 x 9 x 8 x 7 = 4536 Required no of four digit numbers :i,4536.

Example 25 : How many 3-letter words can be made using the letters of the word "DELH1".

Solution: No. ofletters in the word 'DELHI' =5 No. of 3-letter words =5Pa = 5 x 4 x 3 = 60 Required no. of words is 60. Example 26: How many 3-letter words can be made using the letters of the word "ORIGINAL", if (a) 'N is always included in all words ? (b) 'N is always excluded in all words ? (c) the word starts with 'N ? Solution: No. ofletters in the word 'ORIGINAL' is 8 and all letters are distinct. (a) No. of ways to arrange 'N = 3 After arra,nging 'N two places and 7 letters are left. No. of ways to arrange the remaining 2 places =7P2

15

PERMUTATION AND COMSINA nON

No. of words in which N is always included =3)( 7P2 =3)( 7 )( 6 =126 Required number of words is 12'6. (b) If 'N'

is excluded, No. of ways in which the retnaining 7 letters' can be arranged taking three at a time =7Pa =7 x 6 )( 5 =210 Required number of words = 210

(c)

If the 3-letter word starts with 'N, No. of ways to arrange 'N =1 No. of ways to fill the,remaining 2 places =7P2 .. No. of 3-letter words starting with 'N =1 x 7P2 =1 )( 7 )( 6 =42 :. Required number of words is 42.

Example 27 : In how many ways can the letters of the word 'STRANGE be a"anged so that (a) the vowels are never separated (b) the vowels never come together, and (c) the vowels occupy only the odd places. Solution: No. ofletters of the word 'STRANGE' =7 No. of vowels in the word =2 (A and E) :. No. of consonants in the word = 5 (a)

(b)

We take the two vowels as one unit No. of units to be arranged =6 (5 + 1) No. of ways to arrange the 6 units =6P6 =6! No. of ways to arrange the vowels among themselves =2P2 =2 ! N o. of arrangements in which vowels are never separated = 6.! 2! . . The required number of arrangements = 1440 First we arrange the consonants No. of ways to arrange the consonants = 5P5 = 5 !

xC)(CxC)(CxCX There are 6 places in which vowels can be arranged No. of ways to arrange the vowels =6P2 No. of arrangements in which the vo~els never come together =5 ! x 6P 2 =120 x 6 x 5 =3600 The required number of arrangements is 3600. (c)

lolElolElolElo\

No. of odd places =4

=720 )( 2 =1440

16

FINANCIAL MATHEMATICS

The two vowels are to occupy any two of these 4 odd places. No. of ways to arrange the vowels =4P2 Now the remaining 5 places can be filled with the 5 consonants No. of ways to arrange the consonants = 5 ! The no. of arrangements in which the vowels occupy odd places 4P 2 x 5 ! = 4 x 3 x 120 = 1440 The required number of arrangements is 1440.

L Evaluate (a) 4P 3

(c)

20P 4

(d) 9P9

[Ans. (a) 24, (b) 30, (c) 116280, (d) 362880] 2. Show that lOP 3 = 9P 3 + 3 9P2 3. (a) Find n, if2 np3 = n + IP3 and n;> 2. (b)

4. (a) (b)

Find n, ifnps = 3 nP5' Find r, ifl5Pr_l : ISPr _ 2 = 3: 4 Find r, if5Pr = sPr _ 1

[Ans. n = 5] [Ans. n = 8] [Ans.14] (Ans.4]

5. (a) Show that nPn = 2. nPn _ 2 (b) Show that nPr = n. n -IPr _ 1 6. Eight children are to be seated on a bench. In how many ways can the children be seated? [Ans. 40320] ·7. From among the 20 teachers of a college, one director and one dean are to be appointed. In how many ways can this be done? [Ans. 380] 8. From a pool of 12 candidates, in how many ways can we select president, vice president, secretary and a treasurer, if each of the 12 candidates can hold any office ? [Ans.11880] 9. Six candidates are called for interview to fIll four posts in an office. Assuming that each candidate is fit for each post, determine the number of ways in which the four [Ans.360] posts can be filled? 10. In how many ways can 6 boys and 5 girls be arranged for a group photograph, if the girls are to sit on chairs in a row and the boys are to stand in a row behind them? [Ans.86400] lL When a group photograph is taken, all the 5 teachers should be in the first row and all the 8 students should be in the second row. If the two corners of the second row are reserved for the two talent students, interchangeable only between them, and if the middle seat of the front row is reserved for the principal, how many arrangements are possible? [Ans.34560] 12. In how many ways can 5 boys and 3 t..rls be arranged in a row so that no two girls [Ans.14400] may sit together?

PERMUTATION AND COMBINATION

17

13. There are 8 students appearing in an examination of which 3 have to appear in a Mathematics paper and the remaining 5 in different subjects. In how many ways can they be made to sit in a row, if no two candidates in Mathematics sit next to each other. [Ans.14400] 14. Six men and four women are to sit in a row so that the 'women occupy the even [Ans. 86400] places. Find the number of all possible arrangements. 15. How many signals can be given with 6 flags of different colours such that: (a) exactly three flags can be used for a signal? (b) at most three flags are to be used for a signal ? (c) at least three flags are to be used for a signal ? (d) any number of flags may be used for a signal ? [Ans. (a) 120, (b) 156, (c) 1920, (d) 1956] 16. In how many ways can 9 examination papers be arranged so that the best and the worst papers never come together? [Ans. 282240] 17. A family of 5 brothers and 3 sisters is to be arranged for a photograph in one row. In how many ways can they be seated, if (a) all sisters sit together? [Ans. (a) 4320, (b) 14400] (b) no two sisters sit together? 18. How many 6 digit telephone numbers can be constructed if each number starts with 35 and no digit appears more than once? [Ans. 1680] 19. How many different numbers between 100 and 1000 can be formed using the digits 0,1,2,3,4,5, and 6 assuming that, in any number, the digits are not repeated. How many ofthese will be divisible by 5 ? [Ans. 180, 55 20. Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is used more than once in a number. How many of these numbers will be even? [Ans. 120, 48] 21. How many numbers of six digits can be formed from the digits 1, 2, 3, 4, 5, and 6 (no digit being repeated). How many ofthese are divisible by 5 ? [Ans. 720, 120] 22. How many numbers greater than 4000 can be formed with the digits 2, 3, 4, 5, and 6 when no digit is repeated? [Ans. 192] 23. How many numbers lying between 2000 and 4000 can be formed with the digits 1, 2, 3, 4, 5, 6 ? [Ans. 120] 24. Ritu wa~ts to arrange 3 English, 2 Hindi and 4 French books on a shelf. If the books on the same subject are different, determine: (a) the number of possible arrangements. (b) the number of possible arrangements, if all the books on a subject are to be together. [Ans. (a) 362880, (b) 1728] 25. Find the number of ways in which 8 different books can be arranged on a shelf so that 2 particular books are (a) always together. (b) never together. [Ans. (a) 10080, (b) 30240]

18

FINANCIAL MATHEMATICS

26. Find the number of permutations of 8 things taking 5 at a time in which 2 particular things are always (a) included. (b) excluded. [Ans. (a) 2400, (b) 720] 27. The letters of the word TUESDAY are arranged in a line, each arrangement ending with letter S. How many different arrangements are possible? How many of them start with letter D ? [Ans. 720, 120] 28. In how many ways can the letters of the word 'FRACTION' be arranged so that no two vowels are together? [Ans.14400] 29. How many words can be formed from the letters of the word 'DAUGHTER' so that (a) The vowels always come placed together? [Ans. (a) 4320, (b) 36000] (b) The vowels are never placed together? 30. How many different words can be formed of the letters of the word "COMBINE" so that: (a) vowels always remain together? (b) no two vowels are together? [Ans. (a) 720, (b) 1440, (c) 576] (c) vowels may occupy odd places? 31. In how many ways can the letters of the word 'TOWER' be arranged so that the letters '0' and 'E' occupy only even places? [Ans. 12] 32. How many words can be formed out of the letters of the word 'PECULIAR' beginning into 'P' and ending with 'R' ? How many of them will have 'P and 'R' at end places? [Ans. 720, 1440] 33. How many permutations can be made out of the letters of the word 'TRIANGLE' ? How many of these: (a) (c)

begin with T? begin with T and end with E ?

(b) end with E ?

T and E occupy the end places? [Ans. 40320, (a) 5040, (b) 5040, (c) 720, (d) 1440] (c)

34. How many different words can be formed of the letters of the word 'MALENKOV' so that: the first letter is a vowel? (c) vowels may occupy odd places? (a)

no two vowels are together? (d) vowels being always together? [Ans. (a) 15120, (b) 120, (c) 2880, (d) 4320j (b)

35. How many different words containing all the letters of the word 'LOGARITHM' can be formed? How many of these: (b) end with 'M' ? begin with 'L'? (d) have 'L' and 'M' in end places? (c) begin with 'L' and end with 'M' ? (e) begin with 'LOG' ? (f) end with 'THM' ? (h) have no vowels together? (g) have vowels together? (i) have vowels in the end places? (j) have vowels and consonants in their relative position? (a)

19

PERMUTATION AND COMBINA TlON (k) have vowels in odd places? (m)

(l) have vowels in even places?

have vowels in 2nd, 3 rd and 4th places? [Ans. 362880, (a) 40320, (b) 40320, (c) 5040, (d) 10080, (e) 720, (j) 720, (g) 30240, (h) 151200, (i) 8640, (j) 4320, (k) 43200, (l) 17280, (m) 4320]

PERMUTATIONS WHEN ALL OBJECTS ARE NOT DISTINCT In problems of counting, sometimes repetitions are allowed in the arrangements of objects or distinctions between some of the objects are ignored. The number of permutations of n things taken all at a time, when p things are of one kind, q of second kind, r of third kind and so on is given by

n! p!q!r! ...... For example consider the word 'SEE'. We take first E as El and second E as E2 The word has 3 letters, and the number of permutations in which 3 letters can be arranged is 3 !, i.e., 6 [SE 1E 2, SE2Ev E 1SE 2, E 1E 2S, E 2E 1S, E 2SE 1] Clearly, SE 1E 2 }. C

x+y 1 log -2- = 2'log (xy) = log (xy)1I2

=>

(x + y)2 = 4xy => (x - y)2 = 0

=>

x-y = 0,

x+y

=> -2- =

VxY

i.e., x =y.

Example 12 : Find the value of x : (a) 2 log x + log 4 =log 64

(b) 2 log (x + 4) =log 49

Solution: (a) 2 log x = log 64 -log 4

(~)

=>

2 log x = log 16

=> log x = 2'log 4 2

=>

1 logx='2x2Iog4

=> log x = log 4

=>

x=4

2 log (x + 4) = log 49 1 => log (x + 4) = '2 log 49

=>

1 log (x + 4) = 2'log 72

=> 2 log x = log

1

(b)

=> log (x

1 + 4) = '2 x 2 log 7

=> x+4=7

=>

=>

log (x + 4) = log 7

x=3

Example 13 : Solve the following equation: log (x- 6) + log (x-3) =2 log 2

Solution: log ~ - 6) + log (x - 3) = 2 log 2 "

=>

log (x - 6) (x - 3) = log 22 log (x - 6) (x - 3) = 4 (x - 6) (x - 3) = 4

=> =>

x2 -

x 2 -9x+ 14= 0

=>

(x - 7) (x - 2)

"

=> =>

9x + 18 - 4 = 0

x-7=0 x

=7

=0

orx-2=0 or x = 2.

40

FINANCIAL AlA THElIA TICS

L Write the following in the form of logarithm: (a) 3 5 =243 (b) 8-2 =1164 (c)

1()-6 =0.00001

,

(c)

VI6 =2

2. Write the following in exponential form : (a) log61296 = 4

log625 5 =114 3. Find the value of the following:

(b)

logs 243 = 5

(c)

(d) 10glOO 0.01 = -1

(a) logo a'

(b)

(c) log~

logs 27

2401

4. Inog (x 2 - 4x + 5) = 0, find the value ofx. 5. Find the values of the following: s

(a) 10gl0 V100

(b)

[Ana.x

s

log7V343

1210g10 10 6. Find the value of 2 10gl0 100

=2]

[Ana. (a) 3/2, (b) 1] [ADs. 3]

7. Given 10gl0 2 =0.3010 and 10gl0 3 =0.4771, Evaluate the following: (a) 5/2 log 4 (b) log 600 [Ana. (a) 1.5050, (b) 2.7781] 8. Find the value of log2 [log210g210g2 (65536)]. [Ana. 1] 9. Show that: 4 log 5 + 2 log 4 =4 10. Evaluate the following without using log table: (a) log 7 + log ll-log 77 (b) 210g10 5 + 10gl0 8 -1I210g10 4 9 35 15 1L Show that log 7 + log 6-log"2 = 0 [Ana. (a), (b) 2] 12. Prove that

10 2 5 . 81 7 log "9 - 2 log 24 + 3 log 80 =log 2 . . log 3 + log 5 + log 8 13. Simplify: log 4 + log 30 . 14. Prove that log

[Ans.I]

(~!) + log (::) -log (~:) =log 3

. 81 3 2 9 15. SImply 2 log 8- 4 log 2' + 610g g + log 16 . 16. Solve for x : (a) log x + log (x - 9) = 1

(b)

log 16

17. Solve: Log 2 + log (x + 2) -log (3x - 5) =log 3

[ADs. 0]

=x log 2

[Ana. (a) 10, (b) 4] [Ana. x

;9]

=

LOGARITHM

41

18. Solve for x : logx 4 + lo&: 16 + lo&: 64 = 12 19. Find the value of x solving the equation: log4 (x + 3) + log4 (x - 3) =2

[ADs. 2) [ADs. 5]

COMMON LOGARITHM Logarithms to the base 10 are called common logarithms. The common logarithms are most frequently used for computation. If the base is not indicated in the. logarithm, it is understood that the base is 10. In other words 10glO x can be simply written as, log x. From this moment onwards we discard the subscript 10 from the word 'log'. Standard form of a number. Ifm is a positive real number, we can write m =n x loP, where pEl and 1 ~ n < 10. This form is called the standard form of m. Consider the following examples:

(d)

=2.372 x lOS 425 =4.25 x 1()2 8.241 =8.241 x 100 0.8241 =8.241 x 10-1

(e)

0.006652 = 6.25 x 10"3

(a) 2372 (b) (c)

When a number written in this form, we say that it has been written in the standard form. We have m = n x loP, where n is between 1 and 10. Then we find log m =log (n x loP) ::= log n + p log 10 =log n + p Remarks: l~n

log 1 0

~ ~

log n < log 10 log 11 < 1

This log n is between 0 and 1. :. log n will be in the form ofO.xxxx

Example 14 : Write the following numbers in the standard form : (a) 7321

(b) 7.802

(d) 463.2

(e)

0.3253

Solution: (a) 7321

=7321 1000 x 1000 =7.321 x lOS

:. Standard form = 7.321 x 103 . (b) 7.802

=7.802 x 100.

:. Standard form

=7.802 x 100 •

(c)

0.0037

42

FINANCIAL AfATHEAfAnCS (c)

0.0037 = (0.0037 x 1000) x 10-3 =3.7 x 10-3 :. Standard form = 3.7 x lO-S.

(d) 463.2

(e)

2 =463.2 100 x 100 =4.632 x 10

:. Standard form =4.632 x 102 • 0.3253 =(0.3253 x 10) x 10-1 =3.253 :. Standard form = 3.253 x 10-1 .

X

10-1

CHARACTERISTIC AND MANTISSA

We have noticed: log m =log n + p Here p is an integer and 1 :s; n < 10. When log m has been expressed as p + log n, where p is an integer and 0 :s; log n < 1, we say thatp is the 'characteristic' of log m and that log n is the 'mantissa'. Mantissa is never negative and is always less than 1. If we can find the characteristic and the mantissa of log m we have to just add them to get log m.

To find the characteristic If m is a positive number, the characteristic of log m is given by the following two rules: Rule 1. If m > I, count the number of digits in the integral part of m and subtract 1 from it to get the characteristic. Rule 2. If m < 1, count the number of zeroes immediately after the decimal point and add 1 to it. The number so obtained with a negative sign gives the characteristic. Note: If the characteristics is negative, say -3, it can be denoted as 3, when we add with the mantissa. Consider the following examples: (a) log 327.5 = log 3.275 x 102 = 2 + log 3.275 Comparing with log m =P + log, we get the characteristics oflog 327.5 is 2. (b) m

=0.0003275

log m =log 0.0003275 =log (3.275 x 10"4) =-4 + log 3.275 :. Characteristic is -4. (written as 4) To find the mantissa of logarithm We know that log m =p + log n where log n is the mantissa. The mantissa of log m is calculated from the table of logarithms. For finding the mantissa of the logarithm of a number, we disregard the decimal point in the given number. We follow the procedure given below to find the mantissa of the logarithm of a number: 1. Locate the first two digits of the given number in the left-hand column headed by N. If the given number has only one digit, we replace it by a two-digit number obtained by adjoining zero to the right of the number. Thus, 5 is to be replaced by 50 for finding the mantissa ... 2. Locate the entry in the row obtained earlier in the column headed by the third digit in the number whose logarithm is being found out.

43

LOGARnHM

3. Locate now the entry in the same row, and in the column headed by the fourth digit in the given number in the table of proportional parts, and add the entry now obtained to the entry obtained earlier. 4. The sum of the two entries obtained is the mantissa of the logarithm of the given number.

Consider the following example: Let m =0.0002745 =2.745 x 10-' log m =- 4 + log 2.745 From the table oflogarithm, mantissa oflog 2.745 = 0.4386 Thus, log 0.0002745 =4.4386

Example 16: Using log table, find the logarithms of the following numbers: (a) 1270

(b) 0.00001379

(c) 2.307

Solution: (a) 1270 =1.270 x lOS log 1270 =3 + log 1.270 Characteristics of log 1270 =3 From the table oflogarithm, log 1.270 is 0.1038 .. log 1270 =3.1038. (b) 0.00001379 = 1.379 x 1Q-6 log 0.00001379 =- 5 + log 1.379 Characteristics of log 0.00001379 is -5. From the table oflogarithm, log 1.379 =0.0792 log 0.00001379 =5.0792 (c)

2.307

=2.307 x 100 log 2.307 =0 + log 2.3027

Here, the characteristics is 0 From table oflogarithm, log 2.307 =0.3630. (d) 0.0075

=7.5 x 10--3

log (0.0075) =-3 + log 7.5 From the table oflogarithm, log 7.5 =0.8751 log 0.0075 =3.8751 (e)

750 =7.5 x 1()2 log 750 =2 + log 7.5 Characteristic oflog 750 is 2 From the table oflogarithm, log 7.5 =0.8751 .. log 750 =2.8751.

(d) 0.0075

(e) 750

. FINANCIAL MATHEIIA7ICS

44

ANTILOGARITHM Antilogarithm is used to find the number whose logarithm is given. The number m, whose logarithm is x, is called the antilogarithm ofx and is written as m =antilogx. Thus, if log m

=x, then m =antilog-,.

Antilogarithms are calculated by using the antilogarithm tables. The following procedure is followed to find antilogarithm of a given logarithm: 1. While finding antilogarithm, we take into consideration only the mantissa (i.e., the positive decimal part).

2. In the antilogarithm tables, in the first column (left-hand side), we locate the row containing the first two digits in the number. 3. In this row, and the column headed by the third digit in the number, we locate the entry. 4. In this row, and the column headed by the fourth digit in the number in the tables of proportional parts, we locate the entry. 5. We add the two entries obtained earlier. Then we insert the decimal point at a suitable place so that the rule of characteristic is followed. This gives us the antilogarithm of the given number.

Ezample 18 : Fin~ the antilogarithm of the each of the following : (a) 0.0905

(b) 2.051

(c)

2.35

Solution. : (a)

0.0905 = 0 + 0.0905 From the table of Antilogarithm, antilog (0.09054)

= 1.230

antilog (0.0905) = 1.230 x 10° = 1.230 (b)

2.051 = 2 + 0.051 From the table of antilogarithm, antilog (0.051) = 1.125

:. antilog (2.051) • (c) 2.35 = -2 + 0.35

= 1.125 x 102 =112.5

From the table of antilogarithm, antilog (0.35) = 2.239 :. antilog (2.35) = 2.239 x 10-2 = 0.02239 (d) 1.301 = 1

+ 0.301

(d) 1.301

(e) ~.653

LOGARITHM

45

From the table of antilogarithm, antilog (0.301) = 2.000 :. antilog (1.301) = 2.000 x 101 =20 (e)

3.653 = -3 + 0.653 From the table of antilogarithm, antilog 0.653 = 4.4498 :.antilog {a.653) = 4.498 x 1()-3 =0.004498.

Example 17: Using logarithm, evaluate the following: (a) 4.382 (b) 2.632 3.738 0.0045 (c) "42.36 (e)

(d) (0.724)3 (f) 6.45 x 981.4

0.0064 x 1.507

Solution. : (a) Let

Then

.. .. (b)

Let

..

4.382

x

log x

x 4.382 3.738 x

= 3.738 = log (::~::)

= log 4.382 -log 3.738 = 0.6417 - 0.5726 =0.0691 = antilog (0.0691) =1.172; = 1.172 2.632

= .0045

log x -_ (2.632) .0045

= log 2.632 -log .0045 = 0.4203 - 3.6532 =-1 + 1.4203 x 2.632 0.0045 (c) Let

Then

x

= -1 + 3 + (1.4203 - 0.6532) =2 + 0.7671 =?.7671 = antilog (2.7671) =584.09.

= 584.09 = (42.36)112

log x = x

" 4236

(-3 + 0.6532)

2"1 log (42.36) =2"1 (1.6269) = 0.8134

= antilog (0.8134) =6.507. = 6.507 .

46

FINANCIAL MATHEMAncS (d)

Let Then

= (0.724)3 = log (0.724)3

x log x

= 3 log (0.724) =3(1.8597) = 3 (-1 + 0.8597) =3(-0.1403) =- 0.4209

=-1 + 0.5791 =1.5791 = antilog (1.5791) =0.3794 )( 10-1 =0.3794.

x (0.724)3 (e)

Let

= 0.3794 = .0064)( 1.507 = log .0064 + log 1.507

x log x

= 3.8062 + 0.1780 =-3 + 0.8062 + 0.1780 =3.9842 x = antilog (3.9842) =.009642. . . 0.0064)( 1.507 =0.009642

= 6.45)( 981.4

(f) Let x

= log (6.45)( 981.4)

log x

= log 6.45 + log 981.4 =0.8096 + 2.9919 =3.8015 . . 6.45

x

x = antilog (3.8015) = 6331.0. 981.4 = 6331.0

Example 18 : Using logarithm evaluate 7.93 x 0.0054 0.0981

Solution: 7.93 x .0052 .0981

Let

x

=

Then log

x

= I og [ .0981 = log (7.93) + log (.0052) -log (0.981) = 0.8993 + -3.7160-2.9917 = 0.8993 + (-3 + 0.7160)-(-2 + 0.9917)

7.93

x

.0052]

= (-3 + 2) + (0.8993 + 0.7160 - 0.9917)

. = -1 + (1.6153 - 0.9917) =I = antilog (1.6236) =0.4204.

+ 0.6236 = (1.6236)

x

.

Example 19 : Fmd the value of · L t S o I ution: e x

=

(25.36)2 x 0.4569

(25.36)2 x 0.4569 847.5

847.5

LOGARITHM

47

Then Iogx

0.4569] =Iog [ (25.36)2 847.5 X

= log (25.36)2 + log (0.4569) -log (847.5) = 2 log (25.36) + log (0.4569) -log (847.5) = 2(1.4041) + (1.6599) - (2.9282) =(2.8082 + 1.6599) = 2.4681- 2.9282 =-0.4601

2.9282

= 1 + 0.5399 = 1,5399 x

= antilog (1.5399) =0.3466.

Example 20: Using logarithm, Evaluate S I' Le o ution : tx

3.279 x (1.207)112 (120000)113

3.279 x (1.207)112

=

(120000)1/3

Taking log of both sides, we get log x

= log {

3.279 x (1.207)1I2} (120000)113

= log 3.279 + 2"1 log (1.207) -

1 slog (120000)

= 0.5157 + 2"1 x 0.0817 - S1 (5.0792) = 0.5157 + 0.04085 - 1.69306 =0.55655 -

1.69306 =-1.1365

= -2 + 2 -1.1365 = 2.8635

x = antilog (2.8635) = .073030.

'#33immi-1 L Write the following in standard form : (a) 2348 (b) 23.48 (e) 0.0002348 (d) 2.348 (e) 234800 (f) 0.02348 3 [Ans. (a) 2.348 x 10 , (b) 2.348 x 101, (e) 2.348 x 10"4, (d) 2.348 x 100, (e) 2.348 x lOS, (f) 2.348 x 10-2] 2. Write the following numbers in decimal form : (a) 7.317 x 10"-4 (b) 7.317 x 101 (d) 7.317 x 100 (e) 7.317 x 10-2

(e)

7.317 x 1()4

[Ans. (a) 0.0007317, (b) 73.17, (e) 73170, (d) 7.317, (e) 0.07317] 3. Using log table find the logarithms of the following numbers: (a) 12.70 (b) 0.0012 (e) 431.5 (d) 1.123 [Ans. (a) 1.1038, (b) 3.0792, (e) 2.6350, (d) 0.0504]

48

FINANCIAL MATHEMAncS

4. Find the antilogarithm of each of the following: (a) 2.5428 (b) 0.752 (d) 1.301 (c) 6.123 [Ana. (a) 0.03489, (b) 5.649, (c) 1327000, (d) 0.2000] 6. Using logarithm table, find the value of each of the following: (a) 0.0001 x 0.027 x 38.9 x 50.2 (b) 30.30 x 4.5 x 5.2 (c) 2.432 x 37.81 x 5.38 [Ana. (a) 0.0005272, (b) 708.9, (c) 494.8] 6. Using logarithm, evaluate the following: (a)

°o~::;

(b)

~i°:':

(c)

(11.34)113

(d)

V1.96

[Ana. (a) 0.06457, (b) 0.0002497, (e) 2.247, (d) 1.401] 7 U' I bl al (23.5)2 (0.523)3 • SlUg og ta e ev uate 1 _ (0.352)2 8. Find the fIfth roof of 78.62 by using logarithm. 1.121 x 3.378 x (0.5678)2 . I ·th 9 • E valuate (0.8123)2 usmg ogan m. 10. E val uate

8.25 )( 4.63 2.18

[Ana. 90.20]

[Ana. 2.394] [Ans.1.85] [Ans. 13.92]

U Using logarithm, find the value of 41.32 x 20.18 12.69

[Ana. 8.106]

3

12. Evaluate V8.34.x (3.12)2 (4.312)3 x V24.3

[Ana. 0.4995]-

Simple Interest INTRODUCTION Every human being irrespective of their profession, deals with money either as a borrower or as a lender. Business organisations implement new ideas through new projects for expansion, diversification or moderisation. The entire operation is based on the concept that money belonging to one may be used by others and can be returned within a designated future date. The question which immediately arises in mind is whether the value of money borrowed today will remain same after one year. Answer is ·'no'. Because money has time value (i.e a rupee today is worth more than a rupee tomorrow). Money can be employed productively to generate returns. For example, suppose that you have deposited Rs. 1,000 in . a saving bank account and the bank pays 6% per annum return. Then, the amount accumulated at the end of the year will be Rs. 1,060. The use of money bears the cost of interest. Interest plays so important role in business that many individuals and agencies are eng~ged directly or indirectly in the business of lending money. The goal of this chapter is to study about the types of interest and the computation of simple interest.

INTEREST Interest is the consideration for the use of invested or borrowed money. For example,. suppose that Mr. X borrows a particUlar sum of money from Mr. Y. Then Mr. X has to pay certain amount to Mr. Y for the use of this money. This amount is called Interest. The money borrowed is called the principal. When Mr. X return his indebtedness to Mr. Y, he has to pay back both the principal and the interest. Interest is of two kind: (i) (ii)

Simple Interest Compound Interest

50

FINANCIAL MATHEMATICS

In case of simple interest, the interest earned in not added back to the principal amount whereas in compound interest, the interest earned is added back to principal, thus to form a new principal for the new term. Interest at the end of every term is calculated and added to the amount at the beginning of that term. This new amount will be treated as tile principal for the new term. In simple interest, principal will remain same for every term. In the succeeding sections of this chapter, we will elaborately study the formulas and application of simple interest. Now let us briefly discuss the difference between simple interest and compound interest by means of an illustration. Suppose that a person invests Rs. 10,000 for 3 years at 8% simple interest per annum. The year by year interest calculation is shown in the following table:

Year

Principal at the beginning of every year

Interest for the year

Amount at the end of every year

1

Rs.10,000

8% of 10,000 = Rs. 800

Rs.10,800

2

Rs.10,000

8% of 10,000 =Rs. 800

Rs.11,600

3

Rs.10,000

8% of 10,000 =Rs. 800

Rs.12,400

Amount accumulated at the end of 3 years is Rs. 12,400, so the total interest earned for the three year taken together is Rs. 2,400 (Rs. 12,400 - Rs. 10,000). Now suppose that the same amount Rs. 10,000 is invested at 8% per annum but compounded annually. Then the calculation of interest is as follows :

I

Year

Principal at the beginning of every year

Interest for the year

Amount at the end of every year

1

Rs.10,000

8% of 10,000 =Rs. 800

Rs.10,800

2

Rs.10,800

8% of 10,800 =Rs. 864

Rs.11,664

3

Rs.11,664

8% of 11,664 =Rs. 933.12

Rs. 12,597.12

Total amount accumulated at the end of 3 years is Rs. 12,597. The total interest earned in this case is Rs. 2,597.12. It should be noted that compound interest is always more than simple interest.

SOME BASIC TERMS (a) Principal Amount. This is the amount about to be invested or loaned. If a person apply for a bank loan of Rs. 20,000, this amount is referred to as the principal amount to be borrowed. Similarly, if XYZ Ltd. purchased a new machine for Rs. 3,00,000, this is the principal amount invested on the machinery. It is denoted by 'P'. (b) Number of Years (t). This is the tirrle period (in years) over which an amount of money is..invested or borrowed. (c) Conversion period (n). This is the number of times over which interest is compounded on a given principal. In many cases, interest is calculated more than once in a year. For example, in personal loans, interest is calculated on monthly basis but in case of

51

SIMPLE INTEREST

savings bank account interest is compounded on half-yearly basis. "n' is determined using the following relation : n = m xt where m is the number of conversions per year. Thus, if certain principal is invested for 10 years and interest is compounded quarterly, then the number of conversion periods will be 40. It should be noted that n equals t, only when interest is compounded annually. (d) Rate of Interest (r). This is the proportionate amount charged for the use of the

principal. It is defined as the interest charged for keeping Rs. 100 for 1 year. So the rate of interest is always expressed as a percentage rate per annum. However, for the purpose of calculation, interest rate is written in equivalent decimal form: r = 7% as r = 0.07; r = 12% as 0.12 and so on. (e) Interest per Conversion Period (i). We have already discussed that the rate of interest (r) is the percentage rate per annum. What will be the interest rate, if the interest is compounded more than once in a year? The interest rate "i' depends upon the number of conversion periods per year "i' is calculated as follows:

.

l

r m

= --

Thus, ifthe interested is calculated at 10% per annum but compounded quarterly then i = 0 1 = 0.025 or 2.5%

4

If the number conversions per year is one, then i

=r.

(f) Accumulated Amount (A). Amount is the sum of the principal and the interest. Some authors use the term 'sum' and denote it by'S'.

A=P+I

SIMPLE INTEREST Simple interest is generally used only for short-term investments or borrowings which are often of less than one year. It is payable on principal only. For example, simple interest on Rs. 1,000 at a rate of 6% per annum will be Rs. 60 every year through out the entire period. Let P be the principal at a simple rate of interest r% per annum for a period of't' years. The interest charged at the end of t years is !I=Pxrxt! The amount at the end of t years is given by : A=P+I ~

~

A=P+Prt 1

A

=P (1 + rt) !

52

RNANCIAL MATHEMATICS

Given any three of the four variable A, P, r and t, we can solve for the fourth

I p=~;r=~~t=~1

t.e.

Remark. For any transaction, if the transaction is to made for less than one year, the time may be given in months, weeks or days. Since in simple interest formula, 't' represents the number of years, the given time, if it is not in years, should be converted into year. K 30 K days =365 years 30 days =365 years K

K weeks =52 years and.

K months

K =12 years

12 weeks

12

=52

years

4 4 months = 12 =1/3 years.

Example 1 : Find the amount at 6% Simple Interest of Rs. 1,200 (a) in 2 years. (b) in nine months. Solution: Given principleP =Rs. 1,200

Rate of Interest r

=6% =0.06

(a) In 2 years: Amount A

=P (1 + rt) = 1,200 (1 + 0.06 x 2) = 1,200 (1 + 0.12) = 1,200 (1.12) = Rs. 1,344

.. The amount is Rs. 1,344.

(b) In 9 months: t

9

3

=9 months =12 year =4" years A = 1,200 (1 + 0.06 x 3/4)

= 1,200 (1 + 0.045) = 1,200 (1.045) =Rs. 1,254 .. Amount is Rs. 1,254

Example 2: Mr. X borrowed Rs. 7,500 on 2@hMarch 1966 from a corporate bank at a rate of 8% p.a. If he wanted to clear the account on 'Ph June 1966 then what amount would he have to pay to bank ? Solution:

P r

= Rs. 7,500

= 8% i.e. 0.08

73 t = 73 days or 365 yr.

Amount

A

= P (1 + rt)

1

=5" yr.

53

SIMPLE INTEREST

= 7,500 (1 + 0.08 x ~) = 7,500 (1 + 0.016)

= 7,500 (1.016) =Rs. 7,620 . . The amount is Rs. 7,620. Example 3 : What time will be required for a sum of money to double itself at 5% simple interest?

Solution: Let P be the principal. Given that the sum is double of its principal :.

S

=2 P

... (1)

Let t be the time, r S =P(1 +rt) (1)

=5% or 0.05 ... (2)

=(2) ~ 2P =P (1 + rt) ~ 2=1+rt ~

rt= 1

Substituting r =0.05, we get 0.05 t =1 ~ t =20 years . . 20 years are required to get the sum doubled. Example 4 : Mr. Ram deposited Rs. 1O,0{)0 in a saving bank account which pays 10% simple interest. He makes two more deposits of Rs. 15,000 each, the first at the end of 3 months and the second in 6 months. How much will be in the account at the end of the year, if he makes no other deposits and no withdrawals during this time ?

Solution: In case of the deposit ofRs. 10,000 principle, P time, t rate of interest, r Amount is given by

= Rs.10,000

= 1 yr. = 8%:: 0.08

= p(1 + rt) 'AI = 10,000 (1 + 0.08 x 1) -= 10,000 (1.08) =Rs. 10,800 A

Here

Two deposit of Rs. 15,000 are made at the end of 3 months and 6 months respectively. In case of 3 months, the amount will remain in the account for remaining 9 months. 9 3 .. .t = 12 =i years P

= Rs. 15,000

A2 =' 15,000 (1 + 0.08 x'~) A2 = 15,000 (1.06) = Rs. 15,900

54

FINANCIAL MATHEMATICS

In case of 6 months, the amount will remain for 6 months (1 year - 6 months)

t

6

1

= 12 ='2 years

P = Rs. 15,000. A3 = 15,000 (1 + 0.08 x 112) A3 = 15,000 (1.04) = Rs. 15,600 Total amount = Al + A2 + A3 = 10,800 + 15,900 + 15,600 = Rs. 42,300 .. At the end of the year, Rs. 42,300 will remain in the account.

Example 5 : A shopkeeper borrowed Rs. 20,000 from two money tenders. For one loan he paid 12% simple interest and for the other 14% simple interest per annum. After one year, he paid Rs. 2,560 as total interest. How much did he borrow from each money lender?

Solution: Let the shopkeeper borrows Rs. P from money lender 1 at 12%. Then amount borrowed from money lender 2 with be Rs. (20,000 - P) at 14%. Interest is given by I = Prt Interest paid to money lender 1, h = P x 0.12 Interest paid to money lender 2, 12 = (20,000 - P) x 0.14 Given that total interest paid is Rs. 2,560. .. I} + 12 = 2,560 ~ 0.12 P + 0.14 (20,000 -P) = 2,560 ~ 0.12 P + 2,800 - 0.14 P = 2,560 ~

0.02P= 240

~

P = 12,000 20,000 - P = 8,000

..

.

Money borrowed at 12% = Rs. 12,000 Money borrowed at 14% = Rs. 8,000 Example 6 : What percentage of simple interest must a person get on his investment of Rs. 25,000, if he wants his investment to grow to Rs. 26,500 in 6 months?

Solution: Here P A

= Rs. 25,000 = Rs. 26,500

6 1 t = 12 = '2 years We know that I =A-P Here I = 26,500 - 25,000 = Rs. 1,500

55

SIMPLE INTEREST

Simple interest I is given by I = Prt I r = Pt 1,500 Here r = -25-,0-0-'-0-x-(-l/-2) = 0.12 = 1?% . . Required simple interest rate

=12%.

L Find the simple interest and amount on an investment of Rs. 5,000 for 3 years, ifthe interest is calculated at 4% simple interest per annum. [Ans. Rs. 600, Rs. 5,600] [Ans.2.5%] 2. At what rate will Rs. 1,500 yield Rs. 25 simple interest in 8 months? 3. What principal wi~l amount to Rs. 645 in years at 5% simple interest? [Ans. Rs. 600]

Ii

4. What time will be required for a certain sum of money to double itself at 10% simple interest? [Ans. 10 years] 5. How much should an investor deposit now in a Bank to get Rs. 20,000 in 3 months, if bank calculates 9% simple interest? [Ans. Rs. 19,560] 6. How long will Rs. 3,000 take to amount to Rs. 3,300 at 4% simple interest ?[Ans. 2.5 years] 7. Ram borrowed Rs. 5,000 from Shy am at 8% simple interest. After 9 months again he borrowed Rs. 2,000, promising to return the entire indebtedness at the end of two [Ans. Rs. 8,000] years. What amount Shy am will get from Ram? 8. Suresh borrowed Rs. 830 from Vikas at 12% rate of interest for 3 years. He then added some more money to the borrowed sum and lend it to Deepak for the same time at 14% simple interest. If Suresh gains Rs. 93.90 in the whole transaction, find the sum lent by him to Deepak. [Ans.935] 9. A certain amount of money was invested at 8% simple interest and after 9 months an equal amount was invested at 10% simple interest. Find the period in which the amount in each case becomes Rs. 52,000. How much money was invested in each 3 case? [Ans. 3"4 yrs. , Rs. 40,000)

Compound Interest INTRODUCTION The second method of calculating interest is the compound interest method where the interest earned by an invested amc;mnt of money (principal) is reinvested so that it too earns interest. In this concept, the interest earned on the initial principle becomes a part of principal for the next period. At the beginning of every period, the principal will be the sum of the principal of the previous period and the interest earned on that principal. For instance, suppose that interest is converted annually. Then the principal for third year will be the sum of the principal for the second year and the interest for second year. Thus, in case of compound interest, interest is converted into principal and hence there is interest on interests. In this chapter, we study the method of computation of compound interest, present value and future value of investments, and etc.

COMPOUND INTEREST An understanding of compound interest is very important for every common man, since every one has to think of investing money. Let us see an example to know how the concept of compound interest works out. Assume that Mr. A deposits Rs. 5,000 in a savings bank account for 3 years at 10% per annum compounded annually.

Principal at the beginning of first year =Rs. 5,000 Interest for first year = 5,000

x

0.1

= Rs. 500

:. Amount at the end of first year =(Rs. 5,000 + 500) =Rs. 5,500 This is the principal for the second year. Interest for second year = 5,500

x

0.1

= Rs. 550

:. Amount at the end of second year = (Rs. 5,500 + 550) =Rs. 6,050.

57

COMPOUND INTEREST'

This is the principal for third year. Interest for third year = 6,050 x 0.01 =Rs. 605 :. Amount at the end of third year =(Rs. 6,050 + 605) =Rs. 6,655 This is the amount Mr. A will receive at the end of third year against the investment of Rs.5,000. The total amount due at the end of the last period is called the compound amount. The difference between the compound amount and the original principal is called compound interest. Compound Interest = Compound Amount - Initial principal Interest may be compounded annually, half-yearly, quarterly, monthly or at any other regular periods of time. This time between two successive computation of interest is called the conversion period. The number of times over which the interest is compounded in one year is known as frequency of conversion. For example, if the interest is compounded quarterly, then the conversion period will be 3 months and the frequency of conversion is 4, i.e., 4 times in a year. It should be noted that regardless of the frequency of conversion, interest rates are generally quoted as an annual nominal rate. So if the interest is compounded more than once in a year, then the stated annual interest rate' should be converted into the rate per compounding (or conversion) period. As we have discussed earlier in the preceeding chapter, the rate per conversion period is found by dividing the annual nominal rate by the number of conversion periods in a year. To understand conversion period and rate per conversion period more precisely, go through the following table, in which the rate per conversion period equivalent to a nominal rate of 12% compounded is given. Conversion period

Frequencies of conversion (m) Rate per conversion period i =(r I m) %

Annually

1

12% =0.12

Semi-annually

2

12% _ 0.12_ 0 06 2 - 2 -.

Quarterly

4

Monthly

12

12% _ 0.12 _ 0 01 12 - 12 - .

Weekly

52

12% _ 0.12 _ O' 0023 52 - 52 - .

Daily

365

12% 0.12 365 = 365 =0.00333

I

12% _ 0.12 _ 0 03 4 -

4 -

.

FORMULA FOR COMPOUND INTEREST Let P be the principal earning interest compounded m times a year at a rate of i per period. Let "n' be the number of conversion periods.

58

FINANCIAL MATHEMATICS

Then, The amount at the end of first conversion period A 1 =P+Pi=P(I+i) The amount at the end of second conversion period A2 = P (1 + i) + P (1 + i) i =P (1 + i) (1 + i) =P (1 + i)2 The amount at the end of third conversion period = P (1 + i)2 + P (1 + i)2 i = P (1 + i)2 (1 + i) = P (1 + i)3

The amount at the end of the nth period =P (1 + i Y' He.1ce, the compound amount A of a principal P at the end of n conversion periods at the rate of interest of i per conversion period is given as follows:

1A =P (1 + i)n I, n =m x t, i = : Where Principal amount invested or borrowed A - Compound Amount accumulated r - Rate of interest per annum - Rate of interest per conversion period m - Number of conversion periods per year n - Total umber of conversion periods t -- Number of year. P -

It should be noted that the principal amount P is invested for t years at annual rate of r and interest is compounded m times in a year. So the above formula for compound amount can also be given by

Compound interest may be obtained by using the following formula: Compound Interest I =A - P

Example 1 : Find the compound interest earned from Rs. 16,000 for 3 year at 12~ per annum?

Solution: Here, P rate of interest, r

= Rs.16,000 1 = 12~ =0.125

no. of periods, n = 3 Interest compounded annually Amount

A

= P (1 + i)"

.., =r

COMPOUND INTEREST

59

A = 16,000 (1 + 0.125)3 = 16,000 (1·423828)

= Rs. 22,781.25 Now Compound Interest I

=A - P =22,781.25 -16,000 =Rs. 6,781.25

:. Interest paid is Rs. 6,781.25

Example 2 : If Rs. 1, '750 is invested at 9% per annum interest for 10 years and interest is compounded half-year, find the amount and interest. Solution: P = Rs. 1,750 and r = 9% = 0.09 Interest is calculated half-yearly. i = ~ = 0.045 and n = 10 'I( 2 = 20

Amount Let log Now, Interest,

A A x

= P (1 + i)lI = 1,750 (1.045)20 = (1.045)20

x = 20 log 1.045 = 20 x 0.0191 x = antilog (0.03820) =2.41 A = 1,750 x 2.41 = 4,217.50

=0.3820

1= S-P = 4,217.50 -1,750

=Rs. 2,467.50 .. The amount is Rs. 4,217.50, and interest is Rs. 2,467.50 Example 3 : To what amount Rs. 10,000 accumulate in 6 year, if invested at 8% compounded quarterly? [1.02)24 = 1.60B4J. Solution: P = Rs. 10,000, and r Rate of interest as per conversion period

=

8% =0.08

,. = 4"r =40.08 =0.02 and n =6 x 4. 24 Amount

A A

= P (1 + i)lI = 10,000 (1 + 0.02)24 =10,000 (1.02)24 = 10,000 x 1.6084 =Rs. 16,084

. . The amount is Rs. 16,084.

Example 4 : Find the rate of interest, if the sum of money will double itself in 10 years by investing at compound interest. Solution: Let P be the principal. Let the rate of interest b~ r.

60

RNANCIAL MATHEMAncS

Interest is compounded annually, r = i and n = 10 Given that the amount will be doul>led in 10 years.

i.e.

= 2P

A

But A 2P

= P (1 + i)n = P (1 + i)lO

=>

2 =(1 + i)lO

Taking log on both sides log 2 = 10 log (1 + i) I

og

(1

.) - log 2 _ 0.0310 - 0 0301 + £ - 10 - 10 - .

1 + i . = antilog (0.0301) 1 + i = 1.072 i = 0.072 = 7.2% :. The required rate of interest is 7.2%

Now,

Example 5 : At what rate per annum compound interest will Rs. 5,000 amount to Rs. 9,035 in 5 years if the interest is calculated quarterly?

Solution: P = Rs. 5,000, and A = Rs. 9,035 Let r be the rate of interest per annum. Interest is calculated quarterly. i = .A

~ andn= 4x5=20

= P (1 + i)n

9,035 = 5,000' (1 + i)20 (1 + J')20 (1 +

_ -

9,035 5,000

0 20 = 1.807

Taking log on both sides 20 log (1 + i) = log 1.807 1.807 _ 0.2570 - 0 0129 I og (1 + £') -- log 20 - 20 - . 1 + i = antilog (0.0129) = 1.030 i = 1.030 - 1 = 0.030 .

£

r

= 4r => r =4.~,. = 4 x 0.030 = 0.012

r = 12% . . The required rate is 12%.

61

COMPOUND INTEREST

Example 6 : In how many years an amount will triple itself at 11% compound half yearly? Solution: Let P be the principal. r = 11%=0.11

Interest is compounded haIfyearly. i =

Given, But,

~= 0.055

A = 3P A 3P

= P (1 + i)" = P (1 + i)"

3P = P (1 + 0.055)" 3 = (1.055)"

Taking log on both sides n log (1.055) = log 3 log 3 n = log 1.055 0.4771 n

Now, No. of years t

= 0.0233 = 20.555

=nl2

t = 10.28 approx.

.'. Number of years is approximately 10.28 years.

Example 7 : The difference between compound interest and simple interest on a certain sum of money invested for 3 years at 6% per annum is Rs. 110.16. What is that amount invested? Solution: Let P be the principal. Interest is compounded annually. .. t = n = 3 and i = r = 6% = 0.06 Simple interest II = Prt .. II = P x 0.06 x 3 = 0.18P Compound interest 12 =A- P I 2 =P(I+i)n-P 12 = P (1.06)3 - P Given that 12 - II = 110.16 => P (1.06)3 -P - 0.18 P =110.16 => 1.191016P-P-0:18P= 110.16 =>

0.011016 P = 110.16

62

FINANCIAL MATHEMATICS

P = Rs. 10,000 .. The amount deposited is Rs. 10,000.

Example 8 : The amounts for a certain sum with compound interest at a certain rate in two years and in three years are Rs. 8,820 and Rs. 9,261 respectively. Find the rate and sum. Solution: Let P be the principal amount and i be the rate of interest. Amount A = P (1 + iY" At the end of two years, A =Rs. 8,820 .. . 8,820 = P (1 + i )2 At the end of three years, A =Rs. 9,261 " 9,261 =P (1 + i)3 Dividing (2) by (1), we get 9,261 P (1 + i)3 8,820 = P (1 + i)2

... (1) ... (2)

1.05 = 1 + i => i = 0.05 = 5% Now, substituting i = 0.05 in equation (1) we get, 8,820 =P (1.05)2 =>

8,820 =P x 1.1025 => P =Rs. 8,000 .. Required rate is 5% and principal is Rs. 8,000.

=>

Example 9 : A certain sum of money is invested at 4% p.a. compound annually. The interest for 2nd year is Rs. 25. Find the interest for:Jrd year. Solution: Let P be the principal. A = P(l + iY" Amount at the end of 1st year =P (1.04) Amount at the end of 2nd year =P (1.04)2 Interest for 2nd year =Amount the at the end of 2nd year - Amount at the end of 1st. Given interest for second year is Rs. 25. 25 = P (1.04)2 - P (1.04) => P (1.04) (1.04 - 1) = 25 ... (1) => P (1.04) (0.04) = 25 rd Amount at the end of 3 year = P (1.04)3 Interest for 3rd year = Amount at the end of 3rd year - Amount at the end of 2nd year :. Interest for 3 rd year = P (1.04)3 - P (1.04)2 = P (1.04)2 [1.04 -1] = P (1.04)2(0.04)

COMPOUND INTEREST

63

=P (1.04) (1.04) (0.04) =25 x 1.04

[using (1)]

= Rs. 26.

. . Interest for third year is Rs. 26.

Example 10 : Mr. X wants to make an investment of Rs. 20,000 in one of the two banks that fetch the return after 6 years. Bank A offers 8% interest compounded annually and Bank B offers 7.8% compounded semi-annually. Which Bank should be chosen, so that Mr. X will get maximum return ?

Solution: Amount is given by A = P (1 + i)n In case of Bank A : P

= Rs. 20,000, and r = 8% =0.08

Interest is compounded annually.

..

~

= r =0.08 and n = 6 = 20,000 (1.08)6

At At = 20,000

Amount

x

1.586874

At = Rs.31,737.48 In case of Bank B : P

= Rs. 20,000, and r = 7.8% =0.078

Interest is compounded semi-annualy.

.. Amount

0.078 = 0 .039 an d n = 6 x 2 = 1 2 = 2"r = -2A 2 = 20,000 (1.039)12 A2 = 20,000 x 1.582656 A2 = Rs. 31,653.12 .

~

Since AI> A 2 , Mr. X should invest in Bank A to get maximum return.

Example 11 : Lal deposited an amount of Rs. 50,000 in two different Banks A and B, dividing the amount into two investments. Bank A calculates -interest at a rate of 7% per annum and Bank B calculates interest at the rate of 6% per annum convertible semiannually. At the end of 3 years, he received Rs. 10,632.35 as the return on his investment. What amount he has deposited in both Banks ?

Solution: Let Rs. X and Rs. Y be the amounts deposited in Bank A and Bank B respectively. Total investment is Rs. 50,000 X + Y = 50,000 Compound interest is given by I = A - P where A

... (1)

=P (1 + i)n

64

FINANCIAL MATHEMATICS

Interest earned from Bank A : =>

11

= X(1.07)3_X

11

= 1.225043 X-X

11

= 0.225043X

Interest earned from Bank B : 12 = (1.03)6 Y - Y

12 = 1.194052 Y - Y 12 = 0.194052 Y Total interest earned I =11 + 12 :. 1= 0.225043 X + 0.194052 Y It is given that total interest received is Rs. 10,632.35 ... (2) :. 0.225043 X + 0.194052 Y = 10632.35 Solving (1) and (2) we get, X = Rs. 30,000 and Y =Rs. 20,000. Rs. 30,000 and Rs. 20,000 are the' sums deposited in Bank A and Bank B respectively.

INTEREST COMPOUNDED CONTINUOUSLY If a principal of P is invested for t years at an annual rate of r and interest is

compound~d m times a year, then the interest rate per conversion period i is ~ and there are mt conversion periods. The compound amount A at the end of t years becomes -

A=p(l+~rt m, It is well known that if the interest is compounded more frequently, the compound amount will become larger for a fixed principal, time period and annual interest rate. One may think that the compound amount will be increased by indefinitely increasing the frequency of compounding, i.e., m -+ 00. But there is an upper limit on the compound interest that can be achieved in this way. When interest is so computed that the iIlterest per year gets larger and larger as the number of compounding periods increases continuously, i.e., 'm -+ 00, we say that interest is compounded continuously. Thus, in case of continuous compounding of interest, the amount A is given by

. mt

A

= PLim m .....

oo

A

= PLim m-+

oo

(1

+

~) m

r )~xrt ( 1+-

m

COMPOUND INTEREST

Let as m -

x 00,

x-

Til,

= --: r

00.

A

= Pert

Thus, the compound amount A for a principal amount P after t years at an annual interest rate of r compounded continuously is given by

I A=Pert I Remark. Value of e =2.718; log e =0.4343 Example 12: A person deposits Rs: 5,000 in a bank which pays interest at 11% p.a. compounded continuously. How much amount will be in his account after 10 years? (e l .l = 3.0042) P = Rs. 5,000, r =11% =0.11, t =10 years S=P.e rt = 5,000. eO.ll )( 10 ::: 5,000 . el.1 = 5,000 x 3.0042 = 15,021 . . The amount is Rs. 15,021.

Solution: Amount

Example 13 : What is the amount after 5 years, if Rs. 10,000 is invested at 8% compounded (a) annually? (b) semi-annually? (c) quarterly? (d) monthly? and (e) continuously? ,

Solution: P =Rs. 10,000, r =8% =0.08 and t =5 years (a) lfinterest is compounded annually: Amount A = P (1 + i)'l i = r = 0.08 and n =t = 5 A = 10,000 (1 + 0.8)5 =10,000 x 1.469328 =Rs. 14,693.28 Amount

66

FINANCIAL MATHEMATICS

:. The amount is Rs. 14,693.28

(b) If interest is compounded semi-annually: . r 0.08 , = "2= 2=0.04

n Amount

= 2 x 5 =10

A= 10,000 (1.04)10

= 10,000 x\1.480244 = Rs. 14,802.44 :. The amount is Rs. 14,802.44.

(c) If interest is compounded quarterly: ; - !... - 0.08 _ 0' 02 --4-4-·

n = 4 x 5 =20 A = 10,000 (1.02)20 Amount = 10,000 x 1.485947 = Rs. 14,859.47 .• The amount is Rs. 14,859.47 (d) If interest is compounded monthly: . r 0.08 00 , = 12 = 12 = O. 667

n Amount

A

= 12 x 5 =60 = 10,000 (1.00667)60 .=. 10,000 x 1.490142 =Rs. 14901.42

:. The amount is Rs. 14,901.42

(e) Ifinterest is compounded continuously: Amount A = Pert A = 10,000 eO.08x5 A = 10,000 x eM A = 10,000 x 1.4918 :::: E,s. 14,918 . .. The amount is Rs. 14,918. Example 14 : An amount of Rs. 5,000 is invested at an annual rate of 9% compounded continuously. How long it will take to this principal to amount to Rs. 8,580 ?

Solution: Compound AmountA =P e t Given that P = Rs. 5,000, r = 9% =0.09, A Substituting these values, 8,850 = 5,000 e 0.09 t 8,580 eo.o9t = 5 000 ~ eo.09t = 1. 716 ,

= Rs. 8,580

COMPOUND INTEREST

67

Taking log on both sides, log eo.o9t = log 1.716 0.09 x t x log e = log 1.716 0.09 x t x 0.4343 = 0.2345 0.2345

Example 15 : A sum of Rs. 4,000 is deposited in a savings bank account that fetches Rs. 885.60 as compound interest at a rate. of r% per annum, compounded continuously after 4 years. Find r.

Solution: Principal P = Rs. 4,000 Compound InterestA - P = Rs. 885.60 .. A = 4,000 + 885.60 =Rs. 4,885.60 Given that t =·4It is to find the rate of interest r. Compound Amount A = Pert Substituting the values, we get 4,885.60 = 4,000 e 4r 4r 4,885.60 4r ~ e = 4,000 ~ e =1.2214 Taking log on both sides, log e4r = log 1.2214 4r log e = log 1.2214 4 )( r x 0.4343 = 0.0867 r = 0.05 =5% .. The required rate of interest is 5%.

PRESENT VALUE Every one plans for future. One may think of his son's education, or daughter's marriage for the future of his children. When making such kinds of future plans, we must know well in advance what amount should be invested to receive the certain designed sum in the future. This original principal to be invested is called present value or the capital value of the desined amount. Thus, if money is worth 'i' per conversion period, the present value of the compounded amount A due in 'n' conversion periods is the principal 'P' which is invested now at the rate of'i' per period. We derive the formula from the compound interest formula. We know that A ~

=P (1 + i'r

P=A (1 +irn

68

FINANCIAL AfATHEAfAncS

Thus the present value of amount A due n conversion periods hence at the rate of'i' per period is given by

I .P =A (1 + i~ I I The quantity (1 + ii", denoted by 'v', is called the discount factor. v represents the present value of Re. 1 due n periods hence at the rate of i per period.

I

It should be noted that P and A represent the value of the same obligation at different dates. P is the present value of a given obligation while A is the future value of the same obligation. The difference between the two amounts is known as compound interest or compound discount. .

i.e., compound discount:;: A - P

we

To derive the formula for the present value in case of interest compounded continuously, solve the formula for the compound amount. CompOund amount is given by A=Perl

~

P-=:Aert

Thus, the present value of compound amount A due rate of r compounded continuously is given by

a~

the end of t years at the annual .

I P=Ae-rt I Example 16 : Find the present value of Rs. 300 at rate of interest of 6% per annum payable 5 years hence.

Solution: Hence P = Rs. 300 i

= r =6% =0.06

n :;: 5

Present value is given by P

= A (1 + i)"-n

P :;: 300 (1.06)-5:;: 300 (0.74726)

=Rs. 224.18 The required present value is Rs. 224.18. Example 17 : How much should be invested at 5% per annum so that after 3 years the

amount will be Rs. 8,000, when the interest is compounded (a) annually? (b) half-yearly? (c) quarterly? and (d) continuously? $olution: Here P = Rs. 8,000, r = 5% =0.05, t =3 years. (a) In case of interest compounded annually: i = ~ = 0.05 and n = t = 3 Present value P = A (1 + i)-n

COMPOUND INTEREST

69

p = 8,000 (1.05)-3 P = 8,000 (0.86384) = Rs. 6,910.70 .. Amount to be invested is Rs. 6,910.70

(b) In case of interest compounded semi-annually:

!. - 0.05 _ 0 025 2 - .

= 2-

n Present value

P

= 2xt=2x3=6 = 8,000 (1.025~

P = 8,000 (0.8623) = Rs. 6,898.40

:. Amount to be invested is Rs. 6,896.40

(c) In case of interest compounded quarterly: . r 0.05 , = 4"= 4=0.0125 ~

Present value

= 4 x t = 4 x 3 = 12

P = 8,000 (1.0125)-12 P = 8,000 (0.86151) = Rs. 6,892.10

:. Amount to be invested is Rs. 6,892.10

(d) In case of interest compounded continuously: Present value P = A e-rl P = 8,000 e-O·06 II 3 = 8,000 x e-O. 16 P = 8,000 x 0.86071 =Rs. 6,885.68 ., Amount to be invested is Rs. 6,885.68.

Example 18: What is the present value of Rs. 15,000 due after 5 years from now if the interest is comgounded continuously at the rate 'of interest of 6% ?

= Rs. 15,000,

Solution: Here

P

Present value

'A P = A e-rt = rl e

t = 5 years, r = 6% = 0.06

Substituting the values,

P Let

x

15,000

= eO.06116 = eO.3

p_ 15,000 =>

-

eO. 3

log x = 0.3 log e .= 0.3 x 0.4343

=

log x 0.13029 x = antilog (0.13029) x

= 1.3496 15,000

P = 1.3496 = Rs. 11114.40

The present value is Rs. 11,114.40.

70

FINANCIAL MATHEMAnCS

VARYING RATE OF INTEREST In the previous sections, we have assumed that the rate of interest is fiXed through out the duration of the transaction. But practically the interest rates' change frO!!! time to time due to the policy of the Banks or any other financial institution or the decisions taken by the Government. Thus, a bank that charges the rate of interest at 8% per annum, may increase it to 9% or reduce to 7% in the future. The compound amount at changing rates is calculated as follows: Suppose that a principal o{Rs. P is deposited in a bank and the bank compounds interest at 'ii' per conversion period for nl periods. Then the compound Amount will be P (1 + ilf l • We assume that suddenly the interest rate is increased to i2 per conversion 'period and the deposit is withdrawn at the end of another n2 period. :., The compound amount at the end ofnl + n2 periods will be

A

=P (1 +

il)nl (1 + i 2 )nz

In this manner, we can calculate the compound amount, if the interest rate gets changed any number of times.

Example 19 : A person deposited Rs. 1,000 in a bank at 5% compounded annually. After 5 years the rate of interest increased to 6% and after four more years, the rate of interest was further increaS'ed to 7%. The money wa~ withdraw at the end of 12 years. What 'amount he • will get when he withdraws his deposit after 12 years ?

Solution: For first.5 years i = r

Amount after 5. years A A

=5% = 0.05, P = Rs. 1,000,

n =5

= P (1 + i)n = 1,000 (1 + .05)5 = 1,000 (1.05)5

A = 1,000 (1.27628) = Rs. 1276.28

After 5 years, for next 4 years

=r

= 6% = 0.06, P =

Rs. 1,276.28, n = 4

Amount after 9 years A

= 1276.28 (1 + 0.06)4 = 1276.28 (1.06)4 =1276.28 x 1.26248 =Rs. 1611.28

.

After 9 years i = r = 7% = 0.07 Amount is withdrawn at the end of 12 years

. . n =3, P

=Rs. 1,611.28 Amount after 12 years, A = 1611.28 (1 + 0.07)3

71

COMPOUND INTEREST

= 1611.28 (1.07)3 = 1611.28 x 1.225 = Rs. 1973.83 .. The amount the end of 12 years is Rs. 1973.83.

Example 20 : Mr. X deposited Rs. 10,000 in State Bank of India. The bank calculates the interest at 8% p.a. compounded semi-annually. After 5 years, he again deposits an amount of Rs. 8,000 in his account. The bank has increased the rate of interest from 8% to 10% at the end of 8th year. If he withdraws the total amount at the end of 10 years, what amount will have been accumulated in his account ?

Solution: Incp.se of deposit of Rs. 10,000 : Rate of interest is 8% for first 8 years and 10% for next two years. For first 8 years, P = Rs. 10,000, r = 8% = 0.08 Interest compounded semi-anually,

..

i = b 08 = 0.04, and n = 8 x 2,= 16

2

Amount is given by A = P (1 + iY" .. Amount at the end of 8 years A = 10,000 (1 + 0.04)16 A = 10,000 (1.04)16 A = 10,000 (1.872792) = Rs. 18,727.92 This is the principal for 9th year .. For next two years P = Rs. 18,727.92, r = 10% = 0.10, n = 2 x 2 = 4 . _ 0.10_ , - 2 - 005 . Amount at the end of 10th year A = 18,727.92 (1 + 0.05)4 = 18,727.92 (1.05)4 = 18,727.92 x 1.2155 = Rs. 22,763.90 In case of deposit of Rs. 8,OOO/or 5 years starting from (Ph year: Interest rate is 8% for first 3 years, and 10% for next two years. For first three years P = Rs. 8,000, r =8% = 0.08, n = 3 x 2 = 6 . - 0.08 -004 ,- 2 "

Amount at the end of 8th year A = 8,000 (1 + .04)6 = 8,000 (1.04)6 = 8,000 (1.2597) = Rs. 10,017.69

... (1)

72

FINANCIAL AlA THEAIA TICS

This is the principal for 9th year. .. For next two years

P = Rs. 10,077.69, r = 10% = 0.10, n = 2 x 2 = 4 . _ 0.10 _ 005 ,- 2 - . Amount at the end of 10th year A = 10077.69 (1 + 0.05)4= 10077.69 (1.05)4

= 10077.69 (1.2155) =12249.43

... (2)

Total Amount = (1) + (2) = Rs. 22,763.90 + Rs. 12,249.43 =Rs. 35,013.33 :. Mr. X will receive Rs. 35,013.33 at the end of 10 years.

Example 21 : To what sum will Rs. 2,000 amount in 8 years if invested at 6% effective rate for the first 2 years, at ~% compounded semi-annually for the next 8 years at 6% compounded continuously thereafter ?

Solution: For first 2 years:

P

= Rs.2,000

Interest is compounded annually. r = i = 6% = 0.06, and n = 2 Amount is given by A = P (1 + i'Y' A = 2,0004:1. + 0.06)2 = 2,000 (1.06)2

A

= Rs.2,247.20

For next 8 more years : P = Rs. 2,247.2 and r = 6% = 0.06 Interest is calculated semi-annually. ..

i

= i =0.03, and n = 3 x 2 =6

Amount at the end of 5th year

A

= 2,247.20 (1.03)6 =Rs. 2683.27

For last 3 years: P = Rs. 2,683.27 and r = 6% = 0.06

Interest is calculated continuously. Amount at the end of 8th year A

~

Perl

= 2683.27 x eO•06 3 = 2683.27 x eO•18 = 2683.27 x 1.1972 =Rs. 3212.42 Ie

Amount at the end of 8th year is Rs. 3,212.42.

COMPOUND INTEREST

73

L Find the compound amount and compound interest of Rs. 1,00,000 invested for 6 years at 8% compounded quarterly. [ADs. Rs. 1,60,843, Rs. 60,843] 2. !fRs. 1,000 is invested at an annual rate of interest of 15%, what is the amount after 5 years if the interest is compounded monthly? [ADs. Rs.2107.18] S. Find the compound amount of Rs. 2,000 for four years at 6% converted (a) annually, (b) semi-annually, (c) quarterly, and (d) monthly. ' [~. (a) Rs. 2524.95, (b) Rs. 2,533.54, (c) Rs. 2,537.97, (d) Rs. 2,540.97] 4. An amount of Rs. 50,000 is deposited into a savings bank account that pays at the rate of 6% p.a. What amount will be there in the account, if the interest is compounded (a) annually, (b) quarterly, and (c) monthly? [ADs. (a) Rs. 1,60,300, (b) Rs. 1,62,000, (c) Rs. 1,65,500] 5. Find the compound interest for a sum of Rs. 700 invested for 15 years at 8% compounded semi-annually. [Ans. Rs. 1570.37] 8. In how many years an investment of Rs. 5,000 will amount to Rs. 7,000, if it is [ADs. 4.25 years] invested at 8% compounded quarterly? 7. At what rate per cent per annum compound interest will Rs. 2,000 amount to Rs. 3,000 in 3 years, if the interest is reckoned half-yearly? [ADs. 14%] 8. In how many years will an amount double itself at 12.2% compounded annually? [ADs. 6.02 years] 9. At what rate per cent will Rs. 40,000 yield Rs. 13,240 compound interest in 3 years? , [ADs. 10%] 10. How long will it take for a principal to double if the money is worth 12% compounded monthly? [ADs. 5.84 years] lL A sum of money is put at compound interest for two years at 20% per annum. It would fetch Rs. 482 more, if the interest is payable half-yearly than if it is payable [ADs. Rs. 20,000] yearly. Find the sum. 12. The difference between simple interest and compound interest on a sum for 3 years at 5% per annum is Rs. 76.25. Find the sum. [Ans. Rs. 10,000]

18. The difference between simple interest and compound interest on a sum of money invested for 4 years at 5% per annum. is Rs. 150. find the sum. [Ans. Rs. 9677 .40] 14. A person aged 22 has contributed Rs. 25 to his provident fund during the current month. What will this particular contribution amount to by the time he retires at age 60, assuming a rate of interest 7% p.a. ? [ADs. Rs. 326.98] 15. The compound interest on Rs. 8,000 in two years at certain rate is Rs. 820 and in three years, it is Rs. 1,261 at the same rate. Find the rate of interest. [Ans.5%] 18. Mr. X deposited Rs. 10,000 in a savings account for 3 years offering progressive rate of interest. Bank pays 10% per annum compounded semi-annually for the first year, 12% per annum compounded quarterly for the second year and 13% per annum compounded continuously for the third year. Find the amount at the end of 3 years. [ADs. Rs. 14,131.14]

74

FINANCIAL MATHEMATICS

17. If Rs. 2,000 is deposited in a savings account that earns interest at an annual rate of 6% compounded continuously, what is the value of the account after 3 years? [ADs. Rs. 2,107.18] 18. At what rate will a principal amount triple itself in 12 years, if the interest is • compounded continuously? [ADs. 9.15%] 19. How much should be invested at 6% per annum so that after 4 years, the amount will be Rs. 25,000, when the interest is compounded continuously? [ADs. Rs. 19,655] 20. How long will it take for Rs. 4,000 to account to Rs. 7,000, if it is invested at 7% compounded continuously? ' [ A D s . 8 years]

.

21. In how many years a principal amount P will become double, if money is worth 6% compounded continuously? [ADs. 11.5 years] 22. Mr. A deposited Rs. 5,000 in a bank which calculates interest continuously. After 10 years, the amount accumulated was Rs. 15,020. At what rate bank calculated interest? [ADs. 11%] 23. The' population of a town is 8,00,000. During the first year, the population increased by 25%. During the second year, the population increased by 20%. During the third year, the population increased by 10%. Find the population ofthe town after 3 years. [ADs. 13,20,000] 24. A sum of Rs. 2,000 is invested at a rate of interest of 5% per annum. After 7 years, the rate of interest was changed to 5% per annum convertible half-yearly. After a further period of 3 years, the rate was again changed to 6% J>er annum convertible quarterly. What is the accumulated value at the end of 15 years from [ADs. Rs. 4,395.61] commencement? 25. What is the present value of Rs. 500 due at the end of 10 years, the rate of interest being 5% p.a. for the first four years from now and 6% convertible half-yearly for the next 6 years? [ADs. Rs. 288.51] 26. A sum of Rs. 1,000 is due at the end of 10 years 6 months. The present interest rates are 7% per annum but it is expected that there will be a fall in the" rates after 6 years bringing down the rate to 6% per annum. Find the present value of the sum of money under these assumption. [ADs. Rs. 512.65] 27. The difference between the accumulated values of a sum of money accumulated for 10 years at an effective rate of 4% per annum and the accumulated value ofthe same sum of money over the same period at 4% per annum payable quarterly is Rs. 150.• Find the sum. [ADs. Rs. 17,401] , 28. A promises to pay B a sum of Rs. 200 at the end of 3 years and another Rs. 400 at the , end of 5 years from now. What immediate cash payment should B accept in lieu of the above payments, if interest is compounded at 5% per annum? [ADs. Rs. 486.18)

29. Mr. Singh wants to accumulate Rs. 50,000 after 20 years for the purpose of his daughter's marriage. How much money should be set a side now, if interest is [Ans. Rs. 18,394] compounded continuously at an annual rate of 5% ?

75

COMPOUND INTEREST

30. Find the present value of Rs. 2,500 due after 4 years, if the interest rate is 6% per annum compounded (a) annually, (b) quarterly, (c) monthly, and (d) continuously. [ADs. (ci) Rs. 2,137, (b) Rs. 1,974.72, (c) Rs. 1,980.98, (d) Rs. 1,965.40] 31. A person invests a certain sum of money in a bank paying interest 5% effective and plans to receive Rs. 5,000 in 5 years. What is the compound discount of this investment? [ADs. Rs. t082.4] 32. What is the year zero deposit at 8% per annum compounded quarterly to obtain a compound amount of Rs. 1,00,000 after 5 years? How much interest will be earned during this period? [ADs. Rs. 67,297.15, Rs. 32,702.85] 33. Find the compound interest (compounded annually) on Rs. 10,000 for four years at 10% for the first year, 12% for the second year, 14% for the third year and 15% for [Ans. Rs. 6,uiI.52] . the fourth year. 34. Which yields the higher rate of interest---a fixed deposit in a bank which giv~s Rs. 1,629 after 5 years for every Rs. 1,000 deposited or a National savings certificate which gives Rs. 1,901 after 6 years for Rs. 1,000? [ADs. FD : 10.24%, NSC : 11.28%] 35. Mr. A has a right to receive an amount of Rs. 1,000 at the end of 12 years from now. This right has been sold to Mr. B for a present value calculated at the rate of 8% per annum. The money thus received was invested by Mr. A in a deposit account at 9% per annum payable half-yearly. After 8 years the account had to be closed and then Mr. A invested the amount available at 6% per annum in another bank. How much has Mr. A gained or lost in. this transaction, as at the end of 12 years? [ADs. Gain ofRs. 13.90]

•

Nominal and Effective Rates of Interest --~---

-

-

-

~---~

INTRODUCTION The value of money inves~ed at an annual rate of r per annum compounded more than once in a year will be more than the value of money at r per annum compounded annually. This is because interest compounded more than one period will itself earn interest during subsequent period. For example, value of Rs. 100 at 6% per annum compounded semiannually is 100 (1.03)2, i.e. Rs. 106.09, while the value of the same principal at 6% compounded annually is 100 (1.6) i.e. Rs. 106. So the effective return is Rs. 6.09. In this chapter, we discuss the concept of effective rate and how effective r~te is applied in comparing different investment alternatives.

NOMINAL AND EFFECTIVE RATE In transactions involving compound interest, the stated annual rate of interest is called nominal rate of interest. The actual percentage by which money grows during a year is called effective rate of interest. In other words, the effective rate is the simple interest rate that is equivalent to the nominal compound interest rate. Effective rates are also called annual yields or true interest rates. It should be noted that in case of effective rate of interest, interest is compounded only once in a year. Let us understand the concept of nominal and effective rate of interest with a suitable example. Let Rs. 100 be invested for one year at 10% compounded quarterly. Here 10% is the nominal rate of interest.

n

. NOMINAL AND EFFECTIVE RATES OF INTEREST

The amount after one year is A

010)4 = 100 ( 1 + T =100 (1.025)4 =Rs. 110.38

.. The actual interest earned on Rs. 100 is Rs. 10.38 in one year. We say that the effective rate in this case is 10.38%. Now at 10.38% per annum rate of interest, compounded annually, A =100 (1 + 0.1038) =100 (1.1038) =Rs. 110.38. Thus it is obvious that for a principal of Rs. 100, the amount at a rate of 10.38% compounded annually is equivalent to the amount at a rate of 10% compounded quarterly. , We therefore say that the effective rate' is 10.38% and the corresponding nominal rate is 10% (compounded quarterly).

.

RELATION BETWEEN EFFECTIVE RATE AND NOMINAL RATE Let r e be the effective rate of interest and r be the corresponding nominal rate compounded 'm' times in a year. Let P be the principal invested. Let i be the rate of interest per conversion period. . r t=Therefore m At re effect rate of interest, the amount after one year is given by Al

=P (1 + re)

... (1)

At r compounded m time in a year, the amount after one is given by

A2=p(1+~r

... (2)

The amount at re effective is equivalent to the amount at r compounded m times a year. ..

Al

~

P(1 + re)

= A2 = p( 1 + ~

re =

r

r-

(1 + ~ 1

Thus, the relation between nominal rate and effective rate is given by

Using this relation, we can find the effective rate equivalent to)he--no-minal rate r compounded m times a year, i.e. equivalent to rate i per conversion period.

78

FINANCIAL MATHEMATICS

In the above relation, interest is compounded 'm' times a year at r nominal rate of interest. Now we derive the relation between the nominal rate and effective rate, if the nominal rate is r compounded continuously. If the interest is compounded continuously at r per annum nominal rate, the effective rate of interest is given by re re

= ~~ [ (1 + ~ = m_oo Lim

re

= ~~ [ (1 + ~

re

=[ ~~oo

+

fr-

1

~ )~r -1' 1

[Lim (1 + !... )(~)

re

=

~

re

= [Lim x-o

~

re = er - l

1]

!...)m -1 m

(1 +

(1

r~

.m

m-oo

(1 + x)1!%

r-

1

r-

1

Let x

=-mr

. as m _oo,x-O

Thus the relation between the effective rate and nominal rate in case of nominal rate r compounded continuously is given by-

I re =er -11 The nominal rate r compounded continuously and equivalent to a given effective rate, r e , is called the force of interest.

Remark 1 : If the conversion period is one year, at r nominal rate of interest, that is, if

m

=1, then the effective rate equals nominal rate.

Thus,

re

= (1+~r

re

=

(l+ff

re

=

r.

-1

-1=I+r-l=r

m=

1

Remark 2 : The effective rate of interest depends only on the nominal rate r and the number of conversion periods in a year 'm' but is independent of the principal P. Remark 3 : The effective rate can be a useful guide for comparing alternative investment opportunities available to an individual.

79

NOMINAL AND EFFECTIVE RATES OF INTEREST

Example 1 : Find the effective rate of interest corresponding to 8% nominal rate compounded quarterly.

Solution: Nominal rate of interest, r= 8% = 0.08 Interest is compounded quarterly. .. No. of conversions per year m

=4 and mr = 40.08 = 0.02 = 4

Effective rate of interest re = (1 +

~

r

-1

re =. (1 + 0.02)4 - 1 = (1.02)4 r 1 = 1.0824 - 1 = 0.824 = 8.24% The required effective rate is 8.24%. Example 2 : Find the effective rate equivalent to the normal rate of interest 6% compounded continuously.

Solution: In case of continuous compound interest, the effective rate is given by

=

er -1 r = 6% = 0.06 re = eo.6 -1

re Here

re = 1.0618 - 1 = 0.0618 = 6.18% :. The effective rate of interest is 6.1%. Example 3 : Find the effective rate of interest equivalent to the nominal rate 9% converted (a) semi-annually, (b) quarterly, (c) monthly and (d) continuously.

Solution:

(a) In case of Semi-annual conversion: The effective rate re equivalent to the nominate r converted m time a year is given by re

=

r

(1 + ~ -1

Given that r = 9% = 0.09 .. Here, interest is compounded semi-annually, m

=2

re

=

(1 + 0~9r -1

re - (1.045)2 - 1 re = 1.0920 - 1 = 0.920 = 9.2% . . The effective rate is 9.2%.

80

FINANCIAL MATHEMAncS

(b) In case of quarterly conversion: m=4

re

r-1

(1

= + 0~9 = (1.0225)4 - 1 = 1.0931 - 1 =0.0931 =9.31%

re re :. The effective rate is 9.31%

(c) In case of monthly conversion: m = 12 2 .. re = ( 1 + - 1

°i~9

r

re = (1.0075)12 -1 re = 1.0938 - 1 =0.0938 =9.38% The effective rate is 9.38%. (d) In case of interest converted continuously: The effective rate of interest r e equivalent to the nominal rate r compounded continuously is given by re = er -1 Here re = eO.09 - 1 re = 1.0942 - 1 = 0.0942 = 9.42% The effective rate is 9.42%.

Example 4: Find the force of interest corresponding to the effective rate 6%.

Solution: The force of interest r corresponding to the effective rate re is given y re = er -1 Here re = 6% = 0.06 0.06 = er -1 er = 1.06 Taking log on both sides we get r log e = log 1.06 r (0.4343) = .0253 0.0253 r = 0.4343 = 0.583 = 5.83% The force of interest is 5.83%.

Example 5 : Find the nominal rate compounded quarterly which is equivalent to the effective rate 6%.

NOMINAL AND EFFECTIVE RATES OF INTEREST

81

Solution: The relation between the nominal rate and effective rate is given by

=

re

+

~

r-1

re = 6% = 0.06

Here

0.06 =

"

=>

(1

(1

and m=4

(1+~r-1

+~r = 1.06

Taking log on both sides, we get 4 log ( 1 + ~)

= log 1.6

=>

4 log (1 +

~) = 0.0253

=>

log ( 1 +

~) = .0063

=>

1+ ~ 1+

=>

= antilog (0.0063)

r

4" = 1.0146 r

4" = 0.146

r == 0.0584 = 5.84% . . The nominal rate is 5.84%.

=>

Example 6 : Which is the better investment from the stand point of the investor: 6.2% compounded semi-annually or 6% compounded monthly.

Solution: The effective rate (r e) corresponding to the nominal rate r is given by re ::

r

(1 + ~ -1

In case of 6.2% compounded semi-annually: m == 2, amd r =6.2% = 0.062 Here

re

=

(1 +

0.~62 r

- 1

= (1 + 0.031)2 - 1 =(1.031)2 -

1

= 1.06296 - 1 == 0.6296 = 6.3%

In case of interest compounded monthly: r

= 6% =0.06

and m == 12

82

FINANCIAL MATHEMATICS

re

=

0.06)12 -1 ( 1 + 12

= (1.005)12 - 1 = 1.06168 - 1 = 0.06168

= 6.17%.

The effective rate of interest corresponding to the nominal rate 6.2% compounded semiannually is greater. :. The first investment (6.2%) is better investment. Example 7 : When is a better investment : 7.8% compounded semi-annually or 8% compounded annually. Solution: In case of 7.8% compounded semi-annually: The effective rate 're' corresponding to the nominal rate r is given by

re

(1 +

~r-1

r = 7.8% = 0.078 and m = 2

Here .,

=

re = (1 + .0;8f -1

re = (1.039)2 - 1 => re = 1.0795 -1 = 0.0795 = 7.95%. In case of 8% compounded annually: Here r = 8% If the nominal rate r is compounded annually, then the effective rate re =>

=r

re = r = 8% Here, the effective rate of interest corresponding to the nominal rate of interest 8% compounded annually is greater than the effective rate corresponding to the nominal rate 7.8% compounded semi-annually. :. 8% compounded annually is better investment. Example 8 : Find the nominal rate compounded monthly equivalent to 5% compounded semi-annually. Solution: Let r be the nominal rate compounded monthly. The effective rate 'r e', corresponding to the nominal rate converted m times a year is given by

re

=

r

(1 + ~ -1

.. In case of r compounded monthly m

= 12

and re

= (1 + ;2

r1

... (1)

83

NOMINAL AND EFFECTIVE RATES OF INTEREST

In case 5% compounded semi-annually, r

=>

= 5% =0.05

and m

(1 + 0~5 r-1

re

~

re

= (1.025)2 -

=2

1

... (2)

Given that (1) is equivalent to (2) ..

( 1 + ;2 f2 - 1

=>

( 1 + ;2) 12

= (1.025)2 -

1

= (1.025)2

Taking log on both sides, we get 12 log ( 1 +

~2)

=>

12 log ( 1 + ;2 )

=>

log ( 1 + ;2)

=>

r 1+ 12

=>

r 1+ 12 r 12

= 2 log 1.025

= 2 x 0.0107 =

2x 0.107 12

= antilog (0.0018)

= 1.004 =

0.004 => r

= 0.048 =4.8%

:. The required nominal rate is 4.8%. Example 9 : What nominal rate compounded quarterly will be equivalent to 10% compounded continuously ?

Solution: Let r be the nominal rate compounded quarterly. The effective rate r e, corresponding to the nominal rate r converted m times a year is given by

re

=

(1 +: r-1

In case of r compounded quarterly m

=4

re

=

r-

(1 + i 1

... (1)

84

FINANCIAL MATHEMATICS

The effective rate re, corresponding the nominal rate r converted continuously is given by

re = er -1 In case of 10% compounded continuously r = 10% =0.10 .. re = eO. 10 - 1 Given that (1) is equivalent to (2)

(1+~r-1

.. ::;.

(1

... (2)

= eO.1o-1

+~r =

eO. 1O

Taking log on both sides, we get 4 log

(1+~) = 0.10 x loge

=>

4 log ( 1 + ~)

=>

log ( 1 + ~) r

1 + 4"

= 0.10 x 0.4343 =

0.0!343 =0.0109

= antilog (0.0109)

r r 1 + 4" = 1.0254 => 4" = 0.0254 r

= 0.1016 = 10.16%

The nominal rate is 10.16%.

Example 10: A money-lender charges interest at the rate of 10 paise per rupee per month, payable in advance. What effective rate of interest does he charge per annum?

Solution : The money lender charges interest at the rate of 10 paise per rupee per month, payable in advance. Therefore, 10 paise may be treated as interest on 90 paise for one month. ..

The interest rate per month

i.e.

·

t

10 1 =90 =9" .

1

r

12 9 The effective rate re corresponding to the nominal rate r is given by re Here

=

:r

(1 + -1

m = 12

NOlllNAL AND EFFECnVE RATES OF INTEREST

=>

re + 1

85

= ( 190 )12

Taking log on both sides we get log (1 + re)

= 12 log (~O) = 12 [log 10 -log 9] = 12 (1 - 0.9542)

log (1 + r,) => log (1 + re) => log (1 + re) = 12 (0.0458) => log (1 + re) = 0.5496 => 1 + re = antilog (0.5496) => 1 + re = 3.545 re = , 2.545 =254.5% The effective rate is 254.5% =>

Example 11 : A deposited an amount in a bank for 10 years at the effective rates of interest 3% per annum for 4 years, 4% per annum for next 4 year and 5% per annum for the last two years. B deposited the same amount in an another bank at a constant rate of interest compounded semi-annually. After 10 years if both A and B get same accumulate amount, at what rate of interest B deposited his money ?

Solution: Let Re. 1 be the principal. The amount is given by A

=

P (1 + i)n

For A, the Amount after 10 years is

Al

= 1 (1.03)4 (1.04)4 (1.05)2

... (1)

Let r be the rate of interest for B. Interest compounded semi-annually, For B, the amount after 10 years is

A2 = 1 (1 + i)20 = (1 + i)20 To get same accumulated amount,

(1)

= (2)

=>

(1

+ i)2o

= (1.03)4 (1.04)4 (1.05)2

=>

(1

+ i)20

= 1.4516

Taking log on both sides we get 20 log (1 + i)

= log (1.4516)

0.1620 . log (1 + l) = 20 = 0.0081

... (2)

86

FINANCIAL MATHEMA TICS

i

= antilog (0.0081) = 1.019 = 0.019 =1.9%

r

= 2 x i =2 x 1.9 =3.8%

1+i 1+i

Rate of interest for B is 3.8%.

L Find the effective rate of interest corresponding to 12% nominal rate compounded monthly. [ADs. 12.68%] 2. What is the effective rate which is equivalent to a nominal rate 10% per annum compounded semi-annually? [ADs. 10.25%] 3. Find the effective rate equivalent to the nominal rate 5% compounded continuously. [Ans. 5.13%] 4. Find the effective rate equivalent to the nominal rate 8% compounded continuously. [ADs. 8.33~] 5. Find the effective rate-equivalent to the nominal rate of 7%, converted (a) quarterly, • (b) monthly and (c) continuously. [Ans. (a) 7.19%, (b) 7.22%, (c) 7.25%] 6. Find the effective rate equivalent to the nominal rate 6% converted (a) monthly, (b) continuously. [ADs. (a) 6.16%, (b) 6.18%] 7. Find the force of interest corresponding to the effective rate 8%.

[ADs. 7.69%]

8. What annual rate compounded continuously is equivalent to an effective rate of 5% ? [ADs. 5.82%] 9. Find the force of interest corresponding to the effective rate 9%.

[ADs. 8.61%]

10. Find the effective rate of interest corresponding to nominal rate of 5% per annum compounded (i) semi-annually, (ii) quarterly, and (iii) monthly. [ADs. (i) 5.06%, (ii) 5.09%, (iii) 5.12%] 1L A certain bank offers an interest rate of 6% per annum compounded annually. A competing bank compounds its interest continuously. What nominal rate should the competing bank offer so that the effective rates of the two banks will be equal ? [ADs. 5.82%] 12. Find the nominal rate compounded monthly equivalent to the effective rate 12% per

annum.

[Ans. 11.4%]

13. Which is a better investment from the stand point of the investor: 4% per annum compounded quarterly or 4.1% effective? [ADs. 4.1% effective] 14. Which is better from the stand point of investor: 6.1% converted quarterly or 6% converted continuously? [Ans. 6.1 quarterly] 15. Which is a better investment: 5.5% compounded semi-annually or 5% compounded monthly? [ADs. 5.5 semi-anually]

,

.

NOMINAL AND EFFECnVE RATES OF INTEREST

87

16. Bank 'A' offers interest at an annual rate of 9.1% compounded semi-annually, and bank 'B' offers interest at 9% compounded monthly. Which Bank offers the better deal? [Ans. Bank B] 17. Find 'the nominal rate compounded quarterly equivalent to 6% compounded semi, annually. . [Ans. 5.92%] 18. What nominal rate of. interest compounded monthly will ,be equivalent to 8% compounded continuously? [Ans.8.04%] 19. Find the compound interest compounded semi-annually equivalent to 9% compounded monthly. [Ans.9.18%] 20. A money lender charges interest at the rate of 5 paise .per one rupee per quarter, [Ans.22.8%] payable in advance. What effective rate does he change per annum? 21., A money lender charges interest at the rates of 10 rupees per 100 rupees per half year, payable in advance. What effective rate of interest does he charge per annum? [Ans. 23.5%] 22. A deposits an amount in a bank for 16 years at the effective rates of interest 3% per annum for 10 years, 4% per annum for 4 years and 5% per annum for the last two years. B deposits the same amount in another bank at a constant force of interest. After 16 years if both A and B get same accumulated amount, at what rate of interest B deposited his money? [Ans.3.43%] 23. What do you mean by nominal rate of interest? How does it differ from effective rate of interest ? Establish the relationship between the nominal and effective rate of interest (a) when compounded m times a year and (b) when compounded continuously? When will nominal rate be equal to effective rate of interest?

Equation. of Value ------------

~=-==-~

------=--=--=-=--=-==--=_--==--=-_-.J

INTRODUCTION We have so far' discussed the kinds of interest as well as the method bf converting nominal value into effective rate. By using effective rate, we generally compare different investment alternatives. We have also seen that money has time value. A sum of money today will not be the same in value tomorrow. The question now is how to find the value of a certain financial at a future date which is equivalent to some other obligation(s). Answer is the equation of value. In this chapter, we study the concept of equation of value in a detailed manner.

EQUATION OF VALUE Money has different values at different times. In business transactions, different sums are flowing at different points of time. Such flows can be equated at a particular point of time. An equation of value is an equation which states that the sum of the values, on a given date, of one set of obligations is the same as the sum of values, on this date, of another set of obligations. The date chosen for comparing the values of two sets of obligations is called the focal date or comparison date. The equation of value is given by,; Sum of the values ('~ one set of } {sum ofthe values of another set of obligations (old ebligations) on focal date = obligations (new obligations) on focal date Let'Rs. x be a non-interest bearing debt which is due at some specified time. The value of this debt at the due date is of course Rs. x. Suppose that money is worth i per period. If the debt is not repaid on due date and paid 'n' periods after due date, then the value of the money is of course more than Rs. x and is found by multiplying x by (1 + i)'t. But if the debt is

89

EQUATION OF VALUE

repaid 'n' periods before due date, then the value of the money is less than Rs. x and is found by multiplying x by (1 + iT". Thus, per period is x (1 + f)". Value ofRs. x, n periods before due date at at the rate i per period isx (1 + xT".

(a) Value ofRs. x, n periods after due date at the rate i (b)

Remark: (a) Focal date is decided on by the lender and the borrower. In case of only one new obligation, the due date is the focal date. If there are more than one new obligations, then focal date is generally the due date of the last obligation. (b) The flows of money involved may be involving simple interest or compound interest. Simple interest is used only if it is mentioned in the problem, otherwise it is treated as compound interest. (c) It should be noted that if two set of obligations have equal values on one date, then they will have equal values on any other date, provided rate of interest is same for each obligations. (d) If the interest is calculated on the basis of simple interest, the conversion should be based on the methods of conversion which we have discussed in the chapter: Simple Interest. (e) If the problem is to find the focal date, or the number of periods of conversion (months/years), we assume that the focal date is today .

.

Example 1: A man borrowed Rs. 10,000 from a money lender at 9% simple interest and agreed to make two equal payments, one due in 6 months and the other in 12 months. Find the payment at the end of 12 months.

Solution: Let Rs. x be the amount of equal payment. Focal date is 12 months. r

=9% =0.09

The old obligations and new obligations are shown in the following table:

Old Obligations Rs. 10,000 now

Value of each at focal date 10,000 (1.09)

New Obligations Rs. x at the end of 6 months Rs. x at the end of 12 months

Value of each at focal date x (1 + (0.09) x 1/2) x

The equation of value is given by : Sum of the values of old Obligations} _ {sum of the values of new obligations at focal date - at focal date :. The equation is: 10,000 (1.09) = x (1.045) + x ~ 10,900 = 2.045 x ~ x = Rs. 5,330.07 .. Payment at the end of 12 years is Rs. 5,330.07.

90

FINANCIAL MATHEMATICS

Example 2 : A man owes Rs. 2,000 due in 2 months, Rs. 1,000 due in 5 months and Rs. 1,BOO due in 9 months. He wishes to discharge his obligations by two equal payments due in 6 and 12 months respectively. Find the equal payments if money is worth 6% simple interest and at the end of 1 year is the agreed focal date.

Solution: Let Rs. x be the equal payments. The focal date is 1 year. Rate of interest is 6% simple interest. The old obligations and new obligations are shown in the following table : Value of each at focal date

Old Obligations

Rs. 2,000 due in 2 months 2,000 [ 1 + (0.06 x

~~) ]

New Obligations Rs. x due in 6 months

Rs. 1,000 due in 5 months 1,000 [ 1 + ( 0.06 x ;2) ] Rs. x in 1 year

Value of each at focal date x [ 1 + (0.06 x ;2) ] x

Rs. 1,800 due in 9 months 1,800 [ 1 + ( 0.06 x ;2) ] The equation of value is given by : Sum.ofthe values of old Obligations} _ {sum of the values of new obligations at focal date - at focal date :. The equation is 2,000 [ 1 + (0.06 x

~~) ] + 1,000 [ 1 + ( 0.06 x

;2)]

=x [ 1 + (o.o~ x ;2) ] + x 2,000 (1 + 0.05) + 1,000 (1 + 0.035) + 1,800 (1 + 0.015) =x (1 + 0.03) + x + 1,800 [ 1 + (0.06 x ;2) ]

(2,000 x 1.05) + (1,000 x 1.035) + 1,800 (1.015) =X (1.03) + x 2,100 + 1,035 + 1,827 = 2.03 x 2.03 x =4,962 x = 2,444.33 The installment amount is Rs. 2,444.33. Example 3 : What single payment 5 years hence will discharge the debt of Rs. 800 and Rs. 500 in 3 years and 9 years respectively, ifth~ money is worth 6% compounded quarterly?

Solution: Let Rs. x be the amount due in 5 years The focal date = 5 years r

= 6%

..

.

£

=4"r =40.06 =o. 015

91

EQUATION OF VALUE

The old obligations and new obligations are shows in the following table:

800 due in 3 years

Value of each at focal date 800 (1.015)2 x 4

500 due in 9 years

500 (1.015)- 4 x 4

Old Obligations

New Obligations X

Value of each at focal date

at the end of 5 years X

The equation of value is given by : Sum of the values of old ObligatiOns} _ {sum of the values of new obligations at focal date - at focal date The equation is : 800 (1.015)8 + 500 (1.015)-16 => =>

=X

x = 900.8 + 394.94 x =Rs. 1,295.74 The required payment is Rs. 1,295.74. ..

Example 4 : A debt of Rs. 5,000 due in five years is to be repaid by a payment of Rs. 2,000 now and a second payment at the end of 6 years. How much should the second payment be if the rate of interest is 6% compounded quarterly ?

Solution: Let Rs. x be the second payment. The focal date is 6 years. Rate of interest r = 6% = 0.06. Interest is compounded quarterly. ..

; - !:.. - 0.06 4 -_ 0 .015 ~-4-

The old obligations and new obligations are shown in the following table.:

Old Obligations

Value of each at focal date

Rs. 5,000 due in 5 years

5,000 (1.015)4

New Obligations Rs. 2,000 now

Value of each at focal date 2,000 (1.015)6 x 4

Rs. x at the end of 6 years The equation of value is given by : Sum of the values of old obligations } _ {sum of the values of new obligations at focal date - at focal date 5,000 (1.015)4 = 2,000 (1.015)6 x 4 + X =>

=>

5,306.82 = 2,859.01 + x x =5,306.82 - 2,859.01 x == 2,447.81 The second payment is Rs. 2,447.81

x

92

FINANCIAL AlA THEMAncS

Example 5 : Mr. X owes Rs. 1,000 due in 1 year and Rs. 3,000 due in 4 years. He agrees to pay Rs. 2,000 today and the reminder in 2 years. How much he pay at the end of 2 years if the money is worth 5% compounded semi-annually ?

Solution: Let Rs. x be the final payment due at the end of 2 years. The focal date is 2 years. r

=5% =0.05

Interest is calculated half-yearly ..

i

= 0~5 =0.025

The old obligations and the new obligations are shown in the following table.

Old Obligations

Value of each at focal date

Rs. 1,000 due in 1 year

1,000 (1.025)1" 2

Rs. 2,000 today

Rs. 3,000 due in 4 years

1,000 (1.025)- 2" 2

Rs. x due in 2 years.

New Obligations

Value of each at focal date 2,000 (1.025)2" 2 x

The equation of value is given by : Sum of the values of old Obligations} _ {sum of the values of new obligations at focal date - at focal date The required equation .. 1,000 (1.025)2 + 3,000 (1.025)-4

=2,000 (1.025)4 + x

1,050.625 + 2,717.852 = 2,207.626 +x 3,768.477 = 2,207.626 + x x =Rs. 1,560.85 . . The amount to be paid is Rs. 1,560.85.

Example 6 : A debt of Rs. 2,000 due in 2 years and Rs. 3,000 due in 7 years is to be repaid by a single payment of Rs. 1,000 now and 2 equal payments which are due 1 year from now and 4 years from now. If the interest rate is 6% compounded annually, how much will be the equal payments ?

Solution:

Let Rs. x be the equal payment. The focal date is 4 years i = r = 6% = 0.06 The old obligations and now obligations are shown in the following table:

I

Old Obligations

Value of each at focal date

Rs. 2,000 due in 2 years

2,000 (1.06)2

Rs. 1,000 now

1,000 (1.06)4

Rs. 3,000 due in 7 years

3,000 (1.06)-3

Rs. x in 1 year

x (1.06)3

Rs. x due in 4 years

x

New Obligations

Value of each at focal date ._---

I

EQUATION OF VALUE

93

The equation of value is given by: Sum of the values of old Obligations} _ {sum of the values of new obligations at focal date - at focal date . . The required equation 2,000 (1.06)2 + 3,000 (1.06)-3 =1,000 (1.06)4 + X (1.06)3 + X => =>

2,247.2 + 2,518.86 = 1,262.48 + 1.19lx + x 2.191 x = 3,503.58 x

= 1,599.01

The equal payment is Rs. 1,599.01.

Example 7 : Mr. X agrees to pay Rs. 800 due in 2 years without interest and Rs. 300 due in 9 years with 6% annual effective rate of interest. He wishes to repay these debts in 2 equal installments due 4 years and 5 years respectively. If money worth 4% converted semiannually, how much should each installment be ?

Solution: Let Rs. x be the amount of equal installments. The focal date is 5 years. r = 4% = 0.04 Interest is compounded semi-annually. i

=~= 0.02.

The old obligations and new obligations are shown in the following table:

Old Obligations Rs. 800 due in 2 years

Value of each at focal date 800 (1.02)3 x 2

Value of each at focal date x (1.02)1 x 2 Rs. x due in 4 years New Obligations

Rs. 300 due in 9 years with 300 (1.06)9 (1.02)4 x 2 Rs. x due in 5 years 6% effective rate interest [i.e. 300 (1.06)9]

x

The equation of value is given by: Sum of the values of old Obligations} _ {sum of the values of new obligations at focal date - at focal date .. The equation is 800 (1.02)6 + 300 (1.06)9 (1.02)-8 = X (1.02)2 + X => => =>

800 (1.12616) + 30 (1.68948) (0.85439) =1.0404x + x 900.93 + 432.59 = 2.0404 x 1,333.52 =2.0404 x

x

=Rs. 653.56

The required amount is Rs. 653.56.

94

FINANCIAL MATHEMAnCS

Example 8 : Mr. X secured two loans from a bank: one for Rs. 8,000 due in 3 years and another one for Rs. 15,000 due in 6 years, both at an interest rate of 10% per annum compounded semi-annually. The bank has agreed to allow the two loans to be consolidated into one loan payable in 5 years at the same interest rate. What amount will Mr. X be required to pay the bank at the end of 5 years?

Solution: Let Rs. x be the amount to be repaid by Mr. X. The focal date is 5 years. r = 10% = 0.10 Interest is compounded semi-annually. . ..

l

r

0.10

='2 =-5- =0.05

The old obligations and the new obligations are shown in the following table:

Old Obligations

Value of each at focal date

Rs. 8,000 due in 3 years @ 8,000 (1.05)6(1.05)4 10% p.a. compounded semiannually i.e. 8,000 (1.05)6

New Obligations Rs. x at the end of5 years

Value.of each at focal date x

Rs. 15,000 due in 6 years @ 15,000 (1.05)12 (1.05)-2 10% p.a. compounded semiannually i.e. 5,000 (1.05)12 The equation of value is given by: Sum of the values of old Obligations} _ {sum of the values of new obligations at focal date - at focal date :. The equation is: 8,000 (1.05)6 (1.05)4 + 15,000 (1.05)12 (1.05)-2 => 8,000 (1.05)10 + 15,000 (1.05)10 = X => 8,000 (1.62889) + 15,000 (1.62889) = x 13,031.12 + 24,433.35 = x x = 37,464.47 Mr. X has to pay Rs. 37,464.47.

=X

Example 9 : A debt of Rs. 1,500 due in 3 years without interest, Rs. 2,000 due in $ years with 8% p.a. compounded half-yearly and Rs. 5,000 due in 7 years with 7% annually effective rate of interest is to be repaid by 3 equal installments due 3 years, 4 years, 5 years, respectively. If the money is worth 6% convertible quarterly, how much should be each installment ?

Solution: Let Rs. x be the amount of"each installments. The focal date is 5 years r = 6% = 0.06

EQUATION OF VALUE

95

Given that interest is compounded quarterly

..

0.6 ,. = 4'r = "4 = O. 0 15

The old objection and new obligations are shown in the following table:

Old Obligations

Value of each at focal date

New Obligations

Value of each at focal date

Rs. 1,500 due in 3 years 1,500 (1.015)2 x 4 without interest

Rs.x due in 3 years

x1.015)2x 4

Rs. 2,000 due in 5 years at 8% 2,000 (1.04)10 compounded half yearly interest [i.e. (2,000 (1.04)10]

Rs. x due in4 years

x (1.015)1 x 4

Rs. 5,000 due in 7 years at 7% 5,000 (1.07)7 (1.015)-2 x 4 Rs. x due in 5 , years effective rate [i.e. 5,000 (,1.07)7

x

The equation of value is given by : Sum of the values of old Obligations} _ {sum of the values of new obligations at focal date - at focal date 1,500 (1.015)8 + 2,000 (1.04)10 + 5,000 (1.07)7 (1.015)-8 =x (1.015)8 + x (1.015)4 + x 1,689.73 + 2,960.48 + 7127.35 = 1.1264x + 1.0613x + x 11,777.56 = 3.1877x 11,777.56 x = 3.1877 x = Rs. 3,694.69 '" . .. The required amount is Rs. 3,694.69

Example 10 : A debt of Rs. 3,000 which is due 6 years from now is instead to be paid off by 3 payments: Rs. 500 now, Rs. 1,500 in 3 years and final payment ofRs. 475 at the end ofn years. The rate of interest is 6% effective. Find the value of n. .

Solution: Let the focal date be today i = r = 6% = 0.06 The old obligations and new obligations are shown in the following table: Old Obligations

Value of each at focal date

Rs. 3,000 due in 6 years

3,000 (1.06~

New Obligations

Value of each at focal date

500 now 1,500 in 3 years 475 in en' years

500 1,500 (1.06)-3 475 (1.06)-n

The equation of value is given by : Sum of the values of old Obligations} _ {sum of the values of new obligations at focal date - at focal date

96

FINANCIAL MATHEMATICS

. . The equation is 3,000 (1.06)-6

=500 + 1,500 (1.06)-3 + 475 (1.06)-'1

=

3,000 x 0.705

=500 + 1,500 x 0.8396 + 475 (1.06)-n

=>

2,115 = 500 + 1,259 + 475 (1.06)-'1 475 (1.06)-n = 356

=

356

=475

(1.06)-'1

=

{1.06)n

=>

(1.06)n

475

=356

= 1.3343

Taking log on both sides n log 1.06 = log 1.3343 n =

log 1.3343 log 1.06 0.1252

n = 0.0253 n

= 4.95 ..

5 years

The value of n is 5 years

1 A man borrowed Rs. 10,000 from a money lender at 8% simple interest and agreed to pay Rs. 5,000 of loan in 6 months. What payment one year from now will settle the debt? [Ans. Rs. 5,600] 2. Mr. X purchased a television for Rs. 6,000. He paid Rs. 500 cash down and agreed to pay the balance at 5% simple interest. If he paid Rs. 3,000 three months after purchase and Rs. 1,500 six months later, what final payment one year after the date of purchase will discharge his obligations? Assume that the focal date is at the end of 12 months. [Ans. Rs. 1,125] 3. Ram borrows Rs. 50,000 now and agrees to repay Rs. 10,000 in 2 months and Rs. 15,000 in 6 months. What final payment should he make at the end of 18 months to settle down his indebtness, at 12% simple interest, assuming that the focal date is today? [Ans. Rs. 31,829.27] 4. At 5% simple interest, find the value of the following obligations : Rs. 2,000 due today, Rs. 5,000 due in 6 months with interest 6% per annum and Rs. 1,000 due in 1 year with interest at 8% per annum. (a) use today as the focal date. (b) use 1 year from today as the focal date. [Ans. (a) Rs. 17,310.10, (b) Rs. 18,178.75]

5. Mr. X owes Rs. 500 due in 2 months, Rs. 1,000 due in 5 months and Rs. 1,500 due in 8 months. He agrees to pay two equal payments, one due in 6 months and the other

EQUATION OF VALUE

97

due in 10 months. Find the payment, if money is. worth at 6% simple interest and at the end of 10 months is the agreed focal date. [Ans. Rs. 1,514.85] 6. X owes Y Rs. 1,000 due in 6 months without interest and Rs. 2,000 with interest for 1 .. 1"2 years at 4% due in 9 months. Yagrees to accept 3 equal payments, one due today, another in 6 months and the third in 1 year. Find the equal payments using 1 year from today as focal date, if money is worth 5% to Y. [Ans. Rs.1,031.38] 7. A debt of Rs. 4,000 due 3 years hence and another Rs. 10,000 due 8 years hence are to be repaired by a single payment 4 years hence. If the rate of interest is 6% per annum effective, how much is this payment? [Ans. Rs. 12,160.94] 8. A debt of Rs. 200 due 2 years hence and another of Rs. 500 due 7 years hence are to be paid off by a single payment 3 years hence. If the rate of interest is 5% per annum effective, how much is this payment? [Ans. Rs. 621.35] 9. A debt ofRs. 10,000 due in 4 years is to be repaid by a payment ofRs. 3,000 now~d a second payment at the end of 6 years. How much should be the second payment, if the rate or interest is 8% compounded quarterly? [Ans.. Rs./ 6,891.28] 10. A debt of Rs. 30,000 which is due 6 years from now is to repaid by three payments: Rs. 5,000 now, Rs. 15,000 in 3 years and a final payment at the end of 5 years. If the interest rate is 6% compounded annually, how much is the final payment? [Ans.4,756.76] 1l A man owes Rs. 10,000 due in 1 year and Rs. 30,000 due in 4 years. He agrees to pay Rs. 20,000 today and the remainder in 2 years. How much must he pay at the end of 2 years if money is worth 5% compounded semi-annually? [Ans. Rs. 15,608.4] 12. A loan of Rs. 50,000 due 5 years from now and Rs. 50,000 due 10 years from now is to be repaid by a payment of Rs. 20,000 in 2 years, a payment of Rs. 40,000 in 4 years and a final payment at the end of 7 years. If the "interest rat~ is 5% compounded [Ans. Rs. 26,486.4] annually, how much is the final payment?

13. Mr. X agrees to pay Rs. 20,000 due in 3 years without interest and Rs. 40,000 due in 6 years with rate of interest 4% per annum compounded semi-annually. He wishes to repay these debts in 2 equal installments at the end of 1 year and 3 years respectively. If money is worth 4% effective, what equal payment Mr. X has to repay? [Ans. Rs. 31,273] 14. A owes B two sums of money: Rs. 1,000 plus interest at 7% compounded annually which is due in 5 years and Rs. 2,000 + interest at 8% compounded semi-annually which is due in 7 years. If both debts are paid off by a single payment at the end of 6 years, find the amount of payment if the money is worth 6% compounded quarterly. [Ans. Rs. 4,751.73] 15. A man owes Rs. 8,000 due in 2 years with out interest and 3,000 due in 9 years with 6% annual effective rate of interest. He wishes to repay these debts in two equal

98

FINANCIAL MATHEMATICS

installments due 4 and 5 years hence, respectively. If money is worth 4% converted semi-annually how much should each installment be ? [~. Rs. 6535.58] 1& A debt of Rs. 10,000 which is due today is to be tePaid by 4 equal yearly payments. If the interest rate is 5% compounded quarterly, how much should be each installment be: if the first installment is given today? (b) . if the first installment is given 1 year from today? [ADs. (a) Rs. 4,029.32, (b) Rs. 3,860.40]

(a)

17. The sum of Rs. 2,000, 3,000 and 4,000 are due at the end· of 2, 4 and 8 years respectively. It is proposed to replace this series of payments by a single sum of Rs. 9,000 payable at the end of en' years. The rate of interest 10% p.a. effective. Find the value of n. [ADs. 5.04 years] 18. Mr. X browses Rs. 50,000 due 5 years from now and Rs. 50,000 due 10 years from now. This obligation are to repaid by the following agreements : a payment of Rs. 20,000 in 2 years, a payment of 40,000 in 4 years and a payment of Rs. 26,500 in en' years. If the interest rate is 5% compounded annually, find the values of n ? [ADs. 7 years]

./

Discount INTRODucnON It is a common practice in business that money lenders require the borrowers eO make a note bearing no interest until after maturity and discount this note immediately, giving the borrov;er only the net proceeds. In other words, amount borrowed is given to the borrower after deducting the discount at a certain rate of discount. Discount plays a major role in' financial transactions. In this chapter, we discuss the method of calculating discount, compare nominal and effective rate of discounts, and establish the relation between rate of interest and rate of discount.

DISCOUNT A discount is ,a deduction allowed on a financial obligation. When the value of an obligation is known at some future data, the proeess of finding its value at some earlier date is known as discounting. The rate of discount is the per cent of the maturity value charged as bank discount for a discount period of unit length of time. One year is generally taken as the unit length of time. A special case of discounting an obligation is that of finding its present value when its maturity value is known. The present value (the money received by . the borrower) is commonly referred to as the net proceeds. Consider the following example. Suppose a man requests a loan of Rs. 300 from a bank, promising to repay the loan in one year. If the bank charges 10% annual discount, the bank deducts Rs. 30 (10% ofRs. 300) from Rs. 300 and pays only Rs. 270 to the borrower, although the borrower is said to have obtained a loan of Rs. 300. The amount Rs. 270 is the present value of Rs. 300 or the net proceeds and Rs. 30 is the bank discount. At the end of the year, the borrower has to repay Rs. 300. Here the borrower is said to pay the interest in advance.

100

FINANCIAL MATHEIIAncS

Thus, the above transaction is described as follows : The borrower receives Rs. 300 but immediately return Rs. 30 to the bank as 10% interest (in advance) on Rs. 30 and repays the loan amount Rs. 300 at the end of the year.

SIMPLE DISCOUNT Simple discount is computed in much the same way ~s simple interest, with the exception that it is based on the amount \,ather than the principal. Let P = proceeds/present value (amount received by the borrower). d = discount rate per year. t = time in years that proceeds will be held. A = the value of obligation (amount to be paid at the end of t years). By Definition, simple discount (D is computed as follows: Simple discount, D =Amount x Discount rate x Time :. Simple discount, D =Adt The prqceeds received by the borrower is given by : P = Amount - Simple discount P = A-Adt

I P = A(l-dt) I

Example 1 : Find the simple discount and the present value of Rs. 2,000 loan for six months at 8%. Solution: Given A =Rs. 2,000, t Simple Discount is given by D

=6 months = 112 years, and d =8% =0.8

= Adt

D = 2,000 x 0.8 x 1/2 D = Rs.80. Present Value is given by P = A (l-dt) P = Rs. 2,000 (l - 0.08 x 1/2) P = 2,000 (l - 0.04) =2,000 (0.96) =Rs. 1,920 P = Rs.l,920 Alternatively, P =A - D .. P = 2,000 - 80 =Rs. 1,920

PRESENT VALUE AT DISCOUNT RATE The present value of an obligation discounted as bank discount is given by

IP

= A(l-d)"

I

DISCOUNT

where

101

= The value of obligation at the end of 'n' = Present Value of obligation. = Discount rate per period.

A

P

period.

d n = Number of periods. Clearly, the discount is given by

I D=A-P\ Example 2 : Find the present value and discount on Rs. 3,000 due' in 4 years at 8% discount rate, discounted annually. Solution: The present vale is given by

= A (I-d)" = Rs. 3,000

P

Here A The sum is discounted annually

..

n

..

P

~

P

= 4 and, d = 8% = 0.08 = 3,000 (1 - 0.8)4 = 3,000 (0.92)4 =3,000 x 0.71639 = 2,149.18

P The preset value is Rs. 2,149.18. The discount is given by D D

= A:"'P = 3,000 -

2,149.18

D = 850.82 .. Discount is Rs. 850.82

Example 3 : Find the present value and discount on Rs. 2,000 due til 4 years at 8% discount rate, convertible half-yearly. Solution: The present value is given by

= A (1 -d)" A = Rs. 2,000

P

Here

Discount'Rate

t

= 8% =0.08

= 4 years

Sum is discounted half-yearly . n = 4 x 2 =8 and d

= 20.08 =o.04

P = 2,000 (1 - 0.04)8 P = 2,000 (0.96)8 =2,000 (0.72139) P

= 1,442.78

102

FINANCIAL MATHEMATICS

Thus; the present value is Rs. 1,442.78. The discount is given by

= A-P

D

Here

D = 2,000 -1,442.78

= 557.22

D

The discount is Rs. 557.22

Example 4 : If the present value of Rs. 6,000 due in 2 years at a certain nominal rate of discount, convertible quarterly is Rs. 5,536.47. Find the rate of discount. Solution: Given A = Rs. 6,000 Let d be the rate of discount. Present value P No. of years

= Rs. 5,536.47 =2

The sum is discounted quarterly ..

n=2x4=8

. d Rate of discount per quarter =4" The present value is given by P = A (l-d)n Here 5,536.4 7 =>

= 6,000 ( 1 -

r

~

r

~ = 0.922745

(1-

Taking log on both sides we get, 8 log ( I, =>

~) = log (0.922745)

d' 8 log ( 1- 4") = -1 + 0.9650 =-0.0350

=>

log ( 1 _ ~) =

=>

log ( 1 -

~) =

1 -

-O.~350 = -0.004375 -1 + 1 - 0.004375 =-1 + 0.995625

~ = antilog (-1 + 0.9956)

1- ~

= 0.9900

103

DISCOUNT

d 4

= 0.01

d = 0.4=4% The rate of discount is 4%.

DISCOUNT CONVERTIBLE CONTINUOUSLY We now derive the formula for the present value of an obligation discountable at a rate of discount d continuously. Suppose that a loan of Rs. A is repayable after t years and the discount is convertible m times a year. Let d be the nominal rate of discount. Then the present value of Rs. A due in t years at the nominal rate of discount d, convertible 'm' times a year is given by d P = A ( 1-;-

}mlCt

, In case of continuous discounting, the present value P is given by

t

P

= !:~~[A (1-!ft]

P

= A[ m-..CIO Lim (1_E£)mt] m

P= !:~ (1 _! tmt ~)] P= A [ !:i~oo (1-! t ilC -4t] A[

P Let

lC ( -

=A[[ ~!D. (1 + (:) ),l"'r]

x

d = -m

asm-co

P

=A [

Lim (1 + x)~

m-O

r

x-O t

P = Ae-d' , Thus, the present value of an obligation of Re. A due in t years at d nominal rate of discount convertible continuously is given by

I P= Ae~1 Example 5 : What is the present value of Rs. 2,000 due after 5 years from now if the discount is convertible continuously at a discount rate of 8%. Solution: The present value of Rs. A due in t years at d rate of discount, convertible continuously is given by P

= Ae-4t

104

FINANCIAL MATHEMA TICS

Here

A P P p

= Rs. 2,000, d = 8% = 0.08 and t=5 .. = 2,000 e-O· 08 x 5 => = 2,000 e-O· 4O => = 2,000 x 0.67032 = 1,340.64 " The present value is Rs. 1,340.64.

Example 6 : At what rate of discount convertible continuously the present value Rs. 3,000 due in 7 years will be Rs. 2,123 ?

Solution: Let d be the rate of discount, convertible continuously. The present value of Rs. A due in t years at d discount convertible continuously is given by P'= A e-dt Here P = Rs. 2,123, A = Rs. 3,000 and t = 7 years .. 2,123 = 3,000 e-7d 2,123 e-7d = 3000 = 0.7077 => , ,

Taking log on both sid~s we get -7 d log e = log 0.7077 d = _ log (0.7077) 7loge

d

= - (-0.1502)

7 x 0.4343 0.152 d = 3.0401 = 0.0494 = 4.94% The rate of d,iscount is 4.94%.

NOMINAL AND EFFECTIVE RATE OF DISCOUNT In situations where discount is converted more frequently than once each year, the stated rate of discouht if) called a nominal rate of discount and the rate of discount actually obtained during the year is called the effective rate of discount. In the later case, the rate of discount is always convertible annually. Consider the following example: The discount on a maturity value of Rs. 100 due in one year at a rate of 8% per annum is Rs. 8 and the proceeds are Rs. 92 (i.e. Rs. 100 - Rs. 8). Suppose that the same rate of discount is convertible semi-annually. Then the present value ofRs. 100 due in one year would be 100 (1- 0.04)2 =92.16. :. The actual discount obtained on this obligation of Rs. 100 at the end of one year is Rs. (100 - 92.16) =Rs. 7.84. We say that the effective rate in this case is 7.84%. Now at 7.84% rate of discount, convertible per annum, the present value we get is P = 100 (1- 0.0784) = 100 (0.9216) = Rs. 92.16.

Thus for a maturity value of Rs. 100 after one year, the net proceeds at a 7.84% rate of

105

DISCOUNT

discount, convertible annually is equivalent to the net proceeds discounted at a rate of 8%, convertible half-yearly. Therefore, we say that the effective rate of discount is 7.84% and the corresponding nominal rate of discount is 8% convertible semi-annually. Let de denote the effective rate of discount corresponding to the nominal rate of discount d, convertible m times a year. Let A be the maturity amount at the end of 1 year. At de effective rate of discount, the present value is given by Pl

= A (l-d e )

... (1)

!r

..

At d nominal rate of discount, convertible m time a year, the present value is given by P2

=A

(1 -

(2)

Equating the equations (1) and (2), we get

!r !r

(1de = 1- (1-

1-de

=

Thus, the relation between the nominal rate of discount and effective rate of discount is given by

Now, we derive the relation between the nominal rate of discount, convertible continuously and the effective rate of discount. In case of nominal rate d convertible continuously, the effective rate de is given by

de

=

!:~ [1- (l-!r]

1- (l-ELr d tx-d -d de = 1- m-oo (1-m de = 1- !:~oo [(l-!r~]-d -d _! )dlm ] de = 1_!:~ [(1

de

=

Lim

m-loOO

m

Lim

-1

~

-d

=> de = 1- e Thus, the effective rate of discount equivalent to the nominal rate of discount, convertible continuously is given by

I de = 1- e-d I

106

FINANCIAL IfATHEIfAncS

Remark 1 : The nominal rate of discount d compounded continuously and equivalent to a given effective rate of discount re is called the force of discount. Remark 2 : If the conversion period is one year at d nominal rate of discount, then the effective rate of discount will be equal to the nominal rate of discount. If the conversion period is one year, then m = 1 de de

1- (1-!r = 1-(1-1f

=

=l-(1-d)=d

.. ' de = d . Remark 3 : The effective rate of discount (de) depends only on the nominal rate of discount (de) and the number of conversion periods in a year (m). It is independent of the maturity value A. Example 7 : Find the effective rate of discount equivalent to the nominal rate of discount 9% converted monthly.

Solution: Here nominal rate of discount d = 9% = 0.09 Discount is converted monthly ..

m

= 12

The effective rate equivalent to the nominal rate of discount d is given by

Here

1- (l-!r

de

=

de

=1-

(1 -

°i~9 f2

de = 1 - (0.9925)12 = 1- 0.9136 = 0.0864 de = 8.64% The effective rate is 8.64%.

Example 8 : Find the effective rate of discount equivalent to the nominal rate 5% convertible continuously.

Solution : The effective rate of discount equivalent to a nominal rate of discount convertible continuously is given by de = 1- e-d d = 5%,= 0.05 Here de = 1 - e-O·05 = 1 - 0.95123 = 0.04877 de = 4.88% The effective rate is 4.88%

107

DISCOUNT

Example 9 : Find the nominal rate of discount, c.onvertible semi-annually equivalent to the effective rate of discount 6%.

Solution : The effective rate of discount de equivalent to a nominal rate of discount is given by de Here

m

0.06 (1 -

1-(1-!r

= =2

and de

=6% =0.06

= 1- (1-~r

~ r = 1 - 0.06 =0.94

Taking log on both sides we get 2 log ( 1 -

~) = log 0.94 =log (9.4 x 10-10 )

~ 2 log (1 - ~) = -1 + 0.9731 =-0.~269 ~ ~

log (1- ~) 1-

= -0.01345 =-1 + 0.98655

~ = antilog (-1 + 0.98655)

1- d2" = 9.6939

x 10-1 =0.96939

d

2" = 0.03061 d = 0.03061 x 2 =0.06121 =6.12%

.. The nominal rate is 6.12%.

Example 10 : What nominal rate of discount convertible continuously, will be equivalent to the effective rate of discount 7% ?

Solution : The effective rate of interest, equivalent to a nominal rate of discount d, convertible continuously is given by de = 1-e-

. . The scrap value is Rs. 3,486.78.

Example 2 : A machine is being depreciated in such a way that the value of the machine at the end of any year is 90% of the value at the beginning of the year. The actual cost of the machine is Rs. 20,000. Calculate scrap value if the machine, if the estimated useful life of the machine is 6 years. Solution: The value of the machine at the end of any year is 90% of the value of the machine at the beginning of the year. That is, 10% of the value is the depreciation for a year. ..

r=10%=0.10

Number of years n = 6 Cost ofthe machine C =Rs. 20,000 The scrap value of an asset worth Rs. Cat r% rate of depreciation at the end ofn years is given by S=C(1-r)n :. Here S = 20,000 (1 - 0.10)6

=20,000 (0.90)6 =20,000 (0.531441) S = 10,628.82

S

.. The scrap value is Rs. 10,628.82.

Example 3. A machine is purchased for Rs. 10,000. Depreciation is calculated at 8% per annum for first 3 years and after that 10% per annum for next 7 years, depreciation being calculated on the diminishing value. Find the value of the machine after 10 years. Solution: Value of an asset worth C at the end of nl + n2 years with depreciation rates of rl% for nl years and r2% for n2 years is given by S = C (1- rl)

Here

nl

nl

(1 - r2)n2

= 3 years and

n2

= 7 years

117

DEPRECIATION

and rl = 8% = 0.08 and r2 = 10% = 0.10 The cot of the machine C =Rs. 10,000 .. S = 10,000 (1- 0.08)3 (1- 0.10)7 S = 10,000 (0.92)3 (0.90)7

= 10,000 (0.778688) (0.4782969) S = 3,724.44

S

.. Value of the machine after 10 years is Rs. 3,724.44.

Example 4. A machine costing Rs. 80,000 depreciates at a constant rate of 8% per annum. The life of the machine is estimated to be 15 years. (a) What is the value of the machine after 7 years ? (b) What is the depreciation charge for 10th year?

Solution:

(a) Value of the machine after n years is given by S

= C (1- r)lI

Rate of Depreciation r = 8% = 0.08 Cot of the machine C = Rs.80,000 n = 7 years S = 80,000 (1- 0.08)7 S = 80,000 (0.92)1 =80,000 (0.5578466) S = 44627.73 .. The value of the machine after 7 years is Rs. 44,627.73.

=>

(b) Depreciation charge for kth year =Value of the machine at the end of(k _l)th yearValue of the machine at the end of kth year.

Here

k

= 10 years

.'. Depreciation charge for 10th year = Value of the machine at the end of 9th year - Value of the machine at the end of 10th year. Value of the machine at the end of 9th year =80,000 (1- 0.08)9 =80,000 (0.92)9

=80,000 (0.47216136) =37,772.91 Value ofthe machine at the end of the lOthyear =80,000 (1- 0.08)10 =80,000 (0.92)10 = 80,000 (0.43438845) =34,751.08 :. The depreciation charge for 10th year = 37,772.91- 34,751.08 = 3,021.83

The required depreciation charge is Rs. 3,021.83.

118

FINANCIAL AlA THEMA TICS

Example 5. A machine depreciates at the rate of 8% of its value at the beginning of a year. The machine was purchased for Rs. 1,00,000 and the scrap realised when sold was Rs. 43,440. Find the number of years the machine was used.

Solution: Let n be the number of years the machine was used. The scrap value at the end of the life of an asset is given by S =C(l-r)" Here scrap value S = Rs. 43,440 Cost of the machine C = Rs.1,OO,OOO Rate of depreciation r = 8% =0.08 Substituting the given values we get 43,440 = 1,00,000 (1 - 0.08)" 43,440 ~ (0.92)" = 1,00,000 = 0.4344 Taking log on both sides, we get n log 0.92 = log 0.4344 log 0.4344 ~ n= log 0.92

n

= 1.63789 =-0.36211 =10 1.96379

-0.03621

n = 10 years . . The life of the machine is 10 years. Example 6. A machine' depreciates at the rate of 7% of its value at the beginning 01 a year. If the machine was purcha3ed for Rs. 8,500, what is the minimum number of years at' the end of which the worth of the machine will be less than or equal to half of its original cost price?

Solution: The depreciated value is given by S

= C(1-r)"

Cost of the machine C ~ Rs. 8,500 Rate of depreciation r = 7% =0.07 According to the given condition, 8,500 ~ ~

'1

(1 - 0.07)n

:SO

.2 (8,500)

0.07)"

:SO

'2

(1 -

1

(0.93)" :SO 0.5 Taking log on both sides we get n log (0.93) :SO log 0.5

DEPRECIATION

119 =>

log (0.5) n ~ log (0.93)

=>

n ~

1.69897

_

1.96848

-0.30103

= -0 03152 = 9.6 years

.

The minimum number of years =9.6 years.

"

Example 7. A machine costing Rs. 5,600 will depreciate to a scrap value of Rs. 1,951 in 10 years. Given that the depreciation is calculated using diminishing balance method, find the rate of annual depreciation. Solution: Let r be rate of annual depreciation. The scrap value is given by S = C (1-rY' Here, Scrap value S = Rs. 1,,951 Cost of machine C = Rs. 5,600 No. of years n = 10 Substituting these value in the formula, we get 1951 = 5,600 (1- r)10 (1 -

=>

r )10

= !::~~ =0.3484

Taking log on both sides we get 10 log (1- r) =log 0.3484 _ ) _ log (0.3484) I og (1 r => 10 log (1-

=>

rr = I.5:~08 = -O'i~792 = -0.045792 = 1.954208

=>

(1- r)

=>

1- r

=>

= antilog (1.954208)

= 0.8999

r = 1 - 0.8999 = 0.1001 = 10% The rate of depreciation is 10%.

Example 8. A machine depreciates at the ratio of 8% per annum for the first 3 years and then 6% per annum for the next four years. [fthe value of the machine is Rs. 8,000 initial:y, lind the average rate of depreciation and the depreciated value of the machine at the end of the 7th year. Solution: The scrap value at two different ratio of depreciation is given by Here,

S

= C (1 -

rl

= 8% = 0.08 and n1 = 3 = 6% = 0.06 and n2 = 4

r2

r1)nl (1 - r2)n

cost of machine C =Rs. 8,000

120

FINANCIAL MATHEMAnCS

S = 8,000 (1 - 0.08)3 (1 - 0.06)4 S = 8,000 (0.92)3 (0.94)4 S = 8,000 (0.78688) (0.780749) = 4,863.68 The depreciate value of the machine is Rs. 4,863.68. Let r be the average rate of depreciation. At r% rate of depreciation Rs. 8,000 depreciate to Rs. 4,863.68 is 7 years . . 8,000 (1- r)7 = 4,863.68 - 0 6079 (1 - r )7 -- 4,863.68 8,000 - . 6 Taking log on both sides 7 log (1- r)

= 1.78390 =-0.2161

log (1 - r) =

- 0.2161 7 = -0.03087 = 1.96913

= antilog (1.96913) 1 - r = 0.93132 r = 1 - 0.93132 = 0.06868 = 6.87% The average rate is 6.87%. 1- r

Example 9. A machine depreciates each year by 10% of its value at the beginning of the year. At the end of 4th year, its value is Rs. 1,31,220. Find its original value.

Solution: Let C be the original value. The scrap value is given y S :-: C (1-rY'

Depreciation rate r

= Rs.1,31,200 = ~.O% =0.10

No. of years

=4

Here Scrap value

1,31,220 1,31,220

=

C

S n.

= C (1- 0.1)4 = C (0.90)4 =

1,31,20 1,31,22 (0.90)4 = 0.6561 = Rs. 2,00,000.

. . Original value of the machine is Rs. 2,00,000. Example 10. XYZ Ltd. purchases a machinery which costs Rs. 20,000 now. The useful life of the machine is 9 years and the annual rate of depreciation is 6% per annum, depreciation being L'alculated on diminishing balance method. After 8 years the existing machine has to be replac~d by a new one which will cost 25% more than the initial cost of the machine. What amount the company will require at the end of 9th year to replace the existing machine by a new one ?

DEPRECIATION

121

Solution: Cost of machine C Life of machine n Rate of depreciation r The scrap value is given by

= Rs. 20,000 = 9 years

= 6% =0.06

8 =C (1- rY' 8 =20,000 (1 - 0.06)9 8 = 20,000 (0.94)9 8 = 20,000 (0.5729948) = 11,460

. . The value of the machine at the end of 9th year is Rs. 11,460. The cost of new machine at the end of 9th year = Initial cost + 25% of the initial cost

=20,000 + 25% of 20,000 =20,000 + 5,000 =Rs. 25,000 The amount required to rePlace} the old machine

The cost of

= new machine

} -

{scrap value of the old machine

:. The required amount is Rs. 13,540 =Rs. 25,000 - Rs. 11,460 =Rs. 13,540.

STRAIGHT LINE METHOD In the diminishing balance method, the depreciation for one year will not be the same as that of the previous year. The depreciation was calculated on constant percentage. By straight line method, the annual depreciation of an asset is found by dividing the total depreciation by the number of years in its estimated useful life. Thus, by straight line method depreciation per year is calculated as follows:

C-8

D=-n

In this method, the depreciation for every year will be unique through out the life of the asset.

Example 11. A machine costing Rs. 30,000 is expected to have a useful life of 5 years and a final scrap value of Rs. 10,000. Using straight line method, find the annual depreciation

and construct the depreciation schedule.

Solution: Using straight line method the annual depreciation is given by: C -8 D=-n C = Rs. 30,000,8 =Rs. 10,000 and n = 5 years Here

D =3,000 ~ 10,000 =4,000 .. The annual ~epreciation is Rs. 4,000. The book value at the beginning of the 1st yp.al' is P'i;. 30,000. Depreciation for 1st year is Rs.4,000.

122

FINANCIAL MA THEMA TICS

.. The book value at the beginning of the 2nd year is Rs. 26,000 (Rs. 30,000 - Rs. 4,000). Depreciation for second year is Rs. 4,000. :. The book value at the beginning 3rd year is Rs. 22,000 (Rs. 26,000 - Rs. 4,000) In this manner, we calculate the book value of the machine at the beginning of every year. These are shown in the following table:

Year

Bank value at the beginning of the year

Depreciation for that year

Accumulated Depreciation

1 2 3

Rs.30,000 Rs.26,000 Rs.22,000 Rs.18,000 Rs.14,000

Rs.4,000 Rs.4,000 Rs.4,000 Rs.4,000 Rs.4,000

Rs. 4,000 Rs. 8,000 Rs.12,000 Rs.16,000 Rs.20,000

4

5

Value of machine at the end of the year Rs. Rs. Rs. Rs. Rs.

26,000 22,000 18,000 14,000 10,000

Example 12. A machine costing Rs. 50,000 has a useful life of 4 years and the machine has no scrap value at the end of its life. Using the straight line method, find the annual depreciation.

Solution: Using straight line method, the annual depreciation is given by

c-s

D=--

n

Hence C =Rs. 50,000, S =0 and n .. ..

D ~he

= 50,0~0 -

=4 years

0 50,~00

= 12,500

annual depreciation is Rs. 12,500.

SUM-OF-THE-YEARS-DIGITS METHOD By this method, a greater fraction of the cost of the asset is depreciated in earlier years of the life of the asset. The fraction of the asset to be depreciated each year is determined by putting the digit of the year in reverse order over the sum of the digits of the life period. For Example if the life of the asset is 4 years, then the depreciation for each year is calculated in the ratio 4 : 3 : 2 : 1. In other words, the depreciation is calculated in the order ;0 ';0 :0 and ;0'

Example 13. A machine which costs Rs. 20,000 is expected to have a useful life of5 years and a scrap value of Rs. 5,000. Find the annual depreciation and form a depreciation schedule using sum-of-the-years-digits method.

Solution: Cost of the machine C =Rs. 20,000 Scrap value No. of years

S =Rs. 5,000 n=5

DEPRECIA TlON

123

. . Total depreciation for five years = Rs. 20,000 - Rs. 5,000 = Rs. 15,000. The annual depreciation is calculate in the ratio 5: 4 : 3 : 2 : 1 "th d 5 4 3 21 £.e. In e or er 15' 15' 15' 15' 15 Now we prepare the depreciation schedule

Years 1 2 3 4 5

Fraction ofASBets to be depreciated

Annual depreciation

5 15 -4 15 -3 15 -2 15 -1 15

5 15 x 15,000 =Rs. 5,000

Rs. 5,000

4 15 x 15,000 =Rs. 4,000

Rs. 9,000

3 15 x 15,000 =Rs. 3,000

Rs.12,000

2 15 x 15,000 =Rs. 2,000.

Rs.14,000

1 15 x 15,000 =Rs. 1,000

Rs.15,000

-

Accumulated depreciation

L A machine, the life of which is estimated to be 15 years, costs Rs. 40,000. Calculate the scrap value at the end of its life, depreciation on the diminishing balance system [ADs. Rs. 8,236] being calculated at 10% per annum. 2. A machine costing Rs. 5,000 depreciates at a constant rate of 5%. If the estimated [ADs. Rs. 2,316] useful life of the machine is 15 years, determine its scrap value. 3. A machine, the life of which is estimated at 10 years, costs Rs. 6,000. Calculate the scrap value at the end of its life, depreciation on the reducing balance system being [ADs. Rs. 2,904] charged 7% per annum. 4. A machine is be.ing depreciated in such a way that the value of the machine at the end of any year is 95% of the value at the beginning of the year. The actual cost of the machine is Rs. 15,000. Calculate the scrap value of the machine, if the estimated [Ans. Rs. 9,951.31] useful life of the machine is 8 years. 5. An asset is purchased for Rs. 10,000. It is depreciated at a constant rate of 6% for the first 4 years and after that 10% for the next 6 years. Find the value of the asset after [Ans. Rs. 4,149.76] a period of 10 years. 6. A machine depreciates at the rate of 10% pel' annum for the first two years and then 7% per annum fo. the next three years, depreciation being calculated on the diminishing value. If the value of the machine is Rs. 10,000 initially, find the depreciated value of the machine at the end of the 5th year? [ADs. Rs. 6,515.60] 7. An item costing Rs. 50,000 depreciates at a constant rate of 8% per annum. The useful life of the machine is 10 years. What is the depreciation charge for the 8th year? [Ans. Rs. 2,231.39]

124

FINANCIAL MATHEMATICS

8. A machine costing Rs. 30,000 depreciates at a constant rate of 12% per annum, depreciation being calculate on the diminishing value. What is the value of the machine after 5 years? (b) What is the depreciation charge for the 7th year? [Ans. Rs. 15,832, (b) Rs. 1,671.86] 9. A machine costing Rs. 5,000 depreciates at a constant rate of 5% per annum. What is the depreciation charge for the 5th year? [Ans. Rs. 203.50] (a)

10. An item being depreciated in such a way that the value of the item at the end of any year is 90% of the value at the beginning of the year. The cost of the item is Rs. 40,000 and it was sold eventually as waste material for Rs. 15,000 at the end of its life. Obtain the number of years the item was in use. [Ans. 9.3 years] 1L A machine depreciates at the 10% of its value at the beginning of the year. The machine was purchased for Rs. 44,000 and the scrap value realised when sold was Rs. 25,981.56. Find the number of years the machine was used. [Ans. 5 years] 12. A machine depreciated at the rate of 8% of its value at the beginning of a year. If the machine was purchased for Rs. 15,000, what is the minimum number of complete years at the end of which the worth of the machine will not exceed (215)th of its original value. [Ans. 11 years] 13. A machine depreciates at the rate of 10% per annum rate of depreciation. If the purchasing price of the machine is Rs. 10,000, what will be the minimum number of completed years at the end of which the worth of the machine will be less than or equal to quarter of its original cost price? [Ans. 13.16 years] 14. A machine worth Rs. 12,000 is depreciated at the rate of 10% per annum. It was sold eventually as waste metal for Rs. 200. Find the number of years during which the machine was in use. [Ans. 38.84 years] 15. An article, the life of which is estimated to be 10 years, costs Rs. 10,000. The scrap value realised at the end of its life is Rs. 3,483.37. If the depreciation is calculate on the diminishing balance method, what is the rate of annual depreciation? [Ans.10%] 16. A machine costing Rs. 8,000 would reduce to Rs. 2,000 is 8 years. Find the rate of

yearly depreciation, given that the depreciation is calculated using diminishing balance method. [Ans. 15.91%] 17. An asset costing Rs. 2,000 will depreciate to a scrap value of Rs. 160 in 10 years. Find the rate of depreciation. [Ans. 22.33%] 18. A machine depreciates at the rate of 10% per annum for the first two years and then 7% per annum for the next three years, depreciation being calculated on diminishing value. If the value of the machine be Rs. 10,000 initially, find the average rate of depreciation and the depreciate value of the machine at the end of 5 th year. [Ans. Rs. 6,515.29, 8.2%] 19. What the average rate of depreciation equivalent to 9% annual rate of depreciation for 3 years and 7% annual rate of depreciation for next 2 years, if the depreciation is calculated for 5 years. [Ans.8.22%]

DEPRECIATION

125

20. A machine depreciates each year by 10% of its value at the beginning of the year. At the end of 2nd year, its value is Rs. 5,536.47. Find its original value. [Ans. Rs. 6,000] 21. The value of a machine depreciates at the rate of 11 % annually. If its present value [Ans. Rs. 54,527.22] is Rs. 38,440, find its value three years age. 22. A machine costing Rs. 60,000 has a useful life of 5 years. The scrap value is Rs. 20,000. Using straight line method, find the annual depreciation and construct a schedule for depreciation. [Ans. Rs. 8,000] 23. An asset costing Rs. 12,000 is expected to have a useful life of 6 years and a scrap value ofRs, 3,000. Find the annual depreciation charge using straight line method. [Ans. Rs. 1,500] 24. A machine costing Rs. 10,000 is expected to have a useful life of 5 years. It is assumed that the scrap value is nil. Using straight line method, find the annual depreciation charge. • [Ans. Rs. 2,000] 25. A machine costing Rs. 25,000 is expected to have a useful life of 4 years. It is assumed that the scrap value at the end of 4th year is Rs. 5,000. find the annual depreciation and prepare the depreciation schedule using sum-of the years-digit method. [Ans. Rs. 8,00, Rs. 6,000, Rs. 4,000 and Rs. 2,000] 26. An asset costs Rs. 3,000. The useful life of the asset is 5 years and there is no scrap value. Find the annual depreciation and prepare a depreciation schedule using sumof-the-years-digit method. [Ans. Rs. 1,000, Rs. 800, Rs. 600, Rs. 400, Rs. 200] 27. A computer whose cost is Rs. 4,40,000 will depreciate to a scrap value of Rs. 24,000 in 5 years. (a) If the reducing balance method of depreciation is used, find the depreciation rate. What is the book value of the computer at the end of the third year? (c) How much more would the book value be at the end ofthe third year, if straight line method of depreciation has been used ? [Ans. (a) 44.11%, (b) Rs. 76,816.59, (c) Rs. 1,13,583]

(b)

28. A company buys a computer for Rs. 1,25,000 and houses it in a specially constructed suite at a cost ofRs. 20,000. (a) If the computer depreciates at 25% (reducing balance) and the suite appreciates 5% compound, what is the book value of the suite and the computer after 5 years, (b) Taking computer and suite together and using the reducing balance method, what is the overall depreciation rate? [Ans. (a) Rs. 55,188.71, (b) 17.57%] 29. ABC Ltd. purchases a machine which cost Rs. 12,000 now. The useful life of the machine is 10 years. The rate of depreciation is 6%, depreciation being calculate on diminishing balance method. What is the scrap value of the machine at the end of its life? If after 10 years, the existing machine has to be replaced by a new machine which cost 20% more than the cost of the old machine, what amount will be required at the end of the 10th year to replace the old machine by a new one? [Ans.7,936.62]

Bills of Exchange INTRODUCTION Companies due to competition in the market sell their goods on credit and give some time to the buyer to pay the amount before the expiry of some specified time period. This period is known as credit period. To ensure that the seller or creditor will receive his amount on or before the expiry of credit period, he writes a formal instruction to purchaser or debtor to make the payment of the certain specified amount on or before the expiry of credit period to him or to· anyone else according to his instruction. This formal document is known as the Hll of exchange. In this chapter, we discuss the method of discounting a bill of exchange.

BILL OF EXCHANGE A bill of exchange is defined as follows : "A bill of exchange is a written document or undertaking by the debtor to the creditor for paying a certain sum of money on a specified future date." In otherwords, a bill of exchange is a negotiable instrument by which one person undertakes to pay another person a certain sum of money at a future date. Bill of exchange is always drawn by the creditor ordering the debtor to pay the amount specified in it at the end of a specified period. A bill of exchange contains three parties viz. drawer, drawee and payee. The drawer is the person who writes or prepares the bill. The creditor is the drawer of the bill. The drawee is the person who has to pay amount specified in the bill. The debtor is the drawee of the bill. The payee is the person to whom the payment is to be made. In many cases, the drawer and payee are the same. The drawer or the payee who is in possession of the bill is called the polder of the bill.

BILLS OF EXCHANGE

127

Bills are generally prepared by the drawer and sent to the drawee for his acceptance. If the bill is accepted, the drawee writes the word accepted on the bill, with his signature and date. When such an acceptance is written on the bill, it becomes a bill of exchange. Cheque, bank draft and etc., are examples of bills of exchange. A cheque is a bill of exchange drawn on a banker and payable on demand. A bank draft is a bill of exchange drawn by one bank on another and payable on demand. The place of payment should be specified 0 the bill of exchange. If it is not mentioned, then payment should be made at the address of the drawer written in the bill. A bill of exchange must be written on a stamped paper of the court. If it is written in an ordinary paper, it should bear stamps. Stamps are fixed according to the amount of the bill. The fixed period after which the payment of the bill is made is called the term of the bill. If the payment of a bill is to be made after 3 months, we say that the term of the bill is 3 months. DUE DATE AND LEGALLY DUE DATE

The date of writing the bill must be clearly mentioned on the bill, because due date is to be calculated on the basis of this date. The date on which the payment of the bill is due is called the date of maturity or due date. It should be noted that the date of acceptance of a bill may not be the same as the date on which the bill is drawn or prepared.

There are two kinds of bill of exchange. Bill of exchange after date (b) Bill of exchange after right.

(a)

In bill of 3xchange after date, the date of maturity is counted from the date of drawn of the bill. On the other hand, in case of bill of exchange after sight, the date of maturity is counted from the date of acceptance of the bill. Three days are generally added to the due date or maturity date to get the legally due date. These three days are called days of grace. For example, consider a bill drawn on 27th November 2001 for a term of 3 months. We have 3 months from 27 th November 2001 and 3 days of grace. Therefore the maturity date is 27th February 2002. The legally due date is 2nd March, 2002. BANKER'S DISCOUNT AND TRUE DISCOUNT

The drawer or the payee of a bill of exchange is entitled to receive the payment after a fixed period of time. In otherwords, he has to wait for the payment till the bill becomes legally due. If he needs the money immediately, he can get the bill discounted from a bank or can endorse it to the third party to settle his dues. Discountmg a bill of exchange means encashing the bill before the due date form a bank. For discountil1g the bill, the drawer or the holder of the bill transfers the possession and also the ownership ofthe bill to the bank.

128

FINANCIAL MATHEMA TICS

If the drawer or holder of a bill presents the bill for cash payment to a banker or a bill broker, the banker will deduct some amount form the bill. In fact, the banker charges certain rate of interest on the sum due. This interest is known as banker's discount. Thus Banker's Discount is defined as follows: "The interest on the bill value or the face value (i.e. the amount mentioned in the bill)." If a bill of face value of Rs. P is discounted at the rate of r per annum for t years, then the banker's discount at simple interest is the interest on Rs. P at the rate of r per annum for t years. Thus the banker's discount is given by

I Banker's Discount =Prt I Note: 't' is the time period between the date of discounting and the legally due date. If this time period is in months, weeks, or days it should be covered into years. True discount is defined as follows: "True discount is the interest on the present value of a bill of exchange." If a bill of face value Rs. P is discounted at the rate of r per annum for t years, then the P

interest on its present value V = 1 + rt is the true discount. .. True discount

=Interest on V at the rate of r per annum for t years

vrt

. d· TT Prt .. True Iscount = = 1 + rt Thus the true discount is given by t · T rue D Iscoun

= 1Prt + rt

The difference between the banker's discount and the true discount is called banker's gain. Thus the bankers gain is given by Banker's gain = Prt -lP:~t Remark:

Prt Banker gain =Pvt- - -

1+ rt

=Prt

[1 __ 1] 1 + rt

=Prt[~] 1 + rt =

(~)rt .1 + rt

. . Banker's gain =(True discount) rt

BILLS OF EXCHANGE

129

Thus Banker's gain is the interest on true discount. It is also obvious that the banker's gain is greater than the true discount.

Example 1: Find the banker's discount and true discount on a bill of Rs. 20,000 due 4 months at 5% per annum.

Solution: Present· Value of AmountP = Rs. 20,000 No. of Years t = :2 = 0.3333 years Rate ofInterest r = 5% = 0.05 Banker's Dis.count = Prt = 20,000 x 0.05 x 0.3333 = 333.30

:. Banker's Discount is Rs. 333.30 True Dibcount is given by

· Prt True Olscount =l+rl .

333.30

333.30

True 01Rcount =.1 + 0.01667 = 1.01667 = 327.83 :. True Discount is Rs. 327.83. The Banker's Discount is Rs. 333.30 and The True Discount is Rs. 327.83.

Example 2 : Mr. X received from Mr. Y acceptance for Rs. 8,000 on 1st June 2007 for 3 months. He got the acceptance discounted at 6% per annum at his bank after one month. How much amount was received by him from the bank after discounting the bill ?

Solution: Present Value of Amount P = Rs. 8,000 (3 -1)

2

No. of Years t =}2= 12 = 0.167 years Rate of interest r =6% = 0.06 Banker's Discount is given by Bank's Discount = Prt

=8,000 x 0.06 x 0.167 =80.16 . . Banker's Discount is Rs. 80.16 Amount Received by X after discount = Rs. (8,000 - 80.16) =Rs. 7,919.84 The Net Amount Received by X after discount is Rs., 7,919.84.

Example 3 : Find the True Discount and Banker's gain on a bill for Rs. 2,550 due in 4 months hence at 7% per annum. .

Solution: Present Value of Amount P = Rs. 2,550

130

FINANCIAL MATHEMAncS

No. of Years t

4 =12 =0.3333 years

Rate of Interest r = 7% = 0.07 Ban~er's Discount is given by Banker's Discount =Prt =2,550 x 0.07 x 0.3333 =59.49 Banker's Discount is Rs. 59.49 Now, True Discount is given by Prt True Discount 1 + rt _ 2,550 x 0.07 x 0.3333 - 1 + 0.07 x 0.3333 59.49 -1 + 0.0233

=58.13 True Discount is Rs. 58.13. Now, Banker's gain is given by , Bank's gain =Banker's Discount - True Discount = Rs. (59.49 - 58.13) =Rs. 1.36 The Banker's gain is Rs. 1.36.

Example 4: A bill of Rs. 45,000 drawn on March 27,2008 at 8 months was discounted on August 18, 2008 at 6% per annum. How much did the banker charge and what did the holder receive ?

Solution : Date of Drawing the bill =March 27, 2008 Duration or term of the bill =8 months Due Date is give by = 27 March 2008 + 8 months + 3 days of Grace =27 November 2008 + 3 days of Grace = 30 November, 2008 Date of Discounting the bill is August 18, 2008 Now, Number of days form the date of discounting to the due date is calculate as follows: ~3 days + 30 days + 31 days + 30 days =104 days [

For AUgust] [For September] [For October] [For November] Month Month Month Month

Present Value of Amount P

=Rs. 45,000

=104 da

ys

BILLS OF EXCHANGE

131

No. of Years t

104

=365 = 0.285 years

Rate of Interest r = 6% = 0.06 Banker's discount is given by Banker's discount

=Prt = 45,000 x 0.06 x 0.285 =769.50

Banker'oS Discount is Rs. 769.50 Discounted value of the bill is given by Discounted value of the bill

=Present Value of Amount -

Banker's Discount

=P -Prt =Rs. (45,000 -769.50) =Rs. 44,230.50 Discounted value of the Bill is Rs. 44,230.50.

Example 5 : The Bankers gain on a bill due 6 months hence at 8% per annum is Rs. 50. Find the face value of the bill.

Solution : Let the face value of the bill be Rs: P. 6 =12 =0.5 years Rate of Interest r =8% = 0.08 Banker's gain =Rs. 50 No. of Years t

Banker's gain is given by Banker's gain = Banker's Discount - True Discount P Banker's gain = Prt -1 rt

2

Now, substituting the values, we get P x 0.08 x 0.5 P x 0.08 x 0.5 - 1 + 0.08 x 0.5 0.04 P -

=50

°i~~: =50

0.0416P - 0.04 P = 52 0.0016P = 52 P = 32,500 Hence, the face value of the bill is Rs. 32,500.

Example 6: The difference between the True and bankers discount on a bill due after 4 months at 7% interest is Rs. 500. Find (i) The True Discount

132

FINANCIAL MATHEMATICS

(ii) The Banker's Discount (iii) The face value of the bill.

Solution: Let Rs. P be the face value of the bill. 4 =12 =0.3333 years Rate of interest r =7% =0.07

No. of Years t

. . Banker's Discount is given by Banker's Discount = Prt

=P

x

0.07

x

0.3333

=0.023331P

... (1)

Now, True Discount is given by

Prt True Discount = - 1 + rt P x 0.07 x 0.3333 =1 + 0.07 x 0.3333 0.023331P

= 1.023331 Now, Given that

Banker's discount - True Discount =Rs. 500 0.023331P So, 0.023331P - 1.023331 == 500 0.023875335P - 0.023331P = 511.67 ~ 0.000544335P = 511.67 ~ P =9,39,991 ., The face value of the bill is Rs. 9,39,991. Now putting P = Rs. 9,39,991 in eq. (i) and (ii)

~

Banker's discount is given by Banker's discount = Prt =9,39,991 x 0.07 x 0.3333 =21,930.93 :. The Banker's Discount is Rs. 21,930.93 True Discount is given by · True D Iscount

= 1Prt + rt

0.023331 x 939991 1.023331 =21,430.92 . . True Discount is Rs. 21,430.92. =

... (2)

133

BILLS OF EXCHANGE

Example 7: A bill of exchange drawn on February 4, 2004 a.t 4 months after date was discounted on March, 26, 2004 at 8% per annum. If the banker's discount is Rs. 400, find the face value of the bill . Solution: Let the face value of the bill be Rs. P Rate of Interest r =8% =0.08 Now, nate of Drawing the bill is February 4,2004 Tenn of the bill =4 months Legally due date of the bill =7th June, 2004 nate of Discounting the bill =March 26, 2004 Number of days from date of discounting to the due date as follows: =5 days + 30 days + 31 days + 7 days =73 days [March] [April] [May] [June] Thus, we get 73 1 No. of years t 365 ="5 =0.2 year

=

Banker's Discount is given by Banker's Discount = Prt =P )( 0.2 )( 0.08 =0.016P Given that Banker's Discount is Rs. 400. .. 0.016P = 400 400 P = 0.016 =2,500 Hence, the face vahle of the bill is Rs. 2,500.

Example 8: The banker's discount calculated for one year is 26 times his gain. Find the rate of interest. Solution : Let the rate of interest be r per rupee per annum and the face value of the bill beRs.P. . Banker's Discount is given by Banker's Discount =Prt . . Banker's Discount for. one year =Pr Banker's gain =Prt -

1~~

Banker's gain for one year =Pr -

t:

r

It is given that for one year, Banker's Discount is 26 times the banker's gain.

..

Pr =·26

(pr- /: r)

134

FINANCIAL MATHEMATICS

1 = 26 (1 - 1 1+r 25r

~r)

= 26r =1 1

r = 25 = 0.04

Hence, the required rate oflnterest =

(i5

'x 100

)% =4%

:. The required Rate of interest is 4% per annum.

Example 9: A bill of Rs. 1,500 drawn on May 3 for 6 months was discounted on August 25 for a cash payment of Rs. 1,479, find the rate of interest charged by the bank.

Solution.: Bill discounted for a cash payment ofRs. 1,479. Date of drawing the bill is May 3. Term of the bill is 6 months. Due date of the bill is November 6. Date of discounting is August 29. .. Number of -days from the date of discounting (i.e. August 29) to due date (i.e. November 10) calculated as follows: =

6 days [ A!:st]

+

30 days

+ 31 days +

6 days

= 73 days

[sep~:b~r] [o:~er] [No!~ber]

Present value of amount P = Rs. 1,500 73 No. of years t = 365 = 0.2 yF..ars. Let the interest be r per rupee per annum. Banker's Discount is given by :. Banker's discount =Prt = 1,500 x r x 0.2 =300r It is given that the bill was discounted for a cash payment of Rs. 1,479. .. Banker's Discount = Rs. [1,500 -1,479] =Rs. 21 From (i) and (ii), we get 300r = 21 21 r = 300 = 0.07 r

= 7%

So, the rate of Interest is charged by the Bank is 7% per annum.

... (i)

... (ii)

135

BILLS OF EXCHANGE

Example 10 : A bill for Rs. 21,900 drawn on 10 July, 2006 for 6 months was discounted for Rs. 21,720 at 5% per annum. On what date was the bill discounted? Solution : Rate of interest r =5% =0.05

Face value of the billP =Rs. 21,900 Discount value of the bill =Rs. 21,720 Banker's Discount Face value of the bill- Discount value of the bill = Rs. (21,900 - 21,720) = Rs. 180 Date of drawings the bill is July 10, 2006 Duration of the bill = 6 months :. Due Date of the bill is January 13, 2007 Let the bill be discounted 't' years prior to its due date at 5% per annum. Banker Discount is given by Banker's Discount =Pre Rate of InterelSt r =5% =0.05 :. Banker's Discount =21,900 )( 0.05 )( t 1,095t From (i) and (ii), we get l,095t = 180 180 t = 1,095 years

=

=

t t

...(i)

... (ii)

180

=1095 )( 364 days =60 days

Thus, the bill was discounted 60 days prior to its legally due date i.e. 13 January, 2007. Hence, the bill was discounted on 14th November 2006.

Example 11: [fthe banker's gain on a bill is

(~)th of the banker's discount; t.l,.e rate of

interest being 10% per annum (simple), find the expired period of the bill.

Solution : Let 't' years be the expired period of the bill and Rs. P be the face valu~ ''If the bill. Rate ofInterest r = 10% = 0.10 Banker's discount is given by Banker's Discount =Pre =Pt)( 0.10 Now, True Discount is given by True Discount

... (i)

=1~~t Pt)( 0.10

=1 + t )( 0.10

... (ii)

136

FINANCIAL MATHEMATICS

Now, Banker's gain .is given by Banker's gain =Banker's Discount - True Discount Pt x 0.10 =Pt x 0.10 - 1 + t x 0.10

=

(Pt x 0.10) x (1 + t x 0.10) - (Pt x 0.10) (1 + t x 0.10)

.. .(iii)

It is given that

Banker's gain =~ (Banker's discount) ~

(Pt x 0.10) x (1 + t x 0.10) - (Pt x 0.10) _! p' 010 (1 + t x 0.10) - 9x tx .

~

8 (1 + 0.1t) = 9

~ t

10 ='8 =1.25 year

Hence, the expired period of the bill is 1.25 year.

Example 12: A man holds a bill for Rs. 9,600 which is due for payment.after 9 months. After 4 months, however, he sells the bill to a broker who charges 5% per annum. The man then invests the discounted value of the bill in a security where rate of Interest is such that he does not suffer any loss on discounting the bill, find the rate of interest percent per annum of the security. Solution : Since the bill is' due for payment after 9 months and the bill is discotmted after 4 months, the bill has 5 months to run for maturity. Banker's discount is given by Banker's Discount =Prt

=(9,600 x 0.05 x 0.4167) ,

[.; P

:2

=Rs. 9,600, t =

years, r

=1~0 ]

=200.02 . . Banker's Discount ;~ Rs. 200.02 Thus, the Discounted value of the bill =Rs. (9,600 - 200.02) =Rs. 9,399.98 Let this discounted value -qfthe bill be invested in security at r per annum. Interest earned on security for 5 months =9,399.98 x r x 0.4167 Since the man does not suffer any loss in discounting the bill and then investing the discounted value in the security. Therefore, Interest on discounted value in the security =Banker's Discount ~ 9,399.98 x r x 0.4167 = 200.02 200.02 r = 0.4167 x 9,399.98 =0.0511 = 5.11% The rate of interest on the security is 5.11%.

BILLS OF EXCHANGE

137

Example 13: A Banker discounts a bill for a certain amount which has 32 days to run before it matures legally at 5%. The discounted value of the bill is Rs. 7,268. Find the face value of the bill ? Solution : Let the face value of the bill be Rs. P Rate of interest r =5% =0.05 32 No. of years t =365 =0.0877 years Banker's Discount is given by Banker's Discount =Prt = P x 0.5 x 0.0877 =0.004385P Given that the discounted value =Rs. 7,268 ~ P - 0.004385P = 7,268 => 0.995615P = 7,268 => P = 7,300 :. The face value of the bill is Rs. 7,300.

Example 14: The Banker's Discount and True Discount on a Certain sum of money due 3 months hence are Rs. 515 and Rs. 500 respectively. Find the sum of money and rate of interest? Solution : Let the face value be Rs. P and the rte of interest be r per rupee per annum. Given that 3 1 No. of years t = 12 = 4" years =0.25 years. Banker's discount is given by Banker's discount =Prt =Pr x 0.25 Now, True Discount is given by . Prt Pr x 0.25 True DIscount = 1 + rt = 1 + 0.25r It is given that banker's discount = .Rs. 515 and True Discount = Rs. 500.

Prx 0.25

..

Pr x 0.25

=515 and 1 + 0.25r =500

..

515 1 + 0.25r

=500

=>

1 + 0.25 r

=> =>

0.25r

515

=500 =1.03 =0.03

r = 0.12 = 12%

:.rate of interest is 12% Now, P x 0.12 x 0.25 = 515

138

FINANCIAL MATHEMATICS

P

= Rs. 17166.67

The sum of money is Rs. 17,166.67 Example 15 : A merchant has a 3 months bill for Rs. 4,000 which his broker discounts at 5%. Find the rate of interest, if at least he should earn on the discounted value so that he may

not suffer any loss by discounting the bill.

Solution : Let the required rate of interest be r. Broker's Discount is given by Banker's Discount =Prt where Present value of amount P =4,000 Rate ofinterestr =5% =0.05 3 No. of year t = 12 = 0.25 years

Broker's discount =4,000 x 0.05 x 0.25 =50 Broker's discount is Rs. 50 Discounted value =Rs. (4,000 - 50) =Rs. 3,750. Then, The Broker's Discount =Interest on Discounted Value. ~

50=3,750xO.25xr

50

r

50

=3950 x 0.25 =987.5 =0.0506 =5.06%

Hence, the required rate of interest is 5.06%. Example 16: Find the rate of percent at which the true discount on a bill legally due in 9 months will be exactly the same as the banker's discount at 7% per annum?

Solution : Let Rs. P be the face value of the bill legally due in 9 the interest rate of per annum. 9 3 No. of years t =12 =4" =0.75 years

nionth~

time and r be

True Discount is given by True Discount

Prt

= 1 + rt P x r x 0.75 =1+0.75xr 0.75 Pr = 1 + 0.75r

Now, we find Banker's Discount on a bill of Rs. P due in 9 months at 7%. Banker's Discount is given by -"'- Banker's Discount =Prt [.: r =P x 0.07 x 0.75

=0.0525P

=7% =0.07]

BILLS OF EXCHANGE

139

It is given that, True Discount =Banker's Discount 075Pr 0.0525P =1 + 0.75r 0.0525P (1 + 0.75r) =0.75Pr 0.0525P + 0.039375Pr =0.75 Pr 0.0525P = 0.75Pr - 0.039375Pr 0.0525P = 0.710625 Pr 0.0525 = 0.710625r 0.0525 r = 0.710625 = 0.07387 r

[Divide both sides by P]

= 7.39%

.. The rate of percent is 7.39%.

L Find the banker's discount and true discount on a bill of Rs. 12,200 due 4 months at 5% per annum. [ADs. Rs. 203.33, Rs. 200] 2. Find the true discount and banker's gain on a bill for Rs. 1,550 due 3 months hence at 6% per annum. [Ans. Rs. 22.91, Rs. 0.34] 3. A bill of Rs. 80,000 drawn on May 27, 2008 at 6 month is discounted on August 8, 2000 at 6% per annum. How much does the banker charge and what does the holder receive? [ADs. Rs. 1,499.19, Rs. 78,500.81] 4. A bill of exchange for Rs. 1,000 was drawn on the 3rd April, 2000 payable 3 months after date. It was discounted on 15th April, 2000 at 4~ per annum. What was the discounted value of the bill? [ADs. Rs. 990.45] 5. The banker's gain on a bill due 4 months hence at 4% per annum is Rs. 50. Find the sum. [Ans. Rs. 2,85,000] 6. The true discount and banker's gain on a certain bill of exchange due after a certain time are respectively Rs. 700 and Rs. 17.50. Find the face value of the bill. [ADs. Rs. 28,700] 7. The difference between the true and banker's discount on a bill due after 3 months at 5% interest is Rs. 100. Find (a) the true discount, (b) the banker's discount and (c) the face value of the bill. [Ans. (a) Rs. 8,000; (b) Rs. 8,100; (c) Rs. 6,48,000] 8. If the banker's gain on a bill due 6 months hence at 8% per annum is Rs. 600, find the true discount, banker's Discount and the amount of the bir~ ? [Ans. Rs. 15.000;Rs. 15,600; Rs. 3,90,000] 9. A bill of exchange drawn on February 4, 2007 at 6 months after date was discounted on March 27, 2000 at 10% per annum. If the banker's discount is Rs. 600, find the face value of the bill. [ADs. Rs. 16,466.25] 10. A bill was drawn on 25 th July, 2008 at 7 months after sight and was accepted 0 presentation on 8th August, 2008. It was discounted on the 30th August, 2008 at 7% per annum interest to realise Rs. 7,920. Find the value of the bill. [ADs. Rs. 8,224.42]

140

FINANCIAL 'MATHEMATICS

1L The banker's discaunt calculated for 2 year is 28 times his gain. Find the ·.rate of interest. [ADs. 1.82%] 12. The banker's discount and the true discount on a bill due 8 months hence are Rs. 500 and Rs. 400 respectively. Find the rate , percent and the-amount of the bill. [ADs•. 37.5%; Rs. 2,000] 13. A bill of Rs. 10,000 drawn on May 7 -for 7 months was discounted on August 19, following, for a cash payment of Rs. 9,880. Find the rate of interest charged by the bank. [Ans. 3.88%] 14. A bill for Rs. 4,650 was drawn on 8 March, 2008.at 7 months. It was discounted on 18 May, 2008 and the holder of the bill received Rs. 4,497. What rate of interest did the baker charge? . [ADs. 5%] 15. A bill is drawn for Rs. 5,050 on June 20 at five months. It is discounted on September 11 at 5% per annum. How much does the holder of the bill received and what is the banker's gain in the transaction? [ADs. Rs. 4990.50; Rs. 50] • th 16. A bill for Rs. 2,000 drawn on 20 Dec. 1995 at 6 moths is discounted on 11th April, 1996. If the payment made by the broker is Rs. 1,978, pnd the rate of interest. Find also the rate the banker gets on this money.. . [ADs. 5.5%, 5.56%] 17. A bill for Rs. 1,800 drawn on 24th September 2008 due 6 months hence was encashed in a bank at 5% for Rs. 1,782. Find out the date on which it was encashed '1 [ADs. 13 January 2009]

.

18. The difference between banker's discount and true discount on a certain sum for 3 years 4 months at 5% per annum is Rs. 12.50. Find the sum. [ADs. Rs. 525]

19. The banker's discount and True discount on a certain sum of money for 20 months are Rs. 64 and Rs. 60 respectively at the same rate. Find the sum and the rate per cent. [Ans. Rs. 960, 4%] 20. A bill ofRs. 2,525, drawn on January 13, 2007 at 8 months credit was discounted on 5th July at 5% per annum. Find the banker's discount and gain in this transaction. 21.

22. 23.

24.

25.

[Ans. Rs. 25.25; Re. 0.25] A bill of exchange ofRs. 2,000 drawn on January 6,2008 at 4 months was discounted on March 28, 2008. If the banker's discount at 5% per annum is Rs. 1,008, find the face value of the bill. [Ans. Rs. 1,75,200] A bill of exchange of Rs. 2,000 drawn on 14 March for 8 months is discounted on 2nd August at 5% per annum. How much does the banker charges? [ADs. Rs. 29.32] th A bill of exchange of Rs. 2,000 drawn on 15 March, 2008 for 6 months is discounted on 15th May, 2008 at 5% per annum. How much does the bank charge? [ADs. Rs. 34.52] st A received from B acceptance for Rs. 30,000 on 1 March, 2007, at 4 months. A got the acceptance discounted at 6% per annum at his bank after 1 month. How much was received by A from the bank after discounting the acceptance? [ADs. Rs. 29,550] The holder of a ill for Rs. 17,850 received Rs. 357 less than the amount of the bill by having it discounted at 5% per annum. For how many more days the bill has still to run ? [ADs. 146 days]

BILLS OF EXCHANGE

141

26. Rajesh draws a bill on Ramesh for Rs. 6,400, payable after 5 months. Ramesh accepts it and returns it to Rajesh. The bill is discounted by Rajesh from a bank for Rs. 6,000 on the same day. Find the rate at which the bill was discounted by the bank. [Ans. 15% per annum] 27. A bill of exchange drawn on January 4,2005 at 5 months was discounted on March 26,2005. If the banker's discount at 3% be Rs. 603.60, find the face value of the bill. [Ans. Rs. 1,00,600]

28. A bill of exchange for Rs. 27,335 was drawn on 7th April, 2008 payable 7 months after date. It was discounted on the 20th May, 2008 at 7% per annum. What was the discounted value of the bill ? [Ans. Rs. 26,422.84] 29. The true discount and banker's gain on a certain bill of exchange due after a certain time are respectively Rs. 50 and Re. 0.50. Find the face value of the bill. [Ans. Rs. 5,050] th 30. A bill was drawn on 15 May, 1999 at 8 months after date and was discounted on August 25,1999 at 6% per annum. If the banker's gain on the basis of simple interest is Rs. 18, for what sum was the bill drawn. [Ans. Rs. 32,000]

31. What is the face value of a bill discounted at 5% per annum 73 days earlier than the date of maturity, the banker's gain being Rs. 10 only? [Ans. Rs. 1,01,000] 32. One bill ofRs. 5,000 due on June 13 and another for Rs. 4,000 due on August 25 ,are both discounted with a banker on April 1. If the difference between the two discounts is Rs. 30, find the rate of interest at which the discount is calculated. [Ans. 5% p.a.] 33. The banker's discount and the true discount on a bill due 4 months hence are Rs. 210 and Rs. 200 respectively. find the rate of per cent and the amount of the bill. ~Ans. 15%, Rs. 4,200] 34. A bill was drawn on April 14, 2008 at 8 months after date and was discounted on July 24, 2008 at 5% per annum. If the banker's gain on the basis of simple interest is [ADs. Rs. 51,000] Rs. 20, for what sum was the bill drawn? 35. Calculate the banker's gain on a bill of Rs. 18,500 due in 7 months at 9% per annum. [ADs. 48.45] 36. A bill for Rs. 50,000 drawn on January 18, 2002 at 9 months was discounted at a bank on January 18,2002 at 9 months was discounted at a bank on which the date of discounting is March 28, 2002, the rate of interest being 6% per annum. How much did the holder receive? [Ans. Rs. 48,299] 37. If the banker's gain on a bill due 3 months hence at 5% per annum is Rs. 1,500. Find the True Discount, Banker's Discount and the amount of the bill. [ADs. Rs. 40,000; Rs. 40,500; Rs. 32,40,000]

Immediate Annuity INTRODUCTION TO ANNUITY In our daily life, we see a lost of transactions taking place, like sales or purchase of goods, renting or hiring, borrowing or lending of money and etc. Some of these transactions involving money are sometimes done by a single payment at a future date. But not many people are in a position to deposit a large sum of money at one time in an account. They make the payments in instalments over a period of time. For example, a customer, instead of paying the entire price of Rs. 20,000 for a refrigerator, many pay a part of it, say Rs. 5,000 initially and remaining in equal monthly instalments. These instalment are so determined that they compensate the seller for his waiting time and are based on compound interest rate. If a depositor makes equal deposits at regular intervals, he is contributing to an annuity. So by an annuity, we mean a sequence of equal periodic payments made at equal intervals of time. The payments may be generally made weekly, monthly, quarterly, half-yearly or yearly. For example, premium for a life insurance policy may be made in equal instalments periodically, say semi-annually. The equal monthly payment that a retired person receives in the form of pension is another example of annuity. Even bank loans are repaid in equal instalments, say monthly, over a period of time. All these transactions have a common thing equal payments at equal intervals of time. In this chapter, we discuss various kinds of annuities and study in detail the present value and future value concepts of immediate annuity. ANNUITY: DEFINITION

An annuity is defined as follows: "An annuity is a sequence of equal payments made at equal intervals of time."

IMMEDIATE ANNUITY

143

The term annuity is derived from the Latin word 'annum'. Although this word indicates that payments must be annual, its meaning in practice, has been broadened to include any periodic payment of a fixed amount made at a regular interval which may be more or less than one year. The person obligated to make such payments is called annuitator where as the person entitled to receive such payments is called annuitant. SOME BASIC TERMS (i) Period Payments

The size of each payment of an annuity is called the periodic payment or periodic rent of the annuity. For example, if a sequence of payments of Rs. 2,000 is made at the beginning of every month, then we say that it is annuity with periodic payment of Rs. 2,000. (ii) Payment Date

The date oolby which a periodic payment is to be made is called payment date. For example,. if a bank asks a borrower to pay the monthly instalment against a personal loan on 5th of every month this date will be the payment date. (iii) Payment Period

The period of time elapsing between two successive payment dates of an annuity is called its payment period or payment interval or rent period. . For example, if a sequence of payments is to be made on 1st January every year, then it is an annuity with payment period of one year. The payment period of an annuity can be a year, an half-year, a quarter, a month, a week and etc.

Term

(iv)

The total time from the beginning of the first payment period to the end of the last payment period is called the term of an annuity. For example, if a series of payments was made on 1st January every year from 1990 to 1999, it is an annuity with a 10-year term and payment period of one year. It should be noted that the term of an annuity can also be a certain number of months. For example, if a loan is to be repaid in 6 equal monthly instalments, then it is an annuity with 6-months term and payment period of one month. (v)

Amount or Future Value

It is the sum of all the payments made and the interest earned on them at the end of the term of the annuity. In other words, the amount of an annuity is the sum of the terminal or future values 0 f each of the periodic payments. Here, the term terminal value means the sum of the payments made and the interest earned on that payment. (vi)

Present Value or Capital Value

The present value or capital value of an annuity is the sum of the present values of all the payments.

144

FINANCIAL MATHEMATICS

In other words, the present value of an annuity is the amount of money that must be invested now to purchase the payments due in future. TYPES OF ANNUITIES

There are many types of annuities which are classified based on different factors. (i)

Annuities based on their Term

Following are the types of annuities which are classified based on the term of the annuity: (a) Annuity Certain (b) Contingent Annuity (c) Perpetual Annuity (a) Annuity Certain

An annuity payable for a certain fixed term is called Annuity certain. This kind of annuity begins and ends on certain fixed dates. The term of an annuity is fixed and so its payments extend over a fixed period oftime. The payment for hike-purchase of durables like Television, Refrigerator, Car, etc. are usually annuity certain, because the buyer knows the specified dates on which the instalments are to be made and the number of instalments to be made. Repayment of bank loan and Bank recurring deposits are also annuity certain. (b) Contingent Annuity

Contingent annuities one whose payments continue for a period of time which depends on the happening of an event the date of which cannot be accurately foretold. If the contingency did not happen at all, then the periodic payments should be paid till the end of designated term. For example, payment of life insurance premium is contingent annuity. Whenever the insured .person dies, the payment of premium is stopped. (c) Perpetual Annuity

A perpetual annuity, simply perpetuity is an annuity whose payments continue forever (infinite number of years). In perpetuity, the beginning date is known but the terminal date is not known, because the term of this annuity is infinite. For example, in endowment funds of trusts the principal amount is not touched and the interest earned is used for welfare activities. Since the principal is not touched, a certain sum is received periodically forever. (ii)

Annuities based on Payment Date

Following are the types of annuities classified based on the payment interval and time of payment: (a) Ordinary or Immediate Annuity (b) Annuity Due (c) Deferred Annuity.

IMMEDIATE ANNUITY

145

(a) Immediate Annuity An ordinary annuity or immediate annuity is an annuity, in which every periodic payment is made at the end of the corresponding payment period. Even though, the first payment is made at the end of the first payment period, the term of an ordinary annuity begins with the beginning of the first payment period and ends on the day 'of the last payment. For example, repayment of a bank loan is an ordinary annuity. It should be noted that an ordinary annuity is an annuity certain but the converse may not be true.

(b) Annuity Due If each payment of annuity is made at the beginning of the corresponding payment period is called annuity due. Here, even though the last payment is made at the beginning of the last payment period, the term of an annuity due ends at the end of the last payment period. The term of an annuity due begins on the day, the first payment is made. For example, a saving scheme of five years in which equal deposits are made at the beginning of each year is an annuity due. In this annuity every payment is an instalment, the first payment earns interest for five years, the second for four years and the last for one year. Payment of premium for a life insurance policy is the most precise example for annuity due. In life insurance, the first premium is to be made at the commencement of the policy.

(c) Deferred Annuity Deferred stands for postponed. A deferred annuity is an annuity in which the first payment is postponed for a period of time whi~h is equivalent to a certain number of payment periods. In deferred annuity, the term begins after the expiry of this postponed period. This period is called period of deferment. It is understood from the above discussion that in deferred annuity the term is known but postponed for a certain number of intervals. A deferred annuity can either be a deferred ordinary annuity or a deferred annuity due. But it is gener~y an ordinary annuity. For example, when a house building loan is given by an employer to an employee, generally the repayment in equal instalments by the employee does not start thereof but may begin after some time, say five years after the loan is sanctioned. The interval of five years is called the deferred period.

(iii) Annuities based on Payment Period and Conversion Period Annuities based 0:1 payment period and conversion period are classified as follows: Simple Annuity (b) Complex Annuity

(a)

146

FINANCIAL MATHEMAnCS

(a) Simple Annuity

An annuity in which the payment period coincides with the interest conversion period is called simple annuity. For example, consider an annuity of Rs. 1000 payable at the end of every quarter, interest being calculated at 8% per annum cotrtpounded quarterly. Here payment is made at the end of every quarter and intere~t is also compounded on quarterly basis. This kind of annuity is called simple annuity. (b) Complex Annuity

. An annuity in which payment period differs from the interest conversion period is called a complex annuity. For example, suppose that a man deposits Rs. 500 at the end of every month in an account and the bank calculates interest at 6% per annum compounded half-yearly. This kind of annuity is called complex annuity. In this case, payment period is one month whereas the interest conversion period is six months. In other words, six periodic payments are made in one interest conversion period. (iv)

Annuities based on the Period Payment

Annuities based on the amount of periodic payment are classified as follows: (a)

Uniform Annuity

(b)

Variable Annuity

(a) Uniform Annuity

If the periodic payment are all equal through out the term of the annuity, then the annuity is called uniform annuity. It is also known as fixed annuity.

Annuities are generally uniform, since equal payment are made at equal time interval. For example, repayment of a personal loan is a uniform annuity. The debtor is to repay the loan in equal instalments. (b) Varial)le Annuity

It the periodic payment of an annuity changes every year or every payment period, then the annuity is called variable annuity.

Penson plan is an example of variable annuity. Certain additional features may also be added to the plan. The annuity may increase as a result of vested bonuses, if the plan is a participating one. CHARACTERISTICS OF AN ANNUITY

• • •

Annuity is a recurring payment of fixed amount, i.e., a periodic payment. Annuity is payable at an equal interval of time, either annually, semi-annually, quarterly or the like. Annuity is paid against a lump sum received at present or a compound amount receivable after a certain period.

IMMEDIATE ANNUITY

•

•

• • • •

147

Annuity is paid either to discharge an existing obligation instalmentwise or to create a fund to be accumulated in future. If the annuity is paid against a lump sum received at present, then every payment will contain interest for the outstanding principal and the sum of all instalments (periodic payments) will be more than the present value. The difference is the total interest paid. If the compound amount (Future value) is receivable after a certain term, then every periodic payment will bring interest and the sum of the instalments will be less than the future value. The difference is the total interest received. Annuity is paid either at the beginning or at the end of each period. In annuity, every payment is a case of compound interest. To find the future v~lue (or present value) of an annuity, future or present. value of every periodic payment is calculated on the basis of compound interest and the sum is found out. Annuity is paid on the basis of a contract by one party called annuitator to another party called annuitant.

DIFFERENCE BETWEEN COMPOUND INTEREST AND ANNUITY

•

In case of compound interest, a lump sum (principal) is paid only once. But incase of annuity, a series of equal period~c payments are payable (or receivable) at equal interval of time. • The purpose of investment under compound interest system is to crate a heavy fund at the end of a certain term to meet an obligation. But the purpose of investment under annuity system is to create a heavy fund through periodic contribution of small amounts to meet an obligation. • Compound interest system operates without any element of annuity system, but annuity system cannot operate with out the element of compound interest system. • Annuities can be classified into different types based on the payment date, payment interval and etc. But there is no such classification in case of compound interest. • In case of compound interest system, the rate of interest may vary from time to time. But in case of annuity system, interest is generally compounded at an agreed fixed rate of interest. • In case of multiple investments under compound interest, the investments need not to made at equal time intervals. But instalments should be made at equal time intervals in the annuity system. Note: In this chapter, we will study the concept of ordinary annuity in .detail and succeeding chapters are devoted for annuity due, differed annuity and perpetuity.

•

ORDINARY (OR IMMEDIATE) ANNUITY

As we have discussed earlier in this chapter, in an ordinary annuity, the periodic payments are made at the end of each payment period. We consider ordinary annuities that are certain and simple with equal periodic payments. In this chapter, we study the annuities that are subject to the following conditions : (a) The periodic payments are made at equal time interval.

148

FINANCIAL MATHEMATICS

The payment period coincides with the interest conversion period. (c) The periodic payments are made at the end \of every payment period.

(b)

Remark : If the payment date (i.e. beginning or end of the payment period) is not mentioned, it should be then assumed that payments are made at the end of the payment periods and the annuity is ordinary annuity. AMOUNT OF AN ORDINARY ANNUITY

The amount or future value of an annuity is the sum of all the payments made and the interest earned on them at the end of the term ofthe annuity. In other words, future value of an ordinary annuity is the sum of the compound amounts of all the payments accumulated at the end of the term. For example, consider an ordinary annuity of Rs. 1,000 payable at the end of every year for 3 years at 8% effective rate of interest. Three-year term

14

~I

Payment periods

1

1

0

1

3

2

t

t

Rs. 1,000

t

Rs. 1,000 Payments

Rs. 1,000

The payment of Rs. 1,000 made at the end of 1at year will earn interest for 2 years and this amount will be accumulated to 1,000 (1.08)2, i.e., Rs. 1,166.40. The second payment ofRs. 1,000 made at the end of 2nd year will earn interest for 1 year, since this amount remains in the account for one completed year. At the end of the term of3 years, this payment will be accumulated to Rs. 1,000 (1.08), i.e., Rs. 1,080.00. The last payment, i.e., the third payment will earn no interest, because as soon as the payment is made, the term ends. Therefore, the amount of this payment will remain same, i.e., Rs. 1,000. . . The amount of this annuity A = 1,166.40 + 1,080 + 1,000 = Rs. 3,246.40 This is shown in the following figure:

o

Payment periods 1 2 I Rs. 1000 Rs.1000

3 I Rs.1000

Payments -: Rs. 1,000.00 Rs. 1,000 (1.08)

=Rs. 1,080.00

L -_ _ _ _ _ _ _. .

, - - - - - - - - - - - -__ Rs. 1,000 (1.08)

2

=Rs. 1,166.40

Amount =Rs. 3,246.40

149

'MMEDIATE ANNUITY

Now we try to derive the formula for future value of an ordinary annuity in a general problem. Let Rs. R be the size of the periodic payment of an ordinary annuity. Let r be the annual rate of interest and m be the number of conversion periods per year. :. The rate of interest per conversion period is given by i =rlm

Let 'n' be the total number of conversion periods. The first payment Rs. R, made at the end of the first period, earns interest for (n - 1) periods, hence its compound amount would be R (1 + i)n-l. Similarly, the second payment of Rs. R earns interest for (n - 2) years, the third for (n - 3) years and so on. The last payment of Rs. R made at the end of the term does not earn any interest and its value is simply Rs. R. This is illustrated in the following figure:

o

1

R

2

R

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

3

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

R

1

n

I

I

I

R

R

R

I I

I Amount of 1It instalment, Amount of 2 nd instalment,

n- 2n-

L-

:

R R (1 + i) R (1 ! i)2

:+

R (1 i)n-3 R (1 + i)n-2 R (1 + i)n-l

Al A2

= R(1+i)n-l = R(1+i)n-2

Amount of(n - 2}1ld instalment A n - 2 = R (1 + i)2 Amount of(n -1)Btjnstalment A n - 1 = R (1 + i) Amount of nth instalment An =R The Amount of this annuity is given by

=:: '"* '"*

,A =Al +A2 +Aa + ...... A n - 2 + A n - 1 + An A = R(l + iY' -1 + R (1 + iY' - 2 + ...... + R (1 + i)2 + R (1 + i)l + R

=R + R (1 + i) + R (1 + i)2 + ...... + R (1 + i)n-2 + R (1 + i)n-l A =R [1 + (1 + i) + (1 + i)~ + ...... + (1 + i)n-2 + (1 + i)n-l]

A

A

=R

(1 { (1

+ i)n-l } + i)-1

A =R [(1 +

r"-1 .: 1+r+r2 ... r"-1=--1r-

iT - 1 ]

Thus, amount of an ordinary annuity of Rs. R payable at the end of every period for 'n' periods at a rate of i per period is given by

150

FINANCIAL MATHEMATICS

Remark: When the periodic payment is Re. 1, then amount of annuity is given by (1 + i)n - 1

i

. This quantity is denoted by Smi'

S = R. smi where smi =

(1

+ i)n-1 .

,

smi is read as"s sub n at the rate i". The values of smi for various n and i are given in table III. CONTINUOUS COMPOUNDING

The amount A of an ordinary annuity in which Rs. R are paid each year for t (= n) years at the rate of interest of r per annum compounded continuously is given by

I

n(=t)

A=

~

I A =;

Rertdt

(e

rt - 1)

I

TOTALINTERESTEARNED

A is the amount at the end of the term and Rs. R is the periodic payment. If there are n payments, then the total payments paid =nR. We knovT,that the amount is the sum of total payments and the interest· accumulated on every payment. .. The total interest earned is given by

II=A-rilll Example 1 : Find the amount of an ordinary annuity of Rs. 5,000 payable at the end of each year for:3 years at 8% per annum compounded annually.

Solution : Amount A of an ordinary annuity of Rs. R per period for n periods at the rate of i per period is given by A Here,

=R

((1 + it -1 )

periodic payment R = Rs. 5,000 rate of interest r =8% =0.08 Interest is compounded annually i = r = 0.08 and n = t = 3 years

IMM~DIA TE ANNUITY

151

A

=5,000 (1.~~~38-

A

=5,000 (1.25~~~: -

A

=5,000 (0.2:'~~12 ) =5,000 (3.2464)

1) 1)

A =16,232 .. The amount is Rs. 16,232 AlternativelY, A =R siili

Here

A =5,000 Smo.os A = 5,000 (3.2464000) = 16,232 The amount is Rs. 16,232

Example 2 : Find the amount of an ordinary annuity if payment of Rs. 500 is made at the end of every quarter ~or 10 years at the rate of 8% per annum compounded quarterly. Solution : The amount A of a ordinary annuity of Rs. R per period for n periods at the rate of i per period is given by A

Here

=R [(1 + iT - 1 ]

Periodic payment R = Rs. 500 Rate of interest r = 8% = 0.08 Interest is compounded quarterly

. r 0.08 :. '=4'=4=0.02 A"= 500

Let then

[(1.0;~~~ -

and

1]

x = (1.02)40

log x =40 log 1.02 log x =40 x 0.0086 = 0.344 x = antilog (0.344) = 2.2080

.. A:: 500

[2.2~~012-

1]

=500 [1.~.~~0 ] = 500 (60.~) A =30,200

A

.. The amount is Rs. 30,200,

n=4x 10 =40

152

FINANCIAL MATHEMATICS

Example 3: A person decides to put aside Rs. 100 at the end of every month in a money market fund that pay interest at the rate of 8% compounded monthly. After making 12 deposits, how much money does he have ? Solution : Since the payments are made at the end of every month, it is an ordinary annuity. The amount of an ordinary annuity of Rs. R per period for n periods at a rate of i per period is given by A

=R

[(1 + iT - 1]

Here, R =Rs. 100 Rate of interest r =8% =0.08 Interest is compounded monthly.

..

0.08 =0.0066 7 an d ,. - 2"r =12 n = 12

n

=12

A - 100 [(1.00667)12 - 1 ] 0.00667 -

I

A - 100 1.0830425 - 1 ] 0.00667 A = 100

[Oo~~~~:~5] = 100 (12.4501)

A =1245.01 The amount is Rs. 1,245.

Example 4: At 6 months interval, Mr. X deposited Rs. 100 in a savin.gs A I c which carries interest at 10% p.a. compounded semi-annually. The first deposit was made when his son was 6 months old and last deposit was made when his son was 8 years old. The money remained in the account and was presented to his son when he was 10 years old. What much did he received? Soluti!Jn. Rs. 100 was deposited at 6 months interval for 8 years at 10% i.e. R = Rs. 100 Rate of interest r = 10% = 0.10 Interest is compounded semi-annually. i

= ~ =0.05

and

n

= 8 x 2 = 16

The amount of an ordinary annuity is given by

A

=R

[(1 + it -1]

.. Amount at the end of 8 years is

IMMEDIATE ANNUITY

153

= 100 [(1.05)16 -1]

A

0.05

Let

= (1.06)16 = 16 log 1.05

x log x

= 16 x 0.0212 = 0.339 x = antilog (0.339) = 2.183 A

= 100 [2.183 -1] 0.05

= 100 [10~::] = 100 (23.66) = 2,366 This will remain in the account for 2 more years. .. n = 2 x 2 =4 and i =0.05 In case of compound interest, the amount is given by A

= P (1 + iY'

.. Amount at the of 10th year is : A = 2,366 (1.05)4

= 2,366 x 1.2155 = 2,875 .. The amount is Rs. 2,875.

Example 5: If a bank pays 6% interest compounded quarterly, what equal deposits have to be made at the end of each quarter for 3 years to have Rs. 1,500 at the end of 3 years? Solution: Given that Amount A =Rs .. 1,500. Since the payment are made at the end of quarter, it is the case of ordinary annuity. Rate of interest r = 6% =0.06 i

= 0~6 =0.015

and n.= 3 x 4

Let Rs. R be the equal payment. The amount of an ordinary annuity is given by

A

=R

[(1 + iT - 1 ]

1 500 - R [(1.015)12 - 1] , 0.015 1500 - R (1.1956 -1 ) 0.015 , 1,500

= R x 13.04

=12

154

FINANCIAL MATHEMATICS

R - 1,500 - 115 - 13.04.. The quarterly payment is Rs. 115.

Example 6: A company set aside a sum of Rs. 4,500 at the end of each year for 7 years to payoff a debenture issue of Rs. 40,000. If the fund accumulates at 9% compounded annucilly, find the surplus after full redemption of the debenture issue.

Solution: Here, periodic payment R =Rs. 4,500. Interest is compounded annually. .. i = r = 9% = 0.09 and n = t = 7. Since the sum R is set aside at the end of every year, it is an ordinary annuity.

The amount of an ordinary annuity is given by

A

= R [(1 +

it -

1]

A - 4500 [

= 4,500 [0.8~~~39 ] =41,401.96

· . Amount Accumulated =Rs. 41,401.96 Value of debenture =Rs. 40,000 · . The surplus =Rs. (41,401.96 - 40,000) =Rs. 1,401.96 · . The surplus is Rs. 1,401.96

Example 7 : Mr. X deposits Rs. 5,000 at the end of every 6 months into his saving bank account. The bank calculates interest at a rate of 11% per annum compounded semi-annually.

(a) What account will be accumulated at the end of 12 years? (b) What is the amount he will earn as interest ?

Solution: (a)

Since the deposit is made at the end of every 6 months, it is an ordinary annuity. Given that, Rate of interest,

R r

= Rs.500

= 11% =0.11

Interest compounded semi-annually.

..

.

z

0.11 = '2r =2 =0.055

an

d

n == 2 x 12 = 24

IMMEDIATE ANNUITY

155

;

Amount of an ordinary annuity. is given by A

= R [(1 + iT - 1 ]

- 5000 [(1.055)24 -1) A , 0.055 A A A

= 5,000 [3.6~~:5-

1]

= 5,000 [ 2.61459] 0.55 =5,000 [47.53B] = 2,37,690

.. The amount after 12 years is Rs. 2,37,690 (b) The total interest earned is given by

I = A-nR I = Rs. (2,37,690 - 24 x 5,000) I = (2,37,690:- 1,20,000) =Rs. 1,17,690 .. The interest earned is Rs. 1,17,690.

Example 8 : Find the number of years for which an annuity of Rs. 1,500, payable per annum accumulates to Rs. 30,000 at the rate of9% effective.

Solution : Since the payment date is not mentioned, it is assumed to be an ordinary annuity. Periodic Payment, R = Rs. 1,500 Accumulate amount, A = Rs. 30,000 Rate of interest i = r = 9% effective =0.09 The problem is to find 'n' The amount of an ordinary annuity is given by A

= R [(1 + iT - 1 )

30,000

= 1,500 [ (1·~~9-

(l.09)n -1 =

1]

30,0~~;00.09 =1.B ,

(1.09)n = 2.B Taking log on both sides, we get n log 1.09 = log 2.B _ log (2.0B) _ 0.4472 _ 11 957 n - log (1.09) - 0.374 - . .. The number of years is approximately 12 years.

156

FINANCIAL MATHEllAncS

Example 9: A deposits annually Rs. 200 15th for 10 years, the first deposit being made one year from now; and after 10 years the annual deposit is enhanced to Rs. 300 per annum. Immediately after depositing the 15th payment he closes his account. What is the amount payable to him, if the interest is allowed at 6% effective ?

Solution: Since the deposits are made at the end of every year, it is an ordinary annually. This problem contain two annuities : An annuity of Rs. 200 for 10 years and thereafter an annuity of Rs. 300 for 5 years.

oI o

1

2

3

4

5

6

200

200

200

200

200

200

I

I

I

I

I

I

Period in years 7 B 9 I

I

I

200

200

200

10

11

200

300 300 300 300 300

I

I

12 I

13 I

14 I

15 I

Payments,(in Rupees)

Consider the annuity ofRs. 200 for 10 years: Periodic Payment R = Rs. 200 Rate of interest r = 6%=0.06 Interest is compounded annually. .. i = r = 0.06 and n = t = 10 The amount of an ordinary annuity is given by A = RSmi A = 200 BiOIo,06 A = 200 (13.1807949) = 2,636.16

Since the annuity ends at the end of 15th year, this amount will remain in the account for 5 more years, earning compound interest. .. The amount accumulate at the end of 15th year = 2,636.16 (1.06)5 A = P (1 + iY' = 2,636.16 (1.338226) = Rs.3,527.78

... (1)

Now consider the annuity ofRs. 300 for 5 years starting form 11th year: R = Rs. 300 and n = 5 Amount at the end of 15th year is = 300 s51o.06 = 300 (5.6370930) = Rs. 1,691.13 The required amount after 15 year is (1) + (2) = Rs. 3,527.78 + Rs. 1,691.13 = Rs. 5,218.91 ., The total amount is Rs. 5,218.91

... (2)

157

'MMEDIATE ANNUITY

Example 10 : An account fetches interest at 5% per annum compounded continuously. An individual deposits Rs. 1,000 each year in the account. Find how much will ~e in the account after 5 years.

Solution : Giv~n that R Rate of interest r No. of years t rt

= Rs. 1,000 = 5% =0.05 =5 = 0.05 x 5 =0.25

Interest is compounded continuously. The amount of annuity is given by A

= ~r [en -1]

[ 0.26 -1] A -- 1,000 0.05 e A A

= 20,000 [1.284 - 1] = 20,000 (0.284) =5,680

. . The amount is Rs. 5,680.

Example 11 : Find (a) How much should be deposited in a bank each year in order to accumulate Rs. 50,000 in 6 years, if the interest is calculated at a rate of 6% per annum compounded continuously ? (b) How long it will take for an annuity of Rs. 5,000 to amount to Rs. 91,091 at a rate of 11% per annum compounded continuously?

Solution: (a) Let Rs.

R be deposited each year.

Accumulated amount A

=Rs. 50,000

Rate of interest r = 6% = 0.06 Number of years t :. rt

=6 years

=0.06 x 6 =0.36

Amount of an a annuity is given by

A 50,000 ~

Let

3,000 x log x

= Rr [ert·-l] R

= 0.06 [e036 - 1] = R (eO.36 - 1) = eO.36 = 0.36 log e

158

FINANCIAL MATHEMATICS

log x = 0.36 x 0.4343 =0.156348 x = antilog (0.1563) = 1.4332 3,000 = Rs. (1.4332 - 1) 3,000 = R (0.4332) => R =6,925.21 :. The periodic deposit =Rs. 6,925.21 (b)

Let 't" be the number of years Given that R =Rs. 5,000 and A =Rs. 91,091 Rate of interest r =11% =0.,011 Interest is compounded continuously. Amount of the annuity is given by

.. =>

=

A

= !!r [en -

1]

91,091

= 5,000 [

OUt _

0.11 e

1

= 2.004002

eO. 111

= 3.004002

eO. 1lt -

1]

Taking long on both sides we get O.l1t log e = log 3.004002 t = log 3.004002 0.11 log e 0.47769 t = 0.11 x 0.4343 t = 10 .. The r~quired number of years is 10 years.

L Find the amount of an ordinary annuity of Rs. 1,000 payable at the end of each year for 3 years at a rate of 10% per annum compounded annually. [Ans. Rs. 3,310] 2. Find the amount of an ordinary annuity of Rs. 3,000 at the rate of 5% per annum compounded annually for 8 years. [Ans. Rs. 28,628] 3. Find the amount of the following annuities: (a)

Rs. 200 per year for 5 years at 8% per year compounded annually.

(b)

Rs. 500 payable at the end of each year for 14 years at 5% effective rate of interest. [Ans. (a) Rs. 1,173; (b) Rs. 9,810

4. Find the amount of an annuity consisting of payments of Rs. 600 at the end of every three months for 3 years at the rate of 8% compounded quarterly. [Ans. Rs. 8,047.25] 5. Find the amount of an ordinary annuity, if payment ofRs. 200 is made at the end of [Ans. Rs. 9,023.10] every month for 3 years at the rate 18% compounded monthly.

159

IMMEDIATE ANNUITY

6. For the purpose of his daughter's marriage, a man deposits Rs. 3,000 at the end of each 6 month period in a fund paying 8% per year compounded semi-annually. Find the amount accumulated at the end of 18 years. [Ans. Rs. 2,32,795] 7. Find the amount of an ordinary annuity of 12 monthly payments of Rs. 1,000 that earn interest at 12% per year compounded monthly. [Ans. Rs. 12,682.50] 8. A man deposited Rs. 500 at the end of each year for 5 years. He made his first payment at the end of 1990 and the last payment at the end of 1994. How much should have been there on 311t December 1994, if the rate of interest was 10% compounded annually? [Ans. Rs. 30,525.50]

9. A man plans to deposit a sum of Rs. 750 in a savings· account at the end of this month and the same amount at the end of each following months. To what sum will the investment grow at the end of 5 years, if the rate of interest is 5% per annum - compounded monthly? [Ans. Rs 51,004.56] 10. Find the amount of the following ordinary annuities: (a)

Rs. 1,000 a year for 5 years at 7% per year compounded annually.

(b)

Rs. 500 per quarter for 10 years at 8% per year compounded quarterly.

(c)

Rs. 4,000 each six months for 15 years at -5% per year compounded semiannually.

Rs. 40 each month for five years at 6% per year compounded monthly. [Ans. (a) Rs. 5,750.74; (b) Rs. 30,200.99; (c) Rs. 1,75,610.81; (d) Rs. 2,790.80] 1L A person wishes to deposit Rs. 2,500 per year in a savings account which earns interest of 10% per year compounded annually. Assume the first deposit is made at the end ofthis current year and addition deposits at the end of each following year. (d)

(a) To what sum will the investment grow at t~e time of the 20 th deposit? (b)

How much interest will be earned?

[Ans. (a) Rs. 1,43,187.50; (b) Rs. 93,187.50]

12. A c~mpany sets aside a sum of Rs. 5,000 annually for 10 years to payoff a debenture issue ofRs. 60,000. If the fund accumulates at 5% per annum compounded annually, find the surplus after full redemption of the debenture issue. [Ans. Rs. 2,889.46] 13. A company set aside a sum of Rs. 4,000 at the end of each year for 7 years to payoff a debenture issue· of Rs. 35,000. If the surplus after full redemption of the debenture issue. [Ans. Rs. 2,948.68] 14. An annuity consists of equal payments at the end of each month for 2 years is to be purchased for Rs. 2,000. If the interest rate is 6% per annum compounded monthly, [Ans. Rs. 78.64] how much is each payment? 15. For the purpose of education of his child, Ram want to accumulate Rs. 50,000 by making equal payments at the end of each quarter for next 5 years. What will be the size or these investments, if money is worth 6% converted quarterly? [Ans. Rs. 2,162.29] 16. Mr. Shyam deposits an amount of Rs. 2,000 at the end of every 3 months into his savings bank account. The bank calculates interest at the rate of 7% per annum compounded semi-annually.

160

FINANCIAL ..ATHEliA ncs

(a) What amount will be accumulated at the end of 10 years in his account? (b) What is the total interest earned from these deposits? [Ana. (a) Rs. 1,14,468.27; (b) Rs. 34,468.27]

17. How much money must be deposited at the end of each quarter if the objective is to accumulate Ra. 5,00,000 after 5 years? Assume interest is earned at the rate of 10 per cent per year compounded quarterly. How much interest will be earned ? [An,s. Rs. 19,573.56; Rs. 1,08,528.08] 18. A company must accumulate Rs. 12,000 at the end of 10 years to replace certain component of its machine. What sum must the company invest at the end of ~ach year, assuming interest is earned at the rate of 4% per year compounded annually? [Ana. Rs. 1,002.08] 19. If a bank pays interest at the rate of 5% per annum compounded annually, what equal deposits have to be made at the end of ever year to have Rs. 1,00,000 at the end of 10 years. How much interest will be earned on these deposits ? [Ana. Rs. 7,950.46; Rs. 20,495.40] 20. To save for his son's education, Mr. Singh deposited Ra. 1,000 at the end of each 6 months period into a savings account paying 4% interest compounded semi-annually. The first deposit was made when his son was 6 months old and the last deposit was made when his son was 21 years old. The money was kept in the account and was presented to his son on his 25th birth day. How much did he receive? [Ana. Ra. 75,996.43] 2.L An amount of Rs. 1,500 is deposited in a savings bank account at the end of every 3 months for a period of 8 years and no payment is made there after. But the accounts will be closed at the end of loth year. Assuming that the bank calculates interest ala rate of 6% per annum compounded quarterly, calculate the amount which will be [Ana. Rs. 68,752.58] received at the time of closure of the account? 22. What amount should be set aside at the end of each year to accumulate Rs. 1,48,970 at the end of 8 years at 5% effective. [Given (1.05)8 =1.468] [Ana. Rs. 16,915.60] 23. A deposits annually Ra. 200 for 10 years, the first deposit being made one year from now. After 10 years the annual deposit is enhanced to Rs. 300 per annum,. At the end of 20th year, he closes his account. What is the amount payable to him at the end of 20th year, if the interest is allowed at 9% effective? . [Ana. Rs. 8,659.19] 24. A sum of Rs. 500 is deposited in a bank at the end of every 6 months. What is the amount to the credit of the depositor at the end of 10 years, if interest is credited to the account at the rate of 6% for the first five years and 8% thereafter? [Ana. 14,487.72] 25. A person deposits Rs. 2,187 at the end of ever 3 months in a fund created to fulfil his obligation ofRs. 50,000 which is due after a certain number of years. If the money is worth 6% per annum compounded quarterly, find the number of years. [Ana. 5 years] 26. The amount of an annuity of Rs. 500 payable at the end of each year for 12 years is Rs. 8,433. If the interest is compounded at 6% effective, find the number of years. [Ana. 10 years]

IMMEDIATE ANNUITY

161

27. Mr. X has decided to invest Rs. 500 at the end lof each year. He did so for 7 years. Due to some unavoidable situations, he could not make the payments for the next 4 years. He again invested Rs. 500 per annum for the next 4 years beginning from the end of the 12th year. Find the amount to his credit at the end of the 15th year assuming interest at effective rate of 9% per annum. [Ans. Rs. 11,452.75] .

.

28. An annuity is payable for 15 years certain, the first payment falling due at the end of first year. The annuity is payable at the rate ofRs. 5,000 per annum during the Brst 10 years and Rs. 3,000 per annum during the remaining 5 years. Calculate the future value of the annuity on the basis of interest at 4% per annum. [Ans. Rs. 8~i,285.29] 29. If Rs. 500 is deposited each year in a savings bank account paying 5.5% per annum compounded continuously, how much will be in the account after 4 years? [Ans. Rs. 2,237.27] 30. A bank pays interest at the rate of 8% per annum compounded continuously. Find how much should be deposited in the bank each year in order to accumulate Rs. 10,000 in 5 years. [Ans. 1,626.67] 31. An account fetches interest at 10% per annum compounded continuously. A man deposits Rs. 600 at the end of each year in his account. If he wishes to accumulate a sum ofRs. 4,932, how many consecutive deposits he has to make in his account? [Ans.6] 32. Find the amount of an annuity of Rs. 1,200 at the end of every year for 3 years at a rate of 5% per annum compounded continuously. [Ans. Rs. 3,883.35] 33. How much need to be saved each year in a savings account paying 6% per annum compounded continuously in order to accumulate Rs. 6,000 in three years? [Ans. Rs. 1,825.56]

PRESENT VALUE OF AN ORDINARY ANNUITY We have so far studied annuity problems in which payments were made with the objective of obtaining a certain lump sum at a future date. In this section, we shall study problems in which a lump sum is invested at year zero at a certain rate of compound interest so that a series of equal payments can be obtained over some future period of time. The lump sum investment is generally known as the present value of the ordinary annuity. The present value of an ordinary annuity is the sum of the present values of each instalment. For example, suppose that a person purchases a personal loan from a bank at present at 6% per annum compounded annually and agrees to pay Rs. 2,000 at the end of every year for 3 years. This is shown in the following figure: No. of years

Loan ~

0

I

1

2

I

3

I

Rs.2,Ooo Rs.2,ooO Rs.2,OOO Annual payments

The present value of the first instalment of Rs. 2,000 will be Rs. 2,000 (1.06)-1, i.e., Rs.l,886.79.

162

FINANCIAL AfATHEAfAricS

The present value of the second instalment of Rs. 2,000 will be Rs. 2,000 (1.06)2, i.e., Rs.1,779.99. The' present value of the third (final) instalment of Rs. 2,000 will be Rs. 2,000 (1.06)-3, i.e., Rs. 1,679.24. :.' The Present Value of this annuities P =Rs. 1,886.79 +Rs. 1,779.99 + Rs. 1,679.24 =Rs. 5,346.02 P =Rs. 5,346 (approximately) This is shown in the following figure:

o

123

I

I

I

As. 2,000

2,000 (1.06)-' = 1,886.79

Rs. 2,000

I

Rs. 2,000

I

.---.J

I

=1,779.99 2,000 (1.06)-3 =1,679.24 . - - - - - - - - - - ' Present Value =5,346.02

2,000 (1.06)-2

- - - - - - 1 :.

.

Thus, we say that a loan of Rs. 5,346 at a rate of 6% per annum compounded annually is repaid by three annual instalments of Rs. 2,000. The benefits are receivable in the form of either money such as Ipan, etc. or asset such as house, etc. at the beginning of the annuity. And this is repaid through a certain number of equal instalments. Therefore every intalment contains some certain amount of interest. Therefore, the present value of an annuity will be always less than the sum of the all periodic payments. N:ow we derive the formula for the present value of an ordinary annuity. Consider an annuity of n payments of Rs. R each, where the interest rate is 'i' per period. The first payment of Rs. R is made at the end of first period. So the present value of this Rs. R is calculated for one period. Its present value is R(l + i)-I. The second payment of Rs. R is made after 2 payment periods and the present value of this Rs. R is R (1 + i)-2 and so on. This is illustrated in the following figure: 1

2

3 ...... n-1 n-1 n

ORRR ......

RRR

R(1+i)_1~~ ~

R(1 +i)-2 R (1 + i)-3

R (1 + i)-(n-

2>_-----.....

R (1 + i)-(n-

1>_---------'

R (1 + i)-n

IMMEDIATE ANNUITY

163

Thus, The present value of 1st instalment, PI =R (1 + i)-I The present value of 2nd instalment, P 2 =R (1 + i)-2 The present value of (n -

l)st

instalment, P n-l =R(1 + i)-

smi

=

=>

8mi

=

=>

smi =

=>

.. Siili

=(1 +~). [(1 + i)" i

=>

8mi

=(1 + i)

1

(1 + i)" (1 + i) - 1

1

-i

i (1 + i)" (1 + i) - 1 - i ~

i (1 + i)" (1 + i) - (1 + i)

1]

smi

Thus, the relation between smi and smi is given by

8mi

=(1 + i)

smi

[using (1)]

183

ANNUITY DUE

Remark: Using the above relation, the amount of an annuity due of Rs. R, payable at the beginning of every period for n periods at the rate of i per period can also be given as follows: -=-R-(1-+-i)-S-ii1-i--'\ A

rl-

Example 5 : An amount of Rs. 1,500 is deposited in a bank account at the beginning of every year for a period of 12 years and no payment will be made thereafter. The account holder plans to close the account at the end of 16 years. The bank calculates interest at the rate 71/ 2 per annum compounded annually. What amount he will receive at the time of closure of the account ?

Solution: Since the payments are made at the beginning of every year, it is an annuity due. Period payment R =Rs. 1,500. number of payment periods n =12 rate ofinterest r =i 7~ % =0.075 We first calculate the amount at the end of 12 years, since the annuity is only for 12 year. Amount of an annuity due is given by

=

A

=R [(1 + i~ + 1 -

1_ 1]

A-I 500 [(1.075)13 -1 - , 0.075

_ 1 500 [2.560413 -1 0.075

A- ,

A =1.,500 [20.805507 - 1] A =1,500 (19.805507) =29,708.26 .. The amount at the end of12 years is Rs. 29,708.26. This amount will be in the account for four more years earning co~pound interest. :. Principal at the beginning of 13th year P =Rs. 29,708.26 Compound Amount is given by ,A =P(1 + i)'l i =0.075 and n =4 Here A =29,708.26 (1.075)4 A =29,708.26 (1.335469) =39,674.46 .. , The amount at the 'end of 16 years is Rs. 39,674.46. Example 6 : An annuity consisting of equal payments at the beginning of each quarter for 3 years is to be accumulated to Rs. 7,000. [fthe interest rate is 8% compounded quarterly, how much is each payment ?

184

FINANCIAL MATHEMATICS

Solution: Since the payment are made at the beginning of each quarter; it is an annuity due. Let Rs. R be the quarterly payment Amount after 3 years A = Rs. 7,000 rate of interest r =8% =0.08 Interest is compounded quarterly

; =!..4-- 0.08 _ 0 02 4 - .

..

•

and n = 3 x 4 = 12 Amount of an annuity due is given by "(1 ~ i)n+ 1 - 1 A=R [ . . £

1]

7000 - R [(1.02)13 -1 , 0.02

1]

1]

7 000 - R [1.2936066 - 1 , 0.02 7,000 =R [13.68033] 7,000 R = 13.68033 = 511.68

.. TPe quarterly payment is Rs. 511.68.

Example 7 : Find the amount of an annuity due of Rs. 1,850 payable at the beginning of every year for 15 years, if the irtterest is compounded at a rate of 7% effective for 10 years and 7~ effective thereafter. Solution: Since the payments are made at the beginning of every year, it is an annuity due Periodic payment R =Rs. 1,850 The annuity in the problem is shown in the following figure: 0 I R 1111

1 I R

2

I R

3 I R

4 I R

5 I R

6

7

I R

I R

at 7% effective

8 I R

9

I R

10 I R

11 I R

_'III

12 I R

13 I R

7.5% effective

14 I R

15 I

-,

Since the interest rate is changed after 10 years, the given annuity can be split into two : one is an annuity due of Rs. 1,850 for 10 years at 7% effective and another one is an annuity due ofRs. 1,850 for 5 years at 7.5% effective.

Consider the annuity due of Rs. 1,850 for 10 years at 7% effective: Periodic payment R = Rs. 1,850 No. of payment periods n =10 Rate of interest i =r =7% =0.07

ANNUITY DUE

185

Amount of an annuity is given by

=R [sn+ 1Ii -1] A = 1,850 [ smO.07 -

A

1] A =1,850 [15.7835993 - 1] A = 1,850 (14.7835993) = 27,349.66 This is the amount at the end of 10 years. Since the annuity due ends at the end of 15 year, this amount will earn interest for 5 years at 7.5% effective. :. Principal for 11th year, P =Rs. 27,349.66 No. of periods =5 'years Amount is given by A =P (1 + iY' A = 27349.66 (1.075)5 A = 27349.66 (1.43562933) ... (1) Amount at the end of 15 years =Rs. 39,263.97 Consider the annuity due of Rs. 1,850 for last 5 year at 7.5% effective: Periodic payment R =Rs. 1,850 , Rate of interest i =r = 7.5% = 0.075 No. of periods n =5 Amount of this annuity is given by

A

= 1,850 ( s610.075 -

1)

=1,850 (7.2440203 - 1) A =1,850 (6.2440203) =11,551.44

A

... (2)

.. The amount of this annuity is Rs. 11,551.44 Thus the amount ofthe annuity due ofRs. 1,850 for 1,5 years is given by (1) + (2) =39,263.97 + Rs. 11,551.44 :. The required amount is Rs. 50,815.41.

Example 8 : A person deposits Rs. 2,000 per annum for 6 years, the first deposit is made at year zero. After 6 years the annual deposit is enhanced to Rs. 2,500. He closes his account at the end of 1()th year. What is the amount payable to him, if the interest is calculated at 5% effective? Solution: Since the payments are made at the beginning of every year, it is an annuity due. The given annuity consists of two annuities : one is Rs. 2,000 payable for 6 years and another one is Rs. 2,500 payable thereafter for 4 years. This is shown in the following figure:

o

1 2 3 4 5 6 7 8 9 I I I I I I I I I I 2,000 2,000 2,000 2,000 2,000 2,000 2,500 2,500 2,500 2,500 1«

10

I

186

FINANCIAL MATHEMATICS

Consider the annuity of Rs. 2,000 for 6 years at 5% effective Amount at the end of 6 years = 2,000 ( 87)0.05 - 1) =2,,000 (8.1420085 -1) =2,000 (7.1420085) =14,284.02 .. Amount at the end of 6 years is Rs. 14,284.02 This amount will earn compound interest for 4 more years. .. Amount of this annuity at the end of 10 years is A = 14,284.02 (1.05)4 =17326.32 A :;: Rs. 17,326.32 Now Consider the annuity due ofRs. 2,500 for last four years. Amount of this annuity due at the end of 10 years is

=2,500 ( 8510.05 - i) A =2,500 (5.5256313 A =Rs. 11,314.08

... (1)

A

1) =2,500 (4.52556313) ... (2)

.. The amount of the given ~uity due is (1) + (2) =Rs. 17,362.32 + Rs. 11,314.08 =Rs. 28,676.40 :. The required amount is Rs. 28,676.40 Alternative method: The given annuity can also be expressed as follows: 0

1

I

I

2,000 2,000

2 I

3

4

5

6

7

8

9

10

I

I

I

I

I

I

I

I

2,000 2,000 2,000 2,000 2,000 2,000 2,000 2,000 +500 +500 +500 +500

. . Amount oBhe given annuity due =Amount of annuity due of Rs. 2,000 for 10 years + Amount of annuity due ofRs. 500 for 4 years.

=2;000 ( 8rno.05 - 1) + 500 ( 8510.05 - 1) =2,000 (14.2067872 - 1) + 500 (5.5256313 =2,000 (13.2067872) + 500 (4.5256313)

1)

=26,413.57 + 2,262.82 =28,676.39 . . The required amount is Rs. 28,676.39

I §f!i 3jtS

m1--1

L Find the amount at the end of 10 years of an annuity of Rs. 1,500 payable at the beginning of each year until 10 payments have been made of money is worth 8% effective. [ADs. Rs. 15,645.49]

ANNUITY DUE

187

I

2. Find the amount of an annuity due of Rs. 800 payable at the beginning of each year , for 12 years, if money is 'Worth 6% effective. [Ans. Rs. 14,305.71] 3. A person deposits Rs. 2,000 in an account at the beginning of each quarter for 5 years. Ifthe money is worth 7% per annum compounded quarterly, what amount will be accumulated at the end of fifth year? [Ans. Rs. 48,232.78]

4. An annuity consists of 48 monthly payments of Rs. 600, the first being made at the beginning of first month and the subsequent payments at the interval of one month each. If the interest is calculated at the rate of 12% per annum compounded monthly, what is the amount ofthis annuity at the end offour years? [Ans. Rs. 37,100.90] 5. Find the amount of an annuity due to payment of Rs. 500 is made at the beginning of each quarter for 10 years at the rate of 8% per annum compounded quarterly.

[Ans. Rs. 30,805.01] 6. Find the amount of annuity of Rs. 1,600 payable at the beginning of each month for 3 years at the rate of 15% compounded monthly. [Ans. Rs. 73,087.12] 7. Find the amount of the following annuities: (a)

Rs. 1,100 payable at beginning of every year for 9 years at the rate of 6.5% effective. . payable at the beginning of every 6 months for 10 years at the rate of 6% per annum compounded semi-annually.

(b) Rs. 700 (c)

Rs. 500 payable at the beginning of every 3 months for 6 years at the rate of 7% per annum compounded quarterly.

(d)

Rs. 300 payable at the beginning of every month for 6 years at the rate of 6% per . . annum compounded monthly. [Ans. (a) Rs. 13,743.86; (b) Rs. 19,373.54; (c) Rs. 15,013.73; (d) Rs. 26,052.27]

8. For the purpose of their daughter's marriage, a couple decided to deposit Rs. 2,500 in a fund at the beginning of every three months, starting from the girl's fourth birthday. Last payment was made on the 17th birthday. No payment was made thereafter and accumulated amount was withdrawn on the 21 st birthday of the girl. What would have been the amount withdrawn, if the interest was calculated at a rate of 10% per annum compounded quarterly? [Ans. Rs. 4,01,022.88]

9. A person decides to put aside Rs. 150 at the beginning of,every month in a money market fund that pays interest at the rate of 9% compounded monthly. How much money does he have at the end of one year? [Ans. Rs. 1,890.21] 10. What amounL~hould be set aside at the beginning of every year to accumulate a sum of Rs. 1,50,000 at the end of 14 years, if the interest is compounded at a rate of 6.5% per annum compounded annuity? [Ans. Rs. 6,470.49] 11 For his son's higher education, X wants to save a certain amount at the beginning of every three months from now. He estimates that the cost of education would be Rs. 2.5 lakhs. What amount he has to save at the beginni~g of every quarter to accumulate this cost which is required at the end of 12 years from now? Money is [Ans. Rs. 3,308.52] worth 7% compounded quarterly.

188

FINANCIAL MATHEMATICS

12. An annuity of equal payments at the beginning of every 6 months for 8 years is to be calculated for Rs. 20,000. If the interest rate is 18% per annum compounded halfyearly, how much is each payment? [ADs. Rs. 555.96]

13. An amount of Rs. 2,100 is deposited in a savings bank account at the beginning of every 6 month period for a period of 10 years and no payment will be made thereafter. The account holder plans to close the account at the end of 15 years. If the bank calculates interest at the rate of 9% compounded half-yearly, what amount he will receive, at the time of the closure of the account? [ADs. 1,06,913.54] 14. Mr. X wants to purchase a house when he retires after 16 years from now. The price of the house is Rs. 3,00,000 now and it is expected that the price of the house after 16 yeas would be 25%' more than the current price. What amount X has to set aside at the beginning of every quarter, the first payment due now, if money is worth 11% compounded quarterly? [Ans. Rs. 2,146.42] 15. An amount of Rs. 3,600 is deposited at the beginning of every year for 16 years. Interest is calculated at a rate of 5.5% per annum compounded annually. The amount was kept in the account and withdrawn at the end of 20 years. What would [Ans. Rs. 1,15,937.94] have been the amount at the end of 20 years. 16. Find the number of years for which an annuity 'of Rs. 544.25, payable at the beginning of each 6 months accumUlates to Rs. 1,500, if the interest is calculated at 6% compounded semi-annually. [ADs. 10 years] 17. A bank pays interest at a rate of 8% compounded quarterly. At the beginning of each quarter a sum of Rs. 2,000 is deposited into a savings account. After a certain number of years, the amount is accumulated to Rs. 49,600. Find the number of years. [ADs. 5 years] 18. A sum of Rs. 3,800 is deposited in an account at the begin~g of every quarter for 7 years. If the bank calculates interest at 6% per annum compounded quarterly for 5 years and at 7% per annum compounded quarterly thereafter, find the amount of this annuity. [ADs. Rs. 1,35,360.82] 19. Mr. X has decided to invest Rs. 2,400 at the beginning of every 6-month period. He did so for 5 years. Due to some financial problems, he could not make the payment for next 3 years. He again invested Rs. 3,600 per six months, for the next four years from the beginning of the 9th year. Find the amount to his credit at the at the end of the 12th year assuming interest at the rate of 9% per annum compounded semi[ADs. Rs. 92,362.38] annually. I 20. An annuity is payable for 14 years certain, the first payment falling due at the beginning of first year. The annuity is payable at the rate of Rs. 1,800 per annum during the first 8 years and Rs. 1,500 per annum during the remaining years. Calculate the future value of this annuity of money is worth 5.5% effective. [ADs. Rs. 36,355.53] 21. At the beginning of every year, Rs. 650 is deposited into a savings account that pays interest at a ratio of 5% effective. After 7 years, the bank raises the interest rate from 5% to 6%. If the account is to be closed at the end of 12 years, what amount [ADs. Rs. 11,320.37] would be received?

-189

ANNiIITY DUE

PRESENT VALUE OF AN ANNUITY DUE The present value of an annuity due is the sum of the preset values of all the payments. . Consider an annuity due of Rs. R payable for n consecutive payment periods, where the interest rate is i per period and each payment is due at beginning of the corresponding payment period. The first payment of Rs. R is made at the beginning of the first period, i.e., at the beginning of the term. Therefore, its present value is,Rs. R itself. The second payment of Rs. R is made at the beginning of the second payment period, i.e., immediately after the first payment periods. Therefore, the present value of this R is -R (1 + i)-I and so on. ' This is illustrated in the following figure: 0

I

R

1

2

R

R

I

I

......

n-2

n-1

R

R

I

R R (1 + 1)-1 R (1 + 1)-2

:

R(1

_ _ _ _ _ _ _ _ _ _ _ _...J

+~- A 2 ,

••• ,A,J.

• • •

Within the sample space, there exists an event B, which actually occurs. The analytical goal is to compute a conditional probability of the form: p(Ak I B). At least one of the two sets of probabilities are known: (i) P(Ak n B) for each A k, and (ii) peAk) and pCB I A k) for each Ak

Example 11 : In recent years, it has rained only 5 days each year. The weatherman has predicted rain for tomorrow. When it actually rains, the weatherman correctly forecasts rain 90% of the time. When it doesn't rain, he incorrectly forecasts rain 10% of the time. What is the probability that it will rain tomorrow? Solution: The sample space is defined by two mutually-exclusive events - it rains or it does not rain. Additionally, a third event occurs' when the weatherman predicts rain. Event AI: It rains tomorrow. EventA2: It does not rain tomorrow. Event B: The weatherman predicts rain. In terms of probabilities, we calculate the following: p(Al) = 5/365 =0.0136985 [It rains 5 days out of the year.] p(A2) = 360/365

=0.9863014 [It doe~ not rain 360 days out of the year.]

=0.9 [When it rains, the weatherman predicts rain 90% of the time.] I A 2 ) = 0.1 [When it does not rain, the weatherman predicts rain 10% ofthe time.]

p(B I AI) p(B

The problem is to find p(AI I B). The answer can be determined from Bayes' theorem, as shown below. p(A I ) p(B I AI) peAl I B) =p(A I ) p(B I AI) + p(A2) p(B I A 2 ) (0.014)(0.9)

p(A 1 I B)

=(0.014)(0.9) + (0.986)(0.1) =0.111

.. The required probability is 0.111

RANDOM VARIABLE AND PROBABILITY DISTRIBUTION

Random Variable When the numerical value of a variable is determined by a chance event, that variable is called a random variable. A variable is random, if it takes on different values as a result of

288

FINANCIAL MATHEMAnCS

the outcomes of a random experiment. A random variable is also known as a chance variable or a stochastic variable. Random variables can be discrete or continuous. If a random variable takes only a limited number of values. which can be listed in such a manner as O. 1. 2 •....• it is a discrete random variable. Discrete random variables take on only integer values. Suppose. for example. that we.flip three coins ,and count the number of heads. The number of heads results from this random process is O. 1. 2. or 3. Therefore. the number of heads is a discrete random variable. Similarly. number of defective items in a sample. number of printing mistakes in each page. and number of phone calls received by a customer care centre are some examples of discrete random variable. If a random variable takes any value with in a certain range. it is a continuous random variable, For example, the average heig~t of a randomly selected group of boys is a continuous random variable, because it could be a non-integer. Any variable involving measurement of height, weight, time, volume. or etc is essentially a continuous random variable.

Probability Distribution A probability distribution is a table or a listing that links each possible value that a random variable can assume with its probability of occurrence. A probability distribution can be either discrete probability distribution or continuous probability distribution. Generally, we use a capital letter to represent a random variable and a lower-case letter, to represent one ofits values. For example. • •

•

X represents the random variable X. p(X) represents the probability ofX. p(X = r) refers to the probability that the random variable X is equal to a particular value, denoted by r. As an example. P(X = 1) refers to the probability that the random variable X is equal to'''1: "

The probability distribution of a discrete random variable is known as discrete probability distribution. It can always be represented by a table. For example, suppose you flip a coin two times. This simple exercise can have four possible outcomes: HH. HT, TH, and TT. Now, let the variable X represent the number of heads that result from the coin flips. ' The variable X can take on the values 0, 1, or 2; and X is a discrete random variable. The table below shows the probabilities associated with each possible value of the X. The probability of getting 0 heads is 0.25; 1 head, 0.50; and 2 heads. 0.25. Thus. the table below, which associates each outcome with its probability, is an example of a probability distribution for a discrete random variable.

Number of heads, X 0 1

2

Probability, p(X) 0.25 0.50 0.25

The following are discrete probability distributions: •

Binomial distribution

THEORYOFPROBAB~nY

289

•

• • •

Hypergeometric diStribution Multinomial distribution Poisson distribution

The probability distribution of a continuous random variable is knoWn as continuous probability distribution. It is represented by an .equation, called the probability density function. With a continuous diStribution, there are an infinite number of values between any two data points. All probability density functions satisfy the following conditions: • • •

The random variable Y is a function of X; that is, y =fix). The value of y is greater than or equal to zero for all values of x. The total area under the curve of the function is equal to one.

Since the continuous random variable is defined over a continuous range of values; the/~ graph of the density functio~ will also be continuous over that range. The area bounded by the curve of the density function and the x-axis is equal to 1. The probabUity that a random variable assumes a value between a and b is equal to the area under the density function bounded by a andb. For example, consider the probability density function shown in the graph below. Suppose we want to know the probability that the random variable X is less than or equal to a. The probability that X is less than or equal to a is equal to the area under the curve bounded by a and minus infinity as indicated by the shaded area. The following are discrete probability distributions: • • • •

a Figure 1

Normal distribution Student's t distribution Chi-square distribution F distribution ,

In this chapter, we discuss the commonly used probability distributions, namely, Binomial distribution, Poisson distribution, and Normal distribution.

Expected V~ue . The mean of the discrete random variable X is also called the expected value of X. Notationally, the expected value of X is denoted by E(X). We use the following formula to compute the expected value of a discrete random variable. E(X)

=px =I

[Xi xp(Xi )]

where Xi is the value of the random variable for outcome i, px is the mean of random variable X, and p(Xi) is the probability that the random variable will be outcome i.

Remark: Suppose you h':lve two variables. If X and 'Y are random variables, then E(X + Y)

=E(X) + E(Y)

and

E ()( - Y)

=E(X) -

E(Y)

290

RNANCIAL MATHEMATICS

where E(X) is the expected value (mean) of X, E(Y) is the expected value ofY, E(){ + Y) is the expected value of the sum of X and Y, and E(X - Y) is the expected value of the difference between X and Y. Example 12 : In a recent little league softball game, each player went to bat 4 times. The number of hits made by each player is described by the following probability distribution.

Number of hits, X

0

1

2

3

4

Probability, p(X)

0.10

0.20

0.30

0.25

0.15

What is the expected value of the probability distribution?

Solution: The given distribution is a discrete probability distribution. The expected value of a discrete random variable is computed as follows. E(X) = fIX = }: [Xi x p(Xi) ] E(X) = 0 x 0.10 + 1 x 0.20 + 2 x 0.30 + 3 x 0.25 + 4 x 0.15 = 2.15 . . The expected value is 2.15. Example 13 : The table on the right shows the joint probability distribution between two random variables - X and Y. (In a joint probability distribution table, numbers in the cells of the table represent the probability that particular values ofX and Yoccur together.)

x

0

1

2

3

0.1

4

0.1

0.2 0.2

0.2 0.2

Y

What is the expected value of the sum ofX and Y?

Solution: The given distributions are discrete probability distributions. First we find the expected value of the discrete random variable X E(X) = fIX = }: [Xi x p(Xi ) J E(X) = 0 x (0.1 + 0.1) + 1 x (0.2 + 0.2) + 2 x (0.2 + 0.2) = 0 + 0.4 + 0.8 = 1.2 Next, we find the expected value ofthe discrete random variable Y

E(Y) = pY = }: [Yi x p(Yi )] E(Y) = 3 x (0.1 + 0.2 + 0.2) + 4 x (0.1 + 0.2 + 0.2) E(Y) =(3 x 0.5) + (4 x 0.5) = 1.5 + 2 =3.5 And finally the mean of the sum of X and Y is equal to the sum of the means. I

E(X + Y)

=E(X) + E(Y) =1.2 + 3.5 =4.7

Remark: A similar approach is used to find differences between means. The difference between X and Y is E(X - Y) =E(X) - E(Y) = 1.2 - 3.5 =-2.3; and the difference between Y and X is E(Y - X) = E(y) - E(X) = 3.5 - 1.2 = 2.3

THEORY OF PROBABILITY

291

BINOMIAL DISTRIBUTION A binomial experiment (also known as a Bernoulli trial) is a statistical experiment that has the following properties: •

The experiment consists of n repeated trials.

•

Each trial can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure.

•

The probability of success, denoted by p, is the same on every trial.

•

The trials are independent; that is, outcome on other trials.

t~e

outcome on one trial does not affect the

We consider the following st,atistical experiment of tossing a coin 2 times and count the number of times the coin lands on heads. This is a binomial experiment because: •

The experiment consists of repeated trials. We flip a coin 2 times.

•

Each trial can result in just two possible outcomes - heads or tails.

•

The probability of success is constant - 0.5 on every trial.

•

The trials are independent; that is, getting heads on one trial does not affect whether we get heads on other trials. /

\

The following are some helpful notations in binomial probability distribution: r: The number of successes that result from the binomial experiment. n: The number of trials in the binomial experiment. p: The probability of success on an individual trial. q: The probability of failure on an individual trial. (This is equal to 1 - p.) A binomial random variable is the number of successes r in n repeated trials of a binomial experiment. The probability distribution of a binomial random variable is called a binomial distribution (also known as a Bernoulli distribution). The binomial probability refers to the probability that a binomial experiment results in' exactly r successes. Suppose a binomial experiment consists of n trials and results in x successes. If the probability of success on an individual trial is p, then the binomial probability is: p(X = r) =nCr xpr xpn-r The binomial distribution has the following properties: • • •

The mean of the distribution (p,x) = np. The variance. (ox 2 ) of the distribution = npq. The standard deviation (ox) of the distribution

=Vnpq.

A cumulative binomial probability refers to the probability that the binomial random variable falls within a specified range (e.g., is greater than or equal to a stated lower limit and less than or equal to a stated upper limit). In this case, we calculate the all the individual probabilities of successes and add them.

292

RNANCMLMA7HEMAnCS

"' E~ple

13 : Suppose a die is tossed 5 times. What is the probability of getting exactly 2

fours? \

Solution: This is a binomial experiment in which the number of trials (n) is equal to 5. 'Let X be the event of success of getting 4 a~d p be the probability of getting exactly 4. The probability of success on a single trial is p

1

="6 =

0.167.

q =·1-p = 1- 0.167 = 0.833

The number of successes, r =2 The binomial probability of success is given by : p(X =r)

=nCr xpr xpn-r

p(X =2) =5C2 X (0.167)2 p(X =2) =0.161

X

(0.833)3

.. The required probability is 0.161.

Example 14 : The probability that a student is accepted to a prestigious college is 0.3. If 5 students from the same school apply. What is the probability that at most 2 are accepted?

Solution: Let X be the event that a student is accepted to a prestigious college. Given that p =0.3 and therefore q =1- p =1 - 0.3 =0.7 Number of students (trials), n = 5 The binomial probability of success is given by: p(X =r)

=nCr xpr xpn-r

.

The probability that at most 2 students are accepted = p(X s r) To find the probability that at most 2 students are accepted, we compute 3 individual probabilities p(X = 0), p(X = 1), and p(X = 2) and using the binomial formula, we find the sum of all these probabilities. p(X S r) = p(X = 0). + p(X = 1) + p(X = 2) p(X s 2) =5CO X (0.167)0 x (0.833)5 + 5C1 x (0.167)1 x (0.833)4 + 5C2 x (0.167)2 x (0.833)2 p(K.s 2) =0.1681 + 0.3601 + 0.3087 =0.8369 . . The required probability is 0.8369.

Example 15 : If the probability of producing a defective item is 0.10, find the mean·and standard deviation for the distribution of defective items in a total of 400 items produced?

Solution: Let p be the probability of producing a defective item. Given thatp = 0.10 ~ q Number of items n = 400

=1-p =0.90

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The mean of the distribution~) =np. p,,, =400 x 0.10 =40 The standard de.viation of the distribution 0" =Vnpq 0" =V400 x 0.1 x 0.9 =6

POISSON DISTRIBUTION Poisson distribution is a discrete probability distribution which was developed by a French mathematician, Simeon Denis Poisson in 1837. Poisson distribution may be expected in cases where the probability of success of an event is very small. For example, aircraft accident, serious flood, and etc are some rare events and the probability of occurrence of such events is considerably small. The probability distribution of a Poisson random variable is called a Poisson distribution. A Poisson random variable is the number of successes that result from a Poisson experiment. A Poisson experiment is a statistical experiment in which the experiment results in outcomes that can be classified as successes or failures and the probability that a success will occur in an extremely small. The following are some helpful notation in the Poisson distribution. • •

•

e: A constant equal to approximately 2.71828. (Actually, e is the base of the natural logarithm system.) p, : The mean number of successes that occur in a specified region. r : The actual number of successes that occur in a specified region.

Formula: Suppose we conduct a Poisson experiment, in which the average number of successes Within a given region is p,. Then, the Poisson probability is: e-JS p,r r.

p(X=r)=-,-

where r is the actual number of successes that result from the experiment. The Poisson distribution has the following properties: • •

The mean of the distribution i~ equal to p,. The variance is also equal to p,.

Example 16 : The average number of homes sold by the Acme Realty company is 2 homes per day. What is the probability that exactly 3 homes will be said tomorrow?

Solution: This is a Poisson experiment in which p, =2; since 2 homes are sold per day, on average and r = 3; since we want to pnd the likelihood that 3 homes will be sold tomorrow. We know that e =2.71828 The Poisson probability is given by: e-JSpl

p(X= r) =

r.,

..

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p (X -

3) _ 2.71828- 2 x 23 _ 0.13534 x 8 - 0 8 3! 6 -.1

.. The probability of selling 3 homes tomorrow is 0.180. Example 17 : Suppose the average number of defective items produced by a machine in a day is 5. What is the probability that the number of defective items will be fewer than 3 on the next day? .

Solution: This is a Poisson experiment in which f.L day, on average.

= 5; since 5 items are produced per

Here r = 0, 1, or 2; since we want to find the likelihood that the number of defective' items will be fewer than 3. To solve this problem, we need to find the probability p(X < 3).

p(X < 3) = p(X = 0) + p'{X ='1) + P(X = 2) The Poisson probability is given by: p(X= r)

=e-1Jr.pt I e-5 x 50

p(X < 3) (X p