AN INTRODUCTION TO THE MATHEMATICS OF FINANCE

BY

A. H. POLLARD M.Sc, M.Sc. (Econ.) , Ph.D., F.I.A.

Professor of Economic Statistics, Macquaric University

PERGAMON PRESS S Y D N E Y · L O N D O N · E D I N B U R G H · N E W Y O R K

T O R O N T O · O X F O R D · PARIS · B R A U N S C H W E I G

Pergamon Press (Australia) Pty Limited, 19a Boundary Street, Rushcutters Bay, NSW 2011 A. Wheaton & Company, A Division of Pergamon Press, Hennock Road, Exeter, EX2 8RP Pergamon Press Ltd, Headington Hil l Hall , Oxford OX3 OBW Pergamon Press Inc., Maxwell House, Fairview Park, Elmsford, New York 10523 Pergamon of Canada Ltd, 75 T h e East Mall, Toronto , Ontario M8Z 2L9, Canada Pergamon Press G m b H , 6242 Kronberg/Taunus , Pferdstrasse 1, Frankfurt-am-Main, West Germany Pergamon Press SARL, 24 rue des Ecoles, 75240 Paris, Cedex 05, France

© 1968 Pergamon Press Australia Second edition 1977

Typeset in Australia by Press Linotype Co. Pty Ltd Printed in Hong Kong by Wing King T o n g Co. Ltd

Pollard, Alfred Hurlstone, 1916— An introduction to the mathematics of finance

ISBN 0 08 021796 6 1. Business mathematics—Problems, exercises etc. I. T i t l e

510. 2465

All rights reserved. N o part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means: electronic, electrostatic, magnetic tape, mechanical, photo-copying, recording or otherwise, without permission in writing from Pergamon Press (Australia) Pty Limited.

P R E F A C E

Many people who know little about this subject consider it to be very simple. They think of it as a small chapter in a school ari thmetic book. I have heard it described as "nothing more than simple geometrical progression." Th is is a correct description; bu t the fact remains that in the difficult actuarial examinations it has frequently been a s tumbling block for honours graduates who have easily cleared the hurdles of probabili ty and of mathematical statistics. I t can therefore be a most difficult subject calling for clear thinking and ingenuity of a high order.

Th i s volume, however, is intended as a simple, non-mathematical introduction to the subject, and it omits, as far as possible, unnecessary complications. I t aims to give people who will be concerned with problems of growth— accountants, economists, investment officers, demographers— a working knowledge of the subject to enable them to solve the rout ine day-by-day problems.

Dur ing many years in the financial world I was frequently amazed at the lack of knowledge of these techniques among people whom one would expect to be experts in it. This book was written primarily to provide an elementary intro-duction to the subject for students of the School of Economic and Financial Studies at Macquarie University.

Mathematics of Finance is a subject where the lecture content is small but by comparison the practical content is large. T h e student should be warned that unless he solves a large number of problems he will not develop the ability or confidence to deal with new situations. One other warning should be given—in this subject more than any other, the student must ensure that, before a t tempt ing a new chapter, he has mastered all previous chapters. Failure to do this will inevitably lead to trouble.

A. H. P O L L A R D . Macquarie University. May, 1968.

C H A P T E R O N E

SIMPLE INTEREST

Interest (I) is the income earned from investing capital; it is the amount paid for the use of money.

The rate of interest (i) is the amount contracted to be paid for the use of one uni t of capital ($1) for one uni t of time (1 year) . It is often expressed in the form "per cent per annum." Thus , with interest at 6% per annum, i = 0.06.

T h e sum of money invested is called the principal (Ρ ) , and the sum of interest and principal owing at any time is called the amount (A) .

W h e n the interest for any period is charged on the principal only it is called simple interest.

Notes and commercial bills, both of which are in common use, take the form of a writ ten promise to pay a certain sum of money on a specified date. Th is date is called the date of maturity and the sum is called the maturity value. T h e interval between the date of issue and the date of maturi ty is called the term.

Commercial bills before their maturi ty date are regularly sold on the market for an amount a little less than their maturi ty value. Th i s difference is called the discount. T h e rate of discount for a given period is usually expressed as a percentage per annum of the maturity value. T h e amount of money obtained on the sale of a commercial bill is called the proceeds.

Problem 1: Find the simple interest on $250 for 6 months at 6% p.a.

Solution: $250 would earn $250 X .06 in one year. Hence it would

earn $250 X .06 X i or $7.50 in 6 months. I = $7.50.

S I M P L E I N T E R E S T

Problem 2: If $500 is loaned at 3 % p.a. for 150 days, how much

interest is earned and what is the amount at the end of the period? Solution:

$500 earns $500 X .03 in one year. Hence it earns $500

150 χ .03 X or $6.16 in 150 days.

365

/ . I = $6.16 and A = $506.16.

Problem 3:

I paid a friend $1010 for the $1000 he loaned me 50 days ago. W h a t rate of simple interest did I pay? Solution:

He earned $10 for 50 days. At this rate he would have

365 earned $10 X or $73.00 for a full year.

50 73.00

.·. i = = .073 or 7 .3% p.a. 1000

Problem 4: 190 days after borrowing money a man pays back $1000.

How m u d i was borrowed if the $1000 payment includes principal and simple interest at 8% per annum? Solution:

Let the principal = P. 190

T h e n the interest for 190 days = Ρ X .08 X 365

= .041644P the amount = 1.041644P

i.e. 1.041644P = 1000

1000 Ρ - = $960.02

1.041644

S I M P L E I N T E R E S T 3

Problem 5: A bill for $785 due on 14th June is sold on the 28th March,

the discount rate being 7% p.a. Wha t are the proceeds? Solution:

T h e number of days to run is determined as follows:— in March 3 in April 30 in May 31 in June 14

78 days 78

Hence the discount = 785 X .07 X 365

= $11.74 .'. the proceeds = 785 — 11.74

= $773.26

EXERCISES

1-1 A 'bill for $500 is payable at the end of 90 days or 2 % off for cash. If the cash price is considered as a 90-day investment, what simple interest rate does it earn?

1-2 W h a t principal will amount to $1020 in 6 months it the simple interest rate is 4 % p.a.?

1-3 How many days will it take for $600 to earn $4 if it is invested at 5 % p.a. simple interest?

1-4 Find the simple interest on $900 at 4 ^ % p.a. from April 1 to May 16.

1-5 Find the simple interest rate which will cause $2400, invested on May 31, to amount to $2454 on October 18.

1-6 How much should be invested at 6% p.a. simple interest on May 15 so that the amount will be $2550 on September 12?

4 SIMPLE I N T E R E S T

1-7 An investor pays $147 for a note matur ing for $150 in 4 months. Wha t is the discount rate? Wha t simple interest rate does he earn on his investment?

1-8 A note for $400 matures on September 20. It is sold at 4 % p.a. discount on August 21. Find the proceeds of the sale.

1-9 I wish to obtain $1000 from a bank as a 30-day loan. How much should I ask for if the bank charges 6% p.a. interest in advance?

1-10 Find the maturi ty value of the following note and the date of maturi ty:

August 10, 1967. Ninety days after date I promise to pay to O. W. Holmes or order the sum of $300 and simple interest at 5 % per annum.

(Signed) L. S. Dee. If the note is sold on a 6% p.a. discount basis on September 9, 1967, find the proceeds of the sale.

1-11 A 3-month note expressed as being for $300 plus interest at 5 % p.a. is sold 60 days before maturity. T h e pur-chaser wants 6% p.a. simple interest on his investment. How much discount is charged, and what are the proceeds of the sale?

C H A P T E R T W O

COMPOUND INTEREST—Annual Compounding

When the interest earned is periodically added to the prin-cipal and the total amount of principal and interest earns interest for the following period, then the difference between the final amount and the original principal invested is called compound interest.

For the purposes of this chapter we shall assume that interest is added to the principal—a process called compounding—at intervals of 12 months.

If Ρ represents the principal at the beginning of the first year, i the rate of interest per annum—then the interest earned in the first year is Pi and the amount at the end of the first year is Ρ + Pi or Ρ (1+ i) . Th i s latter amount earns interest for the second year equal to Ρ (1 + i) i so that the amount at the end of the second year is Ρ (1 + i) (1 + i) or Ρ (1 + i)

2 and so on.

T h u s after η periods the final compound amount S = P ( l + i)

n

There are four quantit ies in this equat ion—the principal P, the rate of interest i, the term η and the amount S. Given any three, the fourth may be obtained from the equation. Tables of (1 + i)

n for various values of i and η are published

for this purpose. (See tables following Page 68.)

Problem 1: Find the amount if $500 is invested for 10 years at 6% per

annum, interest being compounded annually. Solution:

S = 500 (1.06) 10

= 500 X 1.79085 from the tables = $895.42

6 C O M P O U N D I N T E R E S T

Problem 2: Find the amount which $100 will accumulate to over 200

years if invested at 6% per annum, interest being compounded annually. Solution:

S = 100 (1.06) 2 00

= 100 (1.06) 1 00

(1.06) 1 00

= 100 Χ 339.30208 Χ 339.30208 = $11,512,205 or $11.51 million

[N.B. I t is necessary to split (1.06) 2 00

into factors since the tables do not give values for η > 100.]

Problem 3: Find the amount which $500 will accumulate to over 12

years at 5.2% p.a. compounding annually. Solution:

S = 500 (1.052) 12

= 500[ (1.79586 + | of (1.90121 — 1.79586)]

= 500(1.79586 + .04214) = 500 X 1.83800 = $919

(N.B. Tables are not published at 5.2% interest and we have estimated the value from the 5 % and the 5 J % tables by the above process called interpolation. We could alternatively have evaluated S from the above equat ion by logarithms.)

Problem 4: I am to be given $10,000 in 15 years' time. W h a t is the

present value at 5 % p.a., interest being compounded annually? Solution:

10,000 = Ρ (1 .05)1 5

10,000 Ρ =

(1 .05)1 5

= 10,000 Χ .48102 = $4,810

C O M P O U N D I N T E R E S T 7

1

[N.B. The re are tables of (usually written vn)

(1 + i)n

prepared for determining present values. (See Page 68 onwards.]

Problem 5: Find the amount which $10,000 will accumulate to if in-

vested for 15 years 3 months at 6% p.a. compounded annually.

Solution: S = 10,000 (1.06)

1 5*

= 10,000[2.39656 + J (2.54035 — 2.39656)] = 10,000 X 2.43251 = $24,325 by interpolat ion.

We could alternatively have used compound interest tables for 15 years and added 3 months ' simple interest on the amount at the end of 15 years.

Thus , S = 10,000 (1 .06)1 5 + 3 months ' simple interest

= 10,000(2.39656) + 3 months ' simple interest = 23,965.6 + 23,965.6 X .015 = $24,325

Or, by logarithms log S = log 10,000 + 1 5 1 log 1.06

= 4 + \b\ X 0.025306 = 4.38592

. · . S = $24,320

Problem 6: $1500 amounts to $2400 in 9 years. W h a t is the compound

rate of interest?

Solution: T h e rate of interest i is given by

1500(1 + i )° = 2400 that is by (1 + i)» = 1.6

From the tables we note that (1.05) 9 = 1.55133

(1.055) 9 = 1.61909

Hence the answer to our problem is between 5 % and 5 | % .

8 C O M P O U N D I N T E R E S T

It is near enough to work out the answer by proport ion. At 5 % we obtain 1.55133 At i we obtain 1.6 At 5 | % we obtain 1.61909

Hence 1.6 — 1.55133

51 — 5 1.61909 — 1.55133

= 0.7183 = 5 + ι X 0.7183 - 5.36% p.a.

Problem 7: With interest at 6% p.a. compound, how long will it take

$100 to amount to $150?

Solution: 100(1.06)* = 150 Λ (1.06)* = 1.5 By interpolation from the tables

1.5 —1.41852 η = 6 -|-

1.50363 — 1.41852

= 6.957 years OR, by logarithms,

η log 1.06 = log 1.5

log ί.5 η =

log 1.06

0.17609

0.025406 = 6.9585 years

T h e difference between the two approximations is less than half a day.

C O M P O U N D I N T E R E S T ο

EXERCISES

2-1 How long will it take $100 to accumulate to $200 at 6 % p.a. compound?

2-2 An investment costs $75 and pays $100 ten years later. Wha t is the effective compound interest yield?

2-3 At a given rate of interest $1000 amounts to $1500 in 10 years. Wha t will it amount to after 6 years?

2-4 My only assets consist of the following debts:

W h a t is the value of my assets at the present time? (Interest 6% p.a. compound.)

2-5 Assuming any amounts received are invested to earn 6% p.a. compound, what will the value of my assets in Exercise 2-4 be in:—

Amount Due at the end of 150 200 300 200 500

1000 600

1 year 2 years 3 years 4 years 5 years

10 years 12 years

(1) 3 years? (2) 5 years? (8) 8 years? (4) 12 years?

C H A P T E R T H R E E

ANNUITIES—CERTAIN

An annuity is a series of payments at fixed equal intervals of time, e.g. payments of rent, life assurance payments, pen-sions, instalment payments. Some annuities are only payable if a person is alive (e.g. life assurance premiums or pensions) , others are payable for a fixed term, no mat ter what happens. T h e latter are called annuities-certain.

We shall deal in this chapter only with annuities—certain and only with annuities where the payments are made annually. When the payments are made at the beginning of the intervals the annuity is called an annuity-due; when they are made at the end of the intervals the annuity is called an immediate annuity or an ordinary annuity.

T h e amount which an immediate annuity of 1 per annum will, with compound interest, accumulate to at the end of η years is written S^j. By accumulating each payment (from the last backwards) with interest to the end of η years

$ Π = 1 + ( 1 + 0 + (1 + 0 2

+ · · · · + 0 + 0n _1

(1 + t ) » — 1 = summing the geometrical progression.

i Tables of are published for various values of η and i.

(See Page 68 onwards.) T h e amount of an annuity-due at the end of η years may

be obtained in the same way. I t is, however, the same as an immediate annuity for η + 1 years minus the final payment and hence its amount must equal

— 1

A N N U I T I E S C E R T A I N 11

T h e present value of an η year immediate annuity of 1 per

annum discounted at i% p.a. compound is writ ten a^\

1 1 1

a^= + + + 1 + i (1 + i)

2 (1 + i)

n

1 — (1 + i)~n 1 — v

n

— or

/ i Tables of a^ are published for various values of η and i.

(See Page 68 onwards.) T h e present value of an annuity-due may be obtained in

the same way. It is clear, however, that it is equal to the first payment plus a η—1 year immediate annuity. Hence its present value equals 1 + a^zn.

If payments are to be made for ever, the annuity is called a perpetuity. T h e present value of a perpetuity, from the above

1 1 expression equals — and of a perpetuity-due 1 + —.

i i Problem 1:

I have $1000 invested at 6% per a n n u m compound and I propose to invest $200 on the same terms at the end of each year for the next 12 years. W h a t will my investment then amount to? Solution:

Amount = 1000 (1 .06)1 2 + 2 0 0 S ^

= 1000 X 2.01220 + 200 X 16.8699 = $5,386.18

Problem 2: A local council is paying off a debt by payments of $5000

at the end of each year. Interest is being charged at 5 J % . W h a t is the outs tanding debt if it has just made a payment and there are still to be made 15 further payments of $5000 and a final payment of $2000 at the end of the 16th year? Solution:

2000 Debt outs tanding = 5000αϊη +

(1 .055)1 6

= 5000 X 10.0376 + 2000 X 0.42458 = $51,037

12 A N N U I T I E S C E R T A I N

Problem 3: On my 16th birthday I was given an annuity of $500 per

year for 10 years, the first payment to be made on my 21st birthday. W h a t was the gift worth at 5 % p.a. compound interest-Solution:

This equals an annuity of $500 payable on each birthday from my 17th to my 30th inclusive (i.e. 14 payments) minus an annuity of $500 on each bir thday from my 17th to my 20th (i.e. 4 pavments) . Hence the present value equals

5 0 0 ^ — 5 0 0 ^ = 500 (9.8986 — 3.5460) = $3,176

This is called a deferred annuity.

Problem 4: A house is purchased for $12,500. A deposit of $4250 is

required. W h a t annual payment is required to pay off the balance over 15 years at 6 | % p.a. compound? Solution:

T h e loan = 12,500 — 4250 = $8250. If X is the annual repayment, then the present value of

the repayments equals X a j i n

= 9.4027 X 9.4027 X = 8250

8250 X = = $877.41

9.4027

Problem 5: If no payments were made for the first five years, what

annual payments would then be required to pay off the house by the end of 15 years? Solution:

Let Y be the annual payment. T h e n 8250 = Y ( λ ι ΊΠ — a n )

8250

Y = = $1,572.33 9.4027 — 4.1557

A N N U I T I E S C E R T A I N 13

Problem 6:

If only interest is paid dur ing the first 5 years, what subse-quent annual repayments would be required to pay off the house by the end of 20 years-Solution:

Let Ζ be the required annual payment. T h e n 8250 = 8250 X 0.065«^ + Ζ (a^ — a^)

8250 — 8250 X 0.065 X 4.1557 Ζ -

11.0185 — 4.1557

- $877.41 (N.B. Th is is the same answer as for Problem 4. Th is is

because, by paying interest only, the loan outs tanding at the end of 5 years is still $8250 which then has to be paid oft over 15 years as in Problem 4.)

Problem 7: A loan of $75,000 is to be repaid by equal annua l instalments

of $6000. How long at 6% p.a. compound will it take to repay the loan? Solution:

Let η — the term of the loan. T h e n 75,000 = βΟΟΟα^

75,000 = = 12.5 at 6 %

6000

From the tables it can be seen that η lies between 23 and 24 years. By interpolation:

η — 23 12.5 — 12.3034

24 — 23 12.5504 — 12.3034

.1966

η = 23 + 23.8 years .2470

It is usual in such cases to pay $6000 at the end of each of the first 23 years and a final payment at the end of the

14 A N N U I T I E S C E R T A I N

24th year. Wha t would this final payment be? Π X is the final payment then we have:

1 75,000 = 6 0 0 0 ^ + X

(1 .06)2 4

. · . X = (75,000 — 6000 X 12.3034) 4.04893 = $4,776

Problem S: A loan of S1000 is to be repaid by 12 annual instalments

of S100, the first being in a year's time. Wha t compound rate of interest is being charged? Solution:

If i is the rate of interest 1000 = 100«τη at rate /

.*. αγο] = 10 at rate i From the tables, / lies between 2 | % and 3 % . By interpolat ion:

3 — i 9.9540 — 10

3 — 2 i 9.9540 — 10.2578

—.0460

—.3038 .

. · . s — i = 0.1514 Χ ι i = 3 — .076 ' = 2.924 = 2.92% p.a.

EXERCISES

3-1 A man buys a home for $12,000. He sells his old home for $7000 and uses this as a deposit. He borrows the balance at 7% p.a. compound interest. Wha t ' amoun t , payable at the end of each year would pay off the loan over 10 years?

3-2 If, in the previous question, he were to pay $500 at the end of each year, how long would it take to pay off the loan and what would be the amount of the fractional instalment at the end of the final year?

A N N U I T I E S C E R T A I N 15

3-3 W h a t is the purchase price of a house which can be bought for $4000 cash plus $400 at the end of each year for 20 years? (Interest 6% p.a. compound.)

3-4 If I deposit $100 in the bank now; $100 regularly at 12 monthly intervals for the next 5 years and $200 regularly at 12 monthly intervals for the following 10 years, what is (i) the accumulated value at the end of the 15 years?; (ii) the present value of the payments? (Assume 5 % p.a. compound interest.)

3-5 Find to the nearest $ the present value at 5 % p.a. compound of an annui ty of 16 annual payments of $11,000 each, the first one being due in 5 years' time. Use two methods.

3-6 On the day of his son's bir th a man wishes to deposit with a Trus t Company a lump sum sufficient to pay him $100 on each of the four birthdays that he might expect to be at the university. Assuming 6% p.a. compound interest and that

fhe first such birthday is

his 17th, what sum is required? 3-7 If instead of paying the Trus t Company a lump sum

he decides to make equal annual payments starting on the day of his son's bir th and on each birthday there-after up to, and including, the 16th, what annual pay-ments should he make? (Interest 6% p.a. compound.)

3-8 If money is worth 6% p.a. which is the cheaper—to purchase a property for $30,000 cash or to pay $15,000 cash and four annual payments of $8000, the first due at the end of 5 years?

3-9 A city can make yearly payments of $25,000 into a savings fund paying interest yearly at 3i-% p.a. How many full deposits and what final deposit will be required to accumulate to $500,000?

3-10 Wha t annual rate of interest must be earned for deposits of $500 at the end of each year to accumulate to $10,000 in 15 years?

3-11 A man agreed to repay a debt by 10 equal annual payments of $1000. He failed to make the first four payments and then inherited some money. He then offered to l iquidate the debt. How much money would then be required? (Interest 6% p.a. compound.)

C H A P T E R F O U R

THE USE OF COMPOUND INTEREST TABLES

Four compound interest tables have been described and. each applies to a particular situation. Let us consider each of these in turn.

1. Tables of (1 + i) » If we require the amount to which $1 will accumulate with

compound interest for η years at rate i per annum, the answer is (1 + ι )

n and is found in the tables—thus:

at 5 % p.a. $1 will amount to $1.27628 i n ' 5 years $1 will amount to $2.65330 in 20 years $100 will amount to $265.330 in 20 years

(1 + ί) η _ ι 2. Tables of or

i

If instead of having a single $1 accumulating with compound interest, we have a numbei η (say) of $1 payments being made at equal intervals of one year, then by means of S^-| tables we can find the amount to which they will accumulate with compound interest up to the date of the last payment. [We could, of course, consider each payment separately and use the (1 + i)

n tables and add all the figures. Th is would be

a long procedure and S-^ tables are introduced to save this labour.]

Thus , assuming compound interest at 5 % p.a. and using (1 + i)

n tables:

$ $1 invested now will in one year be worth ... 1.05 $1 invested in one year will then be worth .... 1.00

To ta l $2.05

T H E U S E O F C O M P O U N D I N T E R E S T T A B L E S 17

Or, using S-^ tables, This is a 2-year annuity of $1 p.a. which, as at the date

of the last payment will be worth Sj] or $2.05.

( ϋ )

$ Similarly, using (1 + i)

n tables

$1 invested now will in 2 years' time be worth (1.05)

2 1.1025

$1 invested in 1 year's time will 1 year later be worth (1.05) 1.05

$1 invested in 2 years' time will then be worth $1 1.00

To ta l $3.1525

Or, using S , t i tables, This is a 3-year annuity of $1 p.a. which, as at the date

of the last payment will be worth or $3.1525.

$ 1.27628

(iii) 1.21551 Similarly, taking each payment separately and using 1.15763

(1 + i)n tables at 5 % , a six-year annuity of $1 1.10250

p.a. will, as at the date of the sixth payment, be 1.05000 worth— 1.00000

$6.80192

Bat, it is much simpler to use tables which tell us immediately that the value of a six-year annuity of $1 p.a. as at the date of the last payment is S(7I

0 1 $6.80192.

Note the words—"As at the date of the last payment." If we require the value at any other time we have to make the appropriate adjustment.

Any problem—where we are concerned to find the amount to which given deposits will accumulate (with interest) as at some future time—can be solved using (1 + i)

n and S-^-\

tables with appropriate juggling. Here are some examples— described graphically for both brevity and simplicity. Pay-ments are indicated by a line and an amount and the time at which we wish to know the accumulated amount is rep-resented by V

18 T H E U S E O F C O M P O U N D I N T E R E S T T A B L E S

(i)

V equals

1 $1 $1 $1 $1 $1 $1

minus

equals Si-\ — Sf] = 5.5256 — 1 = $4.5256

equals

$1 $1 $1 $1

minus

$1 $1 equals — Sj-j

= 6.8019 — 3.1525 = $3.6494

(»)

T H E U S E O F C O M P O U N D I N T E R E S T T A B L E S 19

(iii)

$ 1 $ 1

equals

$ 1 $ 1 $ 1

(iv)

$4 $2

plus

minus

$ 1 $ 1

equals 5j] + Sj] — 2S~n = 6.8019 + 3.1525 — 2 = $7.9544

$2

equals

plus

plus

minus

$ 1 $ 1

I $ 1 $ i

I | 1 | 1

$2

equals S.7, + $Π + 2 (1 .05)3 — 2 S n

= 6.8019 + 4.3101 + 2.3153 — 2 = $11.4273

$2

20 T H E U S E OF C O M P O U N D I N T E R E S T T A B L E S

3. Tables of vn or (1 + i)~

n

These are "present value" tables which give the value noxo of an amount of $1 which will not become clue until η years have elapsed. T h u s „ ^

at 5 % p.a. $

$1 due in 1 year's time is now worth 95238 S I due in 2 year's time is now worth 90703 $1 due in 20 year's time is now worth 37689 S100 due in 20 years' time is now worth 37.689

1 — vn

4. Tables of a-^\ or i

These are "present value" tables which give the value now of a series of equal annual future payments of §1 at yearly intervals, the first payment being in one year s tune.

Thus , if we are to receive 6 annual payments of $1, the first being in one year's time, their present value—taking each payment separately, using v

n

tables and adding is . .

To ta l $5.07570

But this value is obtained immediately from the a-^-\ tables because it is the present value of a 6-year annuity of $1, the first payment being in one year's time, i.e. a^-\or $5.07570

Note the words—"The first payment being in one year s time." If we require the value at any time other than one year before the first annuity payment, we must make the appropria te adjustmen t.

Any problem—where we are concerned to find the present value of a series of future payments—can be solved using v

n

and a-^ tables with appropriate juggling. Here are some examples.

$ .95238 .90703 .86384 .82270 .78353 .74622

T H E U S E O F C O M P O U N D I N T E R E S T T A B L E S 21

(0

|1 |1 $1

(ϋ)

equals

plus 1 $1

$1 I I

equals 1 + a^-\ = 1 + 4.3295 = $5.3295

equals

minus ^ $1 $1

equals a-^\ — aj-[ = 4.3295 — 2.7232 = $1.6063

(iii)

η u

equals

$2 $2

plus

minus

$2

Ϊ1 $1

$2

equals 1 + 2aj] — aj\ = 1 + 8.6590 —~ 1.8594 = $7.7996

22 T H E U S E O F C O M P O U N D I N T E R E S T T A B L E S

(iv)

^ . equals v

$2 $2 $2 $4 $1 $1 $1 $1 $l $1

plus

plus

$1 $1 $1

$2

plus

equals ατ-\ + ατ\ + 2t/3 + 2

= 4.3295 + 2.7232 + 1,7277 + 2 = $10.7804

We can work out the value of a series of payments at any time as a combination of these four tables. T h u s :

(i)

i I i I i

equals

$

plus

I $1 $1 $1

fl $1

equals + αγ\ = 4.3101 + 2.7232 = S7.0333

T H E U S E O F C O M P O U N D I N T E R E S T T A B L E S 23

(»)

I I I I ^ I I equals I I I

$1 $1 $2 $3 $3 $1 $1 $1 $1 $1

plus

$1

plus

plus

equals + $7] + + a

T\

= 5.5256 + 3.1525 + 2.0500 + 1.8594 = $12.5875

EXERCISES

4-1 Wri te clown the formula for the value as at 1/1/68 of each of the following payments.

Payments (a) $100 due 1/1/68. (b) $100 due 1/1/69. (c) $100 due 1/1/73. (d) $100 due 1/1/80. (e) $100 due 1/1/68 and 1/1/69. (f) $100 due 1/1/69 to 1/1/72 incl.

(g) $100 due 1/1/68 to 1/1/72 incl. (h) $100 due 1/1/70 to 1/1/75 incl.

I

$1 $1

24 T H E U S E O F C O M P O U N D I N T E R E S T T A B L E S

(i) S100 due 1/1/68 to 1/1/70 plus $100 due 1/1/72 to 1/1/75 inch

(j) $200 due 1/1/68 to 1/1/72 plus $100 due 1/1/73 to 1/1/75 incl.

(k) $100 due 1/1/69 to 1/1/72 plus $300 due 1/1/73 to 1/1/76 incl.

(1) $100 due 1/1/68 to 1/1/71 plus $900 due 1/1/72 plus $300 due 1/1/73 to 1/1/76 incl.

(m) $100 due 1/1/69 to 1/1/71 plus $300 due 1/1/72 to 1/1/74 plus $200 due 1/1/75.

(n) $100 due 5^ years after 1/1/68.

4-2 Wha t is the value as at 1/1/76 of each of the above payments? You may obtain further practice, if you wish, by working out the value of the above payments on 1st January of any year between 1968 and 1976, or of any year before 1968 or of any year after 1976. (Solutions are given on Page 59 for the value as at 1/1/72.)

C H A P T E R FIVE

LOANS REPAYABLE BY EOUAL ANNUAL INSTALMENTS WHEN INTEREST IS CHARGED

ONLY ON THE AMOUNT OF PRINCIPAL FROM TIME TO TIME OUTSTANDING

Amortization is a term used for the process of paying off a debt (i.e. both principal and interest) by a series of pay-ments. T h e payments are usually equal; they cover the interest due u p to the date of payment and any balance is used to repay par t of the debt.

I t is usually necessary for accounting purposes to know how much of each payment is required for interest and how much is therefore available as a repayment of principal . T h e amount of the debt outstanding at any time can then be presented in the form of a loan repayment schedule.

Problem 1: A loan of $10,000 at 5 % per a n n u m compound is to be

repaid by 5 equal annual instalments. Prepare a loan repay-ment schedule.

Solution: W e first find the amount of the annual payment, X (say).

10,000 = Xaji at 5 %

10,000 X = = $2309.75

4.3294767

(N.B. Tables with a large number of decimal places are necessary to obtain the required accuracy )

T h e information we require in the schedule is as follows:

26 L O A N S R E P A Y A B L E B Y E Q U A L A N N U A L I N S T A L M E N T S

Repayment schedule for a loan of $10,000 at 5% p.a. repayable by 5 equal annual instalments of $2309.75

Year number

Principal outstanding

at beginning of year

(1)

Interest due at

5%

(2)

Principal contained

in payment

(3)

Principal outstanding

at end of year

(4)

1 10,000.00 500.00 1809.75 8190.25 2 8,190.25 409.51· 1900.24 6290.01 3 6,290.01 314.50 1995.25 4294.76 4 4,294.76 214.74 2095.01 2199.75 5 2,199.75 110.00 2199.75 0

Th i s information may be filled in line by line commencing with $10,000.00 principal outstanding on the left hand side of line one and ending with 0 on the right hand side of line 5. Column (2) is 5% of the principal outstanding at the beginning of the year (Column 1) ; the principal contained in each payment (Column 3) is obtained by subtracting the interest (Column 2) from the annual payment of $2309.75 and Column 4 is obtained by subtracting Column 3 from Column 1.

(N.B. We should end up with 0 at the bottom of Column 4. Th i s is a check on the arithmetic.)

Problem 2:

A loan of $50,000 at 5 | % p.a. compound interest is being repaid by 30 equal annual payments. Find, without prepar ing a schedule,

(a) the principal outstanding immediately after the 10th payment has been made;

(b) the interest contained in the 11th payment; (c) the principal contained in the 11th payment; (d) the principal outstanding immediately after the 11th

payment has been made.

L O A N S R E P A Y A B L E B Y E Q U A L A N N U A L I N S T A L M E N T S 27

(N.B. Th i s information would be contained in the repay-ment schedule on the line opposite year 11, but can be obtained quite independently and hence may be used as a spot check on the accuracy of the schedule.)

Solution: First calculate the instalment, X.

50,000 = Xam at 5 J %

50,000 X = = $3440.26

14.53374517

(a) Immediately after the 10th payment has been made the amount then outstanding is repaid by, and therefore equal in value to, 20 equal annual payments of $3440.26. Hence the amount then outstanding equals

3440.26öö7T| at h\%

= 3440.26 X 11.9503825 = $41,112.42

(b) T h e interest due at the time of the 11th payment equals 5 J % on the amount then outstanding, namely $41,112.42, and hence equals

41,112.42 X .055 = $2261.18

(c) T h e balance of the payment, namely 3440.26 — 2261.18 = $1179.08

is available for part repayment of principal.

(d) T h e principal outstanding after the 11th payment is therefore

41,112.42 — 1,179.08 = $39,933.34 As a check this may also be calculated as in (a)

3440.26αττη at H% = 3440.26 Χ 1Ϊ.6076535 = $39,933.34 as before.

L O A N S R E P A Y A B L E B Y E Q U A L A N N U A L I N S T A L M E N T S

EXERCISES

1 A loan of $81.00 is to be repaid by ten equal annual instalments of principal and interest which is at the rate of 5 % per annum.

(i) Wha t is the annual instalment? (ii) Draw up a schedule showing the amount of prin-

cipal and the amount of interest contained in each instalment, and the principal still outstanding after each payment.

(iii) As a check, calculate independently the amount outs tanding after the 4th and 7th repayment.

I A loan of $50.00 is to be repaid by annual instalments of $10.00 each, including principal and interest. In-terest is at 6% p.a. These $10.00 payments will carry on for a number of years and will be followed by a final smaller payment at the end of the last year.

(i) How many payments of $10.00 are required? (ii) Wha t is the amount of the final payment?

(iii) Draw up a loan repayment schedule. (iv) As a check, calculate independently the amount

outs tanding after the second and fourth re-payments.

28

C H A P T E R SIX

SINKING FUNDS

A Si?ikino- Fund is the name çiven to a fund into which periodic payments are made and productively invested, usually so as to accumulate to a definite amount at the end of a specified period. Problem 1:

T h e sum of $10,000 will be needed at the end of 10 years. If 5 % compound interest can be earned on a sinking fund, what amount must be placed into it at the end of each year? Solution:

X S-m = 10,000 at 5 %

10,000 X = = $795.05

12.5779

Problem 2: How much money would be in the fund at the end of 7

years?

Solution: Amount = 795.05 X Sr\ at b%

= 795.05 X 8.1420 = $6,473.30

Problem 3: What would the answers be to Problems 1 and 2 if the

amount were placed in the fund at the beginning of each year?

Solution:

Χ(5 ΐη — 1) = 10,000 at 5 %

10,000 .*. Yearly deposit = = $757.19

13.2068

Amount in fund = 757.19 (S 8 — 1) at 5 % = 757.19 X 8.5491 = $6,473.30

(Explain why this last answer is the same as that to Problem 2.)

30 S I N K I N G F U N D S

Sometimes a lender is not prepared to accept an equal annual instalment covering both principal and interest in repayment of the debt. Because of the difficulty of investing small irregular amounts he insists on interest payments only being made and the full loan being repaid in a lump sum at the end of a specified period. In such cases the borrower usually prepares for the ult imate repayment of the loan by setting up a sinking fund. T h e interest earned on the sinking fund is usually less than that which he has to pay on the loan. T h annual expense of the debt is the expression used for the sum of the interest payment and the sinking fund payment.

Problem 4: A council borrows $10,000 for 10 years at 6% p.a., and sets

up a sinking fund at 5 % p.a. to repay the loan. Wha t is the annual expense of the loan? Solution:

T h e annual interest charge = $600.00 T h e sinking fund payment = 795.05

(See Problem 1)

.'. Annual expense of loan = $1,395.05

Problem 5: A council wishes to borrow $5000 for 10 years. Bank A

will lend at 6% p.a. and accept equal annual repayments of principal and interest. Bank Β will lend at 5 % p.a. provided interest only is paid annually and the principal is repaid at the end of the 10 years. T h e council can accumulate a sinking fund at 3 % p.a. Which is the cheaper method of borrowing and by how much? Solution:

T h e annual cost under Bank A terms

5000

5000 = $679.34

7.3601

S I N K I N G F U N D S 31

T h e annual cost under Bank Β terms 5000

= 5000 X .05 + at 3 %

5000 = 250 + = $686.15

11.4639 Bank A's terms are the better by $6.81 per annum.

In each of the following problems we purchase a machine for a given amount (say $1000) and after a given period (say every 10 years) it requires replacement at a given cost (say $800). We have to find the annual cost at (say) 5 % per annum compound interest of owning the machine—that is, the level annual amount (X) payable for ever which is equivalent to the above irregular payments and which will therefore result in us owning such a machine for ever.

There are several ways of solving this problem, all producing the same answer.

Method 1: If we pay $1000 immediately and set up a sinking fund to

accumulate to $800 every 10 years we will then be in a position to buy the machines when they are required. T h e annual cost of this is the interest we lose each year (for ever) on the $1000 plus the annual sinking fund payment.

800 . · . X = 1000 X 0.05 + at 5 % p.a. = $113.60

Method 2: We could pay $200 now out of our own funds and borrow

$800 now, and again every 10 years and pay off these loans by equal annual payments spread over the 10 years between the loans. T h e annual cost is then the interest we lose (for ever) on the $200 plus the annual payment (for ever) to repay the loans.

800 / . X = 200 X 0.05 + at 5 % p.a. = $113.60 as before

«ÎÔJ

32 S I N K I N G F U N D S

Method 3: Alternatively we could equate the present value of the

irregular payments to the present value of a perpetuity of X per annum.

1 1 1 T h u s Χ X = 1000 + 800 ( )

1 0 + 800 ( )

2 0 +

0.05 1.05 1.05

X = 1000 X 0.05 + 800 X 0.05 at 5 % p.a. 1 — v™

= $113.60 as before

It is an interesting exercise to show tnat these formulae are mathematically identical. T h e proof follows simply, using the fact that

1 — xf1

(im =

We shall use Method 1 in each of the following examples.

Problem 6: W h a t is the cost of owning (not running) a hot water

service which costs $200 originally and needs replacing every 10 years at a cost of $175. (Assume that money is worth 5 % per annum.)

Solution: Annual cost = interest charge -f sinking fund payment

175 = 200 X .05 + at 5 %

175 = 10 +

12.5779 = $23.91

S I N K I N G F U N D S 33

Problem 7: r

Which machine is cheaper in the long run—one which costs $1200 and must be replaced at the end of 10 years at a cost of $800 or one which cost $1500 and must be replaced at the end of 15 years at a cost of $1000? (Assume 5 % interest.)

Solution: T h e annual costs allowing for both interest and sinking fund

payments are 800 1000

1200 X .05 + and 1500 X .05 + at 5 %

i.e. $123.60, and $121.34 T h e second machine is the cheaper.

Problem S: A machine costing $15,000 will last 35 years at which time

it can be sold as scrap for $1000. How much could a company afford to pay for an alternative machine which will last only 20 years and have no scrap value? (Assume money is worth 4 ^ % per annum.)

Solution: Let X = the required amount . Annual cost of first machine

14,000 = 15,000 X .045 + at \ \ %

S35I = $846.79

Th i s equals the annual cost of the second machine when

X 846.79 = Χ X .045 + at \ \ %

i.e. X( .045 + .031876) = 846.79 Λ X = $11,015

T h e company can therefore pay up to $11,015 for the alternative machine.

34 S I N K I N G F U N D S

Problem 9: Which is the more economical of the following machines?

Λ Β Cost $6,000 $52,000 Life 20 years 25 years Scrap Value $1,500 $7,000 Repairs $250 p.a. $350 p.a. Operat ing Expenses $4,800 p.a. $6,800 p.a. Ou tpu t 1 uni t 2 units

(Assume 5 % interest.)

Solution:

T h e effective annual cost of A per uni t of ou tpu t

4500 6000 X .05 + + 250 + 4800 at 5 %

= $5,486 T h e effective annual cost of Β per uni t ou tpu t

45,000 • = h (52,000 X .05 + .+ 350 + 6800) at 5 %

= $5,346 Therefore Β is the more economical.

(N.B. Operat ing expenses would normally be spread throughout the year and not payable in a lump sum at the end of the year. T h e method of adjusting for this will be considered later.)

S I N K I N G F U N D S 35

EXERCISES

6-1 A company pays $1000 each year into a bank sinking fund earning 6% p.a. compound interest. After making five payments the rate of interest granted by the bank on the fund is reduced to 4 % p.a. T h e company therefore decides to increase its future deposits to $1200. Wha t amount is in the fund after 15 payments alto-gether have been made?

6-2 A company now has $1000 in a sinking fund that earns interest at 5 % p.a. W h a t annual deposit is required to cause the fund to amount to $10,000 in 10 years' time?

6-3 A council has to pay off two loans, one for $50,000 due in 10 years and one for $10,000 due in 12 years. A sinking fund is to be set up into which equal annual deposits will be paid for 12 years to repay both loans. If the fund earns interest at 5 % per annum, what will the annual deposit be? Find the amount of the fund at the end of 8 years. Will there be enough in the fund after 10 years to meet the $50,000 payment?

6-4 A mine is expected to produce a net income of $50,000 per year for 20 years and will then be worthless. An investor contemplat ing purchasing the mine wishes to earn 10% per a n n u m on his investment. He intends to set u p a sinking fund which will re turn h im his money at the end of the 20 years. If the sinking fund earns 5 % interest, how much should he pay for the mine?

C H A P T E R SEVEN

EQUATIONS OF VALUE

If interest is at 6% per annum, §10,000 now is of the same value at 10,000 (1.06)

10 or $17,908.5 clue in 10 years' time.

Conversely 817,908.5 due in 10 years' time is (at 6% interest) worth SI0,000 at the present time.

Let us consider the value of each of these two payments in four years' time. T h e S 10,000 payable now will, in four years' time, be worth 10,000 (1.06)

4 or $12,624.8; the payment of

S 17,908.5 which will then be due six years later will then be worth (at 6% interest) SI7.908.5 (1.06) -° = $12,624.8 also. T h a t is. at any point in time, at 6°/0 interest, the two payments ire of equal value.

T h e same holds true for any two equivalent sets of pay-ments. For example—again at 6% interest—$1000 due in three years' time plus S2000 due in six years' time is equal ir value to S3585.43 due in eight years' time since . . .

1000 (1.06)·"· + 2000 (1.06) 2 = 3585.43

T h e P.V. of the first two payments = 1000 (1.06)-

3 + 2000(1.06)-« = 2249.54

and the P.V. of the third = 3585.43(1.06)-« = 2249.54 also. T h e values of the two sets of payments at any point of time,

at 6% interest, are equal. If we have two such sets of payments which, at some rate

of interest, are of equal value, then any equation which expresses this equality of value at any time is called an equation of value.

AMOUNT OF ONE PAYMENT UNKNOWN

Problem 1:

In re turn for $700 at the end of 12 years, a man agrees to pay $200 at the end of two years, $300 at the end of eight years and a further sum at the end of 14 years. If interest is at 4 % p.a., what should this final sum be?

E Q U A T I O N S O F V A L U E 37

Solution: Let X be the final payment, then, equat ing the present

values of the two sets of equivalent payments, we obtain 200 (1.04)-

2 4- 300 (1.04)-

8 + X ( 1 . 0 4 ) -

1 4 = 700 (1 .04) -

1 2

700 (1 .04) -1 2 — 300 (1.04)-

8 — 200 (1.04)

Λ X = (1.04) -

1 4

= $57 to the nearest $1 I t would have been equally possible to equate the two sets

of payments at the end of 14 years, giving 200 (1 .04 )

1 2 + 300(1 .04 )6 + X = 700 (1 .04)

2

or X = 700 (1.04) 2 — 300 (1.04)

6 — 200 (1.04)

12

= $57 as before. T h e calculation has thus been considerably simplified, and

it is well worth spending a little time selecting the best point of time to equate the payments.

TIME OF ONE PAYMENT UNKNOWN

Problem 2:

When should $3000 be paid to be equivalent at 4 % interest to $1000 due in five years' time and $2000 due in 10 years? Solution:

If η is the required number of years' time, then the equat ion of value is

3000(1.04)-" = 1000 (1.04)-5 + 2000 (1 .04) -

1 0

that is (1.04)-» = 0.72435

From tables (1.04) -

8 = 0.73069

and (1.04)-» = 0.70259 .'. by interpolat ion

η — 8 .72435 — .73069 η = 8.226 years' time

9 — 8 .70259 — .73069

(Alternatively: T a k i n g logarithms

log 0.72435 T.859948 .140052

η log 1.04 .01'

8.222 years' time)

.017033 .017033

η

38 E Q U A T I O N S O F V A L U E

RATE OF INTEREST UNKNOWN

Problem 3:

A paint ing bought 10 years ago for $200 has just been sold for $285. Wha t is the interest re turn on the investment? Solution:

T h e equat ion of value is 200(1 + i)10

= 285 i.e. ( 1+ = 1.425 From* the tables it can be seen that i is between 3^% and

Ί / ο · ·

By interpolat ion

i — .035 1.425 — 1.41060

.04 — .035 1.48024 — 1.41060

.01440 .*. i = .035 + .005 X

.06964

= .03603 i.e. a little over 3 .6% per annum.

Problem 4: A $100 debenture is purchased in the market for $90 imme-

diately after an interest payment. Interest is payable annually at 5 % (of the face value) and the debenture is due to mature in six years' time. Wha t is the expected net rate of re turn on the investment if income tax is payable at 50 cents in the dollar? Solution:

This means that for $90 the investor receives $2.50 per annum for six years plus $100 at the end of six years. T h e equat ion of value at the date of purchase is therefore . . .

90 = 2 . 5 0 0 7 ] + ΙΟΟϊ;6

at some unknown rate of interest which we wish to determine. A rough guess at the answer may be obtained by noting that for his $90 he obtains $2.50 each year plus a capital profit of

$10 over six years—a return of roughly 2.50 + or $4.17 6

per year. T h e rate of interest should therefore be about 4.17

or 4 .6% 90

/0

E Q U A T I O N S O F V A L U E 39

T o determine the effective net re turn more accurately, we calculate the right hand side of the equat ion of value above at two rates of interest and interpolate.

At 4 i % , R H S = 2.50 X 5.1579 + 76.790 = 89.684. Th is gives a little less than $90 so the actual rate is below

At 4 % , R H S = 2.50 X 5.2421 4- 79.031 = 92.137. Therefore, by interpolat ion

i — .04 90 — 92.137

.045 — .04 89.684 — 92.137

giving i = .04436 or 4.436% p.a.

Problem 5: A has an obligation to pay Β $100 in a year's time and

$300 in three years' time. Β says he will accept in lieu (i) $362.80 or (ii) four payments of $95.18, the first immediately and the other three at intervals of one year. Which is the better for A of these two options? Solution:

Under opt ion (i) the interest rate involved is given by the equat ion

362.80 = lOOi; + 300t;3

and may be shown to be approximately 4 % p.a. At this rate of interest (4%) the present value of the payments under option (ii) is

95.18 (1 + a$ = $359.31 Hence option (ii) is the better opt ion for A. (Alternatively, we could have worked out the rate of

interest involved under opt ion (ii) and found it was about 5%—hence the better option for A.)

Problem 6: A feasibility study is being carried out to determine whether

a new project is likely to prove worthwhile. Estimates of the amount required to be spent on plant and working capital are set out in Column 2. Estimates of the net cash receipts— that is, profits after income tax and after allowing for tax savings on the original investment and tax savings on depre-ciation, plus scrap value of plant—are shown in Column 3.

4 0 E Q U A T I O N S O F V A L U E

What effective net rate of interest is expected on the investment?

(i) Year

(2) Investment

(3) Net Cash Receipts

0 34 0 1 50 6 2 5 13 3 11 4 12 5 13 6 11 7 Q

10 Q

Ο 9

Ο 8

10 6 11 5 12 5 13 5 14 5 15 20

16 (inc. scrap value) —4 (tax)

of the investment At 7% At 6\%

Solution:

(Discounted Cash Outflow) T h e P.V. of net receipts

(Discounted Cash Inflow)

Difference

= 85.1

= 84.5

—0.6

85.4

85.9

+0.5 T h e P.V.s are equal at rate i where

i — 64 0 — 0.5 — 0.5

7 — 6 i — .6

/ = 6.7%

0.5 1.1

T h e project thus provides for a re turn of invested capital plus a 6.7% net return.

(This, in brief, is the method of appraisal of investment projects described as the Discounted Cash Flow technique.)

E Q U A T I O N S O F V A L U E 41

EXERCISES

7-1 W h a t amount payable in 10 years' time would be equivalent to $500 payable in five years' time plus $1000 payable in 15 years' time? (Assume 6% p.a. compound interest.)

7-2 A man owes $1000 payable in three years' time and $1000 payable in 13 years' time. He would like to settle his debt by making a $2000 payment. If interest is 5 % p.a. compound, when should he make the payment?

7-3 A company is prepared to invest in debentures if it earns 7% p.a. or more. W h a t is the highest price it should pay for $100 debentures which pay interest yearly at 5%, which have just made an interest payment and which mature in six years' time? (You may ignore income tax.)

7-4 A man buys a $100 7% debenture which matures in nine years' time. T h e interest is payable yearly and has just been paid. W h a t yield does he obtain (ignor-ing tax) if he pays $104 for the debenture?

7-5 Wha t should the price in Exercise 7-3 be if income tax were 40%?

7-6 Wri te down the equat ion of value which would apply to exercise 7-4 if income tax were 4 0 % and capital gains tax 50%.

C H A P T E R E I G H T

COMPOUNDING MORE FREQUENTLY THAN ANNUALLY

If $1000 is invested, and interest is compounded monthly at a rate of 1% per month, then the interest added to the account at the end of each month is as follows:—

First month : $1000 X .01 = $10 Second month : $1010 X .01 = $10.10 T h i r d month : $1020.10 X .01 = $10.20 and so on.

Using 1% tables of (1 + i)n we can see that at the end

of 12 months the $1000 would have accumulated to 1000 (1.01)

1 2 = $1,126.83

which is the same amount as $1000 would have accumulated to at 12.683% with interest compounded annually.

Hence 1% per month compounding monthly is equivalent to 12.683% per annum compounding annually. A rate of 1% per month compounding monthly is usually referred to as a nominal rate of 12% per annum convertible monthly, and 12.683%, is spoken of as the effective rate of interest. T h e effective rate i corresponding to a nominal rate of / per a n n u m convertible m times a year is thus . . .

/ 1 + i = (1 + — )

m

m

i or i = (1 +—)

m — 1

m

Problem 1: Find the effective rate of interest corresponding to 8% p.a.

convertible quarterly. Solution:

8% p.a. convertible quarterly equals 2 % per quarter. T h e effective rate is 8.24% p.a. since (1.02)

4 = 1.0824.

C O M P O U N D I N G M O R E F R E Q U E N T L Y T H A N A N N U A L L Y 43

Since (1.01) 1 2

= 1.12683, (1.01) 6 0

= (1.12683}5

Hence $1000 invested at* a nominal rate of 12% per annum convertible monthly will accumulate at the end of 5 years (or 60 months) to the same amount as $1000 would accumu-

late to after 5 years at 12.683% p.a. with interest added annually. T h e two rates are thus equivalent.

If we have an annuity of $1000 per a n n u m payable at the end of each year with interest at 12% p.a. convertible monthly, then each payment at the end of η years' time will have accumulated to the same amount as it would have accumulated to at a rate of 12.683% p.a. with interest added annually. Hence at 12% p.a. convertible monthly, the annuity at the end of η years would have accumulated to ΙΟΟΟ τ̂η at a rate 12.683% p.a.

T h e same reasoning applies to the present value of an annuity. T h e P.V. of an annuity of $1000 p.a. for η years, the first payment being due in a year's time, at a rate of 12% p.a. convertible monthly is therefore lOOOa^at 12.683% p.a.

Problem 2: Find the P.V. of an annual annui ty of $100 for 10 years,

first payment in one year's time, at a nominal rate of 6% p.a. convertible monthly. Solution:

T h e effective rate is thus 6.168% p.a. T h e P.V. is thus the P.V. of an annuity of $100 for 10 years

at 6.168% p.a. which (by substi tuting in the formula) equals

.06 (1 + )

1 2 = (1.005)

1 2 = 1.06168

12

1 _ (1.06168) -10 1 _ (1 .005) -1 20

100- = 100 .06168 .06168

1 — 0.5496327 100- = $730.17

.06168

By interpolation, using ajö] tables at 6% and 6£% we obtain $730.27. (N.B. Th is is an approximat ion—the first method is accurate.)

44 C O M P O U N D I N G M O R E F R E Q U E N T L Y T H A N A N N U A L L Y

Problem 3: Find to the nearest $ the amount of an annual annui ty of

$200 after 12 years, first payment in one year's time, at a nominal rate of 6% convertible quarterly. Solution:

.06 (1 + )

4 = (1.015)

4 = 1.06136

4 T h e required amount

(1.06136) 12

— 1 = 200

.06136

(1 .015)4 8 — 1

= 200 .06136

1.043478 = 200

.06136

= $3,402

By interpolation, using S~^\ tables at 6% and 6£% we obtain $3401. (N.B. Th is again is an approximat ion—the first method is accurate if taken to sufficient decimal places.)

Often, in practice, annuities are payable more frequently than annually—say, monthly, half yearly, etc. Usually in such cases interest is added with the same frequency. As an example, consider a monthly deposit of $10 for 8 years at 12% p.a. convertible monthly, the first deposit being in one month 's time. It will amount at the end of 8 years to . .

10 (1.01) 9 5

+ 10 (1.01) 9 4

+ 10 (1.01) »3' + . . . + 10

(1.01) 9G

— 1 = 10 = lOSçjôj at 1% = $1,599

.01

Thus , where annuities are payable monthly and also interest is convertible monthly, the 07η and S-^ tables may be used where η is the number of monthly payments and the rate of

interest is the monthly rate or — of the nominal annual rate.

C O M P O U N D I N G M O R E F R E Q U E N T L Y T H A N A N N U A L L Y 45

Problem 5; What will be the value in 10 years' time of 4 payments of

$100 and 16 payments of $200 made at half-yearly intervals commencing now? Assume interest at 8% p.a. convertible half yearly.

Solution: This is equivalent to 20 payments of $100 commencing now

plus 16 payments of $100 commencing in 2 years' time. Remembering that the payments are made at the beginning

of each half-yearly period, the amount equals 100 ( % ! — 1) + 100 (5τη— 1) at 4%,

= $5,366.67

Problem 6: A person buys a home for $12,000. He has $4000 deposit

and is able to borrow $8000 from an insurance company pro-vided he takes out a 20-year endowment assurance policy for $8000 and repays the loan over 20 years by equal monthly instalments of principal and interest at 6% p.a. convertible monthly. T h e annual premium for the policy at his age is

Problem 4: Find the present value of an annuity of '$50 per quarter for

9 years at 10% p.a. convertible quarterly.

Solution: (i) Assume first payment in a quarter 's time.

P.V. = 5 0 ^ at 2 i % = 50 X 23.55625 = $1,177.81

(ii) Assume first payment is made immediately. T h e P.V. is then that of one payment made immediately and 35 subsequent quarterly payments, namely

50 + 5 0 ^ at 2\% = $1,207.26

(N.B. T o calculate aj=^ from the tables at the end of this book, remember that a-^ — a-^ + v

2*ajô\ ·)

46 C O M P O U N D I N G M O R E F R E Q U E N T L Y T H A N A N N U A L L Y

$5.40 per $100. Because of loss of interest and extra adminis-trative expenses the company adds 5 % to the premium for monthly payments.

(i) W h a t is his monthly commitment to the insurance company?

(ii) Wha t is the amount of loan outstanding after 15 years?

(iii) If the company expects to pay annual compound bonuses at 2 % based on the amount of policy and declared bonuses, what is the amount of the insurance cover at the end of 15 years?

Solution:

(i) T h e monthly premium for $100 would be

1

5.40 X 1.05 X — = 0.4725 12

T h e monthly premium for $8000 is therefore 0.4725 X 80 = $37.80

T h e amount to repay loan plus interest

8000 = — at \ %

= $57.31

(To calculate ατ^\ from the tables in this book you need to use the obvious fact that

His total monthly commitment is therefore 37.80 + 57.31

= $95.11 (ii) T h e amount still owing after 15 years equals the then

present value of his future payments on the loan. This equals

57.31«ΐπη at i % = $2964

(iii) T h e amount of policy plus bonuses after 15 years = 8000 (1.02)

lr»

= $10,767

C O M P O U N D I N G M O R E F R E Q U E N T L Y T H A N A N N U A L L Y 47

EXERCISES

8-1 Wha t is the effective annual rate of interest correspond, ing to . . .

(i) 6% p.a. convertible half-yearly? (ii) 10% p.a. convertible quarterly?

(iii) 9% p.a. convertible half-yearly?

8-2 Wha t is the present value of a ten-year annui ty of $100 a year, the first payment in one year's time, at a nominal rate of interest of 6% p.a. convertible half-yearly?

8-3 Find the amount of the above annuity at the end of 10 years.

8-4 Find the present value of a 10-year annuity of $100 per half year, first payment in 6 months ' time at a rate of 8% p.a. convertible half-yearly.

8-5 Find the amount which $10 per month will accumulate to over 3 years at a rate of 12% p.a. convertible monthly. T h e first payment is payable . . .

(i) in one month 's time (ii) immediately.

8-6 For purchase of a home a man borrows $10,000 at 6% p.a. convertible monthly. He is to repay the loan over 20 years by monthly instalments. W h a t is the amount of the instalment? Wha t amount of loan is still out-standing after 10 years (i.e. immediately after the 120th payment) ? How much principal is repaid in the 121st instalment?

C H A P T E R N I N E

CONTRACTS AT "FLAT" RATES OF INTEREST

No doubt on the grounds of simplicity, it is the custom in many sections of the retail trade, in the hire purchase industry and elsewhere, to charge what are called "flat" rates of interest —i.e. where the interest charge is based on the original amount of the loan and does not reduce as principal repay-ments are made.

T h e system works as follows:— An article is bought for $78. Cash is not available and so

a "sale on terms" is permitted. T h e terms are repayment over 12 months with interest at "10% Hat." This means that the interest charge is $7.80 per year for (in this case) 12 months, or $7.80 which is added to the original cost of the article, making the total debt $85.80. Th i s is to be paid oft evenly over 12 months and hence the monthly payment is $85.80 divided by 12 or $7.15 per month .

We are interested in two aspects of this type of transaction:—

(1) the real or effective rate of interest that is being charged;

(2) the outstanding debt at any stage—that is:— (i) the theoretical payment which should be made if

at any stage the debtor wishes to pay off his outstanding indebtedness; or from another point of view

(ii) the amount which the lender should treat, at any stage, as the value of his asset, i.e. the amount of principal still owing to him.

Τ lie real rate of interest being charged $78 is in fact being paid off by 12 monthly instalments of

$7.15. Hence, if / is the monthly rate of interest, 78 7.15rt77|

giving ι — .015 per month = ( i .015)

1- — 1 = .1956 per annum

or 19.56% per annum

C O N T R A C T S A T " F L A T " R A T E S O F I N T E R E S T 49

The outstanding debt

Although the principle is simple, the actual calculation is complicated. For this reason, very early in the history of these types of contract, a simple, approximate rule was intro-duced called "the rule of 78."

I t recognizes the fact that as the principal outs tanding at the beginning of the 12 months is a max imum ($78) and a min imum at the end ($zero) then the interest content of each instalment must be a max imum in the first instalment and a min imum in the last instalment. The rule assumes that the amount of interest contained in eacli instalment reduces linearly. T h a t is, it assumes that the amount of interest in the second last instalment is double that in the last; the amount in the third last instalment is three times that in the last and so on. Hence the interest payments are assumed to be proport ional to 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1. Since the total of these figures = 78 the fraction of the total interest which has to be paid (namely $7.80), which is paid in the various instalments is . . .

12 11 10 9 8 7 6 5 4 3 2 1

78? 78' 78' 78' Ί8 78* 78* Ί8 78* 78 Ί8 78

T h e actual interest payments, according to the rule of 78, in this case are therefore . . .

1.20, 1.10, 1.00, .90, .80, .70, .60, .50, .40, .30, .20, .10 which total $7.80. T h e balance of each payment is principal. Principal payments therefore are . . .

5.95, 6.05, 6.15, 6.25, 6.35, 6.45, 6.55, 6.65, 6.75, 6.85, 6.95, 7.05

totalling $78.00 which repays the full loan. (It is because 12 + 11 + 10 + . . . + 1 total 78—the 12

months repayment case—that the rule received its name.) Tables have been prepared, based on this assumption that

the interest payment reduces uniformly, and are in regular use for determining the outs tanding debt at any time and the amount of interest which has actually been received and can be included in profits. I t is a reasonably good approxima-tion for loans u p to five years' durat ion. It does however always overstate the interest payments in the earlier years and understates them in the later years. Its use therefore

50 C O N T R A C T S A T " F L A T " R A T E S O F I N T E R E S T

tends to lead companies to overstate their profits in the earlier years and they then find it tougher going in the later years. Its use in longer term contracts (over 5 years) would lead to a serious early overstatement of profits.

We now describe the correct method—curiously called the actuarial method. There is noth ing "actuarial" about it; it is simply the correct method.

We assume the real rate of interest being charged, /, has been calculated (as in the previous sect ion).

As soon as the fourth instalment (say) has been made, the outs tanding debt is the present value of the remaining instalments namely . . .

7 .15081 at rate \ \ %

= $53.52

T h e principal which has been repaid in the first four instalments is therefore

78.00 — 53.52 = $24.48

T h e total of the interest payments in the first four instal-ments is therefore

4 X 7.15 — 24.48 = $4.12

These figures compare with $24.40 and $4.20 obtained above under the rule of 78.

I n practice the debtor is often (for good reason) asked to pay more than the figure calculated in this way to discharge his indebtedness.

Problem: A loan of $300 for 5 years at 10% "flat" is repayable by

monthly instalments.

(i) Wha t is the monthly instalment? (ii) W h a t is the real rate of interest on the loan?

(iii) Assuming payments are made on the due dates, fill in (accurately) the details required in the following table.

C O N T R A C T S A T " F L A T " R A T E S O F I N T E R E S T 51

Year Interest paid during Principal repaid Outstanding debt at the year during the year the end of the year

1

2 3 4

5

(iv) Record in brackets in the table the results obtained by the "rule of 78."

Solution:

300 + 5 X 30 (i) Instalment = = $7.50 per month

5 X 12

(ii) 300 = 7.50öööi a t r a te 2

Pe r

month .'. i = .01440 per mon th

= (1.01440) 12

— 1 = .1872 or 18.72% p.a. (iii) Outs tanding debt at end of

Year 1 2 3 4 5

7 .50a^ 7 .50«a6 i 7.50^24] 7 . 5 0 ^ 0 (at rate of interest 1.44%).

T h e principal repaid during each year may be determined by subtracting the amount of loan outstanding at the end of each year from the amount of loan outs tanding at the beginning of the year; by subtracting the amounts so obtained from $7.50 X 12 or $90.00 the interest paid each year is found.

Hence we obtain the following:—

Year Interest paid during the year

Principal repaid during the year

Outstanding debt at the end of the year

1 2 3 4 5

$48.64 (53.61) $40.91 (41.80) $31.73 (30.00) $20.83 (18.20)

$7.89 (6.39)

$41.36 (36.39) $49.09 (48.20) $58.27 (60.00) $69.17 (71.80) $82.11 (83.61)

$258.64 (263.61) $209.55 (215.41) $151.28 (155.41)

$82.11 (83.61) zero zero

To ta l $150.00 (150.00) $300.00 (300.00)

C O N T R A C T S A T " F L A T " R A T E S O F I N T E R E S T

) According to the "rule of 78" the interest payments are in proport ion to

60, 59, 58, 57, . . . . , 3, 2, 1. By adding the first and last, the second and second

60 χ 61 last, etc. we see that these figures total

2 or 1830. T h e total interest payments are $150. Hence (on the "rule of 78" assumption) the interest payments in the years 1 to 5 are . . .

(60 + 59 + 58 + 57 + . . . + 49)

1830

(48 + 47 + 46 + 45 + . . . + 37)

1830

(36 + 35 + 34 + 33 + . . . + 25)

1830

(24 + 23 + 22 + 21 + . . . + 13)

1830

( 1 2 + 11 + 10 + 9 + . . . + 1)

X 150 = 53.61

X 150 = 41.80

X 150 = 30.00

X 150 = 6.39 1830

(Check that these amounts total $150.) T h e principal repaid column is obtained by sub-tracting each of these amounts from $90 (i.e. 12 X 7.50) and the outs tanding debt column by subtracting the principal repaid figures progres-sively from $300.

52

C O N T R A C T S A T " F L A T " R A T E S O F I N T E R E S T 53

In practice, ta'bles are available from which the results under the "rule of 78" may be written down. I t is useful, however, to realize the assumption on which it is based and the degree of inaccuracy involved. Note that the error was small in the case of the one-year contract but in the above case of a five-year contract the interest in the first instalment wa

ç 10% overstated.

R E M E M B E R : T h e "actuarial" method is accurate or correct. T h e "rule of 78" is a useful approximation in the case of short term loans.

EXERCISES

9-1 A $200 refrigerator is sold "on easy terms." These terms are:—

(i) No deposit. (ii) Interest 10% "flat."

(iii) Monthly repayments over 5 years.

W h a t is the true rate of compound interest which the customer is paying? At the end of 2 years— (a) How much interest would he have paid? (b) How much principal would he have repaid?

9-2 A loan of $400 for 5 months at 9% "flat" is repayable by 5 equal monthly instalments.

(i) W h a t is the monthly instalment? (ii) Draw u p a schedule, based on the assumption

of the "rule of 78," showing the interest con-tained in each instalment, the principal contained in each instalment and the pr incipal outs tanding immediately after each month ' s payment .

(iii) Find the real monthly compound interest ra te involved and determine the corresponding effec-tive annual rate.

(iv) Draw u p a schedule as in (ii) bu t giving the the true figures.

SOLUTIONS TO EXERCISES

1-1 I = $10 Ρ = $490 10 365

/. i = Χ = 8.28% p.a. 490 90

1-2 Ρ + .04Ρ Χ i = 1020 Ρ = $1000

1-3 Let χ = number of days χ

600 Χ .05 Χ = 4 365

.·. χ = 48.7

.'. it will take 49 days

1-4 N u m b e r of days = 29 + 16 = 45 45

/ . I = 900 X .045 X = $4.99 365

1-5 Number of days = 30 + 31 + 31 + 30 + 18 = 140 140

2400 X i = 54 365

i = 5.87% p.a.

1-6 Let Ρ = the amount required Number of days = 16 + 30 + 31 + 31 + 12 = 120

120 .". Ρ + Ρ X .06 Χ = $2550

365 Ρ = $2500.67

3

1-7 d = Χ 3 = 6% p.a. 150 3

i = Χ 3 = 6.12% p.a. 147

S O L U T I O N S T O E X E R C I S E S 55

2-1 100 (1.06)» = 200 .*. (1.06)» = 2

η — 11 2 — 1.89830 By interpolat ion

12 — 11 2.01220 — 1.89830

η — 11.89 years = 11 years 326 days

30

1-8 D = 400 X .04 X = $1.32 365

.*. Proceeds = 400 — 1.32 = $398.68

1-9 Let Ρ = amount asked for. T h e n Ρ minus the interest in advance equals $1000.

30 . · . Ρ — Ρ Χ .06 Χ = 1000

365 Ρ = $1004.96

1-10 Date of maturi ty = 8th November 90

M.V. = 300 + 300 X .05 X = $303.70 365

Number of days to maturi ty = 21 + 31 + 8 = 60 60

Proceeds = 303.70 — 303.70 X .06 X 365

= $300.70

1-11 M.V. = 300+ 300 X .05 X \ = $303.75 If Ρ = proceeds

60 Ρ + Ρ X .06 X = 303.75

365 Ρ = $300.78 and discount = $2.97

56 S O L U T I O N S T O E X E R C I S E S

2-2 75(1 + i )1 0

= 100 (1 + i)

1 0 = 1.33333

i — 2.75 1.33333 — 1.31165 By interpolation =

3 — 2.75 1.34392 — 1.31165 /. i = 2.92% p.a.

2-3 T h e rate of interest i is such that 1000 (I + ί)ΐο = 1500 giving i = 4.1358% p.a.

T h e amount = 1000 (1.041358) 6

= $1,275.35

Amount η υ»(6%) Value 150 1 .94340 141.51 200 2 .89000 178.00 300 3 .83962 251.89 200 4 .79209 158.42 500 5 .74726 373.63

1000 10 .55839 558.39 600 12 .49697 298.18

Tota l $1,960.02

2-5 As the rate of interest earned on the investment (6%) is the same as that used in working the present value in exercise 2-4, we may simply add interest to that present value. Therefore the values will be . . .

(1) 1960.02 (1.06)3 = $2334.42

(2) 1960.02 (1.06) 5 = $2622.95

(3) 1960.02 (1.06)8 = $3123.97

(4) 1960.02 (1.06) 1 2

= $3943.95

5000 3-1 Instalment = at 7 % p.a.

= $711.89

SOLUTIONS T O E X E R C I S E S 57

3-2 Let η — number of years required T h e n ôJOfl^, = 5000 at 7 % p.a. 17 years is not enough .'. η — 18 years In 18 years the loan plus interest

= 5000 (1 .07 )1 8 = 16,899.66

Just before the 18th payment the $500 payments plus interest amount to

500 (Sïïïi — 1) = 16,499.52

Λ Final instalment = $400.14

3-3 P.P. = 4000 4- 4 0 0 a 2 Ö | at 6 % p.a. = $8587.97

3-4 (i) Value = 100 4- 100 S^\ at 5 % p.a. = $3623.54

(ii) P.V. = 100 4- 200ατη— lOOa^at 5 % p.a. = $1742.98

[N.B. As a check, 1742.98 (1.05) 1 5

should equal 3623.54 —which it does.]

3-5 First method

P.V. = 11,000(0201 — βη) = $98,079 Second method P.V. = ΙΙ,ΟΟΟν^ϊβΐ = $98,079 Third method P.V. = 11,000Stö|T/

20 = $98,079

3-6 Let Ρ be the su i n r equ i red .

Ρ = 100(0201 — flïe|) = $136.40

3-7 Let X be the annual payments.

X ( l + «Tïïl) = 100(tf2 o 1— α π π ) = $136.40 X = $12.28

3-8 P.V. of Second Alternative = 15,000 4- 8000 (α^— ατ|) = $36,958

Hence the cash alternative is the cheaper.

58 SOLUTIONS TO EXERCISES

3-9 25,000 Vi = 500,000 at i\% p.a. = 20 at 31% p.a.

Hence 15 full payments are required. T h e final payment (the 16th)

= 500,000 — 25,000 ( S ^ — 1) = $724.26

3-10 500 Sm = 10,000

%i = 20 i — 3 i 20 — 19.2957

By interpolation, = 4 — 31 20.0236 — 19.2957

giving i — 3.98% p.a.

3-11 Amount required = 1000 (Sn 4- a^l) at 6% p.a. = $9,292

4-1 Value as at 1/1/68:— Formulae

(a) $100, (b) lOOv, (c) lOOi;5, (d) 100t;12, (e) 100 + 100v or 100(1 4- a^), (f) lOOa^j, (g) 100(1 4- a~r\), (h) 100 (ατη — α π ) , (i) 100(1 4- a^) — ΙΟΟι;» or 100(1 4- ατη) + 100(0^ — α η ) , (j) 100(1 4- αη) + 100(1 4- α ^ ) , (k) SOOûf, — 2 0 0 ^ , (1) 300(1 4- aj\) 4- 600t/4 — 200(1 4- flgj), (m) 300α η — 200«^ — ΙΟΟτ;7, (η) 100νδ* = 100[^ 5 — \{v* — v«) ] .

4-2 Value as at 1/1/76:— Formulae

(a) 100(1 4- ï)8, (b) 100(1 4- i ) 7 , (c) 100(1 4- if,

(d) 100^ , (e) 100(1 4- i) 8 + 100(1 4- ï) 7, (f) 100 (Sin

- S t t ) , (g) 100(5^ - S^) , (h) 100(5^ - 1) , (i) 100 ( S ^ — 1) — 100(1 4- i)

5, (j) 2 0 0 ( 5 ^ 1 — 1) —

100 (Sr,— 1), (k) lOOSj! 4- 2005η, (1) 1005^1 4- 200%! + 600(1 4- i)

4, (m) 1 0 0 ( 5 ^ — 1) 4- 200 ( S ^ — 1) —

100(1 4- i), (n) 100(1 + i) *ï = 100 [(1 4- i)2 +

f [ ( l 4- i)» — (1 + 0 2 3

S O L U T I O N S T O E X E R C I S E S 59

Value as at 1/1/72:— Formulae

(a) 100(1 + i) \ (b) 100(1 + i)\ (c) lOOr/, (d) lOOr;8,

(e) 100 ( 5 η - 5 η ) , (f) 1005η, (g) 1 0 0 % (h) 100 (5η + αη), (i) 100(5η — 5η) + 100(1 + α η ) , (j) 2005η + ΙΟΟα^, (k) 1005η + 300α η, (1) 100 ( 5 η — 1 ) + 9 0 0 + 300α π, (m) 1005η + 200 + 300#η + 200t;

3, (n) 100t;

1*

= 100 ν — \ (ν — υ2)

5-1 (i) 81

Annual instalment = = $10.49

(iii)

Year Number

Principal outstanding at beginning

of year

Interest due at

5%

Principal contained

in payment

Principal outstanding

at end of year

1 81.00 4.05 6.44 74.56

CM 74.56 3.73 6.76 67.80 3 67.80 3.39 7.10 60.70 4 60.70 3.04 7.45 53.25 5 53.25 2.66 7.83 45.42 6 45.42 2.27 8.22 37.20 7 37.20 1.86 8.63 28.57 8 28.57 1.43 9.06 19.51 9 19.51 .97 9.52 9.99 10 9.99 .50 9.99 —

(Ü) Amounts outstanding = 10.49Α7Π $53.24

and 10.49αη = $28.57

60 S O L U T I O N S T O E X E R C I S E S

5-2 (i) 50 = lOfl^at 6% .'. 6 payments of $10 and a 7th smaller payment

are required.

(Ü) Year

N u m b e i Principal

outstanding at beginning

of year

Interest due at

6%

Principal contained

in payment

Principal outstanding

at end of year

1 50.00 3.00 7.00 43.00 2 43.00 2.58 7.42 35.58

CO

35.58 2.13 7.87 27.71 4 27.71 1.66 8.34 19.37 5 19.37 1.16 8.84 10.53 6 10.53 .63 9.37 1.16 7 1.16 .07 1.16 —

(iii) Hence final payment = = 1.16 + .07 = $1.23.

(iv) Amounts outs tanding = 50 (1.06)

2 — lOSn = $35.58

and 50 (1.06)

4 — IOSji = $19.38

6-1 At the end of 5 years the amount in the fund is lOOOS^-jat 6% p.a. Th is amount and subsequent instal-ments accumulate at 4 % p.a. Hence the amount in the fund after 15 years is . . .

1000 &f% (1.04) 1 0

+ 1200 Sjrf% = $22,752

6-2 Let X = annual deposit required T h e n 1000 (1.05)

10 + XSjy = 10,000 at 5 % p.a.

.'. X = $665.54

S O L U T I O N S T O E X E R C I S E S 61

6-3 If X is the annua l deposit

X S ^ l = 50,000(1.05)2 + 10,000 at 5 %

.'. X = $4,091.50 T h e amount in the fund at the end of 8 years

= 4091.50 S t i = $39,070 T h e amount in the fund just before the $50,000 payment

= 4091.50 STZ) = $51,462 Hence the payment can be met.

6-4 If the purchase price = P, the amount available for the sinking fund = 50,000 — 0.1 P.

Hence (50,000 — 0 . 1 P ) 5 ^ = Ρ at 5 % p.a. . · . Ρ = $383,898—say $383,900

7-1 T h e simplest equat ion of value is as at 10 years' time, i.e. 500 (1.06)

5 4- 1000 (1.06)-

5 = X

X = $1416.37

7-2 T h e equation of value as at 3 years' time is 1000 4- 1000(1.05)-

10 = 2000(1.05)-"

i.e. (1.05)-" = .80696 η = 4.402

that is, 7 years 147 days from now.

7-3 T h e equation of value is . . . Ρ = 5aj] 4- 100t;

6 at 7 %

= $90.47

7-4 T h e equation of value is . . . 104 = la^ + IOOt;

9

at some unknown rate of interest. T h e R .H .S . = 103.328 at 6^% p.a. T h e R .H .S . = 106.802 at 6% p.a. By interpolation, R .H .S . = 104 at 6.40% p.a. T h e yield therefore is 6.40% p.a.

62 S O L U T I O N S T O E X E R C I S E S

7-5 T h e net income would then be $3 p.a. and the new equation of value . . .

Ρ = 3 « , n + 100v« at 7% = $80.93

7-6 104 = 4.2ατπ + 102v° (N.B. T h e "capital gain" is here a loss of $4 on

maturi ty which is an allowable deduction for tax purposes. Hence he saves $2 in tax in 9 years' time, so he then effectively receives $102—i.e. $100 from the debenture and $2 tax saving.)

8-1 (i) (1.03) 2 — 1 = 6.09% p.a.

(ii) (1.025) 4 — 1 = 10.38% p.a.

(iii) (1.045) 2 — 1 = 9.20% p.a.

8-2 Effective rate of interest = (1.03) 2 — 1 = 6.09% p.a.

By interpolation, at 6.09% p.a. = 7.3293 .'. P.V. of annuity = $732.93

8-3 This could also be obtained by interpolation, but it is simpler to use the result of Exercise 8-2. Amount = 732.93 (1.03)

2 0 = $1323.75.

8-4 P.V. = lOOfl^ at 4 % = $1359.03

8-5 (i) Amount = 10%^ at 1% = $430.77

[Remember: %η = S l T| + (1 + 0 1 8

S t s | 3

(ii) Amount = 1 0 ( 5 ^ — 1) at 1% = $435.08

10,000 8-6 Instalment = at - J %

«270)

= $71.64

S O L U T I O N S T O E X E R C I S E S 63

Amount outstanding after 120th payment = 71.64απη at \ % = $6452.86

Interest in 121st instalment = 6452.86 X .005 = $32.26

.'. Principal included in 121st instalment = 71.64 — 32.26 = $39.38

200 X 1.5 Monthly repayment

: • = $5.00

60

T h e equation of value is 200 = 5flsöl i.e. = 40

at some unknown monthly rate of interest. By interpolation, the monthly interest rate = 1.4402

.*. i = (1.014402) 1 2

— 1 = 18.727 0 Pa-

(b) Principal outs tanding at end of 2 years = 5^361 at 1.4402% = $139.70 .*. Principal so far repaid = 200 — 139.70 = $60.30

(a) .". interest so far paid = 24 X 5 — 60.30 = $59.70

(i) Principal = 400 5

To ta l interest = 400 X .09 X - = $15.00 12

415 Monthly instalment = = $83.00

5

(ii) On the assumption of "rule of 78" the interest payments are propor t ional to 5, 4, 3, 2, 1—i.e. the

9-1

9-2

64 S O L U T I O N S T O E X E R C I S E S

I n s t a l m e n t I n t e r e s t i n P r i n c i p a l i n P r i n c i p a l o u t s t a n d i n g

N u m b e r I n s t a l m e n t I n s t a l m e n t a f t e r p a y i n g i n s t a l m e n t

1 5 78.00 $322.00

2 4 79.00 $243.00

3

CO

80.00 $163.00

4 C

M 81.00 $82.00

5 1 82.00 —

(iii) Equat ion of value is . . .

400 = 83tfrr i.e. « n = 4.8193 which is satisfied by a rate of 1.24% .'. Effective annual rate = (1 .0124)

12 — 1

= 15.9% p.a.

(iv) T h e interest in each instalment is obtained by mult iplying the principal outs tanding by the rate of interest, namely .0124, and hence the full table is constructed.

I n s t a l m e n t

N u m b e r

I n t e r e s t i n

I n s t a l m e n t

P r i n c i p a l i n

I n s t a l m e n t

P r i n c i p a l o u t s t a n d i n g

a f t e r p a y i n g i n s t a l m e n t

1 4.96 78.04 $321.96 2 3.99 79.01 $242.95 3 3.01 79.99 $162.96 4 2.02 80.98 $81.98 5 1.02 81.98 —

interest contained in each of the monthly payments

is . . .

5 4 3 2 1 — of 15, — of 15, — of 15, — of 15 and — of 15 15 15 15 15 15

i.e. $5, $4, $3, $2, and $1 . Hence the schedule is as follows:—

65

A P P E N D I X I

TEST ON CHAPTERS ONE TO FIVE

(i) How many days are there from 31st May to 12th August? (i)

(ii) A 3 months ' bill for $1000 plus 4 % interest is due to mature on 12th August. W h a t is the matur i ty value? (ii) .

(iii) If it is to be discounted on 1st J u n e at a discount rate of 6% p.a., what would be the proceeds of the sale? (iii)

Answers

2. W h a t is the value as at 1st January 1968 at 5 % p.a. compound interest of:—

(i) $100 due on 1/1/75? (i) (ii) 5 annual payments of $100 due on

1/1/69 to 1/1/73 inclusive? (ii) (iii) 4 annual payments of $100 due on

1/1/68 to 1/1/71 inclusive? (iii) (iv) 4 annual payments of $100 due on

1/1/72 to 1/1/75 inclusive? (iv) (v) 4 annual payments of $100 due on

1/1/69 to 1/1/72 inclusive and 3 annual payments of $200 due on 1/1/73 to 1/1/75 inclusive? (v)

(vi) 4 annual payments of $200 due on 1/1/69 to 1/1/72 and 3 annual payments of $100 due on 1/1/73 to 1/1/75 inclusive? (vi)

3. If the following investments earn interest at 6% per annum compound interest, how much will each of them have accumulated to by 30th June , 1975:—

(i) $100 invested on 30/6/70? (i) (ii) 5 annual payments made on 30/6/71

to 30/6/75 inclusive, of $100 each? (ii)

1.

66 A P P E N D I X

Answers

(iii) 7 annual payments made on 30/6/68 to 30/6/74 inclusive, of $100 each? (iii)

(iv) 4 annual payments made on 30/6/67 to 30/6 /70 inclusive, of $100 each? (iv)

(v) 3 annual payments made on 30/6/67 to 30/6 /69 of $200 each and 5 annual payments made on 30/6/70 to 30/6 /74 inclusive, of $100 each? (v)

(vi) 3 annual payments made on 30/6/67 to 30/6/69 of $100 each and 5 annual payments made on 30/6/70 to 30/6/74 inclusive, of $200 each? (vi)

4. Work out, by linear interpolation from the tables provided, the values of: —

5. A loan of $60.00 is to be repaid by annual instalments of $12.00 each, including principal and interest. Interest is at 6% p.a. These $12.00 payments will carry on for a number of vears and will be followed by a final smaller payment at the end of the last year.

(i) How many payments of $12.00 are required? (i)

(ii) Wha t is the amount of the final

(iii) Fill in the following loan repayment schedule.

(i) % a t 5 .1% p.a.

(ii) T h e present value at 5 % p.a. of $100 due in 6 years 2 months ' time. (Ü)

payment? (Ü)

(i) . .

A P P E N D I X 67

Year Number

Principal outstanding

at beginning of year

Interest contained

in instalment

Principal contained

in instalment

Principal outstanding

at end of year

(iv) Wri te down below the formula you would use to check independently the amounts outs tanding after the second and fourth repayments.

(a)

(b) Make the check calculations (b)

COMPOUND INTEREST TABLES

^ per cent

η ( 1 + 1 ) η im a—» «I

1 1.00500 .99502 1.0000 0.9950

2 1.01003 .99007 2.0050 1.9851

3 1.01508 .98515 3.0150 2.9702

4 1.02015 .98025 4.0301 3.9505

5 1.02525 .97537 5.0503 4.9259

6 1.03038 .97052 6.0755 5.8964

7 1.03553 .96569 7.1059 6.8621

8 1.04071 .96089 8.1414 7.8230

9 1.04591 .95610 9.1821 8.7791

10 1.05114 .95135 10.2280 9.7304

It 1.05640 .94661 11.2792 10.6770

12 1.06168 .94191 12.3356 11.6189

13 1.06699 .93722 13.3972 12.5562

14 1.07232 .93256 14.4642 13.4887

15 1.07768 .92792 15.5365 14.4166

16 1.08307 .92330 16.6142 15.3399

17 1.08849 .91871 17.6973 16.2586

18 1.09393 .91414 18.7858 17.1728

19 1.09940 .90959 19.8797 18.0824

20 1.10490 .90506 20.9791 18.9874

21 1.11042 .90056 22.0840 19.8880

22 1.11597 .89608 23.1944 20.7841

23 1.12155 .89162 24.3104 21.6757

24 1.12710 .88719 25.4320 22.5629

25 1.13280 .88277 26.5591 23.4456

50 1.28323 .77929 56.6452 44.1428

75 . 1.45363 .68793 90.7265 62.4136

100 1.64667 .60729 129.3337 78.5426

COMPOUND INTEREST TABLES 69

I per cent

η 5ίΠ a—%

1 1.00750 .99256 1.0000 0.9926

2 1.01506 .98517 2.0075 1.9777

CO 1.02267 .97783 3.0226 2.9556

4 1.03034 .97055 4.0452 3.9261

5 1.03807 .96333 5.0756 4.8894

6 1.04585 .95616 6.1136 5.8456

7 1.05370 .94904 7.1595 6.7946

8 1.06160 .94198 8.2132 7.7366

9 1.06956 .93496 9.2748 8.6716

10 1.07758 .92800 10.3443 9.5996

11 1.08566 .92109 11.4219 10.5207

12 1.09381 .91424 12.5076 11.4349

13 1.10201 .90743 13.6014 12.3423

14 1.11028 .90068 14.7034 13.2430

15 1.11860 .89397 15.8137 14.1370

16 1.12699 .88732 16.9323 15.0243

17 1.13544 .88071 18.0593 15.9050

18 1.14396 .87416 19.1947 16.7792

19 1.15254 .86765 20.3387 17.6468

20 1.16118 .86119 21.4912 18.5080

21 1.16989 .85478 22.6524 19.3628

22 1.17867 .84842 23.8223 20.2112

23 1.18751 .84210 25.0010 21.0533

24 1.19641 .83583 26.1885 21.8891

25 1.20539 .82961 27.3849 22.7188

50 1.45296 .68825 60.3943 41.5664

75 1.75137 .57098 100.1833 57.2027

100 2.11108 .47369 148.1445 70.1746

7 0 C O M P O U N D I N T E R E S T T A B L E S

1 per cent η (1+0 η yn 5

»1 α—ι

1 1.01000 .99010 1.0000 0.9901

2 1.02010 .98030 2.0100 1.9704

CO

1.03030 .97059 3.0301 2.9410 4 1.04060 .96098 4.0604 3.9020 5 1.05101 .95147 5.1010 4.8534

6 1.06152 .94205 6.1520 5.7955 7 1.07214 .93272 7.2135 6.7282 8 1.08286 .92348 8.2857 7.6517 9 1.09369 .91434 9.3685 8.5660

10 1.10462 .90529 10.4622 9.4713

11 1.11567 .89632 11.5668 10.3676

12 1.12683 .88745 12.6825 11.2551 13 1.13809 .87866 13.8093 12.1337 14 1.14947 .86996 14.9474 13.0037 15 1.16097 .86135 16.0969 13.8651

16 1.17258 .85282 17.2579 14.7179 17 1.18430 .84438 18.4304 15.5623 18 1.19615 .83602 19.6147 16.3983 19 1.20811 .82774 20.8109 17.2260 20 1.22019 .81954 22.0190 18.0456

21 1.23239 .81143 23.2392 18.8570

22 1.24472 .80340 24.4716 19.6604

23 1.25716 .79544 25.7163 20.4558 24 1.26973 .78757 26.9735 21.2434 25 1.28243 .77977 28.2432 22.0232

50 1.64463 .60804 64.4632 39.1961

75 2.10913 .47413 110.9128 52.5871

100 2.70481 .36971 170.4814 63.0289

C O M P O U N D I N T E R E S T T A B L E S 71

1\ per cent η ( l + f ) n yn η 1

1 1.01250 .98765 1.0000 0.9877

2 1.02516 .97546 2.0125 1.9631 3 1.03797 .96342 3.0377 2.9265 4 1.05095 .95152 4.0756 3.8781 5 1.06408 .93978 5.1266 4.8178

6 1.07738 .92817 6.1907 5.7460 7 1.09085 .91672 7.2680 6.6627 8 1.10449 .90540 8.3589 7.5681 9 1.11829 .89422 9.4634 8.4623

10 1.13227 .88318 10.5817 9.3455

11 1.14642 .87228 11.7139 10.2178

12 1.16075 .86151 12.8604 11.0793 13 1.17526 .85087 14.0211 11.9302 14 1.18995 .84037 15.1964 12.7706 15 1.20483 .82999 16.3863 13.6005

16 1.21989 .81975 17.5912 14.4203 17 1.23514 .80963 18.8111 15.2299 18 1.25058 .79963 20.0462 16.0295 19 1.26621 .78976 21.2968 16.8193 20 1.28204 .78001 22.5630 17.5993

21 1.29806 .77038 23.8450 18.3697 22 1.31429 .76087 25.1431 19.1306 23 1.33072 .75147 26.4574 19.8820 24 1.34735 .74220 27.7881 20.6242 25 1.36419 .73303 29.1354 21.3573

50 1.86102 .53734 68.8818 37.0129

75 2.53879 .39389 123.1035 48.4890

100 3.46340 .28873 197.0723 56.9013

7 2 C O M P O U N D I N T E R E S T T A B L E S

1\ per cent η (1+0

n vn 1

η \ 1 1.01500 .98522 1.0000 0.9852 2 1.03023 .97066 2.0150 1.9559

OS 1.04568 .95632 3.0452 2.9122

4 1.06136 .94218 4.0909 3.8544 5 1.07728 .92826 5.1523 4.7826

6 1.09344 .91454 6.2296 5.6972 7 1.10984 .90103 7.3230 6.5982 8 1.12649 .88771 8.4328 7.4859 9 1.14339 .87459 9.5593 8.3605

10 1.16054 .86167 10.7027 9.2222

11 1.17795 .84893 11.8633 10.0711 12 1.19562 .83639 13.0412 10.9075 13 1.21355 .82403 14.2368 11.7315 14 1.23176 .81185 15.4504 12.5434 15 1.25023 .79985 16.6821 13.3432

16 1.26899 .78803 17.9324 14.1313 17 1.28802 .77639 19.2014 14.9076 18 1.30734 .76491 20.4894 15.6726 19 1.32695 .75361 21.7967 16.4262 20 1.34686 .74247 23.1237 17.1686

21 1.36706 .73150 24.4705 17.9001 22 1.38756 .72069 25.8376 18.6208 23 1.40838 .71004 27.2251 19.3309 24 1.42950 .69954 28.6335 20.0304 25 1.45095 .68921 30.0630 20.7196

50 2.10524 .47500 73.6828 34.9997

75 3.05459 .32738 136.9728 44.8416

100 4.43205 .22563 228.8030 51.6247

C O M P O U N D I N T E R E S T T A B L E S 73

2 per cent η ( 1 + i ) η %1

1 1.02000 .98039 1.0000 0.9804

2 1.04040 .96117 2.0200 1.9416

3 1.06121 .94232 3.0604 2.8839 4 1.08243 .92385 4.1216 3.8077 5 1.10408 .90573 5.2040 4.7135

6 1.12616 .88797 6.3081 5.6014

7 1.14869 .87056 7.4343 6.4720

8 1.17166 .85349 8.5830 7.3255

9 1.19509 .83676 9.7546 8.1622 10 1.21899 .82035 10.9497 8.9826

11 1.24337 .80426 12.1687 9.7868

12 1.26824 .78849 13.4121 10.5753

13 1.29361 .77303 14.6803 11.3484

14 1.31948 .75788 15.9739 12.1062

15 1.34587 .74301 17.2934 12.8493

16 1.37279 .72845 18.6393 13.5777

17 1.40024 .71416 20.0121 14.2919

18 1.42825 .70016 21.4123 14.9920

19 1.45681 .68643 22.8406 15.6785

20 1.48595 .67297 24.2974 16.3514

21 1.51567 .65978 25.7833 17.0112

22 1.54598 .64684 27.2990 17.6580

23 1.57690 .63416 28.8450 18.2922 24 1.60844 .62172 30.4219 18.9139

25 1.64061 .60953 32.0303 19.5235

50 2.69159 .37153 84.5794 31.4236

75 4.41584 .22646 170.7918 38.6771

100 7.24465 .13803 312.2323 43.0984

7 4 C O M P O U N D I N T E R E S T T A B L E S

2\ per cent η (1+0 η yn a—. ,Γ|

1 1.02500 .97561 1.0000 0.975b

2 1.05063 .95181 2.0250 1.9274

3 1.07689 .92860 3.0756 2.8560

4 1.10381 .90595 4.1525 3.7620

5 1.13141 .88385 5.2563 4.6458

6 1.15969 .86230 6.3877 5.5081

7 1.18869 .84127 7.5474 6.3494 8 1.21840 .82075 8.7361 7.1701

9 1.24886 .80073 9.9545 7.9709 10 1.28008 .78120 11.2034 8.7521

11 1.31209 .76214 12.4835 9.5142 12 1.34489 .74356 13.7956 10.2578

13 1.37851 .72542 15.1404 10.9832 14 1.41297 .70773 16.5190 11.6909 15 1.44830 .69047 17.9319 12.3814

16 1.48451 .67362 19.3802 13.0550 17 1.52162 .65720 20.8647 13.7122 18 1.55966 .64117 22.3863 14.3534 19 1.59865 .62553 23.9460 14.9789 20 1.63862 .61027 25.5447 15.5892

21 1.67958 .59539 27.1833 16.1845

22 1.72157 .58086 28.8629 16.7654

23 1.76461 .56670 30.5844 17.3321 24 1.80873 .55288 32.3490 17.8850

25 1.85394 .53939 34.1578 18.4244

50 3.43711 .29094 97.4843 28.3623

75 6.37221 .15693 214.8883 33.7227

100 11.81372 .08465 432.5486 36.6141

C O M P O U N D I N T E R E S T T A B L E S 75

3 per cent

η (1+0 η χ/η a—.

1 1.03000 .97087 1.0000 0.9709

2 1.06090 .94260 2.0300 1.9135

3 1.09273 .91514 3.0909 2.8286

4 1.12551 .88849 4.1836 3.7171

5 1.15927 .86261 5.3091 4.5797

6 1.19405 .83748 6.4684 5.4172

7 1.22987 .81309 7.6625 6.2303

8 1.26677 .78941 8.8923 7.0197

9 1.30477 .76642 10.1591 7.7861

10 1.34392 .74409 11.4639 8.5302

11 1.38423 .72242 12.8078 9.2526

12 1.42576 .70138 14.1920 9.9540

13 1.46853 .68095 15.6178 10.6350

14 1.51259 .66112 17.0863 11.2961

15 1.55797 .64186 18.5989 11.9379

16 1.60471 .62317 20.1569 12.5611

17 1.65285 .60502 21.7616 13.1661

18 1.70243 .58739 23.4144 13.7535

19 1.75351 .57029 25.1169 14.3238

20 1.80611 .55368 26.8704 14.8775

21 1.86029 .53755 28.6765 15.4150

22 1.91610 .52189 30.5368 15.9369

23 1.97359 .50669 32.4529 16.4436

24 2.03279 .49193 34.4265 16.9355

25 2.09378 .47761 36.4593 17.4131

50 4.38391 .22811 112.7969 25.7298

75 9.17893 .10895 272.6309 29.7018

100 19.21863 .05203 607.2877 31.5989

76 C O M P O U N D I N T E R E S T T A B L E S

3\ per cent η (1+0 » Xjn a—.

1 1.03500 .96618 1.0000 0.9662

2 1.07123 .93351 2.0350 1.8997 3 1.10872 .90194 3.1062 2.8016

4 1.14752 .87144 4.2149 3.6731

5 1.18769 .84197 5.3625 4.5151

6 1.22926 .81350 6.5502 5.3286

7 1.27228 .78599 7.7794 6.1145 8 1.31681 .75941 9.0517 6.8740 9 1.36290 .73373 10.3685 7.6077

10 1.41060 .70892 11.7314 8.3166

11 1.45997 .68495 13.1420 9.0016

12 1.51107 .66178 14.6020 9.6633 13 1.56396 .63940 16.1130 10.3027 14 1.61869 .61778 17.6770 10.9205 15 1.67535 .59689 19.2957 11.5174

16 1.73399 .57671 20.9710 12.0941 17 1.79468 .55720 22.7050 12.6513 18 1.85749 .53836 24.4997 13.1897 19 1.92250 .52016 26.3572 13.7098 20 1.98979 .50257 28.2797 14.2124

21 2.05943 .48557 30.2695 14.6980 22 2.13151 .46915 32.3289 15.1671 23 2.20611 .45329 34.4604 15.6204 24 2.28333 .43796 36.6665 16.0584 25 2.36324 .42315 38.9499 16.4815

50 5.58493 .17905 130.9978 23.4556

75 13.19855 .07577 348.5300 26.4067

100 31.19141 .03206 862.6117 27.6554

COMPOUND INTEREST TABLES 77

4 per cent

η (1+ i ) η 5n-l

1 1.04000 .96154 1.0000 0.9615

2 1.08160 .92456 2.0400 1.8861

3 1.12486 .88900 3.1216 2.7751

4 1.16986 .85480 4.2465 3.6299

5 1.21665 .82193 5.4163 4.4518

6 1.26532 .79031 6.6330 5.2421

7 1.31593 .75992 7.8983 6.0021

8 1.36857 .73069 9.2142 6.7327

9 1.42331 .70259 10.5828 7.4353

10 1.48024 .67556 12.0061 8.1109

11 1.53945 .64958 13.4864 8.7605

12 1.60103 .62460 15.0258 9.3851

13 1.66507 .60057 16.6268 9.9856

14 1.73168 .57748 18.2919 10.5631

15 1.80094 .55526 20.0236 11.1184

16 1.87298 .53391 21.8245 11.6523

17 1.94790 .51337 23.6975 12.1657

18 2.02582 .49363 25.6454 12.6593

19 2.10685 .47464 27.6712 13.1339

20 2.19112 .45639 29.7781 13.5903

21 2.27877 .43883 31.9692 14.0292

22 2.36992 .42196 34.2480 14.4511

23 2.46472 .40573 36.6179 14.8568

24 2.56330 .39012 39.0826 15.2470

25 2.66584 .37512 41.6459 15.6221

50 7.10668 .14071 152.6671 21.4822

75 18.94525 .05278 448.6314 23.6804

100 50.50495 .01980 1237.6237 24.5050

7 8 C O M P O U N D I N T E R E S T T A B L E S

4h per cent

η ( l + ι ) « •s,71 ni

1 1 .04500 .95694 1 .0000 0 . 9 5 6 9

2 1 .09203 . 91573 2 . 0 4 5 0 1.8727

3 1 .14117 .87630 3 .1370 2 . 7 4 9 0

4 1 .19252 . 83856 4 . 2 7 8 2 3 .5875

5 1 .24618 .80245 5 .4707 4 . 3 9 0 0

6 1 .30226 . 7 6 7 9 0 6 . 7 1 6 9 5 . 1 5 7 9

7 1 .36086 . 73483 8 .0192 5 .8927

oc 1.42210 . 70319 9 . 3 8 0 0 6 . 5 9 5 9

9 1 .48610 . 67290 10.8021 7 .2688

10 1.55297 . 64393 12 .2882 7 .9127

11 1 .62285 . 61620 13 .8412 8 . 5 2 8 9

12 1 .69588 . 58966 15 .4640 9 . 1 1 8 6

13 1 .77220 .56427 17 .1599 9 . 6 8 2 9

14 1 .85191 .53997 18.9321 10 .2228

15 1 .93528 . 51672 20 .7841 10 .7395

IG 2 . 0 2 2 3 7 .49147 2 2 . 7 1 9 3 11 .2340

17 2 . 1 1 3 3 8 .47318 2 4 . 7 4 1 7 1 1.7072

18 2 . 2 0 8 4 8 . 45280 26 .8551 1 2 . 1 6 0 0

19 2 . 3 0 7 8 6 . 43330 2 9 . 0 6 3 6 12 .5933

2 0 2.11 171 . 41464 3 1 . 3 7 1 4 13 .0079

21 2 . 5 2 0 2 4 . 39679 33 .7831 13 .4047

22 2 . 6 3 3 0 5 . 37970 3 6 . 3 0 3 4 13 .7844

2 3 2 . 7 5 2 1 7 .36335 3 8 . 9 3 7 0 14 .1478

24 2 .87601 . 34770 4 1 . 6 8 9 2 14 .4955

25 3 . 0 0 5 4 3 . 33273 4 4 . 5 6 5 2 14 .8282

5 0 9 . 0 3 2 6 4 .11071 178 .5030 19 .7620

75 2 7 . 1 4 7 0 0 . 03684 5 8 1 . 0 4 4 4 2 1 . 4 0 3 6

100 8 1 . 5 8 8 5 2 . 0 1 2 2 6 1 7 9 0 . 8 5 6 0 2 1 . 9 4 9 8

C O M P O U N D I N T E R E S T T A B L E S 79

5 per cent

η

1 1.05000 .95238 1.0000 0.9524

2 1.10250 .90703 2.0500 1.8594

3 1.15763 .86384 3.1525 2.7232

4 1.21551 .82270 4.3101 3.5460

5 1.27628 .78353 5.5256 4.3295

6 1.34010 .74622 6.8019 5.0757

7 1.40710 .71068 8.1420 5.7864

8 1.47746 .67684 9.5491 6.4632

9 1.55133 .64461 11.0266 7.1078

10 1.62889 .61391 12.5779 7.7217

11 1.71034 .58468 14.2068 8.3064

12 1.79586 .55684 15.9171 8.8633

13 1.88565 .53032 17.7130 9.3936

14 1.97993 .50507 19.5986 9.8986

15 2.07893 .48102 21.5786 10.3797

16 2.18287 .45811 23.6575 10.8378

17 2.29202 .43630 25.8404 11.2741

18 2.40662 .41552 28.1324 11.6896

19 2.52695 .39573 30.5390 12.0853

20 2.65330 .37689 33.0660 12.4622

21 2.78596 .35894 35.7193 12.8212

22 2.92526 .34185 38.5052 13.1630

23 3.07152 .32557 41.4305 13.4886

24 3.22510 .31007 44.5020 13.7986

25 3.38635 .29530 47.7271 14.0939

50 11.46740 .08720 209.3480 18.2559

75 38.83269 .02575 756.6537 19.4850

100 131.50126 .00760 2610.0252 19.8479

80 C O M P O U N D I N T E R E S T T A B L E S

5\ per cent η (1+0

n 5, τ ι

CL—-ι

n\ 1 1.05500 .94787 1.0000 0.9479

2 1.11303 .89845 2.0550 1.8463

3 1.17424 .85161 3.1680 2.6979

4 1.23882 .80722 4.3423 3.5052 5 1.30696 .76513 5.5811 4.2703

6 1.37884 .72525 6.8881 4.9955

7 1.45468 .68744 8.2669 5.6830

8 1.53469 .65160 9.7216 6.3346

9 1.61909 .61763 11.2563 6.9522 10 1.70814 .58543 12.8754 7.5376

11 1.80209 .55491 14.5835 8.0925

12 1.90121 .52598 16.3856 8.6185

13 2.00577 .49856 18.2868 9.1171 14 2.11609 .47257 20.2926 9.5896

15 2.23248 .44793 22.4087 10.0376

16 2.35526 .42458 24.6411 10.4622 17 2.48480 .40245 26.9964 10.8646 18 2.62147 .38147 29.4812 11.2461 19 2.76565 .36158 32.1027 11.6077 20 2.91776 .34273 34.8683 11.9504

21 3.07823 .32486 37.7861 12.2752 22 3.24754 .30793 40.8643 12.5832 23 3.42615 .29187 44.1118 12.8750 24 3.61459 .27666 47.5380 13.1517 25 3.81339 .26223 51.1526 13.4139

50 14.54196 .06877 246.2175 16.9315

75 55.45420 .01803 990.0764 17.8539

100 211.46864 .00473 3826.7025 18.0958

C O M P O U N D I N T E R E S T T A B L E S 81

6 per cent η (1+0 » ytl Snl

1 1.06000 .94340 1.0000 0.9434

2 1.12360 .89000 2.0600 1.8334

GO

1.19102 .83962 3.1836 2.6730 4 1.26248 .79209 4.3746 3.4651 5 1.33823 .74726 5.6371 4.2124

6 1.41852 .70496 6.9753 4.9173 7 1.50363 .66506 8.3938 5.5824 8 1.59385 .62741 9.8975 6.2098 9 1.68948 .59190 11.4913 6.8017

10 1.79085 .55839 13.1808 7.3601

11 1.89830 .52679 14.9716 7.8869

12 2.01220 .49697 16.8699 8.3838

13 2.13293 .46884 18.8821 8.8527 14 2.26090 .44230 21.0151 9.2950

15 2.39656 .4172/ 23.2760 9.7122

16 2.54035 .39365 25.6725 10.1059

17 2.69277 .37136 28.2129 10.4773 18 2.85434 .35034 30.9057 10.8276 19 3.02560 .33051 33.7600 11.1581 20 3.20714 .31180 36.7856 11.4699

21 3.39956 .29416 39.9927 11.7641

22 3.60354 .27751 43.3923 12.0416

23 3.81975 .26180 46.9958 12.3034

24 4.04893 .24698 50.8156 12.5504

25 4.29187 .23300 54.8645 12.7834

50 18.42015 .05429 290.3359 15.7619

75 79.05692 .01265 1300.9487 16.4558

100 339.30208 .00295 5638.3681 16.6175

82 C O M P O U N D I N T E R E S T T A B L E S

6\ per cent η (1+0 η ytl η 1

1 1.06500 .93897 1.0000 0.9390

2 1.13423 .88166 2.0650 1.8206

CO 1.20795 .82785 3.1992 2.6485

4 1.28647 .77732 4.4072 3.4258 5 1.37009 .72988 5.6936 4.1557

6 1.45914 .68533 7.0637 4.8410 7 1.55399 .64351 8.5229 5.4845 8 1.65500 .60423 10.0769 6.0888 9 1.76257 .56735 11.7319 6.6561

10 1.87714 .53273 13.4944 7.1888

11 1.99915 .50021 15.3716 7.6890 12 2.12910 .46968 17.3707 8.1587 13 2.26749 .44102 19.4998 8.5997 14 2.41487 .41410 21.7673 9.0138 15 2.57184 .38883 24.1822 9.4027

16 2.73901 .36510 26.7540 9.7678 17 2.91705 .34281 29.4930 10.1106 18 3.10665 .32189 32.4101 10.4325 19 3.30859 .30224 35.5167 10.7347 20 3.52365 .28380 38.8253 11.0185

21 3.75268 .26648 42.3490 11.2850 22 3.99661 .25021 46.1016 11.5352 23 4.25639 .23494 50.0982 11.7701 24 4.53305 .22060 54.3546 11.9907 25 4.82770 .20714 58.8877 12.1979

50 23.30668 .04291 343.1797 14.7245

75 112.51763 .00889 1715.6559 15.2479

100 543.20127 .00184 8341.5580 15.3563

C O M P O U N D I N T E R E S T T A B L E S 83

7 per cent

η (1+0 n χ/η S

nl

1 1.07000 .93458 1.0000 0.9346

2 1.14490 .87344 2.0700 1.8080 3 1.22504 .81630 3.2149 2.6243 4 1.31080 .76290 4.4399 3.3872 5 1.40255 .71299 5.7507 4.1002

6 1.50073 .66634 7.1533 4.7665

7 1.60578 .62275 8.6540 5.3893 8 1.71819 .58201 10.2598 5.9713 9 1.83846 .54393 11.9780 6.5152

10 1.96715 .50835 13.8164 7.0236

11 2.10485 .47509 15.7836 7.4987 12 2.25219 .44401 17.8885 7.9427 13 2.40984 .41496 20.1406 8.3577 14 2.57853 .38782 22.5505 8.7455 15 2.75903 .36245 25.1290 9.1079

16 2.95216 .33873 27.8881 9.4466 17 3.15882 .31657 30.8402 9.7632 18 3.37993 .29586 33.9990 10.0591 19 3.61653 .27651 37.3790 10.3356 20 3.86968 .25842 40.9955 10.5940

21 4.14056 .24151 44.8652 10.8355

22 4.43040 .22571 49.0057 11.0612 23 4.74053 .21095 53.4361 11.2722 24 5.07237 .19715 58.1767 11.4693 25 5.42743 .18425 63.2490 11.6536

50 29.45703 .03395 406.5289 13.8007

75 159.87602 .00625 2269.6574 14.1964

100 867.71633 .00115 12381.6618 14.2693

84 C O M P O U N D I N T E R E S T T A B L E S

S per cent

η (1 + ί ) η τ/η Sa

1 1.08000 0.92593 1.0000 0.9259 2 1.16640 0.85734 2.0800 1.7833

CO 1.25971 0.79383 3.2464 2.5771

4 1.36049 0.73503 4.5061 3.3121 5 1.46933 0.68058 5.8666 3.9927

6 1.58687 0.63017 7.3359 4.6229 7 1.71382 0.58349 8.9228 5.2064 8 1.85093 0.54027 10.6366 5.7466 9 1.99901 0.50025 12.4876 6.2469

10 2.15893 0.46319 14.4866 6.7101

11 2.33164 0.42888 16.6455 7.1390 12 2.51817 0.39711 18.9771 7.5361 13 2.71962 0.36770 21.4953 7.9038 14 2.93719 0.34046 24.2149 8.2442 15 3.17217 0.31524 27.1521 8.5595

16 3.42594 0.29189 30.3243 8.8514 17 3.70002 0.27027 33.7502 9.1216 18 3.99602 0.25025 37.4502 9.3719 19 4.31570 0.23171 41.4463 9.6036 20 4.66096 0.21455 45.7620 9.8181

21 5.03383 0.19866 50.4229 10.0168 22 5.43654 0.18394 55.4568 10.2007 23 5.87146 0.17032 60.8933 10.3711 24 6.34118 0.15770 66.7648 10.5288

6.84848 0.14602 73.1059 10.6748

50 46.90161 0.02132 573.7702 12.2335

75 321.20452 0.00311 4002.5567 12.4611

100 2199.76130 0.00045 27484.5160 12.4943

C O M P O U N D I N T E R E S T T A B L E S 85

9 per cent η ( 1 + / ) η vn a*!

1 1.09000 0.91743 1.0000 0.9174 2 1.18810 0.84168 2.0900 1.7591 3 1.29503 0.77218 3.2781 2.5313 4 1.41158 0.70843 4.5731 3.2397 5 1.53862 0.64993 5.9847 3.8897

6 1.67710 0.59627 7.5233 4.4859 7 1.82804 0.54703 9.2004 5.0330 8 1.99256 0.50187 11.0285 5.5348 9 2.17189 0.46043 13.0210 5.9952

10 2.36736 0.42241 15.1929 6.4177

11 2.58043 0.38753 17.5603 6.8052 12 2.81267 0.35554 20.1407 7.1607 13 3.06581 0.32618 22.9534 7.4869 14 3.34173 0.29925 26.0192 7.7862 15 3.64248 0.27454 29.3609 8.0607

16 3.97031 0.25187 33.0034 8.3126 17 4.32763 0.23107 36.9737 8.5436 18 4.71712 0.21199 41.3013 8.7556 19 5.14166 0.19449 46.0185 8.9501 20 5.60441 0.17843 51.1601 9.1285

21 6.10881 0.16370 56.7645 9.2922 22 6.65860 0.15018 62.8733 9.4424 23 7.25787 0.13778 69.5319 9.5802 24 7.91108 0.12641 76.7898 9.7066 25 8.62308 0.11597 84.7009 9.8226

50 74.35752 0.01345 815.0836 10.9617

75 641.19089 0.00156 7113.2321 11.0938

100 5529.04078 0.00018 61422.6754 11.1091

86 C O M P O U N D I N T E R E S T T A B L E S

10 per cent η (1 + 0 η vn S*

1 1.10000 0.90909 1.0000 0.9091 2 1.21000 0.82645 2.1000 1.7355

1.33100 0.75132 3.3100 2.4869 4 1.46410 0.68301 4.6410 3.1699 5 1.61051 0.62092 6.1051 3.7908

6 1.77156 0.56447 7.7156 4.3553 7 1.94872 0.51316 9.4872 4.8684

CO

2.14359 0.46651 11.4359 5.3349 9 2.35795 0.42410 13.5795 5.7590

10 2.59374 0.38554 15.9374 6.1446

11 2.85312 0.35049 18.5312 6.4951 12 3.13843 0.31863 21.3843 6.8137 13 3.45227 0.28966 24.5227 7.1034 14 3.79750 0.26333 27.9750 7.3667 15 4.17725 0.23939 31.7725 7.6061

16 4.59497 0.21763 35.9497 7.8237 17 5.05447 0.19785 40.5447 8.0216 18 5.55992 0.17986 45.5992 8.2014 19 6.11591 0.16351 51.1591 8.3649 20 6.72750 0.14864 57.2750 8.5136

21 7.40025 0.13513 64.0025 8.6487 22 8.14028 0.12285 71.4027 8.7715 23 8.95430 0.11168 79.5430 8.8832 24 9.84973 0.10153 88.4973 8.9847 25 10.83471 0.09230 98.3471 9.0770

50 117.39085 0.00852 1163.9085 9.9148

75 1271.89538 0.00079 12708.9538 9.9921

100 13780.61234 0.00007 137796.1236 9.9993

C O M P O U N D I N T E R E S T T A B L E S 87

11 per cent η ( 1 + 1 ) Λ vn

1 1.11000 0.90090 1.0000 0.9009 2 1.23210 0.81162 2.1100 1.7125

1.36763 0.73119 3.3421 2.4437 4 1.51807 0.65873 4.7097 3.1024 5 1.68506 0.59345 6.2278 3.6959

6 1.87042 0.53464 7.9129 4.2305 7 2.07616 0.48166 9.7833 4.7122

CO

2.30454 0.43393 11.8594 5.1461 9 2.55804 0.39093 14.1640 5.5370

10 2.83942 0.35218 16.7220 5.8892

11 3.15176 0.31728 19.5614 6.2065 12 3.49845 0.28584 22.7132 6.4924 13 3.88328 0.25751 26.2116 6.7499 14 4.31044 0.23200 30.0949 6.9819 15 4.78459 0.20900 34.4054 7.1909

16 5.31089 0.18829 39.1899 7.3792 17 5.89509 0.16963 44.5008 7.5488 18 6.54355 0.15282 50.3959 7.7016 19 7.26334 0.13768 56.9395 7.8393 20 8.06231 0.12403 64.2028 7.9633

21 8.94917 0.11174 72.2651 8.0751 22 9.93357 0.10067 81.2143 8.1757 23 11.02627 0.09069 91.1479 8.2664 24 12.23916 0.08171 102.1742 8.3481 25 13.58546 0.07361 114.4133 8.4217

50 184.56483 0.00542 1668.7712 9.0417

75 2507.39877 0.00040 22785.4434 9.0873

100 34064.17527 0.00003 309665.2297 9.0906

88 C O M P O U N D I N T E R E S T T A B L E S

12 per cent η (l + i ) n vn

1 1.12000 0.89286 1.0000 0.8929 2 1.25440 0.79719 2.1200 1.6901 3 1.40493 0.71178 3.3744 2.4018 4 1.57352 0.63552 4.7793 3.0373 5 1.76234 0.56743 6.3528 3.6048

6 1.97382 0.50663 8.1152 4.1114 7 2.21068 0.45235 10.0890 4.5638 8 2.47596 0.40388 12.2997 4.9676 9 2.77308 0.36061 14.7757 5.3283

10 3.10585 0.32197 17.5487 5.6502

11 3.47855 0.28748 20.6546 5.9377 12 3.89598 0.25668 24.1331 6.1944 13 4.36349 0.22917 28.0291 6.4235 14 4.88711 0.20462 32.3926 6.6282 15 5.47357 0.18270 37.2797 6.8109

16 6.13039 0.16312 42.7533 6.9740 17 6.86604 0.14564 48.8837 7.1196 18 7.68997 0.13004 55.7497 7.2497 19 8.61276 0.11611 63.4397 7.3658 20 9.64629 0.10367 72.0524 7.4694

21 10.80385 0.09256 81.6987 7.5620 22 12.10031 0.08264 92.5026 7.6446 23 13.55235 0.07379 104.6029 7.7184 24 15.17863 0.06588 118.1552 7.7843 25 17.00006 0.05882 133.3339 7.8431

50 289.00219 0.00346 2400.0182 8.3045

75 4913.05573 0.00020 40933.7987 8.3317

100 83522.26582 0.00001 696010.5845 8.3333

C O M P O U N D I N T E R E S T T A B L E S 89

13 per cent

η (1 + 0 η vn 5η

1 1.13000 0.88496 1.0000 0.8859 2 1.27690 0.78315 2.1300 1.6681 3 1.44290 0.69305 3.4069 2.3612 4 1.63047 0.61332 4.8498 2.9745 5 1.84244 0.54276 6.4803 3.5172

6 2.08195 0.48032 8.3227 3.9976 7 2.35261 0.42506 10.4047 4.4226 8 2.65844 0.37616 12.7573 4.7988 9 3.00404 0.33289 15.4157 5.1317

10 3.39457 0.29459 18.4197 5.4262

11 3,83586 0.26070 21.8143 5.6869 12 4.33452 0.23071 25.6502 5.9176 13 4.89801 0.20417 29.9847 6.1218 14 5.53475 0.18068 34.8827 6.3025 15 6.25427 0.15989 40.4175 6.4624

16 7.06733 0.14150 46.6717 6.6039 17 7.98608 0.12522 53.7391 6.7291 18 9.02427 0.11081 61.7251 6.8399 19 10.19742 0.09806 70.7494 6.9380 20 11.52309 0.08678 80.9468 7.0248

21 13.02109 0.07680 92.4699 7.1016 22 14.71383 0.06796 105.4910 7.1695 23 16.62663 0.06014 120.2048 7.2297 24 18.78809 0.05323 136.8315 7.2829 25 21.23054 0.04710 155.6196 7.3300

50 450.73593 0.00222 3459.5071 7.6752

75 9569.36814 0.00010 73602.8318 7.6915

100 203162.87491 0.00000 1562783.6532 7.6923

90 C O M P O U N D I N T E R E S T T A B L E S

14 per cent

η ( 1 + 0 η t/n

1 1.14000 0.87719 1.0000 0.8772 2 1.29960 0.76947 2.1400 1.6467 3 1.48154 0.67497 3.4396 2.3216 4 1.68896 0.59208 4.9211 2.9137 5 1.92542 0.51937 6.6101 3.4331

6 2.19497 0.45559 8.5355 3.8887 7 2.50227 0.39964 10.7305 4.2883 8 2.85259 0.35056 13.2328 4.6389 9 3.25195 0.30751 16.0853 4.9464

10 3.70722 0.26974 19.3373 5.2161

11 4.22623 0.23662 23.0445 5.4527 12 4.81791 0.20756 27.2707 5.6603 13 5.49241 0.18207 32.0887 5.8424 14 6.26135 0.15971 37.5811 6.0021 15 7.13794 0.14010 43.8424 6.1422

16 8.13725 0.12289 50.9804 6.2651 17 9.27646 0.10780 59.1176 6.3729 18 10.57517 0.09456 68.3941 6.4674 19 12.05569 0.08295 78.9692 6.5504 20 13.74349 0.07276 91.0249 6.6231

21 15.66758 0.06383 104.7684 6.6870 22 17.86104 0.05599 120.4360 6.7429 23 20.36159 0.04911 138.2970 6.7921 24 23.21221 0.04308 158.6586 6.8351 25 26.46192 0.03779 181.8708 6.8729

50 700.23299 0.00143 4994.5213 7.1327

75 18529.50639 0.00005 132346.4742 7.1425

100 490326.23804 0.00000 3502323.1289 7.1428

C O M P O U N D I N T E R E S T T A B L E S 91

7 5 per rent

η (1 + 0 * im

1 1.15000 0.86957 1.0000 0.8696 2 1.32250 0.75614 2.1500 1.6257 3 1.52088 0.65752 3.4725 2.2832 4 1.74901 0.57175 4.9934 2.8550 5 2.01136 0.49718 6.7424 3.3522

6 2.31306 0.43233 8.7537 3.7845 7 2.66002 0.37594 11.0668 4.1604 oc 3.05902 0.32690 13.7268 4.4873 9 3.51788 0.28426 16.7858 4.7716

10 4.04556 0.24719 20.3037 5.0188

U 4.65239 0.21494 24.3493 5.2337 12 5.35025 0.18691 29.0017 5.4206 13 6.15279 0.16253 34.3519 5.5831 14 7.07571 0.14133 40.5047 5.7245 15 8.13706 0.12289 47.5804 5.8474

16 9.35762 0.10687 55.7175 5.9542 17 10.76126 0.09293 65.0751 6.0472 18 12.37545 0.08081 75.8364 6.1280 19 14.23177 0.07027 88.2118 6.1982 20 16.36654 0.06110 102.4436 6.2593

21 18.82152 0.05313 118.8101 6.3125 22 21.64475 0.04620 137.6316 6.3587 23 24.89146 0.04017 159.2764 6.3988 24 28.62518 0.03493 184.1678 6.4338 25 32.91895 0.03038 212.7930 6.4641

50 1083.6574 0.00092 7217.7163 6.6605

75 35672.86835 0.00003 237812.4557 6.6665

100 1174313.46715 0.00000 7828749.7810 6.6667

92 C O M P O U N D I N T E R E S T T A B L E S

per cent

η ( 1 + ί ) « «71

1 1.17500 0.85106 1.0000 0.8511 2 1.38063 0.72431 2.1750 1.5754 3 1.62223 0.61643 3.5556 2.1918. 4 1.90613 0.52462 5.1779 2.7164 5 2.23970 0.44649 7.0840 3.1629

6 2.63164 0.37999 9.3237 3.5429 7 3.09218 0.32340 11.9553 3.8663 8 3.63331 0.27523 15.0475 4.1415 9 4.26914 0.23424 18.6808 4.3758

10 5.01624 0.19935 22.9500 4.5751

11 5.89409 0.16966 27.9662 4.7448 12 6.92555 0.14439 33.8603 4.8892 13 8.13752 0.12289 40.7858 5.0121 14 9.56159 0.10459 48.9234 5.1167 15 11.23487 0.08901 58.4850 5.2057

16 13.20097 0.07575 69.7198 5.2814 17 15.51114 0.06447 82.9208 5.3459 18 18.22559 0.05487 98.4319 5.4008 19 21.41506 0.04670 116.6575 5.4475 20 25.16270 0.03974 138.0726 5.4872

21 29.56617 0.03382 163.2353 5.5210 22 34.74025 0.02879 192.8015 5.5498 23 40.81979 0.02450 227.5417 5.5743 24 47.96325 0.02085 268.3616 5.5951

56.35682 0.01774 316.3247 5.6129

50 3176.09388 0.00032 18143.3936 5.7125

75 178994.62739 0.00001 1022820.7279 5.7143

100 10087572.30888 0.00000 57643264.62222 5.7143

C O M P O U N D I N T E R E S T T A B L E S 93

20 per cent

η ( 1 + 0 η yn «τη

1 1.20000 0.83333 1.0000 0.8333 2 1.44000 0.69444 2.2000 1.5278 3 1.72800 0.57870 3.6400 2.1065 4 2.07360 0.48225 5.3680 2.5887 5 2.48832 0.40188 7.4416 2.9906

6 2.98598 0.33490 9.9299 3.3255 7 3.58318 0.27908 12.9159 3.6046 8 4.29982 0.23257 16.4991 3.8372 9 5.15978 0.19381 20.7989 4.0310

10 6.19174 0.16151 25.9587 4.1925

11 7.43008 0.13459 32.1504 4.3271 12 8.91610 0.11216 39.5805 4.4392 13 10.69932 0.09346 48.4966 4.5327 14 12.83919 0.07789 59.1959 4.6106 15 15.40702 0.06491 72.0351 4.6755

16 18.48843 0.05409 87.4421 4.7296 17 22.18611 0.04507 105.9306 4.7746 18 26.62333 0.03756 128.1167 4.8122 19 31.94800 0.03130 154.7400 4.8435 20 38.33760 0.02608 186.6880 4.8696

21 46.00512 0.02174 225.0256 4.8913 22 55.20614 0.01811 271.0307 4.9094 23 66.24737 0.01510 326.2369 4.9245 24 79.49685 0.01258 392.4842 4.9371 25 95.39622 0.01048 471.9811 4.9476

50 9100.43815 0.00011 45497.1908 4.9995

75 868147.36931 0.00000 4340731.8466 5.0000

100 82817974.52183 0.00000 414089867.6092 5.0000