Jacques I. Mathematics for Economics and Business

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example. We work with smaller and smaller time periods until the numbers settle down to a

ﬁxed value. However, it turns out that there is a special formula that can be used to compute

this directly. The future value, S, of a principal, P, compounded continuously for t years at an

annual rate of r% is

where e is the number

2.718 281 828 459 045 235 36 (to 20 decimal places)

If r = 12, t = 1 and P = 10 then this formula gives

S = $10e

12×1/100

= $10e

0.12

= $11.27

which is in agreement with the limiting value obtained in the previous example.

S = Pe

rt/100

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200

Advice

The number e and the related natural logarithm function were first introduced in Section

2.4. If you missed this section out, you should go back and read through this work now

before proceeding. The link between the number e and the above formula for continuous

compounding is given in Practice Problem 23 at the end of this section. However, you may

prefer to accept it without justification and concentrate on the applications.

Example

A principal of $2000 is invested at 10% interest compounded continuously. After how many days will the

investment ﬁrst exceed $2100?

Solution

We want to save a total of $2100 starting with an initial investment of $2000. The problem is to determine

the number of days required for this on the assumption that the interest rate is 10% compounded con-

tinuously. The formula for continuous compounding is

S = Pe

rt/100

We are given that

S = 2100, P = 2000, r = 10

so we need to solve the equation

2100 = 2000e

10t/100

for t. Dividing through by 2000 gives

1.05 = e

0.1t

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As explained in Section 2.4, equations such as this can be solved using natural logarithms. Recall that

if M = e

then n = ln M

If we apply this deﬁnition to the equation

1.05 = e

0.1t

with M = 1.05 and n = 0.1t then

0.1t = ln(1.05)

= 0.048 790 2

and so t = 0.488 to three decimal places.

The variable t which appears in the formula for continuous compounding is measured in years, so to

convert it to days we multiply by 365 (assuming that there are 365 days in a year). Hence

t = 365 × 0.488 = 178.1 days

We deduce that the amount invested ﬁrst exceeds $2100 some time during the 179th day.

3.2 • Compound interest

201

Practice Problems

(1) A principal, $30, is invested at 6% interest for 2 years. Determine the future value if the interest is

compounded

(a) annually (b) semi-annually (c) quarterly

(d) monthly (e) weekly (f) daily

(2) Use the formula

S = Pe

rt/100

to determine the future value of $30 invested at 6% interest compounded continuously for

2 years. Confirm that it is in agreement with the results of part (1).

5 Determine the rate of interest required for a principal of $1000 to produce a future value of $4000

after 10 years compounded continuously.

Example

A principal of $10 000 is invested at one of the following banks:

(a) Bank A offers 4.75% interest, compounded annually.

(b) Bank B offers 4.70% interest, compounded semi-annually.

(d) Bank D offers 4.6% interest, compounded continuously.

For each bank, tabulate the value of the investment at the end of every year, for the next 10 years. Which of

these banks would you recommend?

EXCEL

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Solution

In Figure 3.2 the numbers 0 to 10 have been entered in the ﬁrst column, together with appropriate head-

ings. The initial investment is the same for each bank, so the value 10 000 is typed into cells B4 to E4.

For banks A, B and C, the future values can be worked out using the formula

10 000 1 +

for appropriate values of r and n.

(a) In Bank A, the interest rate is 4.75% compounded annually, so at the end of year 1, the investment is

10 000(1 + 0.0475)

The reason for writing it to the power of 1 is so that when we enter it into Excel, and copy down the

ﬁrst column, the power will automatically increase in accordance with the changing years. We type

=10000*(1+0.0475)^A5

in cell B5, and then click and drag down to cell B14.

(b) Bank B offers a return of 4.7% compounded semi-annually, so that at the end of year 1, the investment is

10 000(1 + 0.047/2)

2×1

In general, after t years, the investment is

10 000(1 + 0.0475/2)

2×t

so in Excel, we type

=10000*(1+0.0475/2)^(2*A5)

in cell C5, and copy down this column in the usual way.

100

Mathematics of Finance

202

Figure 3.2

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=10000*(1+0.0465/4)^(4*A5)

in cell D5, and copy down the column.

(d) Bank D offers a return of 4.6% compounded continuously, so after t years the future value is given by

10 000e

4.6t/100

= 10 000e

0.046t

The corresponding values are calculated in column E by typing

=10000*EXP(0.046*A5) in cell E5 and copying down the column.

The completed spreadsheet is shown in Figure 3.3. The amounts have been rounded to 2 decimal places

by highlighting cells B5 through to E14, and using the Decrease Decimal icon on the toolbar.

Figure 3.3 shows that there is very little to choose between these banks and that, in practice, other issues

(such as any conditions or penalties attached to future withdrawals from the account) may well inﬂuence

our recommendation. However, from a purely monetary point of view, we should advise the investor to put

the money into Bank B, as this offers the greatest return. Notice that Bank B is not the one with the highest

rate of interest. This example highlights the importance of taking into account the frequency of com-

pounding, as well as the actual rate of interest.

3.2 • Compound interest

203

Figure 3.3

Given that there are so many ways of calculating compound interest, people often ﬁnd it

difﬁcult to appraise different investment opportunities. What is needed is a standard ‘bench-

mark’ that enables an individual to compare different forms of savings or credit schemes on an

equal basis. The one that is commonly used is annual compounding. All ﬁrms offering invest-

ment or loan facilities are required to provide the effective annual rate. This is often referred to

as the annual percentage rate, which is abbreviated to APR. The APR is the rate of interest

which, when compounded annually, produces the same yield as the nominal (that is, the

stated) rate of interest. The phrase ‘annual equivalent rate’ (AER) is frequently used when

applied to savings. However, in this book we shall use APR for both savings and loans.

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Although the aim of this chapter is to investigate the mathematics of ﬁnance, the mathem-

atical techniques themselves are more widely applicable. We conclude this section with two

examples to illustrate this.

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204

Example

Determine the annual percentage rate of interest of a deposit account that has a nominal rate of 8% com-

pounded monthly.

Solution

The APR is the overall rate of interest, which can be calculated using scale factors. If the account offers a

return of 8% compounded monthly then each month the interest is

==0.67%

of the amount invested at the beginning of that month. The monthly scale factor is

1 +=1.0067

so in a whole year the principal gets multiplied by

(1.0067)

= 1.0834

which can be written as

1 +

so the APR is 8.34%.

8.34

100

0.67

100

Practice Problem

6 Determine the annual percentage rate of interest if the nominal rate is 12% compounded quarterly.

Example

A country’s annual GNP (gross national product), currently at $25 000 million, is predicted to grow by 3.5%

each year. The population is expected to increase by 2% a year from its current level of 40 million. After how

many years will GNP per capita (that is, GNP per head of population) reach $700?

Solution

The per capita value of GNP is worked out by dividing GNP by the size of the population. Initially, this is

= $625

During the next few years, GNP is forecast to grow at a faster rate than the population so this value will

increase.

25 000 000 000

40 000 000

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The scale factor associated with a 3.5% increase is 1.035, so after n years GNP (in millions of dollars) will be

GNP = 25 000 × (1.035)

Similarly, the population (also in millions) will be

population = 40 × (1.02)

Hence GNP per capita is

=× = 625 ×

We want to ﬁnd the number of years required for this to reach 700, so we need to solve the equation

625 ×

= 700

for n. Dividing both sides by 625 gives

= 1.12

and after taking logs of both sides we get

log

= log(1.12)

n log = log(1.12) (rule 3 of logs)

so that

n ==7.76

We deduce that the target ﬁgure of $700 per capita will be achieved after 8 years.

log(1.12)

log(1.035/1.02)

1.035

1.02

1.035

1.02

1.035

1.02

1.035

1.02

1.035

1.02

(1.035)

(1.02)

25 000

25 000 × (1.035)

40 × (1.02)

3.2 • Compound interest

205

Example

A ﬁrm decides to increase output at a constant rate from its current level of 50 000 to 60 000 during the next

5 years. Calculate the annual rate of increase required to achieve this growth.

Solution

If the rate of increase is r% then the scale factor is 1 + so after 5 years, output will be

50 000 1 +

To achieve a ﬁnal output of 60 000, the value of r is chosen to satisfy the equation

50 000 1 +

= 60 000

Dividing both sides by 50 000 gives

1 +

= 1.2

100

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The difﬁculty in solving this equation is that the unknown, r, is trapped inside the brackets, which are raised

to the power of 5. This is analogous to the problem of solving an equation such as

= 5.23

which we would solve by taking square roots of both sides to ﬁnd x. This suggests that we can ﬁnd r by

taking ﬁfth roots of both sides of

1 +

= 1.2

to get

1 +=(1.2)

1/ 5

= 1.037

Hence r = 3.7%.

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Mathematics of Finance

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Practice Problem

7 The turnover of a leading supermarket chain, A, is currently $560 million and is expected to increase

at a constant rate of 1.5% a year. Its nearest rival, supermarket B, has a current turnover of $480 mil-

lion and plans to increase this at a constant rate of 3.4% a year. After how many years will supermar-

ket B overtake supermarket A?

Annual percentage rate The equivalent annual interest paid for a loan, taking into

account the compounding over a variety of time periods.

Compound interest The interest which is added on to the initial investment, so that this

will itself gain interest in subsequent time periods.

Continuous compounding The limiting value when interest is compounded with ever-

increasing frequency.

Future value The ﬁnal value of an investment after one or more time periods.

Principal The value of the original sum invested.

Simple interest The interest which is paid direct to the investor instead of being added to

the original amount.

Key Terms

Practice Problems

8 A bank offers a return of 7% interest compounded annually. Find the future value of a principal of

$4500 after 6 years. What is the overall percentage rise over this period?

9 Find the future value of $20 000 in 2 years’ time if compounded quarterly at 8% interest.

10 Midwest Bank offers a return of 5% compounded annually for each and every year. The rival BFB

offers a return of 3% for the first year and 7% in the second and subsequent years (both compounded

annually). Which bank would you choose to invest in if you decided to invest a principal for (a) 2 years;

(b) 3 years?

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11 The value of an asset, currently priced at $100 000, is expected to increase by 20% a year.

(a) Find its value in 10 years’ time.

(b) After how many years will it be worth $1 million?

12 How long will it take for a sum of money to double if it is invested at 5% interest compounded

annually?

13 A piece of machinery depreciates in value by 5% a year. Determine its value in 3 years’ time if its

current value is $50 000.

14 A principal, $7000, is invested at 9% interest for 8 years. Determine its future value if the interest is

compounded

(a) annually (b) semi-annually (c) monthly (d) continuously

15 A car depreciates by 40% in the first year, 30% in the second year and 20% thereafter. I buy a car

for $14 700 when it is 2 years old.

(a) How much did it cost when new?

(b) After how many years will it be worth less than 25% of the amount that I paid for it?

16 Find the future value of $100 compounded continuously at an annual rate of 6% for 12 years.

17 How long will it take for a sum of money to triple in value if invested at an annual rate of 3% com-

pounded continuously?

18 If a piece of machinery depreciates continuously at an annual rate of 4%, how many years will it take

for the value of the machinery to halve?

19 A department store has its own credit card facilities, for which it charges interest at a rate of 2% each

month. Explain briefly why this is not the same as an annual rate of 24%. What is the annual

percentage rate?

20 Determine the APR if the nominal rate is 7% compounded continuously.

21 Current annual consumption of energy is 78 billion units and this is expected to rise at a fixed rate of

5.8% each year. The capacity of the industry to supply energy is currently 104 billion units.

(a) Assuming that the supply remains steady, after how many years will demand exceed supply?

(b) What constant rate of growth of energy production would be needed to satisfy demand for the

next 50 years?

22 The population of a country is currently at 56 million and is forecast to rise by 3.7% each year. It is

capable of producing 2500 million units of food each year, and it is estimated that each member of

the population requires a minimum of 65 units of food each year. At the moment, the extra food

needed to satisfy this requirement is imported, but the government decides to increase food produc-

tion at a constant rate each year, with the aim of making the country self-sufficient after 10 years.

Find the annual rate of growth required to achieve this.

23 If a principal, P, is invested at r% interest compounded annually then its future value, S, after n years

is given by

S = P 1 +

(a) Use this formula to show that if an interest rate of r% is compounded k times a year then after

t years

S = P 1 +

100k

100

3.2 • Compound interest

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(b) Show that if m = 100k/r then the formula in part (1) can be written as

S = P 1 +

rt / 100

e = lim

m→∞

1 +

to deduce that if the interest is compounded with ever-increasing frequency (that is, continu-

ously) then

S = Pe

rt/100

24 (Excel) The sum of $100 is invested at 12% interest for 20 years. Tabulate the value of the invest-

ment at the end of each year, if the interest is compounded

(a) annually (b) quarterly (c) monthly (d) continuously

Draw graphs of these values on the same diagram. Comment briefly on any similarities and differ-

ences between these graphs.

25 (Excel) A department store charges interest on any outstanding debt at the end of each month. It

decides to produce a simple table of APRs for its customers, based on a variety of monthly rates. Use

a spreadsheet to produce such a table for monthly interest rates of 0.5%, 0.6%, 0.7%, ..., 3%. Plot

a graph of APR against monthly rate and comment briefly on its basic shape.

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section 3.3

Geometric series

Consider the following sequence of numbers:

2, 6, 18, 54,...

One obvious question, often asked in intelligence tests, is what is the next term in the sequence?

All that is required is for you to spot the pattern so that it can be used to generate the next term.

In this case, successive numbers are obtained by multiplying by 3, so the ﬁfth term is

54 × 3 = 162

the sixth term is

162 × 3 = 486

and so on. Any sequence in which terms are calculated by multiplying their predecessor by

a ﬁxed number is called a geometric progression and the multiplicative factor itself is called a

geometric ratio. The sequence above is a geometric progression with geometric ratio 3. The

reason for introducing these sequences is not to help you to answer intelligence tests, but rather

to analyse compound interest problems. You may well have noticed that all of the problems

given in the previous section produced such a sequence. For example, if a principal, $500, is

invested at 10% interest compounded annually, then the future values in successive years are

500(1.1), 500(1.1)

, 500(1.1)

,...

which we recognize as a geometric progression with geometric ratio 1.1.

Objectives

At the end of this section you should be able to:

 Recognize a geometric progression.

 Evaluate a geometric series.

 Calculate the total investment obtained from a regular savings plan.

 Calculate the instalments needed to repay a loan.

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