Jacques I. Mathematics for Economics and Business

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In real terms the house has fallen in value by over $2000.

To adjust the price of the house in 1993 we ﬁrst need to divide by 1.035 to backtrack to the year 1992,

and then divide again by 1.071 to reach 1991. We get

= 90 213

In real terms there has at least been some gain during 1993. However, this is less than impressive, and from

a purely ﬁnancial point of view, there would have been more lucrative ways of investing this capital.

For the 1994 price, the adjusted value is

= 93 476

and, for 1990, the adjusted value is

72 000 × 1.107 = 79 704

Table 3.9 lists both the nominal and the ‘constant 1991’ values of the house (rounded to the nearest thou-

sand) for comparison. It shows quite clearly that, apart from the gain during 1991, the increase in value has,

in fact, been quite modest.

106 000

1.023 × 1.035 × 1.071

100 000

1.035 × 1.071

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Table 3.9

Year

1990 1991 1992 1993 1994

Nominal house price 72 89 94 100 106

1991 house price 80 89 87 90 93

Practice Problem

8 Table 3.10 shows the average annual salary (in thousands of dollars) of employees in a small firm,

together with the annual rate of inflation for that year. Adjust these salaries to the prices prevailing at

the end of 1991 and so give the real values of the employees’ salaries at constant ‘1991 prices’.

Comment on the rise in earnings during this period.

Table 3.10

Year

1990 1991 1992 1993 1994

Salary 17.3 18.1 19.8 23.5 26.0

Inﬂation 4.9 4.3 4.0 3.5

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3.1 • Percentages

191

Annual rate of inﬂation The percentage increase in the level of prices over a 12-month

period.

Index numbers The scale factor of a variable measured from the base year multiplied

by 100.

Laspeyre index An index number for groups of data which are weighted by the quantities

used in the base year.

Nominal data Monetary values prevailing at the time that they were measured.

Paasche index An index number for groups of data which are weighted by the quantities

used in the current year.

Real data Monetary values adjusted to take inﬂation into account.

Scale factor The multiplier that gives the ﬁnal value in percentage problems.

Time series A sequence of numbers indicating the variation of data over time.

Key Terms

Practice Problems

9 Express the following percentages as fractions in their simplest form:

(a) 35% (b) 88% (c) 250% (d) 17

/2% (e) 0.2%

10 Calculate each of the following:

(a) 5% of 24 (b) 8% of 88 (c) 48% of 4563 (d) 112% of 56

11 Write down the scale factors corresponding to

(a) an increase of 19%

(b) an increase of 250%

(d) a decrease of 43%

12 Find the new quantities when

(a) £16.25 is increased by 12%

(b) the population of a town, currently at 113 566, rises by 5%

(d) a good priced at £2300 is reduced by 30% in a sale

(e) a car, valued at £23 000, depreciates by 32%

13 A shop sells books at ‘20% below the recommended retail price (r.r.p.)’. If it sells a book for £12.40

find

(a) the r.r.p.

(b) the cost of the book after a further reduction of 15% in a sale

with the manufacturer’s r.r.p.

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14 A TV costs £940 including 17.5% VAT. Find the new price if VAT is reduced to 8%.

15 An antiques dealer tries to sell a vase at 45% above the £18 000 which the dealer paid at auction.

(a) What is the new sale price?

(b) By what percentage can the dealer now reduce the price before making a loss?

16 Find the single percentage increase or decrease equivalent to

(a) a 10% increase followed by a 25% increase

(b) a 34% decrease followed by a 65% increase

Explain in words why the overall change in part (c) is not 0%.

17 Table 3.11 shows the index numbers associated with transport costs during a 20-year period. The

public transport costs reflect changes to bus and train fares, whereas private transport costs include

purchase, service, petrol, tax and insurance costs of cars.

(1) Which year is chosen as the base year?

(2) Find the percentage increases in the cost of public transport from

(3) Repeat part (2) for private transport.

(4) Comment briefly on the relative rise in public and private transport costs during this 20-year period.

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192

Table 3.11

Year

1985 1990 1995 2000 2005

Public transport 100 130 198 224 245

Private transport 100 125 180 199 221

Table 3.12

Quarter

Q1 Q2 Q3 Q4

Output 13.5 1.4 2.5 10.5

18 Table 3.12 shows the number of items (in thousands) produced from a factory production line dur-

ing the course of a year. Taking the second quarter as the base quarter, calculate the associated index

numbers. Suggest a possible reason for the fluctuations in output.

19 Table 3.13 shows government expenditure (in billions of dollars) on education for four consecutive

years, together with the rate of inflation for each year.

(a) Taking 1994 as the base year, work out the index numbers of the nominal data given in the

second row of the table.

(b) Find the values of expenditure at constant 1994 prices and hence recalculate the index numbers

of real government expenditure.

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(Excel) Table 3.15 shows the annual salaries (in thousands of dollars) of four categories of employee

during a 3-year period. In year 0 the firm employed 24, 250, 109 and 7 people of types A, B, C and

D, respectively. Calculate the Laspeyre index of the total wage bill in years 1 and 2 taking year 0 as

the base year. Comment briefly on these values.

3.1 • Percentages

193

Table 3.13

Year

1994 1995 1996 1997

Spending 236 240 267 276

Inﬂation 4.7 4.2 3.4

Table 3.15

Year 0 Year 1 Year 2

Type A 12 13 13

Type B 26 28 29

Type C 56 56 64

Type D 240 340 560

Table 3.14

Year

1234567 8

Index 1 100 95 105 110 119 127

Index 2 100 112 118

22 (Excel) In the Laspeyre index, the quantities used for the weights are those of the base year. If these

are replaced by quantities for the current year, then the index is called the Paasche index. In Problem

21, suppose that the numbers of employees of types A, B, C and D in year 1 are 30, 240, 115 and 8

respectively. For year 2 the corresponding figures are 28, 220, 125 and 20. By adding two extra

columns for the products P

and P

to the spreadsheet of Problem 21, calculate the Paasche

index for years 1 and 2. Compare with the Laspeyre index calculated in Problem 21.

State one advantage and one disadvantage of using the Laspeyre and Paasche methods for the

calculation of combined index numbers.

20 Index numbers associated with the growth of unemployment during an 8-year period are shown in

Table 3.14.

(a) What are the base years for the two indices?

(b) If the government had not switched to index 2, what would be the values of index 1 in years

7 and 8?

(d) If unemployment was 1.2 million in year 4, how many people were unemployed in years 1 and 8?

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section 3.2

Compound interest

Today, businesses and individuals are faced with a bewildering array of loan facilities and

investment opportunities. In this section we explain how these ﬁnancial calculations are car-

ried out to enable an informed choice to be made between the various possibilities available.

We begin by considering what happens when a single lump sum is invested and show how to

calculate the amount accumulated over a period of time.

Suppose that someone gives you the option of receiving $500 now or $500 in 3 years’ time.

Which of these alternatives would you accept? Most people would take the money now, partly

because they may have an immediate need for it, but also because they recognize that $500 is

worth more today than in 3 years’ time. Even if we ignore the effects of inﬂation, it is still

better to take the money now, since it can be invested and will increase in value over the 3-year

period. In order to work out this value we need to know the rate of interest and the basis

on which it is calculated. Let us begin by assuming that the $500 is invested for 3 years at

10% interest compounded annually. What exactly do we mean by ‘10% interest compounded

annually’? Well, at the end of each year, the interest is calculated and is added on to the amount

currently invested. If the original amount is $500 then after 1 year the interest is 10% of $500,

which is

× $500 =×$500 = $50

so the amount rises by $50 to $550.

100

Objectives

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

 Understand the difference between simple and compound interest.

 Calculate the future value of a principal under annual compounding.

 Calculate the future value of a principal under continuous compounding.

 Determine the annual percentage rate of interest given a nominal rate of interest.

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What happens to this amount at the end of the second year? Is the interest also $50? This

would actually be the case with simple interest, when the amount of interest received is the

same for all years. However, with compound interest, we get ‘interest on the interest’. Nearly

all ﬁnancial investments use compound rather than simple interest, because investors need

to be rewarded for not taking the interest payment out of the fund each year. Under annual

compounding the interest obtained at the end of the second year is 10% of the amount invested

at the start of that year. This not only consists of the original $500, but also the $50 already

received as interest on the ﬁrst year’s investment. Consequently, we get an additional

× $550 = $55

raising the sum to $605. Finally, at the end of the third year, the interest is

× $605 = $60.50

so the investment is $665.50. You are therefore $165.50 better off by taking the $500 now and

investing it for 3 years. The calculations are summarized in Table 3.16.

3.2 • Compound interest

195

Table 3.16

End of year Interest ($) Investment ($)

1 50 550

2 55 605

3 60.50 665.50

Example

Find the value, in 4 years’ time, of $10 000 invested at 5% interest compounded annually.

Solution

 At the end of year 1 the interest is 0.05 × 10 000 = 500, so the investment is 10 500.

 At the end of year 2 the interest is 0.05 × 10 500 = 525, so the investment is 11 025.

 At the end of year 3 the interest is 0.05 × 11 025 = 551.25, so the investment is 11 576.25.

 At the end of year 4 the interest is 0.05 × 11 576.25 = 578.81 rounded to 2 decimal places, so the ﬁnal

investment is $12 155.06 to the nearest cent.

Practice Problem

1 Find the value, in 10 years’ time, of $1000 invested at 8% interest compounded annually.

The previous calculations were performed by ﬁnding the interest earned each year and

adding it on to the amount accumulated at the beginning of the year. As you may have discovered

in Practice Problem 1, this can be rather laborious, particularly if the money is invested over a

long period of time. What is really needed is a method of calculating the investment after, say,

10 years without having to determine the amount for the 9 intermediate years. This can be

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done using the scale factor approach discussed in the previous section. To illustrate this, let us

return to the problem of investing $500 at 10% interest compounded annually. The original

sum of money is called the principal and is denoted by P, and the ﬁnal sum is called the future

value and is denoted by S. The scale factor associated with an increase of 10% is

1 +=1.1

so at the end of 1 year the total amount invested is P(1.1).

After 2 years we get

P(1.1) × (1.1) = P(1.1)

and after 3 years the future value is

S = P(1.1)

× (1.1) = P(1.1)

Setting P = 500, we see that

S = 500(1.1)

= $665.50

which is, of course, the same as the amount calculated previously.

In general, if the interest rate is r% compounded annually then the scale factor is

1 +

so, after n years,

S = P 1 +

Given the values of r, P and n it is trivial to evaluate S using the power key on a calculator.

To see this let us rework the previous example using this formula.

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Mathematics of Finance

196

Example

Find the value, in 4 years’ time, of $10 000 invested at 5% interest compounded annually.

Solution

In this problem, P = 10 000, r = 5 and n = 4, so the formula S = P 1 +

gives

S = 10 000 1 +

= 10 000(1.05)

= 12 155.06

which is, of course, the same answer as before.

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Practice Problem

2 Use the formula

S = P 1 +

to find the value, in 10 years’ time, of $1000 invested at 8% interest compounded annually. [You

might like to compare your answer with that obtained in Practice Problem 1.]

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The compound interest formula derived above involves four variables, r, n, P and S.

Provided that we know any three of these, we can use the formula to determine the remaining

variable. This is illustrated in the following example.

3.2 • Compound interest

197

Example

A principal of $25 000 is invested at 12% interest compounded annually. After how many years will the

investment ﬁrst exceed $250 000?

Solution

We want to save a total of $250 000 starting with an initial investment of $25 000. The problem is to deter-

mine the number of years required for this on the assumption that the interest is ﬁxed at 12% throughout

this time. The formula for compound interest is

S = P 1 +

We are given that

P = 25 000, S = 250 000, r = 12

so we need to solve the equation

250 000 = 25 000 1 +

for n.

One way of doing this would just be to keep on guessing values of n until we ﬁnd the one that works.

However, a more mathematical approach is to use logarithms, because we are being asked to solve an equa-

tion in which the unknown occurs as a power. Following the method described in Section 2.3, we ﬁrst divide

both sides by 25 000 to get

10 = (1.12)

Taking logarithms of both sides gives

log(10) = log(1.12)

and if you apply rule 3 of logarithms you get

log(10) = n log(1.12)

Hence

n =

= (taking logarithms to base 10)

= 20.3 (to 1 decimal place)

Now we know that n must be a whole number because interest is only added on at the end of each year.

We assume that the ﬁrst interest payment occurs exactly 12 months after the initial investment and every

0.049 218 023

log(10)

log(1.12)

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12 months thereafter. The answer, 20.3, tells us that after only 20 years the amount is less than $250 000, so

we need to wait until 21 years have elapsed before it exceeds this amount. In fact, after 20 years

S = $25 000(1.12)

= $241 157.33

and after 21 years

S = $25 000(1.12)

= $270 096.21

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Practice Problem

3 A firm estimates that its sales will rise by 3% each year and that it needs to sell at least 10 000 goods

each year in order to make a profit. Given that its current annual sales are only 9000, how many years

will it take before the firm breaks even?

You may have noticed that in all of the previous problems it is assumed that the interest is

compounded annually. It is possible for interest to be added to the investment more frequently

than this. For example, suppose that a principal of $500 is invested for 3 years at 10% interest

compounded quarterly. What do we mean by ‘10% interest compounded quarterly’? Well, it

does not mean that we get 10% interest every 3 months. Instead, the 10% is split into four equal

portions, one for each quarter. Every 3 months the interest accrued is

= 2.5%

so after the ﬁrst quarter the investment gets multiplied by 1.025 to give

500(1.025)

and after the second quarter it gets multiplied by another 1.025 to give

500(1.025)

and so on. Moreover, since there are exactly 12 three-month periods in 3 years we deduce that

the future value is

500(1.025)

= $672.44

10%

In this example we calculated the time taken for $25 000 to increase by a factor of 10. It can

be shown that this time depends only on the interest rate and not on the actual value of the

principal. To see this, note that if a general principal, P, increases tenfold then its future value

is 10P. If the interest rate is 12%, then we need to solve

10P = P 1 +

for n. The Ps cancel (indicating that the answer is independent of P) to produce the equation

10 = (1.12)

This is identical to the equation obtained in the previous example and, as we have just seen, has

the solution n = 20.3.

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Notice that this is greater than the sum obtained at the start of this section under annual com-

pounding. (Why is this?)

This example highlights the fact that the compound interest formula

S = P 1 +

derived earlier for annual compounding can also be used for other types of compounding.

All that is needed is to reinterpret the symbols r and n. The variable r now represents the rate

of interest per time period and n represents the total number of periods.

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3.2 • Compound interest

199

Example

A principal of $10 is invested at 12% interest for 1 year. Determine the future value if the interest is

compounded

(a) annually (b) semi-annually (c) quarterly (d) monthly (e) weekly

Solution

The formula for compound interest gives

S = 10 1 +

(a) If the interest is compounded annually then r = 12, n = 1, so

S = $10(1.12)

= $11.20

(b) If the interest is compounded semi-annually then an interest of 12/2 = 6% is added on every 6 months

and, since there are two 6-month periods in a year,

S = $10(1.06)

= $11.24

since there are four 3-month periods in a year,

S = $10(1.03)

= $11.26

(d) If the interest is compounded monthly then an interest of 12/12 = 1% is added on every month and,

since there are 12 months in a year,

S = $10(1.01)

= $11.27

(e) If the interest is compounded weekly then an interest of 12/52 = 0.23% is added on every week and,

since there are 52 weeks in a year,

S = $10(1.0023)

= $11.27

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In the above example we see that the future value rises as the frequency of compounding

rises. This is to be expected because the basic feature of compound interest is that we get ‘inter-

est on the interest’. However, one important observation that you might not have expected is

that, although the future values increase, they appear to be approaching a ﬁxed value. It can be

shown that this always occurs. The type of compounding in which the interest is added on with

increasing frequency is called continuous compounding. In theory, we can ﬁnd the future

value of a principal under continuous compounding using the approach taken in the previous

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