Business Mathematics in Schools: Sinking FundsAuthor(s): Philip RobinsonSource: Mathematics in School, Vol. 19, No. 1 (Jan., 1990), pp. 34-35Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30214651 .

Accessed: 10/04/2014 16:27

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact support@jstor.org.

.

The Mathematical Association is collaborating with JSTOR to digitize, preserve and extend access toMathematics in School.

http://www.jstor.org

This content downloaded from 86.26.162.215 on Thu, 10 Apr 2014 16:27:39 PMAll use subject to JSTOR Terms and Conditions

business

mathematics

in

school

Sinkg Funds

by Philip Robinson, St Stephen's School, Rome

Following the recommendation of the Cockcroft report to make mathematics in the classroom more relevant to every- day experience, it is rather surprising that so little work is carried out in the area of finance. There are of course the usual problems involving the sale price of a hat, Ranjit's pocket money and how much change a customer should receive, but these types of problem have always existed in one form or another. Indeed, money problems were once of prime importance before the calculator age, when arithmetic was an important social skill for school leavers.

With more people than ever holding bank or building society accounts, more people using credit cards and buying houses, there is quite a lot of scope for work in this area. Interest rates are often presented only in simple interest problems and where compound interest is mentioned, its importance is understated and the questions usually disappointing.

In this article I would like to suggest a few ways in which strong GCSE candidates and perhaps students of statistics, commerce or economics, might learn some important lessons by working with interest rates in a very real way.

Interest Many students are somewhat baffled by the notion of compound interest, despite the fact that it is the universal method of earning interest. Students should understand that simple interest is the stuff of fairy tales and that an interest rate is 'the price of money'.

Once they have covered the fundamental work using the formula

Pn= Po(1 + i)"

where the letters have their usual meaning, able students could be encouraged to construct a Sinking Fund, i.e. a repayment schedule on a loan at a certain rate of interest. This would involve some work on mathematics outside the syllabus and might be further developed to include a short program in BASIC.

Summing a series In order to tackle the problem some fresh ground would have to be broken. The construction of a Sinking Fund involves the summing of terms which form a geometric progression. This may sound daunting but could be tackled in an experimental way, given the lead by a teacher.

Consider the sum of the terms 1, 1.05, (1.05)2, (1.05)3. Students will immediately say that the sum is easily

summed with a pocket calculator and they should then be asked how to deal with a large number of such terms.

The method may easily be demonstrated:

We have

S4 = 1 + (1.05) + (1.05)2 + (1.05)3 (1)

Multiplying by 1.05 gives

(1.05)S4 = 1.05 + (1.05)2 + (1.05)3 +(1.05)4 (2)

(2) - (1) gives

(1.05)4 - 1 0.05S4 = (1.05)4 - 1 and 4 =(1.05) 0.05

By repeating the above steps for further values students will arrive at the result

(1 +)" - 1 Sn= i

where i = 0.05 in this case.

A description of the problem Before setting up a repayment schedule the student must have a clear idea of the problem: a sum of money is loaned from afinancial institution at a rate of interest which we will take as fixed. The loan is repaid by a series

34 Mathematics in School, January 1990

This content downloaded from 86.26.162.215 on Thu, 10 Apr 2014 16:27:39 PMAll use subject to JSTOR Terms and Conditions

of equal payments at fixed intervals. At the end of each interval interest is added to the outstanding sum. This means that the problem is not as simple as dividing the loan by the number of payment periods as some students tend to think.

The most difficult task for students is to set the problem up mathematically. It is necessary to calculate the value of each payment in such a way that the loan is completely cleared at the end of the agreed period.

To begin with it is wise to consider a loan of a100 for a short period. Once the method has been perfected the periods may be extended.

Example: A loan of a100 is taken out at a rate of 5% p.a. for 4 years. What is the annual repayment?

Let the annual repayment be X. After one year the amount outstanding is a100 (1.05).

After the first repayment the amount outstanding is

100(1.05) - X

After two years, immediately after the second repayment, the amount outstanding is

{ 100(1.05) - X} (1.05) - X

After 3 years:

[{ 100(1.05) - X} (1.05) - X] (1.05) - X

After 4 years:

{[{100(1.05) - X}(1.05) - X](1.05) - X}(1.05) - X

Since the loan outstanding after 4 years must be zero, the last expression is equal to zero.

Simplifying we have

100(1.05)4 = X{1 + (1.05)2 + (1.05)2 + (1.05)3}

Using the result from above, we have

100(0.05)(1.05)4 (1.05)4 - 1

With some guidance students may arrive at the general result for the repayment on a sum of money aC for n years at a rate of interest i% per annum,

Ci(1 + i)" X = (1 + )"- 1

Setting up the repayment schedule With the value of X above, we may now set up the repayment schedule. The value of each payment should be a28.20 if the loan is to be cleared at the end of 4 years.

Using a pocket calculator, the following table may easily be constructed:

Year (1) (2) (1)-(2) Amount Repayment

outstanding

1 105.00 28.20 76.80 2 80.64 28.20 52.44 3 55.06 28.20 26.86 4 28.20 28.20 -

The successful construction of a table convinces students of the efficacy of the method and they achieve a lot of satisfaction. Some may wish to go on and construct a larger table with the help of a computer program.

30 INPUT "CAPITAL IS";C 40 INPUT "RATE OF INTEREST IS";R 50 INPUT "NUMBER OF YEARS"; N 60 LET I = R/100 70 LET X= C*I*(1 + I)AN 80 LEY Y=(l + 1)AN -1 90 LET Z = X/Y: LET Z = INT(Z*100)/100

100 PRINT "BALANCE", WITH INTEREST", "PAYMENT"

110 FOR K= 1 TON 120 LET A= C*(1 + I) 130 LET A= INT(100*A)/100 140 LET B=A-Z 150 LET B= INT(100*B)/100 160 PRINT C,A,Z,B 170 LET C = B 180 NEXT K

1000 END

Monthly payments Most loans are paid on a monthly basis. In this case we must count the number of periods as the number of months and given an annual rate of ifiterest we must calculate a monthly rate.

If the annual rate of interest is 12% then the monthly rate is NOT 1% but is determined using the formula for

compound interest. One unit borrowed at 12% will be worth 1.12 units after one year. The monthly rate i, is therefore given by

(1 + i)12= 1.12

i.e.

i= (1.12)1/12 - 1 = 0.0095 or 0.05%

Example: A ten year mortgage of a20 000 is to be paid at 12% by monthly payments.

There are 120 months and the monthly rate is 0.95 %. Each repayment is therefore given by

20000(0.0095) (1.0095)120 (1.0095)120-1 =a280.05

Present values The amount actually repaid over the ten years of the previous problem is a33 606 which seems a lot to many students. If however, we consider the idea of the present value of a sum of money then we may be able to put the figure into perspective.

Consider the following: an investor buys a bond which will mature in one years time and will be worth a100. How much should an investor pay for the bond if he is to make a profit of 12%? Up go the hands and a88 is the most popular answer.

But

100 1.12P = 100= - P= = 89.29.

1.12

Thus the investor should pay a89.29 if he is to make a profit of 12%.

Further, 32p now will be worth a1 after ten years since

1 S 12) a

= 0.32. (1.12)1a

The above problems may be made more challenging by varying interest rates, as indeed the chancellor does from time to time. Rates of exchange is another area to be explored with the inclusion of export sales in problems rather than the usual tourist rate problem.

Mathematics in School, January 1990 35

This content downloaded from 86.26.162.215 on Thu, 10 Apr 2014 16:27:39 PMAll use subject to JSTOR Terms and Conditions