Skip to content Skip to navigation

Connexions

You are here: Home » Content » Finance - Grade 12

Navigation

Lenses

What is a lens?

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

This content is ...

Affiliated with (What does "Affiliated with" mean?)

This content is either by members of the organizations listed or about topics related to the organizations listed. Click each link to see a list of all content affiliated with the organization.
  • FETMaths display tagshide tags

    This module is included inLens: Siyavula: Mathematics (Gr. 10-12)
    By: Siyavula

    Review Status: In Review

    Click the "FETMaths" link to see all content affiliated with them.

    Click the tag icon tag icon to display tags associated with this content.

Recently Viewed

This feature requires Javascript to be enabled.

Tags

(What is a tag?)

These tags come from the endorsement, affiliation, and other lenses that include this content.
 

Finance - Grade 12

Introduction

In earlier grades simple interest and compound interest were studied, together with the concept of depreciation. Nominal and effective interest rates were also described. Since this chapter expands on earlier work, it would be best if you revised the work done in Grades 10 and 11.

If you master the techniques in this chapter, when you start working and earning you will be able to apply the techniques in this chapter to critically assess how to invest your money. And when you are looking at applying for a bond from a bank to buy a home, you will confidently be able to get out the calculator and work out with amazement how much you could actually save by making additional repayments. Indeed, this chapter will provide you with the fundamental concepts you will need to confidently manage your finances and with some successful investing, sit back on your yacht and enjoy the millionaire lifestyle.

Finding the Length of the Investment or Loan

In Grade 11, we used the Compound Interest formula A=P(1+i)nA=P(1+i)n to determine the term of the investment or loan, by trial and error. Remember that PP is the initial amount, AA is the current amount, ii is the interest rate and nn is the number of time units (number of months or years). So if we invest an amount and know what the interest rate is, then we can work out how long it will take for the money to grow to the required amount.

Now that you have learnt about logarithms, you are ready to work out the proper algebraic solution. If you need to remind yourself how logarithms work, go to Chapter (Reference) (on page (Reference)).

The basic finance equation is:

A = P · ( 1 + i ) n A = P · ( 1 + i ) n (1)

If you don't know what AA, PP, ii and nn represent, then you should definitely revise the work from Grade 10 and 11.

Solving for nn:

A = P ( 1 + i ) n ( 1 + i ) n = ( A / P ) log ( ( 1 + i ) n ) = log ( A / P ) n log ( 1 + i ) = log ( A / P ) n = log ( A / P ) / log ( 1 + i ) A = P ( 1 + i ) n ( 1 + i ) n = ( A / P ) log ( ( 1 + i ) n ) = log ( A / P ) n log ( 1 + i ) = log ( A / P ) n = log ( A / P ) / log ( 1 + i ) (2)

Remember, you do not have to memorise this formula. It is very easy to derive any time you need it. It is simply a matter of writing down what you have, deciding what you need, and solving for that variable.

Exercise 1: Term of Investment - Logarithms

Suppose we invested R3 500 into a savings account which pays 7,5% compound interest. After an unknown period of time our account is worth R4 044,69. For how long did we invest the money? How does this compare with the trial and error answer from Chapters (Reference).

Solution

  1. Step 1. Determine what is given and what is required :
    • PP=R3 500
    • ii=7,5%
    • AA=R4 044,69

    We are required to find nn.

  2. Step 2. Determine how to approach the problem :

    We know that:

    A = P ( 1 + i ) n ( 1 + i ) n = ( A / P ) log ( ( 1 + i ) n ) = log ( A / P ) n log ( 1 + i ) = log ( A / P ) n = log ( A / P ) / log ( 1 + i ) A = P ( 1 + i ) n ( 1 + i ) n = ( A / P ) log ( ( 1 + i ) n ) = log ( A / P ) n log ( 1 + i ) = log ( A / P ) n = log ( A / P ) / log ( 1 + i ) (3)
  3. Step 3. Solve the problem :
    n = log ( A / P ) / log ( 1 + i ) = log ( 4 044 , 69 3 500 ) log ( 1 + 0 . 075 ) Remember that : 7 . 5 % = 7 . 5 100 = 0 . 075 = 2 . 0 n = log ( A / P ) / log ( 1 + i ) = log ( 4 044 , 69 3 500 ) log ( 1 + 0 . 075 ) Remember that : 7 . 5 % = 7 . 5 100 = 0 . 075 = 2 . 0 (4)
  4. Step 4. Write final answer :

    The R3 500 was invested for 2 years.

A Series of Payments

i

i

By this stage, you know how to do calculations such as `If I want R1 000 in 3 years' time, how much do I need to invest now at 10% ?'

What if we extend this as follows: `If I want to draw R1 000 next year, R1 000 the next year and R1 000 after three years ... how much do I need to initially put into a bank account earning 10% p.a. to be able to afford to be able to do this?'

The obvious way of working that out is to work out how much you need now to afford the payments individually and sum them. We'll work out how much is needed now to afford the payment of R1 000 in a year (=R1000×(1,10)-1=R909,09)(=R1000×(1,10)-1=R909,09), the amount needed now for the following year's R1 000 (=R1000×(1,10)-2=R826,45)(=R1000×(1,10)-2=R826,45) and the amount needed now for the R1 000 after 3 years (=R1000×(1,10)-3=R751,31)(=R1000×(1,10)-3=R751,31). Add these together gives you the amount needed to afford all three payments and you get R2 486,85.

So, if you put R2 486,85 into a 10% bank account now, you will be able to draw out R1 000 in a year, R1 000 a year after that, and R1 000 a year after that - and your bank account will come down to R0. You would have had exactly the right amount of money to do that (obviously!).

You can check this as follows:

Table 1
Amount at Time 0 (i.e. Now)   = R2 486,85
Amount at Time 1 (i.e. a year later) = 2 486,85(1+10%) = R2 735,54
Amount after withdrawing R1 000 = 2 735,54 - 1 000 = R1 735,54
Amount at Time 2 (i.e. a year later) = 1 735,54(1+10%) = R1 909,09
Amount after withdrawing R1 000 = R1 909,09 - 1 000 = R909,09
Amount at Time 3 (i.e. a year later) = 909,09(1+10%) = R1 000
Amount after withdrawing R1 000 = 1 000 - 1 000 = R0

Perfect! Of course, for only three years, that was not too bad. But what if I asked you how much you needed to put into a bank account now, to be able to afford R100 a month for the next 15 years. If you used the above approach you would still get the right answer, but it would take you weeks!

There is - I'm sure you guessed - an easier way! This section will focus on describing how to work with:

  • annuities - a fixed sum payable each year or each month, either to provide a pre-determined sum at the end of a number of years or months (referred to as a future value annuity) or a fixed amount paid each year or each month to repay (amortise) a loan (referred to as a present value annuity).
  • bond repayments - a fixed sum payable at regular intervals to pay off a loan. This is an example of a present value annuity.
  • sinking funds - an accounting term for cash set aside for a particular purpose and invested so that the correct amount of money will be available when it is needed. This is an example of a future value annuity.

Sequences and Series

Before we progress, you need to go back and read Chapter (Reference) (from page (Reference)) to revise sequences and series.

In summary, if you have a series of nn terms in total which looks like this:

a + a r + a r 2 + . . . + a r n - 1 = a [ 1 + r + r 2 + . . . r n - 1 ] a + a r + a r 2 + . . . + a r n - 1 = a [ 1 + r + r 2 + . . . r n - 1 ] (5)

this can be simplified as:

a ( r n - 1 ) r - 1 useful when r > 1 a ( 1 - r n ) 1 - r useful when 0 ≤ r < 1 a ( r n - 1 ) r - 1 useful when r > 1 a ( 1 - r n ) 1 - r useful when 0 ≤ r < 1 (6)

Present Values of a series of Payments

So having reviewed the mathematics of Sequences and Series, you might be wondering how this is meant to have any practical purpose! Given that we are in the finance section, you would be right to guess that there must be some financial use to all this. Here is an example which happens in many people's lives - so you know you are learning something practical.

Let us say you would like to buy a property for R300 000, so you go to the bank to apply for a mortgage bond. The bank wants it to be repaid by annually payments for the next 20 years, starting at end of this year. They will charge you 15% interest per annum. At the end of the 20 years the bank would have received back the total amount you borrowed together with all the interest they have earned from lending you the money. You would obviously want to work out what the annual repayment is going to be!

Let XX be the annual repayment, ii is the interest rate, and MM is the amount of the mortgage bond you will be taking out.

Time lines are particularly useful tools for visualizing the series of payments for calculations, and we can represent these payments on a time line as:

Figure 1: Time Line for an annuity (in arrears) of XX for nn periods.
Figure 1 (MG12C4_001.png)

The present value of all the payments (which includes interest) must equate to the (present) value of the mortgage loan amount.

Mathematically, you can write this as:

M = X ( 1 + i ) - 1 + X ( 1 + i ) - 2 + X ( 1 + i ) - 3 + . . . + X ( 1 + i ) - 20 M = X ( 1 + i ) - 1 + X ( 1 + i ) - 2 + X ( 1 + i ) - 3 + . . . + X ( 1 + i ) - 20 (7)

The painful way of solving this problem would be to do the calculation for each of the terms above - which is 20 different calculations. Not only would you probably get bored along the way, but you are also likely to make a mistake.

Naturally, there is a simpler way of doing this! You can rewrite the above equation as follows:

M = X [ v 1 + v 2 + v 3 + . . . + v 20 ] where v i = ( 1 + i ) - 1 = 1 / ( 1 + i ) M = X [ v 1 + v 2 + v 3 + . . . + v 20 ] where v i = ( 1 + i ) - 1 = 1 / ( 1 + i ) (8)

Of course, you do not have to use the method of substitution to solve this. We just find this a useful method because you can get rid of the negative exponents - which can be quite confusing! As an exercise - to show you are a real financial whizz - try to solve this without substitution. It is actually quite easy.

Now, the item in square brackets is the sum of a geometric sequence, as discussion in (Reference). This can be re-written as follows, using what we know from Chapter (Reference) of this text book:

v 1 + v 2 + v 3 + . . . + v n = v ( 1 + v + v 2 + . . . + v n - 1 ) = v ( 1 - v n 1 - v ) = 1 - v n 1 / v - 1 = 1 - ( 1 + i ) - n i v 1 + v 2 + v 3 + . . . + v n = v ( 1 + v + v 2 + . . . + v n - 1 ) = v ( 1 - v n 1 - v ) = 1 - v n 1 / v - 1 = 1 - ( 1 + i ) - n i (9)

Note that we took out a common factor of vv before using the formula for the geometric sequence.

So we can write:

M = X [ ( 1 - ( 1 + i ) - n ) i ] M = X [ ( 1 - ( 1 + i ) - n ) i ] (10)

This can be re-written:

X = M [ ( 1 - ( 1 + i ) - n ) i ] = M i 1 - ( 1 + i ) - n X = M [ ( 1 - ( 1 + i ) - n ) i ] = M i 1 - ( 1 + i ) - n (11)

So, this formula is useful if you know the amount of the mortgage bond you need and want to work out the repayment, or if you know how big a repayment you can afford and want to see what property you can buy.

For example, if I want to buy a house for R300 000 over 20 years, and the bank is going to charge me 15% per annum on the outstanding balance, then the annual repayment is:

X = M i 1 - ( 1 + i ) - n = R 300 000 × 0 , 15 1 - ( 1 + 0 , 15 ) - 20 = R 4 792 844 X = M i 1 - ( 1 + i ) - n = R 300 000 × 0 , 15 1 - ( 1 + 0 , 15 ) - 20 = R 4 792 844 (12)

This means, each year for the next 20 years, I need to pay the bank R47 928,44 per year before I have paid off the mortgage bond.

On the other hand, if I know I will only have R30 000 a year to repay my bond, then how big a house can I buy? That is easy ....

M = X [ ( 1 - ( 1 + i ) - n ) i ] = R 30 000 [ ( 1 - ( 1 , 15 ) - 20 ) 0 , 15 ] = R 187 779 , 90 M = X [ ( 1 - ( 1 + i ) - n ) i ] = R 30 000 [ ( 1 - ( 1 , 15 ) - 20 ) 0 , 15 ] = R 187 779 , 90 (13)

So, for R30 000 a year for 20 years, I can afford to buy a house of R187 800 (rounded to the nearest hundred).

The bad news is that R187 800 does not come close to the R300 000 you wanted to pay! The good news is that you do not have to memorise this formula. In fact , when you answer questions like this in an exam, you will be expected to start from the beginning - writing out the opening equation in full, showing that it is the sum of a geometric sequence, deriving the answer, and then coming up with the correct numerical answer.

Exercise 2: Monthly mortgage repayments

Sam is looking to buy his first flat, and has R15 000 in cash savings which he will use as a deposit. He has viewed a flat which is on the market for R250 000, and he would like to work out how much the monthly repayments would be. He will be taking out a 30 year mortgage with monthly repayments. The annual interest rate is 11%.

Solution
  1. Step 1. Determine what is given and what is needed :

    The following is given:

    • Deposit amount = R15 000
    • Price of flat = R250 000
    • interest rate, i=11%i=11%

    We are required to find the monthly repayment for a 30-year mortgage.

  2. Step 2. Determine how to approach the problem :

    We know that:

    X = M [ ( 1 - ( 1 + i ) - n ) i ] X = M [ ( 1 - ( 1 + i ) - n ) i ] (14)

    In order to use this equation, we need to calculate MM, the amount of the mortgage bond, which is the purchase price of property less the deposit which Sam pays up-front.

    M = R 250 000 - R 15 000 = R 235 000 M = R 250 000 - R 15 000 = R 235 000 (15)

    Now because we are considering monthly repayments, but we have been given an annual interest rate, we need to convert this to a monthly interest rate, i12i12. (If you are not clear on this, go back and revise (Reference).)

    ( 1 + i 12 ) 12 = ( 1 + i ) ( 1 + i 12 ) 12 = 1 , 11 i 12 = 0 , 873459 % ( 1 + i 12 ) 12 = ( 1 + i ) ( 1 + i 12 ) 12 = 1 , 11 i 12 = 0 , 873459 % (16)

    We know that the mortgage bond is for 30 years, which equates to 360 months.

  3. Step 3. Solve the problem :

    Now it is easy, we can just plug the numbers in the formula, but do not forget that you can always deduce the formula from first principles as well!

    X = M [ ( 1 - ( 1 + i ) - n ) i ] = R 235 000 [ ( 1 - ( 1 . 00876459 ) - 360 ) 0 , 008734594 ] = R 2 146 , 39 X = M [ ( 1 - ( 1 + i ) - n ) i ] = R 235 000 [ ( 1 - ( 1 . 00876459 ) - 360 ) 0 , 008734594 ] = R 2 146 , 39 (17)
  4. Step 4. Write the final answer :

    That means that to buy a flat for R250 000, after Sam pays a R15 000 deposit, he will make repayments to the bank each month for the next 30 years equal to R2 146,39.

Exercise 3: Monthly mortgage repayments

You are considering purchasing a flat for R200 000 and the bank's mortgage rate is currently 9% per annum payable monthly. You have savings of R10 000 which you intend to use for a deposit. How much would your monthly mortgage payment be if you were considering a mortgage over 20 years.

Solution
  1. Step 1. Determine what is given and what is required :

    The following is given:

    • Deposit amount = R10 000
    • Price of flat = R200 000
    • interest rate, i=9%i=9%

    We are required to find the monthly repayment for a 20-year mortgage.

  2. Step 2. Determine how to approach the problem :

    We are considering monthly mortgage repayments, so it makes sense to use months as our time period.

    The interest rate was quoted as 9% per annum payable monthly, which means that the monthly effective rate =9%12=0,75%=9%12=0,75% per month. Once we have converted 20 years into 240 months, we are ready to do the calculations!

    First we need to calculate MM, the amount of the mortgage bond, which is the purchase price of property minus the deposit which Sam pays up-front.

    M = R 200 000 - R 10 000 = R 190 000 M = R 200 000 - R 10 000 = R 190 000 (18)

    The present value of our mortgage payments X (which includes interest), must equate to the present mortgage amount

    M = X × ( 1 + 0 , 75 % ) - 1 + X × ( 1 + 0 , 75 % ) - 2 + X × ( 1 + 0 , 75 % ) - 3 + X × ( 1 + 0 , 75 % ) - 4 + ... X × ( 1 + 0 , 75 % ) - 239 + X × ( 1 + 0 , 75 % ) - 240 M = X × ( 1 + 0 , 75 % ) - 1 + X × ( 1 + 0 , 75 % ) - 2 + X × ( 1 + 0 , 75 % ) - 3 + X × ( 1 + 0 , 75 % ) - 4 + ... X × ( 1 + 0 , 75 % ) - 239 + X × ( 1 + 0 , 75 % ) - 240 (19)

    But it is clearly much easier to use our formula than work out 240 factors and add them all up!

  3. Step 3. Solve the problem :
    X × 1 - ( 1 + 0 , 75 % ) - 240 0 , 75 % = R 190 000 X × 111 , 14495 = R 190 000 X = R 1 709 , 48 X × 1 - ( 1 + 0 , 75 % ) - 240 0 , 75 % = R 190 000 X × 111 , 14495 = R 190 000 X = R 1 709 , 48 (20)
  4. Step 4. Write the final answer :

    So to repay a R190 000 mortgage over 20 years, at 9% interest payable monthly, will cost you R1 709,48 per month for 240 months.

Show me the money…

Now that you've done the calculations for the worked example and know what the monthly repayments are, you can work out some surprising figures. For example, R1 709,48 per month for 240 month makes for a total of R410 275,20 (=R1 709,48 ×× 240). That is more than double the amount that you borrowed! This seems like a lot. However, now that you've studied the effects of time (and interest) on money, you should know that this amount is somewhat meaningless. The value of money is dependant on its timing.

Nonetheless, you might not be particularly happy to sit back for 20 years making your R1 709,48 mortgage payment every month knowing that half the money you are paying are going toward interest. But there is a way to avoid those heavy interest charges. It can be done for less than R300 extra every month...

So our payment is now R2 000. The interest rate is still 9% per annum payable monthly (0,75% per month), and our principal amount borrowed is R190 000. Making this higher repayment amount every month, how long will it take to pay off the mortgage?

The present value of the stream of payments must be equal to R190 000 (the present value of the borrowed amount). So we need to solve for nn in:

R 2 000 × [ 1 - ( 1 + 0 , 75 % ) - n ] / 0 , 75 % = R 190 000 1 - ( 1 + 0 , 75 % ) - n = ( 190 000 × 0 , 75 % 2 000 ) log ( 1 + 0 , 75 % ) - n = log [ ( 1 - 190 000 × 0 , 0075 2 000 ] - n × log ( 1 + 0 , 75 % ) = log [ ( 1 - 190 000 × 0 , 0075 2 000 ] - n × 0 , 007472 = - 1 , 2465 n = 166 , 8 months = 13 , 9 years R 2 000 × [ 1 - ( 1 + 0 , 75 % ) - n ] / 0 , 75 % = R 190 000 1 - ( 1 + 0 , 75 % ) - n = ( 190 000 × 0 , 75 % 2 000 ) log ( 1 + 0 , 75 % ) - n = log [ ( 1 - 190 000 × 0 , 0075 2 000 ] - n × log ( 1 + 0 , 75 % ) = log [ ( 1 - 190 000 × 0 , 0075 2 000 ] - n × 0 , 007472 = - 1 , 2465 n = 166 , 8 months = 13 , 9 years (21)

So the mortgage will be completely repaid in less than 14 years, and you would have made a total payment of 166,8×× R2 000 = R333 600.

Can you see what is happened? Making regular payments of R2 000 instead of the required R1,709,48, you will have saved R76 675,20 (= R410 275,20 - R333 600) in interest, and yet you have only paid an additional amount of R290,52 for 166,8 months, or R48 458,74. You surely know by now that the difference between the additional R48 458,74 that you have paid and the R76 675,20 interest that you have saved is attributable to, yes, you have got it, compound interest!

Future Value of a series of Payments

In the same way that when we have a single payment, we can calculate a present value or a future value - we can also do that when we have a series of payments.

In the above section, we had a few payments, and we wanted to know what they are worth now - so we calculated present values. But the other possible situation is that we want to look at the future value of a series of payments.

Maybe you want to save up for a car, which will cost R45 000 - and you would like to buy it in 2 years time. You have a savings account which pays interest of 12% per annum. You need to work out how much to put into your bank account now, and then again each month for 2 years, until you are ready to buy the car.

Can you see the difference between this example and the ones at the start of the chapter where we were only making a single payment into the bank account - whereas now we are making a series of payments into the same account? This is a sinking fund.

So, using our usual notation, let us write out the answer. Make sure you agree how we come up with this. Because we are making monthly payments, everything needs to be in months. So let AA be the closing balance you need to buy a car, PP is how much you need to pay into the bank account each month, and i12i12 is the monthly interest rate. (Careful - because 12% is the annual interest rate, so we will need to work out later what the monthly interest rate is!)

A = P ( 1 + i 12 ) 24 + P ( 1 + i 12 ) 23 + . . . + P ( 1 + i 12 ) 1 A = P ( 1 + i 12 ) 24 + P ( 1 + i 12 ) 23 + . . . + P ( 1 + i 12 ) 1 (22)

Here are some important points to remember when deriving this formula:

  1. We are calculating future values, so in this example we use (1+i12)n(1+i12)n and not (1+i12)-n(1+i12)-n. Check back to the start of the chapter if this is not obvious to you by now.
  2. If you draw a timeline you will see that the time between the first payment and when you buy the car is 24 months, which is why we use 24 in the first exponent.
  3. Again, looking at the timeline, you can see that the 24th payment is being made one month before you buy the car - which is why the last exponent is a 1.
  4. Always check that you have got the right number of payments in the equation. Check right now that you agree that there are 24 terms in the formula above.

So, now that we have the right starting point, let us simplify this equation:

A = P [ ( 1 + i 12 ) 24 + ( 1 + i 12 ) 23 + ... + ( 1 + i 12 ) 1 ] = P [ X 24 + X 23 + ... + X 1 ] using X = ( 1 + i 12 ) A = P [ ( 1 + i 12 ) 24 + ( 1 + i 12 ) 23 + ... + ( 1 + i 12 ) 1 ] = P [ X 24 + X 23 + ... + X 1 ] using X = ( 1 + i 12 ) (23)

Note that this time X has a positive exponent not a negative exponent, because we are doing future values. This is not a rule you have to memorise - you can see from the equation what the obvious choice of X should be.

Let us re-order the terms:

A = P [ X 1 + X 2 + ... + X 24 ] = P · X [ 1 + X + X 2 + ... + X 23 ] A = P [ X 1 + X 2 + ... + X 24 ] = P · X [ 1 + X + X 2 + ... + X 23 ] (24)

This is just another sum of a geometric sequence, which as you know can be simplified as:

A = P · X [ X n - 1 ] / ( ( 1 + i 12 ) - 1 ) = P · X [ X n - 1 ] / i 12 A = P · X [ X n - 1 ] / ( ( 1 + i 12 ) - 1 ) = P · X [ X n - 1 ] / i 12 (25)

So if we want to use our numbers, we know that AA = R45 000, nn=24 (because we are looking at monthly payments, so there are 24 months involved) and i=12%i=12% per annum.

BUT (and it is a big but) we need a monthly interest rate. Do not forget that the trick is to keep the time periods and the interest rates in the same units - so if we have monthly payments, make sure you use a monthly interest rate! Using the formula from Grade 11, we know that (1+i)=(1+i12)12(1+i)=(1+i12)12. So we can show that i12=0,0094888=0,94888%i12=0,0094888=0,94888%.

Therefore,

45 000 = P ( 1 , 0094888 ) [ ( 1 , 0094888 ) 24 - 1 ] / 0 , 0094888 P = 1 662 , 67 45 000 = P ( 1 , 0094888 ) [ ( 1 , 0094888 ) 24 - 1 ] / 0 , 0094888 P = 1 662 , 67 (26)

This means you need to invest R166 267 each month into that bank account to be able to pay for your car in 2 years time.

Exercises - Present and Future Values

  1. You have taken out a mortgage bond for R875 000 to buy a flat. The bond is for 30 years and the interest rate is 12% per annum payable monthly.
    1. What is the monthly repayment on the bond?
    2. How much interest will be paid in total over the 30 years?
  2. How much money must be invested now to obtain regular annuity payments of R 5 500 per month for five years ? The money is invested at 11,1% p.a., compounded monthly. (Answer to the nearest hundred rand)

Investments and Loans

By now, you should be well equipped to perform calculations with compound interest. This section aims to allow you to use these valuable skills to critically analyse investment and loan options that you will come across in your later life. This way, you will be able to make informed decisions on options presented to you.

At this stage, you should understand the mathematical theory behind compound interest. However, the numerical implications of compound interest are often subtle and far from obvious.

Recall the example `Show me the money' in "Present Values of a series of Payments". For an extra payment of R29 052 a month, we could have paid off our loan in less than 14 years instead of 20 years. This provides a good illustration of the long term effect of compound interest that is often surprising. In the following section, we'll aim to explain the reason for the drastic reduction in the time it takes to repay the loan.

Loan Schedules

So far, we have been working out loan repayment amounts by taking all the payments and discounting them back to the present time. We are not considering the repayments individually. Think about the time you make a repayment to the bank. There are numerous questions that could be raised: how much do you still owe them? Since you are paying off the loan, surely you must owe them less money, but how much less? We know that we'll be paying interest on the money we still owe the bank. When exactly do we pay interest? How much interest are we paying?

The answer to these questions lie in something called the load schedule.

We will continue to use the earlier example. There is a loan amount of R190 000. We are paying it off over 20 years at an interest of 9% per annum payable monthly. We worked out that the repayments should be R1 709,48.

Consider the first payment of R1 709,48 one month into the loan. First, we can work out how much interest we owe the bank at this moment. We borrowed R190 000 a month ago, so we should owe:

I = M × i 12 = R 190 000 × 0 , 75 % = R 1 425 I = M × i 12 = R 190 000 × 0 , 75 % = R 1 425 (27)

We are paying them R1 425 in interest. We call this the interest component of the repayment. We are only paying off R1 709,48 - R1 425 = R284.48 of what we owe! This is called the capital component. That means we still owe R190 000 - R284,48 = R189 715,52. This is called the capital outstanding. Let's see what happens at the end of the second month. The amount of interest we need to pay is the interest on the capital outstanding.

I = M × i 12 = R 189 715 , 52 × 0 , 75 % = R 1 422 , 87 I = M × i 12 = R 189 715 , 52 × 0 , 75 % = R 1 422 , 87 (28)

Since we don't owe the bank as much as we did last time, we also owe a little less interest. The capital component of the repayment is now R1 709,48 - R1 422,87 = R286,61. The capital outstanding will be R189 715,52 - R286,61 = R189 428,91. This way, we can break each of our repayments down into an interest part and the part that goes towards paying off the loan.

This is a simple and repetitive process. Table 2 is a table showing the breakdown of the first 12 payments. This is called a loan schedule.

Table 2: A loan schedule with repayments of R1 709,48 per month.
Time Repayment Interest Component Capital Component Capital Outstanding
0       R 190 000 00
1 R 1 709 48 R 1 425 00 R 284 48 R 189 715 52
2 R 1 709 48 R 1 422 87 R 286 61 R 189 428 91
3 R 1 709 48 R 1 420 72 R 288 76 R 189 140 14
4 R 1 709 48 R 1 418 55 R 290 93 R 188 849 21
5 R 1 709 48 R 1 416 37 R 293 11 R 188 556 10
6 R 1 709 48 R 1 414 17 R 295 31 R 188 260 79
7 R 1 709 48 R 1 411 96 R 297 52 R 187 963 27
8 R 1 709 48 R 1 409 72 R 299 76 R 187 663 51
9 R 1 709 48 R 1 407 48 R 302 00 R 187 361 51
10 R 1 709 48 R 1 405 21 R 304 27 R 187 057 24
11 R 1 709 48 R 1 402 93 R 306 55 R 186 750 69
12 R 1 709 48 R 1 400 63 R 308 85 R 186 441 84

Now, let's see the same thing again, but with R2 000 being repaid each year. We expect the numbers to change. However, how much will they change by? As before, we owe R1 425 in interest in interest. After one month. However, we are paying R2 000 this time. That leaves R575 that goes towards paying off the capital outstanding, reducing it to R189 425. By the end of the second month, the interest owed is R1 420,69 (That's R189 425×i12×i12). Our R2 000 pays for that interest, and reduces the capital amount owed by R2 000 - R1 420,69 = R579,31. This reduces the amount outstanding to R188 845,69.

Doing the same calculations as before yields a new loan schedule shown in Table 3.

Table 3: A loan schedule with repayments of R2 000 per month.
Time Repayment Interest Component Capital Component Capital Outstanding
0       R 190 000 00
1 R 2 000 00 R 1 425 00 R 575 00 R 189 425 00
2 R 2 000 00 R 1 420 69 R 579 31 R 188 845 69
3 R 2 000 00 R 1 416 34 R 583 66 R 188 262 03
4 R 2 000 00 R 1 411 97 R 588 03 R 187 674 00
5 R 2 000 00 R 1 407 55 R 592 45 R 187 081 55
6 R 2 000 00 R 1 403 11 R 596 89 R 186 484 66
7 R 2 000 00 R 1 398 63 R 601 37 R 185 883 30
8 R 2 000 00 R 1 394 12 R 605 88 R 185 277 42
9 R 2 000 00 R 1 389 58 R 610 42 R 184 667 00
10 R 2 000 00 R 1 385 00 R 615 00 R 184 052 00
11 R 2 000 00 R 1 380 39 R 619 61 R 183 432 39
12 R 2 000 00 R 1 375 74 R 624 26 R 182 808 14

The important numbers to notice is the “Capital Component" column. Note that when we are paying off R2 000 a month as compared to R1 709,48 a month, this column more than double. In the beginning of paying off a loan, very little of our money is used to pay off the capital outstanding. Therefore, even a small increase in repayment amounts can significantly increase the speed at which we are paying off the capital.

What's more, look at the amount we are still owing after one year (i.e. at time 12). When we were paying R1 709,48 a month, we still owe R186 441,84. However, if we increase the repayments to R2 000 a month, the amount outstanding decreases by over R3 000 to R182 808,14. This means we would have paid off over R7 000 in our first year instead of less than R4 000. This increased speed at which we are paying off the capital portion of the loan is what allows us to pay off the whole loan in around 14 years instead of the original 20. Note however, the effect of paying R2 000 instead of R1 709,48 is more significant in the beginning of the loan than near the end of the loan.

It is noted that in this instance, by paying slightly more than what the bank would ask you to pay, you can pay off a loan a lot quicker. The natural question to ask here is: why are banks asking us to pay the lower amount for much longer then? Are they trying to cheat us out of our money?

There is no simple answer to this. Banks provide a service to us in return for a fee, so they are out to make a profit. However, they need to be careful not to cheat their customers for fear that they'll simply use another bank. The central issue here is one of scale. For us, the changes involved appear big. We are paying off our loan 6 years earlier by paying just a bit more a month. To a bank, however, it doesn't matter much either way. In all likelihood, it doesn't affect their profit margins one bit!

Remember that the bank calculates repayment amounts using the same methods as we've been learning. They decide on the correct repayment amounts for a given interest rate and set of terms. Smaller repayment amounts will make the bank more money, because it will take you longer to pay off the loan and more interest will acumulate. Larger repayment amounts mean that you will pay off the loan faster, so you will accumulate less interest i.e. the bank will make less money off of you. It's a simple matter of less money now or more money later. Banks generally use a 20 year repayment period by default.

Learning about financial mathematics enables you to duplicate these calculations for yourself. This way, you can decide what's best for you. You can decide how much you want to repay each month and you'll know of its effects. A bank wouldn't care much either way, so you should pick something that suits you.

Exercise 4: Monthly Payments

Stefan and Marna want to buy a house that costs R 1 200 000. Their parents offer to put down a 20% payment towards the cost of the house. They need to get a moratage for the balance. What are their monthly repayments if the term of the home loan is 30 years and the interest is 7,5%, compounded monthly?

Solution
  1. Step 1. Determine how much money they need to borrow :

    R 1 200 00 - R 240 000 = R 960 000 R 1 200 00 - R 240 000 = R 960 000

  2. Step 2. Determine how to approach the problem :

    Use the formula:

    P = x [ 1 - ( 1 + i ) - n ] i P = x [ 1 - ( 1 + i ) - n ] i (29)

    Where

    P = 960 000 P = 960 000

    n = 30 × 12 = 360 months n = 30 × 12 = 360 months

    i = 0 , 075 ÷ 12 = 0 , 00625 i = 0 , 075 ÷ 12 = 0 , 00625

  3. Step 3. Solve the problem :
    R 960 000 = x [ 1 - ( 1 + 0 , 00625 ) - 360 ] 0 , 00625 = x ( 143 , 0176273 ) x = R 6 712 , 46 R 960 000 = x [ 1 - ( 1 + 0 , 00625 ) - 360 ] 0 , 00625 = x ( 143 , 0176273 ) x = R 6 712 , 46 (30)
  4. Step 4. Write the final answer :

    The monthly repayments =R6712,46=R6712,46

Exercises - Investments and Loans

  1. A property costs R1 800 000. Calculate the monthly repayments if the interest rate is 14% p.a. compounded monthly and the loan must be paid off in 20 years time.
  2. A loan of R 4 200 is to be returned in two equal annual instalments. If the rate of interest of 10% per annum, compounded annually, calculate the amount of each instalment.

Calculating Capital Outstanding

As defined in "Loan Schedules", Capital outstanding is the amount we still owe the people we borrowed money from at a given moment in time. We also saw how we can calculate this using loan schedules. However, there is a significant disadvantage to this method: it is very time consuming. For example, in order to calculate how much capital is still outstanding at time 12 using the loan schedule, we'll have to first calculate how much capital is outstanding at time 1 through to 11 as well. This is already quite a bit more work than we'd like to do. Can you imagine calculating the amount outstanding after 10 years (time 120)?

Fortunately, there is an easier method. However, it is not immediately clear why this works, so let's take some time to examine the concept.

Prospective method for Capital Outstanding

Let's say that after a certain number of years, just after we made a repayment, we still owe amount YY. What do we know about YY? We know that using the loan schedule, we can calculate what it equals to, but that is a lot of repetitive work. We also know that YY is the amount that we are still going to pay off. In other words, all the repayments we are still going to make in the future will exactly pay off YY. This is true because in the end, after all the repayments, we won't be owing anything.

Therefore, the present value of all outstanding future payments equal the present amount outstanding. This is the prospective method for calculating capital outstanding.

Let's return to a previous example. Recall the case where we were trying to repay a loan of R200 000 over 20 years. A R10 000 deposit was put down, so the amount being payed off was R190 000. At an interest rate of 9% compounded monthly, the monthly repayment was R1 709,48. In Table 2, we can see that after 12 months, the amount outstanding was R186 441,84. Let's try to work this out using the the prospective method.

After time 12, there are still 19×12=22819×12=228 repayments left of R1 709,48 each. The present value is:

n = 228 i = 0 , 75 % Y = R 1 709 , 48 × 1 - 1 , 0075 - 228 0 , 0075 = R 186 441 , 92 n = 228 i = 0 , 75 % Y = R 1 709 , 48 × 1 - 1 , 0075 - 228 0 , 0075 = R 186 441 , 92 (31)

Oops! This seems to be almost right, but not quite. We should have got R186 441,84. We are 8 cents out. However, this is in fact not a mistake. Remember that when we worked out the monthly repayments, we rounded to the nearest cents and arrived at R1 709,48. This was because one cannot make a payment for a fraction of a cent. Therefore, the rounding off error was carried through. That's why the two figures don't match exactly. In financial mathematics, this is largely unavoidable.

Formulae Sheet

As an easy reference, here are the key formulae that we derived and used during this chapter. While memorising them is nice (there are not many), it is the application that is useful. Financial experts are not paid a salary in order to recite formulae, they are paid a salary to use the right methods to solve financial problems.

Definitions

Table 4
P P Principal (the amount of money at the starting point of the calculation)
i i interest rate, normally the effective rate per annum
n n period for which the investment is made
i T i T the interest rate paid TT times per annum, i.e. iT= Nominal Interest Rate TiT= Nominal Interest Rate T

Equations

Present Value - simple Future Value - simple Solve for i Solve for n = P ( 1 + i · n ) Present Value - simple Future Value - simple Solve for i Solve for n = P ( 1 + i · n ) (32)
Present Value - compound Future Value - compound Solve for i Solve for n = P ( 1 + i ) n Present Value - compound Future Value - compound Solve for i Solve for n = P ( 1 + i ) n (33)

Tip:

Always keep the interest and the time period in the same units of time (e.g. both in years, or both in months etc.).

End of Chapter Exercises

  1. Thabo is about to invest his R8 500 bonus in a special banking product which will pay 1% per annum for 1 month, then 2% per annum for the next 2 months, then 3% per annum for the next 3 months, 4% per annum for the next 4 months, and 0% for the rest of the year. The are going to charge him R100 to set up the account. How much can he expect to get back at the end of the period?
  2. A special bank account pays simple interest of 8% per annum. Calculate the opening balance required to generate a closing balance of R5 000 after 2 years.
  3. A different bank account pays compound interest of 8% per annum. Calculate the opening balance required to generate a closing balance of R5 000 after 2 years.
  4. Which of the two answers above is lower, and why?
  5. 7 Months after an initial deposit, the value of a bank account which pays compound interest of 7,5% per annum is R3 650,81. What was the value of the initial deposit?
  6. Thabani and Lungelo are both using UKZN Bank for their saving. Suppose Lungelo makes a deposit of XX today at interest rate of ii for six years. Thabani makes a deposit of 3X3X at an interest rate of 0.050.05. Thabani made his deposit 3 years after Lungelo made his first deposit. If after 6 years, their investments are equal, calculate the value of ii and find XX. If the sum of their investment is R20 000, use XX you got to find out how much Thabani got in 6 years.
  7. Sipho invests R500 at an interest rate of log(1,12)log(1,12) for 5 years. Themba, Sipho's sister invested R200 at interest rate ii for 10 years on the same date that her brother made his first deposit. If after 5 years, Themba's accumulation equals Sipho's, find the interest rate ii and find out whether Themba will be able to buy her favorite cell phone after 10 years which costs R2 000.
  8. Calculate the real cost of a loan of R10 000 for 5 years at 5% capitalised monthly. Repeat this for the case where it is capitalised half yearly i.e. Every 6 months.
  9. Determine how long, in years, it will take for the value of a motor vehicle to decrease to 25% of its original value if the rate of depreciation, based on the reducing-balance method, is 21% per annum.

Content actions

Download module as:

PDF | EPUB (?)

What is an EPUB file?

EPUB is an electronic book format that can be read on a variety of mobile devices.

Downloading to a reading device

For detailed instructions on how to download this content's EPUB to your specific device, click the "(?)" link.

| More downloads ...

Add module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks