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Finance - Grade 11

Introduction

In Grade 10, the ideas of simple and compound interest were introduced. In this chapter we will be extending those ideas, so it is a good idea to go back to the Finance chapter and revise what you learnt in Grade 10. If you master the techniques in this chapter, you will understand about depreciation and will learn how to determine which bank is offering the better interest rate.

Depreciation

It is said that when you drive a new car out of the dealership, it loses 20% of its value, because it is now “second-hand”. And from there on the value keeps falling, or depreciating. Second hand cars are cheaper than new cars, and the older the car, usually the cheaper it is. If you buy a second hand (or should we say pre-owned!) car from a dealership, they will base the price on something called book value.

The book value of the car is the value of the car taking into account the loss in value due to wear, age and use. We call this loss in value depreciation, and in this section we will look at two ways of how this is calculated. Just like interest rates, the two methods of calculating depreciation are simple and compound methods.

The terminology used for simple depreciation is straight-line depreciation and for compound depreciation is reducing-balance depreciation. In the straight-line method the value of the asset is reduced by the same constant amount each year. In the compound depreciation method the value of the asset is reduced by the same percentage each year. This means that the value of an asset does not decrease by a constant amount each year, but the decrease is most in the first year, then by a smaller amount in the second year and by an even smaller amount in the third year, and so on.

Depreciation

You may be wondering why we need to calculate depreciation. Determining the value of assets (as in the example of the second hand cars) is one reason, but there is also a more financial reason for calculating depreciation - tax! Companies can take depreciation into account as an expense, and thereby reduce their taxable income. A lower taxable income means that the company will pay less income tax to the Revenue Service.

Simple Depreciation (it really is simple!)

Let us go back to the second hand cars. One way of calculating a depreciation amount would be to assume that the car has a limited useful life. Simple depreciation assumes that the value of the car decreases by an equal amount each year. For example, let us say the limited useful life of a car is 5 years, and the cost of the car today is R60 000. What we are saying is that after 5 years you will have to buy a new car, which means that the old one will be valueless at that point in time. Therefore, the amount of depreciation is calculated:

R 60 000 5 years = R 12 000 per year . R 60 000 5 years = R 12 000 per year .
(1)

The value of the car is then:

Table 1
End of Year 1 R60 000 - 1××(R12 000) = R48 000
End of Year 2 R60 000 - 2××(R12 000) = R36 000
End of Year 3 R60 000 - 3××(R12 000) = R24 000
End of Year 4 R60 000 - 4××(R12 000) = R12 000
End of Year 5 R60 000 - 5××(R12 000) = R0

This looks similar to the formula for simple interest:

Total Interest after n years = n × ( P × i ) Total Interest after n years = n × ( P × i )
(2)

where ii is the annual percentage interest rate and PP is the principal amount.

If we replace the word interest with the word depreciation and the word principal with the words initial value we can use the same formula:

Total depreciation after n years = n × ( P × i ) Total depreciation after n years = n × ( P × i )
(3)

Then the book value of the asset after nn years is:

Initial Value - Total depreciation after n years = P - n × ( P × i ) = P ( 1 - n × i ) Initial Value - Total depreciation after n years = P - n × ( P × i ) = P ( 1 - n × i )
(4)

For example, the book value of the car after two years can be simply calculated as follows:

Book Value after 2 years = P ( 1 - n × i ) = R 60 000 ( 1 - 2 × 20 % ) = R 60 000 ( 1 - 0 , 4 ) = R 60 000 ( 0 , 6 ) = R 36 000 Book Value after 2 years = P ( 1 - n × i ) = R 60 000 ( 1 - 2 × 20 % ) = R 60 000 ( 1 - 0 , 4 ) = R 60 000 ( 0 , 6 ) = R 36 000
(5)

as expected.

Note that the difference between the simple interest calculations and the simple depreciation calculations is that while the interest adds value to the principal amount, the depreciation amount reduces value!

Exercise 1: Simple Depreciation method

A car is worth R240000240000 now. If it depreciates at a rate of 15%15% p.a. on a straight-line depreciation, what is it worth in 5 years' time ?

Solution

  1. Step 1. Determine what has been provided and what is required :
    P = R 240 000 i = 0 , 15 n = 5 A is required P = R 240 000 i = 0 , 15 n = 5 A is required
    (6)
  2. Step 2. Determine how to approach the problem :
    A = 240 000 ( 1 - 0 , 15 × 5 ) A = 240 000 ( 1 - 0 , 15 × 5 )
    (7)
  3. Step 3. Solve the problem :
    A = 240 000 ( 1 - 0 , 75 ) = 240 000 × 0 , 25 = 60 000 A = 240 000 ( 1 - 0 , 75 ) = 240 000 × 0 , 25 = 60 000
    (8)
  4. Step 4. Write the final answer :

    In 5 years' time the car is worth R6000060000

Exercise 2: Simple Depreciation

A small business buys a photocopier for R 12 000. For the tax return the owner depreciates this asset over 3 years using a straight-line depreciation method. What amount will he fill in on his tax form after 1 year, after 2 years and then after 3 years ?

Solution

  1. Step 1. Understanding the question :

    The owner of the business wants the photocopier to depreciate to R0 after 3 years. Thus, the value of the photocopier will go down by 12000÷3=R400012000÷3=R4000 per year.

  2. Step 2. Value of the photocopier after 1 year :

    12 000 - 4 000 = R 8 000 12 000 - 4 000 = R 8 000

  3. Step 3. Value of the machine after 2 years :

    8 000 - 4 000 = R 4 000 8 000 - 4 000 = R 4 000

  4. Step 4. Write the final answer :

    4 000 - 4 000 = 0 4 000 - 4 000 = 0

    After 3 years the photocopier is worth nothing

Salvage Value

Looking at the same example of our car with an initial value of R60 000, what if we suppose that we think we would be able to sell the car at the end of the 5 year period for R10 000? We call this amount the “Salvage Value"

We are still assuming simple depreciation over a useful life of 5 years, but now instead of depreciating the full value of the asset, we will take into account the salvage value, and will only apply the depreciation to the value of the asset that we expect not to recoup, i.e. R60 000 - R10 000 = R50 000.

The annual depreciation amount is then calculated as (R60 000 - R10 000) / 5 = R10 000

In general, the formula for simple (straight line) depreciation:

Annual Depreciation = Initial Value - Salvage Value Useful Life Annual Depreciation = Initial Value - Salvage Value Useful Life
(9)

Simple Depreciation

  1. A business buys a truck for R560 000. Over a period of 10 years the value of the truck depreciates to R0 (using the straight-line method). What is the value of the truck after 8 years ?
  2. Shrek wants to buy his grandpa's donkey for R800. His grandpa is quite pleased with the offer, seeing that it only depreciated at a rate of 3% per year using the straight-line method. Grandpa bought the donkey 5 years ago. What did grandpa pay for the donkey then ?
  3. Seven years ago, Rocco's drum kit cost him R 12 500. It has now been valued at R2 300. What rate of simple depreciation does this represent ?
  4. Fiona buys a DsTV satellite dish for R3 000. Due to weathering, its value depreciates simply at 15% per annum. After how long will the satellite dish be worth nothing ?

Compound Depreciation

The second method of calculating depreciation is to assume that the value of the asset decreases at a certain annual rate, but that the initial value of the asset this year, is the book value of the asset at the end of last year.

For example, if our second hand car has a limited useful life of 5 years and it has an initial value of R60 000, then the interest rate of depreciation is 20% (100%/5 years). After 1 year, the car is worth:

Book Value after first year = P ( 1 - n × i ) = R 60 000 ( 1 - 1 × 20 % ) = R 60 000 ( 1 - 0 , 2 ) = R 60 000 ( 0 , 8 ) = R 48 000 Book Value after first year = P ( 1 - n × i ) = R 60 000 ( 1 - 1 × 20 % ) = R 60 000 ( 1 - 0 , 2 ) = R 60 000 ( 0 , 8 ) = R 48 000
(10)

At the beginning of the second year, the car is now worth R48 000, so after two years, the car is worth:

Book Value after second year = P ( 1 - n × i ) = R 48 000 ( 1 - 1 × 20 % ) = R 48 000 ( 1 - 0 , 2 ) = R 48 000 ( 0 , 8 ) = R 38 400 Book Value after second year = P ( 1 - n × i ) = R 48 000 ( 1 - 1 × 20 % ) = R 48 000 ( 1 - 0 , 2 ) = R 48 000 ( 0 , 8 ) = R 38 400
(11)

We can tabulate these values.

Table 2
End of first year R60000(1-1×20%)60000(1-1×20%)=R60000(1-1×20%)160000(1-1×20%)1 = R48 000,00
End of second year R48000(1-1×20%)48000(1-1×20%)=R60000(1-1×20%)260000(1-1×20%)2 = R38 400,00
End of third year R38 400(1-1×20%)(1-1×20%)=R60000(1-1×20%)360000(1-1×20%)3 = R30 720,00
End of fourth year R30 720(1-1×20%)(1-1×20%)=R60000(1-1×20%)460000(1-1×20%)4 = R24 576,00
End of fifth year R24 576(1-1×20%)(1-1×20%)=R60000(1-1×20%)560000(1-1×20%)5 = R19 608,80

We can now write a general formula for the book value of an asset if the depreciation is compounded.

Initial Value - Total depreciation after n years = P ( 1 - i ) n Initial Value - Total depreciation after n years = P ( 1 - i ) n
(12)

For example, the book value of the car after two years can be simply calculated as follows:

Book Value after 2 years = P ( 1 - i ) n = R 60 000 ( 1 - 20 % ) 2 = R 60 000 ( 1 - 0 , 2 ) 2 = R 60 000 ( 0 , 8 ) 2 = R 38 400 Book Value after 2 years = P ( 1 - i ) n = R 60 000 ( 1 - 20 % ) 2 = R 60 000 ( 1 - 0 , 2 ) 2 = R 60 000 ( 0 , 8 ) 2 = R 38 400
(13)

as expected.

Note that the difference between the compound interest calculations and the compound depreciation calculations is that while the interest adds value to the principal amount, the depreciation amount reduces value!

Exercise 3: Compound Depreciation

The Flamingo population of the Bergriver mouth is depreciating on a reducing balance at a rate of 12% p.a. If there are now 3 200 flamingos in the wetlands of the Bergriver mouth, how many will there be in 5 years' time ? Answer to three significant numbers.

Solution

  1. Step 1. Determine what has been provided and what is required :
    P = R 3 200 i = 0 , 12 n = 5 A is required P = R 3 200 i = 0 , 12 n = 5 A is required
    (14)
  2. Step 2. Determine how to approach the problem :
    A = 3 200 ( 1 - 0 , 12 ) 5 A = 3 200 ( 1 - 0 , 12 ) 5
    (15)
  3. Step 3. Solve the problem :
    A = 3 200 ( 0 , 88 ) 5 = 3 200 × 0 , 527731916 = 1688 , 742134 A = 3 200 ( 0 , 88 ) 5 = 3 200 × 0 , 527731916 = 1688 , 742134
    (16)
  4. Step 4. Write the final answer : There would be approximately 1 690 flamingos in 5 years' time.

Exercise 4: Compound Depreciation

Farmer Brown buys a tractor for R250 000 and depreciates it by 20% per year using the compound depreciation method. What is the depreciated value of the tractor after 5 years ?

Solution

  1. Step 1. Determine what has been provided and what is required :
    P = R 250 000 i = 0 , 2 n = 5 A is required P = R 250 000 i = 0 , 2 n = 5 A is required
    (17)
  2. Step 2. Determine how to approach the problem :
    A = 250 000 ( 1 - 0 , 2 ) 5 A = 250 000 ( 1 - 0 , 2 ) 5
    (18)
  3. Step 3. Solve the problem :
    A = 250 000 ( 0 , 8 ) 5 = 250 000 × 0 , 32768 = 81 920 A = 250 000 ( 0 , 8 ) 5 = 250 000 × 0 , 32768 = 81 920
    (19)
  4. Step 4. Write the final answer :

    Depreciated value after 5 years is R 81 920

Compound Depreciation

  1. On January 1, 2008 the value of my Kia Sorento is R320 000. Each year after that, the car’s value will decrease 20% of the previous year’s value. What is the value of the car on January 1, 2012?
  2. The population of Bonduel decreases at a rate of 9,5% per annum as people migrate to the cities. Calculate the decrease in population over a period of 5 years if the initial population was 2 178 000.
  3. A 20 kg watermelon consists of 98% water. If it is left outside in the sun it loses 3% of its water each day. How much does it weigh after a month of 31 days ?
  4. A computer depreciates at x%x% per annum using the reducing-balance method. Four years ago the value of the computer was R10 000 and is now worth R4 520. Calculate the value of xx correct to two decimal places.

Present Values or Future Values of an Investment or Loan

Now or Later

When we studied simple and compound interest we looked at having a sum of money now, and calculating what it will be worth in the future. Whether the money was borrowed or invested, the calculations examined what the total money would be at some future date. We call these future values.

It is also possible, however, to look at a sum of money in the future, and work out what it is worth now. This is called a present value.

For example, if R1 000 is deposited into a bank account now, the future value is what that amount will accrue to by some specified future date. However, if R1 000 is needed at some future time, then the present value can be found by working backwards - in other words, how much must be invested to ensure the money grows to R1 000 at that future date?

The equation we have been using so far in compound interest, which relates the open balance (PP), the closing balance (AA), the interest rate (ii as a rate per annum) and the term (nn in years) is:

A = P · ( 1 + i ) n A = P · ( 1 + i ) n
(20)

Using simple algebra, we can solve for PP instead of AA, and come up with:

P = A · ( 1 + i ) - n P = A · ( 1 + i ) - n
(21)

This can also be written as follows, but the first approach is usually preferred.

P = A / ( 1 + i ) n P = A / ( 1 + i ) n
(22)

Now think about what is happening here. In Equation Equation 20, we start off with a sum of money and we let it grow for nn years. In Equation Equation 21 we have a sum of money which we know in nn years time, and we “unwind" the interest - in other words we take off interest for nn years, until we see what it is worth right now.

We can test this as follows. If I have R1 000 now and I invest it at 10% for 5 years, I will have:

A = P · ( 1 + i ) n = R 1 000 ( 1 + 10 % ) 5 = R 1 610 , 51 A = P · ( 1 + i ) n = R 1 000 ( 1 + 10 % ) 5 = R 1 610 , 51
(23)

at the end. BUT, if I know I have to have R1 610,51 in 5 years time, I need to invest:

P = A · ( 1 + i ) - n = R 1 610 , 51 ( 1 + 10 % ) - 5 = R 1 000 P = A · ( 1 + i ) - n = R 1 610 , 51 ( 1 + 10 % ) - 5 = R 1 000
(24)

We end up with R1 000 which - if you think about it for a moment - is what we started off with. Do you see that?

Of course we could apply the same techniques to calculate a present value amount under simple interest rate assumptions - we just need to solve for the opening balance using the equations for simple interest.

A = P ( 1 + i × n ) A = P ( 1 + i × n )
(25)

Solving for PP gives:

P = A / ( 1 + i × n ) P = A / ( 1 + i × n )
(26)

Let us say you need to accumulate an amount of R1 210 in 3 years time, and a bank account pays Simple Interest of 7%. How much would you need to invest in this bank account today?

P = A 1 + n · i = R 1 210 1 + 3 × 7 % = R 1 000 P = A 1 + n · i = R 1 210 1 + 3 × 7 % = R 1 000
(27)

Does this look familiar? Look back to the simple interest worked example in Grade 10. There we started with an amount of R1 000 and looked at what it would grow to in 3 years' time using simple interest rates. Now we have worked backwards to see what amount we need as an opening balance in order to achieve the closing balance of R1 210.

In practice, however, present values are usually always calculated assuming compound interest. So unless you are explicitly asked to calculate a present value (or opening balance) using simple interest rates, make sure you use the compound interest rate formula!

Present and Future Values

  1. After a 20-year period Josh's lump sum investment matures to an amount of R313 550. How much did he invest if his money earned interest at a rate of 13,65% p.a.compounded half yearly for the first 10 years, 8,4% p.a. compounded quarterly for the next five years and 7,2% p.a. compounded monthly for the remaining period ?
  2. A loan has to be returned in two equal semi-annual instalments. If the rate of interest is 16% per annum, compounded semi-annually and each instalment is R1 458, find the sum borrowed.

Finding ii

By this stage in your studies of the mathematics of finance, you have always known what interest rate to use in the calculations, and how long the investment or loan will last. You have then either taken a known starting point and calculated a future value, or taken a known future value and calculated a present value.

But here are other questions you might ask:

  1. I want to borrow R2 500 from my neighbour, who said I could pay back R3 000 in 8 months time. What interest is she charging me?
  2. I will need R450 for some university textbooks in 1,5 years time. I currently have R400. What interest rate do I need to earn to meet this goal?

Each time that you see something different from what you have seen before, start off with the basic equation that you should recognise very well:

A = P · ( 1 + i ) n A = P · ( 1 + i ) n
(28)

If this were an algebra problem, and you were told to “solve for ii", you should be able to show that:

A / P = ( 1 + i ) n ( 1 + i ) = ( A / P ) 1 / n i = ( A / P ) 1 / n - 1 A / P = ( 1 + i ) n ( 1 + i ) = ( A / P ) 1 / n i = ( A / P ) 1 / n - 1
(29)

You do not need to memorise this equation, it is easy to derive any time you need it!

So let us look at the two examples mentioned above.

  1. Check that you agree that PP=R2 500, AA=R3 000, nn=8/12=0,666667. This means that:
    i=(R3000/R2500)1/0,666667-1=31,45%i=(R3000/R2500)1/0,666667-1=31,45%
    (30)
    Ouch! That is not a very generous neighbour you have.
  2. Check that PP=R400, AA=R450, nn=1,5
    i=(R450/R400)1/1,5-1=8,17%i=(R450/R400)1/1,5-1=8,17%
    (31)
    This means that as long as you can find a bank which pays more than 8,17% interest, you should have the money you need!

Note that in both examples, we expressed nn as a number of years (812812 years, not 8 because that is the number of months) which means ii is the annual interest rate. Always keep this in mind - keep years with years to avoid making silly mistakes.

Finding ii

  1. A machine costs R45 000 and has a scrap value of R9 000 after 10 years. Determine the annual rate of depreciation if it is calculated on the reducing balance method.
  2. After 5 years an investment doubled in value. At what annual rate was interest compounded ?

Finding nn - Trial and Error

By this stage you should be seeing a pattern. We have our standard formula, which has a number of variables:

A = P · ( 1 + i ) n A = P · ( 1 + i ) n
(32)

We have solved for AA (in Grade 10), PP (in "Present Values or Future Values of an Investment or Loan") and ii (in "Finding i"). This time we are going to solve for nn. In other words, if we know what the starting sum of money is and what it grows to, and if we know what interest rate applies - then we can work out how long the money needs to be invested for all those other numbers to tie up.

This section will calculate nn by trial and error and by using a calculator. The proper algebraic solution will be learnt in Grade 12.

Solving for nn, we can write:

A = P ( 1 + i ) n A P = ( 1 + i ) n A = P ( 1 + i ) n A P = ( 1 + i ) n
(33)

Now we have to examine the numbers involved to try to determine what a possible value of nn is. Refer to your Grade 10 notes for some ideas as to how to go about finding nn.

Exercise 5: Term of Investment - Trial and Error

We invest R3 500 into a savings account which pays 7,5% compound interest for an unknown period of time, at the end of which our account is worth R4 044,69. How long did we invest the money?

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 A P = ( 1 + i ) n A = P ( 1 + i ) n A P = ( 1 + i ) n
    (34)
  3. Step 3. Solve the problem :
    R 4 044 , 69 R 3 500 = ( 1 + 7 , 5 % ) n 1 , 156 = ( 1 , 075 ) n R 4 044 , 69 R 3 500 = ( 1 + 7 , 5 % ) n 1 , 156 = ( 1 , 075 ) n
    (35)

    We now use our calculator and try a few values for nn.

    Table 3
    Possible nn 1 , 075 n 1 , 075 n
    1,0 1,075
    1,5 1,115
    2,0 1,156
    2,5 1,198

    We see that nn is close to 2.

  4. Step 4. Write final answer :

    The R3 500 was invested for about 2 years.

Finding nn - Trial and Error

  1. A company buys two types of motor cars: The Acura costs R80 600 and the Brata R101 700 VAT included. The Acura depreciates at a rate, compounded annually, of 15,3% per year and the Brata at 19,7%, also compounded annually, per year. After how many years will the book value of the two models be the same ?
  2. The fuel in the tank of a truck decreases every minute by 5,5% of the amount in the tank at that point in time. Calculate after how many minutes there will be less than 30l30l in the tank if it originally held 200l200l.

Nominal and Effective Interest Rates

So far we have discussed annual interest rates, where the interest is quoted as a per annum amount. Although it has not been explicitly stated, we have assumed that when the interest is quoted as a per annum amount it means that the interest is paid once a year.

Interest however, may be paid more than just once a year, for example we could receive interest on a monthly basis, i.e. 12 times per year. So how do we compare a monthly interest rate, say, to an annual interest rate? This brings us to the concept of the effective annual interest rate.

One way to compare different rates and methods of interest payments would be to compare the Closing Balances under the different options, for a given Opening Balance. Another, more widely used, way is to calculate and compare the “effective annual interest rate" on each option. This way, regardless of the differences in how frequently the interest is paid, we can compare apples-with-apples.

For example, a savings account with an opening balance of R1 000 offers a compound interest rate of 1% per month which is paid at the end of every month. We can calculate the accumulated balance at the end of the year using the formulae from the previous section. But be careful … our interest rate has been given as a monthly rate, so we need to use the same units (months) for our time period of measurement.

Tip:

Remember, the trick to using the formulae is to define the time period, and use the interest rate relevant to the time period.

So we can calculate the amount that would be accumulated by the end of 1-year as follows:

Closing Balance after 12 months = P × ( 1 + i ) n = R 1 000 × ( 1 + 1 % ) 12 = R 1 126 , 83 Closing Balance after 12 months = P × ( 1 + i ) n = R 1 000 × ( 1 + 1 % ) 12 = R 1 126 , 83
(36)

Note that because we are using a monthly time period, we have used nn = 12 months to calculate the balance at the end of one year.

The effective annual interest rate is an annual interest rate which represents the equivalent per annum interest rate assuming compounding.

It is the annual interest rate in our Compound Interest equation that equates to the same accumulated balance after one year. So we need to solve for the effective annual interest rate so that the accumulated balance is equal to our calculated amount of R1 126,83.

We use i12i12 to denote the monthly interest rate. We have introduced this notation here to distinguish between the annual interest rate, ii. Specifically, we need to solve for ii in the following equation:

P × ( 1 + i ) 1 = P × ( 1 + i 12 ) 12 ( 1 + i ) = ( 1 + i 12 ) 12 divide both sides by P i = ( 1 + i 12 ) 12 - 1 subtract 1 from both sides P × ( 1 + i ) 1 = P × ( 1 + i 12 ) 12 ( 1 + i ) = ( 1 + i 12 ) 12 divide both sides by P i = ( 1 + i 12 ) 12 - 1 subtract 1 from both sides
(37)

For the example, this means that the effective annual rate for a monthly rate i12=1%i12=1% is:

i = ( 1 + i 12 ) 12 - 1 = ( 1 + 1 % ) 12 - 1 = 0 , 12683 = 12 , 683 % i = ( 1 + i 12 ) 12 - 1 = ( 1 + 1 % ) 12 - 1 = 0 , 12683 = 12 , 683 %
(38)

If we recalculate the closing balance using this annual rate we get:

Closing Balance after 1 year = P × ( 1 + i ) n = R 1 000 × ( 1 + 12 , 683 % ) 1 = R 1 126 , 83 Closing Balance after 1 year = P × ( 1 + i ) n = R 1 000 × ( 1 + 12 , 683 % ) 1 = R 1 126 , 83
(39)

which is the same as the answer obtained for 12 months.

Note that this is greater than simply multiplying the monthly rate by 12 (12×1%=12%12×1%=12%) due to the effects of compounding. The difference is due to interest on interest. We have seen this before, but it is an important point!

The General Formula

So we know how to convert a monthly interest rate into an effective annual interest. Similarly, we can convert a quarterly interest, or a semi-annual interest rate or an interest rate of any frequency for that matter into an effective annual interest rate.

For a quarterly interest rate of say 3% per quarter, the interest will be paid four times per year (every three months). We can calculate the effective annual interest rate by solving for ii:

P ( 1 + i ) = P ( 1 + i 4 ) 4 P ( 1 + i ) = P ( 1 + i 4 ) 4
(40)

where i4i4 is the quarterly interest rate.

So (1+i)=(1,03)4(1+i)=(1,03)4 , and so i=12,55%i=12,55%. This is the effective annual interest rate.

In general, for interest paid at a frequency of TT times per annum, the follow equation holds:

P ( 1 + i ) = P ( 1 + i T ) T P ( 1 + i ) = P ( 1 + i T ) T
(41)

where iTiT is the interest rate paid TT times per annum.

De-coding the Terminology

Market convention however, is not to state the interest rate as say 1% per month, but rather to express this amount as an annual amount which in this example would be paid monthly. This annual amount is called the nominal amount.

The market convention is to quote a nominal interest rate of “12% per annum paid monthly" instead of saying (an effective) 1% per month. We know from a previous example, that a nominal interest rate of 12% per annum paid monthly, equates to an effective annual interest rate of 12,68%, and the difference is due to the effects of interest-on-interest.

So if you are given an interest rate expressed as an annual rate but paid more frequently than annual, we first need to calculate the actual interest paid per period in order to calculate the effective annual interest rate.

monthly interest rate = Nominal interest Rate per annum number of periods per year monthly interest rate = Nominal interest Rate per annum number of periods per year
(42)

For example, the monthly interest rate on 12% interest per annum paid monthly, is:

monthly interest rate = Nominal interest Rate per annum number of periods per year = 12 % 12 months = 1 % per month monthly interest rate = Nominal interest Rate per annum number of periods per year = 12 % 12 months = 1 % per month
(43)

The same principle applies to other frequencies of payment.

Exercise 6: Nominal Interest Rate

Consider a savings account which pays a nominal interest at 8% per annum, paid quarterly. Calculate (a) the interest amount that is paid each quarter, and (b) the effective annual interest rate.

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

    We are given that a savings account has a nominal interest rate of 8% paid quarterly. We are required to find:

    • the quarterly interest rate, i4i4
    • the effective annual interest rate, ii
  2. Step 2. Determine how to approach the problem :

    We know that:

    quarterly interest rate = Nominal interest Rate per annum number of quarters per year quarterly interest rate = Nominal interest Rate per annum number of quarters per year
    (44)

    and

    P ( 1 + i ) = P ( 1 + i T ) T P ( 1 + i ) = P ( 1 + i T ) T
    (45)

    where TT is 4 because there are 4 payments each year.

  3. Step 3. Calculate the monthly interest rate :
    quarterly interest rate = Nominal interest Rate per annum number of periods per year = 8 % 4 quarters = 2 % per quarter quarterly interest rate = Nominal interest Rate per annum number of periods per year = 8 % 4 quarters = 2 % per quarter
    (46)
  4. Step 4. Calculate the effective annual interest rate :

    The effective annual interest rate (ii) is calculated as:

    ( 1 + i ) = ( 1 + i 4 ) 4 ( 1 + i ) = ( 1 + 2 % ) 4 i = ( 1 + 2 % ) 4 - 1 = 8 , 24 % ( 1 + i ) = ( 1 + i 4 ) 4 ( 1 + i ) = ( 1 + 2 % ) 4 i = ( 1 + 2 % ) 4 - 1 = 8 , 24 %
    (47)
  5. Step 5. Write the final answer :

    The quarterly interest rate is 2% and the effective annual interest rate is 8,24%, for a nominal interest rate of 8% paid quarterly.

Exercise 7: Nominal Interest Rate

On their saving accounts, Echo Bank offers an interest rate of 18% nominal, paid monthly. If you save R100 in such an account now, how much would the amount have accumulated to in 3 years' time?

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

    Interest rate is 18% nominal paid monthly. There are 12 months in a year. We are working with a yearly time period, so n=3n=3. The amount we have saved is R100, so P=100P=100. We need the accumulated value, AA.

  2. Step 2. Recall relevant formulae :

    We know that

    monthlyinterestrate = Nominal interest Rate per annum number of periods per year monthlyinterestrate = Nominal interest Rate per annum number of periods per year
    (48)

    for converting from nominal interest rate to effective interest rate, we have

    1 + i = ( 1 + i T ) T 1 + i = ( 1 + i T ) T
    (49)

    and for calculating accumulated value, we have

    A = P × ( 1 + i ) n A = P × ( 1 + i ) n
    (50)
  3. Step 3. Calculate the effective interest rate :

    There are 12 month in a year, so

    i 12 = Nominal annual interest rate 12 = 18 % 12 = 1 , 5 % per month i 12 = Nominal annual interest rate 12 = 18 % 12 = 1 , 5 % per month
    (51)

    and then, we have

    1 + i = ( 1 + i 12 ) 12 i = ( 1 + i 12 ) 12 - 1 = ( 1 + 1 , 5 % ) 12 - 1 = ( 1 , 015 ) 12 - 1 = 19 , 56 % 1 + i = ( 1 + i 12 ) 12 i = ( 1 + i 12 ) 12 - 1 = ( 1 + 1 , 5 % ) 12 - 1 = ( 1 , 015 ) 12 - 1 = 19 , 56 %
    (52)
  4. Step 4. Reach the final answer :
    A = P × ( 1 + i ) n = 100 × ( 1 + 19 , 56 % ) 3 = 100 × 1 , 7091 = 170 , 91 A = P × ( 1 + i ) n = 100 × ( 1 + 19 , 56 % ) 3 = 100 × 1 , 7091 = 170 , 91
    (53)
  5. Step 5. Write the final answer :

    The accumulated value is R170,91170,91. (Remember to round off to the the nearest cent.)

Nominal and Effect Interest Rates

  1. Calculate the effective rate equivalent to a nominal interest rate of 8,75% p.a. compounded monthly.
  2. Cebela is quoted a nominal interest rate of 9,15% per annum compounded every four months on her investment of R 85 000. Calculate the effective rate per annum.

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

S i m p l e I n c r e a s e : A = P ( 1 + i × n ) C o m p o u n d I n c r e a s e : A = P ( 1 + i ) n S i m p l e D e c r e a s e : A = P ( 1 - i × n ) C o m p o u n d D e c r e a s e : A = P ( 1 - i ) n E f f e c t i v e A n n u a l I n t e r e s t R a t e ( i ) : ( 1 + i ) = ( 1 + i T ) T S i m p l e I n c r e a s e : A = P ( 1 + i × n ) C o m p o u n d I n c r e a s e : A = P ( 1 + i ) n S i m p l e D e c r e a s e : A = P ( 1 - i × n ) C o m p o u n d D e c r e a s e : A = P ( 1 - i ) n E f f e c t i v e A n n u a l I n t e r e s t R a t e ( i ) : ( 1 + i ) = ( 1 + i T ) T
(54)

End of Chapter Exercises

  1. Shrek buys a Mercedes worth R385 000 in 2007. What will the value of the Mercedes be at the end of 2013 if
    1. the car depreciates at 6% p.a. straight-line depreciation
    2. the car depreciates at 12% p.a. reducing-balance depreciation.
  2. Greg enters into a 5-year hire-purchase agreement to buy a computer for R8 900. The interest rate is quoted as 11% per annum based on simple interest. Calculate the required monthly payment for this contract.
  3. A computer is purchased for R16 000. It depreciates at 15% per annum.
    1. Determine the book value of the computer after 3 years if depreciation is calculated according to the straight-line method.
    2. Find the rate, according to the reducing-balance method, that would yield the same book value as in Item 61 after 3 years.
  4. Maggie invests R12 500,00 for 5 years at 12% per annum compounded monthly for the first 2 years and 14% per annum compounded semi-annually for the next 3 years. How much will Maggie receive in total after 5 years?
  5. Tintin invests R120 000. He is quoted a nominal interest rate of 7,2% per annum compounded monthly.
    1. Calculate the effective rate per annum correct to THREE decimal places.
    2. Use the effective rate to calculate the value of Tintin's investment if he invested the money for 3 years.
    3. Suppose Tintin invests his money for a total period of 4 years, but after 18 months makes a withdrawal of R20 000, how much will he receive at the end of the 4 years?
  6. Paris opens accounts at a number of clothing stores and spends freely. She gets heself into terrible debt and she cannot pay off her accounts. She owes Hilton Fashion world R5 000 and the shop agrees to let Paris pay the bill at a nominal interest rate of 24% compounded monthly.
    1. How much money will she owe Hilton Fashion World after two years ?
    2. What is the effective rate of interest that Hilton Fashion World is charging her ?

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