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# Compound interest

## Compound Interest

To explain the concept of compound interest, the following example is discussed:

### Exercise 1: Using Simple Interest to lead to the concept Compound Interest

I deposit R1 000 into a special bank account which pays a Simple Interest of 7%. What if I empty the bank account after a year, and then take the principal and the interest and invest it back into the same account again. Then I take it all out at the end of the second year, and then put it all back in again? And then I take it all out at the end of 3 years?

#### Solution

1. Step 1. Determine what is given and what is required :
• opening balance, P=R1000P=R1000
• interest rate, i=7%i=7%
• period of time, 1 year 1 year at a time, for 3 years

We are required to find the closing balance at the end of three years.

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

We know that:

A = P ( 1 + i · n ) A = P ( 1 + i · n )
(1)
3. Step 3. Determine the closing balance at the end of the first year :
A = P ( 1 + i · n ) = R 1 000 ( 1 + 1 × 7 % ) = R 1 070 A = P ( 1 + i · n ) = R 1 000 ( 1 + 1 × 7 % ) = R 1 070
(2)
4. Step 4. Determine the closing balance at the end of the second year :

After the first year, we withdraw all the money and re-deposit it. The opening balance for the second year is therefore R1070R1070, because this is the balance after the first year.

A = P ( 1 + i · n ) = R 1 070 ( 1 + 1 × 7 % ) = R 1 144 , 90 A = P ( 1 + i · n ) = R 1 070 ( 1 + 1 × 7 % ) = R 1 144 , 90
(3)
5. Step 5. Determine the closing balance at the end of the third year :

After the second year, we withdraw all the money and re-deposit it. The opening balance for the third year is therefore R1144,90R1144,90, because this is the balance after the first year.

A = P ( 1 + i · n ) = R 1 144 , 90 ( 1 + 1 × 7 % ) = R 1 225 , 04 A = P ( 1 + i · n ) = R 1 144 , 90 ( 1 + 1 × 7 % ) = R 1 225 , 04
(4)
6. Step 6. Write the final answer :

The closing balance after withdrawing all the money and re-depositing each year for 3 years of saving R1 000 at an interest rate of 7% is R1 225,04.

In the two worked examples using simple interest ((Reference) and Exercise 1), we have basically the same problem because PP=R1 000, ii=7% and nn=3 years for both problems. Except in the second situation, we end up with R1 225,04 which is more than R1 210 from the first example. What has changed?

In the first example I earned R70 interest each year - the same in the first, second and third year. But in the second situation, when I took the money out and then re-invested it, I was actually earning interest in the second year on my interest (R70) from the first year. (And interest on the interest on my interest in the third year!)

This more realistically reflects what happens in the real world, and is known as Compound Interest. It is this concept which underlies just about everything we do - so we will look at it more closely next.

Definition 1: Compound Interest

Compound interest is the interest payable on the principal and its accumulated interest.

Compound interest is a double-edged sword, though - great if you are earning interest on cash you have invested, but more serious if you are stuck having to pay interest on money you have borrowed!

In the same way that we developed a formula for Simple Interest, let us find one for Compound Interest.

If our opening balance is PP and we have an interest rate of ii then, the closing balance at the end of the first year is:

Closing Balance after 1 year = P ( 1 + i ) Closing Balance after 1 year = P ( 1 + i )
(5)

This is the same as Simple Interest because it only covers a single year. Then, if we take that out and re-invest it for another year - just as you saw us doing in the worked example above - then the balance after the second year will be:

Closing Balance after 2 years = [ P ( 1 + i ) ] × ( 1 + i ) = P ( 1 + i ) 2 Closing Balance after 2 years = [ P ( 1 + i ) ] × ( 1 + i ) = P ( 1 + i ) 2
(6)

And if we take that money out, then invest it for another year, the balance becomes:

Closing Balance after 3 years = [ P ( 1 + i ) 2 ] × ( 1 + i ) = P ( 1 + i ) 3 Closing Balance after 3 years = [ P ( 1 + i ) 2 ] × ( 1 + i ) = P ( 1 + i ) 3
(7)

We can see that the power of the term (1+i)(1+i) is the same as the number of years. Therefore,

Closing Balance after n years = P ( 1 + i ) n Closing Balance after n years = P ( 1 + i ) n
(8)

### Fractions add up to the Whole

It is easy to show that this formula works even when nn is a fraction of a year. For example, let us invest the money for 1 month, then for 4 months, then for 7 months.

Closing Balance after 1 month = P ( 1 + i ) 1 12 Closing Balance after 5 months = Closing Balance after 1 month invested for 4 months more = [ P ( 1 + i ) 1 12 ] ( 1 + i ) 4 12 = P ( 1 + i ) 1 12 + 4 12 = P ( 1 + i ) 5 12 Closing Balance after 12 months = Closing Balance after 5 months invested for 7 months more = [ P ( 1 + i ) 5 12 ] ( 1 + i ) 7 12 = P ( 1 + i ) 5 12 + 7 12 = P ( 1 + i ) 12 12 = P ( 1 + i ) 1 Closing Balance after 1 month = P ( 1 + i ) 1 12 Closing Balance after 5 months = Closing Balance after 1 month invested for 4 months more = [ P ( 1 + i ) 1 12 ] ( 1 + i ) 4 12 = P ( 1 + i ) 1 12 + 4 12 = P ( 1 + i ) 5 12 Closing Balance after 12 months = Closing Balance after 5 months invested for 7 months more = [ P ( 1 + i ) 5 12 ] ( 1 + i ) 7 12 = P ( 1 + i ) 5 12 + 7 12 = P ( 1 + i ) 12 12 = P ( 1 + i ) 1
(9)

which is the same as investing the money for a year.

Look carefully at the long equation above. It is not as complicated as it looks! All we are doing is taking the opening amount (PP), then adding interest for just 1 month. Then we are taking that new balance and adding interest for a further 4 months, and then finally we are taking the new balance after a total of 5 months, and adding interest for 7 more months. Take a look again, and check how easy it really is.

Does the final formula look familiar? Correct - it is the same result as you would get for simply investing PP for one full year. This is exactly what we would expect, because:

1 month + 4 months + 7 months = 12 months, which is a year.

Can you see that? Do not move on until you have understood this point.

### The Power of Compound Interest

To see how important this “interest on interest" is, we shall compare the difference in closing balances for money earning simple interest and money earning compound interest. Consider an amount of R10 000 that you have to invest for 10 years, and assume we can earn interest of 9%. How much would that be worth after 10 years?

The closing balance for the money earning simple interest is:

A = P ( 1 + i · n ) = R 10 000 ( 1 + 9 % × 10 ) = R 19 000 A = P ( 1 + i · n ) = R 10 000 ( 1 + 9 % × 10 ) = R 19 000
(10)

The closing balance for the money earning compound interest is:

A = P ( 1 + i ) n = R 10 000 ( 1 + 9 % ) 10 = R 23 673 , 64 A = P ( 1 + i ) n = R 10 000 ( 1 + 9 % ) 10 = R 23 673 , 64
(11)

So next time someone talks about the “magic of compound interest", not only will you know what they mean - but you will be able to prove it mathematically yourself!

Again, keep in mind that this is good news and bad news. When you are earning interest on money you have invested, compound interest helps that amount to increase exponentially. But if you have borrowed money, the build up of the amount you owe will grow exponentially too.

#### Exercise 2: Taking out a Loan

Mr Lowe wants to take out a loan of R 350 000. He does not want to pay back more than R625 000 altogether on the loan. If the interest rate he is offered is 13%, over what period should he take the loan.

##### Solution
1. Step 1. Determine what has been provided and what is required :
• opening balance, P=R350000P=R350000
• closing balance, A=R625000A=R625000
• interest rate, i=13% per year i=13% per year

We are required to find the time period(nn).

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

We know from Equation 8 that:

A = P ( 1 + i ) n A = P ( 1 + i ) n
(12)

We need to find nn.

Therefore we convert the formula to:

A P = ( 1 + i ) n A P = ( 1 + i ) n
(13)

and then find nn by trial and error.

3. Step 3. Solve the problem :
A P = ( 1 + i ) n 625000 350000 = ( 1 + 0 , 13 ) n 1 , 785 . . . = ( 1 , 13 ) n Try n = 3 : ( 1 , 13 ) 3 = 1 , 44 . . . Try n = 4 : ( 1 , 13 ) 4 = 1 , 63 . . . Try n = 5 : ( 1 , 13 ) 5 = 1 , 84 . . . A P = ( 1 + i ) n 625000 350000 = ( 1 + 0 , 13 ) n 1 , 785 . . . = ( 1 , 13 ) n Try n = 3 : ( 1 , 13 ) 3 = 1 , 44 . . . Try n = 4 : ( 1 , 13 ) 4 = 1 , 63 . . . Try n = 5 : ( 1 , 13 ) 5 = 1 , 84 . . .
(14)
4. Step 4. Write the final answer : Mr Lowe should take the loan over four years (If he took the loan over five years, he would end up paying more than he wants to.)

### Other Applications of Compound Growth

The following two examples show how we can take the formula for compound interest and apply it to real life problems involving compound growth or compound decrease.

#### Exercise 3: Population Growth

South Africa's population is increasing by 2,5% per year. If the current population is 43 million, how many more people will there be in South Africa in two years' time ?

##### Solution
1. Step 1. Determine what has been provided and what is required :
• initial value (opening balance), P=43000000P=43000000
• period of time, n=2 year n=2 year
• rate of increase, i=2,5% per yeari=2,5% per year

We are required to find the final value (closing balance AA).

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

We know from Equation 8 that:

A = P ( 1 + i ) n A = P ( 1 + i ) n
(15)
3. Step 3. Solve the problem :
A = P ( 1 + i ) n = 43 000 000 ( 1 + 0 , 025 ) 2 = 45 176 875 A = P ( 1 + i ) n = 43 000 000 ( 1 + 0 , 025 ) 2 = 45 176 875
(16)
4. Step 4. Write the final answer :

There will be 45176875-43000000=217687545176875-43000000=2176875 more people in 2 years' time

#### Exercise 4: Compound Decrease

A swimming pool is being treated for a build-up of algae. Initially, 50m250m2 of the pool is covered by algae. With each day of treatment, the algae reduces by 5%. What area is covered by algae after 30 days of treatment ?

##### Solution
1. Step 1. Determine what has been provided and what is required :
• Starting amount (opening balance), P=50m2P=50m2
• period of time, n=30 days n=30 days
• rate of decrease, i=5% per dayi=5% per day

We are required to find the final area covered by algae (closing balance AA).

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

We know from Equation 8 that:

A = P ( 1 + i ) n A = P ( 1 + i ) n
(17)

But this is compound decrease so we can use the formula:

A = P ( 1 - i ) n A = P ( 1 - i ) n
(18)
3. Step 3. Solve the problem :
A = P ( 1 - i ) n = 50 ( 1 - 0 , 05 ) 30 = 10 , 73 m 2 A = P ( 1 - i ) n = 50 ( 1 - 0 , 05 ) 30 = 10 , 73 m 2
(19)
4. Step 4. Write the final answer :

Therefore the area still covered with algae is 10,73m210,73m2

#### Compound Interest

1. An amount of R3 500 is invested in a savings account which pays compound interest at a rate of 7,5% per annum. Calculate the balance accumulated by the end of 2 years.
2. If the average rate of inflation for the past few years was 7,3% and your water and electricity account is R 1 425 on average, what would you expect to pay in 6 years time ?
3. Shrek wants to invest some money at 11% per annum compound interest. How much money (to the nearest rand) should he invest if he wants to reach a sum of R 100 000 in five year's time ?

The next section on exchange rates is included for completeness. However, you should know about fluctuating exchange rates and the impact that this has on imports and exports. Fluctuating exchange rates lead to things like increases in the cost of petrol. You can read more about this in Fluctuating exchange rates.

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