Future Value Annuities and Sinking Funds

A future value annuity is a series of equal payments that grows into a lump sum, such as monthly savings for a deposit. Sinking funds use the same formula to save for replacing equipment. In the exam you may be asked for the payment, the value or $n$ . Only the two standard annuity formulae are examined: payments start one period from now and end after $n$ periods, and no variations are examinable.

The future value formula

For equal payments $x$ made at the end of each period:

$F = \frac{x [ ( 1 + i ) ^{n} - 1 ]}{i}$

where $n$ is the number of payments and $i$ is the rate per period.

Example. R1 500 is paid into an account at the end of every month for 10 years at 9% p.a. compounded monthly:

$F = \frac{1 500 [ ( 1 , 0075 ) ^{120} - 1 ]}{0 , 0075} \approx R 290271, 42$

Finding the payment. Rearrange: $x = \frac{F \cdot i}{( 1 + i ) ^{n} - 1}$ .

Count the payments carefully. The formula assumes the first payment is made one period from now and the last payment at the end of period $n$ . IEB questions keep to that set-up. If money stays in the account after the last payment, grow the annuity's value forward with the compound growth formula $(1 + i)^{k}$ .

Sinking funds

A sinking fund is a savings plan to replace an asset in the future. The steps:

Future cost of the new asset, grown with inflation: $P (1 + i)^{n}$ .
Scrap value of the old asset, with depreciation: $P (1 - i)^{n}$ .
Amount needed = future cost $-$ scrap value.
Monthly payment into the sinking fund with the future value formula.

Example. A machine costs R850 000 now and will be replaced in 6 years.

Future cost with 7% p.a. inflation: $850000 (1, 07)^{6} \approx$ R1 275 620,80
Scrap value with 20% p.a. reducing-balance depreciation: $850000 (0, 8)^{6} \approx$ R222 822,40
Sinking fund target: R1 052 798,40

Monthly payments at 9,5% p.a. compounded monthly for 72 months:

$x = \frac{1 052 798 , 40 \times \frac{0 , 095}{12}}{( 1 + \frac{0 , 095}{12} ) ^{72} - 1} \approx R 10904, 91$

Key terms

Annuity: A series of equal payments made at regular intervals.
Future value: The total value of the payments and interest at the end.
Sinking fund: An account built up by regular payments to replace an asset.
Scrap value: What the old asset is worth when it is replaced.

Common mistakes

Counting the wrong number of payments when payments start now or stop early.
Using the present value formula for savings.
Forgetting to subtract the scrap value in a sinking fund.
Using annual rates with monthly payments.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

4 marks total

routine

Ayanda pays R1 500 into a savings account at the end of every month for 10 years. Interest is 9% p.a. compounded monthly. Calculate the value of the account after 10 years.

Worked solution

Hidden - try the question first

Solution

$i = \frac{0 , 09}{12} = 0, 0075$ and $n = 120$
$F = \frac{1 500 [ ( 1 , 0075 ) ^{120} - 1 ]}{0 , 0075}$
$\approx$ R290 271,42

Question 2

5 marks total

routine

How much must be saved at the end of every month to have R500 000 after 15 years? Interest is 10,5% p.a. compounded monthly.

Worked solution

Hidden - try the question first

Solution

$i = \frac{0 , 09}{12} = 0, 0075$ and $n = 180$
$500000 = \frac{x [ ( 1 + \frac{0 , 105}{12} ) ^{180} - 1 ]}{\frac{0 , 105}{12}}$
$x = \frac{500 000 \times \frac{0 , 105}{12}}{( 1 + \frac{0 , 105}{12} ) ^{180} - 1}$
$\approx$ R1 151,99

Question 3

7 marks total

complex

A printing company buys a machine for R850 000. It will be replaced in 6 years.

The cost of a new machine rises with inflation at 7% p.a.
The old machine depreciates at 20% p.a. on the reducing-balance method and will be traded in.
The company sets up a sinking fund at 9,5% p.a. compounded monthly. Payments start one month from now, and the last payment is made at the end of the 6 years.

(a) Calculate the cost of the new machine in 6 years. (1)

(b) Calculate the trade-in value of the old machine. (1)

Worked solution

Hidden - try the question first

Solution

(a) $850000 (1, 07)^{6} \approx$ R1 275 620,80
(b) $850000 (0, 8)^{6} \approx$ R222 822,40
(c) Sinking fund target $= 1275620, 80 - 222822, 40 =$ R1 052 798,40
$i = \frac{0 , 095}{12}$ and $n = 72$
$1052798, 40 = \frac{x [ ( 1 + \frac{0 , 095}{12} ) ^{72} - 1 ]}{\frac{0 , 095}{12}}$
$x = \frac{1 052 798 , 40 \times \frac{0 , 095}{12}}{( 1 + \frac{0 , 095}{12} ) ^{72} - 1}$
$\approx$ R10 904,91 per month

Question 4

9 marks total

problem-solving

Lerato pays R2 000 into an account at the end of every month for 5 years. Interest is 8% p.a. compounded monthly. She then stops paying, but leaves the money in the account for another year.

(a) Calculate the value of the account at the end of 5 years. (3)

(b) Calculate the value at the end of 6 years. (2)

(c) Lerato wants R200 000 at the end of 6 years. What single extra deposit, made at the end of year 5, would she need? (4)

Worked solution

Hidden - try the question first

Solution

(a) $850000 (1, 07)^{6} \approx$ , $n = 60$
$F = \frac{2 000 [ ( 1 + \frac{0 , 08}{12} ) ^{60} - 1 ]}{\frac{0 , 08}{12}}$
$\approx$ R146 953,71
(b) $146953, 71 (1 + \frac{0 , 08}{12})^{12}$
$\approx$ R159 150,80
(c) Shortfall at end of year 6: $200000 - 159150, 80 = 40849, 20$
The deposit $D$ grows for 12 months: $D (1 + \frac{0 , 08}{12})^{12} = 40849, 20$
$D = \frac{40 849 , 20}{( 1 + \frac{0 , 08}{12} ) ^{12}}$
$\approx$ R37 718,58

FAQs

Can the IEB ask me to calculate the interest rate in an annuity?

Not in an annuity. Solving the annuity formula for $i$ needs trial and error, so questions ask for the payment, the value or $n$ . Interest rates can still be found from the growth and decay formulae.

What if the first payment is made today?

The IEB does not examine variations of the annuity formulae, so exam questions have the first payment one period from now. Read the dates in the question to confirm this, then count the payments.

Are pyramid schemes examined?

They are part of the IEB Grade 12 curriculum: you should be able to analyse an investment option, including a pyramid scheme, and explain why it is not sustainable.

The future value formula

Sinking funds

Key terms

Common mistakes

Practice questions

FAQs

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