Present Value Annuities and Loans

A present value annuity works out how much a series of future payments is worth today. It is the formula behind home loans, car finance and retirement income. Outstanding balances, and what happens when an extra amount is paid into a loan, are the parts candidates find hardest. Only the standard formula is examined: payments start one period from now, and no variations of the annuity formulae are examinable.

The present value formula

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

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

Example. The present value of R4 000 per month for 5 years at 11% p.a. compounded monthly:

$P = \frac{4 000 [ 1 - ( 1 + \frac{0 , 11}{12} ) ^{- 60} ]}{\frac{0 , 11}{12}} \approx R 183972, 14$

Use present value when money is borrowed now and paid back over time, or when a lump sum now is used to fund payments in the future.

Loan repayments

The loan amount is the present value of the repayments. Rearranged for $x$ :

$x = \frac{P \cdot i}{1 - ( 1 + i ) ^{- n}}$

Example. A home loan of R1 200 000 at 11,75% p.a. compounded monthly over 20 years:

$x = \frac{1 200 000 \times \frac{0 , 1175}{12}}{1 - ( 1 + \frac{0 , 1175}{12} ) ^{- 240}} \approx R 13004, 48$

Total interest = total paid $-$ amount borrowed $= 240 x - 1200000$ .

Outstanding balance

The outstanding balance is what is still owed after some payments. The cleanest method is the present value of the remaining payments.

Example. For the loan above, after 5 years (60 payments) there are 180 payments left:

$Balance = \frac{13 004 , 484 75 \dots [ 1 - ( 1 + \frac{0 , 1175}{12} ) ^{- 180} ]}{\frac{0 , 1175}{12}} \approx R 1098229, 91$

Notice that after a quarter of the term, most of the loan is still owed. Early payments are mostly interest.

Use the unrounded payment (stored in your calculator) in the calculation. With the rounded R13 004,48 you get R1 098 229,51, which is 40 cents out.

Extra payments and new instalments

If a lump sum is paid into a loan, it comes straight off the outstanding balance. The new balance is then paid off with a new instalment over the remaining payments.

Example. A car loan of R350 000 at 12% p.a. compounded monthly is repaid with 60 monthly payments starting one month after the loan is granted, so $x \approx$ R7 785,56.

After the 24th payment, 36 payments remain. The balance is

$\frac{7 785 , 556 \dots [ 1 - ( 1 , 01 ) ^{- 36} ]}{0 , 01} \approx R 234403, 69$

An extra R50 000 is paid with the 24th payment, so R184 403,69 is left to repay over 36 months:

$x_{new} = \frac{184 403 , 69 \times 0 , 01}{1 - ( 1 , 01 ) ^{- 36}} \approx R 6124, 84$

Ask yourself: what is still owed right now, and how many payments are left? Those are the $P$ and $n$ for the formula.

Key terms

Present value: The value today of a series of future payments.
Outstanding balance: The amount still owed on a loan at a given time.
Extra payment: A lump sum paid into a loan on top of the normal instalment. It reduces the outstanding balance immediately.
Amortisation: Paying off a loan with regular payments that cover interest and capital.

Common mistakes

Using the future value formula for a loan.
Counting the remaining payments wrongly when finding the outstanding balance.
Subtracting an extra payment from the original loan amount instead of from the outstanding balance at that time.
Rounding the monthly payment before using it for the balance.

Practice questions

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

Question 1

4 marks total

routine

Calculate the present value of payments of R4 000 at the end of every month for 5 years, if interest is 11% p.a. compounded monthly.

Worked solution

Hidden - try the question first

Solution

$i = \frac{0 , 11}{12}$ and $n = 60$
$P = \frac{4 000 [ 1 - ( 1 + \frac{0 , 11}{12} ) ^{- 60} ]}{\frac{0 , 11}{12}}$
$\approx$ R183 972,14

Question 2

5 marks total

routine

The Mokoena family takes out a home loan of R1 200 000 at 11,75% p.a. compounded monthly, to be repaid over 20 years with monthly payments starting one month after the loan is granted.

Calculate the monthly repayment.

Worked solution

Hidden - try the question first

Solution

$i = \frac{0 , 11}{12}$ and $n = 60$
$P = \frac{4 000 [ 1 - ( 1 + \frac{0 , 11}{12} ) ^{- 60} ]}{\frac{0 , 11}{12}}$
$x = \frac{1 200 000 \times \frac{0 , 1175}{12}}{1 - ( 1 + \frac{0 , 1175}{12} ) ^{- 240}}$
$\approx$ R13 004,48

Question 3

7 marks total

complex

Use the home loan in Question 2.

(a) Calculate the outstanding balance immediately after the 60th payment. (4)

(b) Calculate the total interest paid over the full 20 years. (3)

Worked solution

Hidden - try the question first

Solution

(a) Remaining payments: $240 - 60 = 180$
Balance $= \frac{13 004 , 484 75 \dots [ 1 - ( 1 + \frac{0 , 1175}{12} ) ^{- 180} ]}{\frac{0 , 1175}{12}}$ (unrounded payment)
$\approx$ R1 098 229,91
(b) Total paid $= 240 \times 13004, 48$
$= 3121075, 20$
Interest $= 3121075, 20 - 1200000 =$ R1 921 075,20

Question 4

9 marks total

problem-solving

Pieter buys a bakkie with a loan of R350 000 at 12% p.a. compounded monthly. He repays it with 60 equal monthly payments, starting one month after the loan is granted.

(a) Calculate the monthly payment. (2)

(b) Immediately after his 24th payment, Pieter pays an extra R50 000 into the loan. He keeps the same end date. Calculate his new monthly payment. (5)

Worked solution

Hidden - try the question first

Solution

(a) $240 - 60 = 180$
$\approx$ R7 785,56
(b) Payments left: $60 - 24 = 36$
Balance $= \frac{7 785 , 556 \dots [ 1 - ( 1 , 01 ) ^{- 36} ]}{0 , 01} \approx 234403, 69$
New balance $= 234403, 69 - 50000 = 184403, 69$
$x_{new} = \frac{184 403 , 69 \times 0 , 01}{1 - ( 1 , 01 ) ^{- 36}} \approx$ R6 124,84
(c) Saving $= 36 (7785, 556 \dots - 6124, 841 \dots) - 50000$
$\approx$ R9 785,76

FAQs

How do I know whether to use the present value or future value formula?

Present value when the lump sum is at the start (a loan or a retirement fund being drawn down). Future value when the lump sum is at the end (savings).

What is the easiest way to find an outstanding balance?

Find the present value of the payments still to be made, using the unrounded payment.

Are deferred payments examined?

No. The IEB examines only the standard present value and future value formulae, with the first payment one period from now, and says no variations are examinable. Extra payments, outstanding balances and finding $n$ are examined.

The present value formula

Loan repayments

Outstanding balance

Extra payments and new instalments

Key terms

Common mistakes

Practice questions

FAQs

Need a tutor for this subject?