The Logic of “Buying Time”#
A Present Value Annuity is used when you receive a large sum of money now (a loan) and pay it back with regular, equal payments over time.
Timeline: Present Value (Loan)#
Always draw this before touching the formula:
|--------|--------|--------|--------|--------|
T0 T1 T2 T3 ... Tn
↑ P is x x x x
HEREThe loan ($P$) sits at the start. Each payment of $x$ chips away at the debt, but interest keeps growing it. This is why the balance drops slowly at first and faster later — early payments are mostly interest.
1. The Formula#
$$\boxed{P = \frac{x\left[1 - (1+i)^{-n}\right]}{i}}$$| Symbol | Meaning | Watch out for |
|---|---|---|
| $P$ | Present value (loan amount at $T_0$) | The “big pile” at the start |
| $x$ | Regular payment per period | Must match the compounding period |
| $i$ | Interest rate per period | If 15% p.a. compounded monthly: $i = \frac{0.15}{12} = 0.0125$ |
| $n$ | Number of payments | Not years! Monthly for 20 years = $240$ payments |
For the derivation of this formula from the geometric series sum, see The Logic of Annuities.
2. Worked Examples — Loans#
Worked Example 1 — Finding the Monthly Repayment#
You take out a home loan of R1 200 000 at 11.5% p.a. compounded monthly, to be repaid over 20 years. Find the monthly repayment.
Set up: $P = 1\,200\,000$, $i = \frac{0.115}{12} = 0.009583\overline{3}$, $n = 20 \times 12 = 240$
$$1\,200\,000 = \frac{x\left[1 - (1.009583\overline{3})^{-240}\right]}{0.009583\overline{3}}$$$$(1.009583\overline{3})^{-240} = 0.10028$$$$1\,200\,000 = \frac{x(1 - 0.10028)}{0.009583\overline{3}} = \frac{x(0.89972)}{0.009583\overline{3}} = x \times 93.884$$$$x = \frac{1\,200\,000}{93.884} = \boxed{\text{R}12\,781.01\text{ per month}}$$Reality check: Over 20 years you pay $240 \times \text{R}12\,781.01 = \text{R}3\,067\,442$ for a R1 200 000 loan. That’s more than 2.5 times the original loan — the cost of borrowing is enormous.
Worked Example 2 — Finding How Much You Can Borrow#
You can afford R5 000 per month. The bank offers 12% p.a. compounded monthly over 5 years. What is the maximum loan you can take?
Set up: $x = 5\,000$, $i = \frac{0.12}{12} = 0.01$, $n = 60$
$$P = \frac{5\,000\left[1 - (1.01)^{-60}\right]}{0.01}$$$$(1.01)^{-60} = 0.55045$$$$P = \frac{5\,000(0.44955)}{0.01} = \frac{2\,247.75}{0.01} = \boxed{\text{R}224\,775.46}$$3. Balance Outstanding#
A very common exam question: “How much do you still owe after $k$ payments?”
Method: Present Value of Remaining Payments#
The balance outstanding immediately after the $k$-th payment is the present value of the remaining payments:
$$\text{Balance after } k \text{ payments} = \frac{x\left[1 - (1+i)^{-(n-k)}\right]}{i}$$Simply replace $n$ with $(n - k)$ — the number of payments still to come.
Worked Example 3 — Balance Outstanding#
Using the home loan from Example 1 (R1 200 000, 11.5% p.a. monthly, 20 years, payment R12 781.01), find the balance outstanding after 8 years.
Payments made: $8 \times 12 = 96$. Payments remaining: $240 - 96 = 144$.
$$B_{96} = \frac{12\,781.01\left[1 - (1.009583\overline{3})^{-144}\right]}{0.009583\overline{3}}$$$$(1.009583\overline{3})^{-144} = 0.25335$$$$B_{96} = \frac{12\,781.01(0.74665)}{0.009583\overline{3}} = \frac{9\,543.67}{0.009583\overline{3}} = \boxed{\text{R}995\,861.79}$$After 8 years (96 payments totalling R1 226 976.96), you still owe nearly R996 000 on a R1 200 000 loan. This shows how front-loaded the interest is.
4. Deferred Payments (Grace Period)#
Sometimes a loan is granted now but repayments only begin later (e.g., a student loan where payments start after graduation).
Timeline: Deferred Payment#
|--------|--------|--------|--------|--------|--------|
T0 T1 T2 T3 T4 ... Tn
↑ Loan NO PAY NO PAY ↑ First x x
granted payment
Interest accumulates →During the grace period, interest accumulates on the loan. The bank doesn’t wait for free.
The Strategy#
- Grow the loan for the grace period using compound interest: $P_{\text{new}} = P(1+i)^k$ where $k$ is the number of silent periods
- Use $P_{\text{new}}$ as the present value in the annuity formula to find $x$
Worked Example 4 — Deferred Payment#
A student borrows R80 000 at 9% p.a. compounded monthly. The first repayment is made 7 months after the loan is granted. The loan is repaid over 4 years of monthly payments. Find the monthly repayment.
Step 1 — Grace period: The first payment at $T_7$ means interest accumulates for $6$ months (not 7 — the gap before the first payment is $7 - 1 = 6$).
$$P_{\text{new}} = 80\,000(1.0075)^6 = 80\,000 \times 1.04585 = \text{R}83\,668.27$$Step 2 — Annuity: $P_{\text{new}} = 83\,668.27$, $i = 0.0075$, $n = 48$ (4 years of monthly payments)
$$83\,668.27 = \frac{x\left[1 - (1.0075)^{-48}\right]}{0.0075}$$$$(1.0075)^{-48} = 0.69861$$$$83\,668.27 = \frac{x(0.30139)}{0.0075} = x \times 40.185$$$$x = \frac{83\,668.27}{40.185} = \boxed{\text{R}2\,081.78}$$5. The Final Payment#
In practice, the last payment is rarely exactly $x$ due to rounding. To find the actual final payment:
- Find the balance outstanding after the second-to-last payment (i.e., after $n-1$ payments)
- Add one period of interest
- That’s the final payment
Worked Example 5 — Final Payment#
A loan of R50 000 at 12% p.a. compounded monthly is repaid in 36 monthly payments of R1 660.71. Find the actual final payment.
Step 1: Balance after 35 payments (1 remaining):
$$B_{35} = \frac{1\,660.71\left[1 - (1.01)^{-1}\right]}{0.01} = \frac{1\,660.71 \times 0.009901}{0.01} = \frac{16.438}{0.01} = \text{R}1\,643.79$$Step 2: Add one month of interest:
$$\text{Final payment} = 1\,643.79 \times (1.01) = \boxed{\text{R}1\,660.23}$$The final payment (R1 660.23) is slightly less than the regular payment (R1 660.71) — the small difference is due to rounding in the monthly payment.
6. Settling a Loan Early#
If you want to pay off a loan early (e.g., after receiving a bonus), you pay the balance outstanding at that point. No future interest is charged.
Worked Example 6 — Early Settlement#
Using the home loan from Example 1, you receive a R200 000 inheritance after 8 years and want to settle the loan. How much must you pay?
From Example 3, the balance after 8 years is R995 861.79.
$$\boxed{\text{Settlement amount} = \text{R}995\,861.79}$$You save the remaining $144 \times \text{R}12\,781.01 - \text{R}995\,861.79 = \text{R}844\,583.65$ in interest — a massive saving.
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Forgetting the negative exponent | $(1+i)^{-n}$ not $(1+i)^n$ — without the minus, you get a huge wrong answer | The negative makes the bracket shrink below 1 |
| Wrong $n$ for balance outstanding | Using total $n$ instead of remaining payments | $n_{\text{remaining}} = n_{\text{total}} - k$ |
| Grace period count | “First payment at month 4” means interest for 3 months, not 4 | Grace months $=$ first payment month $- 1$ |
| Mixing up $F$ and $P$ formulas | $F$ has $(1+i)^n - 1$; $P$ has $1 - (1+i)^{-n}$ | Ask: “Is the big pile at the start or end?” |
| Rounding the payment | Using a rounded $x$ in later calculations changes the balance | Store $x$ in calculator memory with full precision |
💡 Pro Tips for Exams#
1. “Start or End?”#
The single most important question in finance: Where does the big pile of money sit?
- Start → Present Value (loan/pension) → use $P$ formula
- End → Future Value (savings/sinking fund) → use $F$ formula
2. Balance Outstanding = Fresh Loan#
Think of the balance outstanding as if you were taking a new loan for just the remaining payments. The formula is identical — just use $n_{\text{remaining}}$.
3. Show Your Timeline for Marks#
In CAPS exams, drawing and labelling a timeline correctly can earn you method marks even if your final answer is wrong. Always show $T_0$, payments, and where $P$ or $F$ sits.
⏮️ Future Value | 🏠 Back to Finance | ⏭️ Nominal vs Effective Interest