The Logic of the “Growing Pile”#
A Future Value Annuity is used when you make regular, equal payments into a savings account and want to know the total at the end.
Timeline: Future Value#
Always draw this before touching the formula:
|--------|--------|--------|--------|--------|
T0 T1 T2 T3 ... Tn
x x x x
↑ F is HEREEach payment of $x$ earns compound interest for a different number of periods:
- Payment at $T_1$ earns interest for $(n-1)$ periods
- Payment at $T_2$ earns interest for $(n-2)$ periods
- Payment at $T_n$ earns no interest (it’s deposited at the end)
The first payment earns the most; the last payment earns nothing. The future value $F$ sits at the end — it’s the total “mountain” of money you’ve accumulated.
1. The Formula#
$$\boxed{F = \frac{x\left[(1+i)^n - 1\right]}{i}}$$| Symbol | Meaning | Watch out for |
|---|---|---|
| $F$ | Future value (total at the end) | This is what you’re solving for (usually) |
| $x$ | Regular payment per period | Must match the compounding period |
| $i$ | Interest rate per period | If 12% p.a. compounded monthly: $i = \frac{0.12}{12} = 0.01$ |
| $n$ | Number of payments | Not years! Monthly for 5 years = $60$ payments |
For the derivation of this formula from the geometric series sum, see The Logic of Annuities.
2. Worked Examples — Saving#
Worked Example 1 — Basic Future Value#
You deposit R2 000 at the end of every month into an account earning 9% p.a. compounded monthly. How much will you have after 3 years?
Set up: $x = 2\,000$, $i = \frac{0.09}{12} = 0.0075$, $n = 3 \times 12 = 36$
$$F = \frac{2\,000\left[(1.0075)^{36} - 1\right]}{0.0075}$$$$(1.0075)^{36} = 1.30865$$$$F = \frac{2\,000(0.30865)}{0.0075} = \frac{617.30}{0.0075} = \text{R}82\,307.16$$You deposited $36 \times \text{R}2\,000 = \text{R}72\,000$, so you earned R10 307.16 in interest.
Worked Example 2 — Finding the Payment#
You want to have R500 000 in 10 years. The bank offers 10% p.a. compounded monthly. How much must you deposit each month?
Set up: $F = 500\,000$, $i = \frac{0.10}{12} = 0.008\overline{3}$, $n = 120$
$$500\,000 = \frac{x\left[(1.008\overline{3})^{120} - 1\right]}{0.008\overline{3}}$$$$(1.008\overline{3})^{120} = 2.70704$$$$500\,000 = \frac{x(1.70704)}{0.008\overline{3}} = x \times 204.845$$$$x = \frac{500\,000}{204.845} = \boxed{\text{R}2\,440.90}$$3. The “Start Immediately” Adjustment#
The standard formula assumes the first payment is at $T_1$ (end of the first period). But what if you start saving immediately (at $T_0$)?
Immediate payment timeline:
|--------|--------|--------|--------|--------|
T0 T1 T2 T3 ... Tn
x x x x x
↑ F is HEREEvery payment now sits in the account for one extra period of interest.
Fix: Multiply the standard formula by $(1+i)$:
$$F_{\text{immediate}} = \frac{x\left[(1+i)^n - 1\right]}{i} \times (1+i)$$Worked Example 3 — Immediate First Payment#
$$F = 82\,307.16 \times (1.0075) = \boxed{\text{R}82\,924.46}}$$Same as Example 1 (R2 000/month, 9% p.a. compounded monthly, 3 years), but the first deposit is made immediately.
The extra month of interest on every payment adds R617.30.
4. Sinking Funds#
A sinking fund is a savings plan used by companies to replace expensive assets (machinery, vehicles, equipment) at the end of their useful life.
The Logic#
- The current asset will depreciate (lose value) over its lifetime
- The replacement asset will cost more due to inflation
- The company sells the old asset for its scrap value
- The sinking fund must cover the difference
The Steps#
- Calculate the scrap value of the current asset using reducing balance depreciation: $A = P(1 - i)^n$
- Calculate the future cost of a new asset using inflation/compound growth: $A = P(1 + i)^n$
- Find the shortfall: Future cost $-$ Scrap value $= F$ (target for sinking fund)
- Use the FV formula to find the monthly payment $x$
Worked Example 4 — Sinking Fund#
A machine costs R800 000 today. It depreciates at 15% p.a. on the reducing balance. Inflation on new machines is 8% p.a. The company plans to replace the machine after 6 years.
(a) Find the scrap value of the old machine. (b) Find the cost of a new machine in 6 years. (c) Find the monthly sinking fund payment if the fund earns 11% p.a. compounded monthly.
(a) Scrap value (reducing balance depreciation):
$$A = 800\,000(1 - 0.15)^6 = 800\,000(0.85)^6 = 800\,000 \times 0.37715 = \text{R}301\,719.80$$(b) Future cost (inflation):
$$A = 800\,000(1 + 0.08)^6 = 800\,000(1.08)^6 = 800\,000 \times 1.58687 = \text{R}1\,269\,498.11$$(c) Sinking fund target:
$$F = 1\,269\,498.11 - 301\,719.80 = \text{R}967\,778.31$$Now find the monthly payment with $i = \frac{0.11}{12} = 0.009\overline{1}$, $n = 72$:
$$967\,778.31 = \frac{x\left[(1.009\overline{1})^{72} - 1\right]}{0.009\overline{1}}$$$$(1.009\overline{1})^{72} = 1.92676$$$$967\,778.31 = \frac{x(0.92676)}{0.009\overline{1}} = x \times 101.201$$$$x = \frac{967\,778.31}{101.201} = \boxed{\text{R}9\,562.92\text{ per month}}}$$5. Changing Interest Rates or Payments#
If the interest rate changes partway through, or you change your payment amount, you cannot use a single formula. Split the problem:
- Calculate the accumulated amount at the point of change
- That amount becomes a lump sum that continues to grow
- Apply a new annuity formula for the remaining period
- Add the two parts together
Worked Example 5 — Rate Change#
You save R1 500/month at 8% p.a. compounded monthly for 2 years. The rate then changes to 10% p.a. compounded monthly for the next 3 years (same payment). Find the total after 5 years.
Phase 1 ($i_1 = \frac{0.08}{12}$, $n_1 = 24$):
$$F_1 = \frac{1\,500\left[(1.00\overline{6})^{24} - 1\right]}{0.00\overline{6}} = \text{R}38\,882.47$$Phase 2 — Lump sum growth (R38 882.47 grows for 36 months at new rate):
$$F_{\text{lump}} = 38\,882.47 \times (1.008\overline{3})^{36} = 38\,882.47 \times 1.34935 = \text{R}52\,463.62$$Phase 2 — New annuity ($i_2 = \frac{0.10}{12}$, $n_2 = 36$):
$$F_2 = \frac{1\,500\left[(1.008\overline{3})^{36} - 1\right]}{0.008\overline{3}} = \text{R}62\,888.07$$Total: $F = 52\,463.62 + 62\,888.07 = \boxed{\text{R}115\,351.69}$
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Using years instead of payment periods | 5 years ≠ $n = 5$ when compounding monthly | $n = \text{years} \times m$ (e.g., $5 \times 12 = 60$) |
| Forgetting scrap value in sinking funds | The target is not the full replacement cost | Target $=$ Future cost $-$ Scrap value |
| Using appreciation rate for depreciation | Depreciation uses $(1 - i)^n$, not $(1 + i)^n$ | Losing value → minus; gaining value → plus |
| Not adjusting for immediate payments | Standard formula assumes first payment at $T_1$ | Multiply by $(1+i)$ if first payment is at $T_0$ |
| Rounding $i$ mid-calculation | $\frac{0.09}{12} = 0.0075$ exactly, but $\frac{0.11}{12} = 0.009\overline{1}$ — rounding kills precision | Store $\frac{i_{\text{nom}}}{m}$ in calculator memory |
💡 Pro Tips for Exams#
1. The Timeline is Non-Negotiable#
Draw it. Mark $F$ at $T_n$. Mark every payment. Mark any changes (rate changes, extra deposits, missed payments). If you skip this step, you will make an error on complex questions.
2. Sinking Fund = 3 Calculations#
Always expect three parts: (1) depreciation of old asset, (2) inflation on new asset, (3) annuity payment. Budget your time accordingly — this is a multi-step question worth 6–8 marks.
3. Check Your Answer’s Reasonableness#
Quick mental check: total deposits $= n \times x$. Your future value should be more than this (because of interest). If it’s less, something’s wrong.
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