Simple and compound growth and decay

Simple interest is calculated on the original amount only:

$A=P(1+in)$

Compound interest is calculated on the growing balance:

$A=P(1+i{)}^n$

where $P$ is the principal, $i$ the interest rate per period as a decimal, and $n$ the number of periods.

Example. R15 000 at 9% p.a. simple interest for 4 years: $A=15000(1+0,09\times 4)=$ R20 400.

At 9% p.a. compounded annually instead: $A=15000(1,09{)}^4\approx$ R21 173,72.

Match $i$ and $n$ to the compounding period. Monthly compounding for 4 years means $i=\frac{0,09}{12}$ and $n=48$.

Straight-line depreciation removes the same amount each year, a percentage of the original value:

$A=P(1-in)$

Reducing-balance depreciation removes a percentage of the current value each year:

$A=P(1-i{)}^n$

Example. A car bought for R320 000 depreciates at 18% p.a.

• Reducing balance after 5 years: $320000(0,82{)}^5\approx$ R118 636,75
• Straight line after 5 years: $320000(1-0,18\times 5)=$ R32 000

Finding the rate. If a R480 000 machine is worth R150 000 after 6 years on reducing balance: $(1-i{)}^6=\frac{150}{480}$, so $i\approx 17,62\%$.

When rates change or money is added or taken out, draw a timeline and move every amount forward to the end date one stage at a time.

1. Mark each change on the timeline.
2. Grow the balance to the first change.
3. Add deposits or subtract withdrawals at that point.
4. Continue with the new rate.

Example. R30 000 is invested at 8,4% p.a. compounded monthly. After 2 years R10 000 is withdrawn. After 3 years the rate changes to 9% p.a. compounded quarterly. Value after 5 years:

$A=\left[30000(1,007{)}^{24}-10000\right](1,007{)}^{12}(1,0225{)}^8\approx \text{R}33085,95$

Alternatively, move each amount forward separately and subtract the withdrawal's future value. Both methods give the same answer.