Simple and compound growth and decay

• Simple interest: $A=P(1+in)$. Interest is earned on the original amount only.
• Compound interest: $A=P(1+i{)}^n$. Interest is earned on interest.

Example. R8 000 at 7% p.a. compounded annually for 5 years:

$A=8000(1,07{)}^5\approx \text{R}11220,41$

Finding the rate. R5 000 grows to R7 500 in 6 years:

• Compound: $(1+i{)}^6=1,5\Rightarrow i\approx 6,99\%$ p.a.
• Simple: $7500=5000(1+6i)\Rightarrow i\approx 8,33\%$ p.a.

Populations grow by a percentage of their current size, so they follow the compound formula:

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

Example. A town of 25 000 people grows at 2,3% p.a.:

$A=25000(1,023{)}^{10}\approx 31383\text{ people after 10 years}$

Round populations to whole numbers, and say whether you rounded up or down if the context matters.

The asset loses the same amount each year, a percentage of the original value:

$A=P(1-in)$

Example. A laptop costing R18 000 depreciates at 20% p.a. on the straight-line method. After 3 years:

$A=18000(1-0,2\times 3)=\text{R}7200$

After 5 years the value is R0, and it cannot go lower.

The asset loses a percentage of its current value each year:

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

Example. The same laptop at 20% p.a. on the reducing-balance method:

$A=18000(0,8{)}^3=\text{R}9216$

Reducing balance loses more in the early years and less later. It never reaches R0.