Compound interest and depreciation
Compound interest and depreciation
A repeated percentage change is a geometric model. Use the rate per compounding period and count the same periods in the exponent.
| Situation | Model | Meaning |
|---|---|---|
| Compound interest | \(A=P\left(1+\dfrac{p}{100m}\right)^{mt}\) | \(P\): initial deposit; \(p\): nominal annual percentage rate; \(m\): periods per year; \(t\): years. |
| Annual depreciation | \(V=P\left(1-\dfrac d{100}\right)^t\) | \(d\): percentage lost each year; \(0<d<100\). |
| Interest earned | \(I=A-P\) | Subtract the initial deposit from the accumulated amount. |
Example
At \(6\%\) nominal annual interest compounded quarterly, the period rate is \(1.5\%\). A deposit of \(\$2000\) grows over two years to \(A=2000(1.015)^8\approx\$2252.99\).
A machine worth \(\$12000\) depreciates by \(15\%\) each year. After three years, \(V=12000(0.85)^3=\$7369.50\).
Technology check
In a financial solver, match payments per year and compounding periods to the stated model. For a single deposit with no payments, set the payment to zero. Use opposite signs for money paid out and money received; check the solver result by substituting into the formula.
To find a time or rate, rearrange the same model. For \(A=Pq^n\), \(n=\dfrac{\ln\left(\dfrac AP\right)}{\ln q}\), where \(q>0\) and \(q\ne1\). For annual crediting, test the next integer year when a target must first be reached.
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