Calculating Compound Interest


How Fast can One's Fortune Grow?

Assume an ideal investor with those variables:

Case 1: without monthly money inflow

Initial condition:

\[A_0 = P\]

For each compounding:

\[A_{n+1} = A_{n} \times (1+r/n)\]

Hence, after \(t\) years or compounding \(c=t\times n\) times:

\[A_{t\times n} = P(1+r/n)^{t\times n}\]

Case 1.1: how many years are needed to double an investor's initial fortune?

Assuming an investor is investing low-to-medium risk assets, and luckily he has a stable yearly interest rate of \(r=2\% =0.02\).

For doubling his initial investment, the final amount would be \(A_{t\times n}=2P\).

\[A_{t\times n}=2P = P(1+r/1)^{t\times 1}\]
\[2=(1+r)^t\]
\[t=\frac{\ln(2)}{\ln(1+r)}\]

Thus, it takes the investor \(t\approx 0.69/\ln(1+0.02) \approx 35.00\) years to double his fortune.

Note that the \(\ln\), instead of for example \(\log_{10}\), was used because we can get a simple empirical relationship.

Because \(\ln(1+r)\) approximates \(r\) when the value of r is very small.

The equation can simplify into

$$ t\approx \frac{0.69}{r} \approx \frac{70}{r\%}, $$ which is also called the Rule of 70.

One's fortune doubles per \(\frac{70}{\text{interest percentage point}}\) years.

Case 2: With monthly inflow

The initial asset is still \(P\) but in each month, the investor adds \(w\) into his investment.

For each month, the final amount is

\[ A_{c+1} = (A_{c}+w)\times(1+r) \]

Sequentially, the amount after each month's compounding is

  1. \(P(1+r/n)\)
  2. \((P(1+r/n)+w)(1+r/n) = P(1+r/n)^2 + w(1+r/n)\)
  3. \([P(1+r/n)^2 + w(1+r/n) +w](1+r/n)= P(1+r/n)^3 + w(1+r/n)^2+w(1+r/n)\)
  4. ...

This can be summerized into

\[ A_{t\times n} = P(1+r/n)^{t\times n} + \sum_{d=0}^{t\times n-1}w(1+r/n)^d, \]

where \(n=12\).

So after one year or 12 compounding, one's fortune would be

\[ A_{12} = P(1+r/12)^{12} + w\sum_{d=0}^{11}(1+r/12)^d. \]

References

(TODO: Approximation of sequences?)