0

Here is the question: I got capital X, and I will compound interest (4%) to roll the capital, until the the interest grow > 3000.

So, I start from 6000 (X), here is my example excel: enter image description here

I mark down the cell number on when it reaches more than 3,000, for 6,000, the number is row 67. Each time, I add the capital by 6,000, so, the next round, the X is 12,000; and the row that is bigger than 49. I do all that util the capital rach 60,000. Here is the result:

enter image description here

And here is the curve. enter image description here

So, my question is: When I got 1x capital, it requires 67 times to reach my goal; but when I got 2x capital, it requires 49. Why it is not 30? And I got 10x capital, it doesn't help me to get 10x faster, instead, it is only 67/8 = 8.375 faster.

Why? Thanks.

Ted Wong
  • 111
  • 2
    Why should it be linear? Using the continuous compounding approximation, you are trying to solve some variant of $e^{rt}C_0=C$ where $C_0$ is the starting capital and $C$ is the desired capital. That gives us $t=\frac 1r\times \ln \frac C{C_0}$ which is not linear. – lulu Nov 29 '19 at 12:40
  • the interest is the gain your starting capital right? so with 6000, you reach a total of 9000 already between 12 and 13. So you then stop? – Henno Brandsma Nov 29 '19 at 12:49

1 Answers1

1

With starting capital $C_0$ and interest rate of $r \%$ the capital after $n$ periods is $C_0(1+\frac{r}{100})^n$ so the gain on the original capital is

$$C_0\left((1+\frac{1}{100})^n -1\right)$$

which is linear in the starting capital. Your gain is $>3000\$$ after 11 periods with starting capital $6000$ (so we need $\left((1+\frac{1}{100})^n -1\right) > \frac12$. With starting capital, say $12000\$$, we'd only need that number to be $> \frac14$ which happens much sooner, after $6$ periods. The period you solve for is in the exponent so will give a logarithmic function.

Henno Brandsma
  • 242,131