Time to Double Money Calculator

This calculator will show all the steps involved in computing the amount of time needed to double an initial amount $A_0$) of money. Common wisdom indicates that the higher the interest rate $r$ you get, the shorter it will take to double your money and that is indeed the case.

It will also depend on whether the compounding occurs more frequently that once a year. Indeed, let $k$ be the number of times the money is compounded in a year.

For example, for yearly compounding we have $k = 1$, for bi-yearly compounding we have $k = 2$, for quarterly compounding we have $k = 4$, etc.

Time to double compounded discretely

When you compound a certain amount of $k$ times a year, you have what is called a discrete compounding. For such type of compounding, the amount of money we will have after $n$ years is

$$FV = A_0 \left( 1+\frac{r}{k}\right)^{ k \times n}$$

So, if we wanted to double our initial amount $A_0$, we would need to end up with $2 A_0$ in the account, so that

$$2 A_0 = A_0 \left( 1+\frac{r}{k}\right)^{ k \times n}$$

and canceling $A_0$ from both sides of the equation leads to

$$2 = \left( 1+\frac{r}{k}\right)^{ k \times n}$$

and then applying natural log and solving for $n$ leads to

$$n = \frac{\ln 2}{k \left( 1+\frac{r}{k}\right)}$$

Time to double compounded continously

Something interesting happens for continuous compounding. Indeed, that case is the same as considering that $k \to \infty$, in which case the amount of money we have after $n$ years is (meaning, the future value of our money):

$$FV = A_0 e^{r \times n}$$

So, same as in the discrete compounding case, if we wanted to double our initial amount $A_0$, we would need to end up with $2 A_0$ in the account, so that

$$2 A_0 = A_0 e^{r \times n}$$

and canceling again $A_0$ from both sides of the equation, we will get

$$2 = e^{r \times n}$$

and then applying natural log and solving for $n$ leads to

$$n = \frac{\ln 2}{r)}$$

Observe the very interesting fact that the number of years required to double your initial amount $A_0$ DOES NOT depend on the initial amount, only on the interest rate $r$ and the type of compounding.

In other words, doubling $1 or double $1 million will take the same amount of time, assuming the same interest rate.

Example: Calculation of Time needed to double money

Question: How many years you need to wait until you double your money, if your bank offers you a nominal annual rate of 3.5%, which is compounded monthly?

Solution:

We need to compute the amount of time needed to double a given amount $A_0$

You have stated that the yearly interest rate is $r = 0.035$, and the compounding is done monthly.

The future value after $n$ years is calculated using the following formula:

$$FV = \displaystyle A_0 \left( 1+\frac{r}{k}\right)^{ k \times n}$$

What we want is to double the money, so we need to have $2 A_0$ as the future value after $n$ periods. So then we need $$2 A_0 = \displaystyle A_0 \left( 1+\frac{r}{k}\right)^{ k \times n}$$

which means that we can cancel $A_0$ from both sides of the equation, and we need to find $n$ from the following:

$$2 = \displaystyle \left( 1+\frac{r}{k}\right)^{ k \times n}$$

so then after applying natural log to both sides and solving for $n$ leads to:

$$n = \displaystyle \frac{\ln 2}{12 \ln\left( 1+\frac{0.035}{12}\right)} = \displaystyle \frac{\ln 2}{12 \ln\left( 1+ 0.003\right)} = 19.83$$

Finally, we plug the value we know about the interest rate and compounding period so we get:

Therefore, regardless of the initial amount $A_0$, the number of years required to double the initial investment for a yearly interest rate of $r = 0.035$, and with monthly compounding is $n = 19.83$ years.