# On Compound InterestGitHub iconHow interest rates and time period of investment, effects return

When should you start investing?

Let say, you find an investment which gives 7% annual interest rate, for 30 years[6]. If you put ₹ 1,000 in the investment. How much money would you get back in 30 years? What if put ₹ 100,000, instead of ₹ 1,000?
Investments have risk associated with them. Should your risk appetite change based on the interest rates available to you? Lets say the safe investments available to you[7] are giving interest rates of 2%-3%. Would you consider slightly risky investment options which give on average 1% more interest rate? If safe investments are giving around 15%-16% interest rates, would you still consider slightly risky investments which promise 1% more interest rate?

Questions like these are attempted better with an understanding of compound interest. This article goes through some ways of building mental modals about the topic.

If you would like to avoid maths, click here to see less of it. Although, I would suggest, to try. If you would be ok with some maths, click here.You should still be able to make sense of the article. There are some interactive parts in the article. Numbers which show up like can be clicked. It opens up a slider to change the number. And hovering over charts show some details. Also, you can change currency among the options: .

TL: DR;

Invest early and often, with best real interest rates you can get. Let compounding do the rest.

• Compounding gives better results with time. Invest early.
• If the interest rates are same, you can split your money in multiple investments. The return would be the same as putting them in a single investment with that interest rate. Ex, two ₹ 100 investments, or a single investment of ₹ 200, both will give the same returns.
• Think of compounding as multiplication factor. Invest amount $$A$$ and get back $$x*A$$ after some time. Where $$x$$ depends on time and interest rate. You can split the investment and the net effect is same (without consider risk etc). Ex: $$x*(A1 + A2 + A3)$$.
• Keep the rule of 72 in mind: Divide 72 by interest rate, to get doubling time. Helps with estimating returns, in head. Ex, for 7% interest rate, your amount will double every 10 years. No matter what that amount is.
• 1% change in interest rate matters a lot for low interest rates.

If you are new to the concept of investment and compound interest, think of investment as a way to grow your money. And compounding as the effect of investing back the extra money you get. Its a process where every year the interest earned in the investment, is invested back and earns you interest next time. With 7%, interest rate compounded annually the table below shows the returns.
year 1year 2year 3year 4year 5year 6year 7year 8year 9year 10100.00+ 7.00107.00+ 7.49114.49+ 8.01122.50+ 8.58131.08+ 9.18140.26+ 9.82150.07+ 10.51160.58+ 11.24171.82+ 12.03183.85+ 12.87107.00114.49122.50131.08140.26150.07160.58171.82183.85196.72
If you didn't invest back the interest. Your ₹ 100 would have given returns of ₹ 170 (or 100 + 100 * 0.07 * 10). With compounding its 196.72. Compound interest is a crucial part of personal finance. It grows your money!
In this article we wont consider risk, real interest rate, taxes or other factors that might (and do) effect investments[8].

Throught the article, to simplify the arguments, we assume interest rates to be annual. And sometimes work with fractional years. These are approximations, but help with building intuition.

Lets start the equation for compounding. If you invest amount $$P$$ with an annualy compounded interest $$r$$, in $$n$$ years the total amount $$P'$$ would be: $$P' = { P(1 + r)^n }$$

## Multiple

Its difficult to gain understanding of the growth, with this equation. One thing to note though, is that the return $$P'$$ is a multiple of initial investment $$P$$. All the observations in the article depend on this property. In other words, we can write the equation as:$$P'/P = { (1 + r)^n }$$Which essentially means, for same interest rate and span of time, investment grows the same way, irrespective of how much money you invest initially.

In terms of growh rate, it doesn't matter whether you invest ₹ 100 or ₹ 100,000. They both grow in the same way. What matters is the interest rate. Ex, if the interest rate is 7%, in the first year both of them would grow to 1.07 times the initial value. In the second year, both of them would be approximately 1.14. In the third year, both of them would be approximately 1.22 times the initial value. So, for the same interest rate, the more money you put in, the more returns you get.

With interest rate, in years your initial investment would grow approximately times. In other words, ₹ 1,000 would grow to ₹ 0 and ₹ 1,000,000 would grow to ₹ 0.
Another nice property is that there is not difference[9] between splitting the money versus putting it all in a single investment.
Another nice property is that [10] you can split the money into multiple investments and get the same returns . Ex, you could replace $$P$$ with $$(P_1 + P_2 + ... + P_n)$$$$P' = (P_1 + P_2 + ... + P_n){ (1 + r)^n }$$

This helps with managing risks. Investments have risks associated with them. Some of them might fail (or give low returns). Splitting your investments (assuming the interest rates are similar) wouldn't change the returns (mathematically speaking), and allows for better risk management.

## Rule of 72

$$Approximate\ doubling\ period = { 72 \over Interest\ rate }$$This is a way to approximate the doubling time of investment.
With an interest rate of , in NaN years (which is approximately 72/7) your initial investment will double.
Again (I find it interesting and worth repeating), doubling period doesn't depend on how much money you put in.

## Doubling continues

Note that the Rule of 72 gives us a number to compare. But its only one doubling time. Every doubling time years, your investment doubles.

interest has doubling period of approx NaN years. By year NaN (3 doubling years) , your investment would grow to 8 times the initial investment. And if you wait another NaN years, it will grow to 16 times. In case you are all to exited, finding real interest rate of 7% might not be easy.

How much time do you have? Lets say you are 35 now. You probably would earn till 65? Thats 30 years. Plan before its too late.

## Value of 1%

What difference does a 1% change in interest rate make? Hover over the plot below to see the doubling time for interest rate.

Notice how 1% makes a big difference, when the interest rates are low. This was a big surprise to me. Think about how it effects your investment. Try out different values of interest rate in the example below.

Lets say your investment is giving you interest rate and the management fees is . That 1% is increasing your investments doubling period by approximately NaN years.

Lets say the market around you provides investments with interest rates mostly around 3%. With a slightly higher risk, you could find investments with 4%-5% returns. What would you do? Think about how many years it saves in doubling. On the other hand, what if most investment options are around 15%-16%, would you consider 17%-18% interest rate investments with slightly higher risk?

## Value of interest rate

The chart below show return curves for different interest rates. The higher the interest rate, the steaper the curve.

Here is a way to think about it, in terms of doubling period. Lets compare interest rates of (which has doubling period of years) and (which has doubling period ). Difference between their doubling period is NaN. For your money to grow 8 times, requires 3 doubling periods. In other words, 7% investment would take NaN[11] more years, than 10%, to grow your money by 8 times.

## Value of time

Each of the curves below represents a time period (ex: 40, 30, 20 years). x axis is the interest rate, and y the multiple of the investment. Again, the longer you let your investment grow, the bigger the returns. Note how fast the curve grows for high interest rates and long periods.

With interest rate of 12%, after 40 years, your investment would have grown 100 times!

## Value of maths?

Much of what we figured out, is just maths and looking at curves. Lets play around with it a little more.

### Where does 72 come from?

We will attempt a close enough answer. Lets look at our equation again.$$P'/P = { (1 + r)^n }$$Doubling means $$P' = 2*P$$, or $$P'/P = 2$$. In other we need to solve for:$$2 = { (1 + r)^n }$$Taking natural log on both sides.$$ln(2) = { n * ln(1 + r) }$$Or$$n = { ln(2) \over ln(1 + r) }$$Now here comes a cute little trick. Turns out, for small values of $$r$$, $$ln(1 + r)$$ is approximately $$r$$. Ex, $$ln(1.01)$$ its 0.01, for $$ln(1.2)$$ its 0.18, approximately. So, for most pratical interest rates, we could just use:$$n = { ln(2) \over r }$$$$ln(2)$$ turns out to be 69, so doubling time could be equated to (approximately)$$n = { 69 \over r }$$But we have been using 72! Wiki explains the reasons behind choosing 72 (or 70 or 69).

### Slopes

Lets again start with the compound interest equation $$(1 + r)^n$$. For a curve, slope at a point, is one way of figuring out how fast change is happening at that certain point. A point in the curve represents some point in time and the return at that time. With the slope of the curve, if we assume it to be same for a year around the point, we can approximate the rate of return in that year, by calculating the change in return. Ex, for a straight line, a unit change in x axis, would change the y axis by slope.

Derivative gives us the slope curve. Derivative of the compund interest equation:$${ \frac{d(1 + r)^n}{dn} } = { ln(1 + r)(1 + r)^n }$$We learned earlier that for small values of r, $$ln(1+r)$$ could be replaced with $$r$$. So, we could write the above equations as: $${ r(1 + r)^n }$$Now, lets consider what this means.
Lets take an investment with interest rate . What would be the growth of the investment in year ? The investment would have grown to 1.97 the initial amount, at the starting of 10. The slope of the line at the middle of the year, ex 10.5, is $$r(1+r)^{10.5}$$, which is (as percentage) 14.24%. The actual change between year 10 and 11 is 13.77%.
Play around with different interest rates and year values. The approximation is mostly within 1% of the actual value.
This also gives us a quick way to approximate the percentange change in the year when the amount doubles (or quadruples, or other multiples of 2). At the year of the doubling, $$(1+r)^n$$ is 2. So the rate change that year comes out to be $$r*2$$. Ex, for interest rate 7, it would be 0.07 * 2 or 14%. On the year of the second doubling it would be 0.07 * 4 or 28%.

## What now?

Also, its interesting to note that throught the article, all we did was to look at implications of the equation $$P' = P(1 + r)^n$$.