What is simple interests and interest compounding? Full explanations with examples and easy to use online calculators to experiment with different values of the parameters involved in order to understand the different formulas related to interests compounding are presented. Problems on Compound interests with detailed solutions are also included in this site.

In order to understand compounding, you need to first understand the percentage increase of a quantity.

If P is a quantity that is increased by a percentage rate r, then the new quantity is P + r P

You need to retain the above:

A quantity P increased by a percentage rate r becomes P + r P = P ( 1 + r)

Example 1: 200 increased by 5% becomes200 + (5/100)*200 = 200(1 + 0.05)

I = P r t

and the total amount A in the account after t years is given by

A = P + P r t = P(1 + r t)

Example 1: $100 in a 3% interest rate saving account, not compounded, would earns 100*(3/100)*5 = $15 in 5 years.

- At t = 0 , A = P (the day money is deposited)

- At the end of the first year t = 1 , A = P + r P = P(1 + r)

we have used what was explained above: A quantity P increased by a percentage r becomes P + r P = P ( 1 + r)

- At the end of the second year t = 2 , A = P(1 + r) + r P(1 + r)

we have used what was explained above: A quantity P ( 1 + r) (at the end of year 1) increased by by a percentage r becomes P ( 1 + r) + r P ( 1 + r)

which gives

A = (P + r P)(1 + r) = P(1 + r )(1 + r) = P(1 + r)^{ 2}

- At the end of the third year t = 3 , A = P(1 + r)
^{ 2}+ r P(1 + r)^{ 2}= P(1 + r)^{ 2}(1 + r) = P(1 + r)^{ 3}

- At the end of the fourth year t = 4 , A = P(1 + r)
^{ 3}+ rP(1 + r)^{ 3}= P(1 + r)^{ 3}(1 + r) = P(1 + r)^{ 4}

- By extension of the above, at the end of year t,

A = P(1 + r)

So if an amount P (principal) is invested at the annual rate r and is compounded annually, the total amount A at the end of t years is given by^{ t}

A = P(1 + r)^{ t}

Example 2: $1000 is invested for 3 years, compounded every year, at the rate of 3%. The amount A (rounded to the nearest cent) at the end of 3 years is shown on the calculator below.(click on Enter)

A = P(1 + r)^{ t} = 1000(1 + 0.03)^{ 3} = $1092.73

You may use the calculator to input and experiment with more values for P, r and t and obtain the amount A. Use your own calculator and compare the results.

Yearly rate is r; set a half yearly rate equal to r/2 and compound twice a year as follows:

t = 0 , A = P

At the end of the first 6 months of the year: A = P(1 + r/2)

At the end of the second 6 months of the same year: A = P(1 + r/2) + (r/2) P(1 + r/2) = P(1 + r/2)

We now extend the above to write:

So if an amount P (principal) is invested at the annual rate r and is compounded n times a year , the amount at the end of t years is given by

A = P(1 + r/n)^{ n t}

Example 3: $1000 is invested for 3 years, compounded twice a year (n = 2), at the rate of 3%. The amount A (rounded to the nearest cent) at the end of 3 years is shown on the calculator below.(click on Enter)

A = P(1 + r/n)^{ n t} = 1000(1 + 0.03 / 2)^{ 2×3} = $1093.44

You may use the calculator to input and experiment with more values for P, r, t and n and obtain the amount A. Use your own calculator and compare the results.

To understand the advantage of Compounding more than once a year, Keep P, r and t constant (The same amount invested at the rate r for t years) and increase n. What happens to A?

A = P(1 + r/n)

Let N = n / r , then r / n = 1 / N and n = r N , hence the formula for A becomes

A = P(1 + 1 / N)

Which can be written as

A = P ( (1 + 1 / N)

The question that one may ask is that what if we increase n indefinitely, which means increasing N indefinitely in our formula?

Use the calculator below to increase N ( 2 ,12, 1000, 10000...) and see how the value of the term (1 + 1 / N)

As the number of compounding n increases, N also increases, the term (1 + 1 / N)^{ N } approaches a constant value which is called e and is approximately equal 2.718282... . More rigorously, e is defined as the limit of (1 + 1/N)^{ N} as N approaches infinity.

Hence for continuous compounding (n very large) at the rate r and an initial amount P and after t years A is given by:

A = P e ^{ r t}

Example 4: $1000 is invested for 3 years, compounded continuously, at the rate of 3%. The amount A (rounded to the nearest cent) at the end of 3 years is shown on the calculator below.(click on Enter)

A = P e ^{ r t} = 1000 e ^{ 0.03 × 3} = $1094.17

You may use the calculator to input and experiment with more values for P, r, and obtain the amount A. Use your own calculator and compare the results.

Conclusion: compare the way the same amount of $1000 was compounded in the eaxamples 1,2 and 4 and make a conclusion as to which compounding earns more.