Compounding and Continuous Compounding of Interest
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. What is Percent Increase?
In order to understand compounding, you need to first understand the percentage increase of a quantity.
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)
Simple Interest (not compounded)If P is deposited in a saving account at an interest r, that is not compounded, for t years then the interest I earned after t years is given byI = 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.
Yearly Interest Compounding (Savings Account for Example)An amount of money P (principal) is invested at an annual percentage rate r. What is the total amount of money after t years?
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)
Interest Compounding n Times Per YearHow about compounding more that once a year? Let us say the interest is compounded twice a year (every 6 months) as follows: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)^{ 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)
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?
Continuous CompoundingSo 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 (see above)A = P(1 + r/n)^{ n t} Let N = n / r , then r / n = 1 / N and n = r N , hence the formula for A becomes A = P(1 + 1 / N)^{ N r t} Which can be written as A = P ( (1 + 1 / N)^{ N } ) ^{ r t} 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)^{ N } behaves?
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.
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)
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. |
More References and Links
Compound Interest Problems with Detailed SolutionsPercent Maths Problems
Percent Maths Questions
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