# Does more frequent compounding result in additional return on the investment?

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Yes it does.

Interests with different compounding frequency can be compared with each other by determining their corresponding effective annual rate.

Using the formula of effective annual rate r,

r = [ 1 + (i/n)]^n - 1

r increases exponentially when the number of compounding periods, *n*, increases.

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To illustrate, let's try this problem.

John wants to invest his $2000 on a plan. He receives two offers which are:

A - 10% annual percentage, compounded semi-annually

B - 10% annual percentage, compounded quarterly.

Among the two, which plan will give John a higher interest after 2 years?

Solution:

>> To determine which investment plan is better, calculate the effective annual interest of each.

For A,

r = [1+(i/n)]^n - 1 = [1+ (.10/2)]^2 - 1 = 1.05^2 - 1

r = 1.1025 - 1 = 0.1025 = 10.25%

Effective annual rate of A is 10.25%

For B,

r = [1+(i/n)^n - 1 = [1+(.10/4)]^4 - 1 = 1.025^4 - 1

r = 1.1038 - 1 = 0.1038 = 10.38%

Effective annual rate of B is 10.38%.

*Since effective annual rate of B is greater than A, hence B is a better investment plan. B will yield a higher interest than A.*

>> The other solution is by computing the future value of each plan after 2 years, using the compounding interest formula.

For A,

F = P [ 1 + (i/n)]^(nt) = 2000 [ 1 + (.10/2)]^(2*2)

F = 2000(1.05)^4 = 2000(1.2155) = 2431

So, after 2 years, John money will increase from $2000 to $2431. The total interest earned is $431.

For B,

F = P [ 1+ (i/n)]^(nt) = 2000 [ 1 + (.10/4)]^(4*2)

F = 2000(1.025)^8 = 2000(1.2184) = 2436.8

John money will increase from $2000 to $2436.8. The total interest earned for 2 years is $436.8.

*Hence, plan B will give John a higher interest earned.*

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Take note that in the above sample problem, B has a higher number of compounding periods than A. When solved, B yields a higher interest.

**Thus, an increase in number of compounding periods in a year result to an additional return on investment.**