Introduction to Compounding Example
In this compounding example article, we are going to see various examples to understand the different set of compounding defined in the financial markets. It is difficult to come up with examples or practical situations for each and every variation. Hence restricting the examples across monthly compounding, quarterly compounding, semi-annual compounding, and annual compounding
Examples of Compounding
Below are the Examples of the Compounding in Finance :
Compounding Example-1
The period considered for adding interest along with the principal, in this case, is one month. For example, I have a fixed deposit with the principal of Rs. 10,000 and the rate of interest is 8% per annum (Rate of Interest usually depicts as annum). I am opting for a monthly compounding and not planning to withdraw any amount in between for 3 years. In this case, the interest that will be added to the principal each month. This can be depicted as follows:
Consider,
- Initial principal (p) = 10,000
- Rate of Interest (i) = 10% (or) 0.1
- Compounding Frequency Per Year (f) = 12
- Term (y) = 3 years
- Interest for 1st Month = (10000 * 0.1 * 1) = 1000
For the second month, the principal will be:
- = Initial Principal + Interest of the First Month
- = 10,000 + 1000
- = 11,000
In this way, the principal will be compounded each month and at the end of 3 years, the compounded amount will be Amount:
Solution:
(A) = (Initial Principal*(1 + Rate of Interest (in Decimal)/Compounding Frequency (f))˄(f*Term (y))
- = (10000 * (1+ (0.1/12)) ˄ (12*3)
- = 13481.81842
Compounding Example -2
Let’s have a case that as part of person X financial planning, she is in need of Rs. 1,00,000 in 3 years. This is when her child will be starting her Higher studies. She is checking for a mutual fund yielding 5% interest compounded quarterly. She wanted to know what would be the investment amount to attain the amount
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The Rate of interest is compounding every quarter, so f = 4. Based on the case given, we got all variables except the initial principal (p). hence on applying all values except P in our formula:
Consider,
- (A) = 1,00,000
- Rate of Interest (i) = 5%, (or) 0.05.
- Compounding Frequency Per Year (f) = 4
- Term (y) = 3 years
Solution:
(A) = (Initial Principal*(1 + Rate of Interest (in Decimal)/Compounding Frequency (f))˄(f*Term (y))
- 1,00,000 = (p * (1+ (0.05/4)^(4*3))
- 1,00,000 = (p*(1.0125)^12)
The logic at this step is to move all values except P to the other side.
- 1,00,000/(1.0125)^12 = p
Hence p = 1,00,000/(1.0125)^12
- = 1,00,000/1.160
- = 86150.87
Person X has to invest about Rs. 86150.87
Compounding Example -3
As we are aware, compounding can be done in different frequencies like daily compounding, monthly compounding, quarterly compounding, half-yearly compounding, yearly compounding or continuous compounding. The shorter the compounding frequency, the more the outcome. We can understand this with an example
Sathya wants to invest in two different types of mutual funds for a tenure of 5 years. Mutual fund A has a return of 8% which is compounded quarterly. Mutual fund B has a return of 8% (same as Mutual fund A) which is compounded semi-annually. He invests Rs.10,000 in both mutual funds. We will see how the amount is compounded in both mutual funds:
Mutual fund A
- Initial Principal (p) = 10,000
- Rate of Interest (i) = 8% (or) 0.08
- Compounding Frequency Per Year (f) = 4
- Term (y) = 5 years
Solution:
(A) = (Initial Principal*(1 + Rate of Interest (in Decimal)/Compounding Frequency (f))˄(f*Term (y))
- = (10000 * (1+ (0.08/4)) ˄ (4*5)
- = 14859.47
Mutual fund B
- Initial Principal (p) = 10,000
- Rate of Interest (i) = 8% (or) 0.08
- Compounding Frequency Per Year (f) = 2
- Term (y) = 5 years
Solution:
(A) = (Initial Principal*(1 + Rate of Interest (in Decimal)/Compounding Frequency (f))˄(f*Term (y))
- = (10000 * (1+ (0.08/2)) ˄ (2*5)
- = 14802.44
When the compounding frequency is increased, the return is substantial. So here on a comparison, between mutual fund A and mutual fund B, mutual fund A gives more returns as the compounding frequency is more when compared to mutual fund B.
Compounding Example -4
Let’s now try to apply to the compound to a practical example. In a city, the population as of today is 280000. Based on a survey, we know that there is an increase in populate rate as 5% per year. We want to know the population after 4 years.
How can we do it? First, let’s identify the parameters for compounding here. The population as of today will be equal to the initial principal (p) = 2,80,000. The compounding frequency here will be annual. Hence f = 1.
Consider,
- initial Principal (p) = 2,80,000
- Rate of Interest (i) = 5% (or) 0.05
- Compounding Frequency Per Year (f) = 1
- Term (y) = 4.
Solution:
Let’s apply the compounding formula, to identify the population after 4 years:
(A) = (Initial Principal*(1 + Rate of Interest (in Decimal)/Compounding Frequency (f))˄(f*Term (y))
- = (2,80,000 * (1+ (0.05/1)) ˄ (1*4)
- = 3,40,341
Hence the population after 4 years will be 3,40,341.
Conclusion – Compounding Example
As far as we know, compounding can be applied for many practical examples in various areas like finance, mutual funds, fixed deposits and to identify population. In the financial world, experts prefer to invest more in compounding with more compounding frequencies. It will benefit more when compared to any other rate of interest. This is also flexible in terms of frequency as in many mutual funds customers will allow choosing the frequency based on their ability to pay the amount. The compounded amount will increase, the more the amount is compounding for frequency.
Recommended Articles
This has been a guide to the Compounding Example. Here we understand the power of compounding with the help of practical examples. You may also have a look at the following articles to learn more –
- Fixed Costs Example
- Variable Costing Example
- Quantitative Research Example
- Monopolistic Competition Examples
