EXAMPLE 1.15 - Simple and Compound Interest
Assume an engineering company borrows $100,000 at 10% per year compound interest and will pay the principal and all the interest after 3 years. Compute the annual interest and total amount due after 3 years. Graph the interest and total owed for each year, and compare with the previous example that involved simple interest.
Solution
To include compounding of interest, the annual interest and total owed each year are calculated
by Equation [1.8].
The repayment plan requires no payment until year 3 when all interest and the principal, a total of $133,100, are due. Figure 1–11 uses a cash fl ow diagram format to compare end-of-year (a) simple and (b) compound interest and total amounts owed. The differences due to compounding are clear. An extra $133,100 – 130,000 = $3100 in interest is due for the compounded interest loan.
Note that while simple interest due each year is constant, the compounded interest due grows geometrically. Due to this geometric growth of compound interest, the difference between simple and compound interest accumulation increases rapidly as the time frame increases. For example, if the loan is for 10 years, not 3, the extra paid for compounding interest may be calculated to be $59,374.
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| Figure 1–11 Interest I owed and total amount owed for ( a ) simple interest (Example 1.14) and ( b ) compound interest (Example 1.15). |
A more effi cient way to calculate the total amount due after a number of years in Example 1.15 is to utilize the fact that compound interest increases geometrically. This allows us to skip the year-by-year computation of interest. In this case, the total amount due at the end of each year is
This allows future totals owed to be calculated directly without intermediate steps. The general form of the equation is
where i is expressed in decimal form. Equation [1.10] was applied above to obtain the $133,100 due after 3 years. This fundamental relation will be used many times in the upcoming chapters.
We can combine the concepts of interest rate, compound interest, and equivalence to demonstrate that different loan repayment plans may be equivalent, but differ substantially in amounts paid from one year to another and in the total repayment amount. This also shows that there are many ways to take into account the time value of money.
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