Learning Objectives for Section 3.2

1. Learning Objectives for Section 3.2 Compound Interest After this lecture, you should be able to • Compute compound interest. • Compute the annual percentage yield of a compound interest investment.

2. Compound Interest • Compound interest: Interest paid on interest reinvested. • Compound interest grows faster than simple interest. • Annual nominal rates: How interest rates are generally quoted • Rate per compounding period:

3. Compounding Periods The number of compounding periods per year (m): • If the interest is compounded annually, then m = _______ • If the interest is compounded semiannually, then m = _______ • If the interest is compounded quarterly, then m = _______ • If the interest is compounded monthly, then m = _______ • If the interest is compounded daily, then m = _______

4. Example 1: Suppose a principal of $1 was invested in an account paying 6% annual interest compounded monthly. How much would be in the account after one year? Example

5. Solution • Solution: Using the Future Value using simple interest formula A = P (1 + rt) we obtain: • amount after one month • after two months • after three months After 12 months, the amount is (1.005)12 = 1.0616778. With simple interest, the amount after one year would be 1.06. The difference becomes more noticeable after several years.

6. Graphical Illustration ofCompound Interest The growth of $1 at 6% interest compoundedmonthly compared to 6% simple interest over a 15-year period. The blue curve refers to the $1 invested at 6% simple interest. Dollars The red curve refers to the $1 at 6% being compounded monthly. Time (in years)

7. General Formula: Compound Interest • The formula for calculating the Amount using Compound Interest is Where A is the future amount, P is the principal, r is the annual interest rate as a decimal, m is the number of compounding periods in one year, and tis the total number of years.

8. Simplified Formula: Compound Interest • The formula for calculating the Amount: Compound Interest is To simplify the formula, let We now have,

9. Example 2: Find the amount to which $1,500 will grow if compounded quarterly at 6.75% interest for 10 years. Then, compare it to the amount if the interest was figured using the simple interest formula. Example

10. Example • Find the amount to which $1500 will grow if compounded quarterly at 6.75% interest for 10 years. • Solution: Use • Helpful hint: Be sure to do the arithmetic using the rules for order of operations.

11. Same Problem Using Simple Interest Using the simple interest formula, the amount to which $1500 will grow at an interest of 6.75% for 10 years is given byA = P (1 + rt) = 1,500(1 + 0.0675(10)) = $2,512.50 which is more than $400 less than the amount earned using the compound interest formula.

12. Changing the number of compounding periods per year Example 3: To what amount will $1,500 grow if compounded daily at 6.75%interest for 10 years?

13. Changing the number of compounding periods per year To what amount will $1,500 grow if compounded daily at 6.75%interest for 10 years? Solution: = $2,945.87 This is about $15.00 more than compounding $1,500 quarterly at 6.75% interest. Since there are 365 days in year (leap years excluded), the number of compounding periods is now 365. We divide the annual rate of interest by 365. Notice, too, that the number of compounding periods in 10 years is 10(365)= 3650.

14. Effect of Increasing the Number of Compounding Periods • If the number of compounding periods per year is increased while the principal, annual rate of interest and total number of years remain the same, the future amount of money will increase slightly.

15. Example 4: Suppose a house that was worth $68,000 in 1987 is worth $104,000 in 2004. Assuming a constant rate of inflation from 1987 to 2004, what is the inflation rate? Computing the Inflation Rate

16. Suppose a house that was worth $68,000 in 1987 is worth $104,000 in 2004. Assuming a constant rate of inflation from 1987 to 2004, what is the inflation rate? 1. Substitute in compound interest formula. 2. Divide both sides by 68,000 3. Take the 17th root of both sides of equation 4. Subtract 1 from both sides to solve for r. Solution: Computing the Inflation RateSolution r = 2.53%

17. Example 5: If the inflation rate remains the same for the next 10 years, what will the house be worth in the year 2014? Computing the Inflation Rate(continued )

18. If the inflation rate remains the same for the next 10 years, what will the house be worth in the year 2014? Solution: From 1987 to 2014 is a period of 27 years. If the inflation rate stays the same over that period, r = 0.0253. Substituting into the compound interest formula, we have Computing the Inflation Rate(continued )

19. Example Example 6: If $20,000 is invested at 4% compounded monthly, what is the amount after a) 5 years b) 8 years?

20. Which is Better? Example 7: Which is the better investment and why: 8% compounded quarterly or 8.3% compounded annually?

21. Inflation Example 8: If the inflation rate averages 4% per year compounded annually for the next 5 years, what will a car costing $17,000 now cost 5 years from now?

22. Investing Example 9: How long does it take for a $4,800 investment at 8% compounded monthly to be worth more than a $5,000 investment at 5% compounded monthly?

23. Annual Percentage Yield • The simple interest rate that will produce the same amount as a given compound interest rate in 1 year is called the annual percentage yield (APY). To find the APY, proceed as follows: Amount at simple interest APY after one year = Amount at compound interest after one year This is also called the effective rate.

24. Annual Percentage Yield Example What is the annual percentage yield for money that is invested at 6% compounded monthly?

25. Annual Percentage Yield Example What is the annual percentage yield for money that is invested at 6% compounded monthly? General formula: Substitute values: Effective rate is 0.06168 = 6.168