learning objectives for section 3.2

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

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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. Compound Interest. Compound interest: Interest paid on interest reinvested. - PowerPoint PPT Presentation

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Page 1: Learning Objectives for Section 3.2

1

Learning Objectives for Section 3.2

After this lecture, you should be able to Compute compound interest. Compute the annual percentage yield of a compound

interest investment.

Compound Interest

Page 2: Learning Objectives for Section 3.2

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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: annual nominal rate

# of compounding periods per year

Page 3: Learning Objectives for Section 3.2

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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 = _______

Page 4: Learning Objectives for Section 3.2

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Example

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?

Page 5: Learning Objectives for Section 3.2

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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

2

2 2 3

0.061 1 1 1 0.005 (1.005)

12

0.061.005 1 1(1.005)(1.005) 1.005

12

0.061.005 1 1.005 1.005 1.005

12

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.

Page 6: Learning Objectives for Section 3.2

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Graphical Illustration ofCompound Interest

The growth of $1 at 6% interest compounded monthly compared to 6% simple interest over a 15-year period.

The blue curve refers to the $1 invested at 6% simple interest.

The red curve refers to the $1 at 6% being compounded monthly.

Time (in years)

Dol

lars

Page 7: Learning Objectives for Section 3.2

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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

t is the total number of years.

General Formula: Compound Interest

1mt

rA P

m

Page 8: Learning Objectives for Section 3.2

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The formula for calculating the Amount: Compound Interest is

To simplify the formula, let

We now have,

Simplified Formula: Compound Interest

1mt

rA P

m

1n

A P i

ri and n mt

m

Page 9: Learning Objectives for Section 3.2

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Example

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.

Page 10: Learning Objectives for Section 3.2

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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.

1n

A P i 10(4)

0.06751,500 1

4

$2,929.50

A

A

Page 11: Learning Objectives for Section 3.2

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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 by

A = 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.

Page 12: Learning Objectives for Section 3.2

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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?

Page 13: Learning Objectives for Section 3.2

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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.

10(365)0.0675

1500 1365

A

Page 14: Learning Objectives for Section 3.2

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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.

Page 15: Learning Objectives for Section 3.2

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Computing the Inflation Rate

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?

Page 16: Learning Objectives for Section 3.2

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Computing the Inflation RateSolution

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:

17

17

17

17

104,000 68,000 1

104,0001

68,000

104,000(1 )

68,000

104,0001 0.0253

68,000

r

r

r

r

r = 2.53%

Page 17: Learning Objectives for Section 3.2

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Computing the Inflation Rate(continued )

Example 5: If the inflation rate remains the same for the next 10 years, what will the house be worth in the year 2014?

Page 18: Learning Objectives for Section 3.2

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Computing the Inflation Rate(continued )

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

2768,000(1 0.0253) $133,501A

Page 19: Learning Objectives for Section 3.2

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Example

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

Page 20: Learning Objectives for Section 3.2

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Which is Better?

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

Page 21: Learning Objectives for Section 3.2

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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?

Page 22: Learning Objectives for Section 3.2

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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?

Page 23: Learning Objectives for Section 3.2

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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:

(1 ) 1

1 1

1 1

m

m

m

rP APY P

m

rAPY

m

rAPY

m

This is also called the effective rate.

Amount at simple interest APY after one year = Amount at compound interest after one year

Page 24: Learning Objectives for Section 3.2

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Annual Percentage Yield Example

What is the annual percentage yield for money that is invested at

6% compounded monthly?

Page 25: Learning Objectives for Section 3.2

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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

1 1m

rAPY

m

120.06

1 1 0.0616812

APY