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Section 13.1Simple and Compound Interest

- Goals
- Study simple interest
- Calculate interest
- Calculate future value
- Study compound interest
- Calculate future value
- Compare interest rates
- Calculate effective annual rate

13.1 Initial Problem

- Suppose you discover you are the only direct descendant of a man who loaned the Continental Congress $1000 in 1777 and was never repaid.
- Using an interest rate of 6% and a compounding period of 3 months, how much should you demand from the government?
- The solution will be given at the end of the section.

Simple Interest

- If P represents the principal, r the annual interest rate expressed as a decimal, and t the time in years, then the amount of simple interest is:

Example 1

- Find the interest on a loan of $100 at 6% simple interest for time periods of:
- 1 year
- 2 years
- 2.5 years

Example 1, cont’d

- Solution: We have P = 100 and r = 0.06.
- For t = 1 year, the calculation is:

Example 1, cont’d

- Solution, cont’d: We have P = 100 and r = 0.06.
- For t = 2 years, the calculation is:
- For t = 2.5 years, the calculation is:

Future Value

- For a simple interest loan, the future value of the loan is the principal plus the interest.
- If P represents the principal, I the interest, r the annual interest rate, and t the time in years, then the future value is:

Example 2

- Find the future value of a loan of $400 at 7% simple interest for 3 years.

Example 2, cont’d

- Solution: Use the future value formula with P = 400, r = 0.07, and t = 3.

Example 3

- In 2004, Regular Canada Savings Bonds paid 1.25% simple interest on the face value of bonds held for 1 year.
- If the bond is cashed early, the investor receives the face value plus interest for every full month.
- Suppose a bond was purchased for $8000 on November 1, 2004.

Example 3, cont’d

- What was the value of the bond if it was redeemed on November 1, 2005?
- What was the value of the bond if it was redeemed on July 10, 2004?

Example 3, cont’d

- Solution: If the bond was redeemed on November 1, 2005, it had been held for 1 year.
- The future value of the bond after 1 year is:

Example 3, cont’d

- Solution: If the bond was redeemed on July 10, 2004, it had been held for 7 full months.
- The future value of the bond after 7/12 of a year is:

Example 4

- What is the simple interest on a $500 loan at 12% from June 6 through October 12 in a non-leap year?

Example 4, cont’d

- Solution: The time must be converted to years.
- (30 - 6) + 31 + 31 + 30 + 12 = 128 days
- A non-leap year has 365 days.
- The interest will be:

Question:

What is the simple interest on a $2000 loan at 8% from March 19th through August 15th in a leap year?

a. $65.32

b. $653.15

c. $651.37

d. $65.14

Ordinary Interest

- Ordinary interest simplifies calculations by using 2 conventions:
- Each month is assumed to have 30 days.
- Each year is assumed to have 360 days.

Example 5

- A homeowner owes $190,000 on a 4.8% home loan with an interest-only option.
- An interest-only option allows the borrower to pay only the ordinary interest, not the principal, for the first year.
- What is the monthly payment for the first year?

Example 5, cont’d

- Solution: Use the simple interest formula, measuring time according to ordinary interest conventions.
- The monthly payments are:

Compound Interest

- Reinvesting the interest, called compounding, makes the balance grow faster.
- To calculate compound interest, you need the same information as for simple interest plus you need to know how often the interest is compounded.

Example 6

- Suppose a principal of $1000 is invested at 6% interest per year and the interest is compounded annually.
- Find the balance in the account after 3 years.

Example 6, cont’d

- Solution: We must calculate the interest at the end of each year and then add that interest to the principal.
- After 1 year:
- The interest is:
- The new balance is $1060.00
- We could also have used the future value formula.

Example 6, cont’d

- Solution, cont’d:
- After 2 years the new balance is:
- After 3 years the new balance is:

Example 6, cont’d

- Solution, cont’d: The interest earned each year increases because of the increasing principal.

Example 6, cont’d

- Solution, cont’d: The following table shows the pattern in the calculations for subsequent years.

Compound Interest, cont’d

- If P represents the principal, r the annual interest rate expressed as a decimal, m the number of equal compounding periods per year, and t the time in years, then the future value of the account is:

Example 7

- Find the future value of each account at the end of 3 years if the initial balance is $2457 and the account earns:
- 4.5% simple interest.
- 4.5% compounded annually.
- 4.5% compounded every 4 months.
- 4.5% compounded monthly.
- 4.5% compounded daily.

Example 7, cont’d

- Solution: We have P = 2457 and t = 3.
- We have r = 0.045 with simple interest.
- We have r = 0.045 compounded annually.

Example 7, cont’d

- Solution, cont’d: We have r = 0.045
- Compounded every 4 months:
- Compounded monthly:
- Compounded daily:

Example 7, cont’d

- Solution, cont’d: The results are summarized below.

Example 8

- Find the future value of each account at the end of 100 years if the initial balance is $1000 and the account earns:
- 7.5% simple interest.
- 7.5% compounded annually.

Example 8, cont’d

- Solution: We have P = 1000, t = 100, and r = 0.075.
- With simple interest, the future value is:
- With annually compounded interest, the future value is:

Question:

If you loan your friend $100 at 3% interest compounded daily, how much will she owe you at the end of 1 year?

a. $103.05

b. $103.00

c. $103.33

d. $103.02

Interest, cont’d

- Simple interest exhibits arithmetic growth.
- The same amount is added each year.
- Compound interest exhibits exponential growth.
- The same amount is multiplied each year.

Example 9

- How much money should be invested at 4% interest compounded monthly in order to have $25,000 eighteen years later?

Example 9, cont’d

- Solution: We know F = 25,000,

r = 0.04, m = 12 and t = 18.

- Solve for P, the necessary principal.

Effective Annual Rate

- The effective annual rate (EAR) or annual percentage yield (APY) is the simple interest that would give the same result in 1 year.
- The stated rate is called the nominal rate.
- This provides a basis for comparing different savings plans.
- APY is used only for savings accounts.
- EAR is used in any context.

Example 10

- Find the effective annual rate by computing what happens to $100 over 1 year at 12% annual interest compounded every 3 months.

Example 10, cont’d

- Solution: The balance at the end of 1 year is:
- Since the account increased by $12.55 in 1 year, the EAR is 12.55%.

Effective Annual Rate, cont’d

- If r represents the annual interest rate expressed as a decimal and m is the number of equal compounding periods per year, then the effective annual rate is:
- Note: The same formula is used for APY.

Example 11

- A bank offers a savings account with an interest rate of 0.25% compounded daily, with a minimum deposit of $100.
- The same bank offers an 18-month CD with an interest rate of 2.13% compounded monthly, with deposits less than $10,000.
- Find the effective annual rate for each option.

Example 11, cont’d

- Solution, cont’d: The effective annual rate for the savings account is:
- The EAR for the account is about 0.2503%.

Example 11, cont’d

- Solution: The effective annual rate for the CD is:
- The EAR for the CD is about 2.1509%.

13.1 Initial Problem Solution

- Suppose you discover you are the only direct descendant of a man who loaned the Continental Congress $1000 in 1777 and was never repaid.
- Using an interest rate of 6% and a compounding period of 3 months, how much should you demand from the government?

Initial Problem Solution, cont’d

- We have P = $1000, r = 0.06, m = 4 and t = 223.
- The value of your ancestor’s loan is:

Section 13.2Loans

- Goals
- Study amortized loans
- Use an amortization table
- Use the amortization formula
- Study rent-to-own

13.2 Initial Problem

- Home mortgage rates have decreased and Howard plans to refinance his home.
- He will refinance $85,000 at either 5.25% for 15 years or 5.875% for 30 years.
- In each case, what is his monthly payment and how much interest will he pay?
- The solution will be given at the end of the section.

Simple Interest Loans

- The interest on a simple interest loan is simple interest on the amount currently owed.
- The simple interest each month is called the finance charge.
- Finance charges are calculated using an average daily balance and a daily interest rate.

Example 1, cont’d

- Assuming the billing period is June 10 through July 9, determine each of the following:
- The average daily balance
- The daily percentage rate
- The finance charge
- The new balance

Example 1, cont’d

- Solution: The daily balances are shown below.

Example 1, cont’d

- Solution, cont’d: The average daily balance is:

Example 1, cont’d

- Solution: The daily percentage rate is:
- Solution: The finance charge is the simple interest on the average daily balance at the daily rate:

Example 1, cont’d

- Solution: The new balance is the sum of the previous balance, any new charges, and the finance charge, minus any payments:

287.84 + 144.10 + 4.33 – 150.00 = 286.27

- The new balance is $286.27.

Amortized Loans

- Amortized loans are simple interest loans with equal periodic payments over the length of the loan.
- The important variables for an amortized loan are:
- Principal
- Interest rate
- Term (length) of the loan
- Monthly payment

Amortized Loans, cont’d

- Each payment includes the interest due since the last payment and an amount paid toward the balance.
- The amount paid each month is constant, but the split between principal and interest varies.
- The amount of the last payment may be slightly more or less than usual.

Question:

When paying off an amortized loan, the percent of the monthly payment going toward the interest will:

a. increase as time goes by.

b. decrease as time goes by.

c. remain the same every month.

Example 2

- Chart the history of an amortized loan of $1000 for 3 months at 12% interest with monthly payments of $340.

Example 2, cont’d

- Solution: Monthly payment #1:
- The interest owed is
- The payment toward the principal is

$340 - $10 = $330

- The new balance is $1000 - $330 = $670.

Example 2, cont’d

- Solution, cont’d: Monthly payment #2:
- The interest owed is
- The payment toward the principal is

$340 - $6.70 = $333.30

- The new balance is $670 - $333.30 = $336.70

Example 2, cont’d

- Solution, cont’d: Monthly payment #3:
- The interest owed is
- The remaining balance plus the interest is: $336.70 + $3.37 = $340.07.
- The third and final payment is $340.07.

Example 2, cont’d

- Solution, cont’d: The amortization schedule for this loan is shown below.

Monthly Payments

- The monthly payments for an amortized loan can be determined in one of three ways:
- Using an amortization table.
- Using a formula.
- Using financial software or an online calculator.

Amortization Table

- An amortization table gives pre-calculated monthly payments for common loan rates and terms.
- An example of a table is shown on the next slide.

Example 3

- A couple is buying a vehicle for $20,995.
- They pay $7000 down and finance the remainder at an annual interest rate of 4.5% for 48 months.
- Use the amortization table to determine their monthly payment.

Example 3, cont’d

- Solution: The amount being financed is $20,995 – $7000 = $13,995.
- In the table, find the row corresponding to 4.5% and the column corresponding to 4 years.
- This entry is highlighted on the next slide.

Example 3, cont’d

- Solution, cont’d: The value 22.803486 indicates the couple will pay $22.803486 for each $1000 they borrowed.
- They will pay $319.14 per month.

Example 4

- The couple in the previous example borrowed $13,995 to buy a car and will pay the loan over 4 years.
- If their payments are $340.02, what interest rate are they being charged?

Example 4, cont’d

- Solution: They are paying $340.02 a month for $13,995, or approximately $24.295820 per $1000.
- Look at the amortization table to see to which interest rate this payment amount corresponds.

Example 4, cont’d

- Solution, cont’d: The interest rate for the amortized loan, according to the table, is approximately 7.75%.

Amortization Formula

- If P is the amount of the loan, r is the annual interest rate expressed as a decimal, and t is the length of the loan in years, then the monthly payment for an amortized loan is:

Example 5

- Use the monthly payment formula to determine the monthly car payment for a loan of $13,995 at 4.5% annual interest for 48 months.

Example 6

- Suppose a student accumulated $7800 in student loans which she must now pay over 10 years.
- Determine her monthly payment amount using an interest rate of 3.37%

Question:

Use the amortization formula to determine the amount of the monthly payment for a loan of $30,000 at 5% for 3 years.

a. $527.38

b. $124.00

c. $722.46

d. $899.13

Example 7

- Suppose you can afford car payments of $250 per month.
- If a 3-year loan at 4% interest is available, how much can you finance?

Example 7, cont’d

- Solution, cont’d: Solve for P.
- You can borrow $8468.

Rent-to-Own

- In a rent-to-own transaction, you rent the item at a monthly rate, but after a contracted number of payments, the item becomes yours.
- The difference between the retail price of the item and the total of your monthly payments is the interest.

Example 8

- Suppose you can rent-to-own a $500 television for 24 monthly payments of $30.
- What amount of interest would you pay for the rent-to-own television?
- What annual rate of simple interest on $500 for 24 months yields the same amount of interest found in part (a)?

Example 8, cont’d

- Solution:
- The total of your monthly payments will be 24($30) = $720.
- You will pay $720 - $500 = $220 in interest over the 2 years.

Example 8, cont’d

- Solution:
- Solve the simple interest formula for r:
- The equivalent simple interest rate is:

13.2 Initial Problem Solution

- Home mortgage rates have decreased and Howard plans to refinance his home. He will refinance $85,000 at either 5.25% for 15 years or 5.875% for 30 years.
- In each case, what is his monthly payment and how much interest will he pay?

Initial Problem Solution, cont’d

- The 15-year loan has an interest rate of 5.25%.
- According to the amortization table, the monthly payment per $1000 would be $8.038777.
- Under this loan, Howard’s monthly payment would be $8.038777(85) which is approximately $683.30.

Initial Problem Solution, cont’d

- For the 15-year loan, Howard will pay a total of ($683.30)(12)(15) = $122,994.
- The amount spent on interest is $122,994 - $85,000 = $37,994.

Initial Problem Solution, cont’d

- The 30-year loan has an interest rate of 5.875%, which is not found in the table.
- Using the amortization formula, we find a monthly payment amount of $502.81.

Initial Problem Solution, cont’d

- For the 30-year loan, Howard will pay a total of ($502.81)(12)(30) = $181,011.60.
- The amount spent on interest is $181,011.60 - $85,000 = $96,011.60

Section 13.3Buying a House

- Goals
- Study affordability guidelines
- Study mortgages
- Interest rates and closing costs
- Annual percentage rates
- Down payments

13.3 Initial Problem

- Suppose you have saved $15,000 toward a down payment on a house and your total yearly income is $45,000. What is the most you could afford to pay for a house?
- Assume you pay 0.5% of the value for insurance, you pay 1.5% of the value for taxes, your closing costs will be $2000, and you can obtain a fixed-rate mortgage for 30 years at 6% interest.
- The solution will be given at the end of the section.

Affordability Guidelines

- The 2 most common guidelines for buying a house are:
- The maximum house price is 3 times your annual gross income.
- Your maximum monthly housing expenses should be 25% of your gross monthly income.

Example 1

- If your annual gross income is $60,000, what do the guidelines tell you about purchase price and monthly expenses for your potential home purchase?

Example 1, cont’d

- Solution:
- The purchase price should be no more than 3($60,000) = $180,000.
- The monthly expenses for mortgage payments, property taxes, and homeowner’s insurance should be no more than

Affordability Guidelines, cont’d

- Some lenders allow monthly expenses up to 38% of the buyer’s monthly income.
- We call the 25% level the low maximum monthly housing expense estimate.
- We call the 38% level the high maximum monthly housing expense estimate.

Example 2

- Suppose Andrew and Barbara both have jobs, each earning $24,000 a year, and they have no debts.
- What are the low and high estimates of how much they can afford to pay for monthly housing expenses?

Example 2, cont’d

- Solution: The low estimate is 25% of the total monthly income.
- The high estimate is 38% of the total monthly income.

Question:

If you make $28,000 per year, can you afford to buy a house for $83,000 with monthly housing expenses of $650?

a. Yes, according to the low maximum guideline.

b. Yes, according to the high maximum guideline.

c. No

Mortgages

- A mortgage is a loan that is guaranteed by real estate.
- The interest rate of a fixed-rate mortgage is set for the entire term.
- The interest rate of an adjustable-rate mortgage (ARM) can change.

Mortgages, cont’d

- The finalizing of a house purchase is called the closing.
- Points are fees paid to the lender at the time of the closing.
- Loan origination fees
- Discount charges
- Points and any other expenses paid at the time of the closing are called closing costs.

Example 3

- Suppose you will borrow $80,000 for a home at 6.5% interest on a 30-year fixed-rate mortgage.
- The loan involves a one-point loan origination fee and a one-point discount charge. What are your added costs?
- Note: One point is equal to 1 percent of the loan amount.

Example 3, cont’d

- Solution: Each fee will cost you 1% of $80,000, or $800.
- Your total added fees are $1600.

Annual Percentage Rate

- The annual percentage rate (APR) helps borrowers compare the true cost of a loan.
- The APR includes the annual interest rate, any points, and any other loan processing or private mortgage insurance fees.

Example 4

- Suppose you plan to borrow $80,000 for a home at 6.5% interest on a 30-year fixed-rate mortgage.
- Determine the loan’s APR if there is a 1-point loan origination fee and a 1-point discount charge.

Example 4, cont’d

- Solution: The loan is $80,000 plus points totaling $1600, for a total of $81,600.
- According to the amortization table, the total monthly payment will be 81.6($6.320680) or about $515.77, if the fees were paid monthly instead of at the time of closing.

Example 4, cont’d

- Solution, cont’d: The interest rate that corresponds to a loan of $80,000 at 6.5% for 30 years with a monthly payment of $515.77 is found:
- Divide $515.77 by 80 to find the amount per $1000, which is $6.44713.
- In the table, this value corresponds to a rate between 6.5% and 6.75%.

Example 4, cont’d

- Solution, cont’d: Use linear interpolation to determine the APR.
- The payment is 76.5% of the way from the payment for 6.5% to the payment for 6.75%.
- The APR is about 6.691%

APR, cont’d

- Note that the APR is always greater than or equal to the stated annual interest rate.

Down Payment

- A down payment on a house is the amount of cash the buyer pays at closing, minus any points and fees.
- Traditionally a down payment is 20% of the value, but can be lower.
- The loan to value ratio of a mortgage is the percent of the home’s value that is not paid for by the down payment.
- For example, a down payment of 20% results in an 80% loan to value ratio.

Down Payment, cont’d

- Another guideline for the maximum price you can afford when buying a home is to find your maximum price by dividing the amount you have for a down payment by the percent of the value of the house that amount represents.

Example 5

- If you have $25,000 for a down payment, what is the highest-priced home you can afford if a 20% down payment is required?

Example 5, cont’d

- Solution: The maximum price you can afford to pay is your down payment amount divided by 20%.
- The most expensive house you can afford is one that is selling for $125,000.

Question:

Suppose you have $5000 for a down payment on a house that is selling for $82,000. If the lender requires a 5% down payment for first-time homebuyers, is this house within your price range?

a. Yes b. No

13.3 Initial Problem Solution

- Suppose you have saved $15,000 toward a down payment on a house and your total yearly income is $45,000. What is the most you could afford to pay for a house?
- Assume you pay 0.5% of the value for insurance, you pay 1.5% of the value for taxes, your closing costs will be $2000, and you can obtain a fixed-rate mortgage for 30 years at 6% interest.

Initial Problem Solution, cont’d

- Your total income is $45,000
- You have $15,000 saved for the purchase
- $2000 will be used for closing costs.
- This leaves $13,000 for a down payment.

Initial Problem Solution, cont’d

- The first affordability guideline says you can spend at most 3($45,000) = $135,000 on a house.
- Next, consider your monthly expenses:
- You would be financing $122,000 at 6% for 30 years.
- The monthly mortgage payments, from the table, would be 122($5.995505) = $732.

Initial Problem Solution, cont’d

- The insurance and taxes are 2% of the home’s value annually.
- This adds $225 to the monthly expenses, for a total monthly expense of $957.
- According to the second affordability guideline you can only afford monthly expenses of at most $938.
- The monthly expenses for this house are above your maximum. You cannot afford it.

Initial Problem Solution, cont’d

- A house priced $135,000 is slightly out of your reach, so your options are:
- Wait for interest rates to fall.
- Increase your income.
- Come up with a larger down payment.
- Choose a less expensive house.

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