Deitz J.E., Southam J.L. Contemporary Business Mathematics for Colleges

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274 Part 4 Interest Applications

Given the annual interest of 12%, the monthly rate is 12% 4 12 5 1% per month.

Month 1: $4,000 3 1% 5 $40 Month 3: $2,000 3 1% 5 $20

Month 2: $3,000 3 1% 5 $30 Month 4: $1,000 3 1% 5 $10

Total interest 5 $40 1 $30 1 $20 1 $10 5 $100

Unpaid InterestPrincipal Total New

Month Balance Payment

Payment

Payment Balance

1 $ 4,000 $ 40 $1,000 $1,040 $3,000

2 3,000 30 1,000 1,030 2,000

3 2,000 20 1,000 1,020 1,000

4 001,000 10 1,000 1,010 0

$100 $4,000 $4,100

EXAMPLE F

Carmel Dufault purchases $4,000 worth of merchandise. She will pay interest of 12% on

$4,000 for four months. First, calculate the total amount of interest. Carmel will repay

one-fourth of the interest amount each month. In addition, she will repay the $4,000 in

four equal monthly amounts of $1,000 each. Display the results in a table.

$4,000 3 12% 35$160

$160 4 4 5 $40 per month for interest

Unpaid InterestPrincipal Total New

Month Balance Payment Payment Payment Balance

1 $ 4,000 $ 40 $1,000 $1,040 $3,000

2 3,000 40 1,000 1,040 2,000

3 2,000 40 1,000 1,040 1,000

4 001,000  40 1,000 1,040 0

$160 $4,000 $4,160

A kitchen stove is priced at $600 and is purchased with a $100 down payment. The $500 remaining

balance is paid in two successive monthly payments of $250 each. Compute the total interest using the

following methods:

a. Interest of 1.5% is calculated on the unpaid balance each month (18% annual rate).

Month 1: $500 3 0.015 5 $7.50

Month 2: New balance is $250. $250 3 0.015 5 $3.75

Total interest 5 $7.50 1 $3.75 5 $11.25

b. Simple interest is calculated on the entire $500 for 2 months at 1.5% per month (18% annual rate).

$500 3 0.015 per month 3 2 months 5 $15.00

✔

CONCEPT CHECK 14.4

14.5 Although the interest is calcu-

lated on the unpaid balance,example

E is not “amortization.”Usually amor-

tized means that all of the total pay-

ments are equal and that each month,

principal payments increase and the

interest payments decrease.Amortiza-

tion is covered in examples J,K,and L.

After studying example L,compare its

table to the tables in examples E and F.

(ASSOCIATED PRESS)

14.6 The method of computing and

paying interest in example F may seem

strange when compared to example E.

However,the method in example F is

somewhat similar to bank discounting,

which will be covered in Chapter 15.

In both,the interest/discount amount

is calculated on a total amount at the

beginning.In example F,this amount is

paid over time.In bank discounting,it

is all paid at the very beginning.In fact,

it is deducted from the face value,

leaving the proceeds.

Examples E and F are very similar, but not quite identical. The numbers are the same:

Both purchases are for $4,000; both repay the $4,000 principal in four equal monthly

payments; both use a 12% annual interest rate. The only difference is the method of

calculating the interest. In example E, the total amount of interest is $100; in example F,

it is $160. In example F, it is more expensive to borrow the same money than in example

E. In example F, interest is calculated as if the entire $4,000 were borrowed for 4 months

($4,000 3 0.12 3 4/12). But Carmel repays $1,000 of the money after only 1 month.

The true interest rate, or the effective interest rate, cannot be the same in each exam-

ple because it costs more in example F to borrow the same amount of money for the

same length of time. To calculate the effective interest rate, we use the familiar formula

from Chapter 13, , where I is the amount of interest in dollars, T is the time of

the loan in years, and P is the average unpaid balance (or the average principal) over the

period of the loan. The average unpaid balance is the sum of all of the unpaid monthly

balances divided by the number of months. (Note: The term effective interest rate is also

used in other contexts where a different formula is used to find the effective rate.)

EXAMPLE G

Use the formula to compute the effective interest rates for (a) example E

and (b) example F. In both examples, the time of the loan is T 5 of a year. Using

the preceding tables, for each example, the average unpaid balance is

P 5 . But in example E,

I 5 $100 and in example F, I 5 $160.

a. Example E: T 5 ; P 5 $2,500; I 5 $100; so that

b. Example F: ; P 5 $2,500; I 5 $160; so that

Rule: When the interest is calculated on the unpaid balance each month, the quoted

rate and the effective rate will always be the same. When interest is computed

only once on the original principal, but the principal is repaid in installments,

then the effective interest rate will always be higher than the quoted rate.

The preceding rule is true even when the principal is not repaid in equal installments

each month.

R 5

P 3 T

$160

$2,500 3

$160

$833.33

5 0.1920008, or 19.2%

T 5

R 5

P 3 T

$100

$2,500 3

$100

$833.33

5 0.120000, or 12%

$4,000 1 $3,000 1 $2,000 1 $1,000

$10,000

5 $2,500

R 5

P 3 T

R 5

P 3 T

Chapter 14 Installment Purchases 275

Computing Effective Interest Rates

14.7 One example of a different use of

the term effective rate is in compound

interest (Chapter 16).Consider the

interest rate 9% compounded monthly.

The “effective rate”is the annual rate

that would yield the same amount of

monetary return over one year as 9%

compounded monthly does.The

formula is R 5 (1 1 0.09/4)

2 1,

so that R 5 0.09308,or 9.308%,

compounded annually.

Compute effective rates.

Learning Objective

14.8 Students may be surprised at the

large difference.Compare the loans in

examples E and F with a simple loan of

$4,000 for 4 months at 12%,where the

entire $4,000 and the $160 interest are

repaid at the end.In examples E and F,

the borrowers do not have the entire

$4,000 for the entire 4 months.In ex-

ample F,the interest amount is calcu-

lated as if the borrower had the entire

$4,000,but on average the principal is

only $2,500 for the 4 months.

EXAMPLE H

Look back at example E where Nancy Bjonerud made four equal principal payments of

$1,000 each. Suppose instead that she repays the principal in four monthly payments of

$900, $1,200, $1,100, and $800. As in example E, she will also pay interest each month on

the unpaid balance for that month, which is calculated at an annual rate of 12%. Compute

the interest amount for each month and display the results in a table. Then, compute the

average unpaid balance and the effective interest rate using the formula .

Given annual interest of 12%, the monthly rate is 12% 4 12 5 1% per month.

Month 1: $4,000 3 1% 5 $40 Month 3: $1,900 3 1% 5 $19

Month 2: $3,100 3 1% 5 $31 Month 4: $800 3 1% 5 $8

Total interest 5 $40 1 $31 1 $19 1 $8 5 $98

Unpaid InterestPrincipal Total New

Month

Balance Payment

Payment Payment Balance

1 $4,000 $40 $ 900 $ 940 $3,100

2 3,100 31 1,200 1,231 1,900

3 1,900 19 1,100 1,119 800

4 00,800  8  800  808 0

$98 $4,000 $4,098

INCREASING THE EFFECTIVE RATE

Example F shows how the effective rate in an installment sale can be increased by using a

different method to calculate interest. Of course, a reputable lender should indicate the

true interest rate in the terms of the agreement. But in installment sales, the interest rate

may be only one of several variables in the total cost of purchasing. Any additional fees

to make the installment purchase increase the actual cost of borrowing.

Naturally, some businesses will attempt to attract buyers by offering very low pur-

chase prices, even “guaranteeing to match all competitors’ advertised prices for 30 days.”

Others may offer installment purchases at low or even 0% interest rates and no addi-

tional fees—but they will charge a higher base price. Different consumers are attracted

by different things—some by low prices, some by favorable terms of purchase. For many

consumers, buying is simply an emotional response with very little thought given to

actual costs.

Lenders and sellers “effectively” increase the cost of borrowing money or buying in

installments by charging or suggesting additional fees. If it is a purchase of merchandise,

the lender could require that the merchandise be insured for the term of purchase. Or

the lender could charge a credit application fee.

Consider the following modification to example G, part a, which had an effective rate

of 12%.

R 5

P 3 T

$98

$2,450 3

$98

$816.67

5 0.11999951, or 12%

P 5

$4,000 1 $3,100 1 $1,900 1 $800

$9,800

5 $2,450

R 5

P 3 T

276 Part 4 Interest Applications

EXAMPLE I

Look back at example G, part a, where we used to calculate the effective rate

for example E, with I equal to the total interest charge of $100. Suppose instead that the

lender had charged Nancy the interest of $100, and a loan origination fee of 1% of the

purchase price, and an insurance premium of $1 per month for the term of the loan.

Use the formula to compute the effective interest rate, but let I be the total

finance charge.

The average unpaid balance is still

I 5 Total finance charge 5 Interest 1 Loan origination fee 1 Insurance

Interest only 5 $40 1 $30 1 $20 1 $10 5 $100

Loan origination fee 5 1% of $4,000 5 0.01 3 $4,000 5 $40

Insurance 5 $1 3 4 months 5 $4

Therefore, I 5 $100 1 $40 1 $4 5 $144

Because the interest in example E was paid on the unpaid balance, the effective rate

was 12%, the same as the quoted interest rate. If the same additional finance charges

from example I were applied to example F, the results would be even more dramatic.

R 5

P 3 T

$144

$2,500 3

$144

$833.33

5 0.17280069, or 17.3%

$10,000

5 $2,500

P 5

$4,000 1 $3,000 1 $2,000 1 $1,000

R 5

P 3 T

R 5

P 3 T

Chapter 14 Installment Purchases 277

From Concept Check 14.4, a kitchen stove priced at $600 is purchased with a $100 down payment. The remaining

balance of $500 may be financed over 2 months with either of the following installment payment plans:

Plan 1: Two monthly principal payments of $250 each and a total interest amount of $11.25

Plan 2: Two monthly principal payments of $250 each and a total interest amount of $15.00

Calculate the effective annual rate of each plan, using , where P is the average unpaid monthly balance

and T is year. In each plan, the monthly unpaid balances are $500 in month 1 and $250 in month 2.

The average unpaid balance is so P 5 $375.

Plan 1:

Plan 2:

COMPLETE ASSIGNMENT 14.2.

R 5

(P 3 T)

$15.00

($375 3

)

$15.00

$62.50

5 0.24, or 24% effective annual rate

R 5

(P 3 T)

$11.25

($375 3

)

$11.25

$62.50

5 0.18, or 18% effective annual rate

($500 1 $250)

$750

5 $375,

R 5

(P 3 T )

✔

CONCEPT CHECK 14.5

14.9 The effective rate of 17.3% in ex-

ample I is still not as high as the 19.2%

of example G,part b,because the total

finance charge in example I was $144

and in example G,part b,it was $160.If

the $40 loan origination fee and the $4

insurance fee were included in exam-

ple G,part b,the total finance charge

would be $204 and the effective rate of

borrowing would be 24.48%.

In example E, interest was calculated on the unpaid balance, but the total payments

were different each month: $1,040, $1,030, $1,020, and $1,010. Equal monthly payments

are usually simpler, especially for the borrower. In example F, the total payments were the

same each month, always $1,040. However, the interest was not calculated on the unpaid

balance. In example E, the effective interest rate was equal to the quoted interest rate of

12%. But in example F, the effective rate was much higher, 19.2%.

Taking the best features of each example, consider a loan where the total payments

are equal each month and the interest is calculated on the unpaid balance each month.

Such a loan is said to be amortized; the method is called amortization. (The word amor-

tize is also used in different contexts and there is more than one way to amortize a loan.)

Although possible for purchases of any length of time, amortization is especially relevant

for larger purchases made over longer periods of time. Loans to pay for homes and auto-

mobiles are usually amortized. There may or may not be a down payment.

COMPUTING THE MONTHLY PAYMENT

The basic concept to amortize a loan is to multiply the loan amount by an amortization

payment factor. The product is the amount of the monthly payment. This factor may be

derived from a calculator or computer or from a book of financial tables. When lenders

amortize loans today, they use computers to do the final calculations. Initial calculations,

however, are often made using calculators or tables. Chapter 23 will describe how to

use a calculator to make amortization calculations. In this chapter we will use tables.

Both methods are still used, and both lead to the same results. (You can also go to the

Internet, search on “amortization calculations,” and find Websites that help you do the

calculations.)

Table 14-1 illustrates the concept of tables for amortization payment factors. Actual

tables would have many pages and would be much more detailed. If you study other

courses in business mathematics, accounting or finance, you may use tables that are

slightly different than Table 14-1. In Chapter 23, we will encounter one such table.

Regardless of the exact format of the table, the concepts are the same. And, to repeat,

financial calculators and computers will eventually completely eliminate the need for

any of these tables.

Notice that the title of Table 14-1 is “Amount of Monthly Payment per $1,000

Borrowed.” Therefore, you must first determine the amount of the loan in “thousands of

dollars,” not in the number of dollars. The annual interest rates in Table 14-1 were selected

because they are evenly divisible by 12. This will eliminate the necessity to round off interest

rates when you convert an annual rate into a monthly rate.

278 Part 4 Interest Applications

Amortizing a Loan

Learning Objective

Amortize a loan.

14.10 Instead of all equal payments,

some amortized loans have equal

payments with one larger “balloon”

payment at the end.Part C of assign-

ment 14.3 has a balloon payment.

to Find the Monthly Payment of an Amortized Loan Using

Table 14-1

1. Divide the loan amount by $1,000 to get the number of thousands of

dollars.

2. Locate the amortization payment factor in Table 14-1.

3. Multiply the quotient in Step 1 by the amortization payment factor.

The product is the amount of the monthly payment.

STEPS

Chapter 14 Installment Purchases 279

Table 14-1: Amortization Payment Factors—Amount of Monthly Payment per $1,000 Borrowed

Term of

Annual Interest Rate

Loan 4.5% 6% 7.5% 9% 10.5% 12%

1 month 1003.75000 1005.00000 1006.25000 1007.50000 1008.75000 1010.00000

2 months 502.81425 503.75312 504.69237 505.63200 506.57203 507.51244

3 months 335.83645 336.67221 337.50865 338.34579 339.18361 340.02211

4 months 252.34814 253.13279 253.91842 254.70501 255.49257 256.28109

5 months 202.25561 203.00997 203.76558 204.52242 205.28049 206.03980

6 months 168.86099 169.59546 170.33143 171.06891 171.80789 172.54837

1 year 85.37852 86.06643 86.75742 87.45148 88.14860 88.84879

2 years 43.64781 44.32061 44.99959 45.68474 46.37604 47.07347

3 years 29.74692 30.42194 31.10622 31.79973 32.50244 33.21431

4 years 22.80349 23.48503 24.17890 24.88504 25.60338 26.33384

5 years 18.64302 19.33280 20.03795 20.75836 21.49390 22.24445

10 years 10.36384 11.10205 11.87018 12.66758 13.49350 14.34709

15 years 7.64993 8.43857 9.27012 10.14267 11.05399 12.00168

20 years 6.32649 7.16431 8.05593 8.99726 9.98380 11.01086

25 years 5.55832 6.44301 7.38991 8.39196 9.44182 10.53224

30 years 5.06685 5.99551 6.99215 8.04623 9.14739 10.28613

EXAMPLE J

Find the monthly payment required to amortize a $4,000 loan over 4 months at 12%

(1% per month).

$4,000 4 $1,000 5 4 (thousands)

Find the intersection of the 12% column and the 4-month row in Table 14-1.

The amortization payment factor is $256.28109 per each one thousand dollars.

Multiply the 4 (from Step 1) by the amortization payment factor.

4 3 $256.28109 5 $1,025.12436, or $1,025.12 monthly.

EXAMPLE K

Judith Kranz agrees to purchase an automobile for $18,300. Judith will make a $2,000

down payment and amortize the balance with monthly payments over 4 years at 9%

(0.75% per month). Determine Judith’s monthly payment.

$18,300 2 $2,000 5 $16,300 amount financed

$16,300 4 $1,000 5 16.3 (thousands)

Find the intersection of the 9% column and the 4-year row in Table 14-1. The

amortization payment factor is $24.88504 per thousand.

Multiply the 16.3 (from Step 1) by the amortization payment factor.

16.3 3 $24.88504 5 $405.62615, or $405.63 monthly.

STEP 3

STEP 2

STEP 1

STEP 3

STEP 2

STEP 1

LOAN PAYMENT SCHEDULE

After determining the amount of the monthly payments, a lender can prepare a schedule

of loan payments called an amortization schedule. The payment for the last month is

determined in the schedule, and it may be slightly different from the payment in the

other months.

280 Part 4 Interest Applications

to Create an Amortization Schedule

For each row except the last:

1. Interest payment 5 Unpaid balance 3 Monthly interest rate

2. Principal payment 5 Monthly payment 2 Interest payment

3. New unpaid balance 5 Old unpaid balance 2 Principal payment

For the last row (the final payment):

1. Interest payment 5 Unpaid balance 3 Monthly interest rate

2. Monthly payment 5 Unpaid balance 1 Interest payment

3. Principal payment 5 Unpaid balance

STEPS

EXAMPLE L

Create an amortization schedule for the loan in example J, a $4,000 loan amortized at

12% over 4 months. The interest rate is 1% per month.

Unpaid InterestPrincipal Total New

Month Balance Payment Payment Payment Balance

1 $ 4,000.00 $ 40.00 $ 985.12 $1,025.12 $3,014.88

2 3,014.88 30.15 994.97 1,025.12 2,019.91

3 2,019.91 20.20 1,004.92 1,025.12 1,014.99

4 001,014.99 10.15 1,014.99 1,025.14 0

Totals $100.50 $4,000.00 $4,100.50

Note: In example L, the last monthly payment is 2 cents larger than the others. Because

the interest payments need to be rounded, the final payment usually will be slightly

different from the previous payments.

Since amortization implies that interest is paid on the unpaid balance, the formula

should show that the effective rate is the same as the quoted rate of 12%.

Looking at the table for example L, the average unpaid balance is

or $2,512.45

The total interest paid is I 5 $40.00 1 $30.15 1 $20.20 1 $10.15 5 $100.50.

Therefore,

R 5

P 3 T

$100.50

$2,512.45 3

$100.50

$837.48

5 0.1200029, or 12%

5 $2,512.445,

P 5

$4,000.00 1 $3,014.88 1 $2,019.91 1 $1,014.99

$10,049.78

R 5

P 3 T

The reason that the result was 12.00029% instead of 12% is that all of the payments

were rounded to the nearest cent. You can easily verify that if you round all payments to

five decimal places, R 5 12.00007. However, also be sure to calculate the monthly pay-

ment to five decimal places, or $1,025.12436.

Chapter 14 Installment Purchases 281

A $2,000 purchase is amortized over 2 months at an annual rate of 9%. First use Table 14-1 to calculate the monthly

payment for month 1.Then show the calculations to construct a 2-month amortization schedule. (Remember: In this

problem,month 2 is the last month.)

$2,000 4 $1,000 5 2 (thousands)

Amortization payment factor from Table 14-1 is $505.63200.

2 3 $505.63200 5 $1,011.264, or $1,011.26 for month 1

Month 1 2

Unpaid balance Original principal: $2,000.00 From end of month 1: $1,003.74

Monthly rate 0.09 4 12 5 0.0075

Interest payment $2,000.00 3 0.0075 5 $ 15.00 $1,003.74 3 0.0075 5 $ 7.53

Total payment From above: $1,011.26 $1,003.74 1 $7.53 5 $1,011.27

Principal payment $1,011.26 2 $15.00 5 $ 996.26 $1,003.74

New balance $2,000.00 2 $996.26 5 $1,003.74 $1,003.74 2 $1,003.74 5 $ 0.00

✔

CONCEPT CHECK 14.6

Persons who decide to purchase a home usually borrow the majority of the money.

The amount that is borrowed is usually amortized, and usually for a long time, such as

15, 20, or 30 years. Such a home loan is called a mortgage. The interest rate may be

fixed, which means that it stays the same for the entire length of the loan. Other mort-

gages are variable-rate loans, which permit the lender to periodically adjust the interest

rate depending on current financial market conditions. Whether a borrower decides on a

fixed or variable rate loan depends on several factors, such as how long he or she plans to

remain in that home.

A mortgage loan is still a loan. And amortizing a mortgage is the same as amortizing

any other loan: Look up the amortization payment factor in Table 14-1 and multiply by

the number of thousands of dollars that are borrowed.

Finding the Monthly Payment of

a Home Mortgage

Compute the monthly payment on

a home mortgage.

Learning Objective

282 Part 4 Interest Applications

EXAMPLE M

George and Kathy Jarvis bought a condominium priced at $190,000. They made a

$20,000 down payment and took out a 30-year, 6% mortgage on the balance. Find the

size of their monthly payment.

$190,000 2 $20,000 5 $170,000 amount borrowed

Divide $170,000 by $1,000 to get 170 (thousands)

Find the amortization factor in the 6% column and 30-year row of Table 14-1.

It is $5.99551.

Multiply the 170 from Step 1 by $5.99551 to get $1,019.2367.

The monthly payment will be $1,019.24.

AMORTIZATION SCHEDULE FOR A MORTGAGE

An amortization schedule for a mortgage is computed line by line, just as the amortiza-

tion schedules are for other loans such as the one in example L. However, a 30-year

home mortgage will have 360 lines, one for each month of the loan. This could be about

six or seven pages of paper with three calculations per line, or 1,080 calculations. Today,

these tables are always produced with a computer. You can create an amortization sched-

ule using Excel or you can find several sources on the Internet to do the calculations for

you. However, to review the concept manually, examine example N.

EXAMPLE N

Construct the first three lines of an amortization schedule for the Jarvis’s home mort-

gage loan in example M.

The Jarvis’s $170,000 mortgage has a monthly payment of $1,019.24.

For a 6% annual interest rate, the monthly rate is 6% 4 12 5 0.5%.

For each row, 1. Monthly interest 5 Unpaid balance 3 0.005

2. Principal payment 5 Total payment 2 Monthly interest

3. New balance 5 Unpaid balance 2 Principal payment

Unpaid InterestPrincipal Total New

Month Balance Payment Payment Payment Balance

1 $170,000.00 $850.00 $169.24 $1,019.24 $169,830.76

2 169,830.76 849.15 170.09 1,019.24 169,660.67

3 169,660.67 848.30 170.94 1,019.24 169,489.73

STEP 3

STEP 2

STEP 1

Chapter 14 Installment Purchases 283

A home cost $280,000.The buyers made a down payment of $30,000. Compute the monthly

payment on a 25-year mortgage with an annual interest rate of 7.5%. Use Table 14-1.

The amount borrowed is $280,000 2 $30,000 5 $250,000.

The amortization payment factor from Table 14-1 is $7.38991.

The amount of the loan in thousands is $250,000 4 $1,000 5 250 (thousands).

The monthly mortgage payment is 250 3 $7.38991 5 $1,847.4775, or $1,847.48.

COMPLETE ASSIGNMENT 14.3.

✔

Concept Check 14.7

amortization

amortization payment factor

amortization schedule

annual percentage rate (APR)

average daily balance

average principal

average unpaid balance

credit card

effective interest rate

finance charge

fixed interest rate

installments

mortgage

Truth in Lending Act (TILA)

variable-rate loans

Chapter Terms for Review