FAPP07 ISM 22 - UW-Madison Department of Mathematicsrobbin/141dir/propp/COMAP/Final...420 Chapter 22 Skill Objectives 1. Become familiar with the various types of loans. 2. Become

419

Chapter 22 Borrowing Models

Chapter Outline Introduction Section 22.1 Simple Interest Section 22.2 Compound Interest Section 22.3 Conventional Loans Section 22.4 Annuities Chapter Summary The concepts of simple and compound interest, developed in Chapter 21 in the context of savings, play an equally important role in borrowing. Most loans, whether for buying a car, paying for college, or mortgages for buying a house, involve compound interest.

To better compare terms of different loans, specific terminology has been developed. The nominal rate is stated for any specific period of time, often monthly. The effective annual rate (EAR) is the actual percentage rate for a full year, taking compounding into account. The amortization formula, obtained by summing a geometric series, is used to compute the periodic payments (generally monthly) needed to repay a loan or mortgage.

While conventional mortgages have a fixed interest rate for the life of the loan, the interest rate in an adjustable mortgage is fixed for a much shorter period (often just one year), and then adjusts either up or down, depending upon interest rates in the economy. Since much of the risk of rising rates is on the borrower, such mortgages often come with a lower initial rate, which translates into lower initial payments. For some borrowers, these lower payments allow them to buy a more expensive house than they could afford with a conventional mortgage. However, because of the periodic adjustments, this type of mortgage may turn out to be more expensive in the long run than fixed-rate mortgages.

420 Chapter 22

Skill Objectives

1. Become familiar with the various types of loans.

2. Become familiar with the terminology for loan rates.

3. Use the amortization formula to compute the monthly payment needed to repay a loan or mortgage.

Teaching Tips

1. As in the previous chapter, students will encounter numerous ads for loans, mortgages, and credit cards, which will contain much of the terminology used in this chapter. You can exploit these practical situations to supplement the problems in the text.

2. A significant proportion of the problems require the use of spreadsheets. Since many of the students in such a course will not be familiar with spreadsheets, be careful to avoid these exercises, unless you are prepared to devote class time to teaching this material.

Research Paper Usury was originally defined as a fee charged for the use of money (Latin from usus or “used”). Have students investigate the origins of lending. Students may wish to include a comparison between biblical and Qur’anic injunctions against Usury.

Another research paper could involve having students investigate unfair lending practices such as “predatory lending” and legislation preventing such lending. Students may wish to discuss the Truth in Lending Act. There are many different tactics that could be considered predatory such as lending by making a secured loan when the borrower has no way to repay that loan.

Spreadsheet Project

To do this project, go to http://www.whfreeman.com/fapp7e.

This spreadsheet project allows students to model a mortgage and investigate the effects of extra payments.

Collaborative Learning Actual Loan Cost

As an icebreaker, ask the students to estimate how much the total repayments will be on a variety of loans, such as:

3-, 4-, and 5-year car loans of $10,000 at 8% compounded monthly,

or

a 15-year mortgage of $100,000 at 6% compounded monthly,

or

a 30-year mortgage of $100,000 at 7% compounded monthly.

Their estimates may be far off the mark. Ask them to hold on to their results until they learn how to make these calculations, and then compare them. They may very well be shocked at the total cost, especially in the case of the 30-year mortgage.

Borrowing Models 421

Solutions Skills Check:

1. b 2. c/a 3. a 4. b 5. b 6. a 7. b 8. a 9. a 10. c

11. a 12. b 13. c 14. b 15. c 16. c 17. c 18. c 19. c 20. b

Exercises:

1. (a) For the 11.25%, a cost of $15,529 has an annual yield of ( )0.0551 $10,000

5.62%.$9802

=

For the 5.51%, a cost of $9802 has an annual yield of ( )0.1125 $10,000

7.24%.$15,529

=

(b) Answers will vary but should remark that the first bond locks in the interest rate much farther into the future.

2. For the 5.75%, a cost of $10,318 has an annual yield of ( )0.0575 $10,000

5.57%.$10,318

=

3. We add on interest of 4 × 0.059 × $5000 = $1180.

The monthly payment is $5000 + $1180

= $128.75.48

4. We add on interest of 2 × 0.085 × $3000 = $510.

The monthly payment is $3000 + $510

= $146.25.24

5. To realize the $3000 that you need, you need to borrow $3000

= $4687.50,1 0.09×4−

for which the

monthly payment is $4687.50

= $97.66.48

If you do just the $3000 discounted loan, the lender

gives you just ( )$3000 1 0.09 4 $1920,− × = on which the monthly payment is $1920

$40.00.48

=

6. To realize the $3000 that you need, you need to borrow $3000

$5217.39,1 0.085 ×5

=−

for which the

monthly payment is $5217.39

$86.96. 60

= If you do just the $3000 discounted loan, the lender

gives you just ( )$3000 1 0.085 5 $1725,− × = on which the monthly payment is $3000

$50.00.60

=

7. Answers will vary but should conclude that the add-on loan always has a lower payment.

422 Chapter 22

8. Let the amount needed be P, with annual interest rate r and payment over t years (= 12t months).

Then the monthly payment on the add-on loan is ( ) ( )1

1 .12 12

P rt Prt

t t

+= + The monthly payment

on the discounted loan is 1 1.

12 12 1

Prt P

t t rt− = −

For simplicity, let .a rt= For positive a, we have

( )( )21 1 1 1a a a− = − + < and hence for 0 < a < 1 we have (by dividing both sides of the

inequality by ( )1 a− also 1

1 .1

aa

+ <−

Thus, for an add-on loan and a discounted loan for

which you receive the same amount from the lender, the monthly payment is always less on the add-on loan.

9. All but a tiny amount after 91 months

10. About $410

11. After 203 months (more than 16 years!), the balance is $500.16.

12. $3706.07

13. 294 months (= 24.5 years!), plus a few cents in the 295th month

14. 294 months (plus a few cents in the 295th month)

15. $24, 995 $2000 $22,995

$383.2560 60

− = =

16. Use the amortization formula ( )

,1 1

n

Aid

i−=

− + with A = $7000, 0.085

12 ,i = and n = 48.

( ) ( )0.085

12480.085

12

$7000$172.54

1 1 1 1n

Aid

i− −

×= = =− + − +

17. The amortization formula gives $19.28 per month for each $1000 financed. Either I miscopied, or the difference is probably due to what is considered a month (30 days?) and what method is used to calculate the monthly interest rate.

18. (a) Use the amortization formula ( )

,1 1

n

Aid

i−=

− + with A = $1000, 0.079

12 ,i = and n = 48.

( ) ( )0.079

12480.079

12

$1000$24.37

1 1 1 1n

Aid

i− −

×= = =− + − +

(b) Use the amortization formula ( )

,1 1

n

Aid

i−=

− + with A = $1000, 0.068

12 ,i = and n = 60.

( ) ( )0.068

12600.068

12

$1000$19.71

1 1 1 1n

Aid

i− −

×= = =− + − +

Additional answers will vary.



,1 1

n

Aid

i−=

− + with A = $100,000, 0.065

12 ,i = and

30 12 360.n = × =

( ) ( )0.065

123600.065

12

$100,000$632.07

1 1 1 1n

Aid

i− −

×= = =− + − +


,1 1

n

Aid

i−=

− + with A = $100,000, 0.07125

12 ,i = and

30 12 360.n = × =

( ) ( )0.07125

123600.07125

12

$100,000$673.72

1 1 1 1n

Aid

i− −

×= = =− + − +


,1 1

n

Aid

i−=

− + with A = $100,000, 0.06125

12 ,i = and

15 12 180.n = × =

( ) ( )0.06125

121800.06125

12

$100,000$850.62

1 1 1 1n

Aid

i− −

×= = =− + − +


,1 1

n

Aid

i−=

− + with A = $100,000, 0.0675

12 ,i = and

15 12 180.n = × =

( ) ( )0.0675

121800.0675

12

$100,000$884.91

1 1 1 1n

Aid

i− −

×= = =− + − +

23. Use the amortization formula ( )1 1

,n

iA d

i

− − + =

with d = $632.07, 0.06512 ,i = and

360 5 12 360 60 300.n = − × = − =

( ) ( ) 3000.06512

0.06512

1 11 1$632.07 $93,611.27

ni

A di

−− − +− + = = =

Thus, the amount of equity is $100,000 $93,611.27 $6388.73.− =


,n

iA d

i

− − + =

with d = $673.72, 0.0712512 ,i = and

360 5 12 360 60 300.n = − × = − =

( ) ( ) 3000.0712512

0.0712512

1 11 1$673.72 $94,246.47

ni

A di

−− − +− + = = =

Thus, the amount of equity is $100,000 $94,246.47 $5743.53.− =

424 Chapter 22


,n

iA d

i

− − + =

with d = $850.62, 0.0612512 ,i = and

180 5 12 180 60 120.n = − × = − =

( ) ( ) 1200.0612512

0.0612512

1 11 1$850.62 $76,186.80

ni

A di

−− − +− + = = =

Thus, the amount of equity is $100,000 $76,186.80 $23,813.20.− =


,n

iA d

i

− − + =

with d = $884.91, 0.067512 ,i = and

180 5 12 180 60 120.n = − × = − =

( ) ( ) 1200.067512

0.067512

1 11 1$884.91 $77,066.56

ni

A di

−− − +− + = = =

Thus, the amount of equity is $100,000 $77,066.56 $22,933.44.− =

27. Using Solver in Excel or otherwise, we get an annual rate of 3.68%.


,1 1

n

Aid

i−=

− + with A = $160,000, 0.036

12 ,i = and

30 12 360.n = × =

The monthly payment would be ( ) ( )

0.036123600.036

12

$160,000$727.43.

1 1 1 1n

Aid

i− −

×= = =− + − +

Regarding the difference, answers will vary.


,1 1

n

Aid

i−=

− + with A = $180,000, 0.0675

12 ,i = and

30 12 360.n = × =

The monthly payment would be ( ) ( )

0.0675123600.0675

12

$180,000$1167.48.

1 1 1 1n

Aid

i− −

×= = =− + − +

30. We must first solve ( ) 48

1 15000 128.75 .

i

i

− − + =

Using a spreadsheet or calculator, we have

0.00900257.i = Thus, APR is approximately 12 0.00900257 10.80%.× = The EAR is

( )121 0.00900257 1 11.35%.+ − =


1 13000 146.25 .

i

i

− − + =



( )121 0.01296119 1 16.71%.+ − =



1 13000 97.66 .

i

i

− − + =



( )121 0.01992686 1 26.72%.+ − =


1 13000 86.96 .

i

i

− − + =



( )121 0.02031339 1 27.29%.+ − =

34. APR = 365 × 0.0004932 = 18.00%; EAR = ( )3651 + 0.0004932 1 = 19.72%.− The company

must disclose the APR.

35. The principal is $300 − $54 = $246 and the (simple) interest over the two weeks is $54, so the interest rate is $54

$246100% × = 21.95% for 2 weeks, for an annual rate of 522 ×21.95% = 571%.

For a 365-day year, we get a daily rate of 21.95%14 = 1.57% per day and an annual percentage rate

of 365 × 1.57% = 572%.

36. You pay back d = $354 on amount A = $300 after n = 1 compounding periods. From spreadsheet or otherwise, the solution i to the amortization formula for these amounts is i = 0.18 = 18% for the two-week period. The APR is thus 26 × 18% = 468% (slightly more if done for 14 days on a 365-day/yr basis).

37. The principal is $1500 − $88 = $1412 and the (simple) interest over the week is $88, so the interest rate is $88

$1412100% × = 6.23% for 1 week, for an annual rate of 52 × 6.23% = 324%. For

a 365-day year, we get a daily rate of 6.23%7 0.89%= per day and an annual percentage rate of

365 × 0.89% = 325%.

38. Because it is a discounted loan, you receive amount A = $1500−$88 = $1412 and pay d = $1500 after n = 1 compounding periods. From spreadsheet or otherwise, the solution i to the amortization formula for these amounts is i = 0.062323 = 6.2323% for the 17-day period, or 0.3666% per day. The APR is thus 365 × 0.3666% ≈ 134%.


,1 1

n

Aid

i−=

− + with A = $100,000, 0.08375

12 ,i = and

30 12 360.n = × =

( ) ( )0.08375

123600.08375

12

$100,000$760.07

1 1 1 1n

Aid

i− −

×= = =− + − +

(b) Use the amortization formula ( )1 1

,n

iA d

i

− − + =

with d = $760.07, 0.0837512 ,i = and

360 5 12 360 60 300.n = − × = − =

( ) ( ) 3000.0837512

0.0837512

1 11 1$760.07 $95,387.80

ni

A di

−− − +− + = = =

Thus, the amount of equity is $100,000 $95,387.80 $4612.20.− = Continued on next page

426 Chapter 22

39. continued

(c) Use the amortization formula ( )

,1 1

n

Aid

i−=

− + with A = $95,387.80, 0.070

12 ,i = and

30 12 360.n = × =

( ) ( )0.070

123600.070

12

$95,387.80$634.62

1 1 1 1n

Aid

i− −

×= = =− + − +

(d) Since the difference between the two payments is $760.07 $634.62=$125.45,− it would

take $2000

15.9426$125.45

≈ or 16 months.


,1 1

n

Aid

i−=

− + with A = $180,000, 0.08375

12 ,i = and

15 12 180.n = × =

( ) ( )0.08375

121800.08375

12

$180,000$1759.37

1 1 1 1n

Aid

i− −

×= = =− + − +

(b) Use the amortization formula ( )1 1

,n

iA d

i

− − + =

with A = $160,000, d = $1759.37,

0.0837512 .i =

( )

( )0.0837512

0.0837512

1 1

1 1$160,000 $1759.37

145

n

n

iA d

i

n

−

−

− + =

− + =

≈

Thus, she has been paying for 180 145 35− = months.

(c) Use the amortization formula ( )

,1 1

n

Aid

i−=

− + with A = $174,000, 0.0875

12 ,i = and

30 12 360.n = × =

( ) ( )0.0875

123600.0875

12

$174,000$1368.86

1 1 1 1n

Aid

i− −

×= = =− + − +

41. (a) $100,250

(b) The payment is $648.60. The interest for one-twelfth of the year is 0.07

12$100,000 $583.33,× = so the payment on the

principal is just $65.27.

(c) $583.33 − 250.00 = $333.33.

(d) The inflation adjusted cost is 0.0312

$333.33 $332.50;

1 + = the interest rate for the month is

$332.50$100,000 0.0033250, = which is an annual rate of ( )12 0.0033250 3.99%.= We have left out

any costs (such as realtor’s fee) involved in the sale.


42. (a) Consider the following balance sheet. REVENUE EXPENSES

Equity in house $5,869 Sale of house $125,000

Cost of house (including down payment)

$105,000

Closing costs at purchase $2,000

Mortgage payments

( )60 $665.30× $39,918

Insurance $1,000 Real estate taxes $10,000 Fee to realtor at sale $9,000 Closing costs at sale $500 Moving expense $3,000 TOTALS $130,869 $170,418

The calculation is in current dollars, so does not take into account inflation. We find the equity in the house by calculating how much of the remainder of the $100,000 mortgage

would be paid off in the last 25 years. Use the amortization formula ( )1 1

,n

iA d

i

− − + =

with d = $665.30, 0.0712 ,i = and 25 12 300.n = × =

( ) ( ) 3000.0712

0.0712

1 11 1$665.30 $94,131.24

ni

A di

−− − +− + = = =

The payment is based off of financing $100,000. Thus, the amount of equity is as follows.

$100,000 $94,131.24 $5,868.76− = ($5869, rounded)

During the five years, there was a loss.

Revenue minus expenses $130,869 $170,418 $39,549= − = −

(b) This would be equivalent to making rent payments of $39,549

$659.15.60

=

43. There would be $81.6 million

$2.7230

= million paid annually (including immediately).

2 291 1 1

2.72 2.72 2.72 ... 2.721.07 1.07 1.07 + + + +

This is a geometric series. From Chapter 21, the sum will be ( )301

1.07

11.07

12.72 36.1.

1

−≈

−

Thus, the payoff is approximately $36.1 million.

44. We assume payments at the end of the month.

For men: The Excel command = PV(0.04/12,199,2000,0,0) yields $290,580.05.

For women: The Excel command = PV(0.04/12,235,2000,0,0) yields $325,514.35.

428 Chapter 22

45. Because her payments increase each year exactly as much as inflation, she will receive $50,000 in 2005 dollars every year. So she needs to accumulate 45 × $50,000 = $2.25 million in 2005 dollars. Her effective yield on investment each year, taking into account inflation and using

Fisher’s formula, is 1 1.072

1 1 3.076923077%.1 1.04

r

a

+ − = − =+

Use the savings formula with

A = 2,250,000, i = 0.03076923077, and n = 35 × 4 = 140 quarters.

( ) ( )140 0.030769230774

0.030769230774

1 11 12,250,000

ni

A d di

+ −+ − = ⇒ =

Thus, d = $8997.74 per quarter.

46. The expected Social Security monthly income of $1628 in 2001 dollars is $1628 195177.1 $1792.55× = in

2005 dollars, using the Consumer Price Index values from Table 21.5. Her desired income in the first year of retirement is $50,000/year in 2005 dollars, or

$4166.67/month, or $4166.67−$1792.55 = $2374.12 more than Social Security will provide. To determine how much she must accumulate to have this much monthly for 33 years, we apply the

amortization formula, ( )1 1

,n

iA d

i

− − + =

with d = $2374.12, 0.0612 0.005,i = = and

33 12 396.n = × =

( ) ( ) 3961 1 1 1 0.005

$2374.12 $408,941.180.005

ni

A di

− − − + − + = = =

This is how much her savings must be worth in 2005 dollars; in 2052 dollars, she must have

( )47$408,941.18 1.03 $1,640,629.07.× = We use the savings formula to save up this amount A

over n = 47 12 = 564× months with monthly interest rate 0.0612 0.005.i = =

( ) ( )

( )

5641 1 1 0.005 1

$1,640,629.070.005

$1,640,629.07$1,640,629.07 3131.87597 $523.85

3131.87597

ni

A d di

d d

+ − + − = ⇒ =

= ⇒ = =

The monthly amount to save is $523.85. If instead of the CPI from Ch. 21 we use 3% inflation per year, the expected Social Security

monthly income of $1628 in 2001 dollars is ( )4$1628 1.03 $1832.33× = in 2005. Her desired

income in the first year of retirement is $50,000/year in 2005 dollars, or $4166.67/month, or $4166.67 − $1832.33 = $2334.34 more than Social Security will provide. To determine how much she must accumulate to have this much monthly for 33 years, we apply the amortization formula with payment d = $2334.34, 33 12 396n = × = months with monthly interest rate

0.0612 0.005.i = =

( ) ( ) 3961 1 1 1 0.005

$2334.34 $402,089.130.005

ni

A di

− − − + − + = = =


( )47$402,089.13 1.03 $1,613,139.39.× =

Continued on next page


46. continued We use the savings formula to save up this amount A over 47 12 564n = × = months with monthly interest rate 0.06

12 0.005.i = =

( ) ( )

( )

5641 1 1 0.005 1

$1,613,139.390.005

$1,613,139.39$1,613,139.39 3131.87597 $515.07

3131.87597

ni

A d di

d d

+ − + − = ⇒ =

= ⇒ = =

The monthly amount to save is $515.07. If the 6% APR is interpreted as 6% APY (a topic from Ch. 21), then the monthly interest rate is

1/121.06 1 0.00486755.− = The corresponding divisor is 2971.908212 (instead of 3131.87597), and she must put away $552.05 (using CPI for inflation 2001–2005) or $542.80 (using 3% annual inflation 2001–2005).

47. The expected Social Security monthly income of $1628 in 2001 dollars is $1628 195177.1 $1792.55× = in

2005 dollars, using the Consumer Price Index values from Table 21.5. Reduction of the benefit by 32.5% makes it $1792.55×(1− 0.325) = $1209.97 per month. Her desired income in 2005 dollars is $50,000/year = $4166.67/month, or $4166.67 − $1209.97 = $2956.70 more than Social Security will provide. To determine how much she must accumulate to have this much monthly for 32 years, we apply the amortization formula with payment d = $2956.70, n = 32×12= 384 months, and 0.06

12 0.005.i = =

( ) ( ) 3841 1 1 1 0.005

$2956.70 $504,229.690.005

ni

A di

− − − + − + = = =


( )48$504,229.69 1.03 $2,083,604.08.× = We use the savings formula to save up this amount A

over n = 48×12 = 576 months with monthly interest rate 0.0612 0.005.i = =

( ) ( )

( )

5761 1 1 0.005 1

$2,083,604.080.005

$2,083,604.08$2,083,604.08 3337.37879 $624.32

3337.37879

ni

A d di

d d

+ − + − = ⇒ =

= ⇒ = =

The monthly amount to save is $624.32.

If instead of the CPI from Ch. 21 we use 3% inflation per year (CPI is from Ch. 21), the expected

Social Security monthly income of $1628 in 2001 dollars is ( )4$1628 1.03 $1832.33× = in 2005

dollars. Reduction of the benefit by 32.5% makes it $1832.33×(1− 0.325) = $1236.82. Her desired income is $50,000/year = $4166.67/month, or $4166.67 − $1236.82 = $2929.85 more than Social Security will provide. To determine how much she must accumulate to have this much monthly for 32 years, we apply the amortization formula with payment d = $2929.85, n = 32×12 = 384 months, and monthly interest rate 0.06

12 0.005.i = =

( ) ( ) 3841 1 1 1 0.005

$2929.85 $499,650.750.005

ni

A di

− − − + − + = = =


( )48$499,650.75 1.03 $2,064,682.75.× =

Continued on next page

430 Chapter 22

47. continued We use the savings formula to save up this amount A over 48 12 576n = × = months with monthly interest rate 0.06

12 0.005.i = =

( ) ( )

( )

5761 1 1 0.005 1

$2,064,682.750.005

$2,064,682.75 $2,064,682.75 3337.37879 $618.65

3337.37879

ni

A d di

d d

+ − + − = ⇒ =

= ⇒ = =

The monthly amount to save is $618.65.

If the 6% APR is interpreted as 6% APY (a topic from Ch. 21), then the monthly interest rate is 1/121.06 1 0.00486755.− = The corresponding divisor is 3162.54921 (instead of 3337.37879),

and she must put away $658.84 (using CPI for inflation 2001–2005) or $652.85 (using 3% annual inflation 2001–2005).

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FAPP07 ISM 22 - UW-Madison Department of Mathematicsrobbin/141dir/propp/COMAP/Final...420 Chapter 22 Skill Objectives 1. Become familiar with the various types of loans. 2. Become