213841945-Chapter-5-Time-Value-of-Money-Multiple-Choice-Questions.pdf

May 7, 2018 | Author: Imran Zulfiqar | Category: Compound Interest, Interest, Mortgage Loan, Present Value, Interest Rates
Share
Report this link


Comments



Description

Chapter 5: Time Value of MoneyMultiple Choice Questions 1. What is the total amount accumulated after three years if someone invests $1,000 today with a simple annual interest rate of 5 percent? With a compound annual interest rate of 5 percent? A. $1,150, $1,103 B. $1,110, $1,158 C. $1,150, $1,158 D. $1,110, $1,103 Level of difficulty: Easy Solution: C. Simple interest rate: $1,000 + ($1,000)(5%)(3) = $1,150 Compound interest rate: $1,000(1.05)3 = $1,158 2. Which of the following has the largest future value if $1,000 is invested today? A. Five years with a simple annual interest rate of 10 percent B. 10 years with a simple annual interest rate of 8 percent C. Eight years with a compound annual interest rate of 8 percent D. Eight years with a compound annual interest rate of 7 percent Level of difficulty: Easy Solution: C. A) $1,000 + ($1,000)(10%)(5) = $1,500 B) $1,000 + ($1,000)(8%)(10) = $1,800 C) $1,000(1.08)8 = $1,851 D) $1,000(1.07)8 = $1,718 Therefore, C is the largest. Interest rates in the following questions are compound rates unless otherwise stated 3. Suppose an investor wants to have $10 million to retire 45 years from now. How much would she have to invest today with an annual rate of return equal to 15 percent? A. $18,561 B. $17,844 C. $20,003 D. $21,345 Level of difficulty: Medium Solution: A. PV=$10,000,000/(1.15)45=10,000,000/538.7693=$18,561 Or using a financial calculator (TI BAII Plus), N=45, I/Y=15, PMT=0, FV=10,000,000, CPT PV= –18,561 Solutions Manual 1 Chapter 5 Copyright © 2008 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission is strictly prohibited. 4. Which of the following is false? A. The longer the time period, the smaller the present value, given a $100 future value and holding the interest rate constant. B. The greater the interest rate, the greater the present value, given a $100 future value and holding the time period constant. C. A future dollar is always less valuable than a dollar today if interest rates are positive. D. The discount factor is the reciprocal of the compound factor. Level of difficulty: Medium Solution: B. The greater the interest rate, the smaller the present value, given a $100 future value and holding time period constant. 5. Maggie deposits $10,000 today and is promised a return of $17,000 in eight years. What is the implied annual rate of return? A. 6.86 percent B. 7.06 percent C. 5.99 percent D. 6.07 percent Level of difficulty: Medium Solution: A. FV=PV(1+k)n 17,000=10,000(1+ k)8 8ln(1+k)=ln(1.7), therefore k=6.86% Or using a financial calculator (TI BAII Plus), N=8, PV= –10,000, PMT=0, FV=17,000, CPT I/Y=6.86% 6. To triple $1 million, Mika invested today at an annual rate of return of 9 percent. How long will it take Mika to achieve his goal? A. 15.5 years B. 13.9 years C. 12.7 years D. 10 years Level of difficulty: Medium Solution : C. FV=PV(1+k)n (3)(1,000,000)=1,000,000(1.09) n ln(3)=(n)ln(1.09) n=12.7 years Or using a financial calculator (TI BAII Plus), I/Y=9, PV= –1,000,000, PMT=0, FV=3,000,000, CPT N=12.7 Solutions Manual 2 Chapter 5 Copyright © 2008 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission is strictly prohibited. 7. Which of the following concepts is incorrect? A. An ordinary annuity has payments at the end of each year. B. An annuity due has payments at the beginning of each year. C. A perpetuity is considered a perpetual annuity. D. An ordinary annuity has a greater PV than an annuity due, if they both have the same periodic payments, discount rate and time period. Level of difficulty: Medium Solution: D. The annuity due has a greater PV because it pays one year earlier than ordinary annuity. 8. Jan plans to invest an equal amount of $2,000 in an equity fund every year-end beginning this year. The expected annual return on the fund is 15 percent. She plans to invest for 20 years. How much could she expect to have at the end of 20 years? A. $237,620 B. $176,424 C. $204,887 D. $178,424 Level of difficulty: Difficult Solution: C.  (1  k )20  1  (1  .15)20  1 FV20  PMT   =$ 2,000    2,000(102.4436)  $204,887  k   .15  Or using a financial calculator (TI BAII Plus), N=20, I/Y=15, PV=0, PMT= -2,000, CPT FV=204,887 9. In Problem 8, what is the present value of Jan’s investments? A. $12,625 B. $12,519 C. $14,396 D. $12,396 Level of difficulty: Medium Solution: B.  1   1  1  (1  k )n  1  (1.15)20  PV0  PMT    $2,000   2,000(6.25933)  $12,519  k   .15      Or using a financial calculator (TI BAII Plus), N=20, I/Y=15, FV=0, PMT= –2,000, CPT PV=12,519 10. What is the present value of a perpetuity with an annual year-end payment of $1,500 and Solutions Manual 3 Chapter 5 Copyright © 2008 John Wiley & Sons Canada, Ltd. Unauthorized copying, distribution, or transmission is strictly prohibited. the total (principal and interest) owing will be: P + (n x P x k) = $1500 + (3 x $1500 x 6%) = $1770.500 short for this year’s tuition fees. Ltd.500 C. distribution. The total interest earned is (n x P x k) = (12 x $1200 x 0. there is no “compounding. The local bank pays simple interest at a rate of 0. how much will she pay each year? Topic: Simple Interest Level of difficulty: Easy Solution: As the exact amount of interest owing each year will be paid.200 more than he needs for tuition this year. or transmission is strictly prohibited. with interest due at the end of each year. these payments never reduce the principal owing.5 percent per month. Your sister has been forced to borrow money to pay her tuition this year. 12. Unfortunately. and the loan is for $2. B.400 D. His parents have agreed to loan him the money for three years at a simple interest rate of 6 percent.000 B. $13. $11. Solutions Manual 4 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.5%) = $72. in total. after three years? Topic: Simple Interest Level of difficulty: Easy Solution: A. $12.expected annual rate of return equal to 12 percent? A. but without compounding. How much will he owe. . In one year you will own P x k = $1500 x 6% = $90 of interest. $14. so the loan will never be paid off! 13.” The amount of each annual payment will be P x k = $2500 x 6% = $150. If she makes annual payments on the loan at year end for the next three years. a student finds himself $1. A.500 at a simple interest rate of 6 percent. PV0=PMT/k=$1. Unauthorized copying. After three years.12=$12. Khalil’s summer job has given him $1. How much interest will he owe his parents after one year? B. After a summer of travelling (and not working).500 Level of difficulty: Easy Solution: D.500 Practice Problems 11.500/. How much interest will he earn in one year? Topic: Simple Interest Level of difficulty: Easy Solution: Khalil will be paid interest each month for 12 months. Payment was made with wampum (likely glass beads and trinkets).14. Ltd.01 of interest will be earned. Suppose the Dutch had invested this money back home in Europe and earned an average return of 5 percent per year. History tells us that a group of Dutch colonists purchased the island of Manhattan from the Native American residents in 1626. 380 years later. How much interest will Khalil’s summer savings of $1. FV= -2500. 15.08)1 Or using a financial calculator (TI BAII Plus). calculated as: 1 1 PV0  FV1   $2500   $2314. and has decided to take out a loan. Simple interest? B. which had an estimated value of $24.500 one year from now. How much would this investment be worth today. . $1. 860 . Value = P + (n x P x k) = $24 + (380 x $24 x 5percent) = $480 B. years) 05$ 2 .81 Solutions Manual 5 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. PMT=0. 74 380 16. 704 .200 earn in one year with this online bank account? Topic: Compound Interest Level of difficulty: Easy Solution: The payment of compound interest means that we must compound (or find the future value) of the amount invested (the present value): FV12 months  $1200  (1  0. However.5 percent per month on deposits.200 was the original amount invested. using: A. David has been awarded a scholarship that will pay $2. The amount that can be borrowed is the present value amount. CPT PV=2314. Unauthorized copying. N=1.500. If the interest rate is 8 percent.01 Of this amount.81 (1  k )1 (1  . or transmission is strictly prohibited. I/Y=8. how much can he borrow so that the scholarship will just pay off the loan? Topic: Discounting Level of difficulty: Easy Solution: The future value of the loan (the amount to be repaid) is $2. Compound interest? Topic: Simple and Compound Interest Level of difficulty: Easy Solution: A.005)12  $1274. 602 . FV $ 380 24 ( 10 . A new Internet bank pays compound interest of 0. so $74. he really needs the money today. distribution. 15 percent compounded monthly.548.85  k   0.  0725  For Bank A. would like to donate some money to her alma mater to endow a $4.07) 4  PV3  PMT    $4000     $13. Bank A pays 7.20 percent compounded quarterly. and Bank C pays 7.000 at the end of each year for four years? Topic: Ordinary Annuities Level of difficulty: Easy Solution: Find the present value of the four-year annuity at year 3:  1   1  1  (1  k ) n  1  (1  0. k 1  17 . Bank B pays 7. 86   k  0 .17. How much will Grace have to invest today. at 7 percent. Unauthorized copying.85   3  $11. Stephen. who plans to begin university in three years. The university will manage the funds. a retired librarian. or transmission is strictly prohibited. 142 . Instead. Which bank pays the highest effective annual rate? Topic: Effective Annual Rates Level of difficulty: Easy Solution: 2 0 . .).000 annual scholarship.  07 18.90  (1  k )   (1. and expects to earn 7 percent per year. Grace. to be able to give Stephen $4.07)  19.548. Grace decides that creating a perpetual scholarship is too costly (see Problem 17Error! Reference source not found.25 percent interest compounded semi-annually.07      Now.059.38 %  2 Solutions Manual 6 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. distribution. How much will Grace have to donate so that the endowment fund never runs out? Topic: Perpetuities Level of difficulty: Easy Solution: Present value of the perpetual scholarship payment:   1  1   PV 0   PMT$ 4000   $ 57 . find the present value of this amount today:  1   1  PV0  FV  3  $13. Ltd. she would like to support the education of her favourite grand-nephew. 12  9. Annually B. the effective annual rate will be the same as the quoted rate. .40 %  4 12 0. Jimmie is buying a new car. If Alysha puts $50. 20. k 1  17 . 4 0 . Ltd. Unauthorized copying.84%  4  C.5%  k  1    1  9. Quoted Rates Level of difficulty: Easy Solution: A.06)  $53. set m=12. Calculate the effective annual rate if the compounding occurs: A. With monthly compounding. Daily? Topic: Effective vs. set m=4. For annual compounding. 4  9.000 in a savings account paying 6 percent per year. or transmission is strictly prohibited.  0715  For Bank C. Monthly Topic: Determining Effective Annual Rates Level of difficulty: Easy Solution: A. k 1  17 . how much money will she have in total at the end of the first year if interest is compounded: A.5%  k  1    1  9. 5%  k 1 1  1 9. To check this: m 1 QR  9.5 percent per year for a car loan. 5% m  1  B. Quarterly C. Annually? B.  0720  For Bank B. k  Quoted Rate  6%  FV1 year  PV0 (1  k )  $50.39 %  12 Bank B pays the highest effective annual rate. Monthly? C. distribution.92%  12  21. His bank quotes a rate of 9. With quarterly compounding.000 Solutions Manual 7 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.000  (1. How much will his investment be worth in five years? Topic: Investing Early Level of difficulty: Easy Solution: FV5 years  $550. as such. Suppose KashKow Inc. He thought he could fund his retirement. Mary-Beth is planning to live in a university residence for four years while completing her degree.12 Each share is worth $16.091.67 k 0.800 and must be paid at the start of each school year.000  (1. they are often viewed as entities that will pay dividends to their shareholders in perpetuity. pays a dividend of $2 per share every year.061678)  $53. k  1    1  6.67. Unauthorized copying. 24. 23. . 12  QR  B. distribution. or transmission is strictly prohibited.90  12  365  QR  C.55  365  22. what is the present value of all the future dividends? Topic: Perpetuities Level of difficulty: Easy Solution: The value of any perpetual stream of payments can be valued as a perpetuity: PMT $2 PV0    $16.000 when he was 55 years old. The annual cost for food and lodging is $5.083.000  (1. this is not an ordinary annuity.000  (1  0. an annuity due.1831%  FV1 year  $50. Ltd. k  1    1  6.50 Tony will have less than his friend in five years because he is not adding more savings to his account.10) 5  $885. Tony sold the business for $550. If the discount rate is 12 percent.1678%  FV1 year  $50. Public corporations have no fixed life span. Tony started a small business and was too busy to consider saving for retirement. because this was a lot more than his friend had amassed in his account. but rather. What is the total present value of Mary-Beth’s residence fees if the discount rate (interest rate) is 6 percent per year? Topic: Annuities Due Level of difficulty: Easy Solution: Because the fees are paid at the start of the year.780. Solutions Manual 8 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.061831)  $53. Tony can invest this total sum and earn 10 percent per year. Unauthorized copying. The effectively monthly rate is: m 365 QR f . Ltd.24 12 A. m=365. on average. Gilda invests $20. f=12.24 4 B.303.12%.000 in a mutual fund which has earned 10 percent per year. 3 . Continuous compounding: k  e.11%. compounded semi-annually E.06)  $21.24  1  27.06    25. m = 365: k  (1  )  1  27. compounded every four months D. m=2. 24 percent.25%.44%. k  (1  )  1= (1  )  1 =1.94% m 3 m 2 QR f . 24 percent. Solution: . compounded quarterly C. m = 2: k  (1  )  1  25. m = 3: k  (1  )  1  25. k  (1  )  1= (1  )  1 =1. Calculate the effective annual rates for the following: A.24 12 B.24 12 C. f=12. m = 4: k  (1  )  1  26.96% m 4 m 3 QR f . 24 percent. 4 . k  (1  )  1= (1  )  1 =1.97%. f=12 k  (1  )  1= (1  )  1=2.  1  1  (1  0. If this rate of return continues. m=4.06)4  PV0  $5800     (1  0. 2 E. f=12. or transmission is strictly prohibited. m=3.24 365 A. how much will her investment be worth in: Solutions Manual 9 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.24 3 C. On the advice of a friend. distribution.24 2 D. 365 . Calculate the effective monthly rate for A to D. Level of difficulty: Medium. compounded daily B.02% m 365 m 4 QR f . in recent years.91% m 2 26. .24 12 D. F. compounded continuously F. 24 percent.47  0. 24 percent. . Unauthorized copying.77 10 better off after 10 years.105)  $54. A.10)5  $32.100. calculate how much better off Gilda would be if she invested $20. he wants to throw a big party. C.000  (1  0.62 – $51. You are ($22.20) = $738. A. FV1 year  $20. FV1 year  $20. ten years? Topic: Compound Interest Level of difficulty: Medium Solution: A.000  (1  0.10)10  $51.000  (1  0.000) = $100 better off 1 after one year.948. It has had an average annual return 0.281.281.10)1  $22.210. how much does he have to invest today if he can earn a compound return of 5 percent per year? Topic: Discounting Level of difficulty: Medium Solution: Jon needs $800 in three years. Topic: Compound Interest Level of difficulty: Medium Solution: First find the value of the investment after each period of time. B. FV10 years  $20.874.000  (1  0.406. one year? B.85 27. distribution. You are ($32.00 . FV5 years  $20.000  (1  0.000  (1  0.5 percent greater than the one Gilda’s friend recommended (see Problem 26).105)  $32. 28.85) = $2.000 in this mutual fund. When Jon graduates in three years.000.62 You are ($54.105)  $22.94 .948. Your investment research has turned up an interesting mutual fund. that is the future value amount.210. For each time period from Problem 26. and then compare to the values from Problem 26 to determine how much difference a small change in the interest rate can make. To have this amount available. FV5 years  $20. FV10 years  $20.94 – $32. which will cost $800. five years? C.74 5 better off after five years. The present value equivalent is: Solutions Manual 10 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.00 B.100 – $22. Ltd.20 C. or transmission is strictly prohibited.874. In Problem 28.05)  $575 B. Felix’s parents can only afford to save $3.000 by the time Felix needs the funds in eight years? Assume Felix’s parents can earn 7 percent (compounded annually) on their savings. if the return on the investment will be: A. FV= -800. CPT PMT= 3898. Felix decided that he wanted to attend a very prestigious (and expensive) university. Topic: Ordinary Annuities Level of difficulty: Medium Solution: The future value amount is $40.07)8  1 $40. N=3. Jon will earn $500 5%  $25 per year in interest. PV=0. .71 31. What rate of return would they require on these savings if they must accumulate $40. distribution. FV= -40. Unauthorized copying.07 29. simple interest at 5 percent per year? B. I/Y=7.71  0.000  PMT    PMT  $3898. At the age of 10. How much can he plan to spend on the graduation party in three years. PMT=0. The value of his investment (or the amount available to spend on the party) will be: Value  P  (n  P  k )  $500  (3  500  0. N=8. compound interest at 5 percent per year? Topic: Simple and Compound Interest Level of difficulty: Medium Solution: A.000.000? Topic: Ordinary Annuities (Solving for IRR) Solutions Manual 11 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. or transmission is strictly prohibited. The amount to be saved each year is really the payment on an ordinary annuity:  (1  0. and that each year’s savings are deposited at the end of the year. The interest earned grows (compounds) each year.81 30. How much will his parents have to save each year to accumulate $40. 1 1 PV0  FV3   $800   $691. Ltd.000. suppose Jon had only $500 to invest. the total available in three years is: FV3  PV0  (1  k )3  $500  (1  . CPT PV=691.05)3  $578.05)3 Or using a financial calculator (TI BAII Plus). I/Y=5.07  Or using a financial calculator (TI BAII Plus). which begins in eight years.000 per year for his university education.07 (1  k ) 3 (1  . 11  0. Level of difficulty: Medium Solution: Solve for the interest rate (or internal rate of return) on an ordinary annuity.11  PMT  (1  0.3% because of taxation. how much can John withdraw from the account at the beginning of each year for his four years at university? The account will continue to earn 9 percent per year. but is easily handled with a financial calculator (TI BAII Plus).000 each year. However. Note that we must use a negative sign for the annual payment (savings) or the future value amount. distribution. PMT = 3. Shortly after John was born. FV= –40. John’s withdrawals at the beginning of each year are essentially “payments” on an annuity due:  1  1  (1  0.000.95  0.30)  6.063)4  28. How much will John’s account be worth after 17 years? Topic: Ordinary Annuity (Future Value) Level of difficulty: Medium Solution: Taxation of the interest income has the effect of reducing the rate of return.973. but not both.063    34. his parents began to put money in a savings account to pay for his post-secondary education.3% . Using this rate of return we find the future value of John’s  (1  0. In effect: k  9%  (1  0. They deposit the money into a Registered Education Savings Plan (RESP) account so that no tax is Solutions Manual 12 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. Topic: Annuities Due Level of difficulty: Medium Solution: As in Problem 32. Unauthorized copying. and earn a return of 9 percent per year.2067% 32. Based on the amount accumulated (from your answer in Problem 320).063  33.063)  PMT  $7919. This is quite difficult to do algebraically. John’s parents used a regular savings account to save for his post-secondary education. k  9%  (1  0. N=8. but interest income is taxed at a rate of 30 percent.063)17  1 account to be : FV17  $1000     $28. They save $1. PV=0. .000 per year for 17 years to pay for her university tuition costs. or transmission is strictly prohibited. CPT I/Y=14. the interest income is taxed each year at a rate of 30 percent. Jane’s parents save $1. Ltd.30)  6.973.000. Stephen has learned that his great-aunt (see Problem 18Error! Reference source not found. the government will add 20 percent to any money contributed to an RESP each year.06)17  1 FV17  $1000     $28. .46  PMT  (1  0.06  B. Based on the amount accumulated (from your answer in Problem 340). Including these grants. so Stephen would like to receive his payments at the beginning of each school year. As an incentive to save for higher education. The future value of the account will now be:  (1  0. this amount will be subject to income tax at a rate of 15%.212. or transmission is strictly prohibited. how much will Jane have in her account? Topic: Ordinary Annuity (Future Value) Level of difficulty: Medium Solution: A.06)  PMT  $9217. payable on the interest income. Solutions Manual 13 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. The grant has the effect of increasing the amount saved from $1.) intends to give him $4. Jane would like to withdraw the same amount of money at the beginning of each year of her four-year degree program.36  0. and the account will earn 6 percent per year on any remaining funds.000 each year he is studying at university.000 to $1. distribution. The future value of Jane’s account will be:  (1  0. A.75 36. Ltd.88  0. Jane’s parents used an RESP account to save for her post-secondary education.200.855. Tuition must be paid in advance. This RESP account provides a return of 6 percent per year. The net amount that Jane will have available for tuition is then: 9217.06  35.855. How much will Jane have available for tuition each year? Topic: Annuities Due Level of difficulty: Medium Solution: Each year. Jane can withdraw:  1  1  (1  0.36%  (1  0.06)4  33.15)  $7834.06)17  1 FV17  $1200     $33. How much will Jane’s account be worth when she begins her university studies? B.46  0.06    However. Unauthorized copying. All funds (interest and principal) withdrawn from this account are taxed at a rate of 15 percent. Stephen’s great-aunt has decided to save the money and pay off Stephen’s student loans when he finishes his degree.834. with monthly compounding.497. Ltd.000.  (1  0. Unauthorized copying. set m=12. . How much will his monthly car payments be if he obtains a loan that is amortized over 60 months.497. distribution. to make the four annual (start-of-year) payments? Topic: Annuities Due Level of difficulty: Medium Solution: Find the present value of the four-year annuity due:  1   1  1  (1  k )n  1  (1  0.07  38. Jimmie’s new car (see Problem 20) will cost $29.07)  $14. or transmission is strictly prohibited. and the nominal interest rate is 8.07      Now. find the payment (amount to be saved) for an ordinary annuity with a future value of $16.7083%  m   12  The 60 car payments form an “annuity” whose present value is the amount of the loan (the price of the car): Solutions Manual 14 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.26   3  $11. Rather than give her grand-nephew some money each year while he is studying. The total amount owing at that time will be $16.000  PMT    PMT  $1. find the effective interest corresponding to the frequency of Jimmie’s car payments (f =12).07)4  PV0  PMT  (1  k )  $4000     (1  0.848.07)  37.000.5 percent per year with monthly compounding? Topic: Effective Interest Rates and Loan Arrangements Level of difficulty: Medium Solution: First.85  0.5%  12 kmonthly  1    1  1    1  0.07) 7  1 FV4  $16.26  k   0. How much will she have to save each year until that time if her investments earn a return of 7 percent per year? Topic: Ordinary Annuities Level of difficulty: Medium Solution: Assuming the savings are invested at the end of each year.08  (1  k )   (1. How much will his great-aunt have to invest today at 7 percent.000. discount this amount back three years:  1   1  PV0  FV  3  $14. m 12  QR  f  8. 559.00 The first monthly payment repays $389.977.823.42 8 26. 59 1.11 24.214.97 421.88 395.42 403. k=0.22 37 13.98 170.90 27.98 424.18 590.42 26.98 197.84 7 26.08 397.11 594.138.13 594. Create an amortization schedule for Jimmie’s car loan (see Problem 38Error! Reference source not found. 35 14.083.34 586.024. 40.11 5 27.94 10 25.805.87 415.177.007083)60  $29.218.01 4 27.98 191. What portion of the first monthly payment goes towards repaying the principal amount of the loan? What portion of the last monthly payment goes towards the principal? Topic: Loan Arrangements Level of difficulty: Medium Solution: Use the effective monthly interest rate from Problem 38.98 199.05 24.98 202.44 2 28.392.79 412.000.138.93 418.42 389.94 594.26 400.98 8.01 24. Ltd. What is the present value of this amount? Solutions Manual 15 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. .22 13.56 26.00 594.22 594.69 409.587.76 495.76 .13 9 25.29 25.587.44 594.98 176.610.977.7083% (4) (3) Ending (1) Principal (2) Principal Period Interest Principal Outstanding Payment Repayment =k*(1) = (1)-(4) = (2)-(3) 1 29.82 11 24.96 .73 13.12 3 28.218.024..72 27.98 185. determine how much Jimmie still owes on the car loan after three years of payments on the five-year loan.823.392.79 594.805.000  PMT    PMT  $594.586..66 392.95 594.98 179.98 92.98 182.01 594.79.620.56 406.79 60 590.610..214.12 594.98  0.).559.18 594.76 13 24.98 99.98 173.72 502.007083    39.00 23.19 25.77 594.25 498.43 594.98 205.40 6 27.40 594. distribution.089. Unauthorized copying.84 594.  1  1  (1  0.42 594.76 594.425.56 28. Using the amortization schedule from Problem 39.620.98 4.26 12.64 590.82 594.98 188.79 0.56 of the principal amount of the loan and the last payment repays $590.98 96.32 28. or transmission is strictly prohibited.77 12 24..714.98 194.425.95 36 13.10 27.089. 02 (see Problem 38Error! Reference source not found.152.). How big a down-payment must Jimmie make on the $29. PMT=588.22 (from the amortization table). k=0.000.6667% Note that we used N=60 months.000. PV= -29.22     $10.5 percent with monthly compounding? Topic: Loan Arrangements Level of difficulty: Medium Solution: Use the effective monthly interest rate from Problem 38.000. so the solution is a monthly interest rate.7083% Find the present value of Jimmie’s 36 payments:  1  1  (1  0. Unauthorized copying. or transmission is strictly prohibited. What is the effective annual interest rate on this loan? What would the quoted rate be? Topic: Loan arrangements and Effective Annual Rates Level of difficulty: Medium Solution: The 60 monthly payments form an annuity whose present value is $29. the problem asks for the effective annual rate.30 36   (1  0. however.30% The quoted rate would be: Solutions Manual 16 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.000 that requires 60 monthly payments of $588.000 car if the nominal interest rate is 8. or 36 monthly payments.98   $18.00  $18. the principal outstanding is $13.007083)36  PV0  $594. CPT I/Y = 0.027.006667)12  1  8.05 42.089.98.089. but can only afford monthly payments of $594. Jimmie would like to pay off his car loan in three years (see Problem 38).95  $10. The present value of this amount is:  1  PV0  $13.95  0.847.007083    Therefore.847. Topic: Loan Arrangements and Discounting Level of difficulty: Medium Solution: After three years. Finding the interest rate is most easily done with a financial calculator (TI BAII Plus): N=60. k  (1  kmonthly )12  1  (1  0. Ltd. Jimmie must make a down payment of $29. . Jimmie is offered another loan of $29.152. distribution.007083)  41. The Business Development Bank is willing to loan Su Mei the $25.00  $6935. Unauthorized copying. what is the nominal (annual) interest rate on this loan? Topic: Determining Rates of Return and Effective Interest Rates Level of difficulty: Medium Solution: A. N= 5. With annual compounding.0%  12. PMT=6935. is: QR  m  kmonthly  12  1. With monthly compounding. CPT I/Y = 12% The effective annual interest rate is 12 percent. CPT I/Y = 1. The compounding period matches the payment frequency. PV = -25. To start a new business. Ltd.24. so the nominal rate. What is the effective monthly rate on this loan? B.00% Or simply: QR  m  kmonthly  12  0.0% per year.00% 43. or transmission is strictly prohibited. what nominal rate would the bank quote for this loan? Topic: Determining Rates of Return and Effective Interest Rates Level of difficulty: Medium Solution: Solve the annuity equation to find k.0% is the effective monthly rate. The loan will require monthly payments of $556. A. N= 60.000. 44. There will be 5 x 12 = 60 monthly payments.000. FV=0. Solutions Manual 17 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.000. PV = –25. and months as the time period.000 she needs to start her new company.0% Because we used monthly payments. 1.24   k ?  k    The calculations are most easily done with a financial calculator (TI BAII Plus). 1 1 QR  m  [(1  k ) 12  1]  12  (1  0.006667  8. the nominal rate (or quoted rate) will also be 12 percent per year. With annual compounding. If the bank asks her to repay the loan in five equal annual instalments of $6. PMT=556.11 over five years.24. determine the bank’s effective annual interest rate on the loan transaction. B. distribution. .11.000 from a local bank. or quoted rate.935. The calculations are most easily done with a financial calculator (TI BAII Plus). Su Mei intends to borrow $25. the interest rate:  1  1  (1  k )5  $25.0830) 12  1  8. What is the effective monthly interest rate? C. Unauthorized copying. 46. With 52 weeks in the year.1% – 1=2. For the initial three-year term. Ltd. or transmission is strictly prohibited. and why? Topic: Effective Interest Rates Level of difficulty: Medium Solution: The two loans have the same principal amount. the loan in Problem 43 is a better deal because the effective interest rate charged is lower. fixed-rate mortgages use semi-annual compounding of interest.1% 47. What is the nominal interest rate per week? Per year? B. What is the effective annual interest rate? B. To avoid a visit from the “collection agency. B. .01)12  1  12.872.40 percent. Which is the better deal. How much will Josephine’s monthly mortgage payments be? Topic: Mortgage Loans and Effective Interest Rates Level of difficulty: Medium Solution: A. Scott visits a loan shark for a $750 loan.772. the effective annual rate is: k  (1  kmonthly )12  1  (1  0. Nova Scotia. After losing money playing on-line poker.” he will have to repay $800 in just one week. the nominal rate per year is then 52 x 6. The effective annual interest rate is k  (1  0.67% per week. What is the effective annual interest rate? Topic: Effective Interest Rates Level of difficulty: Medium Solution: A.7% . the effective annual rate was 12%.772. so m=2.000 to purchase her new house in Yarmouth.0667) 52  1  27. This implies a nominal interest rate of 50/750 = 6. For Problem 44. A. so they cannot be compared on that basis. In Canada. You will pay interest of (800–750) = $50 after one week. Providence Bank has offered her a quoted annual rate of 6. one requires monthly payments. making monthly payments. the other annual. In Problem 43.7210  2. We need an effective rate for the same period of time for comparison. The effective annual rate is therefore: Solutions Manual 18 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. Compare the loans in problems 043 and 440. Josephine needs to borrow $180. Clearly.45. She would like to pay off the mortgage in 20 years. distribution.67%.10% 2.67% = 346. A. We can find the effective monthly interest rate from the effective annual rate. or from Providence Bank (see Problem 470)? Can you answer this question without calculating the monthly mortgage payment? Topic: Mortgage Loans and Effective Interest Rates Level of difficulty: Medium Solution: With monthly compounding and payments.005264    48.005264) 240  $180. 49. The monthly payment for the Credit Union mortgage would be $1. If Josephine can get the same interest rate for a second three-year term as she did originally.5024% 12  1  0. is higher than that at the Bank. how much will her monthly payments be now? Topic: Mortgage Loans Level of difficulty: Medium Solution: If we assume that Josephine does not change the amortization period (now 17 years). Josephine’s monthly payments can be computed as:  1  1  (1  0. the effective monthly interest rate is: m 12  QR  f  0. the compounding will occur monthly. We can confirm this by computing the payment for a mortgage with a principal amount equal to the amount now Solutions Manual 19 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. but unlike most Canadian mortgages. The Yarmouth Credit Union will provide Josephine with a mortgage at a rate of 6. f=12.36 percent.24.327.69  0.0636  12 kmonthly  1    1  1    1  0. Unauthorized copying.5264% 1 1 f C.322. then the same interest rate will result in making the same monthly payments.530%  m   12  Even though the quoted rate is lower at the Credit Union than at the Bank (see Problem 470). . Ltd. Assume Josephine chose the Providence Bank option (see Problem 470). or 20 x1 2 = 240 months.000  PMT    PMT  $1. or transmission is strictly prohibited. which.5024%  m   2  B. Should Josephine take out the mortgage loan from the Credit Union. the effective rate is higher. The amortization period is 20 years. k: kmonthly  1  k   1  1  6. as expected.064  k  1    1  1    1  6. With monthly payments. m 2  QR   0. Josephine should take the mortgage loan from the Bank in this case. distribution. m=2 and f=12.38  PMT    PMT  $1322. If his savings earn 10 percent per year.69  0.039  12 kmonthly  1    1  1    1  0. or transmission is strictly prohibited. distribution.38 based on Problem 47). The real-estate market has been quite active. therefore. $374. . A.000. Ltd. starting on his 21st birthday (40 years of saving). Charlie’s bank has offered him a great mortgage rate of 3. that lasts 204 months (17 years):  1  1  (1  0. Unauthorized copying. As a long-time customer.72  0.172.000 available as a down payment. He has $130.003224    In addition. If the loan will be amortized over 25 years.950.72.005264)204  $165.172. and may sell for more than the asking price. Charlie is very eager to purchase this house.3224%  m   2  The present value of the mortgage payments over the amortization period (25 years x 12 = 300 months) is:  1  1  (1  0.000 = $504. what is the most that Charlie can afford to pay for the house? Topic: Mortgage Loans Level of difficulty: Medium Solution: With semi-annual compounding (the norm in Canada) and monthly payments. Ontario is for sale with an asking price of $499. Charlie has $130.72 + $130. at what age will the value of his savings plan be worth $1 million? Topic: Investing Early Level of difficulty: Medium Solution: Solutions Manual 20 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.00 per month on a mortgage loan. the most he can pay for the house is. 51.003224)300  PV0  $1950. and can pay up to $1. The effective monthly rate is: m 2  QR  f  0.000 available for a down payment. will Timmy achieve his goal? B. Timmy sets himself a goal of amassing $1 million in his retirement fund by the time he turns 61. but is concerned that he may not be able to afford it. A lakefront house in Kingston.000 each year. Will Timmy be able to retire before he turns 60? That is.00     $374. so the house will almost certainly attract several offers. He begins saving $3.005264    50.553.90 percent on a one-year term.553. outstanding ($165.553. 327.  (1  k )n  1 FVn  PMT  (1  k )  k   (1  .000)(497. His expected annual expenditure is $36. there always seemed to be a reason not to save money.  (1  0. Unauthorized copying.000.000     $953. or transmission is strictly prohibited. he began to save. A. B.6 percent.126  Solutions Manual 21 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.000 (1.000. Fortunately.10  No. CPT N = 37.10  Yes. Calculate the amount Jack will receive when he retires. FV = –1. In the equation for part A set FV=$1. He works in a local bank and has an annual after-tax income of $45. so this is an ordinary annuity. Timmy’s savings extend right to age 61 (end of each year). If these savings earn 10 percent per year. will Tommy achieve his $1million goal at the desired time? Topic: Investing Early Level of difficulty: Medium Solution: This is an ordinary annuity.000. Tommy’s executive-level job allowed him to save $30.  (1  0.2749)(1. Jack is 28 years old now and plans to retire in 35 years. so he put it off for many years.000 – $36. Timmy will hit the $1 million dollar mark in just over 37 years.777. However. Tommy will not quite achieve his goal by age 61. I/Y = 10.10)15  1 FV15  $30.000     $1.000. Tommy set the same retirement goal as his friend Timmy (see Problem 510).45  0.000.10) 40  1 FV40  $3. . n. 53. This is an annuity due.000 per year.000.000 = $9.0 3 9. distribution.67  0. 52. This is easiest done with a financial calculator (TI BAII Plus). and solve for the number of years. or shortly after his 58 th birthday.1 2 6)  $5. Finally.000 and the rest of his income will be invested at the beginning of each of the next 35 years at an expected annual rate of return of 12. PMT = 3000. Timmy will achieve his goal by a comfortable margin.1. Ltd.126)  (9.174. with just 15 years to retirement. Solution: Annual investment = Annual income – Annual expenditure = $45. Level of difficulty: Difficult.126)35  1  9.3 8 4  .  (1  k )35  1  (1  .475. I/Y=12. CPT PV=79. or transmission is strictly prohibited. In Problem 53.039. PMT= –9. Hit [2nd] [BGN] [2nd] [Set] N=35.039. N=35.000)(7. CPT FV=5. Unauthorized copying. CPT PV= . where Solutions Manual 22 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.000(497.000.165 2).384/(1.039. Ltd. Level of difficulty: Difficult Solution: Now it is an ordinary annuity.165 55.6.2749)  $4.475. Because each annual investment is compounded one year less in an ordinary annuity.474<5. if Jack prefers to invest a lump-sum amount today instead of investing annually.039. how much he will receive when he retires? Explain why this amount is greater or smaller than the result calculated in Problem 53.126)  (9. PMT= -9.6566=$79. I/Y=12. N=35.384.8118311)(1. if Jack invests the same annual amount at the end of each of the next 35 years instead of at the beginning of the years.000.000.126  Or using a financial calculator (TI BAII Plus).126)  $79. So. FV of an ordinary annuity is less than an otherwise identical annuity due.384.6. PMT= -9. I/Y=12.6. Hit [2nd] [BGN] [2nd] [Set] N=35. PMT=0.126) 35  1 FV35  P M T  = 9.000    9. CPT FV= 4.126    Or using a financial calculator (TI BAII Plus).000 (1.79. .126)35=5.474  k   .  1  1  (1  k )n  PV0  PMT  (1  k )  k     1  1  (1. but still expects to receive the same amount of money when he retires. FV=5.165 Or using a financial calculator (TI BAII Plus).039. distribution.165  .384 54. In Problem 53. 4. PV=0. PV=FV/(1+k)n=5.6.384/63. PV=0. Or using a financial calculator (TI BAII Plus).475. how much should he invest today? Level of difficulty: Difficult Solution: There are two ways to get the answer: 1). I/Y=12. FV=0.474.126)35   $9. As expected. N=120.000.12 12 C. PMT=. The difference is due to the compound factor (1+k).01     1  1   (1  . 10-year loan with an interest rate of 12 percent.920 2 .869  =$182. Determine the month-end payment for a $200. N = (10)(12)=120 months  1  1  (1  .2.000 /    . Ltd.01)102  PV0  2. PMT=$2.000. while in the annuity due this amount is invested at the beginning of each year. Unauthorized copying. distribution. Alternatively. A.000  PMT    .9759% 2 Solutions Manual 23 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. kmonthly= (1  )  1 =. Remaining months to pay=120 – 18=102 months  1  1   (1  . CPT PMT=2. we can look at Equation 5-4 and Equation 5-6.000.01    Or using a financial calculator (TI BAII Plus). FV=0. assuming it is a mortgage loan with monthly payments. Calculate the outstanding loan amount after 18 months. PV=-200. or transmission is strictly prohibited.920  . A. I/Y=1. (Assume there is no down payment) B. Solution: PV=200. FV=0. CPT PV=182. 56. N=102.869. Redo part A. compounded monthly. monthly rate=12%/12=1%. C.869 Or using a financial calculator (TI BAII Plus). . I/Y=1. Level of difficulty: difficult.01)120  PMT  200. the annual investment is at each year-end.01)120  200.01    So.869 B. Ltd.000.19  . I/Y=16.218. beginning in five years.16183424)5=$11. compounded quarterly. Level of difficulty: Difficult Solution: Investor A: k=e.749.183424. continuously compounded. FV=0. PV=-200.009759)120  PMT  200. or transmission is strictly prohibited. I/Y=.23. I/Y=16. discount this amount for five years back to today when she is 20. CPT PV=11. distribution.16183424)8  PV25  5. Determine the yearly payment Investor B has to make in order to have the same present value as Investor A. N=120.009759)120  200.009759     1  1  (1  . The rate of return on her investment is 15 percent. N=8. Investor B is 40 years old and he just started to invest at the beginning of every year an equal amount of money starting today.500  =$23. N=5. 1st.15 – 1=16. PMT=$2. The rate of return on his investment is 16 percent. PV=FV/(1+k)5=23.009759    So. Unauthorized copying. PMT=0. She plans to invest $5. PMT=5.000  PMT    . CPT PV=. .183424%. He will invest for 10 years.749.9759. Investor A just turned 20 years old and currently has no investments. CPT PMT=2. consider an ordinary annuity and the present value of the investment when A turns 25 years old is:  1  1  (1  . FV=0.500 at the end of each eight years.749.000 /    .23.19/(1.19.16183424    Or using a financial calculator (TI BAII Plus).183424.218.500.836 Or using a financial calculator (TI BAII Plus).749.3231 Or.  1  1  (1  .19 2nd. FV=.3231 Investor B: Solutions Manual 24 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.836 57. 303. and $1. Investor B has to make an equal amount of $2.16985856)10  11.44. Determine how much they will have when they retire.000 after taxes each month.16985856)  . A.535 + $50.353.303.400 2nd Calculate the FV of their investment when they retire:  (1  . CPT FV=7. $850 in car payments.218.3231. Hit [2nd] [BGN] [2nd] [Set] N=10. PV=11. Paul and Maria want to have enough money to travel around the world when they retire.38 Or using a financial calculator (TI BAII Plus).000 in mortgage payments.16 4 k= (1  )  1=16.535  .1)30  1 FV30  44.700 Yearly available=$(3.3231  PMT  (1. PMT=.000 = $7.535 3rd Calculate the amount they will have when they retire: $7.218. FV=0. They approached a fund manager and decided to invest the rest of their income at the end of each year. How much can Paul and Maria withdraw annually at the beginning of the year for travelling after they retire if they expect to live until they are 90? A.985856% 4  1  1  (1  . CPT PMT= . Unauthorized copying. Their monthly expenditures include $3. I/Y=16. N=30. or transmission is strictly prohibited.400.057.400  =$7. I/Y=10.303. .057.2.38 so that the present value of the two investments is the same. they will sell their cottage for an expected price of $50. Ltd. distribution. Level of difficulty: Difficult Solution: 1st Calculate their yearly income available for investment Monthly income available=$9. B.450 in other expenses.985856. Level of difficulty: Difficult Solution: Solutions Manual 25 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. 58.000 – $850 –$1.000 – $3. They just turned 30 and will retire when they turn 60.1  Or using a financial calculator (TI BAII Plus). They expect to earn a 10 percent expected annual rate of return for each of the next 30 years. They earn a total of $9.057. PV=0.700)(12)=$44.000.450=$3.38 Therefore. When they retire.535 B.16985856    PMT=$2. . 353. n=30  1  1  (1  . $500 $800 For Investment A: PV0    $453.05)3 1 2 Veda would prefer Investment B. PV=7. If the cost of each investment in Problem 59 0 is $1.69  $1157. CPT PMT=709. Which investment should Veda choose? Year Investment A Investment B 1 $0 $200 2 $500 $400 3 $800 $700 Topic: Discounting Level of difficulty: Difficult Solution: Find the present value of the money paid back to Veda by each investment. 60. PMT=709. I/Y=10.7. Unauthorized copying. using the interest rate on the alternative (the bank account) as the discount rate. FV=0.353.535. This is an annuity due problem.1    So.48  362.300 but at different times. .143 Or using a financial calculator (TI BAII Plus). because it has the higher present value.58 (1  0.1)  . k=10%. PV=.1)30  7.05)3 2 For Investment B: 200 400 700 PV0     190.353. she could leave the funds in a bank account paying 5 percent per year. Both investments will ultimately pay $1.535. distribution. or simply leave the money in the bank account? Would her decision change if the investments cost $1. as shown in the table below.05) (1  0.200 each? Topic: Discounting and Determining Rates of Return Level of difficulty: Difficult Solutions Manual 26 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.000. Hit [2nd] [BGN] [2nd] [Set] N=30.05) (1  0.81  604.05) (1  0. Ltd.535  PMT  (1  .143 59. If Veda doesn’t choose one of these investments. Veda has to choose between two investments that have the same cost today. should Veda invest in one of them.98 (1  0.51  $691. or transmission is strictly prohibited.07  $1144. either investment would be preferable to the bank account (returning 5%).200.11  PMT    PMT  $3. . Solution: Observe that the present value of the money returned by both investments. is greater than the $1.000 per year for four years to pay for tuition (see Problem 30). Therefore. Investment B is more preferable. they both have a positive “net present value. distribution. when Felix is due to begin university studies):  1  1  (1  0. the first step involves an “annuity due. First. Unauthorized copying. as the present value of the money returned is less than the cost (they have a negative net present value).44  0 . find the present value of the four annual tuition payments (at time 8. neither would be acceptable.07) 8  1 $33.000 cost (see Problem 59). and the savings amounts.11  0.301.” as the tuition payments occur at the start of each year: Solutions Manual 27 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. However. using the 5 percent discount rate.” On this basis. Topic: Annuities Due Level of difficulty: Difficult Solution: This problem can be solved in the same two step manner used in Problem 61.000 lump sum.07)4 PV8  10. 61. find the annual savings amount:  (1  0.872. How much will Felix’s parents have to invest at the end of each year for the eight years before he begins his studies if their savings earn compound interest at 7 percent per year? Assume the tuition payments also occur at the end of each year.07      This is the amount of savings required at time 8.872. Topic: Ordinary Annuities Level of difficulty: Difficult Solution: We have two separate annuities to consider: the tuition payments. From the perspective of time 0.) Next. Repeat Problem 61 0assuming the tuition payments occur at the beginning of each year (the savings are still invested at the end of each year).000 in Problem 0. Felix will need $10. this is a future value amount (replaces the $40. Ltd. or transmission is strictly prohibited. If the cost of the investments was $1.000     $33.07  62. Instead of a $40. Unauthorized copying.07)  $36.5.243. or transmission is strictly prohibited. I/Y=7. or turn off “BGN” mode for this step. FV= -20.243.54 63. I/Y=0.000. How many years will it take for an investment to double in value if the rate of return is 9 percent. We know the future value and present value amounts. as this is an ordinary annuity. .000 in his savings account.000.000.07)4 PV8  10. FV=0. FV = -20. distribution. or close to 5 years before Roger can afford to buy the car! B. Roger has his eye on a new car that will cost $20.243.68 It will take nearly 58 months.005  n= 14.000. earning interest at a rate of 0. as well as the monthly interest rate. CPT PMT= 3532.005) n   Or using a financial calculator (TI BAII Plus).000     (1  0. (1.07      Or using a financial calculator (TI BAII Plus).16  PMT    PMT  $3532.86 64. PMT=10.16 This answer is used as the future value amount in the second step:  (1  0. Solving the following equation for “n” we get: 15.86.16  0.000.5. Finding the number of time periods (months) is most easily done with a financial calculator (TI BAII Plus).000. A. not an annuity due) N=8.243. Hit [2nd] [BGN] [2nd] [Set] N=4. I/Y = 0. (Remember to clear the calculator. I/Y=7. How long (to the nearest month) will it be before he can buy the car? B.07)8  1 $36. PV = 15. he can save $250 per month? Topic: Determining Holding Periods Level of difficulty: Difficult Solution: A. in addition to his existing savings.  1  1  (1  0. Ltd. CPT N = 57.005) n  1 20. FV= -36. CPT n = 14. CPT PV= –36. How long will it be before Roger can buy the car if.5 percent per month. and compounding occurs: Solutions Manual 28 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. PMT = 250.16. PV=15. He has $15. PV=0.07  Or using a financial calculator (TI BAII Plus).54  0.000   250  .000  (1. 005264)204  PV3 years  $1322.04 So the investment will double in just over 8 years. Ltd. Therefore:  1  1  (1  0. I/Y = 9. after doubling is then $2. How much interest did Josephine pay over the three-year term of her mortgage loan (see Problem 470)? Topic: Mortgage Loans Level of difficulty: Difficult Solution: Josephine has made 3 x 12 = 36 payments. Based on Problem 47. the payment amount is $1. The total amount she has paid over Solutions Manual 29 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. Quarterly: With quarterly compounding. there are 17 years remaining of the original 20-year amortization period. CPT N = 8.005264    66. annually? B.79 years are needed to double the value of the investment. FV = 2. Using a financial calculator (TI BAII Plus). The future value. or transmission is strictly prohibited. and a financial calculator allows us to find: 4 PV = -1. PV = –1. or 17 x 12 = 204 monthly payments remaining.308. CPT N = 7.09 4 k  (1  )  1  9. 65.308% . distribution.000 principal amount does she still owe? Topic: Mortgage Loans Level of difficulty: Difficult Solution: The principal outstanding at any time is the present value of the remaining payments.38  0.69. 9%. and she must renew the loan for another term. A.69. Assume Josephine chose the Providence Bank option (see Problem 47).79 The higher effective rate means that only 7. Annually: With annual compounding. quarterly? Topic: Effective Interest Rates and Holding Periods Level of difficulty: Difficult Solution: Let’s assume the present value of the investment is $1. of $1322. B. 0. I/Y = 9. After three years.322. Unauthorized copying. The three-year term on Josephine’s mortgage is now over.69     $165. FV = 2. A. the effective rate is the same as the quoted rate. . the effective annual rate is.172. How much of the original $180. 616. Charlie has convinced his bank to extend the amortization period on the loan. 36 x $1.69 = $47. with an initial one-year term at 3. she has paid off ($180.616. 67.172.0535  12 kmonthly  1    1  0. what will be Charlie’s new monthly mortgage payments? Topic: Mortgage Loans Level of difficulty: Difficult Solution: There are 24 years. the outstanding principle amount is:  1  1  (1  0.84 – $14.77  PMT     PMT  $2. .77  0.22 was interest.003224) 288  PV1 year  $1. the new mortgage payment will be:  1  1  (1  0. A year has passed since Charlie purchased his house (see Problem 500).004409) 288  $365.00 per month.827. Unauthorized copying.827.84. which will reduce the monthly payments.003224    The new effective monthly interest rate is: 2  0. Assume the original mortgage amount (from your answer in Problem 500) was to be amortized over 25 years.4409% . ($47. How long must the amortization period be for this mortgage so that Charlie can afford to make the payments? Topic: Mortgage Loans Level of difficulty: Difficult Solution: Here we must solve the annuity equation to find the amortization period required: Solutions Manual 30 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.62) = $32.484.38) = $14. Ltd.004409    68.62 of the principal amount of the mortgage loan. distribution. Rather than sell the house.000 – $165.90 percent. or 24 x 12 = 288 months remaining on the mortgage.484. because he can afford to pay only $1.950. or transmission is strictly prohibited. of her total payments. If the interest rate on a one-year term has increased to 5. The new and higher mortgage payments calculated in Problem 670 creates a big problem for Charlie. and the outstanding principal.789.31  0. Therefore. Based on the solution to Problem 0.  2  Using this rate.35 percent. Using the effective monthly interest rate from Problem 50.243.322.950.00     $365. the three years is therefore. This problem can be solved by trial and error.004409)n  $365.77  0. FV=0. N=9. FV=0. The present value of the annual payments can be found with a financial calculator. CPT PV = 42. Its present value is: PMT $3. above the rate the immediate payment is better.000.484. The lottery offers a third alternative for lottery prize payments: Céline can opt to receive annual payments of $3. A. N=9. Unauthorized copying. so our solution is in terms of months. PMT = -6000. At an interest rate below 1. if the interest rate is 5 percent per year? Topics: Ordinary Annuities.000 PV0    $60.484.000 per year for the rest of her life. she can choose an immediate payment of $50. What would the interest rate have to be for Céline to be indifferent about the two alternatives0? C. She will receive a payment of $6.000.004409    This can be done algebraically if you are familiar with the logarithm function:  $365. or transmission is strictly prohibited.000 lump sum payment today. Perpetuities Level of difficulty: Difficult Solution: A. Determining Rates of Return.0.77. Céline has just won a lottery.98  $1950.000 each year for nine years.00   n?  0. This alternative is a perpetuity. PMT = –6. the nine-year annuity would be preferable. CPT N= 397. not years.5675% per year. CPT I/Y = 1.4409.05 Solutions Manual 31 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.93 As this is less than $50.000 . FV=0. Which alternative should she pick if the interest rate is 5 percent? B. (TI BAII Plus). the immediate payment alternative is better.004409  n  Ln1     1 Ln(1  0. As an alternative. I/Y=0.000.000.98 We have used a monthly interest rate. but the task is much easier with a financial calculator. . 69. Ltd. PMT=1950. I/Y = 5.00  An easier method is to use a financial calculator (TI BAII Plus). Our final answer is that Charlie must extend the amortization period to at least 398 months (or 33 years and 2 months) so that he can still afford the monthly payments. (TI BAII Plus).77  $1950. k 0.004409)  397.  1  1  (1  0.646. C. Should Céline choose this option to the $50.484.5675%. PV = 50. B. distribution. PV= -365. 000 annually. Alysha has decided to use her $50.000. 70. The effective monthly interest rate is. N=20. or the $50.000 k 0. Clearly this is preferable to a $50.613. which were supposed to continue for the rest of her life. Alysha wants to sell the house for a high enough price to cover the remaining principal amount on the mortgage. In hindsight.37 (1  0. and Céline’s payments of $3.05)20 C.000 lump sum payment.000 payments (an annuity) can be easily found using a financial calculator (TI BAII Plus). In two years. The bank has offered her a mortgage rate of 5. CPT PV = 37. distribution. A. with an amortization period of 25 years. Solutions Manual 32 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. After 20 years. What is the minimum sale price she should accept? Topic: Mortgage Loans Level of difficulty: Difficult Solution: A.000 to make a down payment on a house.1 percent compounded semi-annually on a two-year term. the remaining payments form a perpetuity. Ltd.0. . If two friends will rent rooms from Alysha for $475 per room. I/Y = 5. the $50. Unauthorized copying. She will live in the house for the next two years while still at university. PMT = -3000. Even after 20 years. What is the present value of the lost payments? B.000 PV0   $22.63.000 lump-sum payment would have been preferable. the lottery company in Problem 69 goes out of business. FV=0.386. PMT $3. and return her down payment. A. How much were these lost payments worth 20 years ago when Céline won the prize? C. this demonstrates that an annuity can be thought of as the difference between two perpetuities! 71. should Céline have chosen the perpetual payments. Had she been able to predict that the lottery company would go out of business. The same result could be found by subtracting the result in part B from that in A. The house she is interested in purchasing costs $280. stop. how much additional money must she pay to meet her monthly mortgage payments? B.000 PV20 years    $60. We must discount the figure from part (a) back 20 years: $60. or transmission is strictly prohibited. payable at the end of each month.000 lump sum? Topic: Perpetuities and Discounting Level of difficulty: Difficult Solution: A. The value of 20 years of $3.05 B. and then sell it when she graduates. and recoup her down payment.89. FV = -1. The present value of these payments is the outstanding value of the mortgage loan. 72.58. I/Y = 0.336. has just declared that its dividend next year will be $3 per share.12      Solutions Manual 33 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. N=300. as determined by our formula. That rate of payment will continue for an additional four years. distribution.000. Ltd. the reduced dividend stream can be thought of as a perpetuity.5674  $20.000 – $50. (TI BAII Plus).000. after which the dividends will fall back to their usual $2 per share (see Problem 230).000.336. Unauthorized copying. Alysha’s two friends will be paying 2 x $475 = $950 in rent. PV = –230.4206.12  (1  0. the value of which can be found with a financial calculator. set FV=$1. I/Y = 0.336. . KashKow Inc. so she will need an additional $1.603 It will take only a little higher return. This is easiest done with a financial calculator (TI BAII Plus). 2  0.00 (1  0.12)5    $2.051  12 kmonthly  1    1  0.000.89. N=276.6%. In two years. 73. 10.4206. N = 15. What is the present value of all the future dividends? Topic: Multiple Annuities Level of difficulty: Difficult Solution: The dividends for the first five years form an ordinary annuity. PMT = 30. PMT = 1350.350.000. Alysha will have made 24 payments. However.000. the value of this perpetuity.81 to make the mortgage payments. or transmission is strictly prohibited.58 + $50. How much will Tommy have to earn on his savings (see Problem 520) to be able to amass $1 million in 15 years? Topic: Investing Early Level of difficulty: Difficult Solution: In the equation for Problem 52.27   5   0.67  0. Starting in year 6. for Tommy to reach his goal.81  16. Alysha would have to sell the house for at least $220.00  1   10. and must be discounted to the present:   1   1  PV0  $3.58.000 = $270. CPT I/Y = 10.4206%  2  The amount of the mortgage loan will be ($280. and solve for the interest rate. To pay off the loan.000) = $230. leaving 276. CPT PMT = 1350. and there will be 12 x 25 = 300 monthly payments. CPT PV = 220.000.81 – $950 = $400.12)   0. B. PV=0. occurs at year 5 (one year before the first $2 dividend). Using the calculator again. Unauthorized copying. plus the same amount one year from now. rent for the first and last month must be paid up front.99  0. she will pay $450 rent. distribution. and will sub-let her apartment for the same amount she pays in rent (see Problem 740).  1  1  (1  0.393. Consider one year. Topic: Annuities Due Level of difficulty: Difficult Solution: 0. this is a three-year annuity due. Mary-Beth (see Problem 24) decides to rent an apartment for the remaining three years of her degree.75 percent compounded monthly? (Assume that rent payments are made at the start of each month.560.3125% 12 Rent payments are typically made at the start of each month (so this is an annuity due).74. In this situation. Solutions Manual 34 Chapter 5 Copyright © 2008 John Wiley & Sons Canada. we would expect 36 monthly rent payments. and again two years from now. compounded monthly. earning 3.003125    Mary-Beth will need to have this sum now.003125)35  PV0  $450     (1  0. kmonthly  0.77  0. Over three years. how much would she need to have in her bank account. but for only eight months of the year. the last month’s rent must be paid up front. Suppose that Mary-Beth plans to return home for four months each summer. or transmission is strictly prohibited. How much money would Mary-Beth need to have in her bank account right now to be sure she will always have enough for rent? The bank account pays 3. In essence. Ltd. or 8 months of rent payments made at the start of each month:  1  1  (1  0. during each of three years.0375 kmonthly   0.3125% .003125    75.003125) 8  PV0  $450     (1  0.003125)  $3. whose value must be calculated using the effective annual interest rate. so the annuity includes only 35 payments.003125)  $450  $15. In other words. She has found a nice location that will cost $450 per month.75 percent interest. . After one year living in a university residence. However.) Topic: Annuities Due and Multiple Annuities Level of difficulty: Difficult Solution: As in Problem 74. the present value of the last month’s rent is $450 because it will be paid today. 0375  k  1    1  3. or 421 payments to surpass it. the insurance company is able to offer two alternatives. option one is an ordinary annuity.532. Option two is much easier to value.785 each month for as long as he lives. the man must live to be at least 100 years old for option one to be a better deal! Solutions Manual 35 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.70 For the first option to be a better deal. PV = –488.532. The first option is to receive $2. Again using the calculator.70. N = 240.038151    76.5. 12  0. or transmission is strictly prohibited.038151) 3  PV0  $3. Hence. I/Y = 0. .5. FV=0. distribution. How long must the man live so that the first option is a better deal? Topic: Ordinary Annuities and Multiple Annuities Level of difficulty: Difficult Solution: It is tempting to view the first option as a perpetuity. PMT = 3500. it includes exactly 240 monthly payments.5% 12 Using a financial calculator (TI BAII Plus).038151)  $10. PMT = 2785. CPT N = 420.99     (1  0. Ltd. A 65-year-old man intends to use his retirement funds to purchase an annuity from a life insurance company.8151%  12   1  1  (1  0.06 kmonthly   0. Unauthorized copying.560. This is just over 35 years. Given the amount of money the man has available to invest. I/Y = 0.500 each month.3 So option one must continue for 420. CPT PV = –488.19  0. the second option is to receive $3. but this would be incorrect as the man will die at some time. 0. The relevant interest rate is 6 percent per year. and the payment will then cease. Thus.3 monthly payments to equal the value of option two. but for only 20 years (payments will be made to his estate if he should die before that time). it must include enough payments so that its present value is at least as great as for option two. with an uncertain number of payments.295. The material provided herein may not be downloaded. mechanical. 2007. used to create derivative works. modified. Ltd. reproduced. Ltd. scanning. electronic. The data contained in this file is protected by copyright. made available on a network. Ltd. recording. Unauthorized copying. or related companies. stored in a retrieval system. distribution. All rights reserved. or otherwise without the prior written permission of John Wiley & Sons Canada. Solutions Manual 36 Chapter 5 Copyright © 2008 John Wiley & Sons Canada.Legal Notice Copyright Copyright © 2004 by John Wiley & Sons Canada. Note: Permission to post online obtained on December 10. photocopying. . This manual is furnished under licence and may be used only in accordance with the terms of such licence. or transmitted in any form or by any means. or transmission is strictly prohibited.
Copyright © 2024 DOKUMEN.SITE Inc.