Solution Manual for Engineering Economics Financial Decision Making for Engineers Canadian 6th edition-stamped

SOLUTIONS MANUAL Engineering Economics Financial Decision Making for Engineers, 6/e Sixth Edition CHAPTER 1 Solutions to Chapter-End Problems A. Key Concepts When to Use Engineering Economics: 1.1 (a) Yes - several quantifiable alternatives exist (b) Yes - if quantifiable (c) No - alternatives are not quantifiable (d) No - since it mainly involves intangible qualifications of the candidate (e) Yes - at least two alternatives are quantifiable (f) Yes - with quantifiable costs and benefits (g) No - hard to quantify (h) Yes - if the effect is quantifiable (i) Yes - quantifiable (j) No - benefits are hard to quantify (k) Yes - quantifiable (l) Yes - if quantifiable 1.2 Five examples: 1) Location of the business: renting vs. buying decision 2) Equipment: leasing vs. buying decision 3) Equipment: choosing one out of several alternatives; analysis of cost savings 4) Viability of the business: analysis of cash flow; ability to recover the capital investment 5) Effect of the corporate tax rate 1.3 Whatever the description for the items listed, it should be clear that all engineering design is founded on controlling the costs incurred. Ethics in Decision-Making: 1.4 There is no right or wrong answer to any of these questions. The student should observe, however, that there often is a difference in how we should behave compared to how we do behave. B. Applications 1 Copyright © 2016 Pearson Canada Inc. Chapter 1 - Engineering Decision Making 1.5 There is no right answer to which they should move into. However, they probably will move into the duplex, on the grounds that emotion has a more direct impact on our decisions than does logic. 1.6 (a) She should probably not buy the cheapest one. She cannot decide on price alone. (b) If everything else is the same, one would expect Karen to buy the cheapest one. 1.7 Ciel can (1) invest time in making sure she has the best information about all of the issues that will affect her future sales, and (2) do sensitivity analysis in her financial calculations. 1.8 It is not clear that Trevor should sell out to Venture Corp. He will get more money that way, but it may be very important to him that his brainchild continue to thrive. What Trevor has to decide is whether avoiding the displeasure he gets from Venture Corp. closing down his company is worth $1 000 000 to him. If it is, he should take the Investco offer, or if not, the Venture Corp. offer. 1.9 Possible uncertainties that associate with the new technology: 1) Is the new technology proven to be viable? 2) Is the performance or quality that are expected from the technology proven to be consistent? 3) Are there any “side effects”? 4) Is this technology going to stay, or will it be replaced quickly by an alternative technology? 5) What market share might Telekom gain? Sensitivity analysis may be able to address the second and third issues by considering a range of performance/quality/side effect issues. The first and fourth issues, however, may be difficult to address by sensitivity analysis since there is no specific quantity that can be varied. C. More Challenging Problems: 1.10 The Toyota Corolla is the right choice when the interest rate is very high, while the BMW is the right choice if the interest rate is very low. 2 Copyright © 2016 Pearson Canada Inc. Chapter 1 - Engineering Decision Making Notes for Case-in-Point 1.1 1) There is no acceptable death rate, but engineering projects proceed even when it is known that deaths will occur. 2) No right answer. This is a matter of judgement. However, deaths by accident or otherwise will occur whether or not a particular project is approved. 3) No right answer. It depends on the circumstances. Notes for Mini-Case 1.1 1) The student should observe that the offending companies usually have a good reason for staying in business, such as supplying employment, supplying a necessary good, or that efforts to clean up are taking time but should eventually be done. 2) No right answer. Fining a company affects the shareholders, who are the real owners of the company. On the other hand, fining the management may be appropriate in large companies where the shareholder are often unaware of the effects of decisions of management. 3) Perhaps, so there is less pollution. Perhaps not, because companies may be forced to close or move to another country, or goods may become too expensive. Some level of pollution is probably acceptable. 4) Some companies do the socially correct thing anyhow, either for profits resulting from having a “green” image, or because of the moral values of management or owners. 3 Copyright © 2016 Pearson Canada Inc. CHAPTER 2 Solutions to Chapter-End Problems A. Key Concepts Simple Interest: 2.1 P = 3000 N = 6 months i = 0.09 per year = 0.09/12 per month, or 0.09/2 per six months P + I = P + PiN = P(1 + iN) = 3000[1 + (0.09/12)(6)] = 3135 or = 3000[1 + (0.09/2)(1)] = 3135 The total amount due is $3135, which is $3000 for the principal amount and $135 in interest. 2.2 I = 150 N = 3 months i = 0.01 per month P = I/(iN) = 150/[(0.01)(3)] = 5000 A principal amount of $5000 will yield $150 in interest at the end of 3 months when the interest rate is 1% per month. 2.3 P = 2000 N = 5 years i = 0.12 per year F = P(1+i)N = 2000(1+0.12)5 = 3524.68 The bank account will have a balance of $3525 at the end of 5 years. 2.4 (a) P = 21 000 i = 0.10 per year N = 2 years F = P(1+i)N = 21000(1+0.10)2 = 25 410 The balance at the end of 2 years will be $25 410. 5 Copyright © 2016 Pearson Canada Inc. $2000 0 1 2 3 4 5 6 Chapter 2 - Time Value of Money (b) P = 2 900 i = 0.12 per year = 0.01 per month N = 2 years = 24 months F = P(1+i)N = 2900(1+0.01)24 = 3682.23 The balance at the end of 24 months (2 years) will be $3682.23. 2.5 From: F = P(1 + i)N P = F/(1 + i)N = 50 000/(1 + 0.01)20 = 40977.22 Greg should invest about $40 977. 2.6 F = P(1 + i)N 50 000 = 20 000(1 + i)20 (1+i)20 = 5/2 i = (5/2)1/20 − 1 = 0.04688 = 4.688% per quarter = 18.75% per year The investment in mutual fund would have to pay at least 18.75% nominal interest, compounded quarterly. Cash Flow Diagrams: 2.7 Cash flow diagram: $6000 $900 2.8 Showing cash flow elements separately: 6 Copyright © 2016 Pearson Canada Inc. 0 1 2 3 4 5 6 7 8 9 10 11 12 $300 0 1 2 3 4 5 6 7 8 9 10 11 12 $200 $10000 0 1 2 3 4 5 6 7 8 9 10 11 12 $15000 $15000 $15000 $15000 Chapter 2 - Time Value of Money $100 $500 Showing net cash flow: $100 $500 2.9 Showing cash flow elements separately: $10000 $20000 Showing net cash flow: 7 Copyright © 2016 Pearson Canada Inc. $10000 $10000 $10000 $10000 $5000 0 1 2 3 4 5 6 7 $5000 $5000 8 9 $5000 10 11 12 Chapter 2 - Time Value of Money $20000 2.10 The calculation of the net cash flow is summarized in the table below. Time Payment Receipt Net 0 20 −20 1 30 30 2 33 33 3 20 36.3 16.3 4 39.9 39.9 5 43.9 43.9 6 20 48.3 28.3 7 53.1 53.1 8 58.5 58.5 9 20 64.3 44.3 10 70.7 70.7 11 77.8 77.8 12 20 85.6 65.6 Cash flow diagram: $53.1 $58.5 $70.7 $77.8 $65.5 $30 $33 $39.9 $43.9 $28.3 $44.3 $16.3 0 1 2 3 4 5 6 7 8 9 10 11 12 $20 2.11 (a) functional loss (b) use-related physical loss (c) functional loss 8 Copyright © 2016 Pearson Canada Inc. Chapter 2 - Time Value of Money (d) time-related physical loss (e) use-related physical loss (f) use-related physical loss (g) functional loss (h) time-related physical loss 2.12 (a) market value (b) salvage value (c) scrap value (d) market value to Liam, salvage value to Jacque (e) book value 2.13 The book value of the company is $4.5 based on recent financial statements. The market value is $7 million, assuming that the bid is real and would actually be paid. 2.14 Since sewing machine technology does not change very quickly nor does the required functionality, functional loss will probably not be a major factor in the depreciation of this type of asset. Left unused, but cared for, the machine will lose some value, and hence time-related loss may be present to some extent. The greatest source of depreciation on a machine will likely be use-related and due to wear and tear on the machine as it is operated. 2.15 A switch will generally not suffer wear and tear due to use, and thus userelated physical loss is not likely to be a big factor. Nor will there likely be a physical loss due to the passage of time. The primary reason for depreciation will be functional loss - the price of a similar new unit will likely have dropped due to development of new technology and competition in the marketplace. 2.16 The depreciation is certainly not due to use related physical loss, or other non-physical losses in functionality. The depreciation is a time-related physical loss because it has not being used and maintained over time. 2.17 (a) BV(1) = 14 000 – (14 000 – 3000)/7 = $12 429 (b) BV(4) = 14 000 – 4(14 000 – 3000)/7 = $7714 (c) BV(7) = 3000 2.18 (a) BV(1) = 14 000(1 – 0.2) = $11 200 (b) BV(4) = 14 000(1 – 0.2)4 = $5 734 (c) BV(7) = 14 000(1 – 0.2)7 = $2936 2.19 (a) d = 1 – (3000/14 000)1/7 = 19.75% (b) BV(4) = 14 000(1– 0.1975)4 = $5806 9 Copyright © 2016 Pearson Canada Inc. Chapter 2 - Time Value of Money 2.20 Spreadsheet used for chart: Year Straight Line Declining Balance 10 Copyright © 2016 Pearson Canada Inc. 14000 12000 10000 8000 6000 4000 2000 0 Straight line Declining balance 0 1 2 3 4 5 6 7 Year Book value ($) Chapter 2 - Time Value of Money 2.21 Spreadsheet used for chart: Year d = 5% d = 20% d = 30% B. Applications 2.22 I = 190.67 P = 550 N = 4 1/3 = 13/3 years i = I/(PN) = 190.67/[550(13/3)] = 0.08 The simple interest rate is 8% per year. 2.23 F = P(1 + i)N 50 000 = 20 000(1 + 0.01)N (1.01)N = 5/2 N = ln(5/2)/ln(1.01) = 92.09 quarters = 23.02 years Greg would have to invest his money for about 23.02 years to reach his target. 2.24 F = P(1 + i)N = 20 000(1 + 0.01)20 = 24 403.80 11 Copyright © 2016 Pearson Canada Inc. 150000 100000 5% 50000 20% 30% 0 0 5 10 Year Book value ($) Chapter 2 - Time Value of Money Greg would have accumulated about $24 404. 2.25 (a) P = 5000 i = 0.05 per six months F = 8000 From: F = P(1 + i)N N = ln(F/P)/ln(1 + i) = ln(8000/5000)/ln(1 + 0.05) = 9.633 The answer that we get is 9.633 (six-month) periods. But what does this mean? It means that after 9 compounding periods, the account will not yet have reached $8000. (You can verify yourself that the account will contain $7757). Since compounding is done only every six months, we must, in fact, wait 10 compounding periods, or 5 years, for the deposit to be worth more than $8000. At that time, the account will hold $8144. (b) P = 5000 r = 0.05 (for the full year) F = 8000 i = r/m = 0.05/2 = 0.025 per six months From: F = P(1 + i)N N = ln(F/P)/ln(1 + i) = ln(8000/5000)/ln(1 + 0.025) = 19.03 We must wait 20 compounding periods, or 10 years, for the deposit to be worth more than $8000. 2.26 P = 500 F = 708.31 i = 0.01 per month From: F = P(1 + i)N N = ln(F/P)/ln(1 + i) = ln(708.31/500)/ln(1 + 0.01) = 35.001 The deposit was made 35 months ago. 2.27 (a) P = 1000 i = 0.1 N = 20 F = P(1 + i)N = 1000(1+0.1)20 = 6727.50 12 Copyright © 2016 Pearson Canada Inc. Chapter 2 - Time Value of Money About $6728 could be withdrawn 20 years from now. (b) F = PiN = 1000(0.1)(20) = 2000 Without compounding, the investment account would only accumulate $2000 over 20 years. 2.28 Let P = X and F = 2X. (a) By substituting F = 2X and P = X into the formula, F = P + I = P + PiN, we get 2X = X + XiN = X(1 + iN) 2 = 1 + iN iN = 1 N = 1/i = 1/0.11 = 9.0909 It will take 9.1 years. (b) From F = P(1 + i)N, we get N = ln(F/P)/ln(1 + i). By substituting F = 2X and P = X into this expression of N, N = ln(2X/X)/ln(1 + 0.11) = ln(2)/ln(1.11) = 6.642 Since compounding is done every year, the amount will not double until the 7th year. (c) Given r = 0.11 per year, the effective interest rate is i = er − 1 = 0.1163. From F = P(1 + i)N, we get N = ln(F/P)/ln(1 + i). By substituting F = 2X and P = X into this expression of N, N = ln(2X/X)/ln(1 + 0.1163) = ln(2)/ln(1.1163) = 6.3013 Since interest is compounded continuously, the amount will double after 6.3 years. 2.29 (a) r = 0.25 and m = 2 ie = (1 + r/m)m − 1 = (1 + 0.25/2)2 − 1 = 0.26563 The effective rate is approximately 26.6%. (b) r = 0.25 and m = 4 ie = (1 + r/m)m − 1 = (1 + 0.25/4)2 − 1 = 0.27443 13 Copyright © 2016 Pearson Canada Inc. Chapter 2 - Time Value of Money The effective rate is approximately 27.4%. (c) ie = er − 1 = e0.25 − 1 = 0.28403 The effective rate is approximately 28.4%. 2.30 (a) ie = 0.15 and m = 12 From: ie = (1 + r/m)m − 1 r = m[(1 + ie) 1/m − 1] = 12[(1 + 0.15)1/12 − 1] = 0.1406 The nominal rate is 14.06%. (b) ie = 0.15 and m = 365 From: ie = (1 + r/m)m − 1 r = m[(1 + ie) 1/m − 1] = 365[(1 + 0.15)1/365 − 1] = 0.13979 The nominal rate is 13.98%. (c) For continuous compounding, we must solve for r in ie = er − 1: r = ln(1 + ie) = ln(1 + 0.15) = 0.13976 The nominal rate is 13.98%. 2.31 F = P(1 + i)N 14 800 = 665(1 + i)64 i = 0.04967 The rate of return on this investment was 5%. 2.32 The present value of X is calculated as follows: F = P(1 + i)N 3500 = X(1 + 0.075)5 X = 2437.96 The value of X in 10 years is then: F = 2437.96(1 + 0.075/365)3650 = 4909.12 14 Copyright © 2016 Pearson Canada Inc. Chapter 2 - Time Value of Money The present value of X is $2438. In 10 years, it will be $4909. 2.33 r = 0.02 and m = 365 ie = (1 + r/m)m − 1 = (1 + 0.02/365)365 − 1 = 0.0202 The effective interest rate is about 2.02%. 2.34 Effective interest for continuous interest account: ie = er − 1 = e 0.0599 − 1 = 0.08318 = 6.173% Effective interest for daily interest account: ie = (1 + r/m)m − 1 = (1 + 0.08/365)365 − 1 = 0.08328 = 8.328% No, your money will earn less with continuous compounding. 2.35 ie(weekly) = (1 + r/m)m − 1 = (1 + 0.055/52)52 − 1 = 0.0565 = 5.65% ie(monthly) = (1 + r/m)m − 1 = (1 + 0.07/12)12 − 1 = 0.0723 = 7.23% 2.36 ie(Victory Visa) = (1 + r/m)m − 1 = (1 + 0.26/365)365 − 1 = 0.297 = 29.7% ie(Magnificent Master Card) = (1 + 0.28/52)52 − 1 = 0.322 = 32.2% ie(Amazing Express) = (1 + 0.3/12)12 − 1 = 0.345 = 34.5% Victory Visa has the lowest effective interest rate, so based on interest rate, Victory Visa seems to offer the best deal. 2.37 First, determine the effective interest rate that May used to get $2140.73 from $2000. Then, determine the nominal interest rate associated with the effective interest: F = P(1 + ie) N 2140.73 = 2000(1 + ie) 1 ie = 0.070365 ie = er − 1 0.070365 = er − 1 r = 0.068 The correct effective interest rate is then: ie = (1 + r/m)m − 1 = (1 + 0.068/12)12 − 1 = 0.07016 The correct value of $2000 a year from now is: F = P(1 + ie) N = 2000(1 + 0.07016)1 = $2140.32 15 Copyright © 2016 Pearson Canada Inc. $50 0 1 2 3 4 5 6 7 8 9 10 11 12 Chapter 2 - Time Value of Money 2.38 The calculation of the net cash flow is summarized in the table below. Investment A Investment B Time Payment Receipt Net Payment Receipt Net 0 2400 −2400 0 −400 −300 −200 −100 0 12 200 250 50 500 600 100 Cash flow diagram for investment A: $250 $2400 Cash flow diagram for investment B: 16 Copyright © 2016 Pearson Canada Inc. $450 $350 $250 $150 $50 $100 0 1 2 3 4 5 6 7 8 9 10 11 12 $100 $200 $300 Chapter 2 - Time Value of Money $550 $400 Since the cash flow diagrams do not include the time factor (i.e., interest), it is difficult to say which investment may be better by just looking at the diagrams. However, one can observe that investment A offers uniform cash inflows whereas B alternates between positive and negative cash flows for the first 10 months. On the other hand, investment A requires $2400 up front, so it may not be a preferred choice for someone who does not have a lump sum of money now. 2.39 (a) The amount owed at the end of each year on a loan of $100 using 6% interest rate: Year Simple Interest Compound Interest 0 100 100.00 1 106 106.00 2 112 112.36 3 118 119.10 4 124 126.25 5 130 133.82 6 136 141.85 7 142 150.36 8 148 159.38 9 154 168.95 .08 17 Copyright © 2016 Pearson Canada Inc. Chapter 2 - Time Value of Money (b) The amount owed at the end of each year on a loan of $100 using 18% interest rate: Year Simple Interest Compound Interest 0 100 100.00 1 118 118.00 2 136 139.24 3 154 164.30 4 172 193.88 5 190 228.78 6 208 269.96 7 226 318.55 8 244 375.89 9 262 443.55 .38 18 Copyright © 2016 Pearson Canada Inc. 180 170 160 150 140 130 120 110 100 Compound interest Simple interest 0 1 2 3 4 5 6 Year 7 8 9 10 600 500 400 Compound interest 300 200 Simple interest 100 0 1 2 3 4 5 6 7 8 9 10 Year Amount owed ($) Amount owed ($) Chapter 2 - Time Value of Money 2.40 (a) From F = P(1 + i)N, we get N = ln(F/P)/ln(1 + i). At i = 12%: N = ln(1 000 000/0.01)/ln(1 + 0.12) = 162.54 years At i = 18%: N = ln(F/P)/ln(1 + i) = ln(1 000 000/0.01)/ln(1 + 0.18) = 111.29 years (b) The growth in values of a penny as it becomes a million dollars: Year At 12% At 18% 0 0.01 0.01 10 0.03 0.05 20 0.10 0.27 30 0.30 1.43 40 0.93 7.50 50 2.89 39.27 60 8.98 205.55 70 27.88 1 075.82 80 86.58 5 630.68 90 268.92 29 470.04 100 835..32 110 2 594..70 120 8 056..79 ..39 ..70 ..48 ..72 ..84 ..83 2.41 From the table and the charts below, we can see that $100 will double in (a) 105 months (or 8.75 years) if interest is 8% compounded monthly (b) 13 six-month periods (6.5 years) if interest is 11% per year, compounded semi-annually (c) 5.8 years if interest is 12% per year compounded continuously Month 8% 11% 12% 0 100.00 100.00 100.00 12 108.30 111.30 112.75 24 117.29 123.88 127.12 36 127.02 137.88 143.33 48 137.57 153.47 161.61 60 148.98 170.81 182.21 72 161.35 190.12 205.44 19 Copyright © 2016 Pearson Canada Inc. Chapter 2 - Time Value of Money 84 174.74 211.61 231.64 96 189.25 235.53 261.17 108 204.95 262.15 294.47 2.42 P(1 – d)n = P – n(P – S)/N 245 000(1 – d)20 = 245 000 – 20(245 000 – 10 000)/30 (1 – d)20 = 88 333.33/245 000 = 0.3605 1 – d = 0.9503 d = 4.97% The two will be equal in 20 years with a depreciation rate of 4.97%. 2.(1 – d)20 = 60 000 (1 – d)20 = 1/13 d = 1 – (1/13)1/20 = 1 – 0.8796 = 0.1204 A depreciation rate of about 12% will produce a book value in 20 years equal to the salvage value of the press. 2.44 (a) BV(4) = 150 000 – 4[(150 000 – 25 000)/10] = 150 000 – 4(12 500) = 150 000 – 50 000 = 100 000 DC(5) = (150 000 – 25 000)/10 = 12 500 (b) BV(n) = 150 000(1 – 0.2)4 = 150 00(0.8)4 = 61 440 DC(5) = BV(4)0.2 = 61 440(0.2) = 12 288 (c) d = 1 – (25 000/150 000)1/10 = 0.1640 = 16.4% C. More Challenging Problems 2.45 The present worth of each instalment: 20 Copyright © 2016 Pearson Canada Inc. 300 280 260 240 220 200 180 160 140 120 100 12% 11% 8% 100 120 Month Deposit ($) Chapter 2 - Time Value of Money Instalment F P Total 665270 Sample calculation for the third instalment, which is received at the end of the second year: P = F/(1 + r/m)N = 100 000/(1 + 0.10/12)24 = 81 941 The total present worth of the prize is $665 270, not $1 000 000. 2.46 The present worth of the lottery is $665 270. If you take $300 000 today, that leaves a present worth of $365 270. The future worth of $365 270 in 5 years (60 months) is: F = P(1 + r/m)N = 365 270(1 + 0.10/12)60 = 600 982 The payment in 5 years will be $600 982. 2.47 The first investment has an interest rate of 1% per month (compounded monthly), the second 6% per 6 month period (compounding semiannually). (a) Effective semi-annual interest rate for the first investment: ie = (1 + is) N − 1 = (1 + 0.01)6 − 1 = 0.06152 = 6.152% Effective semi-annual interest rate for the second investment is 6% as interest is already stated on that time period. (b) Effective annual interest rate for the first investment: ie = (1 + is) N − 1 = (1 + 0.01)12 − 1 = 0.1268 = 12.68% Effective annual interest rate for the second investment: ie = (1 + is) N − 1 = (1 + 0.06)2 − 1 = 0.1236 = 12.36% 21 Copyright © 2016 Pearson Canada Inc. Chapter 2 - Time Value of Money (c) The first investment is the preferred choice because it has the higher effective interest rate, regardless of on what period the effective rate is computed. 2.48 (a) i = 0.15/12 = 0.0125, or 1.25% per month The effective annual rate is: ie = (1 + i)m − 1 = (1 + 0.0125)12 − 1 = 0.1608 or 16.08% (b) P = 50 000 N = 12 i = 0.15/12 = 0.0125, or 1.25% per month F = P(1 + i)N = 50 000(1 + 0.0125)12 = 58 037.73 You will have $58 038 at the end of one year. (c) Adam's Fee = 2% of F = 0.02(58037.73) = 1160.75 Realized F = 58 037.73 − 1160.75 = 56 876.97 The effective annual interest rate is: F = P(1 + i)1 56 876.97 = 50 000(1 + i) i = 56 876.97/50 000 − 1 = 0.1375 or 13.75% The effective interest rate of this investment is 13.75%. 2.49 Market equivalence does not apply as the cost of borrowing and lending is not the same. Mathematical equivalence does not hold as neither 2% nor 4% is the rate of exchange between the $100 and the $110 one year from now. Decisional equivalence holds as you are indifferent between the $100 today and the $110 one year from now. 2.50 Decisional equivalence holds since June is indifferent between the two options. Mathematical equivalence does not hold since neither 8% compounded monthly (lending) or 8% compounded daily (borrowing) is the rate of exchange representing the change in the house price ($110 000 now and $120 000 a year later is equivalent to the effective interest rate of 9.09%). Market equivalence also does not hold since the cost of borrowing and lending is not the same. 2.51 (a) The amount of the initial deposit, P, can be found from F = P(1 + i)N 22 Copyright © 2016 Pearson Canada Inc. Chapter 2 - Time Value of Money with F = $3000, N = 36, and i = 0.10/12. (b) Having determined P = $2225, then we can figure out the size of the deposit at the end of years 1, 2 and 3. If you had not invested in the fixed interest rate investment, you would have obtained interest rates of 8%, 10%, and 14% for each of the three years. The table below shows how much the initial deposit would have been worth at the end of each of the three years if you had been able to reinvest each year at the new rate. Because of the surge in interest rates in the third year, with 20/20 hindsight, you would have been better off (by about $60) not to have locked in at 10% for three years. Year Fixed Interest Rate Varying Interest Rate 2.52 Interest rate i likely has its origins in commonly available interest rates present in Marlee’s financial activities such as investing or borrowing money. Interest rate j can only be determined by having Marlee choose between X and Y to determine at which interest rate Marlee is indifferent between the choice. Interest rate k probably does not exist for Martlee, since it is unlikely that she can borrow and lend money at the same interest rate. If for some reason she could, then k=j. Also, i could be either greater or less than j. 2.53 BV(0) = 250 000 BV(6) = 250 000(1 − 0.3)6 = 29 412.25 The book values of the conveyor after 7, 8, 9, and 10 years are: 23 Copyright © 2016 Pearson Canada Inc. 3200 3000 2800 2600 Fixed interest rate 2400 Varying interest rate 2200 0 1 2 3 Year Deposit amount ($) Chapter 2 - Time Value of Money BV(7) = 29 412.25 − 29 412.25/41 = 22 059.19 BV(8) = 29 412.25 − 29 412.25/42 = 14 706.13 BV(9) = 29 412.25 − 29 412.25/43 = 7353.07 BV(10) = 29 412.25 − 29 412.25/44 = 0 2.54 d = 1 – (S/P)1/n = 1 – (8300/12 500)1/2 = 1– 0.81486 = 0.18514 = 18.514% BVdb(5) = 12 500 (1 – 0.18514)5 = 4470.87 Enrique should expect to get about $4471 for his car three years from now. 24 Copyright © 2016 Pearson Canada Inc. Chapter 2 - Time Value of Money Notes for Case-in-Point 2.1 1) Close, if the appropriate depreciation method is being used. 2) It makes sense because it is a new technology. 3) Because the accounting department is likely using a specific depreciation method that is not particularly accurate in this case. In particular, they may be using a depreciation method required for tax purposes. 4) Bill Fisher is probably not doing anything wrong, but it wouldn’t hurt to check.. 25 Copyright © 2016 Pearson Canada Inc. Chapter 2 - Time Value of Money Notes for Mini-Case 2.1 3) Money will always be lost over the year. If money could be gained, everybody would borrow as much money as possible to invest. Solutions to All Additional Problems Note: Solutions to odd-numbered problems are provided on the Student CD-ROM. 2S.1 You can assume that one month is the shortest interval of time for which the quoted rental rates and salaries apply. Assembling the batteries will require 24 person-months, and the associated rental space. To maximize the interest you receive from your savings, and minimize the interest you pay on your line of credit, you should defer this expenditure till as late in the year as possible. So you leave your money in the bank till December 1, then purchase the necessary materials and rent the industrial space. Assume that salaries will be paid at the end of the month. As of December 1, you have $100 000(1.005)11 = $105 640 in the bank. You need to spend $360 000 on materials and $240 000 to rent space. After spending all you have in the bank, you therefore need to borrow an additional $494 360 against your line of credit. As of December 31, you owe $494 360(1.01) = $499 304 to the bank, and you owe $240 000 in salaries. So after depositing the government cheque and paying these debts, you have 1 200 000 – 499 304 – 240 000 = $460 696 in the bank. This example illustrates one of the reasons why Just-in-Time (“JIT”) manufacture has become popular in recent years: You want to minimize the time that capital is tied up. An additional motivation for JIT would become evident if you were to consider the cost of storing the finished batteries before delivery. Be aware, however, that the JIT approach also carries risks. December is typically a time when labour, space, and credit are in high demand so there is a possibility that the resources you need will be unavailable or more expensive than expected, and there will then be no time to recover. We will look at methods for managing risk in Chapter 12. 2S.2 26 Copyright © 2016 Pearson Canada Inc. Chapter 2 - Time Value of Money We want to solve the equation Future worth = Present worth (1+i) N, where the future worth is twice the present worth. So we have 2 = (1+i) N Taking logarithms on both sides, we get N = log(2) / log(1+i) For small values of i, log (1+i) is approximately i (this can be deduced from the Taylor series). And log(2) = 0.69315. So, expressing i as a percentage rather than a fraction, we have: N = 69.3 / i Since this is only an approximation, we will adjust 69.3 to an easily factored integer, 72, thus obtaining N = 72 / i 2S.3 Gita is paying 15% on her loan over a two-week period, so the effective annual rate is (1.1526 −1)  100% = 3686% The Grameen Bank was awarded the Nobel Peace Prize in 2006 for making loans available to poor investors in Bangladesh at more reasonable rates. 2S.4 Five hundred years takes us beyond the scope of the tables in Appendix A, so we employ the formula P = F / (1+i) N to find the present value of the potential loss. In this case, we have P = $ / (1.05)500 = $0.025, or two-and-a-half cents. This implies that it is not worth going to any trouble to make the waste repository safe for that length of time. This is a rather troubling conclusion, because the example is not imaginary; the U.S., for example, is currently trying to design a nuclear waste repository under Yucca Mountain in Nevada that will be secure for ten thousand years—twenty 27 Copyright © 2016 Pearson Canada Inc. Chapter 2 - Time Value of Money times as long as in our example. It is not clear how the engineers involved in the project can rationally plan how to allocate their funds, since the tool we usually use for that purpose—engineering economics—gives answers that seem irrational. 2S.5 There is no “right” answer to this question, which is intended for discussion in class or in a seminar. Some of the arguments that might be advanced are as follows: One option is to say, “You cannot play the numbers game with human lives. Each life is unique and of inestimable value. Attempting to treat lives on the same basis as dollars is both cold-blooded and ridiculous.” But this really won’t do. Medical administrators, for example, do have a responsibility to save lives, and they have limited resources to meet this responsibility. If they are to apportion their resources rationally, they must be prepared to compare the results of different strategies. To support the point of view that future lives saved should be discounted by some percentage in comparison with present lives, the following arguments might be offered: 1. Suppose we make the comparison fifty years in the future. If we spend our resources on traffic police, we will have saved the lives of those who would have died in accidents, and, because we spent the money that way, the world of fifty years hence will also contain the descendants of those who would have died. So the total number of live humans will be increased by more than fifty. (This argument assumes that creating a new life is of the same value as preserving an existing life. We do not usually accept that assumption; for example, many governments may promote population control by limiting the number of children born, but it would be unacceptable to control the population by killing off the old and infirm. To give another example, if I am accused of murder, it would not be an acceptable defence to argue that I have fathered two children, and have thus made a greater contribution to society than a law-abiding bachelor.) 2. Just as I charge you interest on a loan because of the uncertainty of what might happen between now and the due date—you could go bankrupt, I could die, etc.—so we should discount future lives saved because we cannot anticipate how the world will change before the saving is realized. For example, a cure for cancer could be discovered ten years from now, and then all the money spent on the anti-smoking campaign will have been wasted, when it could have been used to save lives lost in highway accidents. 3. We should not be counting numbers of lives, but years of human life. Thus it is better to spend the money on preventing accidents, because these kill people of all ages, while cancer and heart disease are mostly diseases of later 28 Copyright © 2016 Pearson Canada Inc. Chapter 2 - Time Value of Money life; so more years of human life are saved by the first strategy. 4. The world population is growing. Thus, for humanity as a whole, losing a fixed number of lives can more easily be born in the future than it can now. (This is an extension of the argument that it is worse to kill a member of an endangered species than of a species that is plentiful.) 29 Copyright © 2016 Pearson Canada Inc. CHAPTER 3 Solutions to Chapter-End Problems A. Key Concepts Recognizing Cash Flows: 3.1 Boarding: an annuity paid monthly, at the end of the months May to September, brought to present worth to the beginning of May. Monthly estimates obtained by historical measurements. Breeding: annuity due, with a six-month period over 10 years Heating, water and sewage: monthly annuity, estimated from the average monthly costs over the last few years Food: annuity due over 10 years, estimated by the amount paid for food in previous years 3.2 Rent: monthly annuity for eight periods. Amount known from rental agreement. Bus: weekly annuity. Amount estimated as 5 return trips, or if on monthly bus pass, an annuity due as the cost of the pass. Groceries: weekly annuity, amount estimated or observed weekly average Lunch: weekly annuity, amount estimated or observed weekly average Printing and copying: two single payments, one at the end of four months and one at the end of eight months, or an annuity with a four-month period. Amount estimated or observed from previous years. Christmas presents: single payment, amount as budgeted Christmas extra cash: single payment, amount as estimated Single Disbursements or Receipts: 3.3 P = 100 i = 8% N = 15 Using the formula: F = P(1 + i)N = 100(1 + 0.08)15 = 317.22 28 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis Or using the compound interest tables: F = P(F/P, 8%, 15) = 100(3.1722) = 317.22 About $317 will be in the bank account. 3.4 F = 1 000 000 i = 12% N = 30 Using the formula: P = F/(1 + i)N = 1 000 000/(1 + 0.12)30 = 33 377.92 Using the tables produces a slightly different result due to the number of significant digits in the table: P = F(P/F, 12%, 30) = 1 000 000(0.0334) = 33 400 You should invest about R 33 400. 3.5 1725(F/P, i, 5) = 3450 (F/P, i, 5) = 2 (F/P, 14%, 5) = 1.9254 (F/P, 15%, 5) = 2.0114 Solve for i using linear interpolation: i = 0.14 + (0.15 − 0.14)[(2 − 1.9254)/(2.0114 − 1.9254)] = 0.1487 = 14.87% In 10 years, you will have: F = 1725(F/P, 14.87%, 10) = 1725(1 + 0.1487)10 = $6900 Annuities: 3.6 P = 500(P/A, 0.5%, 1220) + 5000 = 500[(1 + 0.005)240 − 1]/[0.005(1 + 0.005)240] + 5000 = 69 790 + 5000 = 74 790 Morris purchased the house for £74 790. 3.7 Using the capital recovery formula: A = (P − S)(A/P, i, N) + Si = (45 000 − 25 000)(A/P, 0.15, 5) + 25 000(0.15) = 20 000(0.29832) + 3 750 29 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis = 9 716.40 The juicer would have to save a little over $18 400 per year. 3.8 F = 10 000 i = 0.06/12 = 0.005 per month N = 24 A = 10 000(A/F, 0.5%, 24) = 10 000(0.03932) = 393.20 Fred has to save a little over $393 per month. Arithmetic Gradient Series: 3.9 (a) F = 40(F/A, 1%, 24) = 40(26.969) = 1079 (b) F = [30 + 1(A/G, 1%, 24)](F/A, 1%, 24) = [30 + 1(11.010)](26.969) = 1106 3.10 First find the annuity value of the prize by converting the gradient into an annuity: A = A’ + G(A/G, i, N) = 1000 + 1000(A/G, 15%, 20) = 1000 + 1000(5.3651) = 6365.1 Converting the annuity into a present value gives: P = A(P/A, 15%, 20) = 6365.1(6.2593) = 39 841 Therefore, the winning ticket has a present worth of about £39 800. Since 10 000 tickets are to be sold, on average each ticket is worth £39 800/10 000 = £3.98. Geometric Gradient Series: 3.11 First find the growth adjusted interest rate: i = (1 + i)/(1 + g) − 1 = 1.2/1.1 − 1 = 0.0909 = 9.09% We then make use of the geometric series to present worth conversion factor with A = 10 000, g = 10%, i = 0.0909, and N = 10. Since the growth adjusted interest rate is positive, we can make use of the present worth conversion factor. P = A(P/A, g, i, N) = A(P/A, i, N)/(1 + g) 30 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis = 10 000(P/A, 9.09%, 10)/(1 + 0.1) = 10 000(6.3923/1.1) = 58 110 We then use the present worth factor to calculate the worth in 10 years: F = 58 110(F/P, 20%, 10) = 58 110(6.1917) = 359 810 The savings would have accumulated to about $360 000. 3.12 First find the growth adjusted interest rate: i = (1 + i)/(1 + g) − 1 = 1.1/1.2 − 1 = −0.0833 = −8.333% We then make use of the geometric series to present worth conversion factor with A = 10 000, g = 20%, i = −0.0833, and N = 10. Since the growth adjusted interest rate is negative, we cannot make use of the present worth conversion factor. P = A(P/A, g, i, N) = A[(1 + i) N − 1]/[i(1 + i) N][1/(1+g)] = 10 000[(1 − 0.08333)10 − 1]/[−0.08333(1 − 0.08333)10][1/(1 + 0.2)] = 138 710 We then use the present worth factor to calculate the worth in 10 years: F = 138 710(F/P, 10%, 10) = 138 710(2.5937) = 359 790. The savings would have accumulated to about $360 000. Non-standard Annuities and Gradients: 3.13 Method 1: Consider the annuities as separate future payments. Year Present Worth 7 10 000(P/F, 9%, 7) = 10 000(0.54703) = 5470.3 12 10 000(P/F, 9%, 12) = 10 000(0.35553) = 3555.3 17 10 000(P/F, 9%, 17) = 10 000(0.23107) = 2310.7 22 10 000(P/F, 9%, 22) = 10 000(0.15018) = 1501.8 The investment is worth about $12 838 today. Total = 12 838.1 Method 2: Convert the compounding period from yearly, to every five years. This can be done with the effective interest rate formula: ie = (1 + 0.09)5 − 1 = 53.86% The present value at the end of period 2 is then: 31 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis P2 = 10 000(P/A, 53.86%, 4) and so the present worth today is: P0 = 10 000(P/A, 53.86%, 4)(P/F, 9%, 2) = 10 000(1.5253)(0.84168) = $12 838 Method 3: Convert the annuity to an equivalent yearly annuity. This can be done by considering only the first payment as a future value, and finding the equivalent annuity over the five-year period using the sinking fund factor: A = 10 000(A/F, 9%, 5) This yearly annuity is brought to present worth at the end of period 2: P2 = 10 000(A/F, 9%, 5)(P/A, 9%, 20) and so the present worth today is: P0 = 10 000(A/F, 9%, 5)(P/A, 9%, 20)(P/F, 9%, 2) = 10 000(0.16709)(9.1285)(0.84168) = $12 838 Note that each method produces the same amount, allowing for rounding. When you have a choice in methods as in this example, your choice will depend on what you find convenient, or what is the most efficient computationally. 3.14 Amount owed in one year: P(1-year) = 2000(F/P, 0.5%, 12) = 2000(1.0616) = 2123.2 Convert this to an annuity: A = P(1-year)(A/P, 0.5%, 24) = 2123.2(0.04433) = 94.13 Payments will be $94.13 per month. 3.15 Find the effective annual interest rate first in order to determine the effective interest rate for the four-month period: ie = (1+.12/12) 12 − 1 = 0.1268 0.1268 = (1 + i4-months) 3 − 1 i4-months = (1.1268)1/3 − 1 = 0.040604 32 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis Using this interest rate, P = (P/A, 4.0806%, 20)[400 + 100(A/G, 4.0806%, 20)] = 13.5179[400 + 100(8.1904)] = 16 478.91 The present worth is $16 479. When n Goes to Infinity: 3.16 The present worth computations for the full capacity tunnel can be found as follows: First, the $100 000 paid at the end of 10 years can be thought of as a future amount which has an equivalent annuity: A = 100 000(A/F, 8%, 10) = 100 000(0.06903) = 6903 Thus, at 8% interest, $100 000 every 10 years is equivalent to $6903 every year. Since the tunnel will have (approximately) an infinite life, the present worth of the lining repairs can be found using the capitalized cost formula. Added to the initial cost, the total present worth is thus: P(total) = P(initial cost) + P(lining) = 3 000 000 + 6903/0.08 = 3 086 288 Alternatively, convert to a 10-year annuity period, changing the interest rate instead of the payment amount: i = (1 + 0.08)10 − 1 = 1.1589 P = 3 000 000 + 100 000/1.1589 = $3 086 287 3.17 P = A/i = 18 000/0.07 = 257 143 GA must donate a little over $257 000. 3.18 Sum the present worth of the $350 annuity and the 10 000 future value at 5% per period over 202 = 40 periods. P = 350(P/A, 5%, 40) + 10 000(P/F, 5%, 40) = 350(17.158) + 10 000(0.14205) = 7425.80 33 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis I should pay no more than (approximately) $7426 for the bond. 3.19 P = 10 000(P/F, 4%, 18) + (10 0000.09/2)(P/A, 4%, 18) = 10 000(0.49363) + 450(12.659) = 10 632 The bond is worth about $10 632 today. Estimating Unknowns: 3.20 Using linear interpolation: X* = 100 + (200 – 100) [( – )/( – )] = 100 + 100(/) = 100 + 100(0.625) = 162.5 Trenny should be able to afford a 162.5 MW plant. 3.21 From basic principles: F = P(1 + i)N 5000 = 2300(1 + i)7 1 + i = (5000/2300)1/7 i = (5000/2300)1/7 − 1 = 0.1173 = 11.73% Or from the tables: F = P(F/P, i, N) 5000 = 2300(F/P, i, 7) (F/P, i, 7) = 5000/2300 = 2.1739 (F/P, 11%, 7) = 2.0761 (F/P, 12%, 7) = 2.2106 By linearly interpolating between the two, we get: i = 11 + (12 − 11)[(2.1739 − 2.0761)/(2.2106 − 2.0761)] = 11.73% 3.22 You could use either the capital recovery factor or the series present worth factor. Method 1: Using series present worth factor 2000 = 40(P/A, 1%, N) We can solve using the formula: 34 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis 2000 = 40[(1 + 0.01)N − 1]/[0.01(1 + 0.01)N] 0.5 = (1.01N − 1)/1.01N 1 = 1.01N − 0.5(1.01N) 2 = 1.01N ln(2) = N[ln(1.01)] N = ln(2)/ln(1.01) = 69.66 It will take her about 70 months or about 5.8 years. Method 2: Using linear interpolation 2000 = 40(P/A, 1%, N) (P/A, 1%, N) = 50 (P/A, 1%, 70) = 50.169 (P/A, 1%, 65) = 47.627 Linearly interpolating: N = 65 + 5[(50 − 47.627)/(50.169 − 47.627)] = 69.97 The small difference is the error in linearly interpolating. B. Applications 3.23 5000 = (720 000)(P/F, i, 18) = 140 000/(1 + i)18 (1 + i)18 = 140 000/5000 i = 281/18 − 1 = 0.20336 The annual rate of return must be 20.3%. 3.24 This problem requires the solution for N in the Sinking Fund Factor. Starting with the definition, the explicit formula for N can be obtained through some manipulation: A = F(A/F, i, N) = Fi/[(1 + i)N − 1] which leads to: (1 + i)N = (iF + A)/A 35 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis Taking the logarithm of both sides and rearranging gives: N = ln[(iF + A)/A]/ln(1 + i) = ln[(0.1050 000 + 7000)/7000]/ln(1 + 0.1) = 5.655 It will take about 5.7 years for the members to save the $50 000. Note that this problem can also be effectively solved by trial and error calculations. 3.25 (a) 2000 + 350 + 210 = 100(P/A, 3%, N) (P/A, 3%, N) = 2560/100 = 25.6 Solve for N using linear interpolation: N = 45 + (50 − 45)(25.6 − 24.519)/(25.730 − 24.519) = 49.4632 It will take about 50 months to complete her payment. (b) A = 2560(A/P, 3%, 24) = 2560(0.05905) = 151.168 Yoko’s monthly payment will have to be $151. 3.26 (a) Deposit of $15 per week: F = (154)(F/A, 1.5%, 36) = 60[(1.01536 − 1)/0.015] = $2836.56 Deposit of $20 per week: F = (204)(F/A, 1.5%, 36) = 80  (1.01536 − 1)/0.015 = $3782.08 (b) A = 5000(A/F, 1.5%, 36) = 5000[0.015/(1.01536 − 1)] = 105.76 Rinku needs to deposit $105.76/4 = $26.44 per week. 3.27 A = (P – S)(A/P, i, N) + Si = (140 000 – 37 000)(A/P, 14%, 5) + 37 000(0.14) = 133 000(0.2918) + 37 000(0.14) = 43 989.4 The investment would have to save about $43 989 per year over its 5-year life. 3.28 Using linear interpolation: X* = 500 + (800 – 500)[(15 000 – 11 350)/(18 950 – 11 350)] = 500 + 300(3650/7600) 36 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis = 500 + 300(0.48026) = 644.08 Enrique would have to invest about $644 per month. 3.29 The present worth of the mortgage is 260 000 at the end of August, last year. i = 0.12/12 = 0.01 or 1% per month Length of mortgage = 1220 = 240 months Payments are: A = P(A/P, i, N) =260 000(A/P, 1%, 240) = 2862.82 Method 1: Find the future worth of 260 000 at the end of 5 months. FW(mortgage, end of 5 months) = 260 000(F/P, 1%, 5) = 273 262.61 Future worth of 5 payments at end of 5 months: FW(5 payments) = 2862.82(F/A, 1%, 5) = 14603.28 Amount owing at end of January = FW(mortgage) − FW(5 payments) = 273262.61 − 14603.28 = 258659.33 They still owe $258 659. Method 2: Find the worth of 5 payments as of end of August. PW(5 payments) = 2862.82(P/A, 1%, 5) = 13894.52 PW(debt at end of Aug) = 260 000 PW(amount owed) = 2600 000 − 13894.52 = 246105.48 Amount owing at end of Jan = 246105.48(F/P, 1%, 5) = 2586.59 They still owe $258 659. 3.30 A' = 10 000 G = 1000 i = 0.15 N = 6 for gradient to annuity 6 for annuity to present value 2 for future value to present value P = [10 000 + 1000(A/G, 15%, 6)](P/A, 15%, 6)(P/F, 15%, 2) = [10 000 + 1000(2.0971)](3.7844)(0.75614) = 34 616.29 37 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis The software is worth $34 616 today. 3.31 One way to solve this problem is to sum the future worth of each individual cash flow: Balance: F1 = 2400(F/P, 1%, 24) = 2400(1.2697) = 3047.28 Salary: F2 = 120(F/A, 1%, 24) = 120(26.973) = 3236.76 Dividends: F3 = 200(F/P, 1%, 20) = 200(1.2202) = 244.04 F4 = 200(F/P, 1%, 16) = 200(1.1726) = 234.52 F5 = 200(F/P, 1%, 12) = 200(1.1268) = 225.36 F6 = 200(F/P, 1%, 8) = 200(1.0829) = 216.58 F7 = 200(F/P, 1%, 4) = 200(1.0406) = 208.12 F8 = 200 Fee: F9 = [10+1(A/G, 1%, 24)](F/A, 1%, 24) = [10+1(11.024)](26.973) = 567.08 Total future value = F1 + F2 +  + F8 − F9 = 7045.74 Clem will have saved about $7046. 3.32 P = −30 000 − 1600(P/A, 0.5%, 24) + 40 000(P/F, 0.5%, 24) + 2000(P/A, 0.5%, 12) + 2400(P/A, 0.5%, 12)(P/F, 0.5%, 12) − [400 + 40(A/G, 0.5%, 18)](P/A, 0.5%, 18)(P/F, 0.5%, 6) = −30 000 − 1600(22.558) + 40 000(0.88721) + 2000(11.616) + 2400(11.616)(0.94192) − [400 + 40(8.3198)](17.168)(0.97052) = −30 000 − 36 092 + 35 548 + 13 232 + 26 260 − 12 210 = 6678 The present worth of this investment is $6678. Yogajothi should buy the house. 3.33 P = 15 000(P/A, 12%, 8) = 15 000(4.9676) = 74 514 Barnaby Circuit Boards could afford to spend up to about $75 000 on the wave soldering machine. 3.34 This is an arithmetic gradient with A' = 100 000 G = 10 000 per year 38 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis N = 30 years i = 0.08 per year A = A' + G(A/G, 8%, 30) = 100 000 + 10 000(9.1897) = 100 000 + 91 897 = 191 897 P = A(P/A, i, N) = 191 897(P/A, 8%, 30) = 191 897(11.258) = 2 160 376 Yes, it is a good deal. 3.35 A = 100 000 g = 10% per year i = 8% per year N = 30 years i = (1 + i)/(1 + g) − 1 = 1.08/1.1 − 1 = −0.01818 For a negative interest rate, we cannot use the Series Present Worth Factor from tables, so we must use the full formula: P = 100 000[(1 − 0.01818)30 − 1]/[−0.01818(1 − 0.01818)30][1/(1 + 0.1)] = 3 670 261 No, Leon would have to pay more than $1 000 000 more. 3.36 Amount available for annuity = 20 000 − 1200 = 18 800 P = 18 800 i = 0.07/12 per month N = 5512 = 660 months A = P(A/P, i, N) = P[i(1 + i)N]/[(1 + i)N − 1] = 18 800[(0.07/12)(1 + 0.07/12)660]/[(1 + 0.07/12)660 − 1] = 112.10 The approximate daily amount can be calculated from: Daily amount = (112.1012)/365 = $3.68 No, Tina would not have enough money to retire. She would have about $3.68 available to spend each day. 3.37 Construction costs (in $millions): P1 = 20(P/A, 6%, 5)(P/F, 6%, 4) = 20(4.2124)(0.79209) = 66.728 39 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis Maintenance and Repair costs (in $millions): i = (1 + 0.06)/(1 + 0.01) − 1 = 0.0495  5.0% P2 = 2[(P/A, 5%, 35)/(1 + 0.01)](P/F, 6%, 9) = 2(16.374/1.01)(0.5919) = 19.1914 P = P1 + P2 = 85.9193 The present cost of the water supply project is about $86 million. 3.38 P = 10 000(P/A, 9%, 5) + 20 000(P/A, 9%, 10)(P/F, 9%, 5) = 10 000(3.8896) + 20 000(6.4176)(0.649 93) = 122 316 The present worth is €122 316. 3.39 The actual amount loaned is 500 − 45 = $455 45 = 455(A/P, i, N) (A/P, i, 12) = 45/455 = 0.0989 (A/P, 2.5%, 12) = 0.09749 (A/P, 3%, 12) = 0.10046 By linear interpolation: i = 2.5 + 0.5[(0.0989 − 0.09749)/(0.10046 − 0.09749)] = 2.73% per month The effective interest rate is: ie = (1 + 0.02737)12 − 1 = 38.3% per year 3.40 g = 0.01 i = 0.015 (monthly) ie = (1 + 0.015)2 − 1 = 0.03022 (bimonthly) i = (1 + ie)/(1 + g) − 1 = 1.0302/1.01 − 1 = 0.02 We assume that Shamsir’s monthly profit remains the same as the previous month when there is no increase. For example, his cash flows for months 1, 2, 3, and 4 would be $10 000, $10 100, $10 100, and $10 201. Noting that there are actually two sets of geometric gradient series, identical in all aspects but different only in timing of the first cash flow, we can determine the present value as follows: 40 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis P = 10 000(P/A, g, ie, 12) + 1000(P/A, g, ie, 12)(F/P, i, 1) = 10 000[(P/A, i, 12)/(1 + g)][1 + (F/P, 0.015, 1)] = 10 000[(P/A, 0.02, 12)/1.01](1 + 1.0150) = 10 000(10.575/1.01)(2.015) = 210 976 The present value of all his profit over the next 2 years is about $211 000. 3.41 The 6-month interest rate can be calculated from: 300% = 3 = (1 + i)2 − 1 (1 + i)2 = 4 1 + i = 2 i = 1 = 100% Then: P = 5000(P/F, 100%, 10) + 500(P/A, 100%, 10) = 5000(0.00098) + 500(0.99902) = 504.41 I should pay no more than $504 for the bond now. 3.42 A = 5000(0.015) = 75 i = 0.08/4 = 2% P = 75 + 75(P/A, 2%, 25) + 5000(P/F, 2%, 25) = 75 + 75(19.523) + 5000(0.6095) = 4586.88 You would be willing to pay up to $4587 for the bond. C. More Challenging Problems 3.43 (a) It is not necessary to know how long the mortgages will last, as long as they are equal. For the old plan: Using X to represent the quarterly payments, the present value of one three-month payment cycle is calculated. P(quarterly payment) = X/(1 + i)1 = X/1.06 = 0.943X For the new plan: Each payment is now 0.3X. The present value of three months of payments is 41 Copyright © 2016 Pearson Canada Inc. i  i    Chapter 3 - Cash Flow Analysis P = 0.3X(P/A, 2%, 3) = 0.3X(2.8839) = 0.865X The value of three months of the old plan is greater than that of the new plan. As you prefer to minimize the value of the payments, you prefer the new plan. (b) ie = (1 + is) m − 1 = (1 + 0.06)4 − 1 = 26.24% for the old plan ie = (1 + is) m − 1 = (1 + 0.02)12 − 1 = 26.82% for the new plan The new plan has a higher effective yearly interest rate. 3.44 The cash flow series consists of G at the end of period 2, 2G at the end of period 3 and so on up to (N − 1)G at the end of the Nth period. The first step is to convert each period’s gradient amount to its future value: F = G(1 + i)N−2 + 2G(1 + i)N−3 + ... + (N − 2)G(1 + i) + (N − 1)G Multiplying by (1 + i) and subtracting F gives: F(1 + i) = G(1 + i)N−1 + 2G(1 + i)N−2 + ... + (N − 2)G(1 + i)2 + (N − 1)G(1 + i) F(1 + i) − F = G(1 + i)N−1 + G(1 + i)N−2 + ... + G(1 + i) − (N − 1)G Fi = G[(1 + i)N−1 + (1 + i)N−2 + (1 + i)N−3 + ... + (1 + i) + 1] − NG Noting that the amount in the square brackets is the Series Compound Amount Factor: Fi = G(F/A, i, N) − NG Multiplying both sides by the Sinking Fund Factor: Fi(A/F, i , N) = G − NG(A/F, i, N) Ai = G − NG(A/F, i, N) = G[1 − N(A/F, i, N)] 1 Ni  1 N  A = G −  = G −   i[(1+ i)N −1]  (1+ i)N −1  3.45 The present worth of a geometric series is: P = A + 1+ i A(1+ g) (1+ i) 2 + ... + A(1+ g) N−1 (1+ i) N 42 Copyright © 2016 Pearson Canada Inc.    2 2       Chapter 3 - Cash Flow Analysis If we divide and multiply each term in the present worth expression by (1 + g) and then simplify, we get P = A 1 + g + (1+ g) 2 + ... + (1+ g) N  1+ g  1+ i (1+ i) 2 (1+ i) N  Then, substitute the growth adjusted interest rate, i: i�= 1+i -1 so that 1 = 1+ g 1+ g 1+ i� 1+ i into the present worth expression and we get:   A  1 1 1  P =  + 1+ g 1+ i�  � + ... +    � N  1+ i     1+ i     The right hand side is simply the present worth of an annuity where the constant cash flow each period is A/(1 + g) and the interest rate is i. We can write this as  1 + 1 + ... + 1   1+ i�  �  �N   1+ i  1+ i   (P/A,i�,N) P = A       = A  1+ g       (P/A,i�,N)  1 + g (1+ i�) N −1  1 so that: (P/A,g,i,N) = =   1+ g  i�(1+ i�) N  1 + g 3.46 i = (1 + 0.05)/(1 + 0.5) – 1 = – 0.3 = –30% PWsales = 1 456 988(P/A, –30%, 5)/(1.5) = 1 456 988[(0.75 – 1)/(–0.30.75 )]/(1.5) = Selling price = /2 = 8 013 275 Ruby should sell the company for just over 8 million dollars. 43 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis 3.47 The worth of the loan at month 52 is: F(loan) = 80 000(F/P, 1%, 52) = 134 215.11 The future worth of the payments at month 52 is: F(payments) = 2000(F/A, 1%, 51)(F/P, 1%, 1) = 133 537.78 The difference is the amount of the last payment: Last Payment = F(loan) − F(payments) = 677.33 Clarence's last payment will be about $677. 3.48 Only the spreadsheet for part (a) is provided. (a) Capital Amount $50 000 Annual Interest Rate 8.00% No. of years to repay 15 Payment Annual Interest Recovered Unrecovered Periods Payment Received Capital Capital 0 50000 Total 50000 3.49 Once the project is started, the savings effectively reduce the costs by $50 000 to $100 000 per month. Let X = start of project. 250 000(F/P, 1.5%, X) + 50 000(F/A, 1.5%, X) = 150 000(P/A, 1.5%, 24) 5(F/P, 1.5%, X) + (F/A, 1.5%, X) = 3(20.030) = 60.09 At X = 36: LHS = 5(1.7091) + 47.276 = 60.09 At X = 40: LHS = 5(1.814) + 54.2679 = 60.09 44 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis The project can start about 38 months from now. 3.50 (a) Cash flow diagram 0 1 2 3 4 5 6 7 8 $40000 $26000 (b) There will be a total of 9 payments. The payments can be considered in two portions: 1) The salary costs and the constant portion of the other costs are a 9- period annuity due. The equipment and facility cost declines from $26 000 to a constant $14 000. At the end of eight years, the payments are worth: FA = A (F/A, 7%, 9) = (40 000 + 14 000)(11.978) = 646 812 This includes the first payment, and is equivalent to treating the initial payment separate from a standard 8 period annuity: FA= 54 000(F/P, 7%, 8) + 54 000(F/A, 7%, 8) 2) The declining costs can be modelled as an annuity due less an arithmetic gradient, after four years: FB1 = (26 000 − 14 000)(F/A, 7%, 5) = 12 000(5.7507) = 69 008 The gradient’s worth after the same four years is FB2 = 3000(A/G, 7%, 5)(F/A, 7% 5) = 3000(1.8650)(5.7507) = 32 175 45 Copyright © 2016 Pearson Canada Inc. $14000 Chapter 3 - Cash Flow Analysis For both FB1 and FB2, N = 5 to conform with the standard form for an annuity and a gradient (the first cash flow for an annuity is after one year, while the first cash flow for a gradient is after two years). The value of both cash flows after the eight years is the annuity’s value minus the gradient’s value, taken forward to the eighth year: FB = (69 008 − 32 175)(F/P, 7%, 4) = (36 833)(1.3108) = 48 281 The total worth of the project at the end of the eight years is: FA + FB = 646 812 + 48 281 = $695 093 3.51 g = 0.05 per month i = 0.01 per month i = (1 + i)/(1 + g) − 1 = 1.01/1.05 − 1 = −0.0380 P(expenses) = 15 000(P/A, g, i, 12) = 15 000[(P/A, i, 12)/(1 + g)] = 15 000 [(1.03812 − 1)/(0.0381.03812)]/1.05 = 135 642 Let A be the amount of the monthly instalment: P(grant) = A(P/A, 1%, 6)(P/F, 1%, 12) = A(5.7955)(0.88745) = 5.1432A By letting P(expenses) = P(grant), solve for A: 5.1432A = 135 642 A = $26 373 The amount of the monthly instalment must be 26 373 元. 3.52 The present worth computations for the concrete pool are: P = 1 500 000 + 200 000(A/F, 5%, 10)/0.05 = 1 500 000 + 200 000(0.07951)/0.05 = $1 818 040 3.53 a) i 0  0 ; P = lim A(P / A,i 0 , N) = A N→ i 0 46 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis b) P = A + 1+ i A(1+ g) (1 + i) 2 + ... + A(1+ g) n−1 (1 + i) n + ... noting that A(1+ g) n−1 (1 + i) n = A(1 + g) n−2 (1 + i) n−1 (1 + g) (1+ i) and that since g  1, (1+ g)  1 (1+ i) then A(1+ g) n−1 (1 + i) n  A(1+ g) n−2 (1+ i) n−1 Consequently P is the sum of an infinite number of monotonically increasing amounts, and takes on an infinite value.  c) P = lim N A   =  N→ d) i 0  0 ; 1 + g  P = lim A(P / A,i 0 , N) = A N→ i 0 47 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis Notes for Case-in-Point 3.1 1) No right answer – both are to blame. 2) On the one hand, large infrastructure projects need to be financed. On the other hand, there is a moral hazard for politicians. 3) It is probably not reasonable for political leaders to understand the time value of money. But it is reasonable for them to have competent advisors who do. 4) About 340 months or just over 28 years. Notes for Mini-Case 3.1 Oil industry projects tend to be in the range of 100’s of millions to tens of billions of dollars of investment. As one might expect, very capable people think through all aspects of a project – technical, financial, environmental, political – very carefully. Oil companies take planning seriously, and their engineers are very experienced. The $90 billion of cancelled projects in 2009 does not represent incompetence, but rather something fundamental, which is that no matter how careful the planning process might be, one can never fully predict how the future will turn out. There are three key issues that have strongly affected the Canadian oil sands. The first is that crude oil is a commodity, meaning that there are so many sources that no one can easily predict or control its price. Even if the cost of production is absolutely certain, the profitability and hence project viability depends very much on the world price, which itself is subject to many random forces. No-one predicted that the price of oil would rise to a peak of US$145 per barrel in 2008, or drop the same year to below US$50. Both are extremes outside of the planning window imagined earlier in the same decade. For conventional oil production, costs can be controlled to a degree. Exploration efforts can be reduced, for example, when prices are low. But oil sands project costs are much more difficult to cut back on once they are started, with very large capital costs to recover, and with high direct costs per barrel of oil produced. Investment decisions are thus more vulnerable to low oil prices. The second confounding issue with the oil sands concerns social, environmental and political forces at play. Mining oil sands is very dirty work, damaging the physical sites and producing vast quantities of carbon dioxide. Early oil sands projects sometimes completely failed to recognize the costs of managing their environmental impact, and ended up with unexpected costs for administration, 48 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis legal bills, remediation and meeting regulations. Even in recent years where such costs are recognized and taken into account in the planning process, they are very unpredictable because of the inherent nature of social and political intervention. Otherwise benign activities can suddenly become difficult and expensive if they somehow gain media attention. This has happened in several cases in the oil sands. For example, in 2011 international media attention was given to the “Rethink Alberta” movement, which was started by several environmental groups to dissuade tourists from visiting Alberta unless new oil sands developments were stopped. A third problem is a confluence of global trends, which has been compounded by the oil sand’s remote location. Between 2003 and 2008, for example, a key global force was the industrialization of China. China’s industrialization led to an increase in demand for oil and gas worldwide, causing oil companies to attempt to increase supply. This in turn caused a global proliferation of oil and gas developments, creating a tremendous demand for manufacturing capacity, technical expertise and specialized equipment. This demand significantly increased the costs of acquiring the resources necessary to develop the oil sands. In particular, with a tight labour pool and an isolated location, costs of labour in Alberta skyrocketed – to about double previously estimated costs, far beyond even the most pessimistic expectations. Although some trends can be predicted, others defy prediction. Even with the best planning and with the most reliable process and product, the future remains unpredictable. Economic analyses still need to be done, however, despite an uncertain future. There are sophisticated ways to deal with uncertainty about future cash flows, some of which are discussed in Chapter 12. In most cases it makes sense to carry out the economic analysis with a range of possible values for future cash flows. But, as the oil sands example shows, there are situations when even this approach does not account for what the future actually holds. Solutions to All Additional Problems Note: Solutions to odd-numbered problems are provided on the Student CD-ROM. 3S.1 Mr Dashwood must find the least value of N for which 1500 < 100(P/A,5%,N) which is equivalent to finding N such that (P/A,5%,N) > 15 Consulting Appendix A, this first occurs when N = 29 years. 49 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis If the interest rate is 10%, we come to the end of the table in Appendix A before finding a value of N for which (P/A,10%,N) > 15. We therefore go on to look at the capitalized value of the annuity, which is 100 / 0.1, or £1000. Thus, if he can invest his money at 10%, Mr Dashwood can afford to support his widowed halfsister indefinitely for a cost equivalent to a single present payment of £1000. 3S.2 There are several ways of tackling this. One way is to convert the bi-annuity to an equivalent annuity: A1 = A (A/F,i,2) where ¥A1 is the annual payment equivalent to getting a final payment of ¥A at the end of two years. Then we can convert ¥A1 to its present worth using a formula we already know: P = A1(P/A,i,2N) = A(A/F,i,2)(P/A,i,2N) 3S.3 Having drawn a cash-flow diagram, we write down the PW equivalent of each of the costs: PW = 32 000 − 16 000(P/F,15%,8) + 20 000 (P/ A,15%,8) + 600 + 550(P/A,15%,7) – 50(P/G,15%,7) (represent the decreasing insurance costs as an annuity plus a negative arithmetic gradient) = €122 322. Common mistakes are to treat the insurance payments as coming at year’s end (this is what you assume if you calculate the insurance as 600(P/A,15%,7) − 50(P/G,15%,7)) and to treat the labour costs as occurring at year’s end instead of continuously. (It is rare for a company to wait till the end of the year before paying the year’s wages.) 3S.4 The present worth for each of the two cases can be calculated as follows: Option 1: PW = 230 000 (P/A,0.12,10) = 230 000(5.65) = 1 299 500 Option 2 : PW = 1 100 000 + 500 000 (P/F,0.12,3) + 10 000 (P/A,0.12,3) + 20 000 (P/A,0.12,7)(P/F,0.12,3) − 500 000 (P/F,0.12,10) = 1 383 890 So the first option is better. 50 Copyright © 2016 Pearson Canada Inc. Chapter 3 - Cash Flow Analysis It is also possible to do the comparison in terms of equivalent uniform annual cost; this has the advantage that no calculation is needed for Option 1, though it makes the Option 2 calculations a bit harder. A possible variant is to assume that in Option 1, the company pays rent at the beginning of each year rather than at the end. This makes quite a difference—the present Option 1 cost goes up to 1 455 000, so the second option is then better. 3S.5 The money accumulating in the company’s interest-bearing account is a discrete cash flow, continuously compounded. It is convenient to consider a time interval of six months; for this interval, the nominal interest rate is 2%, and the study period is ten six-monthly intervals. We first convert the arithmetic increase in income to an equivalent annuity, A : A = 10 000(A/G,0.02,10) = 10 000(4.3351) = 43 451 So the incoming cash flow is equivalent to the sum of this and the base annuity, ¥100 000. The future worth of the total equivalent annuity is therefore: F = 143 451(F/A,0.02,10) = 143 451(10.9598) = 1 572 194 So if each of the students were to stay in the group, each would have a one-fifth share of this, which is ¥314 439. The student who leaves the group will instead get a one-fifth share of the present value of the total