(Solution Manual) Engineering Economy

Blank,Tarquin, 7

Engineering

Solution Manual

7th Edition

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Solution Manual for Engineering Economy, The downloadable file consists of 322 pages, and the total size of the file is 3.55 MB. Purchase now only for highly discounted price.

Author(s): Blank, Tarquin

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1 Solutions to end-of -chapter problems Engineering Economy, 7th edition Leland Blank and Anthony Tarquin Chapter 1 Foundations of Engineering Economy 1.1 The four elements are cash flows, time of occurrence of cash flows, interest rates, and measure of economic worth. 1.2 (a) Capital funds are money used to finance projects. It is usually limited in the amount of money available. (b) Sensitivity analysis is a procedure that involves changing various estimates to see if/how they affect the economic decision. 1.3 Any of the fo llowing are measures of worth: p resent worth, future worth, annual worth, rate of return, benefit/cost ratio, capitalized cost, payback period, economic value added. 1.4 F irst cost: e conomic ; leadership: non- economic ; taxes: economic ; salvage value: economic ; morale : non- economic ; dependability : non-economic ; inflation : economic ; profit : economic ; acceptance: non- economic; ethics: non- economic; interest rate : economic. 1.5 Many sections could be identified. Some are: I.b; II.2.a and b; III.9.a and b. 1.6 Example actions are: • Try to t alk them out of doing it now , explaining it is stealing • Try to get them to pay for their drinks • Pay for all the drinks himself • Walk away and not associate with them again 1.7 This is structured to be a discussion question; many responses are acceptable. It is an ethical question, but also a guilt -related situation. He can justify the result as an accident; he can feel justified by the legal fault and punishment he receives; he can get ang ry because it WAS an accident; h e can beco me tormented over time due to the stress caused by accidently causing a child’s death. 1.8 This is structured to be a discussion question; many responses are acceptable. Responses can vary from the ethical (stating the truth and accepting the consequences) to unethical (continuing to deceive himself and the instructor and devise some on- the-spot excuse). Lessons can be learned fro m the experience. A few of them are: • Think before he cheats again. • Think about the longer -term consequences of unethical decis ions. • Face ethical -dilemma situations honestly and make better decisions in real time. 2 Alternatively, Claude may learn nothing from the experience and c ontinue his unethical practices. 1.9 i = [(3,885,000 - 3,500,000)/3,500,000]*100% = 11% per year 1.10 (a) Am ount paid first four years = 900,000(0.12) = $108,000 (b) Final payment = 900,000 + 900,000(0.12) = $1,008,000 1.11 i = (1125/12,500)*100 = 9% i = (6160/56,000)*100 = 11% i = (7600/95,000)*100 = 8% T he $56,000 investment has the highest rate of return . 1.12 Interest on loan = 23,800(0.10) = $2,380 Default insurance = 23,800(0.05) = $1190 Set -up fee = $300 Total amount paid = 2380 + 1190 + 300 = $3870 Effective interest rate = (3870/23,800)*100 = 16.3% 1.13 The market interest rate is usually 3 – 4 % above the expected inflation rate. Therefore, Market rate is in the range 3 + 8 to 4 + 8 = 11 to 12% per year 1.14 PW = pres ent worth; PV = present value; NPV = net present value; DCF = discounted cash flow; and CC = capitalized cost 1.15 P = $150,000; F = ?; i = 11%; n = 7 1.16 P = ?; F = $100,000; i = 12%; n = 2 1.17 P = $3.4 million; A = ?; i = 1 0%; n = 8 1.18 F = ?; A = $100,000 + $125,000?; i = 15%; n = 3 1.19 End -of-period convention means that all cash flows are assumed to take place at the end of the in terest period in which they occur. 1.20 fuel cost : outflow ; pension plan contributions : outflow ; passenger fares : inflow ; m aintenance: outflow ; freight revenue: inflo w; cargo revenue: inflow ; extra bag charges: I nflow ; water and sodas: outflow ; advertising : outflow ; landing fees : outflow ; seat preference fees: i nflow . 3 1.21 End -of-period am ount for June = 50 + 70 + 120 + 20 = $260 End -of-period am ount for Dec = 150 + 90 + 40 + 110 = $390 1.22 Month Receipts, $1000 Disbursements, $1000 Net CF, $ 1000 Jan 500 300 +200 Feb 800 500 +300 Mar 200 400 -200 Apr 120 400 -280 May 600 500 +100 June 900 600 +300 July 800 300 +500 Aug 700 300 +400 Sept 900 500 +400 Oct 500 400 +100 Nov 400 400 0 Dec 1800 700 +1100 Net Cash flow = $2,920 ($2,920,000) 1.23 1.24 4 1.25 1.26 Am ount now = F = 100,000 + 100,000(0.15) = $115,000 1.27 Equivalent present amount = 1,000,000/(1 + 0.15) = $869,565 Discount = 790,000 – 869,565 = $79,565 1.28 5000(40 )(1 + i) = 225,000 1 + i = 1.125 i = 0.125 = 12.5% per year 1.29 Total bonus next year = 8,000 + 8,000(1.08) = $ 16,640 1.30 ( a) Early -bird payment = 10,000 – 10,000(0.10) = $9000 (b) Equivalent future am ount = 9000(1 + 0.10) = $9900 Savings = 10,000 – 9900 = $100 1.31 F 1 = 1,000,000 + 1,000,000(0 .10) = 1,100,000 F 2 = 1,100,000 + 1,100,000(0.10) = $1,210,000 1.32 90,000 = 60,000 + 60,000(5)(i) 300,000 i = 30,000 i = 0.10 (10% per year ) 1.33 (a) F = 1,800,000(1 + 0.10) (1 + 0.10) = $2,178,000 (b) I nterest = 2,178,000 – 1,800,000 = $378,000 5 1.34 F = 6,000,000(1 + 0.09) (1 + 0.09) (1 + 0.09) = $7,770,174 1.35 4,600,000 = P(1 + 0.10)(1 + 0.10) P = $3,801,653 1.36 86,400 = 50,000( 1 + 0.20) n log (86,400/50,000) = n(log 1.20) 0.23754 = 0.07918n n = 3 years 1.37 Simple: F = 10,000 + 10,000(3)(0.10) = $13,000 Compound: 13,000 = 10,000(1 + i) (1 + i) (1 + i) (1 + i) 3 = 1.3000 3log(1 + i) = log 1.3 3log ( 1 + i) = 0.1139 log(1 + i) = 0.03798 1 + i = 1.091 i = 9.1% per year 1.38 Minimum attractive rate of return is also referred to as hurdle rate, cutoff rate, benchmark rate, and minimum acceptable rate of return. 1.39 bonds - debt ; stock sales – equity ; retained earnings – equity ; venture capital – debt ; short term loan – debt ; capital advance from friend – debt ; cash on hand – equity ; credit card – debt ; home equity loan - debt . 1.40 WACC = 0.30( 8%) + 0.70( 13%) = 11.5% 1.41 WACC = 10%(0.09) + 90% (0.16) = 15.3% The company should undertake the inventory, technology, and warehouse projects. 1.42 (a) PV(i%,n,A,F) finds the present value P (b) FV(i%,n,A,P) finds the future value F (c) RATE(n,A,P,F) finds the compound interest rate i (d) IRR(first_cell:last_cell) finds the compound interest rate i (e) PMT(i%,n,P,F) finds the equal periodi c payment A (f) NPER(i%,A ,P,F) finds the number of periods n 6 1.43 (a) NPER(8%, -1500,8000,2000) : i = 8%; A = $ -1500; P = $8000; F = $2000; n = ? (b ) FV(6%,1 0,2000,- 9000) : i = 6%; n = 1 0; A = $2000; P = $ -9000; F = ? (c) RATE(10,1000,- 12000,2000) : n = 10; A = $1000; P = $ -12,000; F = $2000; i = ? (d ) PMT (11%,20, ,14000) : i = 11%; n = 20; F = $ 14,000; A = ? (e) PV(8 %,15, -1000,800) : i = 8%; n = 15; A = $ -1000; F = $800; P = ? 1.44 (a) PMT is A (b) FV is F (c) N PER is n (d) PV is P (e) IRR is i 1.45 (a) For built -in functions, a parameter that does not apply can be left blank when it is not an interior one. For example, if there is no F involved when using the PMT function to solve a particular problem, it can be left blank (omitted) because it is an end parameter . (b) When the parameter involved is an interior one (like P in the PMT function), a comma must be put in its position. 1.46 Spreadsheet shows relations only in cell reference format. Cell E10 will indicate $64 more than cell C10. 1.47 Answer is (b) 1.48 Answer is (d) 1.49 Answer is (a) 1.50 Answer is (d) 1.51 Upper limit = (12,300 – 10,700)/10,700 = 15% Lower limit = (10,700 – 8,900)/10,700 = 16.8% Answer is (c) 1.52 Am ount one year ago = 10,000/(1 + 0.10) = $9090.90 Answer is (b) 7 1.53 Answer is (c) 1.54 2P = P + P(n)(0.04) 1 = 0.04n n = 25 Answer is (b) 1.55 Answer is (a ) 1.56 WACC = 0.70( 16%) + 0.30(12%) = 14,8% Answer is (c) 8 Solution to Case Studies, Chapter 1 There is no definitive answer to case study exercises. The following are examples only. Renewable Energy Sources for Electricity Generation 3. LEC approximation uses (1.05)11 = 0.5847, X = P 11 + A 11 + C 11 and LEC last year = 0.1022. X(0.5847) 0.1027 = 0.1022 + ---------------------- (5.052 B)(0.5847) X = $ 2.526 million Refrigerator Shells 1. The first four steps are: Define objective, information collection, alternative definition and estimates, and criteria for decision -making. Obje ctive: Select the most economic alternative that also meets requirements such as production rate, quality specifications, manufacturability for design specifications, etc. Information: Each alternative must have estimates for life (likely 10 years), AOC and other costs (e.g., training), first cost, any salvage value, and the MARR. The debt versus equity capital question must be addressed, especially if more than $5 million is needed. Alternatives: For both A and B, some of the required data to perform a n analysis are: P and S must be estimated. AOC equal to about 8% of P must be verified. Training and other cost estimates (annual, periodic, one -time) must be finalized. Confirm n = 10 years for life of A and B. MARR will probably be in the 15% to 18% per year range. Criteria: Can use either present worth or annual worth to select between A and B. 2. Consider these and others like them: Debt capital availability and cost Competition and s ize of market share required Employee safety of plastics used in processing 9 3. With the addition of C, this is now a make/buy decision. Economic estimates needed are:  Cost of lease arrangement or unit cost, whatever is quoted.  Amount and length of time the arrangement is available. Some non- economic factors may be:  Guarantee of available time as needed.  Compatibility with current equipment and designs.  Readiness of the company to enter the market now versus later 1 Solutions to end-of -chapter problems Engineering Economy, 7th edition Leland Blank and Anthony Tarquin Chapter 2 Factors: How Time and Interest Affect Money 2.1 ( 1) (P/F, 6%, 8) = 0.6274 ( 2) (A/P, 10%,10) = 0.16275 ( 3) (A/G,15%,20) = 5.3651 ( 4) (A/F,2%,30) = 0.02465 ( 5) (P/G,35%,15) = 7.5974 2.2 P = 21,300(P/A,10%,5) = 21,300(3.7908) = $80,744 2.3 Cost now = 142(0.60) = $85.20 Present worth at regular cost = 142(P/F,10%,2) = 142(0.8264) = $117.35 Present worth of savings = 117.35 – 85.20 = $32.15 2.4 F = 100,000(F/P,10%,3) + 885,000 = 100,000(1.3310) + 885,000 = $1,018,100 2.5 F = 50,000(F/P,6%,14) = 50,000(2.2609) = $113,045 2.6 F = 1,900,000(F/P,15%,3) F = 1,900,000(1.5209) = $2,889,710 2.7 A = 220,000(A/P,10%,3) = 220,000(0.40211) = $88,464 2.8 P = 75,000(P/F,12%,4) = 75,000(0.6355) = $47,663 2 2.9 F = 1.3(F/P,18%,10) = 1.3(5.2338) = 6.80394 ($6,803,940) 2.10 P = 200,000(P/F,15%,1) + 300,000(P/F,15%3) = 200,000(0.8696) + 300,000(0.6575) = $371,170 2.11 Gain in worth of building after repairs = (600,000/0.75 – 600,000) - 25,000 = 175,000 F = 175,000(F/P,8%,5) = 175,000(1.4693) = $257,128 2.12 F = 100,000(F/P,8%,4) + 150,000(F/P,8%,3) = 100,000(1.3605) + 150,000(1.2597) = $325,005 2.13 P = ( 110,000* 0.3) (P/A,12%,4) = (33,000)(3.0373) = $100,231 2.14 P = 600,000(0.04)(P/A,10%,3) = 24,000(2.4869) = $59,686 2.15 A = 950,000(A/P,6%,20) = 950,000(0.08718) = $82,821 2.16 A = 434(A/P,8%,5) = 434(0.25046) = $108.70 2.17 F = (0.18 – 0.04) (100)(F/A,6%,8) = 14(9.8975) = $138.57 2.18 F difference = 10,500(F/P,7%,18) - 10,500(F/P,4%,18) = 10,500(3.3799) - 10,500(2.2058) = $12,328 2.19 F = (200 – 90)(F/A,10%,8) = 110( 11.4359) = $ 1,257,949 3 2.20 A = 350,000(A/F,10%,3) = 350,000(0.30211) = $105,739 2.21 (a) 1. Interpolate between i = 12% and i = 14% at n = 15. 1/2 = x/(0.17102 – 0.14682) x = 0.0121 (A/P,13%,15) = 0.14682 + 0.0121 = 0.15892 2. Interpolate between i = 25 % and i = 30% at n = 1 0. 2/5 = x/ (9.9870 - 7.7872) x = 0.8799 (P/G,27%,1 0) = 9.9870 – 0.8799 = 9.1071 (b) 1. ( A/P,13%, 15) = [ 0.13(1 + 0.13) 15 ]/ [(1 + 0.13)15 - 1] = 0.15474 2. (P/G,27%,1 0) = [(1 + 0. 27) 10 – (0.27)( 10) - 1]/[0.2 72(1 + 0.27)10 ] = 9.0676 2.22 (a) 1. Interpolate between n = 60 and n = 65: 2/5 = x/(4998.22 – 2595.92) x = 960.92 ( F/P,14%, 62) = 4998.22 – 960.92 = 4037.30 2. Interpolate between n = 4 0 and n = 48 : 5/ 8 = x/ (0.02046 – 0.01633) x = 0.00258 (A/ F,1%,45 ) = 0.02046 – 0.00258 = 0.01788 (b) 1. ( F/P,14%, 62) = (1+0.14 ) 62 -1 = 3373.66 2. (A/ F,1%,45 ) = 0.01/[(1+ 0.01) 45 – 1] = 0.01771 4 (c) 1. = -FV(14%,62,,1) displays 3373.66 3. = PMT(1%,45,,1) displays 0.01771 2.23 Interpolated value: Interpolate between n = 40 and n = 45: 3/5 = x/(72.8905 – 45.2593) x = 16.5787 (F/P,1 0%,43) = 45.2593 + 16.5787 = 61.8380 Formula value: (F/P,10%,43) = (1+ 0.10) 43 -1= 59.2401 % difference = [(61.8380 - 59.2401)/ 59.2401]*100 = 4.4% 2.24 Interpolated value: Interpolate between n = 50 and n = 55: 2/5 = x/( 14524 – 7217.72) x = 2922.51 (F/ A,15%, 52) = 7217.72 + 2922.51 = 10,140 Formula value: (F/ A,15%, 52) = [ (1+ 0.15 ) 52 -1]/0.15 = 9547.58 % difference = [( 10,140 - 9547.58)/ 9547.58]( 100) = 6.2% 2.25 (a) Profit in year 5 = 6000 + 1100(4) = $10, 400 (b) P = 6000(P/A,8%,5) + 1100(P/G,8%,5) = 6000(3.9927) + 1100(7.3724) = $32,066 2.26 (a) G = (241 – 7)/9 = $26 billion per year (b) Loss in year 5 = 7 +4(26) = $111 billion (c) A = 7 + 26(A/G,8%,10) = 7 + 26(3.8713) = $107.7 billion 2.27 A = 200 – 5(A/G,8%,8) = 200 – 5( 3.0985 ) = $ 184.51 5 2.28 P = 60,000(P/A,10%,5) + 10,000(P/G,10%,5) = 60,000(3.7908) + 10,000(6.8618) = $296,066 2.29 (a) CF 3 = 70 + 3(4) = $82 ($82,000) (b) P = 74(P/A,10%,10) + 4(P/G,10%,10) = 74(6.1446) + 4(22.8913) = $546.266 ($546,266) F = 546.266(F/P,10%,10) = 521.687(2.5937) = $ 1416.850 ($1,416,850) 2.30 601.17 = A + 30(A/G,10%,9) 601.17 = A + 30(3.3724) A = $500 2.31 P = 2.1B (P/F,18%,5) = 2.1B (0.4371) = $917,910,000 917,910,000 = 50,000,000(P/A,18%,5) + G(P/G,18%,5) 917,910,000 = 50,000,000(3.1272) + G(5.2312) G = $14,557,845 2.32 75,000 = 15,000 + G(A/G,10%,5) 75,000 = 15,000 + G(1.8101) G = $33,147 2.33 First find P g (using equation) and then convert to A For n = 1: P g = {1 – [(1 + 0.04)/(1 + 0.10)]1}/(0.10 – 0.04) = 0.90909 A = 0.90909(A/P,10%,1) = 0.90909(1.1000) = 1.0000 For n = 2: P g = {1 – [(1 + 0.04)/(1 + 0.10)]2}/(0.10 – 0.04) = 1.7686 A = 1.7686(A/P,10%,2) = 1.7686(0.57619) = 1.0190 6 2.34 P g = 50,000{1 – [(1 + 0.06)/(1 + 0.10)]8}/(0.10 – 0.06) = $320,573 2.35 P g1 = 10,000{1 – [(1 + 0.04)/(1 + 0.08)]10}/(0.08 – 0.04) = $78,590 P g2 = 10,000{1 – [(1 + 0.06)/(1 + 0.08)]11}/(0.08 – 0.06) = $ 92,926 Difference = $14,336 2.36 P g = 260{1 – [(1 + 0.04)/(1 + 0.06)]20}/(0.06 – 0.04) = 260(15.8399) = $4118.37 per acre -ft 2.37 P = 30,000[10/(1 + 0.06)] = $283,019 2.38 18,000,000 = 3,576,420(P/A,i,7) (P/A,i,7) = 5.0330 From interest tables in P/A column and n = 7, i = 9% per year . Can be solve d using the RATE function = RATE( 7,3576420,18000000). 2.39 813,000 = 170,000(F/P,i,15) 813,000 = 170,000(1 + i) 15 log 4.78235 = (15)log (1 + i) 0.6796/15 = log (1 + i) log (1 + i) = 0.04531 1 + i = 1.11 i = 11 % per year Can be solved using the RATE function = RATE(15,,- 170000,813000). 2.40 100,000 = 210,325(P/F,i,30) (P/F,i,30) = 0.47545 Find i by inte rpolation between 2% and 3%, by solving the P/F equation for i, or by spreadsheet. By spreadsheet function = RATE(30,,100000,- 210325) , i = 2.51% . 7 2.41 (1,000,000 – 1,900,000) = 200,000(F/P,i,4) (F/P,i,4) = 4.5 Find i by interp olation between 40% and 50%, by s olving F/P equation, or by spreadsheet. By spreadsheet function = RATE(4,, -200000,900000), i = 45.7% per year. 2.42 800,000 = 250,000(P/A,i,5) (P/A,i,5) = 3.20 Interpolate between 16% and 18% interest tables or use a spreadsheet. By spreadsheet function, i = 16.99% ≈ 17% per year. 2.43 87,360 = 24,000(F/A,i,3) (F/A,i,3) = 3.6400 Fo r n = 3 in F/A column, 3.6400 is in 20% interest table . Therefore, i = 20% per year. 2.44 48,436 = 42,000 + 4000(A/G,i,5) 6436 = 4000(A/G,i,5) (A/G,i,5) = 1.6090 For n = 5 in A/G column, value of 1.6090 is in 22% interest table . 2.45 600,000 = 80,000(F/A,15%,n) (F/A,15%,n) = 7.50 Interpolate in the 15 % interest table or use a spreadsheet function. By spreadsheet , n = 5.4 years. 2.46 Starting amount = 1,600,000(0.55) = $880,000 1,600,000 = 880,000(F/P,9%,n) (F/P, 9%,n) = 1.8182 Interpolate in 9% interest table or use the spreadsheet function = NPER(9%,, -880000,1600000) to determine that n = 6.94 ≈ 7 years . 2.47 200,000 = 29,000(P/A,10%,n) (P/A,10%,n) = 6.8966 Interpolate in 10% interest table or use a spreadsheet function to display n = 12.3 years . 2.48 1,500,000 = 18,000(F/A,12%,n) (F/A,12%,n) = 83.3333 Interpolate in 12% interest table or use the spreadsheet function 8 = NPER(12%, - 18000,,1500000) to display n = 21.2 years . Time from now is 21.2 - 15 = 6.2 years . 2.49 350,000 = 15,000(P/A,4%,n) + 21,700(P/ G,4%,n) Solve by trial and error in 4% interest table between 5 and 6 years to determine n ≈ 6 years 2.50 16,000 = 13,000 + 400(A/G,8%,n) (A/G,8%,n) = 7.5000 Interpolate in 8% interest table or use a spreadsheet to determine that n = 21.8 years . 2.51 140(0.06 - 0.03) = 12{1 – [(0.97170)] x} 4.2/12 = 1 – [0.97170]x 0.35 - 1 = – [0.97170]x 0.65 = [0.97170]x log 0.65 = (x)(log 0.97170) x = 15 years 2.52 135,300 = 35,000 + 19,000(A/G,10%,n) 100,300 = 19,000(A/G,10%,n) (A/G,10%,n) = 5.2789 From A/G column in 10% interest table, n = 15 years. 2.53 88,146 = 25,000{1 – [(1 + 0.18)/(1 + 0.10) ] n}/(0.10 – 0.18) 3.52584 = {1 – [(1.18)/(1.10) ]n}/(-.08) - 0.28207 = {1 – [(1.18)/(1.10]n} - 1.28207 = – [(1.18)/(1.10]n 1.28207 = [(1.07273]n log 1.28207 = n log 1.07273 0.10791 = n(0.03049) n = 3.54 years 2.54 P = 30,000(P/F,12%,3) = 30,000(0.7118) = $21,354 Answer is (d) 9 2.55 30,000 = 4200(P/A,8%,n) (P/A,8%,n) = 7.14286 n is between 11 and 12 years Answer is (c) 2.56 A = 22 ,000 + 1000(A/G,8%,5) = $23,847 Answer is (a) 2.57 A nswer is (d) 2.58 A = 800 – 100(A/G,4%,6) = $561.43 Answer is (b) 2.59 Answer is (b) 2.60 F = 61,000(F/P,4%,4) = 61,000(1.1699) = $71,364 Answer is (c) 2.61 P = 90,000(P/A,10%,10) = 90,000( 6.1446) = $553,014 Answer is ( d) 2.62 A = 100,000(A/P,10%,7) = 100,000(0.20541) = $20,541 Answer is (b) 2.63 A = 1,500,000(A/F,10%,20) = 1,500,000(0.01746) = $26,190 Answer is (a) 10 2.64 In $1 million units A = 3(10)(A/P,10%,10) = 30(0.16275) = $4.8825 (≈ $4.9 million) Answer is (c) 2.65 75,000 = 20,000(P/A,10%,n) (P/A,10%,n) = 3.75 By interpolation or NPER function, n = 4.9 years Answer is (b) 2.66 50,000(F/A,6%,n) = 650,000 (F/A,6%,n) = 13.0000 By interpolation or NPER function, n = 9.9 years Answer is ( d) 2.67 40,000 = 13,400(P/A,i,5) (P/A,i,5) = 2.9851 By interpolation or RATE function, i = 20.0 % per year Answer is (a) 2.68 P = 26,000(P/A,10%,5) + 2000(P/G,10%,5) = 26,000(3.7908) + 2000(6.8618) = $112,284 Answer is ( b) 2.69 F = [5000(P/A,10%,20) + 1000(P/G,10%,20)](F/P,10%,20) = [5000(8.5136) + 1000(55.4069)](6.7275) = $ 659,126 Answer is (d) 2.70 A = 300,000 – 30,000(A/G,10%,4) = 300,000 – 30,000(1.3812) = $258,564 Answer is (b) 11 2.71 F = {5000[1 - (1.03 /1.10) 20]/(0.10 – 0.03)}(F/P,10%,20) = {5000[1 - (1.03/ 1.10) 20]/(0.10 – 0.03)}(6.7275) = $351,528 Answer is (c) 12 Solution to Case Stud y, Chapter 2 There is no definitive answer to case study exercises. The following are examples only. Time Marches On; So Does t he Interest Rate 1. Situation A B C D Interest rate 6% per year 6% per year 15% per year Simple: 780% per year Comp’d: 143,213% per year C: 2 million = 300,000(P/A,i%,65) (P/A,i%,64) = 6.666667 i = 15% D: 30/200 = 15% per week Simple: 15%(52 weeks) = 780% per year Compound: (1.15) 52 - 1 = 143,213% per year 2. A: Start -- $24 End -- F = 24(1.06) 385 = $132 billion B: Start -- $2000 per year or $20,000 total over 10 years End -- F 32 = A(F/A,6%,10) = $26,361.60 F 70 = F 32(F/P,6%,38) = $241,320 C: Start -- $2 million End -- 300,000(65) = $19.5 million over 65 years F 65 = 300,000(F/A,15%,65) = $17.6 billion (equivalent) D: Simple interest Start -- $200 End -- (0.15)(12)(200) + 200 = $1760 Compound interest Start -- $200 End -- 200(1.15) 52 = $286,627 1 Solutions to end-of -chapter problems Engineering Economy, 7th edition Leland Blank and Anthony Tarquin Chapter 3 Combining Factors and Spreadsheet Functions 3.1 P = 12,000 + 12,000(P/A,10%,9) = 12,000 + 12,000( 5.7590) = $81,108 3.2 P = 260,000(P/A,10%,3) + 190,000(P/A,10%,2)(P/F,10%,3) = 260,000(2.4869) + 190,000(1.7355)(0.7513) = $ 894,331 3.3 (a) P = - 120(P/F,12%,1) - 100(P/F,12%,2) - 40(P/F,12%,3) + 50(P/A,12%,2 )(P/F,12%,3) + 80(P/A,12%,4)(P/F,12%,5) = - 120(0.8929) - 100(0.7972) - 40(0.7118) + 50(1.6901)(0.7118) + 80(3.0373)(0.5674) = $- 17,320 (b) Enter cash flows in, say, column A, and use the function = NPV(12% ,A2:A10)*1000 to display $ -71,308. 3.4 = 22,000( 5.7466)(0.8573) P = 22,000( P/A,8%,8)(P/F,8%,2) = $108,384 3.5 P = 200(P/A,10%,3)(P/F,10%,1) + 90(P/A,10%,3)(P/F,10%,5) = 200(2.4869)(0.9091) + 90(2.4869)(0.6209) = $591.14 3.6 Discount amount = 1.56 – 1.28 = $0.28/1000 g Savings in cost of water used/yea r = [2,000,000,000/1000]0.28 = $560,000 P = 56 0,000(P/A,6%,20) = 56 0,000(11.4699 ) = $6,423,144 3.7 P = 105,000 + 350 + 350(P/A,10%,30) = 105,000 + 350 + 350(9.4269) = $108,649 2 3.8 P = (20 – 8) + (20 – 8)(P/A,10%,3) + (30 – 12)(P/A,10%,5)(P/F,10 %,3) + (30 - 25)(P/F,10% ,9) = 12 + 12( 2.4869) + 18( 3.7908)(0.7513 ) + 5( 0.4241) = $95,228 3.9 2,000,000 = x(P/F,10%,1) + 2x(P/F,10%,2) + 4x(P/F,10%,3) + 8x(P/F,10%,4) 2,000,000 = x( 0.9091) + 2x( 0.8264) + 4x( 0.7513) + 8x( 0.6830) 11.0311x = 2,000,000 x = $181,306 (first payment) 3.10 A = 300,000 + ( 465,000 – 300,000) (F/A,10%,5)(A/F,10%,9) = 300,000 + 165,000(6.1051)(0.07364) = $374,181 per year 3.11 (a) 2,000,000 = 25,000(F/P,10%,20) + A(F/A,10%,20) 2,000,000 = 25,000(6.7275) + A(57.2750) A = $ 31,983 per year (b) Yes. In fact, they will exceed their goal by $459,188 3.12 (a) A = 16,000(A/P,10%,5) + 52,000 + (58,000 – 52,000)(P/F,10%,1)(A/P,10%5) = 16,000(0.26380) + 52,000 + 6000(0.9091)(0.26380) = $57,660 per year (b) Annual savings = 73,000 - 57,660 = $15,340 per year 3.13 (a) A = 8000(A/P,10%,9) + 4000 + (5000 – 4000)(F/A,10% ,4)(A/F,10%,9) = 8000(0.17364) + 4000 + (5000 – 4000)(4.6410)(0.07364) = $5731 per year (b) Enter cash flows in, say, column B , rows 2 through 11, and use the embedded function = - PMT(10%,9,NPV(10%,B 3:B11) + B2) to display $5731. 3.14 (a) 300 = 200(A/P,10%,7) + 200(P/A,10%,3)(A/P,10%,7) + x(P/F,10%,4)(A/P,10%,7) + 200(F/A,10%,3)(A/F,10%,7) 300 = 200(0.20541) + 200( 2.4869)(0.20541) + x( 0.6830)(0.20541) + 200( 3.3100)(0.10541 ) 0.14030x = 300 – 213.03 x = $619.88 (b) Enter cash flows in B3 through B9 with a number like 1 in year 4. Now, set up PMT function such as = - PMT(10%,7,NPV(10%,B3:B9) + B2) . Use Goal Seek to change year 4 such that PMT function displays 300. Solution is x = $619.97. 3 3.15 Am ount owed after first payment = 10,000,000(F/P,9%,1) - 2,000,000 = 10,000,000(1.0900) - 2,000,000 = $8,900,000 Payment in years 2 - 5: A = 8,900,000(A/P,9%,4) = 8,900,000(0.30867) = $2,747,163 per year 3.16 A = [6000(P/F,10%,1) + 9000(P/F,10%,3) + 10,000(P/F,10%,6)](A/P,10%,7) = [6000(0.9091) + 9000(0.7513) + 10,000(0.5645)](0.20541) = $3669 per year 3.17 Find P 0 and then convert to A . In $1000 units, P 0 = 20(P/A,12%,4) + 60(P/A,12%,5)(P/F,12%4) = 20(3.0373) + 60(3.6048)(0.6355) = $198.197 ($198,197) A = 198.197(A/P,12%,9) = 198.197(0.18768) = $37.197 ($39,197 per year) 3.18 A = - 2500(A/P,10%,10) + (700 – 200)(P/A,10%,4)(A/P,10%,10) + (2000 – 300)(F/A,10%,6)(A/F,10%,10) = - 2500(0.16275) + 500(3.1699)(0.16275) + 1700( 7.7156)(0.06275) = $674.14 per year 3.19 A = 1000 + [100,000 + 50,000(P/A,10%,5)](A/P,10%,20) = 1000 + [100,000 + 50,000(3.7908)](0.11746) = $ 35,009 per year 3.20 Payment amount is an A for 10 years in years 0 through 9. Annual amount = 150,000(P/F,10%,1)(A/P,10%,10) + 2(300) = 150,000(0.9091)(0.16275) + 600 = 22,193 + 600 = $22,793 per year 3.21 360,000 = 55,000(F/P,8%,5) + 90,000(F/P,8%, 3) + A(F/A,8%,3) 360,000 = 55,000(1.4693) + 90,000(1.2597) + A(3.2464) 3.2464A= 165,816 A = $51,076 per year 3.22 F = [ 100(F/A,10%,7) + (300 – 100)(F/A,10%,2) ](F/P,10%,2) = [ 100(9.4872) + (300 – 100)(2.1000) ](1.2100) = $1656.15