Solution Manual For Basics Of Engineering Economy, 2nd Edition

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1Solutions to end-of-chapter problemsBasics of Engineering Economy, 2ndeditionLeland Blank and Anthony TarquinChapter 1Foundations of Engineering Economy1.1If the alternative that is actually the best oneis not even recognized as analternative, it obviously will not be able to be selected by using any economicanalysis tools.1.2Non-economic factors are (b), (c), (e) and (f): morale, goodwill,publicacceptanceand aesthetics,respectively.1.3Revenues and costs are examples of cash flows.1.4The analysis techniques that are used in engineering economic analysis are onlyas good as the accuracy of the cash flow estimates.1.5The do-nothing alternative represents theas-isorstatus quocondition.1.6Evaluation criterion refers to the measurement standard or procedure that is usedto determine which alternative is “best”.1.7In engineering economy, the evaluation criterion is financial units (dollars, pesos,etc).1.8Intangible factors are non-economic considerations that may have to be taken intoaccount in identifying the best alternative.1.9Examples of intangible factors are goodwill, morale, convenience, friendship,implementation know-how, aesthetics, public acceptance1.10Time value of money means that there is a certain worth in having money and thatworth changes as a function of time.1.11Interest is a manifestation of the time value of money.1.12The most important fundamental dimension in economic analysis is time becauseof time value of money.1.13The term that describes compensation for “renting” of money is time value ofmoney, which manifests itself as interest.1.14When an interest rate does not include the time period, the time period is assumedto be one year.

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21.15The original amount of money in a loan transaction is known as the principal.1.16Minimum attractive rate of return is the lowest rate of return (interest rate)on aprojectthatcompanies or individuals consider to be high enough to induce themto invest their money.1.17When the yield on a U.S.Government Bond is 3% per year, investors areexpecting the inflation rate to be near zero.1.18(a) Payment = 1,600,000(1.10)(1.10) = $1,936,000(b) Interest = total amt paid–principal= 1,936,000-1,600,000= $336,0001.19(a) Equivalent cost in 1 year = 38,000 + 38,000(0.10)= $41,800(b) Since $41,600 is less than $41,800, the firm should remodel 1ater (i.e. 1yearfrom now).1.20P = 50,000/1.08= $46,296.301.21Rate ofreturn = (1,450,000/10,000,000)(100%)= 14.5% per year1.22Rate of return = (45/966)(100%)= 4.7%1.23Rate of increase = [(29–22)/22](100%)= 31.8%1.24Interest rate = (275,000/2,000,000)(100%)= 13.75%1.25Rate of return = (2.3/6)(100%)= 38.3%1.26F = P(1 + i)2,360,000 = 2,000,000(1 + i)i =0.18(18%)1.27F =P(1 + i)53,000,000 = P(1 + 0.10)P= $48,181,818

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31.28Amount of earnings in year one =400,000,000(0.25)= $100,000,0001.29Amt at end of year 1 = 280,000(1.15) = $322,000< $425,000Amt at end of year 2 = 322,000(1.15) = $370,300< $425,000Amt at end of year 3 = 370,300(1.15) = $425,845> $425,000Time = 3 years1.30MARR is set relative to the cost of capital.1.31The engineer is wrong, unless the MARR is exactly equal to the cost of capital.Usually, theinequality ROR ≥ MARR > cost of capital is used, and the MARR isestablished higher than the cost of capital so that profit, risk and other factors areconsidered.1.32(a) Amt owed; compound interest = 150,000(1.05)(1.05)(1.05)= $173,644Amt owed;simple interest = P + Pni= 150,000 + 150,000(3)(0.055)= 150,000 + 24,750= $174,750Company should accept 5% compound interest rate(b) Difference = 174,750–173,644= $11061.33F = P + Pni850,000 = P + P(4)(0.10)1.4P = 850,000P = $607,1431.34F = P + Pni100,000 = 1000 + 1000(n)(0.1)99,000 = 1000(n)(0.10)n = 990years1.35Follow plan 4, Figure 1.3 as amodel.A is 9900–2000 = $7900B is 7900(0.10) = $790C is7900 + 790 = $8690D is8690–2000 = $6690

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41.36F =P +Pni35,000 = P + P(3)(0.07)1.21P = 35,000P = $28,925.601.37P(1.20)(1.20) = 20,000P = $13,888.891.38(a) Interest rate = 900,000/6,000,000 = 0.15 = 15% per year(b) Total due after 1 year = 6,000,000 + 900,000 = $6,900,000Interest charged for 2ndyear = 6,900,000(0.15) = $1,035,0001.39(a)F = P + Pni2,800,000 = 2,000,000 + 2,000,000(4)(i)Interest = 10% per year(b)F = P(1 + i) (1 + i) (1 + i) (1 + i)2,800,000 = 2,000,000(1 + i)4(1 + i)4=1.4000log(1 + i)4= log1.4004log(1 + i) = 0.146log(1 + i) = 0.0365(1 + i) = 100.0365(1 + i) = 1.0877i =8.77%1.40F = P + P(n)(i)3P = P + P(n)(0.20)n = 10 years1.41(a) F = P + P(n)(i)2P = P + P(4)(i)i = 25%(b)F = P(1 + i) (1 + i) (1 + i) (1 + i)2P = P(1 + i)42 = (1 + i)4log 2= log(1 + i)40.301 = 4 log(1 + i)log(1 + i) = 0.0753(1 + i) = 100.0753i = 18.93%

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51.42(a) Loan A: interest/year = 500,000(0.10) = $50,000Loan B: interest/year = 500,000(0.10) = $50,000The same amount of interest was be paid on each loan(b) There was no difference paid between the two loans1.43All engineering economy problems will involve i and n1.44P = $13,000; F = $26,000; i = 6.8% peryear; n = ?1.45P= $1000; F = $1270; n = 3; i = ?1.46P = $50,000;F = ?;i = 15%; n = 31.47F= $200,000,000; n = 5; i = 12% per year; A = ?1.48A = $225,000; n = 3; i = 15% per year; F = ?1.49F = $400,000; n = 2; i = 20% per year; P = ?1.50P = $225,000; n = 4; i = 15% per year; A = ?1.51P = $1,800,000; n = 6; i = 12% per year; A = ?1.52P = $16,000,000; A = $3,800,000; i = 18% per year; n = ?1.53End-of-period-convention means end-of-interest period.It does not mean end ofthe calendar year.1.54Period refers to interest period.1.55The difference between cash inflows and cash outflows is known as net cashflow.1.56Office supplies = outflow; GPS surveying equipment = outflow;Auctioning of used earth-moving equipment = inflow;Staff salaries = outflow; Fees for services rendered = inflow;Interest from bank deposits = inflow1.57The following items areinflows: salvage value, sales revenues, cost reductionsby subcontractors.The following items areoutflows: income taxes, loan interest, rebates to dealers,accounting services.

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61.58Assuming down is negative: down arrow of $40,000 in year 5; up arrow in year 0identified as P =?; i = 15% per year.1.59Assuming down is negative: down arrow of $10,000 in year 0; up arrows in theamount of $3000 in years 1 through5; i = 10% per year; arrow in year 5identified asF = ?.1.60Assuming down is negative: down arrow in year 0 identified as P=?; i = 15%per year; up arrows in years 1 thru 5 in the amount of $75,000-$30,000 =$45,0001.61(a) FVis F(b) PMT is A(c) NPERis n(d) IRR is i(e) PV isP1.62(a)F = ?; i = 8%; n = 10; A = $2000; P = $-10,000(b)A = ?; i = 12%; n = 30; P = $16,000; F = 0(c)P = ?; i = 9%; n = 15; A = $1000; F = $700(d) n= ?; i =8.5%;A = $5000; P= $-50000; F = $20000(calculator function)1.63(a)FV(i%,n,A,P) finds the future value, F(b) IRR(first_cell:last_cell) finds the compound interest rate, i(c) PMT(i%,n,P,F) finds the equal periodic payment, A(d) PV(i%,n,A,F) finds the present value, P1.64For built-in spreadsheetfunctions, a parameter that does not apply can be leftblankwhen it is not an interior one. For example, ifno Fisinvolved when usingthe PMT function, it can be left blank because it is an endparameter. When theparameterinvolved is an interior one (like P in the PMT function), a comma mustbe put in itsposition.1.65A common moral is a fundamental belief held by virtually all people. A personalmoral is the translation of a common moral into that which an individual believesand uses as guidance for their own decisions and actions.1.66A code of ethics can be used as an evaluation measure for the decision and actionsof an individual who works in the discipline that honors the code.1.67(a) Assuming that Carol’s supervisor is a trustworthy and ethical person himself,going to her supervisor and informinghim of her suspicion is probably thebest of these options. This puts Carol on record (verbally) as questioningsomething she heard at an informal gathering.(b)Another good option is to go to Joe one-on-one and inform him of herconcern about what she heard him say at lunch. Joe may not be aware he is onthe bid evaluation team and the potential ethical consequences if he acceptsthe free tickets from Dryer.

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71.68Explain your own situation and what you think an ethical action might be.1.69Amount now = 10,000 + 10,000(0.10)= $11,000Answer is (c)1.70Answer is (a)1.71Move both cashflows to year 0and solve for i1000(1 + i) = 1345.60/(1 + i)(1 + i)2= 1345.60/1000(1 + i) = 1.16i = 16%Answer is (d)1.72F in year 2 at 20% compound interest = P(1.20)(1.20) = 1.44PFor simple interest, F = P + Pni = P(1 + ni)P(1 + 2i) = 1.44P(1 + 2i) = 1.44Pi = 22%Answer is (c)1.73F = P(1+i)n16,000 = 8000(1 + i)921/9= 1 + i1.08 = 1 + ii = 0.08(8%)Answer is (b)1.74Answer is (c)1.75Let i = compound rate of increase:268 = 200(1 + i)5(1 + i)5= 268/200(1 + i) = (1.34)0.2(1 + i) = 1.0602i = 6.02%Answer is (b)

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81.762P = P + P(n)(0.05)1 = 0.05nn = 20Answer is (d)1.77Only a parameter at the end of the string can be omitted without an entry orcomma.Answer is (d).

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1Solutions to end-of-chapter problemsBasics of Engineering Economy, 2ndeditionLeland Blank and Anthony TarquinChapter 2Factors: How Time and Interest Affect Money2.1(a) (F/P,10%,20) = 6.7275(b) (A/F,4%,8) = 0.10853(c)(P/A,8%,20) = 9.8181(d)(A/P,20%,28) = 0.20122(e)(F/A,30%,15) = 167.28632.2P = 30,000(P/F,10%,8)= 30,000(0.4665)= $13,9952.3F = 15,000(F/P,6%,25)= 15,000(4.2919)= 64,378.502.4(a)F = 885,000 + 100,000(F/P,10%,3)= 885,000 + 100,000(1.3310)= $1,018,000(b) Spreadsheet function is =-FV(10%,3,,100000) + 885000Display is $1,018,0002.5(a)P = 19,000(P/F,10%,7)= 19,000(0.5132)= $9750.80(b)Ifthecalculator function is PV(10,7,0,19000), display is P = $-9750.00(c)Ifthespreadsheet function is =-PV(10%,7,,19000), display is $9750.002.6(a)Total for 7 lots is 7(120,000) = $840,000P = 840,000(P/F,10%,2)= 840,000(0.8264)= $694,176(b)Ifthecalculator function is PV(10,2,0,840000), display is P = $-694,214.88(c)If the spreadsheet function is =-PV(10%,2,,840000), display is $694,214.88

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22.7(a)F = 3000(F/P,10%,12) + 5000(F/P,10%,8)= 3000(3.1384) + 5000(2.1436)= $20,133.20(b)Sum two calculator functionsFV(10,12,,-3000) + FV(10,8,-5000)9,415.29+10,717.94 = $20,133.23(c)If thespreadsheet function is =–FV(10%,12,,3000)–FV(10%,8,,5000), thedisplay is $20,133.232.8(a)P = 30,000,000(P/F,10%,5)–15,000,000= 30,000,000(0.6209)–15,000,000= $3,627,000(b) If the spreadsheet function is =-PV(10%,5,,30000000)–15,000000, thedisplay is$3,627,640The increased decimal accuracy of a spreadsheet function indicates an increasedthe required amount of $640.2.9F =280,000(F/P,12%,2)= 280,000(1.2544)= $351,2322.10A = 12,700,000(A/P,20%,8)= 12,700,000(0.26061)= $3,309,7472.11P = 6000(P/A,10%,10)= 6000(6.1446)= $36,867.602.12(a)A = 60,000(A/P,8%,5)= 60,000(0.25406)= $15,027.60(b)If calculator function is PMT(8,5,-60000,0), the answer is $15,027.39(c)A spreadsheet function of =-PMT(8%,5,60000) displays $15,027.392.13A = 20,000,000(A/P,10%,6)= 20,000,000(0.22961)= $4,592,200

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32.14A = 50,000(A/F,20%,3)= 50,000(0.27473)= $13,736.502.15(a)17,000,000(A/P,i,8) = 2,737,680(A/P,i,8) = 0.16104From interest tables at n = 8, i = 6%per year(b)Calculator function is i(8,-2737680,17000000,0) to obtain i =6.00%(c)If the spreadsheet function is= RATE(8,-2737680,17000000), display is6.00%2.16(a)A = 3,000,000(10)(A/P,8%,10)= 30,000,000(0.14903)= $4,470,900(b)If calculator function is PMT(8,10,-30000000,0), the answer is $4,470,884.66(c)If the spreadsheet function is=-PMT(8%,10,30000000), display isA =$4,470,884.662.17P = 1,400,000(F/P,7%,4)=1,400,000(1.3108)= $1,835,1202.18P = 600,000(P/F,12%,4)= 600,000(0.6355)= $381,3002.19(a) A = 225,000(A/P,15%,4)= 225,000(0.35027)= $78,811(b) Recall amount = 78,811/0.10= $788,110 per year2.20P = 100,000((P/F,12%,2)= 100,000(0.7972)= $79,7202.21F = 65,000(F/P,4%,5)= 65,000(1.2167)= $79,086

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42.22P = (280,000-90,000)(P/A,10%,5)= 190,000(3.7908)= $720,2522.23F = 649(F/P,8%,2)= 649(1.1664)= $7572.24The value of the system is the interest saved on $20 million for 2 years.F = 20,000,000(F/P,8%,2)= 20,000,000(1.1664)= $23,328,000Interest = 23,328,000-20,000,000= $3,328,0002.25P = 2,100,000(P/F,10%,2)= 2,100,000(0.8264)= $1,735,4402.26P = 40,000(P/F,12%,4)= 40,000(0.6355)= $25,4202.27(a)A= 850,000(A/F,18%,5)= 850,000(0.13978)= $118,813(b)Spreadsheet function=PMT(18%,5,,850000)results in a minus sign.2.28P = 95,000,000(P/F,12%,3)= 95,000,000(0.7118)= $67,621,0002.29F = 375,000(F/P,10%,6)= 375,000(1.7716)= $664,3502.30F = 150,000(F/P,8%,8)= 150,000(1.8509)= $277,6352.31(a)P = 7000(P/F,10%,2) + 9000(P/F,10%,4) +15,000(P/F,10%,5)= 7000(0.8264) + 9000(0.6830) +15,000(0.6209)= $21,245.30

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5(b) Three calculator functions are added.-PV(10,2,0,7000)–PV(10,4,0,9000)–PV(10,5,0,15000)Total is 5785.12 + 6147.12 + 9313,82 = $21,246.062.32P = 600,000(0.10)(P/F,10%,2) + 1,350,000(0.10)(P/F,10%,5)= 60,000(0.8264) + 135,000(0.6209)= $133,4062.33P = 8,000,000(P/A,10%,5)= 8,000,000(3.7908)= $30,326,4002.34A = 10,000,000(A/P,10%,10)= 10,000,000(0.16275)= $1,627,5002.35A = 140,000(4000)(A/P,10%,4)= 560,000,000(0.31547)= $176,663,2002.36P = 1,500,000(P/A,8%,4)= 1,500,000(3.3121)= $4,968,1502.37A =3,250,000(A/P,15%,6)=3,250,000(0.26424)= $858,7802.38P = 280,000(P/A,18%,8)= 280,000(4.0776)= $1,141,7282.39A = 3,500,000(A/P,25%,5)= 3,500,000(0.37185)= $1,301,4752.40A = 5000(7)(A/P,10%,10)= 35,000(0.16275)= $5696.252.41(a)F = 70,000(F/P,12%,6) + 90,000(F/P,12%,4)= 70,000(1.9738) + 90,000(1.5735)= $279,781

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6(b) Spreadsheet function is =-FV(12%,6,0,70000)–FV(12%,4,0,90000) toobtain $279,784.332.42F = (458-360)(0.90)(20,000)(P/A,10%,5)= 1,764,000(3.7908)= $6,686,9712.43(a)Let CF4be the amount in year 4100,000(F/P,9%,3) + 75,000(F/P,9%,2) + CF4(F/P,9%,1) = 290,000100,000(1.2950) + 75,000(1.1881) + CF4(1.0900) = 290,000(1.09)CF4= 71.392.50CF4= $65,497.71(b) F in year 5 for 2 known amounts=-FV(9%,3,0,100000)-FV(9%,2,0,75000)P in year 4 of $290,000 minus amount above (assume it’s in cell H9)=-PV(9%,1,0,290000-H9)Answer is $65,495.052.44P = 225,000(P/A,15%,3)= 225,000(2.2832)= $513,7202.45400,000 = 50,000(F/A,12%,n)(F/A,12%,n) = 8.0000From 12% interest table, n is between 5 and 6 years.Therefore, n = 62.46F = P(F/P,10%,n)3P = P(F/P,10%,n)(F/P,10%,n) =3.000From 10% interest tables, n is between 11and 12years.Therefore, n = 12years2.47(a)1,200,000 = 400,000(F/P,10%,n) + 50,000(F/A,10%,n)Solve for n by trial and error:Try n = 5: 400,000(F/P,10%,5) + 50,000(F/A,10%,5)400,000(1.6105) + 50,000(6.1051)949,455< 1,200,000n too lowTryn = 8: 400,000(2.1436) + 50,000(11.4359)1,429,235> 1,200,000n too high

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7Bycontinuedinterpolation, n is between 6 and 7.Therefore, n = 7 years(b) Spreadsheet function=NPER(10%,-50000,-400000,1200000)displays6.672.482,000,000(F/P,7%,n) = 158,000(F/A,7%,n)Solve for n by trial and error(in $ thousands):Try n = 30: 2,000,000(F/P,7%,30) = 158,000(F/A,7%,30)2,000,000(7.6123) = 158,000(94.4608)15,224,600 >14,924,806n too lowTry n = 32: 2,000,000(8.7153) = 158,000(110.2182)17,430,600 >17,414,476n too lowTry n = 33: 2,000,000(9.3253) = 158,000(118.9334)18,650,600 <18,791,447n too highBy interpolation, n is between 32 and 33, and close to 32 years.Spreadsheet function is= NPER(7%,-158000,2000000) to display 32.1 years2.49(a)P = 26,000(P/A,10%,5)+ 2000(P/G,10%,5)= 26,000(3.7908)+ 2000(6.8618)= $112,284(b) Spreadsheet: entereach annual cost in adjacentcells and use the NPVfunction to display P = $112,284Calculators have no function for gradients; use the PV function on each cashflow and add the five P values to get $112,284.552.50A = 72,000 + 1000(A/G,8%,5)= 72,000 + 1000(1.8465)= $73,8462.51(a)84,000 = 15,000 + G(A/G,10%,5)84,000 = 15,000 + G(1.8101)G = $38,119(b) The annual increase of over $38,000 is substantially larger than the first-yearcost of $15,0002.52A = 9000–560(A/G,10%,5)= 9000–560(1.8101)= $7986

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