Solution Manual For Fundamentals Of Engineering Economics, 3rd Edition

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Fundamentals of Engineering Economics, 3rded.Chapter 1 Engineering Economic Decisions1.1)Not providedForThe Wall Street Journal, go to the Front page to find the section on “What’sNews.” This is a section on a brief summary on major headlines of the day’snews. Quickly browse through the news summary to see if there is any newsrelated to business investment. The best places to find the major business news oninvestment are sections on “BUSINESS”, “MARKETS” or “TECH.”1.2)Not providedSome of the well-known business publications are:•Daily Newspapers:oThe Wall Street JournaloThe New York Times (Business Section)oFinancial Times•Weekly or Monthly Magazines:oBusinessWeekoForbesoMoneyoSmart MoneyoFortune

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Chapter 2: Time Value of Money2.1)()(0.06)($2, 000)(5)$600IiP N===2.2)•Simple interest:(1)$6, 000$3, 000(10.08)12.5 years (or 13 years)FPiNNN=+=+=•Compound interest:$6, 000$3, 000(10.07)21.07log 2log 1.0710.24 years (or 11 years)NNNN=+===2.3)•Simple interest:()(0.07)($15, 000)(25)$26, 250IiP N===•Compound interest:25(1)1$15, 000 (1.07)1$66, 411.50NIPi=+−=−=2.4)•A : Simple interest:()(0.06)($10, 000)(15)$9, 000IiP N===•B : Compound interest:15(1)1$10, 000 (1.055)1$12,324.76NIPi=+−=−=B is a better option.

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Fundamentals of Engineering Economics, 3rded.2.5)•Compound interest:5$1, 000(10.065)$1,370.09F=+=•Simple interest:$1, 000(10.068(5))$1,340F=+=The compound interest option is better.2.6)•Loan balance calculation:End of periodPrincipalPaymentInterestPaymentRemainingBalance0$0.00$0.00$10,000.001$1,670.92$900.00$8,329.082$1,821.30$749.62$6,507.783$1,985.22$585.70$4,522.564$2,163.89$407.03$2,358.675$2,358.64$212.28$0.002.7)$5, 000(/, 7%,5)$5, 000(0.7130)$3,565PPF===2.8)$25, 000(/,8%, 2)$25, 000(1.1664)$29,160FFP===2.9)•Alternative 1$100P=•Alternative 2$120(/,10%, 2)$120(0.8264)$99.168PPF===•Alternative 3$170(/,10%,5)$170(0.6209)$105.553PPF===

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Fundamentals of Engineering Economics, 3rded.•Alternative 3 is preferred2.10)$1, 000(/,5%,3)$1, 000(1.1576)$1,157.6FFP===2.11)$500(/,9%,5)$500(0.6499)$324.95FPF===2.12)10.5%i=, two-year discount rate is2(10.105)1.221(22.1%)+=2.13)(a)$6, 000(/, 6%,8)$6, 000(1.5938)$9,563FFP===(b)$1,550(/,5%,12)$1,550(1.7959)$2, 784FFP===(c)$8, 000(/,9%,32)$8, 000(15.7633)$126,106FFP===(d)$12, 000(/,8%,9)$12, 000(1.999)$23,988FFP===2.14)(a)$5,500(/,10%, 6)$5,500(0.5645)$3,105PPF===(b)$7, 000(/,9%,3)$7, 000(0.7722)$5, 405PPF===(c)$22, 000(/,8%,5)$22, 000(0.6806)$14,973PPF===(d)$13, 000(/, 7%,8)$13, 000(0.5820)$7,566PPF===2.15)(a)$8, 000(/,8%,5)$8, 000(0.6806)$5, 445PPF===(b)$10, 000(/,8%, 4)$10, 000(1.3605)$13, 605FFP===2.16)3(10.07)log 3log 1.0716.24 years (or 17 years)NFPPNN==+==2.17)2(10.06)log 2log 1.0611.896 years (or 12 years)NFPPNN==+==2.18)•Rule of 72:72 / 89 years=

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Fundamentals of Engineering Economics, 3rded.2(10.08)log 2log 1.089 yearsNFPPNN==+==2.19)389$1(1.08)$10, 042, 477,894, 213F==2.20)$35, 000(/,9%, 4)$10, 000(/,9%, 2)$35, 000(0.7084)$10, 000(0.8417)$33, 211PPFPF=+=+=2.21)$450, 000(/,5%,5)450, 000(0.7835)$352,575PPF===2.22)•Simple interest (John):(0.1)($1, 000)(5)$500IiPN===•Compound interest (Susan):5(1)1$1, 000 (1.095)1$574.24NIPi=+−=+−=•Susan’s balance will be greater by $74 (or $74.24 to be exact)2.23)123$2, 000$800$1, 000$3, 230.651.11.11.1P=++=2.24)2345$3, 000$3,500$4, 200$6,500$14, 292.81.051.051.051.05P=+++=2.25)$2, 000(/,8%,10)$3, 000(/,8%,8)$4, 000(/,8%, 6)$16, 218FFPFPFP=++=

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Fundamentals of Engineering Economics, 3rded.2.26)$3, 000, 000$2, 400, 000(/,8%,5)$3, 000, 000(/,8%,5)(/,8%,5)$20, 734, 774.86PPAPAPF=++=2.27)$3, 000(/,9%, 2)$4, 000(/,9%,5)$5, 000(/,9%, 7)$7,859.7PPFPFPF=++=2.28)•Method 1:$2, 000(1.05)(1.1)(1.15)$3, 000(1.1)(1.15)$5, 000$11, 451.5F=++=•Method 2:()$6,451.50$5,100$2, 000(1.05)$3, 000 (1.10)(1.15)$5, 000$11, 451.50F=++=2.29)$180, 000$20, 000(/,9%,5)$10, 000(/,9%,3)(/,9%, 6)180, 00020, 000(3.8897)10, 000(0.7722)(0.5963)$184,350.16PAPFX PFXX=−+−+==2.30)532$80, 000$10, 000(1.08)$12, 000(1.08)(1.08)$43, 029.99FXX==++=2.31)432100(1.08)8(1.08)9(1.08)10(1.08)11$93.67XX=++++=This is the minimum selling price. If John can sell the stock for a higher pricethan $93.67, his return on investment will be higher than 8%.

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Fundamentals of Engineering Economics, 3rded.2.32)2345$60, 000$77, 000$65, 000$57, 00045, 000$212,873.891.141.141.141.141.14P=++++=2.33)$5, 000(/, 6%,10)$5, 000(13.1808)$65,904FFA===2.34)(a)$5, 000(/,5%, 7)$5, 000(8.1420)$40, 710FFA===(b)$5, 000(/,5%, 7)(1.05)$5, 000(8.1420)(1.05)$42, 745.50FFA===2.35)(a)$6, 000(/, 6%, 6)$6, 000(6.9753)$41,851.80FFA===(b)$8, 000(/, 7.25%,9)$96,825.60FFA==(c)$15, 000(/,8%, 25)$15, 000(73.1059)$1, 096,588.50FFA===(d)$3, 000(/,9.75%,10)$47, 242.80FFA==2.36)(a)$18, 000(/,5%,13)$18, 000(0.0565)$1, 017AAF===(b)$11, 000(/, 6%,8)$11, 000(0.1010)$1,111AAF===(c)$8, 000(/,8%, 25)$8, 000(0.0137)$109.6AAF===(d)$12, 000(/, 6.85%,8)$1,176AAF==2.37)$250, 000(/,5%,5)$250, 000(0.1810)$45, 250AAF===2.38)$25, 000(/,5%,5)$25, 000(0.1810)$4,525AAF===2.39)()$35, 000$3, 000(/, 6%,)(/, 6%,)11.666610.06111.66660.06log(1.06)log(1.7)9.11 yearsNFANFANNN==+−=⋅==2.40)$10, 000(/,9%,5)$1, 670.92AAF==

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Fundamentals of Engineering Economics, 3rded.2.41)108642$500(1.04)$1, 000(1.04)$1, 000(1.04)$1, 000(1.04)$1, 000(1.04)$1, 000$6, 625.47F=+++++=2.42)(a)$18, 000(/,8%,5)$18, 000(0.2505)$4,509AAP===(b)$4, 200(/,9.5%, 4)$1,310.82AAP==(c)$7, 700(/,11%,3)$7, 700(0.4092)$3,150.84AAP===(d)$23, 000(/, 6%, 20)$23, 000(0.0872)$2, 005.60AAP===2.43)•Equal annual payment amount:$20, 000(/,10%,3)$20, 000(0.4021)$8, 042AAP===•Loan balance calculation:End of periodPrincipalPaymentInterestPaymentRemainingBalance0$0.00$0.00$20,000.001$6,042.00$2,000.00$13,958.002$6,646.20$1,395.80$7,311.803$7,310.82$731.18$0Interest payment for the second year = $1,395.802.44)(a)$9, 000(/, 6%,8)$9, 000(6.2098)$55,888.20PPA===(b)$1,500(/,9%,10)$1,500(6.4177)$9, 626.55PPA===(c)$7,500(/, 7.25%, 6)$35, 475PPA==(d)$9, 000(/,8.75%,30)$52,529PPA==2.45)(a)()()36360.0625 10.0625(/, 6.25%,36)0.0704410.06251AP+==+−(b)()()12512510.09251(/,9.25%,125)10.810640.0925 10.0925PA+−==+2.46)$500(/, 7%,15)(1.07)$500(25.1290)(1.07)$13, 444.02FFA===2.47)$5, 000(/,11%,5)$5, 000(0.2706)$1,353AAP===If you make the first payment on the loan at the end of the second year:

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Fundamentals of Engineering Economics, 3rded.$5, 000(/,11%,1)(/,11%, 4)$5, 000(1.11)(0.3223)$1, 788.78FFPAP===2.48)New equipment: $195,000O&M cost:$30, 000(/,10%,10)$30, 000(6.1446)$184,338PPA===New equipment isn’t worth buying.2.49)250$3, 4600Pi=−+=I = 7.225%2.50)1, 000$10, 0000.1P==2.51)12$5, 000(/,8%,5)$2, 000(/,8%,5)$5, 000(/,8%,5)$2, 000(/,8%,5)(/,8%,5)$5, 000(5.8666)$2, 000(1.8465)(5.8666)$50,998.35FFFFAFGFAAGFA=+=+=+=+=2.52)$5, 000(/,10%,5)$500(/,10%,5)$5, 000(/,10%,5)$500(/,10%,5)(/,10%,5)$5, 000(6.1051)$500(6.8618)(1.6105)$25, 000.04FFAFGFAPGFP=−=−=−=2.53)$100(/,8%,1)$150(/,8%,3)$200(/,8%,5)$250(/,8%, 7)$300(/,8%,9)$350(/,8%,11)$793.83PPFPFPFPFPFPF=+++++=2.54)$30, 000$3, 000(/,8%,10)$30, 000$3, 000(3.8713)$18,386.1AAG=−=−=

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Fundamentals of Engineering Economics, 3rded.2.55)$3, 000(/,12%,8)$600(/,12%,8)$3, 000(4.9676)$600(14.4714)$23,585.64PPAPG=+=+=2.56)(/,9%, 6)$1000(/,9%, 4)$800(/,9%,3)$600(/,9%, 2)$400(/,9%,1)$200(10.0924)$3796.461, 000(/,9%, 4)800(/,9%, 4)200(/,9%, 4)(/,9%, 4)$376.17C PGFPFPFPFPCFPFAPGFPC=++++=→+−∴=2.57)()()1,4040$6, 000(/5%,9%, 40)11.051.09$6, 0000.090.05$116,379.57$116379.57 *(/,9%, 40)$3, 655.412.47PPAFP−=−=−==2.58)(a)() ()()1,77$10, 000, 000(/10%,12%, 7)110.110.12$10, 000, 0000.120.1$35, 620,126PPA−=−−−+=⋅− −=(b)Note that the oil price increases at the annual rate of 5% while the oilproduction decreases at the annual rate of 10%. Therefore, the annual revenuecan be expressed as follows:1111$100(10.05)100, 000(10.10)$10, 000, 000(0.945)$10, 000, 000(10.055)nnnnnA−−−−=+−==−This revenue series is equivalent to a decreasing geometric gradient series withg= -5.5%.NAn

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Fundamentals of Engineering Economics, 3rded.1$10,000,0002$9,450,0003$8,930,2504$8,439,0865$7,974,9376$7,536,3157$7,121,818() ()()1,77$10, 000, 000(/5.5%,12%, 7)110.05510.12$10, 000, 0000.120.055$39, 746, 494.51PPA−=−−−+=⋅− −=(c)Computing the present worth of the remaining series4567(,,,)AAAAat the end of period 3 gives() ()()144$8, 439, 086.25(/,5.5%,12%, 4)110.05510.12$8, 439, 086.250.120.055$23, 782, 713PPA−=−−−+=⋅− −=2.59)2012011201201(1)(2, 000, 000) (1.06)(1.06)1.06(2, 000, 000 /1.06)()1.06(2, 000, 000 /1.06)20(21)(2, 000, 000 /1.06)2$396, 226, 415.1nnnnnnnnnPAinnn−=−−====+=====∑∑∑∑2.60)(a) The withdrawal series would be:

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Fundamentals of Engineering Economics, 3rded.234PeriodWithdrawal11$3,00012$3,000(1.06)13$3,000(1.06)14$3, 000(1.06)15$3, 000(1.06)Equivalent worth of the withdrawal series at period 10, usingi= 8%:Assuming that each deposit is made at the end of each year,the following equivalence must be hold:$13,384(/,8%,10)14.4866$923.88A FAAA===(b) Equivalent present worth of the withdrawal series at 6%15$3, 000(/, 6%, 6%,5)$3, 000$14,150.9410.06PPA===+$14,151=A(F/A,6%,10)=13.1808AA=$1,073.602.61)$1, 000, 000(/, 6%,30)(79.0582)A FAA==A = $12,649 should be set aside on the accounta)$1, 000, 000(/, 6%, 20)(11.4699)A PAA==A = $87,185 / yearb)P=$3,000(P/A1,6%,8%,5)ŹŹ=$3,000⋅1−1+0.06()51+0.08()−50.08−0.06()ŹŹ=$13,383.92

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Fundamentals of Engineering Economics, 3rded.()()11,20201$1, 000, 000(/3%, 6%, 20)11.031.060.060.03$68, 674 / yearAPAA−=−=−=2.62)2323$50$70$50225.521.11.11.11.11.11.11.331140.87154.1473$33.97CCCCCC++=++=→==2.63)[$100(/,10%,8)$50(/,10%, 6)$50(/,10%, 4)](/,10%,8)[$100(11.4359)$50(7.7156)$50(4.6410)](0.4665)$821.70PFAFAFAPF=++=++=2.64)Select (a).2.65)$500(/,10%,1)$300(/,10%,3)(/,10%,1)$800(/,10%,5)$500(0.9091)$300(2.4869)(0.9091)$800(0.6209)$720.42PPFPAPFPFP= −++= −++=2.66)Computing the equivalent worth at period 3 will require only two differenttypes of interest factors.

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Fundamentals of Engineering Economics, 3rded.12$200(/,10%,5)(/,10%,3)$200(3.7908)(1.3310)$1, 009.11(/,10%, 2)(/,10%,3)(/,10%, 2)(1.7355)(1.3310)(1.7355)(4.0455)$1, 009.11/ 4.0455$249.44PPAFPPA PAFPA PAAAAA====+=+===2.67)1,1$200(/,10%, 4)100(/,10%, 2)$200(3.1699)100(1.7355)460.43PPAPA=−=−=2,1(/,10%, 4)(3.1699)4.1699PXX PAXXX=+=+=1,12,1PP=$460.434.1699$110.42XX==2.68)1212$50(/,10%, 4)$35(/,10%, 2)(/,10%, 2)$50(3.1699)$35(1.7355)(0.8264)208.6926(/,10%, 4)(/,10%, 2)(/,10%,1)(3.1699)(1.7355)(0.9091)4.7476$43.96PPAPAPFPC PAC PAPFCCCPPC=+=+==+=+===2.69)

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Fundamentals of Engineering Economics, 3rded.(/,9%,8)$5, 000(/,9%, 2)$5000(11.0285)$5, 000(1.7591)$5000$1, 250.90C FAPACC=+=+=2.70)The original cash flow series isnAn0$01$8002$8203$8404$8605$8806$9007$9208$3009$30010$300 - $5002.71)2(/,12%, 7)(/,12%,1)$1, 200(/,12%,8)400(/,12%, 4)2(4.5638)(0.8929)$1, 200(4.9676)400(3.0373)CC PAPFPAPACC+=−+=−6.075$4, 746.20$781.27CC==2.72)200(1.06)(1.08)(1.12)(1.15)(1.08)(1.12)(1.15)$300(1.15)$1000247.91.3910434510001.39104360.1$258.87XXXX++=++===

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