Solution Manual For Contemporary Engineering Economics, 5th Edition

Preview (16 of 394 Pages)

100%

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 1 preview image

Loading page ...

Page | 1Chapter 2 Understanding Financial StatementsFinancial Statement2.1(a)•Current assets = $150,000 + $200,000 + $150,000 + $50,000 + $30,000 =$580,000•Current liabilities = $50,000 + $100,000 + $80,000 = $230,000•Working capital = $580,000 - $230,000 = $350,000•Shareholder’s equity = $100,000 + $150,000 + $150,000 + $70,000 =$470,000(b) EPS = $500,000/10,000 = $50 per share(c) Par value = $15; capital surplus = $150,000/10,000 = $15;Market price = $15 + $15 = $30 per share2.2(a)Working capital = Current assets – Current liabilities;Working capital requirements = Changes in current assets – Changes incurrent liabilitiesWC req. = (+$100,000 - $20,000) – (+$30,000 - $40,000) = $90,000(b)Taxable income = $1,500,000 - $650,000 - $150,000 - $20,000 = $680,000(c)Net income = $680,000 - $272,000 = $408,000(d)Net cash flow:A.Operating activities = net income + depreciation – W.C. required =$408,000 + $200,000 - $90,000 = $518,000B.Investing activities = equipment purchase = ($400,000)C.Financing activities = borrowed funds = $200,000D.Net cash flow = $518,000 - $400,000 + $200,000 = $318,0002.3(a)168ROE21%800A==240ROE60%400B==

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 2 preview image

Loading page ...

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 3 preview image

Loading page ...

Page | 2ROAA=168+20(1−0.4)1,000=18%ROAB=240+160(1−0.4)2,000=16.8%(b)Because company has higher income but less equity than that of company A.No, it is just one criterion, so we cannot say that. Further investigation mustbe conducted.(c)408ROE34%1200merge==Merge and Acquisition situation between companies A and B .2.4(a)Debt ratio = $18,542,000/$39,572,000 = 46.86%(b)Time-interest-earned ratio: N/A(c)Current ratio = $34,690,000/$14,092,000 = 2.46 times(d)Quick ratio = ($34,690,000-$509,000)/$14,092,000 = 2.43 times(e)Inventory-turnover ratio = $32,479,000/($509,000 + $346,000)/2 =75.97times(f)DSO = ($6,151,000)/($32,479,000/365) = 69.13 days(g)Total-assets-turnover ratio = $32,479,000/$39,572,000 = 0.82 times(h)Profit margin on sales = $4,834,000/$32,479,000 = 14.88%(i)Return on Total assets$4,834, 000$014.89%($39,572, 000$25,347, 000) / 2+==+(j)Return on Common equity$4,834, 00027.19%($21, 030, 000$14,532, 000) / 2==+(k)Price-earnings ratio = $128.24/ ($4,834,000,000/892,110,000) = $23.66(Note: Theaveragetotal number of outstanding shares in year 2009:892.11 M)(l)Book value per share = ($21,030,000,000 – 0)/892,110,000= $23.57

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 4 preview image

Loading page ...

Page | 32.5(a)Debt ratio = $9,498,000/$10,946,000 = 86.77%(b)Time-interest-earned ratio = ($1,941,000 + $308,000)/$308,000= 7.3 times(c)Current ratio = $2,521,000/$3,552,000= 0.71 times(d)Quick ratio = ($2,521,000- $897,000)/$ 3,552,000= 0.46 times(e)Inventory-turnover ratio =$12,822,000($897,000+$924,000) / 2=14.08times(f)DSO = ($1,100,000)/($12,822,000/365) = 31.31 days(g)Total-assets-turnover ratio = $12,822,000/$10,946,000 = 1.17 times(h)Profit margin on sales = $1,148,000/$12,822,000= 8.95%(i)Return on total assets =$1,148,000+$308,000(1−0.4)($10,946,000+$11,397,000 / 2=11.93%(j)Return on common equity =$1,148,000($1,448,000+$2,526,000) / 2=57.78%(k)Price-earnings ratio = $44.65/($1,148,000,000/382,500,000) = $14.88(Note: Theaveragetotal outstanding number of shares in year 2009: 382.5M)(l)Book value per share = $1,448,000,000/382,500,000 = $3.792.6Given R.C.’s EPS = $8 per share; Cash dividend = $4 per share; Book valueper share = $80; Changes in the retained earnings = $24 million; Total debt =$240 million; Find debt ratio = total debt/total assetsNet Income$8EPSX==whereX= the number of outstanding sharesTotal shareholders' equityBook value$80X==

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 5 preview image

Loading page ...

Page | 4Retained earnings = Net income – Cash dividend; Net income = 8Xfrom EPSrelationship and the total cash dividend = 4X, so we rewrite 8X– 4X= $24million, orX= 6 million sharesFrom book value per share, we know that total shareholders’ equity = 80X, or$480 million; Total assets = Total liabilities + Total shareholders’ equity =$240 million + $480 million = $720 millionDebt ratio = $240 million/$720 million = 33.33%2.7(b)2.8(b)2.9(d)2.10(b)

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 6 preview image

Loading page ...

Page | 1Chapter 3: Interest Rate and Economic EquivalenceTypes of Interest3.1•Simple interest:$20,000=$10,000(1+0.075N)(1+0.075N)=2N=10.075=13.33≈14 years•Compound interest:3.2•Simple interest:(0.06)($5, 000)(5)$1,500IiPN===•Compound interest:[(1)1]$5, 000(1.33821)$1, 691NIPi=+−=−=3.3•Option 1: Compound interest with 8%:408,4$)4693.1(000,3$)08.01(000,3$5==+=F•Option 2: Simple interest with 9%:350,4$)45.1(000,3$)509.01(000,3$==×+∴Option 1 is better.$20,000=$10,000(1+0.07)N(1+0.07)N=2N=10.24≈11years

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 7 preview image

Loading page ...

Page | 23.4End of YearPrincipalRepaymentInterestpaymentRemainingBalance0$10,0001$1,638$1,000$8,3622$1,802$836$6,5603$1,982$656$4,5784$2,180$458$2,3985$2,398$240$0Equivalence Concept3.5$18, 000(/,5%,5)$18, 000(0.7835)$14,103PPF===3.6$25, 000(/,8%,3)$25, 000(1.2597)$31, 493FFP===3.7$100(/,10%,10)$200(/,10%,8)$688FFPFP=+=3.82$1, 000(/, , 2)$1, 200$1, 000(1)$1, 2001.219.54%FP iiii=+==−=Single Payments (Use ofF/PorP/FFactors)3.9$180, 000(/, 6%,10)$322,353FFP==3.10(a)$7, 000(/, 6%,5)$9,368FFP==(b)$3, 250(/,5%,15)$6, 757FFP==(c)$18, 000(/,8%,33)$228,169FFP==(d)$20, 000(/,9%,8)$39,851FFP==

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 8 preview image

Loading page ...

Page | 33.11$300, 000(/,8%,10)$138,958PPF==3.12(a)$15,500(/,14%,8)$5, 434PPF==(b)$18, 000(/, 4%,12)$11, 243PPF==(c)$20, 000(/,8%,9)$10, 005PPF==(d)$55, 000(/,11%, 4)$36, 230PPF==3.13(a)$12, 000(/,13%, 4)$7,360PPF==(b)$30, 000(/,13%,5)$55, 273FFP==3.14F=3P=P(1+0.06)Nlog 3=Nlog(1.06)N=18.85≈19 years3.15•F=2P=P(1+0.08)Nlog 2=Nlog(1.08)N=9 years•Rule of 72:72 / 89 years=3.16(a)Single-payment compound amount),,/(NiPFfactors forN9%10%3520.414028.10244031.409445.2593To find)38%,5.9,/(PF, first, interpolate for38=n:N9%10%3827.011238.3965

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 9 preview image

Loading page ...

Page | 4Then, interpolate for%5.9=i:7039.32)38%,5.9,/(=PFAs compared to formula determination4584.31)38%,5.9,/(=PF(b)Single-payment compound amount(P/F,8%,N)factors forN45500.03130.0213Then, interpolate forN=470273.0)47%,8,/(=FPAs compared to the value from the interest formula:0269.0)47%,8,/(=FP3.17(a)44$18(1)$92, 40021.43%ii+==(b)F=$97.8(F/P,21.43%,22)=$7,007billionUneven Payment Series3.18$1,000+$1,0001.1+$1,5001.13=$1,2101.12+X1.14X=$2,9813.192345$25, 000$33, 000$46, 000$38, 000$110,9611.071.071.071.07P=+++=3.20$2, 000(/, 6%,10)$2,500(/, 6%,8)$3, 000(/,6%,6)$11,822FFPFPFP=++=

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 10 preview image

Loading page ...

Page | 53.21$3, 000, 000$2, 400, 000(/,8%,1)$3, 000, 000(/,8%,10)$20, 734, 618PPFPF=+++="Or,$3, 000, 000$2, 400, 000(/,8%,5)$3, 000, 000(/,8%,5)(/,8%,5)$20, 734, 618PPAPAPF=++=3.22$8, 000(/, 6%, 2)$6, 000(/, 6%,5)$4, 000(/,6%,7)$14, 264PPFPFPF=++=Equal Payment Series3.23(a)With deposits made at the end of each year$2, 000(/,8%,15)$54,304FFA==(b)With deposits made at the beginning of each year$2, 000(/,8%,15)(1.08)$58, 649FFA==3.24$10, 000(/, 6%,20)$367,856FFA==3.25(a)$6, 000(/,8%,5)$35, 200FFA==(b)$4, 000(/, 6.25%,12)$68, 473FFA==(c)$9, 000(/,9.45%, 20)$484,359FFA==(d)$3, 000(/,11.75%,12)$71,308FFA==3.26(a)$32, 000(/,8%,15)$1,179AAF==(b)$55, 000(/, 6%,10)$4,173AAF==

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 11 preview image

Loading page ...

Page | 6(c)$35, 000(/, 7%, 20)$853.8AAF==(d)$8, 000(/,11%, 4)$1, 699AAF==3.27$50, 000(/, 6%,10)$3, 793.40AF=3.28$35, 000$2, 000(/, 6%,)(/, 6%,)17.512.3213yearsFANFANN===≈3.29$15, 000(/,11%,5)$2, 408.57A FAA==3.30$5, 000$500(/, 7%,5)(/, 7%,5)$747.51FPA FAA=+=3.31(a)$12, 000(/, 4%, 6)$2, 289.14AAP==(b)$3,500(/, 6.7%, 7)$642.66AAP==(c)$6,500(/,3.5%,5)$1, 439.63AAP==(d)$32, 000(/,8.5%,15)$3,853.47AAP==3.32(a)The capital recovery factor),,/(NiPAforN6%7%350.06900.0772400.06650.0750To find)38%,25.6,/(PA, first, interpolate forN=38:N6%7%380.06750.0759Then, interpolate for%25.6=i;

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 12 preview image

Loading page ...

Page | 7(A/P, 6.25%, 38)=0.0696:As compared to the value from the interest formula:(A/P, 6.25%, 38)=0.0694(b)The equal payment series present-worth factor)85,,/(iAPfori9%10%11.10389.9970Then, interpolate for%25.9=i:8271.10)85%,25.9,/(=APAs compared to the value from the interest formula:8049.10)85%,25.9,/(=AP3.33•Equal annual payment:$50, 000(/,12%,3)$20,817.45AAP==•Interest payment for the second year:End of YearPrincipalRepaymentInterestpaymentRemainingBalance0$50,0001$14,817.45$6,000$35,182.552$16,595.54$4,221.91$18,587.013$18,587.01$2,230.4403.34$10, 000(/,9%,10)$1,558.2AAP==3.35(a)$1, 000(/, 6.8%,8)$6, 017.86PPA==(b)$3,500(/,9.5%,12)$24, 443.44PPA==(c)$1,900(/,8.25%,9)$11, 746.68PPA==

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 13 preview image

Loading page ...

Page | 8(d)$9,300(/, 7.75%,5)$37,378.16PPA==3.36P=$35,000(P/A,12%,10)=$197,758Since $200,000 > $197,758, You should not purchase the equipment.3.37(a)$3,875, 000$3,125, 000(/, 6%,1)...$8,875, 000(/, 6%, 7)$39,547, 241.99PPFPF=+++=(b)P=$1,375,000+$1,375,000(P/A,6%,7)=$9,050,774.48Since $9,050,774.48 > $8,000,000, the prorated payment option is better choice.Linear Gradient Series3.38$10, 000(/,8%,5)$3, 000(/,8%, 5)$10, 000(/,8%,5)$3, 000(/,8%,5)(/, 8%, 5)$91,163.55FFAFGFAAGFA=+=+=3.39$7,500(/,8%,5)$1,500(/,8%,5)$7,500(/,8%,5)$1,500(/,8%,5)(/, 8%, 5)$27, 750.74FFAFGFAPGFP=−=−=3.40$100[$100(/,9%, 7)$50(/,9%, 6)$50(/,9%,4)$50(/, 9%, 2)](/, 9%, 7)$991.32PFAFAFAFAPF=++++=3.41$15, 000$1, 000(/, 8%, 12)$10, 404.25AAG=−=

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 14 preview image

Loading page ...

Page | 93.42$1, 000(/, 6%,5)$250(/, 6%,5)$6,196PPAPG=+=3.43Using the geometric gradient series present worth factor, we can establishthe equivalence between the loan amount $120,000 and the balloon paymentseries as38.684,25$6721.4)5%,9%,10,/(000,120$1111===AAAPAPayment seriesNPayment1$25,684.382$28,252.823$31,078.104$34,185.915$37,604.513.441$6, 000(/,5%, 7%,30)(/, 7%,30)$987, 093.8FPAFP==3.45(a)076,372,21$)7%,12%,10,/(000,000,6$1=−=APP(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 annualrevenue can be expressed as follows:1111$60(10.05)100, 000(10.1)$6, 000, 000(0.945)$6, 000, 000(10.055)nnnnnA−−−−=+−==−This revenue series is equivalent to a decreasing geometric gradient serieswithg= -5.5%. So,1$6, 000, 000(/,5.5%,12%,7)$23,847,897PPA=−=

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 15 preview image

Loading page ...

Page | 10(c)Computing the present worth of the remaining series4567(,,,)AAAAat the endof period 3 gives3446, 000, 000(10.055)5, 063, 451.75$5, 063, 451.75(/,5.5%,12%, 4)$14, 269, 627.82APPA=−==−=3.462012011201(1)(2, 000, 000) (1.06)(1.06)1.06(2, 000, 000 /1.06)()1.06$396, 226, 415nnnnnnnnPAinn−=−−===+===∑∑∑3.47(a)The withdrawal series would bePeriodWithdrawal11$12,00012$12,000(1.08)13$12,000(1.08)(1.08)14$12,000(1.08)(1.08)(1.08)15$12,000(1.08)(1.08)(1.08)(1.08)101$12, 000(/,8%,12%,5)$49,879.14PPA==Assuming that each deposit is made at the end of each year, then:$49, 879.14=A(F/A,12%,10)A=$2, 842.32(b)101$12, 000(/,8%,9%,5)$54, 045.08PPA==$54, 045.08=A(F/A, 9%,10)A=$3, 557.25

Solution Manual For Contemporary Engineering Economics, 5th Edition - Page 16 preview image

Loading page ...

Page | 11Various Interest Factor Relationships3.48(a)(P/F, 8%, 67)=(P/F, 8%,50)(P/F,8%,17)=0.0058(P/F, 8%, 67)=(1+0.08)−67=0.0058(b)),,/(1),,/(NiFPiNiPA−=0394.0)2%,8,/)(40%,8,/()42%,8,/(==FPFPFP0833.00394.0108.0)42%,8,/(=−=PA0833.01)08.1()08.1(08.0)42%,8,/(4242=−=PA(c)4996.1208.0)35%,8,/)(100%,8,/(1),,/(1),,/(=−=−=FPFPiNiFPNiAP(P/A, 8%,135)=(1.08)135−10.08(1.08)135=12.49963.49(a)NNNNiiiiiiNiAFiNiPF)1(11)1(11)1()1(1),,/(),,/(+=+−+=+−+=++=(b)NNNNNNNNiiiiiiiiiiiNiAPNiFP−−+=+−+−++=+−+−=+−=)1()1(1)1()1()1()1(1)1(1)1(),,/(1),,/(

Preview Mode

This document has 394 pages. Sign in to access the full document!

Report

Study Now!

Document Details

Related Documents

ENGR 111 Spring 2025 ScheduleFeb 19

MyOakton Instructions Change Term View Placement I

Computer Assembling Plant Design

Reliability Analysis and Maintenance Strategies for Complex Systems and Components

Analysis and Calculations of Resistances, Wire Sizes, and Magnetomotive Force in DC Machine Field Windings

ECET105 � Digital Fundamentals Homework Assignment #5

INDE 2333 Engineering Statistics I Practice Exam 2

Test Bank for Automotive Service Management, 3rd edition

Solution Manual for Sensors and Actuators: Engineering System Instrumentation, 2nd Edition

Solution Manual For Refrigeration and Air Conditioning Technology, 8th Edition

Company

Explore

Study Tools