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Solution Manual For Contemporary Engineering Economics, 5th Edition

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    Page | 1 Chapter 2 Understanding Financial Statements Financial Statement 2.1 (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 share 2.2 (a) Working capital = Current assets – Current liabilities; Working capital requirements = Changes in current assets – Changes in current liabilities WC 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 = $518,000 B. Investing activities = equipment purchase = ($400,000) C. Financing activities = borrowed funds = $200,000 D. Net cash flow = $518,000 - $400,000 + $200,000 = $318,000 2.3 (a) 168 ROE 21% 800 A = = 240 ROE 60% 400 B = =

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    Page | 2 ROA A = 168 + 20(1 − 0.4) 1,000 = 18% ROA B = 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 must be conducted. (c) 408 ROE 34% 1200 merge = = 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 =75.97 times (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 $0 14.89% ($39,572, 000 $25,347, 000) / 2 + = = + (j) Return on Common equity $4,834, 000 27.19% ($21, 030, 000 $14,532, 000) / 2 = = + (k) Price-earnings ratio = $128.24/ ($4,834,000,000/892,110,000) = $23.66 (Note: The average total 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

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    Page | 3 2.5 (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 = 14.08 times (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: The average total outstanding number of shares in year 2009: 382.5M) (l) Book value per share = $1,448,000,000/382,500,000 = $3.79 2.6 Given R.C.’s EPS = $8 per share; Cash dividend = $4 per share; Book value per share = $80; Changes in the retained earnings = $24 million; Total debt = $240 million; Find debt ratio = total debt/total assets Net Income $8 EPS X = = where X = the number of outstanding shares Total shareholders' equity Book value $80 X = =

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

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    Page | 1 Chapter 3: Interest Rate and Economic Equivalence Types of Interest 3.1 • Simple interest: $20,000 = $10,000(1 + 0.075 N ) (1 + 0.075 N ) = 2 N = 1 0.075 = 13.33 ≈ 14 years • Compound interest: 3.2 • Simple interest: (0.06)($5, 000)(5) $1,500 I iPN = = = • Compound interest: [(1 ) 1] $5, 000(1.3382 1) $1, 691 N I P i = + − = − = 3.3 • Option 1: Compound interest with 8%: 408 , 4 $ ) 4693 . 1 ( 000 , 3 $ ) 08 . 0 1 ( 000 , 3 $ 5 = = + = F • Option 2: Simple interest with 9%: 350 , 4 $ ) 45 . 1 ( 000 , 3 $ ) 5 09 . 0 1 ( 000 , 3 $ = = × + ∴ Option 1 is better. $20,000 = $10,000(1 + 0.07) N (1 + 0.07) N = 2 N = 10.24 ≈ 11years

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    Page | 2 3.4 End of Year Principal Repayment Interest payment Remaining Balance 0 $10,000 1 $1,638 $1,000 $8,362 2 $1,802 $836 $6,560 3 $1,982 $656 $4,578 4 $2,180 $458 $2,398 5 $2,398 $240 $0 Equivalence Concept 3.5 $18, 000( / ,5%,5) $18, 000(0.7835) $14,103 P P F = = = 3.6 $25, 000( / ,8%,3) $25, 000(1.2597) $31, 493 F F P = = = 3.7 $100( / ,10%,10) $200( / ,10%,8) $688 F F P F P = + = 3.8 2 $1, 000( / , , 2) $1, 200 $1, 000(1 ) $1, 200 1.2 1 9.54% F P i i i i = + = = − = Single Payments (Use of F / P or P / F Factors) 3.9 $180, 000( / , 6%,10) $322,353 F F P = = 3.10 (a) $7, 000( / , 6%,5) $9,368 F F P = = (b) $3, 250( / ,5%,15) $6, 757 F F P = = (c) $18, 000( / ,8%,33) $228,169 F F P = = (d) $20, 000( / ,9%,8) $39,851 F F P = =

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    Page | 3 3.11 $300, 000( / ,8%,10) $138,958 P P F = = 3.12 (a) $15,500( / ,14%,8) $5, 434 P P F = = (b) $18, 000( / , 4%,12) $11, 243 P P F = = (c) $20, 000( / ,8%,9) $10, 005 P P F = = (d) $55, 000( / ,11%, 4) $36, 230 P P F = = 3.13 (a) $12, 000( / ,13%, 4) $7,360 P P F = = (b) $30, 000( / ,13%,5) $55, 273 F F P = = 3.14 F = 3 P = P (1 + 0.06) N log 3 = N log(1.06) N = 18.85 ≈ 19 years 3.15 • F = 2 P = P (1 + 0.08) N log 2 = N log(1.08) N = 9 years • Rule of 72: 72 / 8 9 years = 3.16 (a) Single-payment compound amount ) , , / ( N i P F factors for N 9% 10% 35 20.4140 28.1024 40 31.4094 45.2593 To find ) 38 %, 5 . 9 , / ( P F , first, interpolate for 38 = n : N 9% 10% 38 27.0112 38.3965

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    Page | 4 Then, interpolate for % 5 . 9 = i : 7039 . 32 ) 38 %, 5 . 9 , / ( = P F As compared to formula determination 4584 . 31 ) 38 %, 5 . 9 , / ( = P F (b) Single-payment compound amount ( P / F ,8%, N ) factors for N 45 50 0.0313 0.0213 Then, interpolate for N = 47 0273 . 0 ) 47 %, 8 , / ( = F P As compared to the value from the interest formula: 0269 . 0 ) 47 %, 8 , / ( = F P 3.17 (a) 44 $18(1 ) $92, 400 21.43% i i + = = (b) F = $97.8( F / P ,21.43%,22) = $7,007billion Uneven Payment Series 3.18 $1,000 + $1,000 1.1 + $1,500 1.1 3 = $1,210 1.1 2 + X 1.1 4 X = $2,981 3.19 2 3 4 5 $25, 000 $33, 000 $46, 000 $38, 000 $110,961 1.07 1.07 1.07 1.07 P = + + + = 3.20 $2, 000( / , 6%,10) $2,500( / , 6%,8) $3, 000( / ,6%,6) $11,822 F F P F P F P = + + =

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    Page | 5 3.21 $3, 000, 000 $2, 400, 000( / ,8%,1) $3, 000, 000( / ,8%,10) $20, 734, 618 P P F P F = + + + = " Or, $3, 000, 000 $2, 400, 000( / ,8%,5) $3, 000, 000( / ,8%,5)( / ,8%,5) $20, 734, 618 P P A P A P F = + + = 3.22 $8, 000( / , 6%, 2) $6, 000( / , 6%,5) $4, 000( / ,6%,7) $14, 264 P P F P F P F = + + = Equal Payment Series 3.23 (a) With deposits made at the end of each year $2, 000( / ,8%,15) $54,304 F F A = = (b) With deposits made at the beginning of each year $2, 000( / ,8%,15)(1.08) $58, 649 F F A = = 3.24 $10, 000( / , 6%,20) $367,856 F F A = = 3.25 (a) $6, 000( / ,8%,5) $35, 200 F F A = = (b) $4, 000( / , 6.25%,12) $68, 473 F F A = = (c) $9, 000( / ,9.45%, 20) $484,359 F F A = = (d) $3, 000( / ,11.75%,12) $71,308 F F A = = 3.26 (a) $32, 000( / ,8%,15) $1,179 A A F = = (b) $55, 000( / , 6%,10) $4,173 A A F = =

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    Page | 6 (c) $35, 000( / , 7%, 20) $853.8 A A F = = (d) $8, 000( / ,11%, 4) $1, 699 A A F = = 3.27 $50, 000( / , 6%,10) $3, 793.40 A F = 3.28 $35, 000 $2, 000( / , 6%, ) ( / , 6%, ) 17.5 12.32 13years F A N F A N N = = = ≈ 3.29 $15, 000 ( / ,11%,5) $2, 408.57 A F A A = = 3.30 $5, 000 $500( / , 7%,5) ( / , 7%,5) $747.51 F P A F A A = + = 3.31 (a) $12, 000( / , 4%, 6) $2, 289.14 A A P = = (b) $3,500( / , 6.7%, 7) $642.66 A A P = = (c) $6,500( / ,3.5%,5) $1, 439.63 A A P = = (d) $32, 000( / ,8.5%,15) $3,853.47 A A P = = 3.32 (a) The capital recovery factor ) , , / ( N i P A for N 6% 7% 35 0.0690 0.0772 40 0.0665 0.0750 To find ) 38 %, 25 . 6 , / ( P A , first, interpolate for N = 38 : N 6% 7% 38 0.0675 0.0759 Then, interpolate for % 25 . 6 = i ;

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    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 , , / ( i A P for i 9% 10% 11.1038 9.9970 Then, interpolate for % 25 . 9 = i : 8271 . 10 ) 85 %, 25 . 9 , / ( = A P As compared to the value from the interest formula: 8049 . 10 ) 85 %, 25 . 9 , / ( = A P 3.33 • Equal annual payment: $50, 000( / ,12%,3) $20,817.45 A A P = = • Interest payment for the second year: End of Year Principal Repayment Interest payment Remaining Balance 0 $50,000 1 $14,817.45 $6,000 $35,182.55 2 $16,595.54 $4,221.91 $18,587.01 3 $18,587.01 $2,230.44 0 3.34 $10, 000( / ,9%,10) $1,558.2 A A P = = 3.35 (a) $1, 000( / , 6.8%,8) $6, 017.86 P P A = = (b) $3,500( / ,9.5%,12) $24, 443.44 P P A = = (c) $1,900( / ,8.25%,9) $11, 746.68 P P A = =

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    Page | 8 (d) $9,300( / , 7.75%,5) $37,378.16 P P A = = 3.36 P = $35,000( P / A ,12%,10) = $197,758 Since $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, 241.99 P P F P F = + + + = (b) P = $1,375,000 + $1,375,000( P / A ,6%,7) = $9,050,774.48 Since $9,050,774.48 > $8,000,000, the prorated payment option is better choice. Linear Gradient Series 3.38 $10, 000( / ,8%,5) $3, 000( / ,8%, 5) $10, 000( / ,8%,5) $3, 000( / ,8%,5)( / , 8%, 5) $91,163.55 F F A F G F A A G F A = + = + = 3.39 $7,500( / ,8%,5) $1,500( / ,8%,5) $7,500( / ,8%,5) $1,500( / ,8%,5)( / , 8%, 5) $27, 750.74 F F A F G F A P G F P = − = − = 3.40 $100 [$100( / ,9%, 7) $50( / ,9%, 6) $50( / ,9%,4) $50( / , 9%, 2)]( / , 9%, 7) $991.32 P F A F A F A F A P F = + + + + = 3.41 $15, 000 $1, 000( / , 8%, 12) $10, 404.25 A A G = − =

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    Page | 9 3.42 $1, 000( / , 6%,5) $250( / , 6%,5) $6,196 P P A P G = + = 3.43 Using the geometric gradient series present worth factor, we can establish the equivalence between the loan amount $120,000 and the balloon payment series as 38 . 684 , 25 $ 6721 . 4 ) 5 %, 9 %, 10 , / ( 000 , 120 $ 1 1 1 1 = = = A A A P A Payment series N Payment 1 $25,684.38 2 $28,252.82 3 $31,078.10 4 $34,185.91 5 $37,604.51 3.44 1 $6, 000( / ,5%, 7%,30)( / , 7%,30) $987, 093.8 F P A F P = = 3.45 (a) 076 , 372 , 21 $ ) 7 %, 12 %, 10 , / ( 000 , 000 , 6 $ 1 = − = A P P (b) Note that the oil price increases at the annual rate of 5% while the oil production decreases at the annual rate of 10%. Therefore, the annual revenue can be expressed as follows: 1 1 1 1 $60(1 0.05) 100, 000(1 0.1) $6, 000, 000(0.945) $6, 000, 000(1 0.055) n n n n n A − − − − = + − = = − This revenue series is equivalent to a decreasing geometric gradient series with g = -5.5%. So, 1 $6, 000, 000( / , 5.5%,12%,7) $23,847,897 P P A = − =

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    Page | 10 (c) Computing the present worth of the remaining series 4 5 6 7 ( , , , ) A A A A at the end of period 3 gives 3 4 4 6, 000, 000(1 0.055) 5, 063, 451.75 $5, 063, 451.75( / , 5.5%,12%, 4) $14, 269, 627.82 A P P A = − = = − = 3.46 20 1 20 1 1 20 1 (1 ) (2, 000, 000) (1.06) (1.06) 1.06 (2, 000, 000 /1.06) ( ) 1.06 $396, 226, 415 n n n n n n n n P A i n n − = − − = = = + = = = ∑ ∑ ∑ 3.47 (a) The withdrawal series would be Period Withdrawal 11 $12,000 12 $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) 10 1 $12, 000( / ,8%,12%,5) $49,879.14 P P A = = 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) 10 1 $12, 000( / ,8%,9%,5) $54, 045.08 P P A = = $54, 045.08 = A ( F / A , 9%,10) A = $3, 557.25

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    Page | 11 Various Interest Factor Relationships 3.48 (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 ) , , / ( N i F P i N i P A − = 0394 . 0 ) 2 %, 8 , / )( 40 %, 8 , / ( ) 42 %, 8 , / ( = = F P F P F P 0833 . 0 0394 . 0 1 08 . 0 ) 42 %, 8 , / ( = − = P A 0833 . 0 1 ) 08 . 1 ( ) 08 . 1 ( 08 . 0 ) 42 %, 8 , / ( 42 42 = − = P A (c) 4996 . 12 08 . 0 ) 35 %, 8 , / )( 100 %, 8 , / ( 1 ) , , / ( 1 ) , , / ( = − = − = F P F P i N i F P N i A P ( P / A , 8%,135) = (1.08) 135 − 1 0.08(1.08) 135 = 12.4996 3.49 (a) N N N N i i i i i i N i A F i N i P F ) 1 ( 1 1 ) 1 ( 1 1 ) 1 ( ) 1 ( 1 ) , , / ( ) , , / ( + = + − + = + − + = + + = (b) N N N N N N N N i i i i i i i i i i i N i A P N i F P − − + = + − + − + + = + − + − = + − = ) 1 ( ) 1 ( 1 ) 1 ( ) 1 ( ) 1 ( ) 1 ( 1 ) 1 ( 1 ) 1 ( ) , , / ( 1 ) , , / (

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