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Solution Manual For Fundamentals Of Engineering Economics, 3rd Edition

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    Fundamentals of Engineering Economics, 3 rd ed. Chapter 1 Engineering Economic Decisions 1.1) Not provided For The Wall Street Journal , go to the Front page to find the section on “What’s News.” This is a section on a brief summary on major headlines of the day’s news. Quickly browse through the news summary to see if there is any news related to business investment. The best places to find the major business news on investment are sections on “BUSINESS”, “MARKETS” or “TECH.” 1.2) Not provided Some of the well-known business publications are: • Daily Newspapers: o The Wall Street Journal o The New York Times (Business Section) o Financial Times • Weekly or Monthly Magazines: o BusinessWeek o Forbes o Money o Smart Money o Fortune

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    Chapter 2: Time Value of Money 2.1) ( ) (0.06)($2, 000)(5) $600 I iP N = = = 2.2) • Simple interest: (1 ) $6, 000 $3, 000(1 0.08 ) 12.5 years (or 13 years) F P iN N N = + = + = • Compound interest: $6, 000 $3, 000(1 0.07) 2 1.07 log 2 log 1.07 10.24 years (or 11 years) N N N N = + = = = 2.3) • Simple interest: ( ) (0.07)($15, 000)(25) $26, 250 I iP N = = = • Compound interest: 25 (1 ) 1 $15, 000 (1.07) 1 $66, 411.50 N I P i     = + − = −     = 2.4) • A : Simple interest: ( ) (0.06)($10, 000)(15) $9, 000 I iP N = = = • B : Compound interest: 15 (1 ) 1 $10, 000 (1.055) 1 $12,324.76 N I P i     = + − = −     = B is a better option.

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    Fundamentals of Engineering Economics, 3 rd ed. 2.5) • Compound interest: 5 $1, 000(1 0.065) $1,370.09 F = + = • Simple interest: $1, 000(1 0.068(5)) $1,340 F = + = The compound interest option is better. 2.6) • Loan balance calculation: End of period Principal Payment Interest Payment Remaining Balance 0 $0.00 $0.00 $10,000.00 1 $1,670.92 $900.00 $8,329.08 2 $1,821.30 $749.62 $6,507.78 3 $1,985.22 $585.70 $4,522.56 4 $2,163.89 $407.03 $2,358.67 5 $2,358.64 $212.28 $0.00 2.7) $5, 000( / , 7%,5) $5, 000(0.7130) $3,565 P P F = = = 2.8) $25, 000( / ,8%, 2) $25, 000(1.1664) $29,160 F F P = = = 2.9) • Alternative 1 $100 P = • Alternative 2 $120( / ,10%, 2) $120(0.8264) $99.168 P P F = = = • Alternative 3 $170( / ,10%,5) $170(0.6209) $105.553 P P F = = =

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    Fundamentals of Engineering Economics, 3 rd ed. • Alternative 3 is preferred 2.10) $1, 000( / ,5%,3) $1, 000(1.1576) $1,157.6 F F P = = = 2.11) $500( / ,9%,5) $500(0.6499) $324.95 F P F = = = 2.12) 10.5% i = , two-year discount rate is 2 (1 0.105) 1.221(22.1%) + = 2.13) (a) $6, 000( / , 6%,8) $6, 000(1.5938) $9,563 F F P = = = (b) $1,550( / ,5%,12) $1,550(1.7959) $2, 784 F F P = = = (c) $8, 000( / ,9%,32) $8, 000(15.7633) $126,106 F F P = = = (d) $12, 000( / ,8%,9) $12, 000(1.999) $23,988 F F P = = = 2.14) (a) $5,500( / ,10%, 6) $5,500(0.5645) $3,105 P P F = = = (b) $7, 000( / ,9%,3) $7, 000(0.7722) $5, 405 P P F = = = (c) $22, 000( / ,8%,5) $22, 000(0.6806) $14,973 P P F = = = (d) $13, 000( / , 7%,8) $13, 000(0.5820) $7,566 P P F = = = 2.15) (a) $8, 000( / ,8%,5) $8, 000(0.6806) $5, 445 P P F = = = (b) $10, 000( / ,8%, 4) $10, 000(1.3605) $13, 605 F F P = = = 2.16) 3 (1 0.07) log 3 log 1.07 16.24 years (or 17 years) N F P P N N = = + = = 2.17) 2 (1 0.06) log 2 log 1.06 11.896 years (or 12 years) N F P P N N = = + = = 2.18) • Rule of 72: 72 / 8 9 years =

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    Fundamentals of Engineering Economics, 3 rd ed. 2 (1 0.08) log 2 log 1.08 9 years N F P P N N = = + = = 2.19) 389 $1(1.08) $10, 042, 477,894, 213 F = = 2.20) $35, 000( / ,9%, 4) $10, 000( / ,9%, 2) $35, 000(0.7084) $10, 000(0.8417) $33, 211 P P F P F = + = + = 2.21) $450, 000( / ,5%,5) 450, 000(0.7835) $352,575 P P F = = = 2.22) • Simple interest (John): (0.1)($1, 000)(5) $500 I iPN = = = • Compound interest (Susan): 5 (1 ) 1 $1, 000 (1 .095) 1 $574.24 N I P i     = + − = + −     = • Susan’s balance will be greater by $74 (or $74.24 to be exact) 2.23) 1 2 3 $2, 000 $800 $1, 000 $3, 230.65 1.1 1.1 1.1 P = + + = 2.24) 2 3 4 5 $3, 000 $3,500 $4, 200 $6,500 $14, 292.8 1.05 1.05 1.05 1.05 P = + + + = 2.25) $2, 000( / ,8%,10) $3, 000( / ,8%,8) $4, 000( / ,8%, 6) $16, 218 F F P F P F P = + + =

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

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    Fundamentals of Engineering Economics, 3 rd ed. 2.32) 2 3 4 5 $60, 000 $77, 000 $65, 000 $57, 000 45, 000 $212,873.89 1.14 1.14 1.14 1.14 1.14 P = + + + + = 2.33) $5, 000( / , 6%,10) $5, 000(13.1808) $65,904 F F A = = = 2.34) (a) $5, 000( / ,5%, 7) $5, 000(8.1420) $40, 710 F F A = = = (b) $5, 000( / ,5%, 7)(1.05) $5, 000(8.1420)(1.05) $42, 745.50 F F A = = = 2.35) (a) $6, 000( / , 6%, 6) $6, 000(6.9753) $41,851.80 F F A = = = (b) $8, 000( / , 7.25%,9) $96,825.60 F F A = = (c) $15, 000( / ,8%, 25) $15, 000(73.1059) $1, 096,588.50 F F A = = = (d) $3, 000( / ,9.75%,10) $47, 242.80 F F A = = 2.36) (a) $18, 000( / ,5%,13) $18, 000(0.0565) $1, 017 A A F = = = (b) $11, 000( / , 6%,8) $11, 000(0.1010) $1,111 A A F = = = (c) $8, 000( / ,8%, 25) $8, 000(0.0137) $109.6 A A F = = = (d) $12, 000( / , 6.85%,8) $1,176 A A F = = 2.37) $250, 000( / ,5%,5) $250, 000(0.1810) $45, 250 A A F = = = 2.38) $25, 000( / ,5%,5) $25, 000(0.1810) $4,525 A A F = = = 2.39) ( ) $35, 000 $3, 000( / , 6%, ) ( / , 6%, ) 11.6666 1 0.06 1 11.6666 0.06 log(1.06) log(1.7) 9.11 years N F A N F A N N N = = + − = ⋅ = = 2.40) $10, 000( / ,9%,5) $1, 670.92 A A F = =

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    Fundamentals of Engineering Economics, 3 rd ed. 2.41) 10 8 6 4 2 $500(1.04) $1, 000(1.04) $1, 000(1.04) $1, 000(1.04) $1, 000(1.04) $1, 000 $6, 625.47 F = + + + + + = 2.42) (a) $18, 000( / ,8%,5) $18, 000(0.2505) $4,509 A A P = = = (b) $4, 200( / ,9.5%, 4) $1,310.82 A A P = = (c) $7, 700( / ,11%,3) $7, 700(0.4092) $3,150.84 A A P = = = (d) $23, 000( / , 6%, 20) $23, 000(0.0872) $2, 005.60 A A P = = = 2.43) • Equal annual payment amount: $20, 000( / ,10%,3) $20, 000(0.4021) $8, 042 A A P = = = • Loan balance calculation: End of period Principal Payment Interest Payment Remaining Balance 0 $0.00 $0.00 $20,000.00 1 $6,042.00 $2,000.00 $13,958.00 2 $6,646.20 $1,395.80 $7,311.80 3 $7,310.82 $731.18 $0 Interest payment for the second year = $1,395.80 2.44) (a) $9, 000( / , 6%,8) $9, 000(6.2098) $55,888.20 P P A = = = (b) $1,500( / ,9%,10) $1,500(6.4177) $9, 626.55 P P A = = = (c) $7,500( / , 7.25%, 6) $35, 475 P P A = = (d) $9, 000( / ,8.75%,30) $52,529 P P A = = 2.45) (a) ( ) ( ) 36 36 0.0625 1 0.0625 ( / , 6.25%,36) 0.07044 1 0.0625 1 A P + = = + − (b) ( ) ( ) 125 125 1 0.0925 1 ( / ,9.25%,125) 10.81064 0.0925 1 0.0925 P A + − = = + 2.46) $500( / , 7%,15)(1.07) $500(25.1290)(1.07) $13, 444.02 F F A = = = 2.47) $5, 000( / ,11%,5) $5, 000(0.2706) $1,353 A A P = = = If you make the first payment on the loan at the end of the second year:

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    Fundamentals of Engineering Economics, 3 rd ed. $5, 000( / ,11%,1)( / ,11%, 4) $5, 000(1.11)(0.3223) $1, 788.78 F F P A P = = = 2.48) New equipment: $195,000 O&M cost: $30, 000( / ,10%,10) $30, 000(6.1446) $184,338 P P A = = = New equipment isn’t worth buying. 2.49) 250 $3, 460 0 P i = − + =  I = 7.225% 2.50) 1, 000 $10, 000 0.1 P = = 2.51) 1 2 $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) $50,998.35 F F F F A F G F A A G F A = + = + = + = + = 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, 000.04 F F A F G F A P G F P = − = − = − = 2.53) $100( / ,8%,1) $150( / ,8%,3) $200( / ,8%,5) $250( / ,8%, 7) $300( / ,8%,9) $350( / ,8%,11) $793.83 P P F P F P F P F P F P F = + + + + + = 2.54) $30, 000 $3, 000( / ,8%,10) $30, 000 $3, 000(3.8713) $18,386.1 A A G = − = − =

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    Fundamentals of Engineering Economics, 3 rd ed. 2.55) $3, 000( / ,12%,8) $600( / ,12%,8) $3, 000(4.9676) $600(14.4714) $23,585.64 P P A P G = + = + = 2.56) ( / ,9%, 6) $1000( / ,9%, 4) $800( / ,9%,3) $600( / ,9%, 2) $400( / ,9%,1) $200 (10.0924) $3796.46 1, 000( / ,9%, 4) 800( / ,9%, 4) 200( / ,9%, 4)( / ,9%, 4) $376.17 C P G F P F P F P F P C F P F A P G F P C = + + + + = → + − ∴ = 2.57) ( ) ( ) 1, 40 40 $6, 000( / 5%,9%, 40) 1 1.05 1.09 $6, 000 0.09 0.05 $116,379.57 $116379.57 *( / ,9%, 40) $3, 655.412.47 P P A F P − = − = − = = 2.58) (a) ( ) ( ) ( ) 1, 7 7 $10, 000, 000( / 10%,12%, 7) 1 1 0.1 1 0.12 $10, 000, 000 0.12 0.1 $35, 620,126 P P A − = − − − + = ⋅ − − = (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 $100(1 0.05) 100, 000(1 0.10) $10, 000, 000(0.945) $10, 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%. N A n

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    Fundamentals of Engineering Economics, 3 rd ed. 1 $10,000,000 2 $9,450,000 3 $8,930,250 4 $8,439,086 5 $7,974,937 6 $ 7,536,315 7 $ 7,121,81 8 ( ) ( ) ( ) 1, 7 7 $10, 000, 000( / 5.5%,12%, 7) 1 1 0.055 1 0.12 $10, 000, 000 0.12 0.055 $39, 746, 494.51 P P A − = − − − + = ⋅ − − = (c) Computing the present worth of the remaining series 4 5 6 7 ( , , , ) A A A A at the end of period 3 gives ( ) ( ) ( ) 1 4 4 $8, 439, 086.25( / , 5.5%,12%, 4) 1 1 0.055 1 0.12 $8, 439, 086.25 0.12 0.055 $23, 782, 713 P P A − = − − − + = ⋅ − − = 2.59) 20 1 20 1 1 20 1 20 1 (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, 415.1 n n n n n n n n n P A i n n n − = − − = = = = + = = = = = ∑ ∑ ∑ ∑ 2.60) (a) The withdrawal series would be:

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    Fundamentals of Engineering Economics, 3 rd ed. 2 3 4 Period Withdrawal 11 $3,000 12 $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, using i = 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 $923.88 A F A A A = = = (b) Equivalent present worth of the withdrawal series at 6% 1 5 $3, 000( / , 6%, 6%,5) $3, 000 $14,150.94 1 0.06 P P A = = = + $14,151 = A ( F / A ,6%,10) = 13.1808 A A = $1,073.60 2.61) $1, 000, 000 ( / , 6%,30) (79.0582) A F A A = =  A = $12,649 should be set aside on the account a) $1, 000, 000 ( / , 6%, 20) (11.4699) A P A A = =  A = $87,185 / year b) P = $3,000( P / A 1 ,6%,8%,5) Ź Ź = $3,000 ⋅ 1 − 1 + 0.06 ( ) 5 1 + 0.08 ( ) − 5 0.08 − 0.06 ( ) Ź Ź = $13,383.92

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    Fundamentals of Engineering Economics, 3 rd ed. ( ) ( ) 1 1, 20 20 1 $1, 000, 000 ( / 3%, 6%, 20) 1 1.03 1.06 0.06 0.03 $68, 674 / year A P A A − = − = − = 2.62) 2 3 2 3 $50 $70 $50 2 2 5.52 1.1 1.1 1.1 1.1 1.1 1.1 1.331 140.8715 4.1473 $33.97 C C C C C C + + = + + = → = = 2.63) [$100( / ,10%,8) $50( / ,10%, 6) $50( / ,10%, 4)]( / ,10%,8) [$100(11.4359) $50(7.7156) $50(4.6410)](0.4665) $821.70 P F A F A F A P F = + + = + + = 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) $720.42 P P F P A P F P F P = − + + = − + + = 2.66) Computing the equivalent worth at period 3 will require only two different types of interest factors.

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    Fundamentals of Engineering Economics, 3 rd ed. 1 2 $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 $249.44 P P A F P P A P A F P A P A A A A A = = = = + = + = = = 2.67) 1,1 $200( / ,10%, 4) 100( / ,10%, 2) $200(3.1699) 100(1.7355) 460.43 P P A P A = − = − = 2,1 ( / ,10%, 4) (3.1699) 4.1699 P X X P A X X X = + = + = 1,1 2,1 P P = $460.43 4.1699 $110.42 X X = = 2.68) 1 2 1 2 $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 $43.96 P P A P A P F P C P A C P A P F C C C P P C = + = + = = + = + = = = 2.69)

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    Fundamentals of Engineering Economics, 3 rd ed. ( / ,9%,8) $5, 000( / ,9%, 2) $5000 (11.0285) $5, 000(1.7591) $5000 $1, 250.90 C F A P A C C = + = + = 2.70) The original cash flow series is n A n 0 $0 1 $800 2 $820 3 $840 4 $860 5 $880 6 $900 7 $920 8 $300 9 $300 10 $300 - $500 2.71) 2 ( / ,12%, 7)( / ,12%,1) $1, 200( / ,12%,8) 400( / ,12%, 4) 2 (4.5638)(0.8929) $1, 200(4.9676) 400(3.0373) C C P A P F P A P A C C + = − + = − 6.075 $4, 746.20 $781.27 C C = = 2.72) 200(1.06)(1.08)(1.12)(1.15) (1.08)(1.12)(1.15) $300(1.15) $1000 247.9 1.39104 345 1000 1.39104 360.1 $258.87 X X X X + + = + + = = =

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