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Solution Manual for Fundamentals of Engineering Economics 4th Edition / All Chapters 1 - 13 / Full Complete 2023 $12.99   Add to cart
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Solution Manual for Fundamentals of Engineering Economics 4th Edition / All Chapters 1 - 13 / Full Complete 2023

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Solution Manual for Fundamentals of Engineering Economics 4th Edition / All Chapters 1 - 13 / Full Complete 2023

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  • May 26, 2023
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  • fundamentals of engineering economics 4th edition solution manual
  • fundamentals of engineering economics 4th edition
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Chapter 1 Engineering Economic Decisions Fundamentals of Engineering Economics 4th Edition Solution Manual 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 Chapter 2: Time Value of Money 2.1) I  (iP)N  (0.06)($2, 000)(5)  $600 2.2)  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: $20,000  $10,000(1  0.07)N (1  0.07)N  2 N  10.24  11years 2.3)  Simple interest: I  iPN  (0.10)($10, 000)(5)  $5, 000  Compound interest: I  P[(1 i)N 1]  $10, 000(1.6105 1)  $6,105 2.4)  Option 1: Compound interest with 8.5%: F  $4, 500(1 0.085)5  $4, 500(1.5037)  $6, 766.65  Option 2: Simple interest with 9%: $4, 500(1 0.09  5)  $5, 000(1.45)  $6, 525  Option 1 is still better. 2.5)  Compound interest: F  $1, 000(1 0.065)5  $1, 370.09  Simple interest: F  $1, 000(1 0.068(5))  $1, 340 The compound interest option is better. 2.6) End of Year Principal Repayment Interest payment Remaining Balance 0 $15,000.00 1 $4,620.50 $1,200.00 $10,379.50 2 $4,990.14 $830.36 $5,389.36 3 $5,389.35 $431.15 $0 2.7) P  $22, 000(P / F , 5%, 5)  $22, 000(0.7835)  $17, 237.58 2.8) F  $30, 000(F / P, 9%, 3)  $30, 000(1.295)  $38,850.87 2.9) F  $100( F / P,10%,10)  $200( F / P,10%,8)  $688 2.10) F  $250, 000(F / P, 6%,10)  $447, 712 2.11) P  $300, 000(P / F ,8%,10)  $138, 958 2.12) i  10.5% , two-year discount rate is (1 0.105)2  1.221 (or 22.1%) 2.13) (a) F  $5, 000(F / P, 7%, 5)  $7, 013 (b) F  $7, 250(F / P, 9%,15)  $26, 408 (c) F  $9, 000(F / P, 6%, 33)  $61, 565 (d) F  $12, 000(F / P, 5.5%,8)  $18, 416 2.14) (a) P  $25, 500(P / F ,12%,8)  $10, 299 (b) P  $58, 000(P / F , 4%,12)  $36, 227 (c) P  $25, 000(P / F , 6%, 9)  $14, 797 (d) P  $35, 000(P / F , 9%, 4)  $24, 795 2.15) (a) P  $12, 000(P / F ,13%, 4)  $7, 360 (b) F  $30, 000(F / P,13%, 5)  $55, 273 2.16) F  3P  P(1 0.08)N log 3  N log(1.08) N  14.27  15 years 2.17) F  2P  P(1 0.06)N log 2  N log 1.06 N  11.896 years (or 12 years) 2.18) F  2P  P(1 0.06)N  log 2  N log(1.06) N  11.90 years 12 years  Rule of 72:  12 years 2.19) F  $1(1.08)394  $14, 755, 694, 730, 611

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