Lecture No6c3.ppt - Contemporary Engineering Economics 4 th...

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Contemporary Engineering Economics, 4 th edition, ©2007 Interest Formulas for Single Cash Flows Lecture No.6 Chapter 3 Contemporary Engineering Economics Copyright © 2006
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Contemporary Engineering Economics, 4 th edition, © 2007 Types of Common Cash Flows in Engineering Economics
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Contemporary Engineering Economics, 4 th edition, © 2007 Equivalence Relationship Between P and F
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Contemporary Engineering Economics, 4 th edition, © 2007 Single Cash Flow Formula- Compound-Amount Factor Single payment compound amount factor (growth factor) Given : Find : F P F N 0 F P i F P F P i N N ( ) ( / , , ) 1 12% 8 years $5,000 i N P 8 $5,000(1 0.12) $5,000( / ,12%,8) $12,380 F F P
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Contemporary Engineering Economics, 4 th edition, © 2007 Practice Problem If you had $2,000 now and invested it at 10%, how much would it be worth in 8 years? $2,000 F = ? 8 0 i = 10%
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Contemporary Engineering Economics, 4 th edition, © 2007 Solution 8 Given: $2,000 10% 8 years Find: $2,000(1 0.10) $2,000( / ,10%,8) $4,287.18 EXCEL command: =FV(10%,8,0,2000,0) =$4,287.20 P i N F F F P
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