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Unformatted text preview: P = G/i[((1+i)^n1)/((1+i)^n*1)n/(i+1)^n]= (P/G,I,n) Ex. Pump Repair cost for the first year + increase $100/yr over 5 years I=6% P=? P=G(P/G,6%,5) =100(7.9345) = $793.40 Ex. $1000 first year that increase $500/yr for 5 yrs i=8% P=? A=? F=? P = 1000(P/A, 8%,5) + 500(P/G,8%,5) =$7,679 Ex. Liner Savings 1 4 0 5 1000 6 2000 7 3000 11 7000 P=? i=10% P1=1000(P/A,10%,7)+1000(P/G,10%,7) P=P1(P/F,10%,4) Perpetuity P/G = 1/i^2 P=G/i^2 For n = ∞ P= P1(P/F, 10%,4) Ieff= (1+.10)^5 1 = 0.611 61.1% P1 = (200,000/0.611) + (100,000/0.611^2) = $ 792,209 P = 792,209(F/P,10%,3)...
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This note was uploaded on 06/07/2009 for the course CEE 3000 taught by Professor Meyer during the Fall '07 term at Georgia Tech.
 Fall '07
 Meyer

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