November 1st Lecture 3000

November 1st Lecture 3000 - = $162,740 Ex. Traffic =...

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November 1 st Lecture CEE 3000 3 cases * r=i r>i r<i r=i P= C*n/(1+r) r>i w= (1+r)/(1+i) -1 P= C/(1+i)(F/A,w,n) = C/(1+i)((1+w)^n -1)/w) r<i w=(1+i)/(1+r) -1 P= C/(1+r)(P/A,w,n) = C/(1+r)[(1+w)n-1)(w(1+w)n)] Imperpetuity r<i P = C/((1+r)w) W= ((1 +i) – 1)/(1+r) Ex. Computer N=3 Operation cost $60,000 at first year Increase/yr r = 10% i=10% P? r=i P= C*n/(1+r) = 60000(3)/(1+0.10) Ex. R<i W=(1+i)/(1+r) -1 = 0.09524 P=18,000/(1.05)(P/A, 0.9524,10) = 18000/1.05[(1+0.09524)^10 – 1)/( 0.09524*(1+0.09524)^10)]
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Unformatted text preview: = $162,740 Ex. Traffic = $100,000 Estimated $0.05/vehicle Current traffic => 800 cars/day (365) I=5.77% N=10 yrs R=10% each yr W=(1+r)/(1+i)=1 = 0.04 P= 146,000/(1+0.0577)(F/A,4%,10) Ex. Bridge n= heavy maintenance every 5 yrs first occurs at EOY 3 8 -13 -18 $80,000 each time maintenance r _> 25% i=12% Ieff= (1+0.12)^5) -1 = 76.23% W= (1+i)/(1+r)-1 = 0.41 P = 80,000/(1.25*.41) = 256,174 156,174(F/P,12%,2) = $124,500...
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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 Institute of Technology.

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November 1st Lecture 3000 - = $162,740 Ex. Traffic =...

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