October 30th Lecture 3000

October 30th Lecture 3000 - P =...

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October 30 th Lecture CEE 3000 Arithmetic Distribution Ex. Recycling Plant 2000 tons/yr N= 25 yrs I= 10% Construction costs= $200,000 Tipping fee= ? A=200,000(A/P,10%,25) = $22,034 Annual Costs= 60000 + 22034 = $82,034/yr. 2000(fee) = $82,034 Fee= $41.02/ton Ex. Airport Terminal Fee starts @ EOM 3 Nominal i= 5% # pax EOM 3 EOM 24 = 25k pax/m EOM 25 EOM 60= 30k pax/m i= 5% = 5/12 = 0.0042 Prev= 25000*fee*(P/A,0.0042,22)(P/F,0.0042,2) + 30000*fee*(P/A,0.0042,36) (P/F,0.0042,24) = $1,424,569(fee) Pcost= 10,000(P/F,5%,5) = $7,776,550 7,776,550=1424569(fee) Fee= 5.46/pax Arithmetic Distribution F= G/i[((1+i)^n -1)/i-n] = (F/G,I,n) A= G/i-G*n/((1+i)^n-1) = (A/G,I,n)
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Unformatted text preview: P = G/i[((1+i)^n-1)/((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.

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October 30th Lecture 3000 - P =...

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