October_11th_Lecture_3000

# October_11th_Lecture_3000 - P = 15,000(P/A 10 5 20,000(P/A...

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October 11 th Lecture CEE 3000 A = P[(1 + i) n i ]/[(1 + i) n – 1] P = A*[(1 + i) n – 1]/ [(1 + i) n i ] Capitalizing- bring back to the present Ex. N = 10 yrs i = 7% P= \$10,000 A=? A=P(A/F,i,n) A= 10,000[0.142378] Ex. \$10,000 loan Annual payments A= P[(1 + 0.1) 10 (0.1)]/[(1 + 0.1) 10 – 1] = \$1,627 F = P(1 + i) n P = F/(1 + i) n A= P(A/P, 10%, 10) = 10,000(0.162745)= \$1,627 N infinity --- perpetuity When n= P= A/i P = P’*(P/F, i, n) If n = 1 then time = 0 If not n =1 then must discount time Till time = 0 Power company antipolluliton technology N= 10 years Fuel cost savings = \$15,000/yr for first 5 yrs \$20,000/yr for next 5 yrs What is the present value of their savings? I = 10%

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Unformatted text preview: P = 15,000(P/A, 10%, 5) + 20,000(P/A, 10%, 5)(P/F, 10%,5) P = \$103,936 P = 20,000(P/A, 10%,10) – 5000(P/A, 10%, 5) = \$103,936 Interpolation 6% 3.1836 6.17% 7% 3.2149 0.17/1 = x/0.0313 X= 0.0053 3.1836 + x 3.1836 + 0.0053= 3.1889 So F = 1200(3.1889) = \$3,826.70 Ex. Dam construction at t = 3 Dredge every 5 years thereafter n = ∞ \$500,000 i= 10% What is Pcap? A = 500,000(A/F, 10%,5) A = \$81,900/yr P= A/i = 81,900/0.10 = \$819,000(P/F,10%, 3) = \$615,300 C’eff = (1 + 0.5) 5 – 1 = 0.61 or 61% P = A/i = 500,000/0.61 = = \$819,672(P/F,10%, 3) = \$615,300...
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October_11th_Lecture_3000 - P = 15,000(P/A 10 5 20,000(P/A...

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