3232ex5-2d (1) - PE based on just 3 years was -22601 PE(A,...

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Ex 5.2d: Analysis period differs from system life Case 4: Continuous project with unequal system lives Cash flow, MARR =15% Year A B 0 -12,500 -15,000 A salvage year 3 = 2000 1 -5,000 -4,000 B salvage year 4 = 1500 2 -5,000 -4,000 3 -3,000 -4,000 4 -2,500 PE(15%) -22,601 -25,562 Is this meaningful? Now what?
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Least common multiple = 12 years Cash flow, MARR =15% Year A B 0 -12,500 -15,000 1 -5,000 -4,000 2 -5,000 -4,000 3 -3,000 -4,000 4 -2,500 -53,657 -48,534 CHOOSE B -9,899 -8,954 AE computed based on 3 years for A, 4 years for B PE (15%) 12 years AE (15%) 3/4 years Consider A.
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Unformatted text preview: PE based on just 3 years was -22601 PE(A, 12years @ 15%) = -22601 - 22601/(P/F,15,3) - 22601/(P/F,15,6) - 22601(P/F,15,9) =-53657 AE(A, 12 years, 15%) = PE(A,12yr, 15%) * (A/P,15,12) = -9899 also AE(A, 12yr, 15%) = AE(A, 3 yr , 15%) AE(A, 3yr, 15%) = (-12500 - 5000(P/F,15,1) - 5000(P/F,15,2)-3000(P/F(15,3))*(A/P,15,3)=-9899 AEs for projects of differing lengths can be compared because they both are based on the same year. For PEs to be compared, one must consider the entire project cash flow....
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This note was uploaded on 11/19/2010 for the course ME 3232 taught by Professor Lyon during the Fall '10 term at University of New Brunswick.

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3232ex5-2d (1) - PE based on just 3 years was -22601 PE(A,...

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