{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

3232ex5-2d (1)

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

This preview shows pages 1–2. Sign up to view the full content.

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?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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.
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online