Engineering Economics Present Worth

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The discussion focuses on calculating the Net Present Worth (NPW) for three alternatives in an engineering economics problem, with given initial costs, annual benefits, and useful lives. The alternatives include X with a high initial cost but significant annual benefits, Y with a low initial cost and infinite life, and Z with moderate costs and benefits. The participant successfully calculated the NPW for alternative Y but encountered difficulties with X and Z. A typo in the problem was identified, which contributed to the confusion. The discussion emphasizes the importance of accurate data in engineering economic analysis.
MHrtz
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Homework Statement



Three different alternatives shown below are being considered. Assume that alternatives X and Z are replaced at the end of their lives


Alt. X Alt. Y Alt. Z
Initial Cost - 6000 1000 1500

Annual Benefits - 810 125 230

Useful Life - 20 infinite 10

MARR = 12% for all

Find the Net Present Worth for Alt. X, Alt. Y, and Alt. Z.

Homework Equations



capitalized cost A = iP

F = A [((1 + i)n - 1))/i]

P = A [((1 + i)n - 1))/i(1 + i)n]

The Attempt at a Solution



I found the NPW for Alt. Y

P = (125/.12) - 1000 = 41.67

I'm still stuck on X and Z.
 
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Nevermind. One of choices was a typo so I got it right.
 

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