Engineering Economics Present Worth

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SUMMARY

The discussion focuses on calculating the Net Present Worth (NPW) for three alternatives in an engineering economics scenario, specifically alternatives X, Y, and Z. The alternatives have different initial costs, annual benefits, and useful lives, with a Minimum Attractive Rate of Return (MARR) set at 12%. The NPW for Alternative Y was successfully calculated as 41.67, while the calculations for Alternatives X and Z remain unresolved due to a typo in the problem statement. The relevant formulas for capitalized cost and present worth were also provided.

PREREQUISITES
  • Understanding of Net Present Worth (NPW) calculations
  • Familiarity with engineering economics concepts
  • Knowledge of financial formulas including capitalized cost and present worth
  • Basic proficiency in using financial calculators or spreadsheet software
NEXT STEPS
  • Research how to calculate NPW using different cash flow scenarios
  • Learn about the implications of MARR in project evaluation
  • Explore the differences between finite and infinite useful life in economic analysis
  • Study the impact of initial costs on the overall profitability of engineering projects
USEFUL FOR

Engineering students, financial analysts, project managers, and anyone involved in evaluating investment alternatives in engineering projects.

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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