Difficulty: Easy
Correct Answer: All of these
Explanation:
Introduction / Context:Engineering economy compares alternatives on a consistent financial basis over a common study period. A complete analysis requires more than just the initial purchase price; it must consider the time value of money, operating costs, service life, and terminal values. This question checks whether you can identify the full set of inputs for life-cycle cost or present-worth/annual-worth comparisons.
Given Data / Assumptions:
Concept / Approach:Key methods include Present Worth (PW), Annual Worth (AW), and Equivalent Uniform Annual Cost (EUAC). Each requires the initial cost (first cost), annual operating and maintenance (O&M) costs, the asset life (n), and salvage value (SV). Discounting all cash flows at the required rate i ensures a like-for-like comparison across alternatives of differing timing and amounts.
Step-by-Step Solution:
List inputs: First cost (negative at t = 0), O&M costs (annual negatives), revenue or savings (positives), salvage value at end of life (positive), interest rate i.Compute PW or AW for each alternative using consistent i and study period.Select the alternative with the lowest EUAC (for cost-only cases) or highest PW/AW (for benefit cases).Verification / Alternative check:
Perform a sensitivity analysis on life n, i, O&M, and SV to test robustness; check the crossover rate if comparing two options with different cost profiles.Why Other Options Are Wrong:
Any single element (e.g., only first cost) is insufficient and can flip the decision when life-cycle costs or salvage are material.Common Pitfalls:
Ignoring different lives (failing to use common multiple or replacement chain); omitting taxes or escalation where relevant.Final Answer:
All of these
Discussion & Comments