Engineering Economics Flashcards
6 cards from real Fundamentals of Engineering practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 6 Engineering Economics flashcards as text
In straight-line depreciation, what is the annual depreciation for an asset costing $50,000 with a salvage value of $5,000 and useful life of 5 years?
Answer: $9,000
Annual depreciation = (Cost - Salvage) / Life = ($50,000 - $5,000) / 5 = $9,000.
What is the uniform series present worth factor (P/A, i%, n)?
Answer: [(1+i)ⁿ - 1] / [i(1+i)ⁿ]
The P/A factor converts a uniform series of payments A to present worth P: P = A × [(1+i)ⁿ - 1] / [i(1+i)ⁿ].
A benefit-cost (B/C) ratio greater than 1.0 for a public project indicates:
Answer: Benefits exceed costs and the project is economically justified
A B/C ratio > 1.0 means the present worth of benefits exceeds costs, justifying the public expenditure.
What is the payback period for a project with an initial cost of $100,000 and uniform annual net cash flow of $25,000?
Answer: 4 years
Payback period = Initial cost / Annual cash flow = $100,000 / $25,000 = 4 years.
Which cost does NOT change with the level of production or output?
Answer: Fixed cost
Fixed costs remain constant regardless of production level, unlike variable costs which change with output.
If the inflation rate is 4% and the market interest rate is 9%, what is the approximate real interest rate?
Answer: ~4.8%
Using the Fisher equation: real rate = (1.09/1.04) - 1 ≈ 0.048 or 4.8%.