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Engineering Economics Flashcards

6 cards from real EIT 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

  1. What is the present value of $10,000 received in 5 years at a discount rate of 8%?

    Answer: $6,806

    PV = FV/(1+i)ⁿ = 10000/(1.08)⁵ = 10000/1.469 ≈ $6,806.

  2. Which method computes the rate of return at which the net present value of a project equals zero?

    Answer: Internal Rate of Return (IRR)

    The Internal Rate of Return (IRR) is the discount rate at which NPV = 0, representing the project's effective return.

  3. In engineering economics, what does MARR stand for?

    Answer: Minimum Acceptable Rate of Return

    MARR (Minimum Acceptable Rate of Return) is the minimum return a company requires before approving an investment.

  4. Depreciation using the straight-line method distributes the cost:

    Answer: Equally over the useful life

    Straight-line depreciation spreads the cost minus salvage value equally across each year of the asset's useful life.

  5. What is the capital recovery factor (A/P, i, n) used to calculate?

    Answer: Uniform annual payment equivalent to a present sum

    The capital recovery factor converts a present sum P into an equivalent uniform annual payment A over n periods at rate i.

  6. The payback period of a $50,000 investment that generates $10,000/year is:

    Answer: 5 years

    Payback period = Initial Investment / Annual Cash Flow = $50,000 / $10,000 = 5 years.