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Economic Analysis and Engineering Economics Flashcards

7 cards from real CCP practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 7 Economic Analysis and Engineering Economics flashcards as text

  1. What does Net Present Value (NPV) represent in project economic analysis?

    Answer: The present value of all future cash inflows minus the present value of all cash outflows

    NPV is the difference between the present value of all cash inflows and outflows discounted at the required rate of return, measuring whether an investment creates value.

  2. Which discount rate makes the Net Present Value (NPV) of a project exactly equal to zero?

    Answer: The Internal Rate of Return (IRR)

    The IRR is by definition the discount rate at which NPV equals zero, representing the project's true rate of return.

  3. A project requires an initial investment of $120,000 and generates $30,000 in net cash flow each year. What is the simple payback period?

    Answer: 4 years

    Simple payback period = Initial Investment / Annual Cash Flow = $120,000 / $30,000 = 4 years.

  4. The 'time value of money' concept in engineering economics is fundamentally based on which principle?

    Answer: A dollar available today is worth more than a dollar available in the future due to its earning potential

    Time value of money holds that a present dollar can be invested to earn a return, making it worth more than the same dollar received in the future.

  5. What is the Present Worth Factor (P/F, i%, n) formula used in engineering economics?

    Answer: 1 / (1 + i)^n

    The Present Worth Factor (P/F) = 1/(1+i)^n converts a single future value to its equivalent present value by discounting at rate i for n periods.

  6. What is the Benefit-Cost Ratio (BCR) decision rule for accepting a project?

    Answer: Accept if BCR ≥ 1.0

    A BCR ≥ 1.0 means the present value of benefits equals or exceeds the present value of costs, indicating the project is economically justified.

  7. The Capital Recovery Factor (A/P, i%, n) in engineering economics is used to:

    Answer: Convert a present sum into an equivalent uniform series of annual payments

    The Capital Recovery Factor converts a present investment (P) into an equivalent uniform annual payment series (A), such as converting a loan into annual installments.