Engineering Economics Flashcards

What is the future worth of $2,000 invested today at 7% per year compounded annually for 8 years? (F/P, 7%, 8) = 1.7182

Answer: $3,436.40

F = P × (F/P, i, n) = $2,000 × 1.7182 = $3,436.40.

Which of the following best describes the minimum attractive rate of return (MARR)?

Answer: The minimum interest rate a company will accept for any investment

MARR is the minimum return a firm requires before it will invest; projects below this rate are rejected.

A geometric gradient series has a first payment of $1,000 and increases by 5% each year for 4 years at i = 10%. What is the present worth? (P/G, geom formula: P = A₁[(1−(1+g)ⁿ(1+i)⁻ⁿ)/(i−g)])

Answer: $3,397

Using the geometric gradient PW formula with A₁=$1,000, g=5%, i=10%, n=4, PW ≈ $3,397.

A project requires $200,000 now and yields $50,000/year for 6 years. At MARR = 10%, (P/A,10%,6)=4.3553, the net present value is:

Answer: $17,765

NPV = −200,000 + 50,000(4.3553) = −200,000 + 217,765 = +$17,765.

Book value under straight-line depreciation after year k is calculated as:

Answer: Cost − k × (Cost − Salvage) / n

BV_k = Cost − k × annual depreciation = Cost − k × (Cost − Salvage)/n.

Which analysis method is most appropriate when comparing alternatives with unequal lives over a study period longer than both lives?

Answer: Annual worth analysis

Annual worth analysis converts each alternative to an equivalent annual cost/benefit, making unequal-life comparisons straightforward without repeating cycles.

A $500,000 project has expected cash inflows of $150,000/year. The simple payback period is:

Answer: 3.33 years

Simple payback = Initial cost / Annual cash inflow = $500,000 / $150,000 ≈ 3.33 years.