NCEES Engineering Economics 4 — Questions and Answers
Question 1: 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
- $3,436.40 (Correct answer)
- $3,200.00
- $2,980.00
- $4,000.00
Correct answer: $3,436.40
F = P × (F/P, i, n) = $2,000 × 1.7182 = $3,436.40.
Question 2: Which of the following best describes the minimum attractive rate of return (MARR)?
- The minimum interest rate a company will accept for any investment (Correct answer)
- The average return on all company investments
- The prime lending rate set by the Federal Reserve
- The risk-free rate of return on government bonds
Correct 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.
Question 3: 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)])
- $3,397 (Correct answer)
- $4,000
- $3,000
- $3,641
Correct answer: $3,397
Using the geometric gradient PW formula with A₁=$1,000, g=5%, i=10%, n=4, PW ≈ $3,397.
Question 4: 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:
- $17,765 (Correct answer)
- −$17,765
- $50,000
- $100,000
Correct answer: $17,765
NPV = −200,000 + 50,000(4.3553) = −200,000 + 217,765 = +$17,765.
Question 5: Book value under straight-line depreciation after year k is calculated as:
- Cost − k × (Cost − Salvage) / n (Correct answer)
- Cost × (1 − k/n)
- Salvage + k × (Cost − Salvage) / n
- Cost / (n − k)
Correct answer: Cost − k × (Cost − Salvage) / n
BV_k = Cost − k × annual depreciation = Cost − k × (Cost − Salvage)/n.
Question 6: Which analysis method is most appropriate when comparing alternatives with unequal lives over a study period longer than both lives?
- Annual worth analysis (Correct answer)
- Present worth analysis with LCM
- Incremental benefit-cost ratio
- Payback period analysis
Correct answer: Annual worth analysis
Annual worth analysis converts each alternative to an equivalent annual cost/benefit, making unequal-life comparisons straightforward without repeating cycles.
Question 7: A $500,000 project has expected cash inflows of $150,000/year. The simple payback period is:
- 3.33 years (Correct answer)
- 4.00 years
- 2.50 years
- 5.00 years
Correct answer: 3.33 years
Simple payback = Initial cost / Annual cash inflow = $500,000 / $150,000 ≈ 3.33 years.
NCEES Engineering Economics 4
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