NCEES FE Other Disciplines · USA · NCEES
Original practice questions written to the published syllabus · not past papersMoney and time in engineering decisions: discounting, costs, break-even, project appraisal, scheduling and starting a business.
In the exams: NCEES FE Engineering Economics 6–9 questions; NCEES PE engineering economics and project management; ICAR AICE subject 41 Unit 10; PRC ABE project management
How our questions for NCEES FE Other Disciplines spread across the units of Engineering Economics, Project & Business Management. Difficulty: 15 easy, 17 medium, 11 hard.
| NPV, IRR, benefit-cost ratio and payback | 11 | |
| Time value of money | 10 | |
| Depreciation methods | 9 | |
| Break-even analysis | 7 | |
| Replacement analysis | 5 | |
| Fixed and variable costs | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A bank quotes 12 % a year compounded monthly on a tractor loan. What is the effective annual interest rate?
D. 12.68 %
i_eff = (1 + r/m)^m − 1 = (1 + 0.12/12)^12 − 1 = 1.01^12 − 1 = 0.1268, or 12.68 %. 12.00 % is the nominal rate; 12.36 % is the effective rate for semi-annual compounding; 1.00 % is the monthly rate, not an annual one.
At 9 % a year compound interest, how long does it take for a sum of money to double?
D. 8.04 years
Solve 2 = 1.09^n: n = ln 2 / ln 1.09 = 8.04 years. 11.11 years is the doubling time under simple interest, 1/0.09; the rule of 72 gives 72/9 = 8, close to the exact answer. ln 2/0.09 = 7.70 years treats the rate as continuously compounded.
A grading line costs 240,000 and earns net receipts of 60,000 a year. What is its discounted payback period at 10 % a year, interpolating within the year?
B. 5.4 years
Cumulative discounted receipts: 60,000(P/A, 10 %, n) reaches 240,000 between year 5 (227,447) and year 6 (261,316); interpolation gives 5.4 years. 4.0 years is simple payback, 240,000/60,000, which ignores interest; discounted payback is always longer. 3.6 years compounds the receipts instead of discounting them (240,000/(60,000 × 1.1)).
Machine X costs 600,000, lasts 5 years, has salvage 100,000 and costs 80,000 a year to run. Machine Y costs 900,000, lasts 10 years, has salvage 150,000 and costs 60,000 a year to run. At 10 %, which has the lower equivalent annual cost?
A. Y, at about 197,100 a year against about 221,900 for X
EUAC_X = 600,000(A/P,10 %,5) − 100,000(A/F,10 %,5) + 80,000 = 221,899; EUAC_Y = 900,000(A/P,10 %,10) − 150,000(A/F,10 %,10) + 60,000 = 197,059. Y is cheaper. Annual cost compares unequal lives directly under the repeatability assumption, so no extra study period is needed; first cost alone ignores life and running cost.
A tractor costs 800,000, has a salvage value of 80,000 and a recovery period of 8 years. Using double declining balance, what is the depreciation in year 2?
A. 150,000
d = 2/n = 0.25. D1 = 0.25 × 800,000 = 200,000; BV1 = 600,000; D2 = 0.25 × 600,000 = 150,000. 200,000 is year 1; 135,000 wrongly applies the rate to B − S, but declining balance uses book value and ignores salvage until the floor is reached; 90,000 is straight line.
How does depreciation affect the cash flow of a profitable business?
A. It is not itself a cash flow, but it lowers taxable income and therefore the tax paid
Cash leaves when the asset is bought; depreciation only allocates that cost over time. Because tax law allows it as a deduction, it reduces taxable income and tax, so its cash effect is the tax saving, depreciation × tax rate.
A spice-grinding unit has fixed costs of 600,000 a year, sells at 50 a kg and has variable costs of 30 a kg. What is its break-even output?
A. 30,000 kg
Q* = F/(p − v) = 600,000/(50 − 30) = 30,000 kg. 12,000 divides by price and 20,000 by variable cost; 7,500 adds them instead of subtracting. The contribution margin, p − v, is what covers fixed cost.
A spice-grinding unit has fixed costs of 600,000 a year, a price of 50 a kg and variable costs of 30 a kg. If it plans to sell 40,000 kg, what is its margin of safety as a percentage of planned sales?
A. 25.0 %
Break-even = 600,000/20 = 30,000 kg. Margin of safety = 40,000 − 30,000 = 10,000 kg = 10,000/40,000 = 25.0 % of planned sales. 33.3 % divides by break-even output instead; 75 % is break-even as a share of plan; 20 % is the contribution per kg read as a percentage.
A defender's marginal cost of keeping it one more year is 140,000 and rising each year. The best challenger has a minimum equivalent annual cost of 150,000 at its economic life. What should be done?
C. Keep the defender for at least one more year and repeat the comparison next year
The defender is cheaper for the coming year (140,000 < 150,000), so keeping it saves money now. Because its marginal cost is rising, it will soon exceed 150,000, and the comparison should be repeated each year. Book depreciation and novelty are not decision criteria.
A pump still works reliably, but a new model uses 30 % less energy for the same duty. The loss in value of the old pump for this reason is called:
C. Obsolescence, or functional depreciation
Obsolescence is the loss of value caused by better technology or changed requirements, not by wear. Physical depreciation comes from use, corrosion and accidents. It is one of the main reasons assets are replaced before they wear out.
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