Engineering Economics
5–8 of 110 itemsNCEES draws between 5 and 8 of the 110 from this area on any legal paper. Its outline lists 4 sub-topics; the bank holds 27 verified templates against them, which expand to 864 questions.
Sub-topics, from the NCEES outline
- ATime value of money (e.g., equivalence, present worth, equivalent annual worth, future worth, rate of return)
- BCost (e.g., fixed, variable, direct and indirect labor, incremental, average, sunk)
- CAnalyses (e.g., break-even, benefit-cost, life cycle, sustainability, renewable energy)
- DUncertainty (e.g., expected value and risk)
5 worked problems
Real questions from the bank, at the depth the exam asks them. Pick an answer first: the reasoning opens when you do, along with why each wrong answer is tempting.
Question 1 of 5
Equipment costs $500,000, has a salvage value of $80,000 after 15 years, and costs $46,000 per year to operate. At 10% interest, its equivalent uniform annual cost is most nearly:
Show the answer and the reasoning
Why C
Given: P = $500,000, S = $80,000 at year 15, O&M $46,000 per year, i = 10%. Relation: EUAC = P(A/P, i, n) + O&M - S(A/F, i, n); the salvage is a single sum at the end, so it takes the sinking fund factor. Substitute: (A/P) = 0.13147, (A/F) = 0.03147; EUAC = 65,737 + 46,000 - 2,518. Result: EUAC = $109,219.
Why A is tempting
Left the annual operating cost out.
Why B is tempting
Spread the capital and salvage straight-line with no interest.
Why D is tempting
Ignored the salvage value entirely.
Source NCEES FE Reference Handbook 10.6 — Engineering Economics, interest factor tables
Question 2 of 5
Maintenance costs are zero in year 1 and rise by $4,000 each year thereafter through year 6. At 8% interest, the present worth of the gradient is most nearly:
Show the answer and the reasoning
Why C
Given: G = $4,000 per year, starting from zero in year 1; i = 8%, n = 6. Relation: P = G(P/G, i, n), with (P/G) = [(1 + i)^n - 1]/[i^2 (1 + i)^n] - n/[i(1 + i)^n]. Substitute: (P/G, 8%, 6) = 10.5233; P = 4,000(10.5233). Result: P = $42,093.
Why A is tempting
Treated the gradient as a level annuity of G.
Why B is tempting
Multiplied the gradient by the number of years.
Why D is tempting
Kept only the first half of the gradient factor and dropped the n/(i(1+i)^n) correction.
Source NCEES FE Reference Handbook 10.6 — Engineering Economics, interest factor tables
Question 3 of 5
$200,000 has already been spent on a project. Completing it will cost another $480,000, and abandoning it now would recover $140,000 in salvage. The relevant net cost of continuing is most nearly:
Show the answer and the reasoning
Why B
Given: $480,000 to finish; $140,000 salvage if abandoned now. The $200,000 already spent is sunk. Relation: compare the two futures. Continuing costs the money to finish plus the salvage it gives up; abandoning costs nothing more. Substitute: net = 480,000 + 140,000. Result: net cost = $620,000.
Why A is tempting
Ignored the salvage value that would be forgone by continuing.
Why C is tempting
Added the sunk cost and ignored the salvage.
Why D is tempting
Included the money already spent. Sunk costs cannot be recovered and do not affect the decision going forward.
Source NCEES FE Reference Handbook 10.6 — Engineering Economics, interest factor tables
Question 4 of 5
A public project has annual benefits of $75,000, disbenefits of $45,000, an equivalent annual capital cost of $300,000 and operating costs of $26,000. Its benefit-cost ratio is most nearly:
Show the answer and the reasoning
Why B
Given: benefits $75,000, disbenefits $45,000, capital $300,000, O&M $26,000, all annual. Relation: B/C = (benefits - disbenefits)/(capital + O&M). Substitute: B/C = (75,000 - 45,000)/(300,000 + 26,000). Result: B/C = 0.0920, below one, so the project is not justified.
Why A is tempting
Moved the operating cost to the benefits side as a negative benefit. Conventional practice keeps agency costs in the denominator.
Why C is tempting
Left the operating and maintenance cost out of the denominator.
Why D is tempting
Ignored the disbenefits, which are deducted from the benefits side.
Source NCEES FE Reference Handbook 10.6 — Engineering Economics, interest factor tables
Question 5 of 5
A facility faces a 2% annual chance of a $3,250,000 loss. Insurance costs $95,000 per year. On an expected value basis, the annual advantage of buying the insurance is most nearly:
Show the answer and the reasoning
Why A
Given: annual probability 0.02 of a $3,250,000 loss; premium $95,000 per year, paid regardless. Relation: advantage = expected loss avoided - premium. Substitute: 0.02(3,250,000) - 95,000 = 65,000 - 95,000. Result: advantage = $-30,000; negative, so on expected value alone the premium costs more than the risk.
Why B is tempting
Reversed the sign of the comparison.
Why C is tempting
Reported the expected loss without deducting the premium.
Why D is tempting
Added the premium rather than subtracting it.
Source NCEES FE Reference Handbook 10.6 — Engineering Economics, interest factor tables
Four mistakes that cost the question
Each of these lands on an answer that is offered, so it costs the question outright. The minutes are our estimate of the time each one burns on top, against an average of 2.9 minutes a question (320 minutes for 110).
Costs the question and about 2 min
Counting money already spent
A sunk cost is the same under every alternative, so it cannot change the decision. Questions hand you one, such as the price of the old equipment or what a project has cost to date, because adding it in is the most natural slip, and the result is usually a choice. Compare only the cash flows that differ from here on.
Costs the question and about 3 min
Mismatching the interest rate and the period
The rate and n must share a period. Monthly payments at a nominal annual rate need the rate divided by 12 and n in months; a rate compounded more often than yearly has an effective annual value of (1 + r/m)^m − 1. Using the nominal rate as if it were effective lands a few percent off, which is close enough to be offered.
Costs the question and about 3 min
Starting a series or gradient in the wrong year
The P/A factor gives present worth one period before the first payment, and the P/G factor assumes a zero cash flow at the end of year 1 with the first increment at year 2. A series that starts at year 0, or a gradient whose first year is not zero, has to be split or shifted. Unshifted, the answer is one period of interest off.
Costs the question and about 2 min
Treating salvage as a cost, or annualising it with the wrong factor
In an equivalent uniform annual cost, the first cost is spread with A/P and the salvage, a receipt at the end of the life, comes off with A/F. Using A/P on both, or adding the salvage instead of subtracting it, gives answers within a few percent of each other, and each is offered.
Questions about Engineering Economics
Are interest factor tables provided on the FE Civil exam?
Yes. The FE Reference Handbook carries the factor formulas and tables of compound interest factors for a range of rates. Knowing the formulas as well lets you handle a rate the tables do not list and check a table read you are unsure of.
What is the difference between nominal and effective interest?
A nominal annual rate is the periodic rate times the number of periods in a year; the effective rate includes the compounding, (1 + r/m)^m − 1. A 10% nominal rate compounded semiannually is 10.25% effective. Use the effective rate for the period in which the cash flows actually occur.
Which method should I use when a question does not name one?
Present worth, annual worth and future worth rank alternatives the same way when applied correctly, so use whichever needs the fewest factors for the cash flows given. Alternatives with different lives are easiest compared on annual worth. Benefit-cost and rate-of-return comparisons between alternatives have to be made on the increment between them.
How is a benefit-cost ratio set up?
Conventionally as benefits minus disbenefits, divided by costs, all expressed as present or annual worth at the same rate. Disbenefits are losses to the public, so they come off the benefits rather than being added to the costs; the two conventions give different ratios, and a question that gives disbenefits is checking which you use.
How many study hours this area is worth · Using the handbook under the clock