2.5 Engineering Economics
Key Takeaways
- Engineering Economics is 4-6 of 110 FE Civil questions; draw the cash-flow diagram first, then pick a factor from the NCEES Handbook 'Engineering Economics' interest tables.
- Single-payment present worth uses P = F(P/F, i, n) = F/(1+i)ⁿ; uniform-series present worth uses P = A(P/A, i, n).
- Capital recovery converts a present cost to equal annual cost: A = P(A/P, i, n); sinking fund converts a future target to annual deposits: A = F(A/F, i, n).
- Rate of return sets net present worth to zero and solves for i; accept a project when ROR ≥ MARR or net present worth ≥ 0 at the MARR.
- Benefit-cost ratio B/C ≥ 1.0 justifies a public project; straight-line depreciation is d = (cost − salvage)/N per year.
Draw the cash-flow diagram first
Engineering Economics is 4-6 questions, and they are nearly always cash-flow timing problems, not accounting cases. Put time zero on the left, mark the interest rate i per period, place each cost or benefit at its correct date, and assign signs before touching a factor. Initial cost is negative; annual savings positive; annual operating cost negative; salvage positive at end of life. If answer choices differ mainly by sign, the model is right but the direction is wrong; if they differ by about one period, a payment sits at the wrong year.
The standard interest factors
All factors are tabulated in the NCEES FE Reference Handbook 'Engineering Economics' chapter — you read a value from the table for a given i and n; you do not memorize the closed forms, but knowing them speeds checking:
| Need | Convert from | Factor | Closed form |
|---|---|---|---|
| Present worth | Single future F | (P/F, i, n) | 1/(1+i)ⁿ |
| Future worth | Single present P | (F/P, i, n) | (1+i)ⁿ |
| Present worth | Uniform series A | (P/A, i, n) | [(1+i)ⁿ − 1] / [i(1+i)ⁿ] |
| Annual from present | Present P | (A/P, i, n) | i(1+i)ⁿ / [(1+i)ⁿ − 1] |
| Annual from future | Future F | (A/F, i, n) | i / [(1+i)ⁿ − 1] |
| Future from series | Uniform series A | (F/A, i, n) | [(1+i)ⁿ − 1] / i |
A classic error is using (P/A) for a single future salvage value or (P/F) for a repeating annual cost. Another is mixing monthly cash flows with an annual interest rate without converting the period.
Decision methods
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Present worth (PW): move all cash flows to time zero. For mutually exclusive revenue alternatives, choose the largest PW; for equal-service cost alternatives, choose the lowest present cost.
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Annual worth (AW): best when lives differ or costs are naturally annual. Convert capital cost with (A/P), add annual costs, subtract salvage's annual equivalent (A/F).
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Rate of return (ROR): set net present worth = 0 and solve for i. Accept the project when ROR ≥ MARR (minimum attractive rate of return), equivalently when PW ≥ 0 at the MARR.
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Benefit-cost (B/C): used for public projects; accept when B/C ≥ 1.0 (present worth of benefits ÷ present worth of costs).
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Break-even: find the value (production volume, interest rate, life) that makes two alternatives equal in cost or makes PW = 0.
Depreciation
Straight-line depreciation per year is d = (initial cost − salvage value) / N, where N is the depreciable life. Book value declines linearly to the salvage value. The Handbook also lists MACRS and declining-balance, but straight-line is the most commonly tested on FE Civil.
Exam tip: match the interest period to the cash-flow period before using any factor. If a question gives a nominal annual rate compounded monthly, convert to an effective rate or use the monthly i with monthly n.
A culvert project costs $50,000 today and saves $12,000 per year for 5 years, with a $8,000 salvage value at year 5. At i = 8%, which Handbook factors give the net present worth? (P/A, 8%, 5) = 3.993; (P/F, 8%, 5) = 0.6806.
A grader costs $120,000, has a 6-year life, and a $24,000 salvage value. Using straight-line depreciation, what is the annual depreciation and the book value after 2 years?