FE Civil · Chapter 4 · 4–6 exam questions

FE Civil Engineering Economics

This chapter tests your ability to evaluate the economic viability of engineering projects using time-value-of-money principles, cost analysis, and depreciation methods.

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What the FE tests in Engineering Economics

Time Value of Money

Time value of money is the foundation of every infrastructure investment decision. When a city asks whether to replace a bridge now for $8M or do major rehabilitation in 10 years for $12M, you convert both options to present worth at the appropriate discount rate. On the FE, you need the nine standard interest factors and the skill to read factor tables quickly (pp. 229, 232–236).

Cost & Economic Analysis

Civil engineers compare alternatives constantly — asphalt vs. concrete over a 30-year life, pump station upgrade vs. status quo, highway widening vs. new interchange. The FE tests whether you can classify costs correctly, find break-even points, compute benefit-cost ratios, and use decision trees to handle uncertainty (pp. 230–231).

Depreciation & Finance

Depreciation determines the book value of infrastructure assets like treatment plants, fleet vehicles, and heavy equipment. The FE tests straight-line and MACRS methods, the inflation-adjusted interest rate, and basic bond/capitalized-cost calculations (pp. 230–231).

Key Engineering Economics formulas

$F = P(1+i)^n$
Future Worth (Single Payment)FE Handbook p. 229
$P = F(1+i)^{-n}$
Present Worth (Single Payment)FE Handbook p. 229
$A = P\frac{i(1+i)^n}{(1+i)^n - 1}$
Capital RecoveryFE Handbook p. 229
$B/C \geq 1$
Benefit-Cost RatioFE Handbook p. 230
$D_j = \frac{\text{Cost} - \text{Salvage}}{n}$
Straight-Line DepreciationFE Handbook p. 230
$d = i + f + (i \times f)$
Inflation-Adjusted Interest RateFE Handbook p. 230
$PW_{benefits} - PW_{costs} = 0$
Internal Rate of Return (IRR)FE Handbook p. 229

Sample Engineering Economics problems

Q1. A civil engineering firm sets aside 15,000 dollars in a reserve account earning 8% annual interest. If the money is left untouched, what will it be worth in 5 years?

Answer: 22,040 dollars

Explain it simply

This is the most basic time-value-of-money problem: you have money now ( $P$ ) and want to know what it grows to ( $F$ ). Use the compound amount factor $(F/P)$ . Multiply $15{,}000 \times (1.08)^5 = 15{,}000 \times 1.4693 = 22{,}040$ . Answer A (21,000) comes from using simple interest ( $15{,}000 \times 0.08 \times 5 = 6{,}000$ , then adding to 15,000 = 21,000). Answer C uses the wrong number of years or the wrong rate. Answer D overshoots, possibly by applying the rate twice somewhere.

Q2. A public works department takes out a loan of 200,000 dollars for a new dump truck at 10% annual interest. The loan will be repaid in 8 equal annual payments at the end of each year. What is the annual payment amount?

Answer: 37,490 dollars

Explain it simply

You have a present amount (the loan) and need to find equal annual payments that pay it off. This is the capital recovery factor $(A/P)$ . Look up $(A/P,\, 10\%,\, 8) = 0.18744$ in the 10% table and multiply by 200,000. Answer A (25,000) divides 200,000 by 8 and ignores interest entirely. Answer C might use $(A/F)$ instead of $(A/P)$ — that is for a sinking fund, not a loan. Answer D uses the wrong factor or adds interest on top of the payment.

These are 2 of 1,126 problems across all 15 chapters. The full bank, lessons, mastery tracking, and timed exam simulation live inside the app.

Common Engineering Economics mistakes on the FE

Forgetting to draw the cash flow diagram — most errors come from misplacing payments in time.
Using the wrong interest factor table (P/A vs A/P, F/A vs A/F) — always double-check which conversion you need.
Including sunk costs in economic analysis — sunk costs are irrelevant to future decisions.
Confusing MARR (minimum attractive rate of return) with the actual rate of return.
Not converting between nominal and effective interest rates when compounding periods differ.
IRR is the rate where net present worth = 0; accept a project only if IRR ≥ MARR.
For mutually exclusive alternatives, use INCREMENTAL rate-of-return analysis — do not just pick the highest individual IRR.

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