FE Civil · Chapter 4 · 4–6 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.

What the FE tests

The Engineering Economics skills NCEES checks

01

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).

02

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).

03

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).

Reference

Key Engineering Economics formulas

  • F=P(1+i)nF = P(1+i)^nFuture Worth (Single Payment)FE Handbook p. 229
  • P=F(1+i)nP = F(1+i)^{-n}Present Worth (Single Payment)FE Handbook p. 229
  • A=Pi(1+i)n(1+i)n1A = P\frac{i(1+i)^n}{(1+i)^n - 1}Capital RecoveryFE Handbook p. 229
  • B/C1B/C \geq 1Benefit-Cost RatioFE Handbook p. 230
  • Dj=CostSalvagenD_j = \frac{\text{Cost} - \text{Salvage}}{n}Straight-Line DepreciationFE Handbook p. 230
  • d=i+f+(i×f)d = i + f + (i \times f)Inflation-Adjusted Interest RateFE Handbook p. 230
  • PWbenefitsPWcosts=0PW_{benefits} - PW_{costs} = 0Internal Rate of Return (IRR)FE Handbook p. 229
Try it

Sample Engineering Economics problems

Q1A 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?

22,040 dollars

Explain it simply

This is the most basic time-value-of-money problem: you have money now (PP) and want to know what it grows to (FF). Use the compound amount factor (F/P)(F/P). Multiply 15,000×(1.08)5=15,000×1.4693=22,04015{,}000 \times (1.08)^5 = 15{,}000 \times 1.4693 = 22{,}040. Answer A (21,000) comes from using simple interest (15,000×0.08×5=6,00015{,}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.

Q2A 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?

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)(A/P). Look up (A/P,10%,8)=0.18744(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)(A/F) instead of (A/P)(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.

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

Avoid these

Common Engineering Economics mistakes

  • 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.
FAQ

FE Civil Engineering Economics: common questions

How many FE Civil Engineering Economics questions are on the exam?

NCEES includes roughly 4–6 Engineering Economics questions on the computer-based FE Civil exam, which has 110 questions total over about six hours.

What Engineering Economics topics does the FE Civil exam cover?

The FE Civil exam tests Engineering Economics across: Time Value of Money, Cost & Economic Analysis, Depreciation & Finance.

Is FE Civil Engineering Economics hard?

Engineering Economics is very learnable for the FE. It rewards recognizing a handful of standard problem types and applying the right FE Reference Handbook formula — not deep theory. Practicing past-style problems is the fastest way to master it.

Where can I find free FE Civil Engineering Economics practice problems?

FE for Raccoons has free Engineering Economics lessons and practice problems with step-by-step explanations — part of a 1,126-problem bank covering all 15 FE Civil topics, free at fe4raccoons.com.

Study Engineering Economics the smart way

Bite-sized lessons, one-problem-at-a-time practice with instant feedback, and progress that adds up every day you study — built for the FE Civil exam.

Start practicing Engineering Economics