Engineering Economics is 5-8 of the 110 questions on the FE Civil exam and is the most formulaic area on the test. Every problem reduces to placing cash flows on a timeline and applying a factor from the handbook interest tables. Candidates lose these points to sign errors and to picking the wrong factor, not to conceptual difficulty.
The problems below cover the patterns that actually recur: present worth comparison of two alternatives, equivalent uniform annual cost, a gradient series, straight-line and MACRS depreciation, and a benefit-cost ratio on a public project. Each solution shows the timeline first, then the factor notation, then the arithmetic.
Exam weight: NCEES lists Engineering Economics at 5-8 questions of the 110-question FE Civil exam. Work each problem below on paper first, then reveal the worked solution — reading a solution you have not attempted builds recognition, not recall.
What NCEES Tests in Engineering Economics
The specification covers time value of money, cost estimating, economic analyses, and depreciation. In practice that means single payment, uniform series, and gradient factors used to move money between present, future, and annual equivalents. Expect alternative comparison by present worth, future worth, or annual worth, plus rate of return and payback period. Public infrastructure framing is common, so benefit-cost ratio appears more often here than on other FE disciplines.
Depreciation questions center on straight-line and MACRS, with the recovery period tables printed in the handbook. Cost estimating covers escalation, inflation adjustment, and the difference between nominal and effective interest rates when compounding is not annual. Break-even analysis shows up as a crossover point between two alternatives. Numbers are usually clean because the factors do the work.
5 Free Engineering Economics Practice Problems
Each engineering economics problem below comes from the FEExamClick FE Civil question bank, with worked solutions, and matches the style, difficulty and format of the real exam. Attempt each one under a three-minute limit — roughly the pace the exam demands.
Problem 1 — B. Cost (e.g., fixed, variable, direct and indirect labor, incremental, average, sunk)
An asset costs \(\$12{,}000\), has a salvage value of \(\$2{,}000\), and a life of 5 years. Using straight-line depreciation, what is the total depreciation after 3 years?
Answer: C) \(\$6{,}000\)
Step 1 – Identify given information:
| Initial cost (C) | \( \$12{,}000 \) |
| Salvage value (S) | \( \$2{,}000 \) |
| Useful life (n) | \( 5 \text{ years} \) |
Find total depreciation after 3 years using straight-line method.
Step 2 – Straight-line depreciation formula: From the FE Handbook:
$$D = \frac{C - S}{n}$$
Straight-line depreciation spreads the depreciable amount equally over the useful life.
Step 3 – Substitute and solve:
\( D = \frac{12{,}000 - 2{,}000}{5} = \frac{10{,}000}{5} = \$2{,}000\text{/yr} \)
Total depreciation after 3 years:
\( D_{\text{total}} = 3 \times 2{,}000 = \$6{,}000 \)
Step 4 – Answer: The total depreciation after 3 years is \( \$6{,}000 \), matching option (C). Option (B) \(\$4{,}000\) represents only 2 years of depreciation, and (D) \(\$8{,}000\) represents 4 years.
Option (A) is the annual straight-line charge of \(\$2{,}000\)/yr reported as if it were the accumulated depreciation - the classic annual-versus-total confusion.
Correct answer: (C)
Problem 2 — A. Time value of money (e.g., equivalence, present worth, equivalent annual worth, future worth, rate of return)
A precast concrete batching plant costs \(\$120{,}000\). The owner finances it at 7% annual interest for 20 years with uniform annual payments. Its cash flow diagram is shown. What is the annual payment amount?
Use: \(A = P\left[\frac{i(1+i)^n}{(1+i)^n - 1}\right]\)
Answer: D) \(\$11{,}327\)
Handbook: FE Reference Handbook 10.6, Engineering Economics, p. 235 (PDF viewer: page 241). The factor table on that page lists
Capital Recovery — converts to A given P — symbol (A/P, i%, n)
with exactly the formula quoted in the problem. This is a direct table lookup; you do not have to derive anything.
Note on the factor tables. The Handbook prints numerical interest tables on pp. 238–242, but only for i = 0.5, 1, 1.5, 2, 4, 6, 8, 10, 12 and 18%. There is no 7% table, so this factor must be evaluated from the formula rather than looked up.
Step 1 – Read the diagram. One outflow of \(P = \$120{,}000\) at year 0, and 20 equal unknown inflows \(A\) at the ends of years 1 through 20. Interest is 7% per year.
Step 2 – Evaluate the factor.
$$(1.07)^{20} = 3.86968$$
$$(A/P, 7\%, 20) = \frac{0.07(3.86968)}{3.86968 - 1} = \frac{0.27088}{2.86968} = 0.094393$$
Step 3 – Multiply.
$$A = 120{,}000 \times 0.094393 = \mathbf{\$11{,}327}$$
Step 4 – Answer: (D).
Sanity check. Twenty payments of \(11{,}327\) total \(\$226{,}540\) on a \(\$120{,}000\) loan, so about \(\$106{,}500\) is interest over 20 years. That is the right order of magnitude for 7% money held that long, and it confirms the payment must sit well above the \(\$6{,}000\) of bare principal.
Step 5 – Eliminate the distractors. The three wrong options are the three ways to get the interest wrong, and they bracket the answer:
(A) \(\$6{,}000 = P/n = 120{,}000/20\). This repays the principal and charges no interest at all. It has to be too small.
(B) \(\$8{,}400 = P \times i = 120{,}000(0.07)\). This is an interest-only payment: the balance never falls, and after 20 years you still owe the full \(\$120{,}000\).
(C) \(\$14{,}400 = P/n + P \times i\), the sum of (A) and (B). It charges 7% on the original balance every single year, as though nothing were ever paid off. Because the balance actually amortizes, the true payment must fall between (B) and (C) — and \(\$11{,}327\) does.
Problem 3 — D. Uncertainty (e.g., expected value and risk)
A sensitivity analysis on a toll road project yields the following results, plotted in the figure. The base-case NPV is \(\$2{,}500{,}000\).
| Variable | Change | Resulting NPV |
| Pavement material cost | +10% | \(\$1{,}300{,}000\) |
| Toll revenue | -25% | \(\$1{,}000{,}000\) |
| Maintenance cost | +20% | \(\$2{,}100{,}000\) |
Answer: A) Pavement material cost, with a sensitivity ratio of 4.8
Handbook: not covered. The terms "sensitivity ratio", "spider" and "tornado" appear zero times in FE Reference Handbook 10.6, and so does "NPV" (p. 235 calls the quantity \(P\), present worth). The closest heading, Breakeven Analysis on p. 236, states the one-variable-at-a-time idea in words but gives no formula. The stem therefore defines the ratio for you.
Step 1 – Compute each percentage change in NPV from the \(\$2.5\) M base:
$$\text{Material cost:}\quad \frac{2.5 - 1.3}{2.5} = 48\%$$
$$\text{Toll revenue:}\quad \frac{2.5 - 1.0}{2.5} = 60\%$$
$$\text{Maintenance cost:}\quad \frac{2.5 - 2.1}{2.5} = 16\%$$
Step 2 – Divide each by the change in its own variable. This is the step the whole item turns on: the three variables were not probed by the same amount.
$$\text{Material cost:}\quad \frac{48\%}{10\%} = \mathbf{4.8}$$
$$\text{Toll revenue:}\quad \frac{60\%}{25\%} = 2.4$$
$$\text{Maintenance cost:}\quad \frac{16\%}{20\%} = 0.8$$
Step 3 – Answer: the highest ratio is the pavement material cost at 4.8 — (A).
The trap, and what the figure is really showing. The toll revenue caused the biggest NPV drop in raw dollars: \(\$1.5\) M against the material cost's \(\$1.2\) M. Its line reaches the lowest point on the chart. But it took a 25% move to do that, while the material cost did nearly as much damage with a 10% move. Per point of change, the material cost is twice as dangerous.
On a spider diagram this is simply slope. The toll-revenue line goes further; the material-cost line goes steeper. Sensitivity is the steepness, not the reach — so read the gradient of each ray, never the depth of its endpoint.
A caution on the method. A high sensitivity ratio identifies the variable the NPV reacts to most, not necessarily the greatest risk. Risk is sensitivity combined with how much the variable is actually likely to move. A steep but stable input can matter less than a shallow but volatile one — which is why spider charts are a screening tool, not a decision.
Step 4 – Eliminate the distractors:
(B) 2.4 is the toll revenue's ratio, correctly computed. It is the right number for the wrong variable, and it is what you choose if you rank by the size of the NPV drop instead of by the ratio.
(C) 0.8 is the maintenance cost's ratio, also correct. A ratio below 1 means the NPV moves proportionally less than the input, so maintenance is the least threatening of the three, not the most.
(D) 0.21 inverts the material-cost ratio: \(10/48\) instead of \(48/10\). The right variable, the division upside down. A sensitivity ratio above 1 means amplification; getting 0.21 for the most sensitive input should look wrong immediately.
Problem 4 — A. Time value of money (e.g., equivalence, present worth, equivalent annual worth, future worth, rate of return)
Payments occur at the end of each year for 5 years: \(\$1{,}000\) in year 1 increasing by \(\$200\) each year. If \( i = 8\% \), what is the present worth?
Answer: B) \(\$5{,}470\)
Step 1 – Identify given information:
| Base annual payment (A) | \( \$1{,}000 \) |
| Annual gradient (G) | \( \$200\text{/yr} \) |
| Interest rate (i) | \( 8\% \) |
| Number of years (n) | \( 5 \) |
Payments: \(\$1{,}000\); \(\$1{,}200\); \(\$1{,}400\); \(\$1{,}600\); \(\$1{,}800\).
Step 2 – Arithmetic gradient present worth formula: From the FE Handbook, a gradient series is split into a uniform base plus an increasing gradient:
$$P = A(P/A,\, i,\, n) + G(P/G,\, i,\, n)$$
Step 3 – Substitute and solve:
Look up at \( i = 8\%,\; n = 5 \): \( (P/A) = 3.9927 \), \( (P/G) = 7.3724 \)
$$P = 1{,}000(3.9927) + 200(7.3724)$$
\( P = 3{,}993 + 1{,}474 = \$5{,}467 \)
Step 4 – Answer: The present worth is approximately \( \$5{,}470 \), matching option (B). The gradient factor \( (P/G) \) accounts for the increasing portion only; the base amount is handled separately by \( (P/A) \). The gradient starts at zero in year 1 and increases by \( G \) each year.
Correct answer: (B)
Problem 5 — A. Time value of money (e.g., equivalence, present worth, equivalent annual worth, future worth, rate of return)
A project has an NPV of \(\$5{,}385\) over \( 4 \) years at \( i = 10\% \). What is the equivalent annual worth (EAW)?
Answer: A) \(\$1{,}700\)/yr
Step 1 – Identify given information:
| NPV of project | \( \$5{,}385 \) |
| Interest rate (i) | \( 10\% \) |
| Number of years (n) | \( 4 \) |
Step 2 – Equivalent Annual Worth formula: EAW converts a lump-sum NPV into an equivalent uniform annual value using the capital recovery factor:
$$EAW = NPV \times (A/P,\, i,\, n)$$
Step 3 – Substitute and solve:
Look up: \( (A/P,\, 10\%,\, 4) = 0.3155 \)
\( EAW = 5{,}385 \times 0.3155 = \$1{,}699\text{/yr} \)
Step 4 – Answer: The EAW is approximately \( \$1{,}700\text{/yr} \), matching option (A). This means the project is equivalent to receiving \( \$1{,}700 \) per year above the required 10% return. EAW is especially useful when comparing alternatives with different lifespans.
Option (B) is what you get by spreading the principal evenly over four years and adding a full year's interest on the entire present worth each year, instead of applying the capital recovery factor.
Correct answer: (A)
Using the FE Reference Handbook for Engineering Economics
The Engineering Economics section ends with several pages of interest factor tables at standard rates. Learn the factor notation before exam day, because searching for a plain phrase will not find them: the table headers read (P/F, i%, n), (A/P, i%, n), and so on. Read the notation as what you want over what you have. If the problem uses a rate not tabulated, the closed-form equations sit just above the tables.
Four Mistakes That Cost Points
- Mixing period length and interest period When a problem gives an annual rate but monthly payments, the interest rate and the number of periods must both convert. Using 6 percent with 60 monthly periods instead of 0.5 percent per month produces a confidently wrong answer that often appears as a choice. Convert the rate and n together, every time.
- Confusing nominal and effective interest rates Twelve percent compounded monthly is not twelve percent effective. Problems state the compounding frequency for a reason, and the effective annual rate formula is in the handbook. Whenever a problem mentions compounding more often than annually, compute the effective rate before comparing alternatives or the comparison itself is invalid.
- Getting the gradient series timing wrong The arithmetic gradient factor assumes the first gradient increment occurs at the end of period two, not period one. Candidates routinely apply the factor to a series that starts increasing immediately. Draw the cash flow diagram and separate the uniform base amount from the gradient before selecting any factor.
- Comparing alternatives with unequal lives Present worth comparison requires equal study periods. When one alternative lasts 5 years and another 10, you must use the least common multiple with repeated cycles, or switch to equivalent uniform annual cost. Comparing raw present worths across unequal lives is the single most common setup error in alternative selection problems.
Frequently Asked Questions
How many engineering economics questions are on the FE Civil exam?
NCEES specifies 5-8 questions from Engineering Economics out of 110. Because the problems follow a small number of repeating patterns, this is one of the highest return areas per hour of study. Candidates who drill twenty or thirty problems typically move from unreliable to consistently correct on this material.
Do I need to memorize the interest factor formulas?
No. Both the closed-form equations and the tabulated factors are in the handbook. What you must know cold is the notation, so that you can translate a problem into (P/A, i%, n) without hesitating. The lookup takes seconds once you know which factor you need. Deciding which factor you need is the actual skill being tested.
Which depreciation methods appear on the FE Civil exam?
Straight-line and MACRS are the two that matter. Straight-line is a one-line calculation from cost, salvage value, and useful life. MACRS requires the recovery period and the percentage table, both of which are in the handbook. Know the half-year convention in the first year, since that is where MACRS problems most often catch candidates.
How is benefit-cost ratio tested on the FE Civil exam?
Usually on a public works scenario: a bridge, a road improvement, or a drainage project with annual benefits, annual costs, and an initial capital cost. Convert everything to the same basis, typically annual, then take benefits over costs. A ratio above one justifies the project. Watch for disbenefits, which belong in the numerator as a reduction, not added to costs.
Keep Going
Done with engineering economics? Work through the other FE Civil knowledge areas with more free practice problems, review the full FE Civil exam topic breakdown, or plan your preparation with our FE exam study guide and study timeline. Or test yourself against the clock with the free FE Civil practice exam: 20 timed questions in the CBT format.