Engineering economics is a section of every FE discipline, and much of it runs on eight interest factors. Below are the factor formulas in compact notation, a worked example, how much each discipline asks and 25 solved problems.
Updated Oct 29, 2026 · Not affiliated with NCEES.
problems on this page25
interest factors8
questions on FE Civil5–8
The interest factors
Notation: i is the interest rate per period, n the number of periods, P a present amount, F a future amount, A a uniform end-of-period amount and G the gradient, the amount by which a series grows each period. (X/Y, i, n) reads “X given Y”.
A/P is the capital recovery factor and A/F the sinking fund factor. Each pair is a reciprocal: (A/P) = 1/(P/A), (A/F) = 1/(F/A), (P/F) = 1/(F/P). On exam day you look these up in the FE Reference Handbook by name rather than typing symbols into its search box (how the search works).
Worked example: a cost that grows
Pump maintenance costs $1,200 in year 1 and rises by $300 a year through year 10. At 8% a year, what single amount today covers it? Split the series into a uniform base and a gradient.
FE Environmental: 5–8; O&M costs, depreciation, lease/buy/make and projects with unequal lives.
FE Chemical, where the section is called Economics: 4–6; optimal economic life, expected value and unequal lives.
Coming: FE Industrial and Systems gives it 9–14 questions, the most of any FE; FE Other Disciplines 6–9.
The same factors carry into the PE: on PE HVAC&R, economic analysis is part of supportive knowledge.
Where the points go
Timing of the gradient. P/G assumes the first gradient amount falls at the end of year 2. A real cost series rarely starts at zero, so its year-1 amount goes in as a uniform base with P/A, as in the example.
Signs. Write costs and receipts with opposite signs before you sum present worths; a single dropped sign flips an accept/reject answer.
Rounded factors. Factors read from a table are rounded. On a large cash flow two answer options can sit closer together than that rounding, so compute the factor from its formula when they are near.
25 engineering economics problems
0 / 25 answered · 0 correct
Question 1 of 25 · Engineering Economics
A wastewater plant upgrade is scheduled 4 years from today. Its cost today would be $150,000, and that cost escalates at 5% per year. The municipality uses a real MARR of 7%, and general inflation is 3% per year. What is the present worth of the upgrade cost?
Given
Cost today $150,000, escalating 5%/yr
Timing: year 4
Real MARR 7%, inflation 3%
Find
Present worth of the upgrade cost
Solution
Escalate to the year-4 nominal cost and discount at the nominal rate (FE Reference Handbook, Engineering Economics — Inflation):
P = 150,000 × 1.054/1.10214 = 182,326/1.4753 = $123,585
$101,673 discounts today’s cost at the nominal rate but never escalates it; the 5% escalation must be applied first.
$114,434 discounts today’s cost at the real rate, which would be right only if the cost escalated at exactly the general inflation rate.
The nominal rate is 1.07 × 1.03 − 1 = 10.21%, and $150,000 × (1.05/1.1021)4 = $123,585.
$124,531 approximates the nominal rate as 7% + 3% = 10% instead of using the Fisher relation.
$139,096 escalates the cost to nominal dollars but discounts it at the real 7%, mixing nominal cash flow with a real rate.
Question 2 of 25 · Engineering Economics
A laboratory analyzer bought for $80,000 is depreciated straight-line over 5 years to zero salvage. It is sold after 3 years for $40,000, and the tax rate is 25%. What is the after-tax cash flow from the sale?
Given
First cost $80,000
SL depreciation, 5 years, zero salvage
Sold at year 3 for $40,000
Tax rate 25%
Find
After-tax cash flow from the sale
Solution
Tax applies to the gain over book value (FE Reference Handbook, Engineering Economics — Depreciation, Book Value):
BV3 = 80,000 − 3(16,000) = $32,000
Tax = 0.25 × (40,000 − 32,000) = $2,000
After-tax cash = 40,000 − 2,000 = $38,000
$30,000 taxes the whole sale price (25% of $40,000); only the gain over book value is taxable.
$32,000 is the book value after 3 years, not the cash received.
Book value $80,000 − 3 × $16,000 = $32,000, gain $8,000, tax $2,000, so $40,000 − $2,000 = $38,000.
$40,000 is the sale price before the tax on the $8,000 gain.
$42,000 adds the tax on the gain instead of subtracting it, as if the sale produced a loss.
Question 3 of 25 · Engineering Economics
An environmental trust fund pays for long-term monitoring. Costs are $10,000 at the end of the first year and grow by 3% per year forever. The fund earns 5% per year. What endowment is needed today?
$125,000 adds the growth rate to the interest rate: $10,000/(0.05 + 0.03).
$200,000 is a level perpetuity $10,000/0.05; it ignores the growth of the payments.
$333,333 divides by the growth rate alone, $10,000/0.03.
$476,190 discounts the growing-perpetuity value one extra year, $500,000/1.05, although A1/(i − g) already gives the value today.
a growing perpetuity is worth A1/(i − g) = $10,000/0.02 = $500,000.
Question 4 of 25 · Engineering Economics
A machine (the defender) bought 5 years ago now has a market value of $10,000, a remaining life of 3 years, annual O&M costs of $5,000 and zero salvage. With a MARR of 10%, what is the equivalent uniform annual cost (EUAC) of keeping the defender?
Given
Market value $10,000
Remaining life 3 years, zero salvage
O&M $5,000/yr
MARR 10%
Find
EUAC of the defender
Solution
The defender’s first cost is its current market value (opportunity cost) (FE Reference Handbook, Engineering Economics — Factor Table):
(A/P, 10%, 3) = 0.1(1.1)3/(1.13 − 1) = 0.40211
EUAC = 10,000 × 0.40211 + 5,000 = $9,021
$5,000 counts only the O&M cost; keeping the machine also gives up its $10,000 market value.
$6,000 adds only one year’s interest on the market value; the value itself must also be recovered.
$8,021 uses the sinking-fund factor (A/F) instead of the capital-recovery factor (A/P) on the market value.
$8,333 recovers the market value by straight-line ($10,000/3) and ignores interest.
An asset with a cost basis of $50,000 and a 5-year life is depreciated by double declining balance (DDB). The estimated salvage value is $5,000. What is the book value at the end of year 3?
Given
Cost basis $50,000
Life 5 years
Salvage $5,000
Find
Book value at end of year 3 (DDB)
Solution
DDB rate 2/n applied to the start-of-year book value (FE Reference Handbook, Engineering Economics — Depreciation):
$6,480 is the book value after year 4; one DDB step too many.
$9,720 applies the DDB rate to the cost minus salvage ($45,000 × 0.63); DDB works on the full book value and ignores salvage until the floor is reached.
The DDB rate is 2/5 = 40%, so $50,000 × 0.63 = $10,800, still above the $5,000 salvage floor.
$18,000 is the book value after year 2.
$23,000 is straight-line book value: $50,000 − 3 × ($45,000/5).
Question 6 of 25 · Engineering Economics
A municipality sets up a sinking fund to accumulate $100,000 for equipment replacement in 5 years. The fund earns a nominal 6% per year compounded monthly, and deposits are made at the end of each month. What is the required monthly deposit?
$1,933 uses the capital-recovery factor (A/P) instead of the sinking-fund factor (A/F), adding interest on a present sum that does not exist.
$1,667 divides $100,000 by 60 deposits and ignores interest.
$1,478 finds an annual deposit at 6% for 5 years and divides it by 12, ignoring monthly compounding.
i = 0.5% per month and n = 60, so A = $100,000 × 0.005/(1.00560 − 1) = $1,433.
$1,426 is the deposit for beginning-of-month payments (an annuity due); the deposits here are at the end of each month.
Question 7 of 25 · Engineering Economics
Two pump systems are compared at 8% interest. System A costs $15,000, lasts 5 years, has a $2,000 salvage value and O&M of $1,000 per year. System B costs $25,000, lasts 10 years, has a $5,000 salvage value and O&M of $500 per year. What is the EUAC of the more economical system?
$4,226 is System B with its salvage value ignored: $25,000 × 0.14903 + $500.
$4,416 is System A’s EUAC, the more expensive of the two.
Question 8 of 25 · Engineering Economics
Two mutually exclusive filtration upgrades each have a 10-year life, and the MARR is 15%. System A costs $20,000 and saves $6,000 per year; System B costs $30,000 and saves $8,200 per year. Using incremental analysis, which system should be selected, and why?
Given
MARR 15%, n = 10 yr
A: $20,000, saves $6,000/yr
B: $30,000, saves $8,200/yr
Find
Selection and reason
Solution
Analyze the increment B − A (FE Reference Handbook, Engineering Economics — Rate-of-Return, Present Worth):
$19,802 discounts for 4 years instead of 5: $25,000/1.064.
$23,585 discounts for only one year: $25,000/1.06.
Question 11 of 25 · Engineering Economics
A firm compares two production processes. Process A has a life-cycle cost of $1,000,000 for 5,000 units; Process B has a life-cycle cost of $1,260,000 for 6,000 units. What is the incremental cost per additional unit of switching from A to B?
Given
A: $1,000,000 for 5,000 units
B: $1,260,000 for 6,000 units
Find
Incremental cost per additional unit
Solution
Incremental cost = change in total cost/change in output:
(1,260,000 − 1,000,000)/(6,000 − 5,000) = $260 per unit
This is the difference between the two average costs ($210 − $200), not the cost of the extra units.
$200 per unit is Process A’s average cost, $1,000,000/5,000.
$210 per unit is Process B’s average cost, $1,260,000/6,000; it spreads the extra cost over all units.
Incremental cost = ($1,260,000 − $1,000,000)/(6,000 − 5,000) = $260 per added unit.
Question 12 of 25 · Engineering Economics
A company deposits $2,000 at the end of each year for 8 years into a fund earning 5% per year. What is the fund’s value at the end of year 8?
A company invests $150,000 in equipment that saves $40,000 per year for 5 years. With a MARR of 12%, what is the net present worth of the investment?
Given
First cost $150,000
Savings $40,000/yr for 5 years
MARR 12%
Find
Net present worth
Solution
Present worth of the uniform series minus first cost (FE Reference Handbook, Engineering Economics — Factor Table):
NPW = −150,000 + 40,000 × 3.60478 = −$5,809
−$36,515 discounts all $200,000 of savings as a single amount at year 5 instead of as a uniform series.
−$150,000 + $40,000 × (P/A, 12%, 5) = −$150,000 + $40,000 × 3.60478 = −$5,809, so the project misses the MARR.
+$5,809 has the right magnitude but subtracts in the wrong order (PW of savings is less than the investment, not more).
+$50,000 adds undiscounted savings ($200,000) and ignores the time value of money.
Question 16 of 25 · Engineering Economics
A project costs $200,000 to install, yields gross benefits of $60,000 per year for 6 years and has annual operating costs of $15,000. Its salvage value at year 6 is $20,000. With a MARR of 9%, what is the conventional benefit-cost ratio, and should the project be accepted?
This leaves out the salvage value; its present worth ($11,925) belongs in the ratio, as a benefit or as a cost reduction (both give 1.05).
This leaves the operating costs out of the denominator ($281,079/$200,000); O&M is a cost of the project.
Question 17 of 25 · Engineering Economics
A project requires a $200,000 investment at year 0 and has a 5-year life. Annual revenue is $70,000 and annual operating cost is $20,000. Straight-line tax depreciation runs to a $10,000 salvage value at year 5, the tax rate is 30% and the MARR is 12%. What is the after-tax NPW, and should the project be accepted?
Given
First cost $200,000
Revenue $70,000/yr, cost $20,000/yr, 5 years
SL depreciation to $10,000 salvage
Tax 30%, MARR 12%
Find
After-tax NPW and decision
Solution
After-tax cash flow with the depreciation shield (FE Reference Handbook, Engineering Economics — Depreciation, Taxation):
Two mutually exclusive projects have 5-year lives and a MARR of 10%. Project X: first cost $250,000, net benefit $70,000 per year, salvage $20,000. Project Y: first cost $180,000, net benefit $48,000 per year, salvage $10,000. What is the IRR of the increment X − Y, and which project should be chosen?
Given
MARR 10%, n = 5 yr
X: $250,000, $70,000/yr, S = $20,000
Y: $180,000, $48,000/yr, S = $10,000
Find
Incremental IRR and selection
Solution
Increment X − Y (FE Reference Handbook, Engineering Economics — Rate-of-Return):
ΔP = −$70,000, ΔA = +$22,000/yr, ΔS = +$10,000
0 = −70,000 + 22,000(P/A, i, 5) + 10,000(P/F, i, 5) → i = 19.9% > 10% → choose X
This is Y’s own IRR; picking the cheaper project ignores whether the extra investment in X earns the MARR.
−$70,000 + $22,000(P/A, i, 5) + $10,000(P/F, i, 5) = 0 at i = 19.9%, above the 10% MARR.
This drops the $10,000 difference in salvage values from the incremental cash flows.
This is X’s own IRR, not the IRR of the $70,000 increment.
Question 19 of 25 · Engineering Economics
Process equipment costs $250,000 and has a 10-year life with no salvage value. What is the annual straight-line depreciation?
$45,455 is the first-year sum-of-years’-digits charge, $250,000 × 10/55, not straight-line.
This is the first-year double-declining-balance charge (2/10 × $250,000), not straight-line.
$250,000 writes off the entire cost in one year.
Question 20 of 25 · Engineering Economics
Which of the following is a sunk cost in an engineering economic analysis?
Solution
A sunk cost has already been incurred and cannot be recovered by any choice, so it does not affect which alternative is better.
Future maintenance can still change with the decision, so it is a relevant cost.
A past expenditure that cannot be recovered whatever is decided is sunk and is left out of the comparison.
Future price changes affect the alternatives and belong in the analysis.
Warranty claims lie in the future and may differ between alternatives.
Question 21 of 25 · Engineering Economics
An investment of $150,000 today returns $50,000 in year 1, $60,000 in year 2, $70,000 in year 3 and $40,000 in year 4. What is the internal rate of return, to the nearest tenth of a percent?
Given
P = $150,000 at year 0
Inflows $50,000, $60,000, $70,000, $40,000 in years 1–4
Find
IRR (nearest 0.1%)
Solution
Find the rate where NPW = 0 (FE Reference Handbook, Engineering Economics — Rate-of-Return):
NPW(17%) = +$1,618, NPW(18%) = −$1,300
IRR ≈ 17% + 1,618/(1,618 + 1,300) × 1% = 17.6%
9.2% leaves out the year-4 inflow of $40,000 when solving for the rate.
10.0% treats the $220,000 total as one amount received at year 4: (220/150)1/4 − 1.
11.7% is the average annual profit ($70,000/4) over the investment, a simple return that ignores timing.
NPW is +$1,618 at 17% and −$1,300 at 18%; interpolating gives 17% + 1,618/2,918 × 1% = 17.6%.
Question 22 of 25 · Engineering Economics
A capital outlay of $150,000 yields cash inflows of $40,000 in year 1, $50,000 in year 2, $30,000 in year 3 and $30,000 in year 4, with no salvage value. The MARR is 8%. What are the simple payback and the discounted payback?
Discounted payback does not occur within the 4-year life.
Cumulative cash reaches $150,000 at year 4, while the present worth of all four inflows at 8% is $125,770, so the outlay is never recovered.
Simple payback lands exactly at year 4, but the discounted inflows total only $125,770, short of $150,000.
Discounted payback can never be shorter than simple payback when the MARR is positive.
Three years recovers only $120,000; 3.0 years comes from dividing $150,000 by the largest single inflow.
Question 23 of 25 · Engineering Economics
Which type of cost increases in proportion to the number of units produced?
Solution
Cost behavior: variable cost = (cost per unit) × (units); fixed cost does not change with volume over the relevant range.
A sunk cost was spent in the past and does not change with future output.
A fixed cost stays the same over the relevant range of output, such as rent or insurance.
Variable costs, such as raw materials or energy per unit, rise in direct proportion to output.
Overhead is indirect cost allocated to products; much of it is fixed and does not rise unit by unit.
Question 24 of 25 · Engineering Economics
A company can buy equipment for $50,000 or lease it with end-of-year payments of $10,000 for 7 years. The nominal interest rate is 12% compounded monthly. What is the present worth of the lease payments?
Given
A = $10,000 at each year-end, n = 7
r = 12%/yr compounded monthly
Find
Present worth of the lease payments
Solution
Convert to the effective rate for the payment period (FE Reference Handbook, Engineering Economics — Nonannual Compounding):
The effective annual rate is 1.0112 − 1 = 12.68%, and $10,000 × (P/A, 12.68%, 7) = $44,667.
$45,638 discounts the annual payments at the nominal 12% and ignores monthly compounding.
$55,824 uses 6% (the semiannual rate) as if it were the annual rate.
$67,282 applies the 1% monthly rate to annual payments, as if each year were one month.
$70,000 adds the seven payments with no discounting.
Question 25 of 25 · Engineering Economics
Parameter
Value
Original purchase price
$40,000
Market value today
$12,000
Market value next year
$9,000
O&M cost next year
$4,500
A company must decide whether to keep its current asset (the defender) one more year or replace it now. The MARR is 10%. Using the data in the table, what is the marginal cost of keeping the defender for one more year?
Given
Market value today $12,000
Market value next year $9,000
O&M cost next year $4,500
MARR 10%
Original purchase price $40,000
Find
Marginal cost of keeping the defender one more year
Solution
Marginal cost = loss in market value + interest forgone on today’s value + next year’s O&M:
(12,000 − 9,000) + 0.10 × 12,000 + 4,500 = $8,700
The $40,000 purchase price is sunk.
$5,700 leaves out the $3,000 loss in market value during the year.
$7,500 leaves out the interest forgone on the $12,000 that selling now would release.
$8,400 charges interest on next year’s value ($9,000) instead of on today’s value, which is the money tied up.
Yes, on all seven specifications, from 4–6 questions on FE Mechanical and FE Chemical to 9–14 on FE Industrial and Systems. On FE Chemical it is called Economics.
Is a financial calculator allowed for the economics items?
No. Only models on the NCEES list are allowed, and none of them is a financial calculator; the factors are computed from the formulas or read from the Handbook (approved calculators).
Do the interest-factor formulas have to be memorized?
No. They are in the FE Reference Handbook on screen. What saves time is knowing which factor a problem needs and its name, so the search box finds it on the first try.