Economic Analysis Methods

Key Takeaways

  • Present worth (PW) discounts every cash flow to time 0 at the MARR; pick the highest PW, and accept a lone project if PW ≥ 0.
  • Annual worth (AW = PW × A/P) restates a project as a level yearly amount and compares unequal-life alternatives directly, with no least-common-multiple study period needed.
  • Rate of return (ROR/IRR) is the i* that drives PW to zero; accept when ROR ≥ MARR, and use incremental ΔROR — never raw ROR — to rank mutually exclusive options.
  • Benefit-cost ratio B/C = PW(benefits)/PW(costs) governs public projects; B/C > 1 means justified, and ranking also uses incremental ΔB/C.
  • Sunk costs are ignored; only present and future cash flows (and opportunity costs) affect the decision.
  • Straight-line depreciation is (cost − salvage)/life; MACRS uses fixed IRS percentages, ignores salvage, and is tabulated in the FE Reference Handbook.
Last updated: June 2026

Present Worth (PW) Analysis

Discount every cash flow to time 0 using the MARR (Minimum Attractive Rate of Return), then sum: PW=t=0nCFt(1+i)tPW = \sum_{t=0}^{n} CF_t\,(1+i)^{-t}

Decision rules. For a single project, accept if PW ≥ 0. Among mutually exclusive alternatives, choose the highest PW (least negative cost, or greatest net benefit). A subtle but heavily tested rule: when alternatives have different service lives, a direct PW comparison is invalid. You must evaluate over a common horizon — the least common multiple (LCM) of the lives (e.g. a 3-year and a 4-year option are compared over 12 years, repeating each project) — or switch to annual worth, which sidesteps the problem.

Annual Worth (AW) Analysis

Convert the project's entire cash-flow stream to one equivalent end-of-year amount: AW=PW×(A/P,i%,n)AW = PW \times (A/P, i\%, n)

AW is often the cleanest method because it automatically handles unequal lives — comparing $/year requires no LCM study period, provided each alternative repeats indefinitely. Decision rule: pick the alternative with the highest AW (for costs, the least-negative AW). AW also reads naturally to a client: "this design costs $8,400 per year."

Rate of Return (ROR / IRR)

The rate of return i* is the interest rate that makes present worth exactly zero: 0=t=0nCFt(1+i)t0 = \sum_{t=0}^{n} \frac{CF_t}{(1+i^*)^t} For a single project, accept if ROR ≥ MARR. The trap: you cannot simply pick the alternative with the highest individual ROR. A small project can show a huge ROR yet leave money on the table.

Incremental (ΔROR) Analysis

To rank mutually exclusive alternatives correctly:

  1. Order them by increasing first cost.
  2. Compute the incremental cash flow of the larger-cost option minus the smaller (defender vs. challenger).
  3. Find the ROR on that increment, Δi*.
  4. If Δi* ≥ MARR, the extra investment earns its keep — keep the higher-cost option; otherwise keep the lower-cost one. The same incremental logic applies to ΔB/C ratios for public projects.

Benefit-Cost (B/C) Analysis

Standard for government/public-sector work: B/C=PW(benefits)PW(costs)B/C = \frac{PW(\text{benefits})}{PW(\text{costs})}

B/C valueDecision
B/C > 1Justified (benefits exceed costs)
B/C = 1Break-even
B/C < 1Not justified

Watch placement: disbenefits may be subtracted from benefits (numerator) or added to costs (denominator) — the two conventions give different ratios, so follow the problem's instruction.

Break-Even Analysis

Find the quantity Q where two alternatives cost the same (or revenue equals cost). Example: Machine A costs $10,000 with $5/unit variable cost; Machine B costs $20,000 with $3/unit. Setting total costs equal: 10,000 + 5Q = 20,000 + 3Q → 2Q = 10,000 → Q = 5,000 units. Below 5,000 units choose A (lower fixed cost dominates); above 5,000 units choose B (lower variable cost wins).

Capitalized Cost

For assets assumed to last forever (dams, perpetual endowments, roadways with indefinite service), the present worth of a perpetual annual cost A at rate i is the capitalized cost: CC=AiCC = \frac{A}{i} This is the limit of (P/A) as n → ∞. Example: maintaining a structure at $8,000/year forever, at i = 8%, has a capitalized cost of 8,000/0.08 = $100,000. Capitalized cost is the standard way to compare very-long-life public alternatives without an LCM study period.

Choosing the Right Method

SituationBest method
Compare equal-life alternativesPresent worth (highest PW)
Compare unequal-life alternativesAnnual worth (no LCM needed)
Public/government projectBenefit-cost ratio (incremental ΔB/C)
Perpetual-life assetCapitalized cost A/i
Rank by return, mutually exclusiveIncremental rate of return (ΔROR ≥ MARR)

All four worth-based methods give the same accept/reject decision for a single project when applied consistently at the same MARR — PW ≥ 0 is equivalent to AW ≥ 0, to ROR ≥ MARR, and to B/C ≥ 1. They can disagree only when misused for ranking mutually exclusive options, which is exactly why incremental analysis exists.

Cost Terminology

TermDefinition
Sunk costAlready spent and unrecoverable; ignore it in any decision
Opportunity costValue of the best alternative forgone
Fixed costIndependent of output (rent, insurance)
Variable costScales with output (materials, direct labor)
Marginal costCost of one more unit
Life-cycle costAcquisition through operation, maintenance, and disposal

The most-tested item is sunk cost: money already paid for a machine you now own does not change which future option is best — only present and future cash flows (and opportunity costs) do.

Depreciation

Depreciation spreads an asset's cost over its useful life for tax accounting.

Straight-Line (SL)

D=CostSalvageUseful lifeD = \frac{\text{Cost} - \text{Salvage}}{\text{Useful life}} Equal depreciation each year. Example: a $40,000 asset with $4,000 salvage over 6 years gives D = (40,000 − 4,000)/6 = $6,000/year.

MACRS (Modified Accelerated Cost Recovery System)

The U.S. tax-default method. It applies fixed IRS percentages to the original basis for each year of a stated recovery period (3, 5, 7, … years), ignores salvage value, and front-loads deductions (accelerated). The FE Reference Handbook tabulates the MACRS factors. Year-n depreciation is D_n = (rate_n) × (initial cost).

Book Value

BVn=Initial costk=1nDkBV_n = \text{Initial cost} - \sum_{k=1}^{n} D_k Book value is what remains on the books after accumulated depreciation; under SL it declines linearly toward salvage, while under MACRS it drops faster early.

Test Your Knowledge

A public project costs $50,000 now and yields $12,000 of benefits per year for 6 years. At MARR = 10%, the benefit-cost ratio is approximately:

A
B
C
D
Test Your Knowledge

How should a sunk cost be treated in an engineering economic analysis?

A
B
C
D
Test Your Knowledge

An asset costs $40,000 with a $4,000 salvage value after a 6-year life. What is the annual straight-line depreciation?

A
B
C
D
Test Your Knowledge

Two mutually exclusive designs have service lives of 4 and 6 years. To compare them fairly with present-worth analysis, you should:

A
B
C
D