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.
Present Worth (PW) Analysis
Discount every cash flow to time 0 using the MARR (Minimum Attractive Rate of Return), then sum:
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 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: 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:
- Order them by increasing first cost.
- Compute the incremental cash flow of the larger-cost option minus the smaller (defender vs. challenger).
- Find the ROR on that increment, Δi*.
- 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 value | Decision |
|---|---|
| B/C > 1 | Justified (benefits exceed costs) |
| B/C = 1 | Break-even |
| B/C < 1 | Not 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: 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
| Situation | Best method |
|---|---|
| Compare equal-life alternatives | Present worth (highest PW) |
| Compare unequal-life alternatives | Annual worth (no LCM needed) |
| Public/government project | Benefit-cost ratio (incremental ΔB/C) |
| Perpetual-life asset | Capitalized cost A/i |
| Rank by return, mutually exclusive | Incremental 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
| Term | Definition |
|---|---|
| Sunk cost | Already spent and unrecoverable; ignore it in any decision |
| Opportunity cost | Value of the best alternative forgone |
| Fixed cost | Independent of output (rent, insurance) |
| Variable cost | Scales with output (materials, direct labor) |
| Marginal cost | Cost of one more unit |
| Life-cycle cost | Acquisition 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)
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
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.
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:
How should a sunk cost be treated in an engineering economic analysis?
An asset costs $40,000 with a $4,000 salvage value after a 6-year life. What is the annual straight-line depreciation?
Two mutually exclusive designs have service lives of 4 and 6 years. To compare them fairly with present-worth analysis, you should: