2.4 Engineering Economics
Key Takeaways
- Engineering Economics is 3-5 of 110 questions and is almost entirely table-driven: pick the right factor from the FE Handbook, then plug in i and n.
- Single-payment future worth is F = P(1+i)^n; present worth is P = F(1+i)^-n, which is the (P/F, i, n) factor.
- The capital-recovery factor converts a present amount to a uniform series: (A/P, i, n) = i(1+i)^n / [(1+i)^n - 1].
- Compare alternatives over equal lives with present worth, annual worth, or rate of return; the chosen project maximizes worth or has ROR above the MARR.
- Straight-line depreciation is (cost - salvage)/life per year; MACRS uses IRS percentage tables, not a derived formula.
Time value of money
Engineering Economics is 3-5 of 110 questions and among the most learnable areas because the FE Reference Handbook provides every factor. A dollar today is worth more than a dollar later, so cash flows are moved to a common point in time using the interest rate i per period and the number of periods n.
The core relationship is single-payment compounding: a present amount P grows to a future amount F as
F = P(1 + i)^n
Inverting gives the present worth of a future amount, the (P/F) factor:
P = F(1 + i)^-n, so (P/F, i, n) = (1 + i)^-n
NCEES uses standard factor notation. (X/Y, i, n) reads "find X given Y." You locate the numeric factor in the Handbook table for the given i and n, multiply by the known quantity, and read off the unknown.
| Factor | Name | Formula |
|---|---|---|
| (F/P, i, n) | Single-payment compound amount | (1+i)^n |
| (P/F, i, n) | Single-payment present worth | (1+i)^-n |
| (A/P, i, n) | Capital recovery | i(1+i)^n / [(1+i)^n - 1] |
| (P/A, i, n) | Uniform-series present worth | [(1+i)^n - 1] / [i(1+i)^n] |
| (F/A, i, n) | Uniform-series compound amount | [(1+i)^n - 1] / i |
| (A/F, i, n) | Sinking fund | i / [(1+i)^n - 1] |
Here A is a uniform end-of-period annuity payment. The (A/P) capital-recovery factor is the one you use to turn an upfront equipment cost into an equivalent annual cost.
Comparing alternatives
Three equivalent decision methods appear on the FE:
- Present worth (PW): discount all cash flows to time zero; choose the option with the greatest PW (or least PW of cost). Alternatives must be compared over equal lives.
- Annual worth (AW): convert all cash flows to an equivalent uniform annual amount; useful when lives differ because it normalizes per year.
- Rate of return (ROR): the interest rate that makes PW = 0. Accept a project if its ROR exceeds the minimum attractive rate of return (MARR).
Benefit-cost ratio (B/C) is used for public projects: accept when B/C >= 1. Breakeven analysis finds the volume or time where two alternatives have equal cost.
Depreciation
Depreciation spreads an asset's cost over its useful life for accounting and tax purposes.
- Straight-line (SL): equal charge each year, D = (cost - salvage) / useful life.
- MACRS (Modified Accelerated Cost Recovery System): the U.S. tax method using IRS percentage tables by property class. There is no closed-form factor; you read the year's percentage from the table and multiply the original basis. Salvage is ignored under MACRS.
For the FE, recognize which method the stem names and apply the matching rule. Mixing SL with a declining-balance or MACRS expectation is a common distractor.
Inflation and effective rates
A nominal annual rate r compounded m times per year gives an effective annual rate i_eff = (1 + r/m)^m - 1. When inflation matters, distinguish the market (nominal) rate from the real rate; the FE usually states which to use.
You will receive $10,000 in 5 years. At an interest rate of 8% per year, what is its present worth? Use (P/F, 8%, 5).
A machine costs $50,000, has a $5,000 salvage value, and a 10-year life. What is the annual straight-line depreciation?