2.4 Engineering Economics

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.

Here A is a uniform end-of-period annuity payment. The (A/P) capital-recovery factor turns an upfront equipment cost into an equivalent annual cost.

Worked example (present worth, P/F): You will receive $10,000 in 5 years; with i = 8% per year, what is its present worth? P = F(1+i)^-n = 10,000(1.08)^-5 = 10,000(0.6806) = $6,806. The (P/F, 8%, 5) factor is 0.6806, exactly what the Handbook table lists.

Worked example (uniform-series present worth, P/A): A maintenance contract pays $2,000 at the end of each year for 4 years at i = 6%. Its present worth uses (P/A, 6%, 4) = [(1.06)⁴ - 1]/[0.06(1.06)⁴] = (1.26248 - 1)/(0.06·1.26248) = 0.26248/0.075749 = 3.465. So PW = 2,000 × 3.465 = $6,930. The most common time-value error is mismatching the cash-flow timing to the factor: P/A and A/P assume end-of-period payments, so a payment made today (time zero) must be handled separately, not run through the annuity factor.

Comparing alternatives, rate of return, and break-even

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. When choosing among mutually exclusive alternatives by rate of return, you must perform incremental analysis — compare the extra investment of the more expensive option against the extra benefit, and accept the increment only if its incremental ROR exceeds the MARR. Picking the option with the highest standalone ROR is a classic trap, because a small project can post a high percentage return yet leave value on the table compared with a larger one that still clears the MARR.

Worked example (capital recovery, A/P): A $50,000 machine is financed at i = 10% over n = 5 years. The equivalent annual cost uses (A/P, 10%, 5) = i(1+i)^n/[(1+i)^n - 1] = 0.10(1.61051)/(1.61051 - 1) = 0.161051/0.61051 = 0.2638. Annual cost = 50,000 × 0.2638 = $13,190 per year.

Break-even analysis finds the volume or time where two alternatives have equal cost. If fixed cost is FC and the per-unit margin is (price - variable cost), the break-even quantity is Q* = FC / (price - variable cost). Example: with FC = $20,000, price $50, variable cost $30, then Q* = 20,000/(50-30) = 1,000 units.

Depreciation, inflation, and effective rates

Depreciation spreads an asset's cost over its useful life:

• 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; read the year's percentage and multiply the original basis. Salvage is ignored under MACRS.

A nominal annual rate r compounded m times per year gives an effective annual rate i_eff = (1 + r/m)^m - 1. Example: 12% nominal compounded monthly gives i_eff = (1 + 0.12/12)^12 - 1 = (1.01)^12 - 1 = 12.68%. When inflation matters, distinguish the market (nominal) rate from the real rate; the FE usually states which to use. A frequent trap is using the nominal rate directly when compounding is more frequent than annual.

When the compounding period and the payment period differ, the cleanest approach is to find the effective rate per payment period first, then apply the standard factors with n = number of payment periods. For continuous compounding, the effective rate is i_eff = e^r - 1. Keep i and n in the same time units throughout a problem — a monthly cash flow needs a monthly i and a count of months, not an annual rate with a year count. Getting units consistent resolves most engineering-economics mistakes, just as it does in the rest of the FE math areas.