Engineering Economics, Cash Flow, and Decision Methods

Key Takeaways

  • Engineering economics is cash-flow timing: draw the cash-flow diagram first, with the interest rate matching the cash-flow period.
  • Single-payment factors: (F/P, i, n) = (1+i)ⁿ and (P/F, i, n) = (1+i)⁻ⁿ convert between present and future single sums.
  • Uniform-series factors (A/P), (P/A), (A/F), (F/A) link an equal annual amount A to present or future worth.
  • Compare alternatives by Present Worth (NPV), Annual Worth/EUAC, or Rate of Return; use EUAC for unequal service lives.
  • Straight-line depreciation = (cost − salvage)/life; MACRS uses IRS percentage tables with the half-year convention.
Last updated: June 2026

Time Value of Money

Engineering economics is about 4–6 FE questions and is almost entirely cash-flow timing. A dollar today is worth more than a dollar later because it can earn interest. The first step on every problem is to draw the cash-flow diagram: time on the horizontal axis, upward arrows for receipts and downward arrows for disbursements, with consistent signs.

The interest rate i must match the compounding period n. If cash flows are annual but interest compounds monthly, convert using the effective annual rate i_eff = (1 + r/m)^m − 1, where r is the nominal annual rate and m the compounding periods per year. For 12% nominal compounded monthly, i_eff = (1 + 0.01)¹² − 1 = 12.68%.

Single-Payment Factors

The NCEES Handbook lists compound-interest factors in functional notation:

  • (F/P, i, n) = (1 + i)ⁿ — find future worth of a present sum.
  • (P/F, i, n) = (1 + i)⁻ⁿ — find present worth of a future sum.

Example: $1,000 invested at 8% for 5 years grows to F = 1000(1.08)⁵ = 1000(1.469) = $1,469.

Uniform-Series Factors

Many problems involve an equal annual amount A (annuity). The Handbook factors are:

FactorSymbolFormula
Uniform series present worth(P/A, i, n)[(1+i)ⁿ − 1]/[i(1+i)ⁿ]
Capital recovery(A/P, i, n)[i(1+i)ⁿ]/[(1+i)ⁿ − 1]
Uniform series compound amount(F/A, i, n)[(1+i)ⁿ − 1]/i
Sinking fund(A/F, i, n)i/[(1+i)ⁿ − 1]

The capital recovery factor (A/P) converts a present cost into equal annual payments—the basis for loan and EUAC calculations.

Worked Example — Annual Worth

A machine costs $20,000, lasts 5 years with no salvage, at i = 10%. The equivalent annual cost = 20,000·(A/P, 10%, 5). With (A/P) = [0.10(1.1)⁵]/[(1.1)⁵ − 1] = 0.1610/0.6105 = 0.2638, the annual cost = 20,000(0.2638) = $5,276/yr. Adding annual operating cost gives the total EUAC (Equivalent Uniform Annual Cost).

Decision Methods and Depreciation

Three equivalent bases compare alternatives:

  • Present Worth / Net Present Value (NPV): discount all cash flows to time zero. Choose the highest NPV (most positive). A positive NPV means the project earns more than the MARR (minimum acceptable rate of return).
  • Annual Worth / EUAC: convert everything to an equal annual amount. Best for alternatives with unequal service lives, since it normalizes per year automatically.
  • Rate of Return (ROR): the interest rate that makes NPV = 0. Accept if ROR ≥ MARR.

For mutually exclusive choices pick the single best; for independent projects fund every one that meets the MARR within the budget. Carry correct signs: initial investment and operating costs are negative, revenue and salvage are positive.

Depreciation

Straight-line: annual depreciation = (cost − salvage)/useful life; book value declines linearly. MACRS (Modified Accelerated Cost Recovery System, the U.S. tax method) applies IRS percentage tables to the full cost basis (salvage ignored) under the half-year convention—only half a year's depreciation in years 1 and the last. For a $50,000 asset with $5,000 salvage over 9 years, straight-line gives (50,000 − 5,000)/9 = $5,000/yr.

Gradients, Benefit–Cost, and Common Traps

Beyond uniform series, the FE includes a few specialized cash-flow patterns and public-project methods.

Arithmetic gradient (G): when a cash flow increases by a constant amount each period (e.g., maintenance rising $200/yr), the (P/G, i, n) and (A/G, i, n) factors convert the gradient to present worth or to an equivalent uniform series. The first cash flow occurs at the end of year 2, not year 1—a frequent setup error. A geometric gradient grows by a constant percentage each period and uses its own factor.

Benefit–cost ratio (B/C): public-sector projects are often judged by B/C = (present worth of benefits)/(present worth of costs). Accept when B/C ≥ 1.0. Be careful to put benefits in the numerator and all costs (including annual operating cost) in the denominator on a consistent present-worth or annual basis.

Inflation and Real vs. Nominal

The FE may distinguish the market (nominal) interest rate from the real interest rate after removing inflation f: (1 + i_market) = (1 + i_real)(1 + f). Use the rate consistent with whether cash flows are stated in actual (then-current) or constant (today's) dollars.

High-Yield Traps to Avoid

TrapFix
Interest period ≠ cash-flow periodConvert to effective rate first
Wrong sign on costs vs. revenueCosts negative, income positive
Comparing unequal lives by present worthUse EUAC or a common multiple of lives
Gradient starts year 1Gradient's first increment is year 2
Using salvage in MACRSMACRS depreciates full basis
Picking lowest first costCompare total equivalent cost, not initial outlay

Working economics as a fixed sequence—diagram, match the period, choose the basis, apply the factor, state the rule—turns these into reliable points. Because the Handbook supplies every factor formula and many factor tables, the skill being tested is correct setup and interpretation, not memorization.

Test Your Knowledge

How much must you deposit today at 6% annual interest to have $5,000 in 4 years?

A
B
C
D
Test Your Knowledge

Which method is most appropriate for comparing two machines with different service lives?

A
B
C
D
Test Your Knowledge

An asset costs $30,000 with $3,000 salvage and a 9-year life. Straight-line annual depreciation is:

A
B
C
D
Test Your Knowledge

A public project has present-worth benefits of $480,000 and present-worth costs of $400,000. By the benefit–cost criterion the project should be:

A
B
C
D