2.5 Engineering Economics

Time Value of Money

Engineering Economics is 4–6 questions on the FE Civil exam and rests on one idea: money has a time value — a dollar today is worth more than a dollar in the future because it can earn interest. To compare cash flows occurring at different times, you must move them all to a common point in time (usually present worth, time 0) using interest factors. ).

The Interest Factors

The Handbook lists six standard discrete-compounding factors. The four you must command:

Here P = present amount, F = future amount, A = uniform end-of-period amount (annuity), i = interest rate per period, n = number of periods. The companion factors F/A (uniform series compound amount) and A/F (sinking fund) round out the set. Note A/P and P/A are reciprocals, as are F/P and P/F.

Two more relationships round out the toolkit. A uniform gradient G (cash flows that increase by a fixed amount each period) converts to an equivalent annuity with the A/G factor, also tabulated. And the stated nominal annual rate r compounded m times per year gives an effective annual rate i_eff = (1 + r/m)ᵐ − 1; for continuous compounding i_eff = eʳ − 1. The classic trap is using a 12% annual rate with monthly periods without first converting to the 1% monthly rate — match the interest rate's period to the cash-flow period before picking any factor.

Evaluation Methods

Four methods select among alternatives, and a correct analysis gives the same decision:

• Present Worth (PW/NPW): discount every cash flow to time 0; for revenue projects the highest PW wins, and for cost comparisons the least-negative (lowest-cost) PW wins. Compare alternatives over equal study periods.
• Annual Worth (AW/EUAC): convert all cash flows to an equivalent uniform annual amount using A/P; convenient when lives differ.
• Rate of Return (ROR): the interest rate i* at which NPW = 0; accept if i* ≥ the minimum attractive rate of return (MARR).
• Benefit-Cost (B/C): a public-project staple — compute B/C = (PW of benefits)/(PW of costs); accept when B/C ≥ 1. For incremental comparisons, accept the costlier option only if the incremental ΔB/ΔC ≥ 1.

Depreciation & Inflation

Depreciation spreads an asset's cost over its useful life for accounting/tax purposes. Straight-line (SL): D = (initial cost − salvage value)/n — an equal amount each year, with book value declining linearly to the salvage value. S. tax method that front-loads depreciation; the annual deduction = (recovery rate from the IRS/Handbook table) × (initial cost), where MACRS assumes zero salvage and uses a half-year convention so an n-year asset depreciates over n+1 years. You are not expected to memorize the percentages — they are in the Handbook tables — only to apply them.

Inflation erodes purchasing power. The combined (market) interest rate is (1+d) = (1+i)(1+f), where i is the real rate and f the inflation rate. Discount inflated (actual-dollar) cash flows at the market rate, or convert to constant dollars and discount at the real rate — never mix the two.

Replacement and break-even analysis also appear. In replacement studies you compare the equivalent uniform annual cost (EUAC) of keeping a 'defender' asset against acquiring a 'challenger,' and you ignore sunk costs — money already spent does not affect the forward-looking decision. Break-even analysis finds the volume or time at which two alternatives have equal cost; setting their cost equations equal and solving for the unknown gives the break-even point. A frequent error is letting a sunk cost (the original purchase price of equipment you already own) bias a replacement decision.

Worked Example — Present Worth

A culvert upgrade costs $50,000 now and saves $8,000 per year in maintenance for 10 years. At i = 6%, is it worthwhile (find the present worth of savings)?

Step 1 — choose the factor: the savings are a uniform annual series A converted to present worth P, so use P/A at 6%, n = 10.

Step 2 — compute the factor: P/A = [(1.06)¹⁰ − 1]/[0.06(1.06)¹⁰] = [1.7908 − 1]/[0.06 × 1.7908] = 0.7908/0.10745 ≈ 7.360.

Step 3 — present worth of savings: PW = A × (P/A) = 8,000 × 7.360 = $58,880.

Step 4 — net present worth: NPW = 58,880 − 50,000 = +$8,880. Because NPW > 0, the upgrade is economically justified at 6%. (Equivalently, the B/C ratio = 58,880/50,000 = 1.18 ≥ 1.)

Worked Example — Straight-Line vs Declining-Balance Depreciation

An asset costs B = $30,000, has a salvage S = $3,000, and a 5-year life. Compare the first two years under straight-line (SL) and double-declining-balance (DDB).

Straight-line: D = (B − S)/n = (30,000 − 3,000)/5 = $5,400 each year. Book value falls in equal steps to the $3,000 salvage.

Double-declining-balance: the rate is 2/n = 2/5 = 40% applied to the current book value (salvage ignored until the floor is reached). Year 1: D₁ = 0.40 × 30,000 = $12,000 (book value → $18,000). Year 2: D₂ = 0.40 × 18,000 = $7,200 (book value → $10,800).

DDB front-loads the deductions — larger early write-offs (helpful for tax timing) but identical total depreciation over the asset's life. Book value may never be depreciated below salvage, so later DDB years are reduced or switched to SL to land exactly on the $3,000 floor.

Worked Example — Break-Even Analysis

Two paving alternatives must be compared. Method A has a fixed cost of $20,000 plus $15 per square meter; Method B has a fixed cost of $50,000 plus $9 per square meter. Find the area at which the two cost the same.

Set total costs equal: 20,000 + 15·x = 50,000 + 9·x.

Solve for x: 15x − 9x = 50,000 − 20,000 → 6x = 30,000 → x = 5,000 m².

Below 5,000 m², Method A (lower fixed cost) is cheaper; above 5,000 m², Method B (lower variable cost) wins. The break-even point is where the choice flips, and it is the area at which both equations yield $95,000. Break-even logic also applies to production volume, time-to-payback, and lease-versus-buy decisions — set the two cost (or cost-versus-revenue) expressions equal and solve for the unknown.