Engineering Economics: Time Value of Money

Quick Answer: Use present worth (PW) and annual worth (AW) with Handbook factors. NPV > 0 accepts independent projects. Ignore sunk costs in replacement analysis.

Single Payment Factors

[ P = F/(1+i)^n \quad F = P(1+i)^n ]

Example: $500,000 in 10 years at 6% → P = 500,000/1.791 ≈ $279,200 today.

Uniform Series

Capital recovery: ( A = P \times (A/P, i, n) ). Sinking fund: ( A = F \times (A/F, i, n) ).

Example: $200,000 scrubber, 15 yr, 8% → A ≈ $23,360/yr capital recovery.

NPV and Benefit-Cost

[ NPV = \sum \frac{B_t - C_t}{(1+i)^t} ]

Accept if NPV > 0 or B/C > 1 (when benefits and costs separated).

Effective Annual Rate

When compounding is m times per year: ( i_e = (1 + i/m)^m - 1 ). Nominal 12% compounded monthly: ( i_e = (1.01)^{12} - 1 = 12.68% ).

Gradient and Geometric Series

Some environmental costs grow each year (electricity, chemical prices). Present worth of gradient ( G ):

[ P = G \times \frac{(1+i)^n - 1 - ni}{i(1+i)^n} ]

Handbook supplies ((P/G, i, n)) factors — locate before exam day.

Worked Example — UV vs. Chlorine Capital

UV system: P = $400,000, annual O&M $25,000, life 15 yr, MARR 7%.

Chlorine: P = $120,000, annual O&M $55,000 (chemicals + labor), life 15 yr.

Find PW of costs for each (cost-only comparison):

[ PW_{UV} = 400{,}000 + 25{,}000 \times (P/A, 7%, 15) ] [ PW_{Cl} = 120{,}000 + 55{,}000 \times (P/A, 7%, 15) ]

Using (P/A, 7%, 15) ≈ 9.108: PW_UV ≈ 627,700; PW_Cl ≈ 620,940 — chlorine slightly lower on cost alone; add risk, DBPs, safety qualitatively in essay-style stems.

IRR Concept

Internal rate of return is the rate where NPV = 0. If IRR > MARR, project is attractive. FE may ask you to compare two rates without full IRR calculation — use bracketing or calculator IRR function.

Bond and Loan Equivalence

Equal loan payments use capital recovery factor. Bond interest-only with balloon principal uses different cash flow pattern — read stem carefully.

Sinking Fund for Closure

Landfill post-closure care $2M in 30 years, i = 5%:

[ A = 2{,}000{,}000 \times (A/F, 5%, 30) \approx 2{,}000{,}000 \times 0.0151 = $30{,}200/\text{yr} ]

Escrow this annual deposit for regulatory financial assurance.

Time Value of Money — FE Formulas in Practice

Worked Example

A UV disinfection upgrade costs $400,000 now and saves $55,000/year in chemicals for 12 years at MARR = 8%.

$PW = -400{,}000 + 55{,}000(P/A,8%,12)$

$(P/A,8%,12) = \frac{(1.08)^{12}-1}{0.08(1.08)^{12}} \approx 7.536$

$PW \approx -400{,}000 + 414{,}500 = +$14{,}500$ → accept at 8% MARR.

On the Exam: Match cash-flow diagram to the correct factor; mixing $(P/A)$ with $(A/P)$ is the most common arithmetic trap.

Time Value of Money — FE Formulas in Practice

Worked Example

A UV disinfection upgrade costs $400,000 now and saves $55,000/year in chemicals for 12 years at MARR = 8%.

$PW = -400{,}000 + 55{,}000(P/A,8%,12)$

$(P/A,8%,12) = \frac{(1.08)^{12}-1}{0.08(1.08)^{12}} \approx 7.536$

$PW \approx -400{,}000 + 414{,}500 = +$14{,}500$ → accept at 8% MARR.

On the Exam: Match cash-flow diagram to the correct factor; mixing $(P/A)$ with $(A/P)$ is the most common arithmetic trap.

Time Value of Money — FE Formulas in Practice

Worked Example

A UV disinfection upgrade costs $400,000 now and saves $55,000/year in chemicals for 12 years at MARR = 8%.

$PW = -400{,}000 + 55{,}000(P/A,8%,12)$

$(P/A,8%,12) = \frac{(1.08)^{12}-1}{0.08(1.08)^{12}} \approx 7.536$

$PW \approx -400{,}000 + 414{,}500 = +$14{,}500$ → accept at 8% MARR.

On the Exam: Match cash-flow diagram to the correct factor; mixing $(P/A)$ with $(A/P)$ is the most common arithmetic trap.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.

Additional review point: verify assumptions, boundary conditions, and whether the problem is steady-state or transient before selecting an answer.