FE Mechanical: Engineering Economics study guide

Engineering Economics

The FE Mechanical engineering economics items test time value of money, equivalence, project evaluation, depreciation, taxes, and inflation. Most problems reduce to converting cash flows to a common point in time and comparing alternatives on a consistent basis.

Core rules and concepts

Single payment: F = P(1 + i)^n and P = F(1 + i)^(-n).
Uniform series: (F/A, i, n) = ((1 + i)^n - 1)/i; (A/F, i, n) = i/((1 + i)^n - 1); (P/A, i, n) = ((1 + i)^n - 1)/(i(1 + i)^n); (A/P, i, n) = i(1 + i)^n/((1 + i)^n - 1).
Effective annual rate for nominal rate r compounded m times per year: i_eff = (1 + r/m)^m - 1. For continuous compounding, i_eff = e^r - 1.
Net present value: NPV = Σ CF_t/(1 + i)^t. For an independent project, accept if NPV > 0. IRR is the i that makes NPV = 0.
For mutually exclusive alternatives, use incremental analysis or compare NPV; ranking by IRR alone can mislead.
Depreciation: straight-line annual amount = (P - S)/n. MACRS is accelerated and uses a half-year convention for the first year.
Inflation: (1 + i_nominal) = (1 + i_real)(1 + f).

Worked example

A machine costs $10,000, saves $3,000 per year for 5 years, and has a $1,000 salvage value. MARR is 10%. Compute NPV = -10,000 + 3,000(P/A, 10%, 5) + 1,000(P/F, 10%, 5). Using (P/A, 10%, 5) = 3.7908 and (P/F, 10%, 5) = 0.6209: NPV = -10,000 + 11,372.4 + 620.9 = $1,993.3. Since NPV is positive, the machine is economically justified at a 10% MARR.

Traps that cost marks

Treating an annuity due as an ordinary annuity; begin-of-period cash flows need an extra (1 + i) factor.
Mixing nominal and effective interest rates.
Including sunk costs in a decision.
Omitting salvage value or giving it the wrong sign.
Assuming MACRS equals straight-line depreciation.
Choosing the higher IRR when alternatives differ in scale; use incremental NPV.