Engineering Economics for the FE Exam
Engineering economics questions appear on every FE exam discipline. These problems involve comparing the costs and benefits of engineering alternatives using the time value of money. The good news is that these problems follow consistent patterns, and the formulas are in the reference handbook.
The Time Value of Money
A dollar today is worth more than a dollar in the future because money can earn interest. This fundamental concept underlies all engineering economics calculations. To compare cash flows occurring at different times, they must be converted to equivalent values at a common point in time.
Interest and Equivalence
- Simple interest: Interest is calculated only on the original principal. I = P * i * n. Rarely used in engineering economics.
- Compound interest: Interest is calculated on the principal plus accumulated interest. This is the standard for engineering economics.
- Nominal vs. effective interest rate: The nominal rate is the stated annual rate. The effective rate accounts for compounding periods. i_eff = (1 + r/m)^m - 1, where r is the nominal rate and m is the number of compounding periods per year.
Cash Flow Diagrams
Always draw a cash flow diagram before solving an economics problem:
- Horizontal line represents time (divided into periods)
- Upward arrows represent positive cash flows (income, salvage value)
- Downward arrows represent negative cash flows (costs, investments)
- Label each arrow with the amount and the time period
Interest Factors
The reference handbook provides interest factor tables. The six basic factors are:
Single Payment Factors
- Future worth factor (F/P, i, n): F = P * (1 + i)^n. Converts a present amount to a future amount.
- Present worth factor (P/F, i, n): P = F * (1 + i)^(-n). Converts a future amount to a present amount.
Uniform Series Factors
- Uniform series compound amount (F/A, i, n): F = A * [(1 + i)^n - 1] / i. Converts a uniform annual amount to a future amount.
- Sinking fund factor (A/F, i, n): A = F * i / [(1 + i)^n - 1]. Converts a future amount to a uniform annual amount.
- Uniform series present worth (P/A, i, n): P = A * [(1 + i)^n - 1] / [i * (1 + i)^n]. Converts a uniform annual amount to a present amount.
- Capital recovery factor (A/P, i, n): A = P * [i * (1 + i)^n] / [(1 + i)^n - 1]. Converts a present amount to a uniform annual amount (like a loan payment).
Gradient Factors
- Arithmetic gradient: Cash flows that increase by a constant amount G each period. The gradient starts at zero in period 1 and increases by G each subsequent period.
- Geometric gradient: Cash flows that increase by a constant percentage each period.
Comparison Methods
Present Worth (PW) Analysis
Convert all cash flows to their present value at time zero. The alternative with the highest (least negative) present worth is preferred.
PW = -Initial cost + PW of benefits - PW of costs + PW of salvage value
Important: When comparing alternatives with different service lives, use the least common multiple of the lives, or use the annual worth method instead.
Annual Worth (AW) Analysis
Convert all cash flows to an equivalent uniform annual amount. The alternative with the highest (least negative) annual worth is preferred.
AW = -A_capital + A_benefits - A_costs
Advantage: You do not need to use the least common multiple for different service lives. Just compute AW over each alternative's own life.
Rate of Return (ROR) Analysis
Find the interest rate at which the present worth of costs equals the present worth of benefits (NPW = 0). Compare the ROR to the minimum attractive rate of return (MARR). If ROR is greater than or equal to MARR, the investment is acceptable.
For comparing mutually exclusive alternatives, use incremental ROR analysis.
Benefit-Cost Ratio (B/C)
B/C = PW of benefits / PW of costs
If B/C is greater than or equal to 1.0, the project is economically justified. For comparing alternatives, use incremental B/C analysis.
Depreciation
Depreciation allocates the cost of an asset over its useful life for tax purposes:
- Straight-line: D = (P - S) / n, where P is initial cost, S is salvage value, and n is useful life. Equal depreciation each year.
- MACRS (Modified Accelerated Cost Recovery System): Uses IRS-specified percentages for each year. Salvage value is assumed to be zero. The percentages are given in the reference handbook.
Book value at any time: BV = P - accumulated depreciation
Inflation
Inflation reduces the purchasing power of money over time. To account for inflation:
- Real (inflation-free) interest rate: i_r = (i - f) / (1 + f), where i is the market (nominal) interest rate and f is the inflation rate.
- Actual dollars vs. constant dollars: Actual (then-current) dollars include inflation. Constant (today's) dollars remove the effect of inflation.
Common FE Exam Problem Types
- Finding the present worth or future worth of a series of cash flows
- Calculating loan payments using the capital recovery factor
- Comparing two or more alternatives using present worth or annual worth
- Finding the rate of return on an investment
- Calculating depreciation using straight-line or MACRS methods
- Determining the breakeven point between two alternatives
Practice Tips
- Always draw a cash flow diagram
- Practice using the interest factor tables in the reference handbook
- Know which method to use (PW, AW, ROR, B/C) based on the problem setup
- Watch for different compounding periods (monthly vs. annual)
- Remember to adjust for different service lives when using present worth
Engineering economics is one of the most scoreable topics on the FE exam. The problems are formulaic, the tables are in the handbook, and practice makes these problems fast and reliable.