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Engineering Economics, Cash Flow, and Decision Methods

Key Takeaways

  • Engineering economics questions are primarily cash-flow timing problems, so draw the timeline first.
  • Present worth, future worth, annual worth, and rate-of-return methods all convert money to a common basis.
  • The interest period must match the cash-flow period before any factor is used.
  • Salvage value, operating cost, maintenance savings, and initial investment must carry correct signs.
  • Decision methods differ for revenue projects, cost-only alternatives, mutually exclusive choices, and independent projects.
Last updated: May 2026

Draw the cash-flow diagram first

FE Mechanical engineering-economics questions are rarely long accounting cases. They are timeline problems. Put time zero on the left, mark the interest rate per period, place each cost or benefit at the correct date, and assign signs before using a factor. An initial cost is usually negative for a revenue project. Annual savings are positive. Annual operating costs are negative. Salvage value is usually positive at the end of the service life.

If the answer choices differ mainly by sign, the model may be right but the cash-flow direction is wrong. If the choices differ by about one interest period, the payment was probably placed at the wrong end of the year.

Common factors and when to use them

NeedConvert fromFactor idea
Present worthFuture single amountP/F
Present worthUniform annual seriesP/A
Future worthPresent single amountF/P
Annual worthPresent investmentA/P
Annual equivalent of future salvageFuture amountA/F
Sinking fundFuture targetA/F
Loan paymentPresent loanA/P

The FE Reference Handbook provides standard interest factors, but you still choose the right one. A common error is using P/A for a single future salvage value or P/F for a repeating annual cost. Another error is mixing monthly cash flows with annual interest without conversion.

Decision methods

Present worth compares alternatives by moving all cash flows to time zero. For revenue alternatives, choose the option with the largest net present worth if alternatives are mutually exclusive. For cost-only alternatives that provide the same service, choose the lowest present cost or lowest equivalent annual cost.

Annual worth is useful when alternatives have different lives or when costs are naturally annual. Convert capital cost, annual operating cost, and salvage into one annual equivalent. Remember that salvage reduces annual cost because it is value recovered at the end.

Rate of return sets net present worth equal to zero and solves for the interest rate. On the FE exam, this is often a recognition or interpolation problem. If the project net present worth is positive at the minimum attractive rate of return, the project is acceptable by present-worth logic. If an internal rate of return is above the required rate for a conventional cash flow, it is acceptable.

Benefit-cost ratio is common in public projects. Keep benefits in the numerator and costs in the denominator. A ratio above 1 generally supports the project, but mutually exclusive public alternatives may require incremental analysis.

Fast error checks

Use three checks before selecting an answer:

  1. Does the rate period match the cash-flow period?
  2. Did you include salvage and maintenance costs at the correct time?
  3. Is the decision rule correct for revenue, cost-only, independent, or mutually exclusive alternatives?

Economics is one of the best places to gain efficient FE points because the models are standardized. A neat timeline plus correct signs often matters more than memorizing every factor name.

Test Your Knowledge

What is the present worth of $8,000 received 4 years from now at 6% annual interest?

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Test Your Knowledge

A machine costs $18,000, saves $5,200 per year for 5 years, and has $2,000 salvage at 8% interest. What is the approximate net present worth?

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B
C
D
Test Your Knowledge

A cost-only asset has a $24,000 first cost, $4,000 annual operating cost, $6,000 salvage value, 6-year life, and 10% interest. What is the approximate equivalent annual cost?

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