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FE Other Disciplines Engineering Economics Practice Problems

The current FE Other Disciplines specification assigns 6-9 of the 110 exam questions to Engineering Economics. Form A contains 7 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

Free sample problem

A municipal wastewater pumping station incurs a maintenance cost of USD 8{,}000 at the end of year 1. The cost increases by USD 1{,}200 in each following year through the end of year 8. At an interest rate of 7% per year, the present worth of the 8-year maintenance stream is most nearly:

  1. (A)USD 47{,}800
  2. (B)USD 54{,}900
  3. (C)USD 65{,}400
  4. (D)USD 70{,}300
Show worked solution

Answer: (D)

Base series and gradient together discount to a present worth of about USD 70{,}300.

Costs that grow by a constant amount each year separate into a uniform base series of USD 8{,}000 and an arithmetic gradient of USD 1{,}200 whose first increment falls at the end of year 2.

At 7% over eight years the two factors take the values below, with .

Gradient growth supplies almost a third of the obligation, so a reserve sized on the first-year cost alone falls badly short.

FE Reference Handbook — Engineering Economics: Uniform Series Present Worth Factor and Uniform Gradient Present Worth Factor

Why the other choices appear

  • (A)Discounting only the uniform USD 8,000 base, 8,000(5.9713) = 47,770, discards the gradient entirely.
  • (B)Treating the whole stream as a uniform USD 9,200 (the year-2 amount) gives 9,200(5.9713) = 54,936.
  • (C)Using (P/G, 7%, 7) = 14.7149, from counting the seven gradient steps as the factor's period count, gives 47,770 + 17,658 = 65,428.

The complete Form A contains 7 problems in this area and 110 problems overall, with an answer key and full solutions.