FE Civil Engineering Economics Practice Problems

The current FE Civil specification assigns 5-8 of the 110 exam questions to Engineering Economics. Form A contains 5 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

Two booster pump stations are being considered for a water main extension. Station A has a first cost of USD 60,000 and annual operating costs of USD 11,000. Station B has a first cost of USD 95,000 and annual operating costs of USD 5,000. Both stations have 10-year lives with no salvage value. At an interest rate of 8% per year, the equivalent uniform annual cost of the more economical alternative is most nearly:

(A)USD 11,560
(B)USD 14,500
(C)USD 19,160
(D)USD 19,940

Show worked solution

Answer: (C)

At 8% the capital recovery factor is 0.14903; Station B annualizes to USD 19,158 against USD 19,942 for Station A, so the economical alternative costs most nearly USD 19,160 per year.

Equal 10-year lives with no salvage make equivalent uniform annual cost the natural basis; first find the capital recovery factor.

(A / P, 8%, 10) = \frac{0.08 ( 1.08 ) ^{10}}{( 1.08 ) ^{10} - 1} = 0.14903

E U A C_{A} = 60, 000 (0.14903) + 11, 000 = 19, 942

E U A C_{B} = 95, 000 (0.14903) + 5, 000 = 19, 158

FE Reference Handbook — Engineering Economics: Capital Recovery (A/P, i%, n)

Why the other choices appear

(A)Using the sinking fund factor (A/F, 8%, 10) = 0.06903 in place of (A/P) for Station B gives 95,000(0.06903) + 5,000 = 11,558.
(B)Dividing the first cost of Station B by the life without interest gives 95,000/10 + 5,000 = 14,500.
(D)USD 19,942 is the equivalent uniform annual cost of Station A, the less economical alternative.