Question 4 of 6: Engineering Economics

Question 4: Engineering Economics (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two mutually exclusive projects with the investment, operating cost, periodic major maintenance, revenue and service life tabulated below, all evaluated at a discount rate of 10% per year.

Find. The present-value profit of each project and, because the lives are unequal, a comparison made over a common study period, so that the more economical plan can be identified.

Approach. Net the yearly revenue against the yearly operating cost to get a uniform annual series, discount it with the series present-worth factor, subtract the investment and the discounted major-maintenance events to get the present worth over each project's own life, then place the two on a common footing by repeating each project over the 30-year least common multiple of the lives (equivalently, by comparing annual worths).

1. Net the recurring annual cash flow. Operating cost and revenue are both uniform annual series, so they combine before discounting: $$R_A = \$13{,}500 - \$2{,}500 = \$11{,}000 \text{ per year}$$ $$R_B = \$16{,}000 - \$1{,}000 = \$15{,}000 \text{ per year}$$
2. Discount the annual series. With \((P/A, i, n) = \left[1 - (1+i)^{-n}\right]/i\) at \(i = 0.10\): $$(P/A, 10\%, 15) = \frac{1 - 1.10^{-15}}{0.10} = 7.60608, \qquad (P/A, 10\%, 10) = \frac{1 - 1.10^{-10}}{0.10} = 6.14457$$ so the operating surplus is worth \(11{,}000 \times 7.60608 = \$83{,}666.87\) for A and \(15{,}000 \times 6.14457 = \$92{,}168.51\) for B at time zero.
3. Discount the major-maintenance events. A charge "every 5 years" over a 15-year life falls in years 5, 10 and 15; over a 10-year life it falls in years 5 and 10. Using \((P/F, 10\%, n) = 1.10^{-n}\), the factors are 0.62092, 0.38554 and 0.23939, so $$P_{M,A} = 5{,}000\,(0.62092 + 0.38554 + 0.23939) = \$6{,}229.28$$ $$P_{M,B} = 3{,}000\,(0.62092 + 0.38554) = \$3{,}019.39$$
4. Present worth over each project's own life. Assembling the three pieces against the investment: $$PW_A = -70{,}000 + 83{,}666.87 - 6{,}229.28 = \boxed{+\$7{,}437.59}$$ $$PW_B = -50{,}000 + 92{,}168.51 - 3{,}019.39 = \boxed{+\$39{,}149.11}$$ Both projects clear the 10% hurdle rate, so both are acceptable in absolute terms and the question is which to choose.
5. Put the two on a common study period. These present worths are not comparable as they stand, because one buys 15 years of service and the other only 10. Assuming each project is repeatable on the same terms, the least common multiple of the lives is 30 years, over which A runs twice (starting in years 0 and 15) and B three times (years 0, 10 and 20). Each repetition is the same present worth pushed back by one life, so $$PW_A^{30} = PW_A \left[1 + (P/F, 10\%, 15)\right] = 7{,}437.59 \times 1.23939 = \$9{,}218.09$$ $$PW_B^{30} = PW_B \left[1 + (P/F, 10\%, 10) + (P/F, 10\%, 20)\right] = 39{,}149.11 \times 1.53419 = \$60{,}062.06$$
6. Cross-check on annual worth. Dividing each own-life present worth by its own series factor converts it to an equivalent uniform annual profit, which is directly comparable without any repetition argument: $$AW_A = \frac{7{,}437.59}{7.60608} = \$977.85 \text{ per year}, \qquad AW_B = \frac{39{,}149.11}{6.14457} = \$6{,}371.34 \text{ per year}$$ The two routes agree, as they must: B is preferred by a factor of about 6.5 on either measure.

Project B is therefore the more economical plan, and the margin is far too wide to be overturned by any reasonable adjustment to the assumptions. The result is intuitive once the annual worths are in view: B earns a larger net annual surplus ($15,000 against $11,000) from a smaller investment ($50,000 against $70,000), and its shorter life is a benefit rather than a penalty under a repeatability assumption because the capital is recovered and redeployed sooner.