Question 4 of 6: Engineering Economics — present-worth comparison of two projects with unequal lives

Question 4: Engineering Economics — present-worth comparison of two projects with unequal lives (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 unequal service lives, appraised at 10 percent per year.

Find. The present worth of each project, compared over a basis on which the comparison is legitimate despite the unequal lives, and a recommendation as to the more economical plan.

Approach. Collapse revenue and operating cost into one net annual amount, discount it with the uniform-series factor, subtract the discounted major maintenance and the first cost to get the present worth of one life cycle, then repeat each project over the 30-year least common multiple so the two are compared over the same span — with annual worth as an independent cross-check.

1. Reduce each project to a net annual amount. Revenue and operating cost are both uniform and annual, so they combine before any discounting:$$A_A = 23{,}500 - 12{,}500 = \$11{,}000 \text{ per year},\qquad A_B = 26{,}000 - 11{,}000 = \$15{,}000 \text{ per year}$$Project B earns more per year from a smaller investment; the whole question is whether its ten-year life gives it enough time to do so.
2. Discount the net annual series. With \(i = 0.10\) the uniform-series present-worth factor is \((P/A, i, n) = [1 - (1+i)^{-n}]/i\), giving \((P/A, 10\%, 15) = 7.6061\) and \((P/A, 10\%, 10) = 6.1446\):$$P_{A,\text{ann}} = 11{,}000 \times 7.6061 = \$83{,}666.87,\qquad P_{B,\text{ann}} = 15{,}000 \times 6.1446 = \$92{,}168.51$$
3. Discount the major maintenance. These are single payments every five years, so each is brought back with \((P/F, i, n) = (1+i)^{-n}\). Within a 15-year life Project A incurs them at years 5 and 10, and within a 10-year life Project B incurs one at year 5; an overhaul in the final year of service is not spent, because the asset is retired at that instant:$$P_{A,\text{mm}} = 15{,}000\,(0.6209 + 0.3855) = \$15{,}096.97,\qquad P_{B,\text{mm}} = 13{,}000\,(0.6209) = \$8{,}071.98$$
4. Present worth of one life cycle. Combining the first cost, the series and the overhauls,$$PW_A^{15} = -170{,}000 + 83{,}666.87 - 15{,}096.97 = \boxed{-\$101{,}430.09}$$$$PW_B^{10} = -150{,}000 + 92{,}168.51 - 8{,}071.98 = \boxed{-\$65{,}903.47}$$Both present worths are negative. That is not an arithmetic slip and must be reported plainly: at 10 percent neither project recovers its investment. The undiscounted totals confirm the loss is structural rather than an artefact of discounting — Project A returns \(-170{,}000 + 15 \times 11{,}000 - 2 \times 15{,}000 = -\$35{,}000\) and Project B \(-150{,}000 + 10 \times 15{,}000 - 13{,}000 = -\$13{,}000\) even at a zero interest rate, so no discount rate rescues either and neither has a real internal rate of return.
5. Put the two on a common study period. A 15-year and a ten-year present worth cannot be compared directly. Repeating each project over the least common multiple, \(\mathrm{LCM}(15, 10) = 30\) years — Project A twice, Project B three times — and discounting each repetition to time zero:$$PW_A^{30} = PW_A^{15}\left[1 + (1.10)^{-15}\right] = -101{,}430.09 \times 1.2394 = \boxed{-\$125{,}711.65}$$$$PW_B^{30} = PW_B^{10}\left[1 + (1.10)^{-10} + (1.10)^{-20}\right] = -65{,}903.47 \times 1.5342 = \boxed{-\$101{,}108.24}$$
6. Cross-check with annual worth. Under the same repeatability assumption the annual worth of one cycle already represents the alternative indefinitely, so it should rank the projects identically:$$AW_A = \frac{-101{,}430.09}{7.6061} = -\$13{,}335.40\text{/yr},\qquad AW_B = \frac{-65{,}903.47}{6.1446} = -\$10{,}725.49\text{/yr}$$Multiplying each by \((P/A, 10\%, 30) = 9.4269\) reproduces the 30-year present worths above exactly, which is the arithmetic proof that the two methods are the same comparison.
7. Recommend. Over the common 30-year period Project B is worth \(\$24{,}603.41\) more than Project A, so$$\boxed{\text{Project B is the more economical plan}}$$with the essential qualification that it is the smaller loss and not a profit. The annual worths say exactly how far short each falls: Project B would need \(\$10{,}725.49\) per year of extra net revenue to break even at 10 percent, about 41 percent on top of its \(\$26{,}000\) revenue, and Project A would need \(\$13{,}335.40\) per year, about 57 percent on top of its \(\$23{,}500\). If this appraisal is a genuine go/no-go decision, the recommendation is to proceed with neither at these figures.