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 appraised at $i = 10\%$ per year, with unequal service lives of 9 and 12 years.

Find. The present value profit (net present worth) of each alternative and, allowing correctly for the unequal lives, a recommendation as to which plan is the more economical.

Approach. Reduce each project to a net annual benefit, discount that annuity and the periodic maintenance to year 0 over the project’s own life, and then — because the lives differ — convert both to an equivalent annual worth (equivalently, repeat each project over the 36-year least common multiple) before comparing.

1. Reduce each project to a net annual benefit. Revenue and operating cost are both uniform annual series, so they combine into a single annuity for each alternative: $$A_{\text{net},A} = 12{,}000 - 2{,}500 = \$9{,}500\text{/yr}, \qquad A_{\text{net},B} = 16{,}000 - 1{,}000 = \$15{,}000\text{/yr}$$ The major maintenance is not part of this annuity: it is a set of discrete disbursements at years 3, 6 and 9 for A, and years 4, 8 and 12 for B, and each must be discounted individually.
2. Evaluate the discount factors. With $i = 0.10$, $$(P/A, 10\%, 9) = \frac{1-(1.10)^{-9}}{0.10} = 5.759024, \qquad (P/A, 10\%, 12) = \frac{1-(1.10)^{-12}}{0.10} = 6.813692$$ and for the maintenance events, $(P/F,10\%,n) = (1.10)^{-n}$: $$\begin{aligned} \text{A:}&\quad (P/F,3) + (P/F,6) + (P/F,9) = 0.751315 + 0.564474 + 0.424098 = 1.739886\\ \text{B:}&\quad (P/F,4) + (P/F,8) + (P/F,12) = 0.683013 + 0.466507 + 0.318631 = 1.468152 \end{aligned}$$
3. Present value profit over each project’s own life. For Project A, $$PW_A = -60{,}000 + 9{,}500(5.759024) - 12{,}000(1.739886)$$ $$PW_A = -60{,}000 + 54{,}710.73 - 20{,}878.64$$ $$\boxed{PW_A = -\$26{,}167.91}$$ and for Project B, $$PW_B = -70{,}000 + 15{,}000(6.813692) - 17{,}000(1.468152)$$ $$PW_B = -70{,}000 + 102{,}205.38 - 24{,}958.58$$ $$\boxed{PW_B = +\$7{,}246.80}$$ Project B earns a present value profit; Project A returns a present value loss. The negative figure is not an arithmetic slip — it is the correct statement that A does not recover its investment at a 10 % cost of capital, and the answer must say so plainly rather than quietly changing a sign.
4. Test whether A’s loss is structural. Adding A’s cash flow without any discounting at all gives $$9{,}500(9) - 60{,}000 - 12{,}000(3) = 85{,}500 - 60{,}000 - 36{,}000 = -\$10{,}500$$ so Project A loses money even at a zero discount rate, and no interest rate can rescue it; its internal rate of return is negative (about −4.3 %). The same test on B gives $\;15{,}000(12) - 70{,}000 - 17{,}000(3) = +\$59{,}000$, and solving $PW_B(i) = 0$ gives an internal rate of return of about 12.3 %, comfortably above the 10 % hurdle. This is the check that separates a genuinely unprofitable alternative from one that merely fails at the chosen discount rate.
5. Correct for the unequal lives. The two present worths above cover 9 years of service and 12 years of service respectively, so they are not directly comparable; comparing them as they stand is the error the question invites. Under the usual repeatability assumption, convert each to an equivalent annual worth, $$AW = PW \times (A/P, i, n) = \frac{PW}{(P/A,i,n)}$$ $$AW_A = \frac{-26{,}167.91}{5.759024} = -\$4{,}543.81\text{/yr}, \qquad AW_B = \frac{7{,}246.80}{6.813692} = +\$1{,}063.56\text{/yr}$$ Equivalently, repeat each project over the least common multiple of the lives, $\operatorname{lcm}(9,12) = 36$ years, with $(P/A,10\%,36) = 9.676508$: $$PW_A^{36} = -4{,}543.81(9.676508) = -\$43{,}968.21, \qquad PW_B^{36} = 1{,}063.56(9.676508) = +\$10{,}291.59$$ The two routes agree, as they must, and both rank B above A by a margin of about $5,607 per year of service.
6. Recommendation. $$\boxed{\text{Adopt Project B: } PW_B = +\$7{,}247 \text{ (}AW = +\$1{,}064\text{/yr)} \text{ versus } PW_A = -\$26{,}168 \text{ (}AW = -\$4{,}544\text{/yr)}}$$ Project B is the more economical plan and is the only one of the two that is economically justified at all. Project A should be rejected outright rather than merely ranked second: it fails the absolute test as well as the relative one, and if the service it provides is mandatory, then some third alternative must be sought rather than accepting A.