Question 4 of 6: Engineering Economics — present-value comparison of two projects of unequal life

Question 4: Engineering Economics — present-value comparison of two projects of unequal life (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, each described by a single initial outlay, a uniform annual operating cost, a major maintenance event every five years, a uniform annual revenue, and a service life; the discount rate is 10 percent per year. The lives are not equal, which is the feature that governs how the comparison must be made.

Find. The present-value profit of each project and, on a basis that is valid when the lives differ, which of the two is the more economical plan.

Approach. Collapse revenue and operating cost into a single net annual cash flow, discount it with the uniform-series present-worth factor, subtract the present worth of the major-maintenance outlays and the initial investment to obtain each project’s present-value profit over its own life, then — because the lives differ — convert to annual worth so the two are compared on a common basis, and confirm the ranking over the 30-year least common multiple.

1. Reduce each project to a net annual cash flow. Revenue and operating cost are both uniform, so they combine directly: $$\begin{aligned} R_A &= 23{,}500 - 12{,}500 = 11{,}000 \ \text{per year} \\ R_B &= 26{,}000 - 11{,}000 = 15{,}000 \ \text{per year} \end{aligned}$$ Project B earns more each year and costs less to build, so the whole comparison turns on whether the five extra years of life carry Project A.
2. Evaluate the factors at 10 percent. With $i = 0.10$, $$(P/A,i,n) = \frac{(1+i)^n-1}{i(1+i)^n}, \qquad (P/F,i,n) = (1+i)^{-n}$$ which give $(P/A,10\%,15) = 7.60608$, $(P/A,10\%,10) = 6.14457$, and the single-payment factors $(P/F,10\%,5) = 0.62092$, $(P/F,10\%,10) = 0.38554$ and $(P/F,10\%,15) = 0.23939$.
3. Present worth of Project A. Major maintenance falls in years 5, 10 and 15, so $$\begin{aligned} PW_A &= -170{,}000 + 11{,}000\,(P/A,10\%,15) - 15{,}000\,[\,0.62092+0.38554+0.23939\,] \\ &= -170{,}000 + 83{,}667 - 18{,}688 \\ &= \boxed{-105{,}021\ \text{CAD}} \end{aligned}$$ The result is negative: at 10 percent, Project A does not recover its investment.
4. Present worth of Project B. Over a ten-year life the maintenance events fall in years 5 and 10, so $$\begin{aligned} PW_B &= -150{,}000 + 15{,}000\,(P/A,10\%,10) - 13{,}000\,[\,0.62092+0.38554\,] \\ &= -150{,}000 + 92{,}169 - 13{,}084 \\ &= \boxed{-70{,}915\ \text{CAD}} \end{aligned}$$ Project B is also a loss, but a substantially smaller one.
5. Put the two on a common basis, because the lives differ. A raw comparison of the two present worths is not legitimate: one covers fifteen years of service and the other ten. Assuming the service is required indefinitely and each project is repeatable on the same terms, convert each present worth to an equivalent annual worth over its own life using $(A/P,10\%,15) = 0.131474$ and $(A/P,10\%,10) = 0.162745$: $$\begin{aligned} AW_A &= -105{,}021 \times 0.131474 = \boxed{-13{,}808\ \text{CAD per year}} \\ AW_B &= -70{,}915 \times 0.162745 = \boxed{-11{,}541\ \text{CAD per year}} \end{aligned}$$ Project B costs about 2,267 CAD per year less to own and operate, so Project B is the more economical plan.
6. Cross-check over the least common multiple of the lives. Fifteen and ten years repeat together over 30 years — Project A twice, Project B three times. With $(P/A,10\%,30) = 9.42691$, $$\begin{aligned} PW_{A,30} &= -13{,}808 \times 9.42691 = -130{,}166 \\ PW_{B,30} &= -11{,}541 \times 9.42691 = -108{,}797 \end{aligned}$$ The study-period comparison reproduces the annual-worth ranking exactly, as it must, which confirms that the unequal lives have been handled correctly rather than assumed away.
7. Test whether the loss is structural. Evaluating each project undiscounted, at $i = 0$, gives $-170{,}000 + 15(11{,}000) - 3(15{,}000) = -50{,}000$ for Project A and $-150{,}000 + 10(15{,}000) - 2(13{,}000) = -26{,}000$ for Project B. Neither total is positive, so no discount rate could rescue either alternative and neither has a real rate of return; the losses are structural, not an artefact of the 10 percent rate. The answer to “which is the most economical plan” is therefore “the one that loses least”, and that must be said plainly rather than presented as a profit.