11-CS-1 Engineering Economics · December 2015
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — December 2015 — 11-CS-1 Engineering Economics. Open book; non-communicating calculator permitted. Any four of the five questions constitute a complete paper; all questions are of equal value. Fully worked solutions to all five questions follow; standard compound-interest factors are used and minor rounding is immaterial.
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.
The part asks specifically for a rate-of-return method, so the analysis has two stages: first discard any project that cannot earn the 12% MARR on its own, then choose among the survivors by incremental rate of return. Each project's rate of return $i^*$ is the rate that zeroes its own present worth,
Every project's cash-flow sequence has exactly one sign change (outflow at $t=0$, outflows in years 1–4, a single large inflow in year 5), so by Descartes' rule each has a unique real rate of return. Solving each by interpolation between trial rates:
| Project | First cost | Expenses / yr | Year-5 return | Rate of return $i^*$ | Accept vs 12% MARR? |
|---|---|---|---|---|---|
| 1 | $250,000 | $75,000 | $1,000,000 | 14.73% | Yes |
| 2 | $350,000 | $100,000 | $1,450,000 | 16.59% | Yes |
| 3 | $450,000 | $150,000 | $1,900,000 | 14.72% | Yes |
For Project 1, for instance, trial rates of 14% and 15% give present worths of $+\$11{,}888$ and $-\$4{,}235$, and interpolating between them gives $i^*_1 = 14 + \frac{11{,}888}{11{,}888+4{,}235} = 14.74\%$, against the exact 14.73%.
All three clear the MARR, so all three are acceptable — but only one may be chosen, and the standalone rates cannot rank mutually exclusive alternatives. Order them by increasing first cost (1, 2, 3) and test each extra increment of investment in turn. Both increments begin with a cash outflow, so each is a genuine investment and the ordinary "accept if the rate exceeds the MARR" test applies.
| Increment | Extra first cost | Extra expense / yr | Extra year-5 return | Incremental ROR | Decision |
|---|---|---|---|---|---|
| 2 − 1 | $100,000 | $25,000 | $450,000 | 21.06% | > 12% → take it; move to Project 2 |
| 3 − 2 | $100,000 | $50,000 | $450,000 | 8.76% | < 12% → reject; stay with Project 2 |
The extra $100,000 needed to move from Project 1 to Project 2 earns 21.06%, comfortably above the 12% MARR, so it is worth spending. The next $100,000, needed to move from Project 2 to Project 3, earns only 8.76% — that increment would have to be funded at 12% to earn 8.76%, which destroys value. Choose Project 2.
Present-worth cross-check. Using $(P/A,12\%,5)=3.60478$ and $(P/F,12\%,5)=0.56743$:
Project 2 maximizes present worth, agreeing with the incremental result — as it must, since the rejected increment's 8.76% rate is exactly the crossover rate at which Projects 2 and 3 have equal present worth, and the MARR of 12% sits above it.
No. Annual Worth is just Present Worth multiplied by $(A/P,12\%,5)=0.27741$, a single positive constant applied identically to all three alternatives. Multiplying by a positive constant cannot reorder them, so AW selects Project 2 as well ($AW_1 = \$13{,}057$, $AW_2 = \$31{,}150$, $AW_3 = \$24{,}244$). The lives are equal here, so no study-period adjustment is needed either; PW, AW and incremental ROR must all agree.
No. A rate of return is a ratio and carries no information about scale, so a small project can post a high percentage while adding little total value. Mutually exclusive alternatives must therefore be ranked by incremental rate of return, which always agrees with maximum present worth.
This paper supplies its own illustration. Project 1 has a marginally higher standalone rate of return than Project 3 (14.73% against 14.72%), so a naive ranking on standalone ROR would place Project 1 ahead of Project 3 — yet Project 3's present worth ($87,400) is nearly double Project 1's ($47,100), because the same percentage is earned on a much larger investment. Only the incremental test gets the ordering right.
When the result is wanted as a single percentage to compare against the MARR or cost of capital and to communicate to management, and when the MARR is uncertain (the ROR shows the break-even rate). It is well suited to judging a single project's acceptability.
