11-CS-1 Engineering Economics · December 2014
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — December 2014 — 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.
Each investment is a five-year stream: the initial cost at $t=0$, the annual expense at $t=1\ldots5$, and the return at the end of year 5. The net flow in year 5 is therefore the return less that year's expense, and
Step 1 — screen each alternative on its own rate of return. Setting $PW_k(i^*)=0$ and solving (by bisection; linear interpolation between bracketing trial rates gives the same values to a tenth of a point):
| Investment | Initial cost | Expense/yr | Return, yr 5 | IRR $i^*$ | $PW$ at 8% |
|---|---|---|---|---|---|
| 1 | $250,000 | $75,000 | $1,000,000 | 14.73% | +$131,100 |
| 2 | $350,000 | $100,000 | $1,450,000 | 16.59% | +$237,600 |
| 3 | $450,000 | $150,000 | $1,900,000 | 14.72% | +$244,200 |
Each stream has a single sign change (an outflow block, then one large net inflow), so each has exactly one real IRR, and all three clear the 8% MARR. All three are therefore individually acceptable; because they are mutually exclusive, the largest IRR does not settle the choice.
Step 2 — incremental analysis. Rank the alternatives by first cost (1, 2, 3), make the cheapest acceptable one the defender, and test each increment in turn. Every increment here opens with a negative cash flow (an extra $100,000 at $t=0$), so each is an investment rather than a loan and the ordinary accept test applies: take the increment if its rate exceeds the MARR.
Accept the increment, so Investment 2 displaces Investment 1 as the defender. (Cross-check at the MARR: $\Delta PW = +\$106{,}400 > 0$, consistent.)
This increment also clears the MARR, though only just, so it too is accepted. (Cross-check: $\Delta PW = +\$6{,}600 > 0$ at 8%.) No alternative is left to test, so
which is also the alternative with the largest present worth ($244,200) — as a correctly applied incremental rate-of-return analysis must be.
Watch the margin on the last increment. $\Delta(3-2)$ returns 8.76% against a MARR of 8% — a margin of only three-quarters of a percentage point. Had RTCI's MARR been 9% instead of 8%, that increment would be rejected and Investment 2 selected. The recommendation should be reported with that sensitivity attached.
No—PW, AW, and a correctly applied incremental ROR give the same decision; AW would also select Investment 3. (No calculation needed.)
No — and this very paper is the counter-example. Investment 2 has the highest standalone rate of return (16.59%, against 14.73% for Investment 1 and 14.72% for Investment 3), yet part (a) selects Investment 3. A rate of return is a ratio: it measures how hard each dollar works, not how many dollars are put to work, so a smaller alternative can post a higher percentage while adding less total value. Here the extra $100,000 committed in moving from Investment 2 to Investment 3 earns 8.76% — worse than the 16.59% that Investment 2 earns overall, but better than the 8% MARR, and since the alternative use of that money is by definition worth only the MARR, the extra outlay is worth making. That is precisely what the incremental test formalizes, and why it, rather than a ranking by standalone IRR, agrees with maximizing present worth. Ranking by standalone IRR is correct only for independent projects under no capital constraint, which the question's capacity constraint rules out.
