11-CS-1 Engineering Economics · May 2014
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — May 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.
Using $(P/A,12\%,5)=3.60478$ and $(P/F,12\%,5)=0.56743$, the present worth of each is:
All three are profitable at 12%, but the question asks for a rate-of-return method, and for mutually exclusive alternatives that means an incremental analysis. The present worths above are used only as the independent cross-check at the end.
Solving $0 = -C - E(P/A,i,5) + R(P/F,i,5)$ for each stream:
Every alternative clears the 12% MARR, so all three are viable candidates and none is eliminated at this stage. Ranking by these percentages is not a valid selection rule (see part (d)).
Order: 1 ($500,000), 2 ($700,000), 3 ($900,000). Each increment begins with a cash outflow, so each is a genuine investment and the ordinary test applies — accept the increment if its rate of return exceeds the MARR. (Had an increment begun with an inflow it would be a financing flow and the test would reverse.)
| Increment | t = 0 | Years 1–5 | End of year 5 | Rate of return | vs. MARR 12% |
|---|---|---|---|---|---|
| 2 − 1 | −$200,000 | −$50,000/yr | +$900,000 | 21.06% | > 12% → accept: move to 2 |
| 3 − 2 | −$200,000 | −$100,000/yr | +$900,000 | 8.76% | < 12% → reject: stay at 2 |
The extra $200,000 spent to move from Investment 1 to Investment 2 earns 21.06%, comfortably above the MARR, so 2 becomes the current best. The next $200,000, to move from 2 to 3, earns only 8.76% — Blue Star would do better leaving that money in its ordinary 12% opportunities — so Investment 3 is rejected.
Blue Star should select Investment 2. The present-worth column confirms it independently: $PW_2 = \$224{,}600$ is the largest of the three, exactly as a correctly applied incremental rate-of-return analysis requires.
No. All three alternatives share the same 5-year study period, so each one's annual worth is simply its present worth multiplied by the same positive factor $(A/P,12\%,5)=0.27741$. Multiplying every candidate by one positive constant cannot reorder them, so the AW ranking is identical to the PW ranking and AW would also select Investment 2. (Had the lives differed, AW would still agree with PW provided PW were taken over a common study period such as the least common multiple.)
When the result is wanted as a single percentage for easy comparison against the MARR or cost of capital and for communication to management/investors, and when the MARR is uncertain (the ROR shows the break-even rate the project can tolerate). It is well suited to judging a single project's acceptability.
No. For mutually exclusive alternatives, the one with the highest standalone ROR need not maximize value—a smaller project can show a higher percentage yet add less total worth. The correct method is incremental ROR, accepting each increment whose return exceeds the MARR, which coincides with maximizing present worth.
A standard counter-example: a $10,000 project returning 30% adds $3,000 of value, while a $100,000 project returning 18% adds $18,000 — the lower percentage is plainly the better choice when both can be funded to the same limit. This paper happens to be a case where the two rules agree (Investment 2 has both the highest standalone ROR, 16.59%, and the highest present worth), but that coincidence is not something a candidate may rely on: only the incremental test is guaranteed to reproduce the maximum-present-worth answer.
