11-CS-1 Engineering Economics · December 2017
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — December 2017 — 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 names the method, so the answer is built from rates of return, not present worths. Each investment is a conventional cash flow (one outlay at $t=0$, expenses in years 1–5, one receipt at $t=5$), so each has a single rate of return: the $i$ that solves
Step 1 — screen each alternative against the MARR. Solving by trial and interpolation (all figures in millions):
| Investment | Initial cost | Trial brackets | Rate of return | Clears 12%? |
|---|---|---|---|---|
| 2 | $5M | PW(14%)=+0.238, PW(15%)=−0.085 | 14.7% | yes |
| 3 | $7M | PW(16%)=+0.259, PW(17%)=−0.171 | 16.6% | yes |
| 1 | $9M | PW(14%)=+0.437, PW(15%)=−0.164 | 14.7% | yes |
All three are acceptable on their own, so the choice among mutually exclusive alternatives must be made incrementally, in order of increasing first cost: 2, then 3, then 1.
Step 2 — Δ(2→3), the extra $2M of investment. The increment is $-2$ at $t=0$, $-0.5$/yr in years 1–5, $+9$ at $t=5$. Its first cash flow is negative, so this is a genuine investment increment and the ordinary accept test applies (accept if $\Delta ROR > MARR$):
The extra $2M earns 21.1%, well above the MARR, so 3 displaces 2 as the current best.
Step 3 — Δ(3→1), the next $2M. The increment is $-2$ at $t=0$, $-1$/yr in years 1–5, $+9$ at $t=5$ — again an investment increment:
That extra $2M earns only 8.8% — less than the 12% the money could earn elsewhere — so the step to Investment 1 is rejected and Investment 3 stands. Choose Investment 3.
Cross-check by present worth at 12%, using $(P/A,12\%,5)=3.60478$ and $(P/F,12\%,5)=0.56743$:
The maximum present worth is Investment 3, and the incremental present worths agree with the incremental rates: $\Delta PW_{2\to3}=+\$1.30$M (accepted) and $\Delta PW_{3\to1}=-\$0.50$M (rejected). The two methods select the same alternative, as they must.
No—PW, AW, and incremental ROR give the same decision; AW also selects Investment 3.
When the result is wanted as a single percentage to compare against the MARR/cost of capital and communicate to management, and when the MARR is uncertain (the ROR shows the break-even rate). It suits judging a single project's acceptability.
No. A smaller investment can have a higher standalone ROR yet add less total value, because a rate says nothing about the size of the base it is earned on: 20% on $1M adds less wealth than 15% on $10M when the firm's alternative use of capital earns only the MARR. Mutually exclusive alternatives must therefore be chosen by incremental ROR — accepting each extra increment of investment that itself earns more than the MARR — which always coincides with maximum present worth.
On this paper the two happen to agree: Investment 3 has both the highest standalone rate (16.6%, against 14.7% for each of the others) and the highest present worth. That agreement is a coincidence of these particular cash flows, not a rule — had Investment 2 returned, say, $22M instead of $20M, it would have carried the highest standalone rate while Investment 3 still won on incremental analysis. Ranking by standalone ROR is unsafe in general, which is exactly why step 2 of part (a) tests the increments rather than the alternatives.
