23-CS-1 Engineering Economics · May 2016
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — May 2016 — 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 for a rate of return method, so the decision must be reached from rates, not from present worths. Each option's net stream is $-P_0$ now, $-A$ in each of years 1–5, and $+R$ at year 5. Note the three options are ordered by increasing initial investment, which is what an incremental analysis requires.
Step 1 — is each option acceptable on its own? Solving $-P_0 - A(P/A,i^*,5) + R(P/F,i^*,5) = 0$ by trial and interpolation:
All three exceed the 12% MARR, so all three are acceptable and none can be discarded before the incremental step.
Step 2 — incremental analysis. Take the smallest investment (Option 1) as the defender and challenge it with the next larger. Each increment's first cash flow is an outflow (−$200,000 at $t=0$), so each increment is an ordinary investment and the normal accept test applies: take the increment only if its rate is above the MARR.
$21.06\% > 12\%$, so the extra $200,000 invested in Option 2 earns well above the MARR—Option 2 replaces Option 1 as the current best.
$8.76\% < 12\%$, so the further step to Option 3 does not earn the MARR and is rejected. Choose Option 2.
Cross-check by present worth (using $(P/A,12\%,5)=3.60478$ and $(P/F,12\%,5)=0.56743$):
Option 2 has the largest present worth, confirming the incremental result. Note that the increments' rates, not the standalone rates, carry the decision: Option 3 has a respectable 14.72% on its own, and is rejected only because the last $200,000 it requires returns 8.76%.
No. With a common 5-year horizon, Annual Worth is present worth multiplied by $(A/P,12\%,5)$, the same positive constant for all three options, so AW ranks them exactly as PW does—and PW at the MARR always agrees with a correctly executed incremental rate-of-return analysis. AW would therefore also select Option 2. (The methods differ in what they report—a percentage versus a dollar amount per year—not in the decision they lead to.)
No. A smaller option can have a higher standalone ROR yet add less total value, because a rate of return is a percentage and is blind to the size of the investment it is earned on. Mutually exclusive options must therefore be chosen by incremental ROR, which always coincides with maximum present worth. On this paper the two rules happen to agree—Option 2 has both the highest standalone rate (16.59%) and the highest present worth—but that agreement is a coincidence of these numbers, not a rule: had Option 1's return been large enough to push $i^*_1$ past 16.59% while leaving $i^*_{2-1}$ above 12%, Option 2 would still be the correct choice despite no longer having the highest standalone rate.
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.
