11-CS-1 Engineering Economics · May 2017
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — May 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 is worth 12 marks and explicitly says use a rate of return method, so the decision must be reached on rates. For mutually exclusive alternatives that means the incremental procedure: order by increasing first cost (2, 3, 1), check that the cheapest clears the MARR on its own, then buy each successive increment only if that increment itself earns more than 12%. Every alternative here begins with an outflow, so each increment below is an ordinary investment and the normal “accept if the rate exceeds the MARR” test applies (a financing increment, which would open with a receipt, reverses that test).
Step 1 — is the cheapest alternative acceptable on its own? Solve $-5-1.5(P/A,i,5)+20(P/F,i,5)=0$. The expression is $+0.238$ at $i=14\%$ and $-0.389$ at $i=16\%$, so the root lies between; solving gives $i^{*}_{2}=\boxed{14.73\%}$. Since $14.73\% > 12\%$, Investment 2 is acceptable and becomes the current best, and do-nothing is discarded.
Step 2 — increment 2 to 3: $-\$2$M now, $-\$0.5$M/yr for five years, $+\$9$M at year 5.
$21.06\% > 12\%$, so the extra $2M of capital earns well above the hurdle rate: advance to Investment 3.
Step 3 — increment 3 to 1: $-\$2$M now, $-\$1$M/yr, $+\$9$M at year 5.
$8.76\% < 12\%$, so the last $2M fails to earn the MARR and that step is rejected. Bright Star should select Investment 3.
Cross-check by present worth at 12%, using $(P/A,12\%,5)=3.60478$ and $(P/F,12\%,5)=0.56743$ (all figures in millions):
The two routes agree, as they must: $PW_3-PW_2=+\$1.30$M confirms the accepted increment and $PW_1-PW_3=-\$0.50$M the rejected one. Investment 3 is the recommended choice.
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 standalone rate of return measures efficiency per dollar invested, not the amount of value created, so a small alternative can post the highest percentage while a larger one adds more worth. Mutually exclusive alternatives must therefore be ranked by incremental ROR, which always agrees with maximum present worth at the MARR.
On this paper the two criteria happen to coincide: the standalone rates are $i^{*}_{1}=14.72\%$, $i^{*}_{2}=14.73\%$ and $i^{*}_{3}=16.59\%$, so Investment 3 carries both the highest standalone return and the largest present worth, and the given data contain no counter-example to point at. The ranking is nonetheless not driven by size — the smallest alternative (2) edges out the largest (1) on standalone return while creating only $0.94M of value against $1.75M. A genuine counter-example is one keystroke away: raise Investment 2's year-5 return from $20M to $22M and its standalone rate becomes 17.75%, the highest of the three, yet $PW_2=+\$2.08$M still trails $PW_3=+\$2.25$M and Investment 3 still wins the incremental comparison.
