23-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.
Net annual = revenue − expenses: A = $54,000; B = $94,000. For independent projects, accept each whose IRR exceeds the MARR. Solving $Cost = A(P/A,i,10) + S(P/F,i,10)$:
Trial rates bracket each root (present worth at the trial rate):
| Project | PW at lower trial | PW at upper trial | Interpolated IRR |
|---|---|---|---|
| A | +$6,274 at 13% | −$6,191 at 14% | 13 + 6,274/12,465 = 13.5% |
| B | +$21,277 at 9% | −$7,712 at 10% | 9 + 21,277/28,989 = 9.7% |
So invest in Project A only (A clears the 10% hurdle; B does not). Because the projects are independent, each is judged on its own against the MARR and no incremental comparison is needed; B is rejected not because A is better but because B on its own earns less than A & G's 10% cost of capital (its PW at 10% is negative).
This is the incremental IRR of (B − A): extra cost $320,000, extra net $40,000/yr, extra salvage $45,000. Solving $320{,}000 = 40{,}000(P/A,i,10) + 45{,}000(P/F,i,10)$ gives $i \approx \boxed{6.0\%}$ (the increment's PW is +$16,495 at 5% and −$469 at 6%, so the root is 5.97%). The increment opens with an outflow, so it is an ordinary investment. At this rate the two projects have equal present worth (about $123,200 each), so they are economically equivalent. Below ~6% Project B is preferred; above it, Project A.
Taking each project forward to the end of year 10 at the 10% MARR, with $(F/P,10\%,10)=2.59374$ and $(F/A,10\%,10)=15.93742$:
Equivalently, $FW = PW\times(F/P,10\%,10)$ with $PW_A \approx +\$49{,}200$ and $PW_B \approx -\$7{,}700$, so Future Worth gives exactly the same accept/reject signal as Present Worth: $FW_A > 0$ (accept), $FW_B < 0$ (reject). Hence FW analysis also says invest in A only.
Now the projects are mutually exclusive. Since 4% is below the ~6% break-even MARR from part (b), the increment to the larger Project B is justified—so choose Project B. (No new calculation needed: below the crossover rate, B dominates.)
