23-CS-1 Engineering Economics · May 2015
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — May 2015 — 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 projects are ranked by internal rate of return, not by present worth. Each project's cash flow is one outlay at $t=0$, a base annuity plus an arithmetic gradient of expenses in years 1–7, and a single receipt at $t=7$; the present worth at any trial rate $i$ is
Each stream changes sign exactly once (net outflows followed by one large net inflow), so each has a single, unambiguous IRR.
Step 1 — standalone IRR of each project (trial rates, then linear interpolation):
| Project | PW at lower trial rate | PW at upper trial rate | IRR (interpolated) | vs. MARR 8% |
|---|---|---|---|---|
| 1 | 9%: +$153 | 10%: −$12,368 | 9.01% | accept |
| 2 | 9%: +$33,955 | 10%: −$3,815 | 9.90% | accept |
| 3 | 7%: +$3,782 | 8%: −$17,456 | 7.18% | reject |
For Project 1, for example, $i^{*} = 9\% + \dfrac{153}{153+12{,}368}(1\%) = 9.01\%$. Project 3 earns only about 7.2%, below the 8% MARR, so it is not acceptable and drops out; Projects 1 and 2 both clear the hurdle.
Step 2 — incremental IRR on the surviving alternatives. Because only one project may be chosen, the alternatives are mutually exclusive and must be compared incrementally, cheapest first. The defender is Project 1 (first cost $100,000); the challenger is Project 2 ($360,000). The increment 2 − 1 is
Its first cash flow is negative and it has one sign change, so it is an ordinary investment increment and the usual accept test applies (take the increment if its rate exceeds the MARR). Trial rates give $PW_\Delta(10\%)=+\$8{,}553$ and $PW_\Delta(11\%)=-\$14{,}765$, so
The extra $260,000 committed by moving from Project 1 to Project 2 earns about 10.4%, comfortably above the 8% MARR, so the step is worth taking. (For completeness the increment 3 − 1 — $85,000 extra outlay, $15,000 extra first-year expense rising $1,000/yr, $250,000 extra return — earns only 3.76%, which confirms Project 3's rejection by the incremental route as well.)
As a cross-check for part (c), the present worths at the 8% MARR—using $(P/A,8\%,7)=5.20637$, $(P/G,8\%,7)=14.02422$, $(P/F,8\%,7)=0.58349$—are:
The signs agree with the IRR screening (Projects 1 and 2 positive, Project 3 negative) and the largest present worth is Project 2—the same answer the incremental rate-of-return analysis gave.
No. The highest standalone ROR need not maximize value—a smaller project can show a higher percentage yet add less total worth, because a rate of return says nothing about the size of the capital it is earned on. Mutually exclusive alternatives must therefore be compared by incremental ROR (accept each increment whose return exceeds the MARR), which coincides with maximizing present worth. This paper happens to be the benign case: Project 2 has both the highest standalone IRR (9.90% against Project 1's 9.01%) and the highest present worth, so the shortcut and the correct method agree. Had Project 1's standalone IRR been the higher of the two, the incremental test in part (a) would still have selected Project 2, since the extra $260,000 it commits earns 10.37%—more than the 8% the firm requires of it.
No. Present Worth, Annual Worth, and a correctly applied incremental ROR give the same decision at a given MARR; the PW computation above already confirms Project 2 as the best. (No further calculation needed.)
