NivaarExam PrepOfficial exam papers ↗

11-CS-1 Engineering Economics · December 2018

Question 4 of 5: Three Investments — Rate of Return

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2018 — 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 follow; standard compound-interest factors are used and minor rounding is immaterial.

Question 4: Three Investments — Rate of Return (25 marks)

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.

(a) Selection by a Rate-of-Return Method

Each investment is the stream $-C$ at $t=0$, expenses $E_1$ growing by $G$ per year over $t=1\ldots7$, and the return $R$ at $t=7$. With $(P/A,i,7)$, $(P/G,i,7)$ and $(P/F,i,7)$, its worth at any rate is

$$PW(i) = -C - \left[E_1(P/A,i,7) + G(P/G,i,7)\right] + R\,(P/F,i,7)$$

Step 1 — screen each alternative against do-nothing by solving $PW(i^*)=0$ (trial-and-error or interpolation between bracketing rates):

InvestmentStandalone IRRvs MARR 8%PW at 8% (cross-check)
19.01%pass+$13,791
29.90%pass+$74,946
37.17%fail−$17,456

Investment 3 returns only 7.17%, below the 8% MARR, and is rejected outright. Step 2 — compare the survivors incrementally, cheapest first. The defender is Investment 1 ($100,000); the challenger is Investment 2 ($360,000). Differencing the two streams gives

$$\Delta_{2-1}:\quad -\$260{,}000 \text{ at } t=0;\quad -\$45{,}000 \text{ at } t=1 \text{ rising } \$2{,}000\text{/yr through } t=6;\quad +\$943{,}000 \text{ net at } t=7$$

The increment's first cash flow is an outflow, so this is an investment (not a loan) and the ordinary accept test applies: take the increment if its rate exceeds the MARR. Solving $PW_\Delta(i)=0$:

$$i^*_{\Delta_{2-1}} = \boxed{10.36\%} \;>\; 8\% \;\Rightarrow\; \text{the extra }\$260{,}000\text{ is justified}$$

There is no further challenger, so select Investment 2. The present-worth column confirms the same ranking (Investment 2 has the largest PW, and $74{,}946-13{,}791 = \$61{,}155$ is exactly the incremental PW that the 10.36% rate expresses).

(b) Is the Highest-ROR Alternative Always Best?

No. Mutually exclusive alternatives must be ranked by incremental rate of return (or, equivalently, by maximum present worth), never by standalone IRR: a small alternative can post a high percentage return on a small base and still create less total value than a larger one whose extra capital also clears the MARR.

On this paper, however, the two criteria happen to coincide—Investment 2 has both the highest standalone return (9.90%, against 9.01% and 7.17%) and the largest present worth—so the paper itself supplies no conflict. One is easy to construct: raise Investment 1's year-7 return from $600,000 to $650,000 and leave everything else unchanged. Investment 1's standalone IRR rises to 11.02%, above Investment 2's 9.90%, yet the increment 2 − 1 still earns 9.27% > 8%, so Investment 2 remains the correct choice despite the lower individual return.

(c) Different Result Under Future Worth?

No. $FW = PW(F/P,8\%,7)$ multiplies every alternative by the same positive constant, so the ranking is unchanged—Future Worth also selects Investment 2. No calculation is needed.