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23-CS-1 Engineering Economics · December 2019

Question 3 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 2019 — 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.

Question 3: 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

Each investment is a single outlay at $t=0$, an expense series that starts at $A_1$ and rises by an arithmetic gradient $G$ over years 1–7, and one receipt $F$ at $t=7$. Its net present worth at a rate $i$ is

$$PW(i) = -P - A_1(P/A,i,7) - G(P/G,i,7) + F(P/F,i,7)$$

The part asks for a rate of return method, so the rate that makes $PW(i)=0$ is what must be found — first for each alternative on its own (the screening test against do-nothing), then for the increment between the survivors (the selection test).

Step 1 — standalone IRR of each investment. Solving $PW(i)=0$ for each:

InvestmentStandalone IRRPasses MARR = 8%?Present Worth at 8%
19.01%Yes+$13,790
29.90%Yes+$74,950
37.17%No−$17,460

Investment 3 earns only 7.17%, below the 8% MARR, so it is rejected outright. (Equivalently, its present worth at 8% is negative: using $(P/A,8\%,7)=5.20637$, $(P/G,8\%,7)=14.0242$ and $(P/F,8\%,7)=0.583490$, $PW_3 = -185{,}000 - 55{,}000(5.20637) - 3{,}000(14.0242) + 850{,}000(0.583490) = -\$17{,}456$.)

Step 2 — incremental IRR between the survivors. Order the survivors by first cost and test the extra capital. The increment 2 − 1 is

$$\Delta P = -\$260{,}000,\quad \Delta A_1 = -\$45{,}000,\quad \Delta G = -\$2{,}000,\quad \Delta F = +\$1{,}000{,}000$$

Its first cash flow is an outflow, so this increment is an investment and the ordinary accept test applies (accept if its rate exceeds the MARR). Solving $\Delta PW(i)=0$:

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

At the MARR the same increment is worth $\Delta PW(8\%) = 74{,}950 - 13{,}790 = +\$61{,}160$, confirming the accept decision. Select Investment 2.

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

No. A standalone IRR is a percentage on whatever capital the alternative happens to require, and percentages cannot be compared across different investment sizes; the only valid rate-of-return test for mutually exclusive alternatives is whether each increment of capital earns at least the MARR — equivalently, maximise present worth.

On this paper the two criteria happen to agree: Investment 2 has both the highest standalone IRR (9.90%) and the highest present worth, so this exam does not itself supply a counter-example. One is easy to construct from its own data. Raise Investment 1's year-7 return from $600,000 to $700,000 and nothing else changes; Investment 1's standalone IRR rises to 12.86%, far above Investment 2's 9.90%, yet its present worth at 8% is only $+\$72{,}140$, still below Investment 2's $+\$74{,}950$. The incremental test would take Investment 2 anyway, and it would be right to — the higher-percentage alternative adds less total value.

(c) Different Result Under Future Worth?

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