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11-CS-1 Engineering Economics · May 2013

Question 3 of 6: Laundry Systems — Rate of Return

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

Notes on this paper

National Exams — May 2013 — 11-CS-1 Engineering Economics. Open book; non-communicating calculator permitted. Any five of the six questions constitute a complete paper; all questions are of equal value. Fully worked solutions to all six questions are given below. Standard compound-interest factors are used throughout; minor rounding differences are immaterial.

Question 3: Laundry Systems — Rate of Return (20 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 Rate-of-Return (Incremental) Analysis

Net cash flows over the 10-year horizon (savings less costs), with $(P/A,12\%,10)=5.6502$, $(P/F,12\%,10)=0.32197$, $(P/F,12\%,1)=0.89286$:

Step 1 — rate of return of each alternative on its own. Setting each present worth to zero and solving for $i^{*}$ (bisection; linear interpolation between 4-figure factor-table entries gives the same values):

$$\text{A:}\ 1000 = 200\,(P/A,i,10) \;\Rightarrow\; (P/A,i,10)=5.000 \;\Rightarrow\; i^{*}_{A}=15.10\%$$
$$\text{B:}\ 3600 = 750\,(P/A,i,10)+500\,(P/F,i,10) \;\Rightarrow\; i^{*}_{B}=16.99\%$$
$$\text{C:}\ 1250+3750\,(P/F,i,1) = 800\,(P/A,i,10)+1000\,(P/F,i,10) \;\Rightarrow\; i^{*}_{C}=13.40\%$$

All three clear the 12% MARR, so all three are acceptable in isolation and none can be eliminated yet. The systems are mutually exclusive, so the decision must be made on the increments, not by ranking the individual $i^{*}$ values.

Step 2 — order by size of investment, measured by the present worth of the outlays: A $1,000 < B $3,600 < C $1,250 + 3,750(0.89286) = $4,598.

Step 3 — increment B − A: an extra $2,600 now buys an extra $550/yr plus an extra $500 salvage. The first cash flow is an outflow, so this is an ordinary investment increment — accept it if its rate exceeds the MARR:

$$2600 = 550\,(P/A,i,10) + 500\,(P/F,i,10) \;\Rightarrow\; i^{*}_{B-A} = \boxed{17.66\%} > 12\% \;\Rightarrow\; \text{B beats A}$$

Step 4 — increment C − B. Check the sign of the first cash flow before applying any accept test. Because C pays only $1,250 today against B's $3,600, the increment receives $2,350 at $t=0$ and pays $3,700 at $t=1$: it is a financing (borrowing) flow, not an investment.

$$+2350 - \frac{3700}{1+i} + 50\sum_{t=2}^{10}(1+i)^{-t} + \frac{500}{(1+i)^{10}} = 0 \;\Rightarrow\; i^{*}_{C-B} = \boxed{53.1\%}$$

For a borrowing increment the accept test reverses: take the increment only if the rate you pay is below the MARR. Deferring three-quarters of C's price to year 1 costs 53.1%, far above the 12% the firm requires, so the increment is rejected. The present worths agree: $\Delta PW_{C-B} = 243.94 - 798.65 = -\$554.7$.

Decision: the A→B increment is accepted and the B→C increment is rejected, so install System B. Present worth confirms the ranking:

$$PW_A = -1000 + 200(5.6502) = +\$130$$
$$PW_B = -3600 + 750(5.6502) + 500(0.32197) = -3600 + 4237.7 + 161.0 = +\$799$$
$$PW_C = -1250 - 3750(0.89286) + 800(5.6502) + 1000(0.32197) = +\$244$$

On this paper the winner also happens to have the highest individual rate of return, but that is a coincidence of the numbers, not a rule — see part (b). The tell is already visible here: System A returns 15.10% against System C's 13.40%, yet A's present worth (+$130) is barely half of C's (+$244), because A's rate is earned on an investment less than a quarter the size.

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

No. For mutually exclusive alternatives, the option with the highest individual rate of return is not necessarily the best choice: a small, cheap project can show a high percentage return yet generate little total value. The correct procedure is incremental analysis—rank by increasing first cost and accept each increment whose incremental ROR exceeds the MARR—which selects the alternative that maximizes total worth (here, System B).

(c) Would Annual Worth Give a Different Result?

No. Present Worth, Annual Worth, and a correctly applied (incremental) Rate-of-Return analysis always give the same decision for a given MARR. An Annual-Worth comparison would also select System B.