23-CS-1 Engineering Economics · December 2014
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — December 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.
A sunk cost is a cost already incurred that cannot be recovered by any future decision. In replacement analysis it is irrelevant—only future costs and the asset's current market value matter; the original purchase price of an existing asset does not affect the keep-or-replace decision. Counting sunk costs leads to "throwing good money after bad."
The $600 running-in cost is incurred immediately, so it joins the purchase price in the $t=0$ outlay: $P=15{,}000+600=\$15{,}600$. The declining-balance salvage applies to the pump, so $S_n = 15{,}000(0.85)^{n}$, and the O&M stream is an arithmetic gradient with base $1,500 and gradient $660. Hence
Worked example, $n=3$: $S_3 = 15{,}000(0.85)^3 = \$9{,}212$; $(A/P,10\%,3)=0.402115$ and $(A/F,10\%,3)=0.302115$, so capital recovery $=15{,}600(0.402115)-9{,}212(0.302115)=6{,}273-2{,}783=\$3{,}490$; $(A/G,10\%,3)=0.93656$, so the O&M equivalent is $1{,}500+660(0.93656)=\$2{,}118$, giving $EAC(3)=\$5{,}608$. Repeating for each $n$:
| n (yr) | Salvage | Capital cost | O&M equiv. | EAC(n) |
|---|---|---|---|---|
| 1 | $12,750 | $4,410 | $1,500 | $5,910 |
| 2 | $10,838 | $3,828 | $1,814 | $5,642 |
| 3 | $9,212 | $3,490 | $2,118 | $5,608 |
| 4 | $7,830 | $3,234 | $2,412 | $5,646 |
The two components pull in opposite directions: capital recovery falls steadily ($4,410 → $3,234) as the large first cost is spread over more years, while the O&M equivalent rises ($1,500 → $2,412) as the gradient accumulates. Their sum therefore has a minimum, reached at $n=3$ years (EAC $\approx$ $5,608/yr). Since the question states that cash flows and interest rates stay constant over the plant's horizon, that minimum-cost life repeats, so the pump should be replaced every 3 years. The curve is flat near its minimum — $5,642 at $n=2$ and $5,646 at $n=4$, both within $40/yr of the optimum — so a two- or four-year cycle would cost very little more, and non-economic factors (warranty, shutdown scheduling) could reasonably decide between them.
This takes the depreciation basis as the $15,000 purchase price, the same basis the salvage estimate uses in part (b). If the $600 running-in cost were capitalized into the asset's cost instead (basis $15,600), the rate would be $1-(1{,}500/15{,}600)^{1/4} \approx 44.3\%$. Either way, this book-value rate is almost three times the 15% rate the question uses to estimate resale value. Book value and market value are different quantities, and only the market value belongs in the replacement analysis of parts (b) and (c).
