11-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.
