23-CS-1 Engineering Economics · December 2015
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — December 2015 — 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.
$EAC(n) = 49{,}000(A/P,10\%,n) - S_n(A/F,10\%,n) + [\text{annualized O\&M}]$, where the O&M present worth is annualized with $(A/P,10\%,n)$:
| n (yr) | Capital cost | O&M equiv. | EAC(n) |
|---|---|---|---|
| 1 | $22,400 | $17,000 | $39,400 |
| 2 | $18,769 | $19,057 | $37,826 |
| 3 | $14,974 | $21,398 | $36,372 |
| 4 | $14,005 | $24,065 | $38,070 |
The $49,000 market value is the defender's first cost — it is what keeping the cutter costs the plant in forgone sale proceeds. Taking $n = 2$ as a worked example: $(A/P,10\%,2)=0.576190$ and $(A/F,10\%,2)=0.476190$, so the capital cost is $49{,}000(0.576190) - 19{,}875(0.476190) = 28{,}233 - 9{,}464 = \$18{,}769$. The O&M charges are not uniform, so they are first brought to the present, $17{,}000(0.909091) + 21{,}320(0.826446) = \$33{,}074$, and then spread over the two years: $33{,}074(0.576190) = \$19{,}057$. The two together give $EAC(2) = \$37{,}826$.
The EAC is minimized at n = 3 years (≈$36,372), so the old cutter's remaining economic life is 3 years.
The new cutter's EAC ($35,000) is below the old cutter's minimum EAC over its remaining life ($36,372), and also below the old cutter's marginal cost of keeping it even one more year ($39,400). Since it is cheaper to own the new cutter than to keep the old one, yes—replace the old cutter now.
The original $130,000 purchase price (paid 7 years ago) is the sunk cost. It cannot be recovered and is irrelevant to the replacement decision—only the current market value ($49,000) and future costs matter.
