11-CS-1 Engineering Economics · December 2016
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — December 2016 — 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.
Assumptions: present $t=0$ at end of 2016; construction $200M at ends of 2022–2024 ($t=6\text{–}8$); operation 2025–2059 ($t=9$ to $t=43$, 35 years); maintenance $5M/yr flat for the first 6 operating years ($t=9\text{–}14$), then increasing $50,000/yr for the remaining 29 years; savings $55M/yr over operation; salvage $+25M at $t=43$; $i=5\%$.
Time is measured in years from the present, $t=0$ at the end of 2016. Upward arrows are receipts to the province, downward arrows are disbursements; all amounts are in millions of dollars.
Figure 1 — Cash-flow diagram for the Niagara wind farm, end of 2016 ($t=0$) to end of 2059 ($t=43$). Construction: three disbursements of $200M at $t=6,7,8$ (ends of 2022, 2023, 2024). Operation runs $t=9$ to $t=43$ (2025–2059, 35 years): energy savings of $55M/yr, against maintenance of $5M/yr for the first six operating years ($t=9$ to $t=14$) and then rising by $50,000 each year to $6.45M at $t=43$. The scrap value of $25M is received at $t=43$. Only representative arrows are drawn for the two 35-year series.
Construction ($200M at $t=6\text{–}8$): $PW_c = 200(0.746215+0.710681+0.676839)=200(2.133735)=\$426.747$M.
Maintenance (valued first at $t=8$, one period before the first operating payment): a $5M annuity for the whole 35 operating years, $5(P/A,5\%,35)=5(16.374194)=\$81.871$M, plus the escalating part. The increments above $5M run from $t=15$ to $t=43$ — 29 payments of $0.05, 0.10, \dots, 1.45$ — so at $t=14$ they are worth $0.05[(P/A,5\%,29)+(P/G,5\%,29)]=0.05(15.141074+161.912605)=0.05(177.053679)=\$8.8527$M. Bringing that back six years by $(P/F,5\%,6)$ gives $8.8527(0.746215)=\$6.606$M, so maintenance at $t=8$ totals $81.871+6.606=\$88.477$M and
Savings ($55M/yr, $t=9\text{–}43$): at $t=8$, $55(16.374194)=\$900.581$M; $PW_S = 900.581(0.676839)=\$609.548$M.
Salvage: $PW_{sv}=25(P/F,5\%,43)=25(0.122704)=\$3.068$M. Combining:
The savings stream alone ($609.5M) outweighs the construction outlay ($426.7M) by a wide margin; maintenance, even with the escalation, costs less than a tenth of the savings in present terms, and the scrap value is almost immaterial at 43 years' discount.
The future worth is the same equivalence carried forward to the end of the project's life, so it uses the unrounded present worth ($PW = 125.984$M):
(Equivalently, $FW$ could be accumulated term by term — the two routes must agree, because $(F/P,5\%,43)$ multiplies every term identically.)
The present worth is positive at $+126M, so yes — on these figures it is a good investment for the province: at 5% the energy savings comfortably dominate the construction and maintenance costs, and the project would still show a positive present worth if the savings were about 21% lower than forecast.
