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11-CS-1 Engineering Economics · December 2017

Question 2 of 5: Hydropower Plant — Present and Future Worth

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

Notes on this paper

National Exams — December 2017 — 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 2: Hydropower Plant — Present and Future Worth (25 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.

Assumptions: present $t=0$ at end of 2017; construction $200M at ends of 2023–2025 ($t=6\text{–}8$); operation 2026–2060 ($t=9$ to $t=43$, 35 years); maintenance $5M/yr flat for the first 6 operating years then +$50,000/yr; savings $55M/yr over operation; salvage $+25M at $t=43$; $i=5\%$.

(a) Cash-Flow Diagram

Time is measured in years from the present, $t=0$ at the end of 2017. Upward arrows are receipts to the province, downward arrows are disbursements; all amounts are in millions of dollars.

+55 M/yr energy savings (t = 9 to 43) +25 M scrap value (t = 43) -200 M/yr construction (t = 6, 7, 8) maintenance -5 M/yr (t = 9 to 14), then rising 0.05 M/yr 0 6 8 9 14 43 0

Figure 1 — Cash-flow diagram for the hydropower plant, end of 2017 ($t=0$) to end of 2060 ($t=43$). Construction: three disbursements of $200M at $t=6,7,8$ (ends of 2023, 2024, 2025). Operation runs $t=9$ to $t=43$ (2026–2060, 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.

(b) Present Worth (t = 0, i = 5%)

Construction ($200M at $t=6\text{–}8$): $PW_c = 200(0.74622+0.71068+0.67684)=200(2.13374)=\$426.75$M.

Maintenance (value at $t=8$): $5M annuity for 35 yr $=5(16.3742)=\$81.871$M, plus the escalating part (a $0.05M gradient beginning in operation year 7, worth $8.853M at $t=14$, brought to $t=8$: $8.853(0.74622)=\$6.606$M) — total $81.871+6.606=88.477$M; then $PW_M=88.477(P/F,5\%,8)=88.477(0.67684)=\$59.88$M.

Savings ($55M/yr): at $t=8$, $55(16.3742)=\$900.58$M; $PW_S=900.58(0.67684)=\$609.55$M. Salvage: $PW_{sv}=25(P/F,5\%,43)=25(0.122704)=\$3.07$M. Combining:

$$PW = -426.75 - 59.88 + 609.55 + 3.07 \approx \boxed{+\$126.0\text{M}}$$

(c) Future Worth (t = 43, end of 2060)

$$FW = PW\,(F/P,5\%,43) = 126.0(8.1497) \approx \boxed{+\$1{,}027\text{M}}$$

(d) Good Investment?

The present worth is strongly positive (+$126M), so yes—it is a good investment: the energy savings comfortably outweigh the construction and maintenance costs at 5%.