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11-CS-1 Engineering Economics · May 2017

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

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

Notes on this paper

National Exams — May 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 Station — 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.

Timeline: present $t=0$ at end of 2016; construction $75M at ends of 2023–2025 ($t=7\text{–}9$); operation 2026–2060 ($t=10$ to $t=44$, 35 years)—maintenance ($5M rising $0.05M/yr) and savings ($22.5M/yr) both over this period; overhaul $10M in 2045 ($t=29$); salvage $+15M at $t=44$; $i=6\%$.

(a) Cash-Flow Diagram

Set $t=0$ at the end of 2016. Three construction disbursements of $75M fall at $t=7,8,9$ (2023–2025). From $t=10$ (2026) to $t=44$ (2060) two series run together: maintenance starting at $5M and growing by $0.05M each year, and savings of $22.5M each year. One extra disbursement of $10M falls at $t=29$ (2045) and the $15M salvage is received at $t=44$.

0 5 7 9 15 20 25 29 35 40 44 t (years from end of 2016) +22.5 M/yr savings (t = 10 to 44) +15 M salvage (t = 44) -75 M/yr construction, t = 7, 8, 9 -5 M rising 0.05 M/yr (t = 10 to 44) -10 M overhaul (t = 29, year 2045) All amounts in millions of dollars; arrow lengths are schematic, not to scale. i = 6% per year, all flows at year-end.

Figure 1 — Cash-flow diagram, end of 2016 ($t=0$) to end of 2060 ($t=44$). Receipts point up, disbursements down.

(b) Present Worth (t = 0, i = 6%)

Construction ($75M at $t=7\text{–}9$): $PW_c = 75(1.884368) = \$141.33$M.

Maintenance (annuity + gradient, $t=10\text{–}44$): at $t=9$, $5(P/A,6\%,35)+0.05(P/G,6\%,35)=5(14.4982)+0.05(165.743)=\$80.78$M; then $\times(P/F,6\%,9)=0.591898$ gives $PW_M=\$47.81$M.

Overhaul: $PW_{oh}=10(P/F,6\%,29)=10(0.184557)=\$1.85$M. Savings ($22.5M/yr, $t=10\text{–}44$): at $t=9$, $22.5(14.4982)=\$326.21$M; $\times0.591898$ gives $PW_S=\$193.08$M. Salvage: $PW_{sv}=15(P/F,6\%,44)=15(0.077009)=\$1.16$M. Combining:

$$PW = -141.33 - 47.81 - 1.85 + 193.08 + 1.16 \approx \boxed{+\$3.25\text{M}}$$

(c) Future Worth (t = 44, end of 2060)

$$FW = PW\,(F/P,6\%,44) = 3.2529(12.98548) \approx \boxed{+\$42.24\text{M}}$$

(d) Good Investment?

The present worth is positive but small (+$3.25M), so the project is marginally worthwhile at 6%. Because the margin is thin relative to the ~$140M construction cost, the decision is sensitive to the estimates (savings, cost, rate), so a sensitivity check would be prudent before committing.