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23-CS-1 Engineering Economics · May 2016

Question 2 of 5: Solar Power Station — Present and Future Worth

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

Notes on this paper

National Exams — May 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 2: Solar Power 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.

Assumptions: present $t=0$ at end of 2016; construction $150M at ends of 2023–2025 ($t=7\text{–}9$); operation from 2026 to 2060 ($t=10$ to $t=44$, 35 years)—maintenance ($10M rising $0.1M/yr arithmetic) and savings ($45M/yr) both run over this period; salvage $+30M at $t=44$; $i=6\%$.

(a) Cash-Flow Diagram

0 7 9 10 44 construction 150/yr (t = 7–9) savings 45/yr (t = 10–44) salvage 30 (t = 44) maintenance 10 rising 0.1/yr (t = 10–44) t (yr) amounts in millions of dollars; t = 0 at end of 2016; arrows to scale

Figure 1 — Cash-flow diagram, present (end of 2016, $t=0$) to end of 2060 ($t=44$), in millions of dollars. Downward arrows are costs: construction of 150 at $t=7,8,9$ (2023–2025) and maintenance of 10 at $t=10$ rising by 0.1 each year to 13.4 at $t=44$. Upward arrows are benefits: savings of 45 each year from $t=10$ to $t=44$, plus the 30 salvage at $t=44$.

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

Construction ($150M at $t=7\text{–}9$): $PW_c = 150(0.66506+0.62741+0.59190)=150(1.88437)=\$282.66$M.

Maintenance (annuity + arithmetic gradient, $t=10\text{–}44$): worth at $t=9$ is $10(P/A,6\%,35)+0.1(P/G,6\%,35)=10(14.4982)+0.1(165.743)=\$161.56$M; then $\times(P/F,6\%,9)=0.59190$ gives $PW_M=\$95.63$M.

Savings ($45M/yr, $t=10\text{–}44$): worth at $t=9$ is $45(14.4982)=\$652.42$M; then $\times0.59190$ gives $PW_S=\$386.17$M.

Salvage: $PW_{sv}=30(P/F,6\%,44)=30(0.077009)=\$2.31$M. Combining:

$$PW = -282.66 - 95.63 + 386.17 + 2.31 \approx \boxed{+\$10.2\text{M}}$$

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

$$FW = PW\,(F/P,6\%,44) = 10.2(12.9855) \approx \boxed{+\$132\text{M}}$$

(d) Good Investment?

The present worth is positive (+$10.2M), so yes—it is a good investment: the energy savings and salvage outweigh the construction and maintenance costs at the 6% required return.