NivaarExam PrepOfficial exam papers ↗

23-CS-1 Engineering Economics · December 2013

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

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

Notes on this paper

National Exams — December 2013 — 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 are given below; standard compound-interest factors are used and minor rounding is immaterial.

Question 2: 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 2015; construction $75M at ends of 2020–2024 ($t=5\text{–}9$); O&M from 2025 to 2054 ($t=10$ to $t=39$), first payment $5M at $t=10$ growing geometrically at 1%; salvage $+10M at $t=39$; $i=6\%$.

(a) Cash-Flow Diagram

t 0 5 9 10 39 five construction outlays of 75M (t = 5 to 9) O&M: 5M at t = 10, rising 1% per year through t = 39 salvage +10M (t = 39) t = 0 is the end of 2015; amounts in millions of dollars

Figure 1 — Cash-flow diagram, end of 2015 ($t=0$) to end of 2054 ($t=39$). Downward arrows are disbursements, the upward arrow is a receipt; arrow lengths are indicative only and not to scale.

Down-arrows: five $75M construction costs ($t=5\text{–}9$) and the rising O&M series ($t=10\text{–}39$); the up-arrow is the $10M salvage at $t=39$.

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

Construction ($75M at $t=5\text{–}9$):

$$PW_c = 75\sum_{t=5}^{9}(P/F,6\%,t) = 75(0.74726+0.70496+0.66506+0.62741+0.59190) = 75(3.33659)=\$250.24\text{M}$$

O&M (geometric, $A_1=5$M at $t=10$, $g=1\%$, $n=30$). Worth at $t=9$:

$$P_9 = 5\,\frac{1-\left(\frac{1.01}{1.06}\right)^{30}}{0.06-0.01} = 5\,\frac{1-0.23467}{0.05} = 5(15.3065)=\$76.53\text{M}$$
$$PW_{OM} = 76.53\,(P/F,6\%,9) = 76.53(0.59190)=\$45.30\text{M}$$

Salvage: $PW_s = 10\,(P/F,6\%,39)=10(0.10306)=\$1.03$M. Combining (costs negative):

$$PW = -250.24 - 45.30 + 1.03 \approx \boxed{-\$294.5\text{M}}$$

(c) Future Worth (t = 39, end of 2054)

$$FW = PW\,(F/P,6\%,39) = -294.5(1.06)^{39} = -294.5(9.7035) \approx \boxed{-\$2{,}858\text{M}}$$

The station carries a net present cost of about $294.5M (≈$2.86 billion by 2054), as expected when only costs—not the value of the power delivered—are counted.