23-CS-1 Engineering Economics · December 2015
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — December 2015 — 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 2015; construction $10M at ends of 2020–2023 ($t=5\text{-}8$); O&M from 2024 to 2068 ($t=9$ to $t=53$, 45 payments), first $2M growing 1%/yr; overhaul $8M in 2045 ($t=30$); salvage $+15M at $t=53$; $i=6\%$.
Figure 1 — Cash-flow diagram from the present (t = 0, end of 2015) to the end of 2068 (t = 53). Downward arrows are disbursements, the upward arrow is the salvage receipt; only eight of the 45 O&M payments are drawn, and their arrows lengthen to the right because the series grows 1% per year.
Construction ($10M at $t=5\text{-}8$): $PW_c = 10(2.744688) = \$27.45$M.
O&M (geometric, $A_1=2$M, $g=1\%$, $n=45$): worth at $t=8$ is $P_8 = 2\frac{1-(1.01/1.06)^{45}}{0.05}=2(17.7263)=\$35.45$M; then $PW_{OM}=35.45(P/F,6\%,8)=35.45(0.62741)=\$22.24$M.
Overhaul: $PW_{oh}=8(P/F,6\%,30)=8(0.174110)=\$1.39$M. Salvage: $PW_s = 15(P/F,6\%,53)=15(0.045582)=\$0.68$M.
