NivaarExam PrepOfficial exam papers ↗

11-CS-1 Engineering Economics · December 2018

Question 3 of 5: Quebec Bridge — Present and Future Worth

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

Notes on this paper

National Exams — December 2018 — 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 follow; standard compound-interest factors are used and minor rounding is immaterial.

Question 3: Quebec Bridge — 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 ($t=0$ = end 2018): construction $25M at $t=2\text{–}5$ (ends 2020–2023); O&M a geometric series, first $2.5M at $t=6$ (2024) growing 2.8%/yr for 30 years to $t=35$ (2053); salvage $+5M at $t=35$; $i=8\%$.

(a) Cash-Flow Diagram

Cash flows in millions of dollars; t = 0 is the end of 2018. t 0 2 5 6 35 2018 2020 2023 2053 −25 construction, four payments at t = 2, 3, 4, 5 O&M: −2.5 at t = 6, growing 2.8%/yr to t = 35 +5 salvage (t = 35) Down = cost, up = receipt. The only inflow is the salvage; the bridge earns no toll revenue.

Figure 1. Cash-flow diagram, Quebec bridge, end of 2018 to end of 2053 (millions of dollars, i = 8%).

(b) Present Worth (i = 8%)

Construction: $25(P/A,8\%,4)(P/F,8\%,1) = 25(3.312127)(0.925926) = \$76.67$M.

O&M (geometric series, $g=2.8\%$, value at $t=5$): using $\;A_1\dfrac{1-\left(\frac{1+g}{1+i}\right)^{n}}{i-g}$ with $n=30$:

$$PW_{t=5} = 2.5\cdot\frac{1-(1.028/1.08)^{30}}{0.08-0.028} = 2.5\cdot\frac{1-0.227552}{0.052} = 2.5(14.85477) = \$37.137\text{M}$$

to $t=0$: $37.137(P/F,8\%,5)=37.137(0.680583)=\$25.27$M. Salvage: $5(P/F,8\%,35)=5(0.0676345)=\$0.34$M. Combining:

$$PW = -76.67 - 25.27 + 0.34 \approx \boxed{-\$101.6\text{M}}$$

(c) Future Worth (t = 35, end of 2053)

$$FW = PW\,(F/P,8\%,35) = -101.6(14.7853) \approx \boxed{-\$1{,}502\text{M}}$$

The worth is strongly negative because this bridge generates no revenue—it is a pure cost. The −$101.6M present worth is the cost the municipality must justify against non-monetary benefits (congestion relief, safety, regional development), which lie outside the cash-flow model.