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11-CS-1 Engineering Economics · May 2013

Question 2 of 6: Bridge Project — Present and Future Worth

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

Notes on this paper

National Exams — May 2013 — 11-CS-1 Engineering Economics. Open book; non-communicating calculator permitted. Any five of the six questions constitute a complete paper; all questions are of equal value. Fully worked solutions to all six questions are given below. Standard compound-interest factors are used throughout; minor rounding differences are immaterial.

Question 2: Bridge Project — Present and Future Worth (20 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 (stated per the exam's instruction): the present is $t=0$ at the end of 2013; construction outflows of $25M occur at the ends of 2015–2018 ($t=2,3,4,5$); O&M runs from 2019 to 2048 ($t=6$ to $t=35$), with the first payment $2.5M at $t=6$ growing geometrically at 2.8%; salvage $+5M at $t=35$; $i=8\%$.

(a) Cash-Flow Diagram

$M   +5 (salvage, t=35)
  ^                                                     |
0 -+--+--+--+--+--+--+--+---- ... ----+   (years t=0..35)
     |  |  |  |  |  ↓  ↓          ↓
     2  3  4  5   O&M grows 2.8%/yr from 2.5 (t=6) to t=35
   -25 -25 -25 -25   (construction, t=2..5)

Down-arrows are costs (construction $25M at $t=2\text{–}5$; O&M $2.5M rising at $t=6\text{–}35$); the single up-arrow is the $5M salvage at $t=35$.

(b) Present Worth (at t = 0, i = 8%)

Construction ($25M at $t=2,3,4,5$):

$$PW_{c} = 25\big[(P/F,8\%,2)+(P/F,8\%,3)+(P/F,8\%,4)+(P/F,8\%,5)\big]$$
$$= 25\,(0.85734+0.79383+0.73503+0.68058) = 25(3.06678) = \$76.67\text{M}$$

O&M (geometric gradient, $A_1=2.5$M at $t=6$, $g=2.8\%$, $n=30$). Worth at $t=5$:

$$P_{5} = A_1\,\frac{1-\left(\frac{1+g}{1+i}\right)^{n}}{i-g} = 2.5\,\frac{1-(0.951852)^{30}}{0.08-0.028} = 2.5\,\frac{1-0.22755}{0.052} = 2.5(14.855)=\$37.14\text{M}$$
$$PW_{OM} = 37.14\,(P/F,8\%,5) = 37.14(0.68058) = \$25.28\text{M}$$

Salvage: $PW_{s} = 5\,(P/F,8\%,35) = 5(0.06764) = \$0.34$M. Combining (costs negative):

$$PW = -76.67 - 25.28 + 0.34 \approx \boxed{-\$101.6\text{M}}$$

(c) Future Worth (at t = 35, end of 2048)

$$FW = PW\,(F/P,8\%,35) = -101.6\,(1.08)^{35} = -101.6(14.785) \approx \boxed{-\$1{,}502\text{M}}$$

The project has a net present cost of about $101.6M (future cost ≈ $1.5 billion at 2048)—as expected for a public bridge whose benefits (reduced congestion) are not monetized here.