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11-CS-1 Engineering Economics · May 2019

Question 2 of 5: New Bridge — Present and Future Worth

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

Notes on this paper

National Exams — May 2019 — 11-CS-1 Engineering Economics. Three hours; open book; any non-communicating calculator permitted. Any four of the five questions constitute a complete exam paper and each question is of equal value (25 marks). Fully worked solutions to all five follow.

Question 2: New 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. Take $t=0$ at the end of 2018, so the end of calendar year $Y$ is $t=Y-2018$. Construction runs through 2020, 2021, 2022 and 2023, giving four end-of-year outlays of $25M at $t=2,3,4,5$. Operation and maintenance begins in the first year after completion, 2024, so the first $2.5M falls at $t=6$ and the series grows at 2.8% per year until the end of 2053, $t=35$ — that is 30 payments. The $5M salvage is a receipt at $t=35$. Interest is 8% per year.

(a) Cash-Flow Diagram

Cash flows in millions of dollars. Present (t = 0) = end of 2018 / beginning of 2019.2018t = 02023t = 52028t = 102038t = 202048t = 304 × 25 M/yr construction (t = 2–5, years 2020–2023)O&M: 2.5 M at t = 6 (2024), rising 2.8%/yr to t = 35+5 M salvage (t = 35, end of 2053)

Figure 1 — Cash-flow diagram for the bridge, present ($t=0$) at the end of 2018. Downward arrows are costs, the upward arrow at $t=35$ is the salvage receipt. The four red construction arrows stand at $t=2$ to $t=5$; the gold series is the geometric operation-and-maintenance cost, beginning at $t=6$ and growing 2.8% per year through $t=35$.

(b) Present Worth (i = 8%)

  1. Construction. Four equal outlays of $25M at $t=2\ldots5$ form a uniform series whose present value sits at $t=1$, so discount that back one more year:
    $$PW_{c} = 25\,(P/A,8\%,4)(P/F,8\%,1) = 25(3.312127)(0.925926) = \$76.67\text{M}$$
  2. Operation and maintenance. A geometric gradient with $A_1=2.5$, $g=2.8\%$, $i=8\%$ and $n=30$ payments starting at $t=6$; its present value sits at $t=5$:
    $$P_{5} = 2.5\;\frac{1-\left(\dfrac{1.028}{1.08}\right)^{30}}{0.08-0.028} = 2.5(14.854768) = \$37.14\text{M}$$
    $$PW_{om} = 37.136920\,(P/F,8\%,5) = 37.136920(0.680583) = \$25.27\text{M}$$
  3. Salvage. A single receipt 35 years out:
    $$PW_{s} = 5\,(P/F,8\%,35) = 5(0.067635) = \$0.34\text{M}$$
  4. Combine (costs negative, the salvage positive):
    $$PW = -76.67 - 25.27 + 0.34 = \boxed{-\$101.60\text{M}}$$

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

The future worth is the present worth carried forward 35 years at 8%, with $(F/P,8\%,35)=14.78534$:

$$FW = PW\,(F/P,8\%,35) = -101.606(14.78534) = \boxed{-\$1{,}502\text{M}}$$

As a check, building the future worth directly from the cash flows gives the same figure: the construction series is worth $-25(F/A,8\%,4)(F/P,8\%,30)=-1{,}133.6$M at $t=35$, the operation-and-maintenance series $-37.136920(F/P,8\%,30)=-373.7$M, and the salvage adds $+5$M, for $-\$1{,}502$M.

ComponentPresent worth (t = 0)Future worth (t = 35)
Construction, 4 × $25M (t = 2–5)−$76.67M−$1,133.6M
Operation & maintenance, geometric (t = 6–35)−$25.27M−$373.7M
Salvage (t = 35)+$0.34M+$5.0M
Total−$101.60M−$1,502M

The bridge carries no modelled revenue, so both worths are strongly negative: the $101.60M present cost is what the city must weigh against the congestion relief, travel-time savings and redundancy benefits that sit outside this cash-flow model. Note how little the distant items matter — the $5M salvage 35 years out contributes only $0.34M of present value, while the operating series, though each payment is small, accumulates to a quarter of the total cost because it runs for 30 years.