NivaarExam PrepOfficial exam papers ↗

23-CS-1 Engineering Economics · December 2014

Question 2 of 5: Wastewater Plant — Present and Future Worth

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

Notes on this paper

National Exams — December 2014 — 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 2: Wastewater Plant — 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 2014; construction $30M at ends of 2018–2022 ($t=4\text{–}8$); O&M from 2023 to 2052 ($t=9$ to $t=38$), first $2M at $t=9$ growing 1%/yr; salvage $+6M at $t=38$; $i=6\%$.

(a) Cash-Flow Diagram

t = 0 is the end of 2014; amounts in millions of dollars, i = 6% t salvage +6M at t = 38 (end of 2052) 0 4 8 9 38 O&M: 2M at t = 9, rising 1% per year through t = 38 five construction outlays of 30M (t = 4 to 8)

Figure 1 — Cash-flow diagram, end of 2014 ($t=0$) to end of 2052 ($t=38$). Downward arrows are disbursements, the upward arrow is a receipt; arrow lengths are indicative only and not to scale.

The down-arrows are the five $30M construction outlays at the ends of 2018–2022 ($t=4$ to $t=8$), and the 30-payment O&M series that starts at $2M at $t=9$ (the end of 2023, the first year of operation) and grows 1% per year through $t=38$. The single up-arrow is the $6M salvage at $t=38$. Note that nothing at all happens at $t=1,2,3$: the present is the end of 2014, but the first outlay is not until the end of 2018, and it is this four-year gap that the deferred-series discounting in part (b) has to carry.

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

Construction ($30M at $t=4\text{–}8$): $PW_c = 30\sum_{t=4}^{8}(P/F,6\%,t) = 30(3.53678) = \$106.10$M.

O&M (geometric, $A_1=2$M, $g=1\%$, $n=30$): worth at $t=8$ is $P_8 = 2\frac{1-(1.01/1.06)^{30}}{0.05}=2(15.3065)=\$30.61$M; then $PW_{OM}=30.61(P/F,6\%,8)=30.61(0.62741)=\$19.21$M.

Salvage: $PW_s = 6(P/F,6\%,38)=6(0.10924)=\$0.66$M. Combining (costs negative):

$$PW = -106.10 - 19.21 + 0.66 \approx \boxed{-\$124.7\text{M}}$$

(c) Future Worth (t = 38, end of 2052)

$$FW = PW\,(F/P,6\%,38) = -124.655(9.15425) \approx \boxed{-\$1{,}141\text{M}}$$

(The unrounded PW, $-124.655$M, is carried here; multiplying the rounded $-124.7$M instead would overstate the future cost by about $0.4M. Carrying every flow forward to $t=38$ one at a time gives the same $-\$1{,}141.1$M.)

The plant has a net present cost of about $124.7M (≈$1.14 billion by 2052), counting only costs.