16-Civ-B8 Management of Construction · May 2015

Question 4 of 6: Engineering economics — present-worth comparison of two projects

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

Notes on this paper

Paper format. National Exams, May 2015 — 98-Civ-B8 Management of Construction (the paper now catalogued as 16-Civ-B8). Three hours, closed book; one of two approved calculator models permitted. Six questions of equal value (20 marks each); the rubric states that any five constitute a complete paper and that only the first five presented in the answer book will be marked. All six are worked here, because this set is a study resource rather than an exam script. The paper is three calculation questions (1, 3, 4) and three discussion questions (2, 5, 6).

Source note. The two side tables on page 2 — the activity/duration/predecessor list in Question 1 and the trenching-machine production table in Question 3 — are given in full in the Given blocks below. The final activity in the Question 1 table is printed as a two-character label that reads QI; it is a closing activity of one day's duration following V and S, and the answer does not depend on how the label is read.

Reference texts.

Halpin, D. W. and Senior, B. A., Construction Management, 4th ed., Wiley — Ch. 9–11 (network scheduling, activity-on-arrow networks, float), Ch. 4 (project delivery methods), Ch. 13–14 (equipment production and cost estimating).
Hendrickson, C. and Au, T., Project Management for Construction, 2nd ed. — Ch. 8 (construction contracts and risk allocation), Ch. 10 (fundamental scheduling procedures, critical path and float), Ch. 5 (cost estimation, productivity).
Peurifoy, R. L., Schexnayder, C. J., Shapira, A. and Schmitt, R., Construction Planning, Equipment, and Methods, 9th ed., McGraw-Hill — Ch. on trenching and excavating machinery (ladder- and wheel-type trenchers, production tables, operating-efficiency factors).
Fraser, N. M., Jewkes, E. M. et al., Global Engineering Economics: Financial Decision Making for Engineers, Canadian ed., Pearson — Ch. 3 (compound-interest factors), Ch. 4 (present worth), Ch. 5 (comparison of alternatives).
Canadian Construction Documents Committee, CCDC 2 — Stipulated Price Contract, CCDC 14 — Design-Build Stipulated Price Contract; and the Canadian Construction Association Guide to Calling Bids and Awarding Contracts.
Hinze, J. W., Construction Safety, 2nd ed.; WorkSafeBC, Occupational Health and Safety Regulation (Parts 3, 4, 5, 8, 11, 12, 13, 18, 20) and the Canadian Centre for Occupational Health and Safety (CCOHS) fact sheets.
Construction Industry Institute and the U.S. Business Roundtable productivity studies, as summarised in Halpin & Senior Ch. 15 (labour productivity, work sampling, the five-minute rating).

Question 4: Engineering economics — present-worth comparison of two projects (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.

Given. Two mutually exclusive projects of equal nine-year life, each with a single initial outlay, uniform annual revenue and operating cost, and a major maintenance event every third year, all discounted at 10 per cent per year.

Cash-flow data for the two alternatives
Item	Project A	Project B
Initial investment (year 0)	$62,000	$80,000
Yearly operating cost	$1,500	$1,000
Major maintenance, every 3 years	$15,000	$13,000
Yearly revenue	$11,500	$16,000
Life	9 years	9 years
Discount rate	10 % per year, no salvage value stated

Find. The present-worth profit of each alternative and, from the comparison, the more economical plan.

Project A cash-flow diagram: $62,000 invested at year 0, a net $10,000 per year for nine years (revenue $11,500 less operating cost $1,500), and major maintenance of $15,000 in years 3, 6 and 9.

Project B cash-flow diagram: $80,000 invested at year 0, a net $15,000 per year for nine years (revenue $16,000 less operating cost $1,000), and major maintenance of $13,000 in years 3, 6 and 9.

Approach. Net the annual revenue against the annual operating cost, bring the resulting uniform series back with the series present-worth factor, bring each major-maintenance outlay back with a single-payment present-worth factor, subtract the initial investment, and compare the two present worths directly — which is legitimate here because both lives are nine years.

Reduce each project to a net annual cash flow. Revenue and operating cost are both uniform annual series over the same nine years, so they may be combined before discounting: $$A_{\text{A}}=11{,}500-1{,}500=\$10{,}000\ \text{per year}$$ $$A_{\text{B}}=16{,}000-1{,}000=\$15{,}000\ \text{per year}$$ Project B earns half as much again per year but costs $18,000 more to build, which is the trade-off the present-worth calculation must resolve.
Evaluate the compound-interest factors at 10 per cent. The uniform series runs for nine years, and the major maintenance falls in years 3, 6 and 9 — three occurrences in a nine-year life on the literal reading of "every 3 years": $$\left(\frac{P}{A},10\%,9\right)=\frac{1-(1.10)^{-9}}{0.10}=5.7590$$ $$\left(\frac{P}{F},10\%,3\right)=0.7513,\qquad\left(\frac{P}{F},10\%,6\right)=0.5645,\qquad\left(\frac{P}{F},10\%,9\right)=0.4241$$ Summing the three single-payment factors gives the multiplier that converts one maintenance amount into the present worth of the whole maintenance programme: $$\Sigma\left(\frac{P}{F}\right)=0.7513+0.5645+0.4241=1.7399$$
Present worth of Project A. Discounting the net annual series and the maintenance programme and subtracting the initial investment, $$PW_{\text{A}}=-62{,}000+10{,}000(5.7590)-15{,}000(1.7399)$$ $$PW_{\text{A}}=-62{,}000+57{,}590-26{,}098=\boxed{-\$30{,}508}$$
Present worth of Project B. The same three terms with B's data, $$PW_{\text{B}}=-80{,}000+15{,}000(5.7590)-13{,}000(1.7399)$$ $$PW_{\text{B}}=-80{,}000+86{,}385-22{,}619=\boxed{-\$16{,}233}$$
Compare and select. Because the two alternatives have identical nine-year lives and are being evaluated over the same study period, their present worths are directly comparable and the larger value wins: $$PW_{\text{B}}-PW_{\text{A}}=-16{,}233-(-30{,}508)=\boxed{+\$14{,}275\ \text{in favour of B}}$$ Project B is therefore the more economical plan by a little over $14,000 in present-worth terms.
Confirm the choice incrementally. A defensible comparison of mutually exclusive alternatives should also be checked on the increment, because that is the cash flow the decision actually commits: choosing B rather than A means spending an extra $18,000 now to gain an extra $5,000 a year and save $2,000 of maintenance every third year. $$PW_{\text{B}-\text{A}}=-18{,}000+5{,}000(5.7590)+2{,}000(1.7399)=+\$14{,}275$$ The increment is positive, so the extra investment in B earns more than 10 per cent and the same conclusion is reached by both routes.
Report the result honestly — both present worths are negative. Neither alternative recovers its investment at a 10 per cent discount rate, so "the most economical plan" here means the smaller loss, not a profit. For Project A the failure is structural rather than a matter of discounting: even with no discounting at all, the nine years of net income total $90,000 against an outlay of $62,000 plus three maintenance events of $15,000, or $107,000 in all — a loss of $17,000. Project B's undiscounted position is positive at $-80{,}000+135{,}000-39{,}000=+\$16{,}000$, but that surplus is earned too slowly: its internal rate of return is about 4.2 per cent, well below the 10 per cent hurdle. The correct engineering recommendation is that if one of the two must be built, build B; if the 10 per cent rate genuinely represents the organisation's cost of capital and the "do nothing" option is available, neither project should proceed on these figures alone.

Check: "Major maintenance every 3 years" over a nine-year life has been taken literally as three events, in years 3, 6 and 9. Some marking schemes omit the year-9 event on the ground that maintenance performed at retirement buys nothing. That variant gives $PW_{\text{A}}=-\$24{,}147$ and $PW_{\text{B}}=-\$10{,}720$ — both still negative, and B still ahead, this time by $13,427. The ranking, and therefore the answer to the question asked, is unchanged by the assumption. No salvage value is stated for either project, so none has been assumed.

Question 4 — final results
Quantity	Project A	Project B
Net annual cash flow (revenue less operating cost)	$10,000/yr	$15,000/yr
Present worth of the net annual series	$57,590	$86,385
Present worth of major maintenance (years 3, 6, 9)	$26,098	$22,619
Present worth of the project	−$30,508	−$16,233
Advantage of B over A	$14,275 in present worth (confirmed on the increment)
Selection	Project B is the more economical plan; neither is profitable at 10 %
Sensitivity — maintenance in years 3 and 6 only	−$24,147	−$10,720

Topic. Present-worth comparison of mutually exclusive alternatives with equal lives.

Key relations. Series present worth $(P/A,i,N)=\left[1-(1+i)^{-N}\right]/i$; single-payment present worth $(P/F,i,n)=(1+i)^{-n}$; $PW=-P_{0}+A(P/A,i,N)-\sum F_{n}(P/F,i,n)$; the incremental test $PW_{\text{B}-\text{A}}>0$ selects the higher-cost alternative; the internal rate of return is the $i$ at which $PW=0$.

Why this works. Present worth converts every cash flow, whenever it occurs, into an equivalent amount today, so alternatives become comparable as single numbers. The method is only valid when the alternatives are compared over the same study period, which is why the equal nine-year lives here are worth remarking on explicitly — had the lives differed, the analysis would have needed either a least-common-multiple study period with repeated lives or an annual-worth comparison. Combining revenue and operating cost into one net series before discounting is legitimate because both are uniform over the same nine years; the maintenance cannot be folded in the same way because it is a set of discrete future amounts, so each is brought back separately. The incremental check matters because a comparison of totals can conceal what is actually being decided: the decision is not "is B good" but "is the extra $18,000 worth spending," and only the increment answers that.

Common pitfall. The most frequent error is miscounting the maintenance events — taking three events but discounting them all at year 3, or applying $(P/A)$ to a cost that occurs only every third year instead of the three separate $(P/F)$ factors. A subtler error is stopping at "B has the larger present worth, therefore B is profitable"; both present worths are negative here, and a candidate who does not say so has answered only half the question. The third is treating a negative present worth as though it forbade a comparison — it does not, since the alternative with the smaller loss is still the correct choice when the work must be done.

Source. Fraser, Jewkes et al., Global Engineering Economics: Financial Decision Making for Engineers, Canadian ed., Ch. 3 (compound-interest factors), Ch. 4 (present worth), Ch. 5 (comparison of mutually exclusive alternatives, incremental analysis); Hendrickson & Au, Project Management for Construction, Ch. 6.

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