16-Civ-B8 Management of Construction · May 2014

Question 4 of 6: Engineering Economics — maximum justified investment in a new surface

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

Notes on this paper

Paper format. National Exams, May 2014 — 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.

Reference texts.

Halpin, D. W. and Senior, B. A., Construction Management, 4th ed., Wiley — unit-price estimating and crew productivity, precedence diagramming with lags, labour relations, project control.
Hendrickson, C. and Au, T., Project Management for Construction, 2nd ed. — Ch. 5 (cost estimation from productivity data), Ch. 10 (scheduling with activity relationships and resource constraints).
Project Management Institute, A Guide to the Project Management Body of Knowledge (PMBOK Guide) — §6.5 Develop Schedule (leads, lags, critical path, float), §7 Cost Management.
RSMeans, Building Construction Cost Data — front matter on crew tables, daily output, labour-hours per unit and the overhead-and-profit columns.
Fraser, N. M. et al., Global Engineering Economics: Financial Decision Making for Engineers, Canadian ed., Pearson — Ch. 3–4 (present worth, uniform series, deferred annuities).
Canadian Construction Documents Committee, CCDC 2 — Stipulated Price Contract (2020) — GC 6.5 delays and extensions of time; and Society of Construction Law, Delay and Disruption Protocol, 2nd ed. — concurrency, time-slice and as-built but-for analysis.
Hinze, J. W., Construction Safety, 2nd ed.; WorkSafeBC, Occupational Health and Safety Regulation (Parts 4, 8, 11, 18, 20) and the BC Ministry of Transportation Traffic Management Manual for Work on Roadways.

Question 4: Engineering Economics — maximum justified investment in a new surface (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. The do-nothing case costs $3,500 per year indefinitely; resurfacing changes that cost profile for ten years only, after which the two alternatives are identical again.

Given data — annual maintenance cost under each alternative
Period	Do nothing	With new surface	Annual saving
Years 1–5	$3,500	$650	$2,850
Years 6–10	$3,500	$1,100	$2,400
After year 10	$3,500	$3,500	nil
Interest rate	$i = 5\%$ per year

Find. The largest first cost $P$ that the resurfacing can carry and still be economically justified — that is, the present worth of the maintenance savings it produces.

Cash-flow diagram of the incremental (savings) stream. Only the ten years in which the two alternatives differ appear; the identical $3,500 per year thereafter cancels between them and needs no discounting.

Approach. Work with the incremental cash flow — the difference between doing nothing and resurfacing — so the identical post-year-10 maintenance cancels; then value the two uniform savings blocks as a five-year annuity plus a five-year annuity deferred by five years, and set the maximum justifiable investment equal to their combined present worth, the point at which net present worth is exactly zero.

Form the incremental savings stream. Comparing the resurfaced pavement against doing nothing, the annual benefit is the avoided maintenance: $$A_1 = 3500 - 650 = \$2\,850/\text{yr (years 1--5)}, \qquad A_2 = 3500 - 1100 = \$2\,400/\text{yr (years 6--10)}.$$ From year 11 onward both alternatives cost $3,500 per year, so the difference is zero and no perpetuity term is needed — a point worth stating explicitly, because the mention of $3,500 returning after ten years tempts candidates into capitalising a stream that cancels.
Evaluate the two interest factors at 5 %. The five-year uniform-series present-worth factor and the five-year single-payment present-worth factor are $$(P/A,\,5\%,\,5) = \frac{1 - (1.05)^{-5}}{0.05} = 4.3295, \qquad (P/F,\,5\%,\,5) = (1.05)^{-5} = 0.7835.$$
Present worth of the first block. The years 1–5 savings are an ordinary annuity valued at time zero: $$P_1 = A_1 (P/A,\,5\%,\,5) = 2850 \times 4.3295 = \$12\,339.$$
Present worth of the deferred block. The years 6–10 savings are the same shape of annuity, but its own “time zero” sits at the end of year 5, so it must be discounted back one further five-year step: $$P_2 = A_2 (P/A,\,5\%,\,5)(P/F,\,5\%,\,5) = 2400 \times 4.3295 \times 0.7835 = \$8\,141.$$ Forgetting the second factor — treating the $2,400 as though it also began in year 1 — is the single most common error on this question and overstates the answer by about $2,250.
Maximum justifiable investment. The resurfacing is worth doing as long as its first cost does not exceed the present worth of what it saves, so the break-even first cost is $$P_{\max} = P_1 + P_2 = 12\,339 + 8\,141 = \boxed{\$20\,480}$$ (more precisely $20,480.43). At that price the net present worth is exactly zero and the investment earns precisely the 5 % required return; below it the resurfacing is justified, above it the money is better left in the alternative earning 5 %.
Independent cross-check by two other decompositions. Discounting the ten annual savings one at a time, $\sum_{t=1}^{5} 2850(1.05)^{-t} + \sum_{t=6}^{10} 2400(1.05)^{-t}$, reproduces $20,480.43 exactly. So does re-cutting the stream horizontally instead of vertically — a uniform $2,400 for ten years plus an extra $450 for the first five: $$PW = 2400\,(P/A,\,5\%,\,10) + 450\,(P/A,\,5\%,\,5) = 2400(7.7217) + 450(4.3295) = \$20\,480.$$ Three routes to the same figure is the confidence the exam answer should carry.

Check: the calculation prices maintenance savings only, as the question directs. A real resurfacing decision would also weigh the salvage or residual condition of the surface at year 10, user-delay and vehicle-operating-cost savings on a high-traffic section, and the possibility that the do-nothing alternative is not in fact sustainable for ten years. It also assumes end-of-year cash flows and a constant 5 % real rate, and that the $3,500 baseline is itself stable rather than escalating — if maintenance costs escalate, the savings grow and the justifiable investment rises.

Question 4 — results
Quantity	Value
Annual saving, years 1–5	$2,850
Annual saving, years 6–10	$2,400
$(P/A,5\%,5)$ and $(P/F,5\%,5)$	4.3295 and 0.7835
Present worth of years 1–5	$12,339
Present worth of years 6–10	$8,141
Maximum justified investment	$20,480

Topic. Present-worth analysis of an incremental cost-saving stream, including a deferred uniform series, to establish a break-even (maximum justifiable) capital cost.

Key relations. $(P/A,i,n) = \left[1-(1+i)^{-n}\right]/i$; $(P/F,i,n) = (1+i)^{-n}$; a series beginning in year $k+1$ is valued as $A\,(P/A,i,n)(P/F,i,k)$; and the break-even condition $NPW = 0 \Rightarrow P_{\max} = \sum PW(\text{savings})$.

Why this works. Two alternatives can always be compared on the difference between them rather than on their absolute cash flows, and doing so here removes the awkward part of the problem: because maintenance returns to $3,500 in year 11 under both alternatives, every cash flow beyond year 10 is common and cancels, so what looks like a perpetuity problem is really a ten-year one. Within those ten years the savings are two uniform blocks, and the uniform-series factor collapses each block into a single equivalent value at the instant the block begins. Shifting the second block back to time zero needs one single-payment factor, because discounting is multiplicative — five years of discounting followed by an annuity valuation is the same as valuing the annuity and then discounting five years. Setting the investment equal to that present worth defines the price at which the decision-maker is indifferent, which is exactly what “maximum justified investment” means.

Common pitfall. The dominant error is omitting the $(P/F,5\%,5)$ factor on the second block, which credits the later savings as though they were immediate and inflates the answer to about $22,730. The second is being drawn into capitalising the $3,500 that resumes after year 10 — it is identical in both alternatives and must not appear anywhere in the incremental analysis. A third is comparing the $650 and $1,100 figures against each other rather than each against the $3,500 baseline, which is what generates the savings in the first place.

Source. Fraser et al., Global Engineering Economics: Financial Decision Making for Engineers, Canadian ed., Ch. 3 (cash-flow analysis, compound-interest factors, deferred annuities) and Ch. 4 (present worth and break-even comparisons); Hendrickson & Au, Project Management for Construction, 2nd ed., Ch. 6 (economic evaluation of facility investments).

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