07-Str-B2 · December 2018

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

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

Notes on this paper

Paper format: National Exams, December 2018 — 07-Str-B2 Management of Construction. Three hours, closed book, one approved Casio or Sharp calculator permitted. Six questions of equal value (20 marks each); any five constitute a complete paper and only the first five that appear in the answer book are marked. All six are worked below so the paper serves as a complete revision set whichever five a candidate elects.

Reference texts: Hegazy, T., Computer-Based Construction Project Management (Prentice Hall) — activity-on-arrow and activity-on-node networks, forward and backward passes, total and free float, and the contractor cash-flow / overdraft model with mark-up, retention and payment lag; these chapters carry Questions 1 and 3. Hendrickson, C. & Au, T., Project Management for Construction (2nd ed., Carnegie Mellon) — Chapter 5 (cost estimation and unit-cost data), Chapter 8 (construction contracts and the allocation of risk), Chapter 10 (fundamental scheduling procedures) and Chapter 12 (cost control, monitoring and accounting, including project financing). Halpin, D.W. & Senior, B.A., Construction Management (4th ed., Wiley) — delivery systems, crew productivity and quantity take-off, surety bonding and lien law. R.S. Means, Building Construction Cost Data (annual) — the anatomy of a unit-price line (crew, daily output, labour-hours per unit, bare material / labour / equipment / total, and total including overhead and profit) and of the related crew table, behind Question 6. Sullivan, W.G., Wicks, E.M. & Koelling, C.P., Engineering Economy (17th ed., Pearson) — Chapters 4 and 5, the uniform-series present-worth factor and deferred annuities, used in Question 4. Canadian Construction Documents Committee, CCDC 2 Stipulated Price Contract (2020), CCDC 4 Unit Price Contract, CCDC 3 Cost Plus Contract, CCDC 14 Design-Build Stipulated Price Contract, and CCDC 220 Bid Bond, CCDC 221 Performance Bond and CCDC 222 Labour and Material Payment Bond — the Canadian contract and surety machinery behind Questions 2 and 5. Provincial lien statutes — the British Columbia Builders Lien Act (SBC 1997 c.45) and the Ontario Construction Act (RSO 1990 c.C.30, as amended 2018) — supply the Canadian equivalent of the American “mechanics lien” named in Question 5.

Check — how the two printed figures on page 2 were read. Network (Question 1): nine numbered event circles and eleven arrows, every arrow carrying a letter and a duration — A(4) 1→2, B(6) 1→4, C(2) 1→7, D(8) 2→3, E(4) 3→6, F(10) 4→5, G(16) 4→8, H(8) 5→6, I(6) 6→9, J(6) 7→8, K(10) 8→9. There is no dummy arrow on this drawing, so the translation to activity-on-node in part (c) is exact and needs no extra logic. Budget S-curve (Question 3): the six labelled markers fall squarely on months 1 to 6 of the printed axis (ticks 0 to 7). The project therefore runs six months and the budget (cost) at completion is $127,000.

Question 4: Engineering Economics — maximum justifiable investment in a new pavement 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. Two maintenance regimes for the same section of pavement, discounted at 10 per cent per year:

Annual maintenance cost under each alternative
Years	Do nothing ($/yr)	New surface ($/yr)	Saving ($/yr)
1 to 5	7,000	2,500	4,500
6 to 10	7,000	4,000	3,000
after 10	7,000	7,000	0

Find. The largest first cost that could be spent on the new surface today and still be justified — that is, the present worth of the maintenance savings at 10 per cent.

Figure 4.1 — cash-flow diagram of the maintenance savings. Upward arrows are savings relative to doing nothing: $4,500 per year in years 1 to 5 and $3,000 per year in years 6 to 10. Nothing accrues from year 11 onward because both alternatives revert to $7,000 per year. The downward arrow at time zero is the investment being sized.

Approach. The maximum justifiable investment is the first cost that drives the net present worth of the difference between the two alternatives to exactly zero, so it equals the present worth of the savings stream — a five-year uniform series starting immediately, plus a second five-year uniform series deferred by five years.

Reduce the problem to a savings stream. Because the pavement is the same asset under either alternative and maintenance is the only difference, everything except maintenance cancels. Subtracting the new-surface cost from the do-nothing cost year by year gives $$A_1=7{,}000-2{,}500=4{,}500\ \text{dollars/yr in years 1 to 5}$$ $$A_2=7{,}000-4{,}000=3{,}000\ \text{dollars/yr in years 6 to 10}$$ and zero from year 11 onward, since the question states that maintenance returns to $7,000 under both regimes. That last point matters: nothing beyond year 10 enters the calculation at all, so no assumption about the pavement's ultimate life is needed.
Set up the criterion. An investment $P$ is justified while the present worth of what it saves is at least what it costs, so the break-even first cost is $$P_{max}=PW(\text{savings})=A_1\,(P/A,10\%,5)+A_2\,(P/A,10\%,5)(P/F,10\%,5)$$ The second term treats years 6 to 10 as an ordinary five-year annuity whose present worth sits at the end of year 5, then discounts that single amount back five more years.
Evaluate the two factors. With $i=0.10$ and $n=5$, $$(P/A,10\%,5)=\frac{(1+i)^n-1}{i(1+i)^n}=\frac{1.61051-1}{0.10\times1.61051}=3.7908$$ $$(P/F,10\%,5)=\frac{1}{(1+i)^n}=\frac{1}{1.61051}=0.62092$$ so the deferred-annuity factor for years 6 to 10 is $3.7908\times0.62092=2.3538$.
Discount each block of savings. Substituting, $$PW_1=4{,}500\times3.7908=17{,}058.54\ \text{dollars}$$ $$PW_2=3{,}000\times2.3538=7{,}061.34\ \text{dollars}$$ The second block is worth less than half the first even though it runs for the same five years, which is the combined effect of a smaller annual saving and five extra years of discounting.
Add the two blocks. Summing, $$\boxed{P_{max}=17{,}058.54+7{,}061.34=24{,}119.88\approx24{,}120\ \text{dollars}}$$ Spending exactly this amount today makes the two alternatives economically indifferent at 10 per cent; anything less is justified, anything more is not.
Check by direct discounting. Present-worth factors are a convenience, not a separate theory, so the same figure must come out of discounting all ten payments individually: $$PW=\sum_{n=1}^{5}\frac{4{,}500}{1.10^{\,n}}+\sum_{n=6}^{10}\frac{3{,}000}{1.10^{\,n}}=24{,}119.88\ \text{dollars}\ \checkmark$$ As a sanity check on the order of magnitude, the undiscounted savings total $5\times4{,}500+5\times3{,}000=37{,}500$ dollars, so the present worth is 64 per cent of the nominal saving — entirely reasonable for a ten-year stream at 10 per cent.

Check — assumptions carried through Question 4. The investment is treated as a single lump sum at time zero, the savings as end-of-year amounts, and the new surface as having no salvage value and requiring no intermediate rehabilitation within the ten years. The question gives no analysis period, and none is needed: the savings vanish after year 10, so any horizon of ten years or more returns the same present worth. All figures are before tax, as the question implies no tax data.

Question 4 — present worth of the maintenance savings at 10 per cent
Component	Annual amount ($)	Years	Factor	Present worth ($)
First block of savings	4,500	1 to 5	(P/A,10%,5) = 3.7908	17,058.54
Deferred block of savings	3,000	6 to 10	(P/A,10%,5)(P/F,10%,5) = 2.3538	7,061.34
Savings after year 10	0	11 onward	—	0
Maximum justifiable investment in the new surface				24,119.88 (say 24,120)

Topic. Present-worth analysis of a cost-saving investment, with a deferred uniform series.

Key relations. Uniform-series present worth $(P/A,i,n)=\dfrac{(1+i)^n-1}{i(1+i)^n}$; single-payment present worth $(P/F,i,n)=(1+i)^{-n}$; a series running from year $k+1$ to year $k+n$ has present worth $A\,(P/A,i,n)(P/F,i,k)$. The maximum justifiable investment is the $P$ that makes $NPW=0$, that is $P_{max}=PW(\text{benefits})$.

Why this works. Sizing a maximum justifiable investment is the same calculation as a present-worth comparison, viewed from the other end. Rather than asking whether a known first cost pays back, it asks how large a first cost the benefit stream could support before the net present worth turns negative — the break-even point. Because both alternatives serve the identical function on the identical asset, every cost they share cancels and only the maintenance difference survives, which is why the answer can be built entirely from savings without ever writing down the total cost of either regime. The deferred-annuity treatment works because the $(P/A)$ factor always places its present worth one period before the first payment it covers: applied to a series beginning in year 6, it lands at the end of year 5, and a single $(P/F,i,5)$ then brings it to the present.

Common pitfall. The most common error is discounting the second block with $(P/A,10\%,10)-(P/A,10\%,5)$ mis-remembered, or applying $(P/F,10\%,6)$ instead of $(P/F,10\%,5)$ — an off-by-one that shifts the whole block a year and changes the answer by about $640. A second is to keep discounting savings past year 10; the question is explicit that maintenance reverts to $7,000 for both alternatives, so the saving is genuinely zero thereafter and a perpetuity term would be simply wrong. A third is to compare the two alternatives on unequal lives when no lives are given at all — here the maintenance streams are defined only for ten years and that is the entire problem.

Source. Sullivan, Wicks & Koelling, Engineering Economy, Chapter 4 (interest factors, deferred annuities) and Chapter 5 (present-worth method and the break-even or maximum-justifiable-investment criterion).

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