16-Civ-B8 Management of Construction · December 2016

Question 4 of 6: Engineering Economics

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

Notes on this paper

Paper format. 98-Civ-B8 Management of Construction, National Exams December 2016. Three hours, closed book, one approved calculator (Casio or Sharp). Six questions of equal value (20 marks each); any five constitute a complete paper and only the first five presented in the answer book are marked. All six are solved here, because the set is a study resource rather than an examination script.

Reference texts.

Halpin & Senior, Construction Management, 4th ed. — precedence networks with lags, bonding, project control.
Hendrickson, Project Management for Construction, 2nd ed. — scheduling, cost control, earned value, financing of constructed facilities.
Project Management Institute, A Guide to the Project Management Body of Knowledge (PMBOK Guide), 6th ed. — schedule and cost management, earned value.
Fraser et al., Global Engineering Economics, 5th Canadian ed. — present worth, deferred annuities, maximum justifiable investment.
Canadian Construction Documents Committee: CCDC 2 (stipulated price), CCDC 4 (unit price), CCDC 14 (design-build), CCDC 3 (cost-plus), CCDC 23 Guide to Calling Bids and Awarding Contracts; CCDC 220 / 221 / 222 bond forms.
Builders Lien Act (British Columbia, RSBC 1997 c. 45) and the provincial construction-lien / prompt-payment statutes; RSMeans Residential Square Foot Costs.

Question 4: Engineering Economics (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 courses of action on one section of pavement, compared over the ten years during which they differ.

Given data
Quantity	Symbol	Value
Maintenance cost, do nothing (all years)	$M_0$	$7,000 per year
Maintenance cost with new surface, years 1–5	$M_1$	$2,500 per year
Maintenance cost with new surface, years 6–10	$M_2$	$4,000 per year
Maintenance cost with new surface, year 11 onward	—	$7,000 per year (identical to do-nothing)
Interest rate	$i$	5 % per year
Period over which the alternatives differ	$n$	10 years

Find. The largest first cost $P$ that could be paid for the new surface and still leave the investment justified at 5 % — that is, the present worth of the maintenance savings it produces.

Figure 4.1 — cash-flow diagram of the incremental (savings) position: an outlay $P$ at time zero against ten years of maintenance savings, $4,500 per year for five years then $3,000 per year for five years.

Approach. Work with the incremental cash flow between the two alternatives — the annual saving in maintenance — and set the maximum justifiable investment equal to the present worth of that saving, since at exactly that first cost the investment earns precisely the required 5 %.

Form the incremental cash flow. The only difference between the two alternatives is maintenance, and only during the first ten years; from year 11 onward both cost $7,000 per year, so those flows cancel and can be ignored entirely. The annual savings are $$A_1 = 7{,}000 - 2{,}500 = \$4{,}500 \text{ per year, years 1 to 5},$$ $$A_2 = 7{,}000 - 4{,}000 = \$3{,}000 \text{ per year, years 6 to 10}.$$ This cancellation is the key simplification: it turns an infinite-horizon comparison into a ten-year one.
Evaluate the two interest factors at 5 %. The uniform-series present-worth factor and the single-payment present-worth factor for five years are $$(P/A,\,5\%,\,5) = \frac{1-(1.05)^{-5}}{0.05} = 4.3295, \qquad (P/F,\,5\%,\,5) = (1.05)^{-5} = 0.78353 .$$
Discount the first five years of saving. This series starts at year 1, so the uniform-series factor places it directly at time zero: $$P_1 = A_1\,(P/A,\,5\%,\,5) = 4{,}500 \times 4.3295 = \$19{,}482.65 .$$
Discount the deferred second series. The second series runs from year 6 to year 10. The $(P/A)$ factor placed on it returns a value at year 5, not at year 0, so it must then be brought back five more years with the single-payment factor: $$P_2 = A_2\,(P/A,\,5\%,\,5)\,(P/F,\,5\%,\,5) = 3{,}000 \times 4.3295 \times 0.78353 = \$10{,}176.77 .$$ Forgetting the second factor is the classic error in deferred-annuity problems and would overstate the answer by about $2,800.
Add the two components to obtain the maximum justifiable investment. $$P_{\max} = P_1 + P_2 = 19{,}482.65 + 10{,}176.77$$ $$\boxed{P_{\max} = \$29{,}659 \;\;(\text{approximately } \$29{,}660)}$$
Check the result independently. Discounting the ten savings one year at a time, $\sum_{y=1}^{5} 4{,}500(1.05)^{-y} + \sum_{y=6}^{10} 3{,}000(1.05)^{-y}$, reproduces $29,659.42, confirming the factor arithmetic. As a sanity bracket, the undiscounted savings total $5(4{,}500)+5(3{,}000) = \$37{,}500$, so the present worth must be somewhat less than that but clearly more than ten years of the smaller saving alone ($3{,}000 \times 7.7217 = \$23{,}165$) — and $29,659 sits properly between the two.

Check: the answer is a break-even first cost, not a recommendation. Spending exactly $29,659 makes the new surface earn precisely 5 % and leaves the owner indifferent; any price below that is justified and any price above it is not. Two assumptions are built in and should be stated on an exam script: that the maintenance savings occur as end-of-year amounts (the standard convention for annual costs), and that the new surface has no salvage value and requires no other cost or benefit — the question’s phrase “maintenance costs are the only saving” authorises both. It is also worth noting that the surface evidently has a ten-year effective life, since maintenance reverts to $7,000 afterwards, so no replacement or salvage term belongs in the analysis.

Question 4 — results
Quantity	Value
Annual saving, years 1–5	$4,500 per year
Annual saving, years 6–10	$3,000 per year
Present worth of years 1–5	$19,482.65
Present worth of years 6–10	$10,176.77
Maximum justifiable investment at 5 %	$29,659

Topic. Maximum justifiable investment from a deferred, stepped annuity — present-worth analysis of an incremental cash flow.

Key relations. $(P/A,i,n) = \dfrac{1-(1+i)^{-n}}{i}$; $(P/F,i,n) = (1+i)^{-n}$; a series beginning in year $k+1$ is valued by $(P/A,i,n)(P/F,i,k)$. Maximum justifiable investment $P_{\max} = PW(\text{benefits})$, the first cost at which $NPW = 0$.

Why this works. A capital expenditure is justified when the present worth of what it buys is at least the present worth of what it costs. Setting the two equal locates the break-even first cost, which is exactly what “maximum investment that can be justified” asks for. The analysis is legitimate over ten years rather than perpetuity because the two alternatives are identical from year 11 on: identical future flows cancel in an incremental comparison, so they cannot affect the decision no matter how long they run. This cancellation is also why no study-period or repeated-lives adjustment is needed here, unlike the unequal-lives comparisons that appear elsewhere in this subject.

Common pitfall. Applying $(P/A,5\%,5)$ to the second series and stopping there, which values it at the end of year 5 instead of at time zero; and computing the present worth of the new maintenance costs rather than of the savings, which answers a different question entirely. A third, subtler slip is to include years beyond 10 because the do-nothing case “continues forever” — it does, but so does the improved case, at the same $7,000.

Source. Fraser et al., Global Engineering Economics, 5th Canadian ed., Ch. 3–4 (cash-flow analysis, present worth, deferred annuities); Halpin & Senior, Construction Management, 4th ed., Ch. 16.

← Question 3 Next: Question 5 →

Quantity	Symbol	Value
Maintenance cost, do nothing (all years)	\(M_0\)	$7,000 per year
Maintenance cost with new surface, years 1–5	\(M_1\)	$2,500 per year
Maintenance cost with new surface, years 6–10	\(M_2\)	$4,000 per year
Maintenance cost with new surface, year 11 onward	—	$7,000 per year (identical to do-nothing)
Interest rate	\(i\)	5 % per year
Period over which the alternatives differ	\(n\)	10 years