07-Str-B2 · May 2014

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

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

Notes on this paper

National Examinations — May 2014 — 07-Str-B2 Management of Construction. Three hours, closed book; candidates may use one of the two approved calculators (Casio or Sharp). The paper prints six questions of equal value (20 marks each) and states that any five questions constitute a complete paper, only the first five appearing in the answer book being marked. Candidates are urged to submit a clear statement of any interpretive assumptions with their answers. All six questions are worked below, because this set is intended as a study resource rather than as a single exam sitting.

Reference texts: RSMeans, Building Construction Cost Data — the "How to Use the Cost Data" front matter, which defines daily output, labour-hours, bare costs and the Total Incl. O&P column used in Question 1; Halpin, D.W. & Senior, B.A., Construction Management (4th ed., Wiley) — unit-price estimating, crew balancing, labour relations and construction safety; Hegazy, T., Computer-Based Construction Project Management (Prentice Hall) — precedence networks with SS/FS/FF relationships and lags, which is exactly the notation of Question 2; Hendrickson, C. & Au, T., Project Management for Construction (2nd ed., Carnegie Mellon) — scheduling and cost control; Sullivan, W.G., Wicks, E.M. & Koelling, C.P., Engineering Economy (17th ed., Pearson) — present-worth analysis and the maximum-justified-investment problem of Question 4; Canadian Construction Documents Committee, CCDC 2 — Stipulated Price Contract (2020), General Conditions 6.5 (delays) and 6.6 (claims for a change in Contract Price); Goldsmith, I. & Heintzman, T.G., Goldsmith on Canadian Building Contracts (5th ed., Thomson Reuters) — delay and notice law in Canada; AACE International, Recommended Practice 29R-03: Forensic Schedule Analysis — the but-for and windows methods named in Question 5; British Columbia Labour Relations Code, RSBC 1996 c. 244 — certification, bargaining units and the construction-industry provisions behind Question 3; WorkSafeBC, Occupational Health and Safety Regulation (Parts 4, 8, 11, 18 and 20) and the BC Workers Compensation Act — the prime-contractor duty and the traffic-control, fall-protection and hazardous-substance rules behind Question 6.

Check — two readings taken from the printed page. The RS Means extract in Question 1 prints two cells as question marks; both are recovered below from the crew table, and the recovered labour-hour figure is checked against the printed $27 labour column before it is used. Two arrows leave the right-hand edge of activity D and turn vertically to reach E and C; they are read here as ordinary finish-to-start links, which is the only reading consistent with the drawing and with the fact that every unlabelled arrow on the sheet carries no lag.

Question 4: Engineering Economics — maximum justified 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.

Quantity	Symbol	Value
Present annual maintenance cost	$A_0$	$3,500 / year
Maintenance cost with new surface, years 1–5	$A_1$	$650 / year
Maintenance cost with new surface, years 6–10	$A_2$	$1,100 / year
Maintenance cost after year 10	—	$3,500 / year (unchanged)
Interest rate	$i$	5 % per year
Analysis horizon	$n$	10 years

Find. The largest first cost that could be paid for the new surface and still leave the owner no worse off — that is, the present worth of the maintenance savings the surface creates.

Approach. Difference the two maintenance streams year by year to isolate the saving, recognise that the saving is zero after year 10 and so contributes nothing, then discount the two uniform blocks of saving to the present with the uniform-series present-worth factor, shifting the second block back to time zero with a single-payment factor.

Figure 4.1 — cash-flow diagram of the incremental saving. Upward arrows are the annual maintenance savings created by the new surface; the downward arrow at time zero is the investment whose maximum justifiable size is sought. After year 10 the two alternatives cost the same and the diagram is blank.

Isolate the incremental saving. Only the difference between the two maintenance streams is relevant; everything common to both alternatives cancels. Year by year, $$A_1^{\text{sav}}=3500-650=\$2{,}850\ \text{/yr (years 1--5)}$$ $$A_2^{\text{sav}}=3500-1100=\$2{,}400\ \text{/yr (years 6--10)}$$ From year 11 onward both alternatives cost $3,500 a year, so the saving is zero and the analysis need not extend past year 10. This is the key structural observation in the problem: the resurfacing buys ten years of reduced maintenance and nothing else.
Compute the two interest factors. At $i=5\ \%$, $$(P/A,5\%,5)=\frac{1-(1.05)^{-5}}{0.05}=4.32948$$ $$(P/F,5\%,5)=(1.05)^{-5}=0.78353$$ Both are needed because the second block of savings is a five-year annuity that does not begin until the end of year 6.
Discount the first block. The first five savings form an ordinary annuity beginning at the end of year 1, so the uniform-series factor places its present worth directly at time zero: $$P_1=A_1^{\text{sav}}(P/A,5\%,5)=2850\times 4.32948=\$12{,}339.01$$
Discount the second block in two moves. Applying $(P/A,5\%,5)$ to the second annuity values it at the end of year 5, one period before its first payment, so that value must then be brought back five more years: $$P_2=A_2^{\text{sav}}(P/A,5\%,5)(P/F,5\%,5)=2400\times 4.32948\times 0.78353=\$8{,}141.36$$
Add the two present worths. The maximum justified investment is the point at which the present worth of the savings exactly offsets the first cost, leaving $PW=0$: $$P_{\max}=P_1+P_2=12{,}339.01+8{,}141.36=\boxed{\$20{,}480\ \text{(rounded)}}$$
Check by discounting the ten savings individually. Summing $\sum_{k=1}^{10} A_k^{\text{sav}}(1.05)^{-k}$ term by term — five terms of $2,850 and five of $2,400 — returns $20,480.37, confirming both factors and the five-year shift. As a sanity check on magnitude, the undiscounted savings total $5(2850)+5(2400)=\$26{,}250$, so a present worth of about $20,500 implies an average discount of roughly 22 %, which is what a 5 % rate over an average delay of about five years should produce.

Result	Value
Annual saving, years 1–5	$2,850
Annual saving, years 6–10	$2,400
$(P/A,5\%,5)$	4.32948
$(P/F,5\%,5)$	0.78353
Present worth of years 1–5	$12,339.01
Present worth of years 6–10	$8,141.36
Maximum justified investment	$20,480

Check — assumptions stated as the paper invites. End-of-year cash flows are assumed throughout, the resurfacing is assumed to be paid for as a single outlay at time zero, and the new surface is assumed to be fully consumed by year 10 with no salvage or residual condition benefit. Inflation is not treated separately, so the 5 % is a real rate applied to constant dollars. If the new surface in fact left the pavement in better condition at year 10 than the do-nothing case, the justified investment would rise above $20,480.

Topic. Present-worth analysis of an incremental cash-flow stream, and the maximum-justified-investment form of the breakeven question.

Key relations. $(P/A,i,n)=\dfrac{1-(1+i)^{-n}}{i}$; $(P/F,i,n)=(1+i)^{-n}$; a deferred annuity of $n$ payments starting at the end of period $m+1$ has present worth $A(P/A,i,n)(P/F,i,m)$; the maximum justifiable first cost is the $P$ that makes the incremental present worth zero.

Why this works. Comparing two alternatives over a common horizon needs only their differences, because identical cash flows discount identically and cancel. Here the difference is a ten-year stream of maintenance savings that stops abruptly at year 10, and the question “what is the most I should pay” is exactly the question “what present sum is equivalent to that stream at 5 %”. Splitting the stream into two uniform blocks lets the uniform-series factor do the work twice; the single-payment factor then repairs the five-year offset that the second application introduces, because $(P/A)$ always values an annuity one period before its first payment.

Common pitfall. The most frequent error is to apply $(P/A,5\%,10)$ to the second block, or to shift it by six years instead of five — the factor already discounts to one period before the first payment, so the extra shift is $m=5$, not $m=6$. The second is to extend the analysis beyond year 10 on the assumption that the surface goes on saving money; the question states explicitly that maintenance returns to $3,500, so the incremental stream is exactly ten years long. The third is to answer with the undiscounted total of $26,250, which ignores the time value of money entirely and overstates the justifiable investment by nearly 30 %.

Source. Sullivan, Wicks & Koelling, Engineering Economy, 17th ed., ch. 4 (equivalence and the P/A and P/F factors) and ch. 5 (present-worth method, breakeven and maximum justified investment); Hendrickson & Au, Project Management for Construction, ch. 6.

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