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07-Str-B2 · December 2017

Question 3 of 6: Estimating — duration and cost from R.S. Means production data, and the activity time–cost relationship

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

Notes on this paper

Paper format: National Exams, December 2017 — 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 answered in the answer book are marked. All six are worked below so that the paper can be used for revision whichever five a candidate chooses.

Reference texts: Hegazy, T., Computer-Based Construction Project Management (Prentice Hall) — precedence (activity-on-node) networks with lags, forward and backward passes, total and free float, earned-value progress measurement and the time–cost trade-off curve; 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 10 (fundamental scheduling procedures), Chapter 11 (advanced scheduling with lags) and Chapter 12 (cost control, monitoring and accounting), including percent-complete and earned-value reporting. Halpin, D.W. & Senior, B.A., Construction Management (4th ed., Wiley) — labour productivity, crew balance, construction contracts and bonding, and construction safety management. Peurifoy, R.L. & Schexnayder, C.J., Construction Planning, Equipment and Methods (9th ed., McGraw-Hill) — crew productivity and the physical determinants of daily output, behind Question 3. R.S. Means, Building Construction Cost Data (annual) — the structure of a unit-price line: crew, daily output, labour-hours per unit, bare material / labour / equipment, and total including overhead and profit. Sullivan, W.G., Wicks, E.M. & Koelling, C.P., Engineering Economy (17th ed., Pearson) — Chapters 5 and 6, present-worth analysis and the repeatability (common multiple of lives) assumption for alternatives with unequal lives, used in Question 4. AACE International, Recommended Practice 29R-03, Forensic Schedule Analysis, and the Society of Construction Law Delay and Disruption Protocol (2nd ed., 2017) — the delay-analysis taxonomy required by Question 2. Canadian Construction Documents Committee, CCDC 2 — Stipulated Price Contract (2020), CCDC 220 Bid Bond, CCDC 221 Performance Bond, CCDC 222 Labour and Material Payment Bond and CCDC 40 — Rules for Mediation and Arbitration, together with the BC Builders Lien Act holdback provisions — the Canadian contractual machinery behind Questions 2 and 5. Transportation Association of Canada, Manual of Uniform Traffic Control Devices for Canada (MUTCDC) and the BC Ministry of Transportation and Infrastructure Traffic Management Manual for Work on Roadways, with WorkSafeBC's Occupational Health and Safety Regulation (Part 18 Traffic Control, Part 4 lighting and workplace conditions, Part 8 personal protective clothing) — the Canadian rule set behind Question 6.

Check — what the printed R.S. Means line in Question 3 actually contains. On line 04810‑3000 the printed cells are labour-hours 0.092, bare material $3.62, bare labour $2.93, bare total $6.55 and total including O&P $8.45. The CREW, DAILY OUTPUT, UNIT and EQUIPMENT cells are blank on the paper — they are not faint, they carry no ink at all. The self-consistency of the printed row ($3.62 + $2.93 = $6.55) confirms the money columns were read correctly.

Question 3: Estimating — duration and cost from R.S. Means production data, and the activity time–cost relationship (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. A masonry wall requiring 2,000 jumbo units laid in running bond, to be built by a single crew of three skilled bricklayers plus two helpers — five workers — on a standard eight-hour day. From the printed R.S. Means line 04810‑3000, the values available are:

Given data — R.S. Means line 04810-3000 and the site crew
QuantitySymbolValueSource
Quantity of masonry unitsQ2,000 EaQuestion stem
Labour-hours per unitL0.092 L.H./EaPrinted LABOR-HOURS column
Bare materialm$3.62 /EaPrinted MAT. column
Bare labourl$2.93 /EaPrinted LABOR column
Bare equipmentenilEQUIP. cell is blank on the printed line
Bare totalc$6.55 /EaPrinted TOTAL column
Total including overhead and profitC$8.45 /EaPrinted TOTAL INCL O&P column
Crew supplied by the contractorn3 skilled + 2 helpers = 5 workersQuestion stem
Working dayh8 hStandard R.S. Means basis
Modular note—3.00 units per S.F. of wall facePrinted line description

Find. (a) the duration in working days and the total cost of the 2,000-unit wall for the stated five-person crew, distinguishing bare cost from cost including overhead and profit; and (b) a sketch of the typical relationship between the duration of an activity and its direct cost.

Approach. The R.S. Means labour-hours-per-unit figure is a property of the work, not of the crew that does it — a different crew size changes the daily output but not the hours required per unit. So the duration follows from total labour-hours divided by the labour-hours the stated crew supplies per day, and the cost follows from multiplying the quantity by the printed unit prices. The daily output the printed line omits is then recovered as a check, and part (b) is answered with the standard crash-to-normal direct-cost curve.

  1. Convert the quantity into total labour-hours. The labour-hours column states the work content of one unit, so $$H = Q \times L = 2000\ \text{Ea} \times 0.092\ \tfrac{\text{L.H.}}{\text{Ea}}$$ $$\boxed{H = 184\ \text{labour-hours}}$$ This is the whole of the production information the question needs, and it is independent of how many workers are used — that choice affects only how many days the 184 hours take to consume.
  2. Compute the labour-hours the crew supplies each day. A crew of three skilled workers and two helpers is five people, all of whom work the standard eight-hour shift, so $$H_{day} = n \times h = 5 \times 8 = 40\ \text{L.H./day}.$$ R.S. Means counts helpers in exactly this way: labourers and helpers appear in the crew and contribute their full hours to the crew total, because the published labour-hours per unit already averages the mix of skilled and unskilled effort.
  3. Recover the daily output the printed line omits, as a check. Daily output and labour-hours per unit are two ways of saying the same thing, related through the crew's daily hours: $$\text{Daily output} = \frac{H_{day}}{L} = \frac{40}{0.092} = 434.8 \approx 435\ \text{units/day}.$$ So the blank DAILY OUTPUT cell is not information that was lost — for this crew it is determined by the two figures that are printed. It is worth quoting in the estimate because a foreman thinks in units per day, not in labour-hours per unit.
  4. Compute the duration. Dividing the work content by the crew's daily supply, $$D = \frac{H}{H_{day}} = \frac{184}{40} = 4.6\ \text{days}$$ $$\boxed{D = 4.6\ \text{working days} \;\rightarrow\; \text{5 working days scheduled}}$$ The same answer comes from the recovered daily output, $2000/434.8 = 4.6$ days, which closes the loop on Step 3. A schedule cannot buy a fraction of a shift, so the activity is entered on the bar chart as five days; the residual 0.4 day is the crew's float within the last shift and is where the estimator's allowance for scaffolding, clean-up and cure typically goes.
  5. Price the work at bare cost. Multiplying the quantity by each printed bare unit rate, $$\text{Material} = 2000 \times \$3.62 = \$7{,}240, \qquad \text{Labour} = 2000 \times \$2.93 = \$5{,}860,$$ with no equipment cost because the EQUIP. cell is blank and the printed bare TOTAL equals material plus labour. Adding the two components, $$\text{Bare cost} = \$7{,}240 + \$5{,}860 = \$13{,}100,$$ which agrees with pricing the printed bare TOTAL directly, $2000 \times \$6.55 = \$13{,}100$. That agreement is the arithmetic check that the row was read correctly.
  6. Price the work as it would be billed, including overhead and profit. The TOTAL INCL O&P column already contains the subcontractor's payroll burden, overhead and profit, so it is the figure that goes into a bid: $$\text{Cost}_{O\&P} = Q \times C = 2000 \times \$8.45$$ $$\boxed{\text{Cost including O\&P} = \$16{,}900}$$ The markup carried by the line is therefore $\$16{,}900 - \$13{,}100 = \$3{,}800$, or 29.0 % on bare cost. Expressed against the wall face, at three units per square foot the wall is $2000/3.00 = 667$ S.F. and the installed rate is $3.00 \times \$8.45 = \$25.35$ per square foot, which is the form in which the number is usually reported to an owner.
  7. Confirm that the O&P figure is consistent with the R.S. Means convention. In R.S. Means the material component is marked up ten per cent while the labour component is converted at the "including O&P" wage rate, which already contains burden, overhead and profit. Working backwards from the printed $8.45, $$\text{Labour rate}_{O\&P} = \frac{C - 1.10\,m}{L} = \frac{8.45 - 1.10(3.62)}{0.092} = \$48.57\ \text{per L.H.},$$ against a bare labour rate of $l/L = 2.93/0.092 = \$31.85$ per L.H. The ratio of 1.52 is the usual masonry burden-plus-O&P factor, so the printed row is internally coherent and the two cost answers can be relied on. If the contractor's own union or open-shop wage package differs from the R.S. Means average, the duration above is unchanged — the labour-hours are the same — and only the labour money is rescaled by the ratio of the two wage rates.
  8. Sketch the time–cost relationship for an activity (part b). Every activity has a normal duration, at which its direct cost is lowest because crew size, shift length and method are the economical ones, and a crash duration, the shortest technically achievable, reached by adding crew, working overtime, adding shifts or changing method. Between the two the direct cost rises as the duration is shortened, and the rise is convex — the first day bought back is cheap, the last is expensive, because congestion, overtime premium and diminishing returns on added labour compound. Beyond the normal duration the direct cost rises again, since stretching an activity keeps the crew and its equipment mobilised for longer. For scheduling arithmetic the segment between the crash and normal points is idealised as a straight line whose slope is the cost slope, $$\text{Cost slope} = \frac{C_c - C_n}{D_n - D_c}\ \ \left[\tfrac{\$}{\text{day}}\right],$$ the price of buying back one day on that activity. Least-cost crashing then proceeds by shortening the critical activity with the smallest cost slope first.
    Activity duration (days)Direct cost of the activity ($)Crash point(Dc, Cc)Normal point(Dn, Cn)DcDnCcCnlinear idealisation: cost slope = (Cc − Cn) / (Dn − Dc)the $ price of buying back one day on this activitystretching past Dn costsmore again (idle crew)infeasible(technical limit)
    Figure 3.1 — Typical time–cost relationship for a single activity. The solid curve is the true direct cost; the dashed straight line between the crash point and the normal point is the linear idealisation used in crashing calculations, and its gradient is the cost slope. Durations shorter than the crash duration are technically unattainable at any price.
    Note that this curve is the direct cost of one activity. At project level the indirect cost — site overhead, supervision, financing, liquidated damages — falls as the project duration shortens, so the sum of the two is U-shaped and has a minimum at the economical project duration; that optimum, not the crash point, is the target of a time–cost trade-off study.
Question 3 — final results
QuantityResult
Total work content184 labour-hours
Crew supply40 labour-hours per day (5 workers × 8 h)
Recovered daily output≈ 435 units per day
Duration4.6 working days, scheduled as 5 days
Bare material cost$7,240
Bare labour cost$5,860
Bare total cost$13,100
Total cost including overhead and profit$16,900
Markup carried by the O&P column$3,800 (29.0 % on bare cost)
Installed rate on the wall face$25.35 per S.F. (667 S.F. of wall)
Cost slope (part b)(Cc − Cn) / (Dn − Dc), in dollars per day