Question 3 of 6: Estimating

Question 3: Estimating (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 wall requires 2,000 jumbo masonry units. The R.S. Means line 04810-3000 describes jumbo brick 6" × 4" × 12" laid in running bond at a density of 3.00 units per square foot of wall, and prices the assembly per square foot. The crew is three skilled masons plus two helpers, five workers in total, and a standard eight-hour shift is assumed. All rates below are per square foot of wall.

Find. The duration in crew-days and the cost of the wall, both at bare cost and including overhead and profit, and a sketch of the typical time-cost relationship for an activity.

Approach. The Means line is priced per square foot, so the take-off is converted from units to square feet first; the labour-hours per square foot then give the total labour-hours, which the crew's daily labour-hour supply converts into a duration, and the unit rates give the cost directly.

1. (a) Convert the take-off from masonry units to the Means unit of measure. The line is priced per square foot of wall and the description gives the coursing density as 3.00 units per square foot: $$A = \frac{N_{\text{units}}}{n_{\text{units/S.F.}}} = \frac{2{,}000}{3.00} = 666.67 \text{ S.F.}$$ This conversion is the whole difficulty of the question. Every rate in the line — labour-hours, material, labour, and the marked-up total — is per square foot, so pricing 2,000 directly against them would overstate the wall by a factor of three.
2. Convert the area to total labour-hours. Multiplying the area by the labour-hour rate: $$LH = A \times 0.092 = 666.67 \times 0.092 = 61.33 \text{ labour-hours}$$ Note that this is a measure of effort, independent of how many people are put on it; the crew size converts it into elapsed time in the next step.
3. Convert labour-hours to a duration using the crew's daily supply. Five workers on an eight-hour shift supply $$LH_{\text{day}} = 5 \times 8 = 40 \text{ labour-hours per day}$$ so the elapsed duration is $$D = \frac{LH}{LH_{\text{day}}} = \frac{61.33}{40} = \boxed{1.53 \text{ crew-days}}$$ For scheduling purposes this is rounded up to two working days, since a crew is normally engaged in whole days; the half-day of spare capacity on the second day is available for scaffold striking, cleaning down and pointing.
4. Cross-check the crew against the Means daily output. The DAILY CREW OUTPUT cell is blank in the printed table, but it can be recovered from the labour-hour figure, and doing so confirms that the specified crew is the one Means assumed: $$Q_{\text{day}} = \frac{LH_{\text{day}}}{0.092} = \frac{40}{0.092} = 434.8 \approx 435 \text{ S.F. per day}$$ Dividing the area by this output returns the same duration, $666.67 / 435 = 1.53$ days, as it must. A three-mason-plus-two-helper crew laying roughly 435 square feet of jumbo brick a day corresponds to about 1,300 units per day, which is a realistic figure for this unit size.
5. Price the wall at bare cost. Applying the material and labour rates to the same area: $$\begin{aligned} C_{\text{mat}} &= 3.62 \times 666.67 = 2{,}413.33 \\ C_{\text{lab}} &= 2.93 \times 666.67 = 1{,}953.33 \end{aligned}$$ Adding them, and confirming against the printed bare total of 6.55 per square foot: $$C_{\text{bare}} = 6.55 \times 666.67 = \boxed{4{,}366.67 \text{ dollars}}$$ The two routes agree to the cent, which checks the printed line ($3.62 + 2.93 = 6.55$) as well as the arithmetic.
6. Price the wall including overhead and profit. The TOTAL INCL O&P column already contains the subcontractor's overhead and profit on both material and labour: $$C_{\text{total}} = 8.45 \times 666.67 = \boxed{5{,}633.33 \text{ dollars}}$$ The markup is therefore $1,266.67, or 29.0% on bare cost. Expressed per masonry unit, the wall costs $2.18 bare and $2.82 installed — a useful sanity figure to carry into a bid comparison.
7. (b) Sketch the time-cost relationship for an activity. The relationship is drawn below. An activity has a normal duration, at which its direct cost is lowest because the work is resourced in the most efficient way, and a crash duration, the shortest technically achievable time, at which direct cost is highest. Between the two, direct cost rises as duration is shortened, because acceleration is bought with overtime premiums, additional crews working in a congested area, shift work, premium-rate deliveries and additional equipment — all of which reduce output per labour-hour even as they increase output per day. The true relationship is a convex curve, but CPM crashing approximates it by the straight line joining the two points, whose gradient is the cost slope: $$\text{cost slope} = \frac{C_k - C_n}{D_n - D_k}$$ where $C_n$ and $D_n$ are the normal cost and duration and $C_k$ and $D_k$ the crash values. The slope is the price of buying back one day on that activity, and it is what a crashing analysis compares against the daily indirect cost. Durations shorter than the crash duration are technically unattainable at any price, and lengthening the activity beyond its normal duration produces no further direct-cost saving.