07-Str-B2 · Undated paper

Question 1 of 6: Scheduling — least-cost plan, and accelerating to eight days inside a four-worker limit

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

Notes on this paper

Paper format: National Exams, May 2019 — 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-node networks, the forward and backward passes, total and free float, resource profiles and levelling, the time–cost trade-off, and the earned-value formulation with the 20/80 progress convention; these chapters carry Questions 1 and 5. Hendrickson, C. & Au, T., Project Management for Construction (2nd ed., Carnegie Mellon) — Chapter 8 (construction contracts, delivery systems and the allocation of risk), Chapter 10 (fundamental scheduling procedures) and Chapter 12 (cost control, monitoring and accounting), behind Questions 1, 3 and 5. Halpin, D.W. & Senior, B.A., Construction Management (4th ed., Wiley) — delivery-system comparison, bid evaluation and responsibility determination, and construction safety management, behind Questions 3 and 6. Sullivan, W.G., Wicks, E.M. & Koelling, C.P., Engineering Economy (17th ed., Pearson) — Chapters 4, 5 and 6: the uniform-series present-worth factor, single-payment factors, and the comparison of alternatives with unequal lives by repeated study period or by annual worth; this is Question 4. AACE International, Recommended Practice 29R-03, Forensic Schedule Analysis, together with the Society of Construction Law Delay and Disruption Protocol (2nd ed., 2017) — the delay-analysis methods and the excusable / compensable / concurrent taxonomy in Question 2. Canadian Construction Documents Committee, CCDC 2 Stipulated Price Contract (2020), CCDC 5B Construction Management Contract — for Services and Construction, and CCDC 23 A Guide to Calling Bids and Awarding Contracts — the Canadian contract and tendering machinery behind Questions 2 and 3. WorkSafeBC Occupational Health and Safety Regulation (B.C. Reg. 296/97), especially Part 20 (Construction, Excavation and Demolition) and Part 6 (Substance Specific Requirements — asbestos and lead), with the federal Transportation of Dangerous Goods Regulations — the Canadian regulatory frame for Question 6.

Question 1: Scheduling — least-cost plan, and accelerating to eight days inside a four-worker limit (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. Five activities, each priced two ways — a slow machine (Estimate 1) and a fast machine (Estimate 2) — with the “Depends on” column referring to the activity number. Site Preparation is offered on the slow machine only.

Activity data as printed on page 2 (Estimate 1 = slow machine, Estimate 2 = fast machine)
#	Activity	Depends on	Estimate 1 — slow machine			Estimate 2 — fast machine
#	Activity	Depends on	Cost	Duration (d)	Workers/day	Cost	Duration (d)	Workers/day
1	Site Preparation	—	$5,000	4	3	—	—	—
2	Trench 1 Excavation	1	$5,000	4	3	$10,000	2	2
3	Trench 2 Excavation	1	$5,000	4	2	$7,000	2	2
4	Lay Pipe 1 & Backfill	2	$5,000	4	3	$7,000	1	2
5	Lay Pipe 2 & Backfill	3	$5,000	4	3	$9,000	2	2

Find. (a) With every activity on its cheapest option: the project duration, the total cost, the peak crew size, and the network with its critical path marked. (b) The cheapest combination of options that finishes in eight days without ever exceeding four workers on site in a day, its cost, and its bar chart.

Approach. Read the “Depends on” column as the precedence list, run a forward and backward pass on the resulting activity-on-node network for part (a), then in part (b) treat the duration limit as a length budget on each chain to eliminate option mixes before pricing them, and finally confirm the surviving plan against the daily worker profile.

Part (a) — the least-cost plan

Identify the cheapest option for every activity. Estimate 1 costs $5,000 for all five activities, while Estimate 2 costs between $7,000 and $10,000. The fast machine is dearer in every single row, so the cheapest plan puts every activity on Estimate 1: four days each, at the workers per day printed in the slow-machine columns (3, 3, 2, 3 and 3).
Build the network from the “Depends on” column. Activity 1 has no predecessor; activities 2 and 3 both depend on 1; activity 4 depends on 2 and activity 5 on 3. Site preparation therefore opens the job and the work then splits into two independent pipe runs, trench-then-lay, that never rejoin: $$1\rightarrow\{2,3\},\qquad 2\rightarrow 4,\qquad 3\rightarrow 5$$
Forward pass — earliest times. With every duration equal to four days and $$ES_j=\max_{i\in P(j)}\left(EF_i\right),\qquad EF_j=ES_j+d_j$$ site preparation runs from day 0 to day 4; both trenches then run from day 4 to day 8; and both pipe-laying activities from day 8 to day 12. The two chains have identical length, so $$T=\max\left(EF_4,\;EF_5\right)=\max(12,\,12)=\boxed{T=12\ \text{working days}}$$
Backward pass — the critical path. Setting the latest finish of both terminal activities to T = 12 and working back through $LF_i=\min_{j\in S(i)}\left(LS_j\right)$ returns latest times identical to the earliest ones at every node, so the total float $$TF_i=LS_i-ES_i=0\quad\text{for all five activities}$$ Both chains, 1 → 2 → 4 and 1 → 3 → 5, measure 4 + 4 + 4 = 12 days and are therefore both critical. That matters for the resource answer that follows: with no float anywhere, nothing can be slid to flatten the crew profile without pushing the finish date out.

Figure 1.1 — activity-on-node network for the all-cheapest plan. Each box carries ES / EF on the top row, the activity and its duration in the middle, and LS / LF below. Every box is critical (shown in red): the two pipe runs are exactly the same length.

Project cost. Five activities, each on its $5,000 option: $$C=5\times 5{,}000=\boxed{C=\$25{,}000}$$
Required number of workers. Because both chains are critical, all activities run at their earliest start and the crew demand is simply the sum of the concurrent activities' workers per day. Days 1–4 carry site preparation alone at 3 workers/day; days 5–8 carry the two trenches together at 3 + 2 = 5 workers/day; days 9–12 carry the two pipe-laying activities at 3 + 3 = 6 workers/day. The crew that must be available is therefore the peak of that profile, $$W_{\max}=\max\left(3,\;5,\;6\right)=\boxed{W_{\max}=6\ \text{workers/day}}$$ For comparison the work content is $\sum d_i w_i = 12+12+8+12+12 = 56$ worker-days, an average of 56/12 = 4.67 workers/day — so the profile is markedly uneven, and with zero float in the network there is no way to level it without extending the project.

Figure 1.2 — bar chart and daily worker histogram for the all-cheapest plan. The profile steps 3 → 5 → 6 and peaks at six workers per day, half again over the four-worker ceiling imposed in part (b) (dashed line).

Part (b) — eight days with at most four workers per day

Convert the deadline into a budget on each chain. Site preparation is on the critical path of both chains and has only one option, so it consumes four of the eight days no matter what. That leaves $$8-4=4\ \text{days}$$ for each of the two remaining chains, 2 → 4 and 3 → 5, running in parallel.
Screen the option mixes chain by chain. For the first chain, $d_2\in\{4,2\}$ and $d_4\in\{4,1\}$, giving four mixes of length 8, 6, 5 and 3 days; only the all-fast mix, $2+1=3\le 4$, fits. For the second chain, $d_3\in\{4,2\}$ and $d_5\in\{4,2\}$, giving lengths 8, 6, 6 and 4 days; again only the all-fast mix, $2+2=4\le 4$, fits. Every mix that keeps even one slow machine on either chain overruns the eight-day deadline, so the choice is forced: $$\boxed{\text{all four optional activities on Estimate 2 (fast machine)}}$$ There is exactly one feasible combination, which makes it trivially the best one — the “determine the best combination” instruction is answered by the elimination itself, not by a cost comparison among survivors.
Re-run the passes on the accelerated network. Site preparation still occupies days 0–4. Trench 1 (2 d) and Trench 2 (2 d) then run days 4–6, Lay Pipe 1 (1 d) runs day 6–7 and Lay Pipe 2 (2 d) days 6–8, so $$T=4+2+2=\boxed{T=8\ \text{working days}}$$ The chains are no longer equal: 1 → 3 → 5 measures 8 days and is critical, while 1 → 2 → 4 measures 4 + 2 + 1 = 7 days, so Trench 1 and Lay Pipe 1 each carry one day of total float.
Check the four-worker ceiling on the daily profile. Every fast-machine activity needs two workers per day. Days 1–4 carry site preparation at three; days 5–6 carry both trenches at 2 + 2 = 4; day 7 carries both pipe-laying activities at 2 + 2 = 4; day 8 carries Lay Pipe 2 alone at two. The profile is $$\left[3,\,3,\,3,\,3,\,4,\,4,\,4,\,2\right]\ \text{workers/day},\qquad W_{\max}=4$$ which sits exactly on the ceiling and never breaches it. The plan is feasible on both counts, and the one day of float on Lay Pipe 1 provides the only manoeuvring room there is.
Price the accelerated plan. Site preparation stays at its single option; the other four move to Estimate 2: $$C=5{,}000+10{,}000+7{,}000+7{,}000+9{,}000=\boxed{C=\$38{,}000}$$ Buying four days therefore costs $38,000 − $25,000 = $13,000, a 52 % cost premium for a 33 % reduction in duration, or about $3,250 per day saved.

Figure 1.3 — the accelerated network. Only Site Preparation → Trench 2 → Lay Pipe 2 is critical now (red); Trench 1 and Lay Pipe 1 each carry one day of total float.

Figure 1.4 — bar chart for the accelerated plan, answering part (b). Dashed outlines show the one day of float on Trench 1 and Lay Pipe 1. The histogram touches the four-worker limit on days 5, 6 and 7 without exceeding it.

Check — assumptions behind the resource answer. Two readings are taken as given and are stated here because they change the numbers. First, the “Workers/day” column is treated as a crew size that must be present for the whole of an activity's duration and cannot be part-loaded — the standard reading, and the only one under which “four workers/day” is a meaningful constraint. Second, the two pipe runs are treated as physically independent, exactly as the “Depends on” column states: Lay Pipe 2 waits on Trench 2 only, not on Lay Pipe 1. If the site in fact allowed only one pipe crew, both parts would change and part (b) would become infeasible at any price. The part (b) result was additionally confirmed by exhaustive enumeration over all sixteen option combinations and every admissible start vector; exactly one combination satisfies both the eight-day and four-worker constraints.

Question 1 — results
Quantity	(a) All cheapest options	(b) Eight days, four workers/day
Option chosen	Estimate 1 (slow) for all five	Estimate 1 for Site Prep; Estimate 2 (fast) for the other four
Project duration	12 working days	8 working days
Project cost	$25,000	$38,000
Peak crew required	6 workers/day	4 workers/day
Daily worker profile	3, 3, 3, 3, 5, 5, 5, 5, 6, 6, 6, 6	3, 3, 3, 3, 4, 4, 4, 2
Critical path	Both chains: 1–2–4 and 1–3–5	1–3–5 (Site Prep → Trench 2 → Lay Pipe 2)
Total float	Zero on every activity	1 day on Trench 1 and on Lay Pipe 1
Acceleration premium: $13,000 (+52 %) to remove four days (−33 %), about $3,250 per day saved.

Topic. The time–cost–resource trade-off on a small activity-on-node network: choosing among discrete execution options subject to both a deadline and a crew ceiling.

Key relations. Forward pass $ES_j=\max_{i\in P(j)}EF_i$, $EF_j=ES_j+d_j$; backward pass $LF_i=\min_{j\in S(i)}LS_j$, $LS_i=LF_i-d_i$; total float $TF_i=LS_i-ES_i=LF_i-EF_i$; and the resource profile $R(t)=\sum_{i\,\text{active at}\,t} w_i$ evaluated on a chosen schedule.

Why this works. Discrete-option acceleration is not a continuous crashing problem — there is no cost slope to walk down. Because the deadline decomposes into an independent length budget on each parallel chain, the infeasible mixes can be eliminated by inspection before anything is priced, which is far quicker than enumerating and costing all sixteen combinations. The resource check must come last and separately: a plan can satisfy the deadline and still breach the crew ceiling, since duration and resource are different constraints on the same schedule.

Common pitfall. Reporting the average crew (56 worker-days over 12 days, 4.67) as the “required number of workers”. The requirement is the peak of the daily profile, six, because the site must be staffed on its busiest day. A second, subtler trap in part (a) is missing that both chains are critical — a candidate who marks only one loses the observation that no levelling is available without extending the schedule.

Source. Hegazy, Computer-Based Construction Project Management, chapters on CPM calculations, resource management and the time–cost trade-off; Hendrickson & Au, Project Management for Construction, Chapter 10 (Fundamental Scheduling Procedures) and Chapter 11 (Advanced Scheduling Techniques, resource-constrained scheduling).

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