07-Str-B2 · December 2018

Question 1 of 6: Scheduling — activity-on-arrow network, floats, critical path and AON conversion

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

Notes on this paper

Paper format: National Exams, December 2018 — 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-arrow and activity-on-node networks, forward and backward passes, total and free float, and the contractor cash-flow / overdraft model with mark-up, retention and payment lag; 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 8 (construction contracts and the allocation of risk), Chapter 10 (fundamental scheduling procedures) and Chapter 12 (cost control, monitoring and accounting, including project financing). Halpin, D.W. & Senior, B.A., Construction Management (4th ed., Wiley) — delivery systems, crew productivity and quantity take-off, surety bonding and lien law. R.S. Means, Building Construction Cost Data (annual) — the anatomy of a unit-price line (crew, daily output, labour-hours per unit, bare material / labour / equipment / total, and total including overhead and profit) and of the related crew table, behind Question 6. Sullivan, W.G., Wicks, E.M. & Koelling, C.P., Engineering Economy (17th ed., Pearson) — Chapters 4 and 5, the uniform-series present-worth factor and deferred annuities, used in Question 4. Canadian Construction Documents Committee, CCDC 2 Stipulated Price Contract (2020), CCDC 4 Unit Price Contract, CCDC 3 Cost Plus Contract, CCDC 14 Design-Build Stipulated Price Contract, and CCDC 220 Bid Bond, CCDC 221 Performance Bond and CCDC 222 Labour and Material Payment Bond — the Canadian contract and surety machinery behind Questions 2 and 5. Provincial lien statutes — the British Columbia Builders Lien Act (SBC 1997 c.45) and the Ontario Construction Act (RSO 1990 c.C.30, as amended 2018) — supply the Canadian equivalent of the American “mechanics lien” named in Question 5.

Check — how the two printed figures on page 2 were read. Network (Question 1): nine numbered event circles and eleven arrows, every arrow carrying a letter and a duration — A(4) 1→2, B(6) 1→4, C(2) 1→7, D(8) 2→3, E(4) 3→6, F(10) 4→5, G(16) 4→8, H(8) 5→6, I(6) 6→9, J(6) 7→8, K(10) 8→9. There is no dummy arrow on this drawing, so the translation to activity-on-node in part (c) is exact and needs no extra logic. Budget S-curve (Question 3): the six labelled markers fall squarely on months 1 to 6 of the printed axis (ticks 0 to 7). The project therefore runs six months and the budget (cost) at completion is $127,000.

Question 1: Scheduling — activity-on-arrow network, floats, critical path and AON conversion (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. An activity-on-arrow (AOA) network of nine events and eleven activities, each arrow labelled with its letter and duration in working days, as printed in the paper:

Activities read from the printed network (i → j event pairs)
Activity	i → j	Duration (d)	Activity	i → j	Duration (d)
A	1 → 2	4	G	4 → 8	16
B	1 → 4	6	H	5 → 6	8
C	1 → 7	2	I	6 → 9	6
D	2 → 3	8	J	7 → 8	6
E	3 → 6	4	K	8 → 9	10
F	4 → 5	10	Event 1 is the start, event 9 the finish.

Find. The total float of every activity and the critical path; the effect on project duration of a three-day delay to activity F; and the equivalent activity-on-node (AON) network.

Figure 1.1 — the printed activity-on-arrow network, re-traced. The critical chain B–G–K is shown in red; every other arrow carries float.

Approach. Work the network on its events: a forward pass gives each event its earliest time, a backward pass its latest time, and every activity's total float then follows from the one-line identity total float equals late event time of its head, minus early event time of its tail, minus its duration.

Forward pass — earliest event times. An event cannot occur until every activity entering it has finished, so $$E_j=\max_{i}\left(E_i+d_{ij}\right),\qquad E_1=0$$ Taking the events in numerical order (which happens to be a valid topological order here): $$E_2=0+4=4,\quad E_4=0+6=6,\quad E_7=0+2=2$$ $$E_3=E_2+8=12,\quad E_5=E_4+10=16$$ $$E_8=\max(E_4+16,\;E_7+6)=\max(22,\,8)=22$$ $$E_6=\max(E_3+4,\;E_5+8)=\max(16,\,24)=24$$ Finally the finish event collects activities I and K: $$E_9=\max(E_6+6,\;E_8+10)=\max(30,\,32)=32$$ so the project duration is $$\boxed{T=32\ \text{working days}}$$
Backward pass — latest event times. Setting the latest finish equal to the earliest finish, $L_9=E_9=32$, and working backwards through $$L_i=\min_{j}\left(L_j-d_{ij}\right)$$ gives $$L_6=32-6=26,\quad L_8=32-10=22$$ $$L_3=L_6-4=22,\quad L_5=L_6-8=18,\quad L_7=L_8-6=16$$ $$L_4=\min(L_5-10,\;L_8-16)=\min(8,\,6)=6,\qquad L_2=L_3-8=14$$ $$L_1=\min(L_2-4,\;L_4-6,\;L_7-2)=\min(10,\,0,\,14)=0$$ The pass closing exactly on zero at event 1 is the arithmetic check that both passes are consistent.
Total and free float of every activity. Total float is the delay an activity can absorb without pushing the project end date; free float is the delay it can absorb without disturbing any successor's early start: $$TF_{ij}=L_j-E_i-d_{ij},\qquad FF_{ij}=E_j-E_i-d_{ij}$$ Applying both to each arrow gives the schedule table below. For example, activity F runs 4 → 5, so $TF_F=L_5-E_4-10=18-6-10=2$ days, while $FF_F=E_5-E_4-10=16-6-10=0$ — F cannot slip at all without pushing H.
(a) Identify the critical path. The activities with zero total float are B, G and K, and they form an unbroken chain of events 1 → 4 → 8 → 9. Their durations add to the project duration, which confirms the chain: $$6+16+10=32\ \text{days}$$ $$\boxed{\text{Critical path } = \text{B}\rightarrow\text{G}\rightarrow\text{K},\ \ T=32\ \text{days}}$$ Note that C and J carry the largest float (14 days each), and that E has 10 days of total float but 8 days of free float — the two are not the same number, because E shares event 6 with H.
(b) Effect of delaying activity F by three days. F has only two days of total float, so the first two days of the delay are free and the third is not. Re-running the passes with F occupying $10+3=13$ days: $E_5=6+13=19$, hence $E_6=\max(12+4,\;19+8)=27$ and $$E_9=\max(27+6,\;22+10)=\max(33,\,32)=33$$ $$\boxed{\text{Project duration extends by 1 day, from 32 to 33 days}}$$ The delay also moves the critical path: B–F–H–I now measures $6+13+8+6=33$ days and becomes critical, while B–G–K falls back to one day of float. This is the point of the question — consuming an activity's float does not merely “use up slack”, it re-routes where management attention has to go next.
(c) Equivalent AON network. In an activity-on-arrow diagram the predecessors of an activity are exactly the activities entering its tail event, so the translation is mechanical: A, B and C leave the start event and have no predecessor; D follows A (event 2); F and G follow B (event 4); J follows C (event 7); E follows D (event 3); H follows F (event 5); I follows E and H (event 6); K follows G and J (event 8). Because the printed network contains no dummy arrow, no dependency is lost or invented in the conversion, and the early / late times computed above transfer unchanged onto the nodes.

Figure 1.2 — the equivalent activity-on-node network, each box carrying early start / early finish on the top line, the activity and its duration in the middle, and late start / late finish / total float on the bottom line. Critical boxes and links are red.

Question 1 — schedule results
Activity	i → j	Duration	ES	EF	LS	LF	Total float	Free float
A	1 → 2	4	0	4	10	14	10	0
B	1 → 4	6	0	6	0	6	0	0
C	1 → 7	2	0	2	14	16	14	0
D	2 → 3	8	4	12	14	22	10	0
E	3 → 6	4	12	16	22	26	10	8
F	4 → 5	10	6	16	8	18	2	0
G	4 → 8	16	6	22	6	22	0	0
H	5 → 6	8	16	24	18	26	2	0
I	6 → 9	6	24	30	26	32	2	2
J	7 → 8	6	2	8	16	22	14	14
K	8 → 9	10	22	32	22	32	0	0
Project duration 32 days · critical path B–G–K · delaying F by 3 days extends the project to 33 days and makes B–F–H–I critical

Topic. Critical-path analysis on an activity-on-arrow network, and its translation to activity-on-node.

Key relations. Forward pass $E_j=\max_i(E_i+d_{ij})$; backward pass $L_i=\min_j(L_j-d_{ij})$; total float $TF_{ij}=L_j-E_i-d_{ij}$; free float $FF_{ij}=E_j-E_i-d_{ij}$; interfering float $=TF-FF$. Project duration $T=E_n=L_n$.

Why this works. The two passes are simply the longest-path and the shortest-slack recursions on a directed acyclic graph. The forward pass propagates the constraint “an event waits for its slowest feeder”, so $E_n$ is the length of the longest chain from start to finish — the definition of the project duration. The backward pass propagates the complementary constraint “an activity must finish early enough for its successors to still fit”, so $L_i$ is the latest an event can occur without lengthening that longest chain. An activity's total float is the gap between the window its events allow it ($L_j-E_i$) and the time it actually needs ($d_{ij}$); when that gap is zero the activity lies on the longest chain and is critical by construction. Free float differs because it uses the early time of the head event, which asks a narrower question: can this activity slip without disturbing anyone else, as opposed to without disturbing the deadline.

Common pitfall. Two errors dominate. The first is identifying the critical path by adding durations along paths that “look long” instead of reading it off the zero-float column — on this network A–D–E–I sums to 22 and B–F–H–I to 30, so an eyeball estimate easily misses that B–G–K (32) governs. The second is assuming that a delay smaller than an activity's total float is always harmless and one larger than it always costs the full excess; here F's three-day delay exceeds its two days of float by one day, and the project slips by exactly that one day — never by three. Also resist the habit of quoting one float per activity: total and free float answer different management questions and this network makes them differ for A, C, D, E, F and H (for E, 10 days total against 8 free), while for I and J the two happen to coincide.

Source. Hegazy, Computer-Based Construction Project Management, chapters on CPM calculations and network representation; Hendrickson & Au, Project Management for Construction, Chapter 10 (Fundamental Scheduling Procedures), sections on activity-on-branch networks and float calculation.

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