16-Civ-B8 Management of Construction · May 2015

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

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

Notes on this paper

Paper format. National Exams, May 2015 — 98-Civ-B8 Management of Construction (the paper now catalogued as 16-Civ-B8). Three hours, closed book; one of two approved calculator models permitted. Six questions of equal value (20 marks each); the rubric states that any five constitute a complete paper and that only the first five presented in the answer book will be marked. All six are worked here, because this set is a study resource rather than an exam script. The paper is three calculation questions (1, 3, 4) and three discussion questions (2, 5, 6).

Source note. The two side tables on page 2 — the activity/duration/predecessor list in Question 1 and the trenching-machine production table in Question 3 — are given in full in the Given blocks below. The final activity in the Question 1 table is printed as a two-character label that reads QI; it is a closing activity of one day's duration following V and S, and the answer does not depend on how the label is read.

Reference texts.

Halpin, D. W. and Senior, B. A., Construction Management, 4th ed., Wiley — Ch. 9–11 (network scheduling, activity-on-arrow networks, float), Ch. 4 (project delivery methods), Ch. 13–14 (equipment production and cost estimating).
Hendrickson, C. and Au, T., Project Management for Construction, 2nd ed. — Ch. 8 (construction contracts and risk allocation), Ch. 10 (fundamental scheduling procedures, critical path and float), Ch. 5 (cost estimation, productivity).
Peurifoy, R. L., Schexnayder, C. J., Shapira, A. and Schmitt, R., Construction Planning, Equipment, and Methods, 9th ed., McGraw-Hill — Ch. on trenching and excavating machinery (ladder- and wheel-type trenchers, production tables, operating-efficiency factors).
Fraser, N. M., Jewkes, E. M. et al., Global Engineering Economics: Financial Decision Making for Engineers, Canadian ed., Pearson — Ch. 3 (compound-interest factors), Ch. 4 (present worth), Ch. 5 (comparison of alternatives).
Canadian Construction Documents Committee, CCDC 2 — Stipulated Price Contract, CCDC 14 — Design-Build Stipulated Price Contract; and the Canadian Construction Association Guide to Calling Bids and Awarding Contracts.
Hinze, J. W., Construction Safety, 2nd ed.; WorkSafeBC, Occupational Health and Safety Regulation (Parts 3, 4, 5, 8, 11, 12, 13, 18, 20) and the Canadian Centre for Occupational Health and Safety (CCOHS) fact sheets.
Construction Industry Institute and the U.S. Business Roundtable productivity studies, as summarised in Halpin & Senior Ch. 15 (labour productivity, work sampling, the five-minute rating).

Question 1: Scheduling — activity-on-arrow network, critical path and total floats (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. Fourteen activities with fixed durations in days and finish-to-start precedence only — no lags, no leads and no resource limits are stated.

Activity list transcribed from the side table on page 2
Activity	Duration (days)	Predecessors
A	1	—
B	8	A
C	4	A
P	7	A
L	2	B
M	4	C
Q	4	P, C
N	9	P
Y	5	L, Q
F	10	M
J	2	Q
S	2	N
V	5	Y, F, J
QI	1	V, S

Find. The activity-on-arrow (AOA) network, the project duration and the critical path, and the total float of every activity.

Activity-on-arrow network for the fourteen activities. Circles are events, labelled with the event number above the divider and the earliest / latest event times below it. Solid arrows are activities (name and duration in days); dashed arrows are the three dummies. The heavy red chain is the critical path A–C–M–F–V–QI, giving a project duration of 25 working days.

Approach. Convert the precedence list into an arrow diagram, inserting a dummy wherever two activities share only part of a predecessor set; run a forward pass over the events to get earliest event times, a backward pass to get latest event times, and read each activity's total float as the slack left in the interval its arrow must fit into.

Lay out the arrow diagram and place the dummies. In an activity-on-arrow network every activity is an arrow between two events, and an activity may start only when every arrow entering its tail event has finished. A dummy — a zero-duration arrow — is needed wherever a logical dependency cannot be drawn without also creating a false one. Three are required here:
- Q depends on both P and C, but M depends on C alone and N on P alone. Neither C's head event nor P's head event may serve as Q's tail, so a merge event (8) is created and fed by dummies from event 4 (end of C) and event 5 (end of P).
- Y depends on L and Q, while J depends on Q alone. L therefore ends at a separate event (13), and a dummy carries the Q dependency from event 9 into event 13.
Everything else merges without a dummy: Y, F and J all feed V and nothing else, so all three may end at event 16; V and S both feed QI alone, so both may end at event 18. The result is the seventeen-arrow network in the figure above, with thirteen events.
Forward pass — earliest event times. Set the start event to day zero and take, at every event, the largest of the earliest-finish times arriving there: $$E_j=\max_{i\to j}\left(E_i+D_{ij}\right)$$ Working left to right, $E_2=0+1=1$ (after A); $E_3=1+8=9$ (after B); $E_4=1+4=5$ (after C); $E_5=1+7=8$ (after P); $E_7=5+4=9$ (after M). At the merge event 8 the two dummies arrive carrying $E_4=5$ and $E_5=8$, so $E_8=\max(5,8)=8$ and Q cannot start before day 8 even though C finished on day 5. Continuing, $E_9=8+4=12$, $E_{10}=8+9=17$, $E_{13}=\max(9+2,\;12)=12$, and $E_{16}=\max(12+5,\;9+10,\;12+2)=\max(17,19,14)=19$, the governing value being F.
Complete the forward pass and read the project duration. At the last merge, V arrives at $E_{16}+5=24$ and S at $E_{10}+2=19$, so $E_{18}=24$ and the closing activity QI gives $$E_{19}=24+1=\boxed{T=25\ \text{days}}$$ The project therefore takes 25 working days from the start of A to the finish of QI.
Backward pass — latest event times. Set $L_{19}=E_{19}=25$ and work right to left, taking the smallest requirement imposed by the arrows leaving each event: $$L_i=\min_{i\to j}\left(L_j-D_{ij}\right)$$ So $L_{18}=25-1=24$; $L_{16}=24-5=19$; $L_{13}=19-5=14$ and $L_7=19-10=9$; $L_{10}=24-2=22$; and at event 9, which is left by both J and the dummy into 13, $L_9=\min(19-2,\;14-0)=14$. Then $L_8=14-4=10$, and event 5 is left by the dummy into 8 and by N, giving $L_5=\min(10-0,\;22-9)=10$. Event 4 is left by M and by the dummy into 8, so $L_4=\min(9-4,\;10-0)=5$; event 3 gives $L_3=14-2=12$; and finally $L_2=\min(12-8,\;5-4,\;10-7)=1$ with $L_1=1-1=0$. The backward pass returning to exactly zero at the start event is the arithmetic check that the whole traverse is consistent.
Total float of each activity. The total float is the difference between the interval the arrow is allowed to occupy and the time it actually needs: $$TF_{ij}=L_j-E_i-D_{ij}$$ Equivalently, in activity terms, $TF=LS-ES=LF-EF$. Applying this arrow by arrow gives the schedule table below. As a worked example, activity N runs from event 5 (earliest day 8) to event 10 (latest day 22) and needs 9 days, so $TF_N=22-8-9=5\ \text{days}$; activity B runs from event 2 (earliest day 1) to event 3 (latest day 12) and needs 8 days, so $TF_B=12-1-8=3\ \text{days}$.

Assemble the schedule table. Reporting the same result in activity form — earliest and latest start and finish for each activity — is what a construction schedule actually needs, because it tells the superintendent the window inside which each crew may be mobilised.

Forward and backward pass results, all values in working days
Activity	Duration	ES	EF	LS	LF	Total float
A	1	0	1	0	1	0 — critical
B	8	1	9	4	12	3
C	4	1	5	1	5	0 — critical
P	7	1	8	3	10	2
L	2	9	11	12	14	3
M	4	5	9	5	9	0 — critical
Q	4	8	12	10	14	2
N	9	8	17	13	22	5
Y	5	12	17	14	19	2
F	10	9	19	9	19	0 — critical
J	2	12	14	17	19	5
S	2	17	19	22	24	5
V	5	19	24	19	24	0 — critical
QI	1	24	25	24	25	0 — critical

Identify and check the critical path. The zero-float activities form one unbroken chain from start to finish: $$\text{A}\to\text{C}\to\text{M}\to\text{F}\to\text{V}\to\text{QI}$$ Because every link in this network is a plain finish-to-start relationship with no lag, the length of the critical path must equal the plain sum of its durations, and that sum must equal the project duration: $$1+4+4+10+5+1=\boxed{25\ \text{days}}$$ The agreement with the forward pass confirms both traverses. Note how narrowly C beats P into the critical chain: P is the longer activity (7 days against 4), but it feeds only Q and N, whose downstream chains are short, whereas C feeds M and then the 10-day F.

Question 1 — final results
Quantity	Value
Project duration	25 working days
Critical path	A – C – M – F – V – QI
Activities with zero total float	A, C, M, F, V, QI
Total float — P, Q, Y	2 days each
Total float — B, L	3 days each
Total float — N, J, S	5 days each
Dummies required	3 (two into the Q merge event, one from the Q head event into the Y tail event)

Topic. Critical path method on an activity-on-arrow network, and the meaning of total float.

Key relations. Forward pass $E_j=\max(E_i+D_{ij})$; backward pass $L_i=\min(L_j-D_{ij})$; total float $TF_{ij}=L_j-E_i-D_{ij}$; free float $FF_{ij}=E_j-E_i-D_{ij}$; the critical path is the longest chain of arrows through the network and, for an all-finish-to-start network, its durations sum to the project duration.

Why this works. The forward pass answers "how soon can this event possibly happen," so it must take the largest arriving value — an event is not reached until its slowest feeder arrives. The backward pass answers "how late may this event happen without pushing the completion date," so it must take the smallest departing requirement. Total float is what is left over when the time an activity actually needs is subtracted from the widest window it could occupy, between the earliest its predecessors can release it and the latest its successors can accept it. Because that window is defined by events shared with other activities, total float is a shared resource: if B consumes its three days of float, L has none left. Dummies matter for exactly the same reason — merging two activities at a common event asserts that every arrow entering it is a predecessor of every arrow leaving it, so a merge that is only partly true must be split and reconnected with zero-duration arrows.

Common pitfall. Two mistakes dominate. The first is omitting a dummy and merging Q's two predecessors at P's head event, which silently makes C a predecessor of N; the network then reports a 25-day duration for the wrong reason and the floats on the P–N–S chain come out wrong. The second is treating total float as private to one activity and letting two activities on the same chain each absorb "their" float, which consumes the same days twice and pushes the completion date out.

Source. Halpin & Senior, Construction Management, 4th ed., Ch. 9–10 (arrow networks, dummy logic, forward and backward pass, float); Hendrickson & Au, Project Management for Construction, 2nd ed., Ch. 10; PMI, PMBOK Guide, §6.5 Develop Schedule.

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