16-Civ-B8 Management of Construction · December 2016

Question 1 of 6: Scheduling

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

Notes on this paper

Paper format. 98-Civ-B8 Management of Construction, National Exams December 2016. Three hours, closed book, one approved calculator (Casio or Sharp). Six questions of equal value (20 marks each); any five constitute a complete paper and only the first five presented in the answer book are marked. All six are solved here, because the set is a study resource rather than an examination script.

Reference texts.

Halpin & Senior, Construction Management, 4th ed. — precedence networks with lags, bonding, project control.
Hendrickson, Project Management for Construction, 2nd ed. — scheduling, cost control, earned value, financing of constructed facilities.
Project Management Institute, A Guide to the Project Management Body of Knowledge (PMBOK Guide), 6th ed. — schedule and cost management, earned value.
Fraser et al., Global Engineering Economics, 5th Canadian ed. — present worth, deferred annuities, maximum justifiable investment.
Canadian Construction Documents Committee: CCDC 2 (stipulated price), CCDC 4 (unit price), CCDC 14 (design-build), CCDC 3 (cost-plus), CCDC 23 Guide to Calling Bids and Awarding Contracts; CCDC 220 / 221 / 222 bond forms.
Builders Lien Act (British Columbia, RSBC 1997 c. 45) and the provincial construction-lien / prompt-payment statutes; RSMeans Residential Square Foot Costs.

Question 1: Scheduling (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 precedence (activity-on-node) network of eight activities with three lagged relationships, read directly off the printed drawing on page 2 of the paper.

Activity durations and relationships, as printed on the network
Activity	Duration (days)	Predecessor links (type, lag)
B	8	— (start activity)
D	12	B (SS 2)
E	7	D (FS 0)
C	8	D (FS 0)
F	9	D (FS 0), E (FS 0), C (FS 0)
H	10	B (FS 6), C (FS 0)
G	12	F (SS 8)
I	8	F (SS 8), H (FF 3)

Find. (a) the critical path, the project duration and the total float of every activity; (b) the change in project duration if activity F is delayed by three days.

[Figure not reproduced: Figure 1.1 — the printed precedence network redrawn: activity letter above, duration in days below; lag labels in red. See the official exam paper.]

Approach. Run a precedence-diagram forward pass to get early start and early finish for every activity (each relationship type imposes its own inequality on the successor’s early times), take the project duration as the largest early finish, then run the backward pass for late finish and late start and report total float as $LS-ES$.

State the four relationship rules in early-time form. For a link from predecessor $p$ to successor $s$ with lag $L$, the forward pass imposes $$ES_s \ge ES_p + L \quad (\text{SS}), \qquad ES_s \ge EF_p + L \quad (\text{FS}), \qquad EF_s \ge EF_p + L \quad (\text{FF}),$$ and each activity takes the largest value its predecessors demand. Only the FF rule constrains a finish, so it is converted to a start by $ES_s \ge EF_p + L - D_s$.
Forward pass through the network. Starting at $ES_B=0$, B finishes on day 8. The SS2 link releases D two days after B starts, so $ES_D = 0+2 = 2$ and $EF_D = 2+12 = 14$. D releases both E and C at day 14, giving $EF_E = 21$ and $EF_C = 22$. F waits on the latest of its three finish-to-start predecessors, $\max(14,\,21,\,22)$, so $$ES_F = 22, \qquad EF_F = 22 + 9 = 31 .$$ H is pulled by B through FS6 to day $8+6=14$ and by C to day 22, so $ES_H = 22$ and $EF_H = 32$.
Carry the two lagged successors G and I. G hangs off F by SS8, so it may start eight days after F starts rather than after F finishes: $$ES_G = ES_F + 8 = 30, \qquad EF_G = 30 + 12 = 42 .$$ I is pulled by F through SS8 to day 30, and by H through FF3, which requires $EF_I \ge EF_H + 3 = 35$, i.e. $ES_I \ge 35 - 8 = 27$. The governing value is 30, so $ES_I = 30$ and $EF_I = 38$.
Read the project duration. The network has two terminal activities, G and I, and the project cannot finish before the later of them: $$T = \max\,(EF_G,\;EF_I) = \max\,(42,\;38) = \boxed{42 \text{ days}}$$
Backward pass. Setting $LF_G = LF_I = 42$ and working right to left with the mirror-image rules (an SS link constrains the predecessor’s late start, an FF link its late finish), G gives $LS_G = 30$ and I gives $LS_I = 34$. H is held by the FF3 link to $LF_H = LF_I - 3 = 39$. F is held by G to $LS_F = LS_G - 8 = 22$ and by I to $LS_I - 8 = 26$; the binding one is 22, so $LF_F = 31$. Continuing, $LF_C = LS_F = 22$, $LF_E = 22$, $LF_D = \min(LS_E,\,LS_C,\,LS_F) = 14$ and $LS_B = LS_D - 2 = 0$, which returns exactly to zero and confirms the pass.
Tabulate the schedule and take total float as $TF = LS - ES$.
Precedence-network schedule, all times in working days
Activity Duration ES EF LS LF TF

B 8 0 8 0 8 0

D 12 2 14 2 14 0

E 7 14 21 15 22 1

C 8 14 22 14 22 0

F 9 22 31 22 31 0

H 10 22 32 29 39 7

G 12 30 42 30 42 0

I 8 30 38 34 42 4
Identify the critical path. The zero-float activities form one unbroken chain through the network: $$\boxed{\;B \rightarrow D \rightarrow C \rightarrow F \rightarrow G \;=\; 42 \text{ days}\;}$$ Walking it link by link reproduces the duration exactly: B starts at 0, the SS2 lag releases D at day 2, D runs 12 days to day 14, C runs 8 days to day 22, F runs 9 days to day 31 but releases G after only 8 of them at day 30, and G’s 12 days close the project at day 42. Note that the plain sum of the five critical durations is $8+12+8+9+12 = 49$ days — seven days more than the answer, because the SS2 and SS8 links let successors overlap their predecessors. That sum-equals-duration check is valid only on an all-finish-to-start network, and using it here would be wrong.
Part (b): delay activity F by three days. F carries zero total float, so it has no slack to absorb a delay. Pushing its start from day 22 to day 25 pushes the SS8 link with it, and G — the activity that sets the project finish — starts at day 33 and ends at day 45: $$ES_G' = 25 + 8 = 33, \qquad EF_G' = 33 + 12 = 45 .$$ The other terminal activity moves to $ES_I' = 33$, $EF_I' = 41$, which is still not governing, so $$\boxed{T' = 45 \text{ days} \;=\; T + 3 \text{ days}}$$ The whole three-day delay is transmitted to the completion date, and there is no other path able to take over as critical: the nearest competitor, the chain through I, still finishes four days earlier.

Precedence-network schedule, all times in working days
Activity	Duration	ES	EF	LS	LF	TF
B	8	0	8	0	8	0
D	12	2	14	2	14	0
E	7	14	21	15	22	1
C	8	14	22	14	22	0
F	9	22	31	22	31	0
H	10	22	32	29	39	7
G	12	30	42	30	42	0
I	8	30	38	34	42	4

Check: the two readings of “delaying F by 3 days” give different answers, and that is the point of the question. If F is delayed in the sense of starting three days late (the reading taken above), the project slips the full three days to 45. If instead F’s duration grows by three days (9→12) while it still starts on day 22, the project duration does not move: both of F’s successors are tied to F by start-to-start links, so G still starts on day 30, F now finishes on day 34 — still inside its own late finish of 34 — and the project still ends on day 42. A candidate should state which reading is being used; the first is the standard meaning of “delaying an activity” and is the one carried into the Final Results.

Question 1 — results
Quantity	Value
Project duration (as scheduled)	42 days
Critical path	B → D → C → F → G
Total float — B, D, C, F, G	0 days (critical)
Total float — E	1 day
Total float — I	4 days
Total float — H	7 days
(b) Project duration with F delayed 3 days	45 days (a 3-day slip)
(b) Alternative reading: F’s duration extended 3 days	42 days (no slip)

Topic. Precedence diagramming with start-to-start, finish-to-finish and lagged finish-to-start relationships.

Key relations. Forward pass: $ES_s = \max\{ES_p+L\ (\text{SS}),\;EF_p+L\ (\text{FS}),\;EF_p+L-D_s\ (\text{FF})\}$. Backward pass: $LF_p = \min\{LS_s-L\ (\text{FS}),\; LS_s-L+D_p\ (\text{SS}),\;LF_s-L\ (\text{FF})\}$. Total float $TF = LS-ES = LF-EF$; free float $FF = \min(ES_s) - EF_p$ over finish-to-start successors.

Why this works. The forward pass answers “how early can this happen if nothing goes wrong?” by pushing each activity as far left as its most demanding predecessor allows; the backward pass answers “how late can it happen without moving the completion date?” by pulling each activity as far right as its least tolerant successor allows. The difference between the two is by construction the amount of slippage the schedule can absorb, so zero difference marks the chain that governs the duration. Lags do not change that logic — they only change the arithmetic of each inequality — but they do break the intuition that the critical path length equals the sum of its activities, because an SS lag lets a successor overlap its predecessor. On this network the SS2 and SS8 overlaps buy back seven of the 49 days of critical work.

Common pitfall. Two mistakes dominate. The first is treating an SS lag as though it were an FS lag, which here would push G to day $31+8=39$ and give a 51-day project. The second is forgetting that a finish-to-finish link constrains a finish, not a start: converting H→I FF3 into a start constraint requires subtracting I’s own duration, and candidates who skip that step usually report I as critical.

Source. Halpin & Senior, Construction Management, 4th ed., Ch. 11 (precedence networks and lag relationships); Hendrickson, Project Management for Construction, 2nd ed., Ch. 10.

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Activity	Duration	ES	EF	LS	LF	TF
B	8	0	8	0	8	0
D	12	2	14	2	14	0
E	7	14	21	15	22	1
C	8	14	22	14	22	0
F	9	22	31	22	31	0
H	10	22	32	29	39	7
G	12	30	42	30	42	0
I	8	30	38	34	42	4

Activity	Duration	ES	EF	LS	LF	TF
B	8	0	8	0	8	0
D	12	2	14	2	14	0
E	7	14	21	15	22	1
C	8	14	22	14	22	0
F	9	22	31	22	31	0
H	10	22	32	29	39	7
G	12	30	42	30	42	0
I	8	30	38	34	42	4

Activity	Duration	ES	EF	LS	LF	TF
B	8	0	8	0	8	0
D	12	2	14	2	14	0
E	7	14	21	15	22	1
C	8	14	22	14	22	0
F	9	22	31	22	31	0
H	10	22	32	29	39	7
G	12	30	42	30	42	0
I	8	30	38	34	42	4