07-Str-B2 · May 2014

Question 2 of 6: Scheduling — precedence network with SS, FS and FF lags

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

Notes on this paper

National Examinations — May 2014 — 07-Str-B2 Management of Construction. Three hours, closed book; candidates may use one of the two approved calculators (Casio or Sharp). The paper prints six questions of equal value (20 marks each) and states that any five questions constitute a complete paper, only the first five appearing in the answer book being marked. Candidates are urged to submit a clear statement of any interpretive assumptions with their answers. All six questions are worked below, because this set is intended as a study resource rather than as a single exam sitting.

Reference texts: RSMeans, Building Construction Cost Data — the "How to Use the Cost Data" front matter, which defines daily output, labour-hours, bare costs and the Total Incl. O&P column used in Question 1; Halpin, D.W. & Senior, B.A., Construction Management (4th ed., Wiley) — unit-price estimating, crew balancing, labour relations and construction safety; Hegazy, T., Computer-Based Construction Project Management (Prentice Hall) — precedence networks with SS/FS/FF relationships and lags, which is exactly the notation of Question 2; Hendrickson, C. & Au, T., Project Management for Construction (2nd ed., Carnegie Mellon) — scheduling and cost control; Sullivan, W.G., Wicks, E.M. & Koelling, C.P., Engineering Economy (17th ed., Pearson) — present-worth analysis and the maximum-justified-investment problem of Question 4; Canadian Construction Documents Committee, CCDC 2 — Stipulated Price Contract (2020), General Conditions 6.5 (delays) and 6.6 (claims for a change in Contract Price); Goldsmith, I. & Heintzman, T.G., Goldsmith on Canadian Building Contracts (5th ed., Thomson Reuters) — delay and notice law in Canada; AACE International, Recommended Practice 29R-03: Forensic Schedule Analysis — the but-for and windows methods named in Question 5; British Columbia Labour Relations Code, RSBC 1996 c. 244 — certification, bargaining units and the construction-industry provisions behind Question 3; WorkSafeBC, Occupational Health and Safety Regulation (Parts 4, 8, 11, 18 and 20) and the BC Workers Compensation Act — the prime-contractor duty and the traffic-control, fall-protection and hazardous-substance rules behind Question 6.

Check — two readings taken from the printed page. The RS Means extract in Question 1 prints two cells as question marks; both are recovered below from the crew table, and the recovered labour-hour figure is checked against the printed $27 labour column before it is used. Two arrows leave the right-hand edge of activity D and turn vertically to reach E and C; they are read here as ordinary finish-to-start links, which is the only reading consistent with the drawing and with the fact that every unlabelled arrow on the sheet carries no lag.

Question 2: Scheduling — precedence network with SS, FS and FF lags (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. Eight activities and eleven precedence relationships, as printed:

Activity	Duration (days)	Predecessor relationships
B	8	— (project start)
D	12	B, start-to-start, lag 2
E	7	D, finish-to-start, lag 0
C	8	D, finish-to-start, lag 0
F	9	D, E and C, all finish-to-start, lag 0
H	10	B, finish-to-start, lag 6; C, finish-to-start, lag 0
G	12	F, start-to-start, lag 8
I	8	F, start-to-start, lag 8; H, finish-to-finish, lag 3

[Figure not reproduced: Figure 2.1 — the precedence network as printed, redrawn. Each box carries the activity letter over its duration in days. Labelled arrows carry the relationship type and its lag; unlabelled arrows are finish-to-start with zero lag. See the official exam paper.]

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

Approach. Run a forward pass in which each lag type imposes its own inequality on the successor's early start, take the project duration as the largest early finish, run the backward pass with the mirror-image inequalities, and read total float as late start minus early start. Part (b) then follows from whether C carries float.

State the three lag rules before touching a number. In precedence diagramming each relationship constrains a different pair of dates, and mixing them up is the usual source of a wrong answer. For a link from predecessor $i$ to successor $j$ with lag $L$: $$\text{SS: } ES_j\ge ES_i+L\qquad \text{FS: } ES_j\ge EF_i+L\qquad \text{FF: } EF_j\ge EF_i+L$$ Each activity takes the largest value its incoming links demand, and $EF=ES+D$ throughout. Day zero is the instant the project starts, so an activity with $ES=2$ begins at the start of day 3 in calendar terms.
Forward pass. Working in the order B, D, E, C, F, H, G, I — a valid topological order — and applying the rules above: $$ES_B=0,\quad EF_B=8$$ $$ES_D=ES_B+2=2,\quad EF_D=2+12=14$$ $$ES_E=EF_D=14,\quad EF_E=21\qquad ES_C=EF_D=14,\quad EF_C=22$$ $$ES_F=\max(EF_D,\,EF_E,\,EF_C)=\max(14,21,22)=22,\quad EF_F=31$$ $$ES_H=\max(EF_B+6,\,EF_C)=\max(14,22)=22,\quad EF_H=32$$ $$ES_G=ES_F+8=30,\quad EF_G=42$$ For I both an SS and an FF link apply: the SS link gives $ES_I\ge 30$ and the FF link gives $EF_I\ge EF_H+3=35$. Starting at day 30 yields $EF_I=38$, which already satisfies the finish-to-finish requirement, so $ES_I=30$ and $EF_I=38$.
Read the project duration. The project cannot end before every activity has finished, so it is governed by the latest early finish: $$T=\max(EF)=\max(42,\,38)=\boxed{42\ \text{days}}$$ Note that the last activity to finish is G, not I, even though I is drawn furthest down and to the right. Position on the sheet says nothing about criticality.
Backward pass. Setting $LF=42$ for the two terminal activities and working backwards, each predecessor takes the smallest late date its outgoing links allow, using the mirrored rules $LS_i\le LS_j-L$ for SS, $LF_i\le LS_j-L$ for FS and $LF_i\le LF_j-L$ for FF: $$LF_G=42,\ LS_G=30\qquad LF_I=42,\ LS_I=34$$ $$LS_F=\min(LS_G-8,\,LS_I-8)=\min(22,26)=22,\ LF_F=31$$ $$LF_H=LF_I-3=39,\ LS_H=29$$ $$LF_C=\min(LS_F,\,LS_H)=\min(22,29)=22,\ LS_C=14\qquad LF_E=LS_F=22,\ LS_E=15$$ $$LF_D=\min(LS_E,\,LS_C,\,LS_F)=\min(15,14,22)=14,\ LS_D=2$$ $$LS_B=LS_D-2=0,\ LF_B=8$$
Tabulate the four dates and the total float. Total float is the amount by which an activity can slip without pushing the project finish, $TF=LS-ES=LF-EF$:

Activity	$D$	$ES$	$EF$	$LS$	$LF$	$TF$	Critical?
B	8	0	8	0	8	0	yes
D	12	2	14	2	14	0	yes
E	7	14	21	15	22	1	no
C	8	14	22	14	22	0	yes
F	9	22	31	22	31	0	yes
H	10	22	32	29	39	7	no
G	12	30	42	30	42	0	yes
I	8	30	38	34	42	4	no

Identify the critical path. The zero-float activities form the continuous chain $$\boxed{B\xrightarrow{SS2}D\longrightarrow C\longrightarrow F\xrightarrow{SS8}G,\qquad T=42\ \text{days}}$$ Reading the chain link by link reproduces the duration and shows where the time actually goes: the SS2 link contributes 2 days, D contributes its full 12, C its full 8, the SS8 link out of F contributes 8 of F's 9 days, and G contributes its full 12, so $2+12+8+8+12=42$. Two features are worth noting. Only part of F lies on the critical path, because the SS8 link releases G once F has been under way for eight days; that is a normal consequence of start-to-start logic and is why the critical path in a precedence network cannot always be described as a chain of whole activities. And E, despite being an immediate predecessor of the critical activity F, has a day of float, because C finishes one day later than E and is the link that actually governs F.
Part (b): assess a three-day delay to activity C. C carries zero total float, so by definition it has no room to slip; any delay to it moves the project finish day for day. The recomputation confirms it. With C starting on day 17 instead of day 14 and still taking 8 days, $EF_C=25$, so $ES_F=25$ and $EF_F=34$; $ES_H=\max(14,25)=25$; the SS8 link then gives $ES_G=33$ and $EF_G=45$, while I starts on day 33 and finishes on day 41, comfortably satisfying its finish-to-finish tie to H. The controlling finish is therefore G at day 45: $$\boxed{T'=45\ \text{days, i.e. the project is delayed by the full 3 days}}$$ The same answer follows if the phrase is read as C taking 11 days rather than 8, because C's finish is what governs both F and H either way. The delay is not absorbed anywhere: H's seven days of float and I's four days are carried along by the shift unchanged rather than consumed, and neither of those chains was controlling to begin with.

Figure 2.2 — the early-start bar chart. Red bars are the zero-float critical activities B, D, C, F and G; blue bars are E, H and I, whose dashed tails show the one, seven and four days of total float respectively. The project finish is governed by G at day 42.

Result	Value
Project duration	42 days
Critical path	B → D → C → F → G
Total float — B, D, C, F, G	0 days each
Total float — E	1 day
Total float — H	7 days
Total float — I	4 days
(b) Project duration with C delayed 3 days	45 days
(b) Net effect on completion	3-day slip, absorbed nowhere

Topic. Critical path analysis of a precedence (activity-on-node) network containing start-to-start, finish-to-start and finish-to-finish relationships with lags.

Key relations. Forward pass: $ES_j=\max$ over incoming links of $ES_i+L$ (SS), $EF_i+L$ (FS) and $EF_i+L-D_j$ (FF); $EF=ES+D$. Backward pass: $LS_i=\min$ over outgoing links of $LS_j-L$ (SS), and $LF_i=\min$ of $LS_j-L$ (FS) and $LF_j-L$ (FF); $LS=LF-D$. Total float $TF=LS-ES=LF-EF$; free float $FF=\min(ES_{\text{succ}})-EF$.

Why this works. The forward pass answers “how soon can this happen” by taking the most demanding of the constraints pushing an activity later, and the backward pass answers “how late can it happen without moving the end date” by taking the most demanding of the constraints pulling it earlier. Where the two answers coincide the activity has no room to move, and the connected chain of such activities is the critical path. Lags do not change that logic; they only change which pair of dates a link constrains. A start-to-start lag lets a successor overlap its predecessor, so the critical path may pass through part of an activity rather than all of it, and a finish-to-finish lag constrains only the tail of an activity, which is why the FF3 tie from H could bind only I's finish — and here it does not bind at all, because the SS8 link from F already puts I's finish at day 38, three days later than the day 35 the FF link requires.

Common pitfall. The first mistake is to add raw durations along each chain and ignore the lags: the longest such chain, B–D–C–F–G, happens to be the critical one, but it sums to 49 days, seven more than the true 42, because the SS2 link lets D overlap B by six days and the SS8 link lets G start one day before F ends. The second is to treat SS8 as though G waited eight days after F finishes (FS8), which would give a 51-day project. The third is to forget that an FF link reaches a successor's start only through the successor's own duration, $ES_j\ge EF_i+L-D_j$, so an activity with both SS and FF predecessors must be tested both ways, as I was in step 2. Finally, once part (a) is done, part (b) needs no new arithmetic at all: an activity with zero total float transmits its delay in full, and the only reason to recompute is to confirm that no downstream float absorbs it.

Source. Hegazy, Computer-Based Construction Project Management, ch. 5–6 (precedence diagramming, lags, forward and backward passes); Halpin & Senior, Construction Management, 4th ed., ch. 9 (CPM and float); Hendrickson & Au, Project Management for Construction, ch. 10.

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Activity	$D$	$ES$	$EF$	$LS$	$LF$	$TF$	Critical?
B	8	0	8	0	8	0	yes
D	12	2	14	2	14	0	yes
E	7	14	21	15	22	1	no
C	8	14	22	14	22	0	yes
F	9	22	31	22	31	0	yes
H	10	22	32	29	39	7	no
G	12	30	42	30	42	0	yes
I	8	30	38	34	42	4	no

Activity	$D$	$ES$	$EF$	$LS$	$LF$	$TF$	Critical?
B	8	0	8	0	8	0	yes
D	12	2	14	2	14	0	yes
E	7	14	21	15	22	1	no
C	8	14	22	14	22	0	yes
F	9	22	31	22	31	0	yes
H	10	22	32	29	39	7	no
G	12	30	42	30	42	0	yes
I	8	30	38	34	42	4	no

Activity	$D$	$ES$	$EF$	$LS$	$LF$	$TF$	Critical?
B	8	0	8	0	8	0	yes
D	12	2	14	2	14	0	yes
E	7	14	21	15	22	1	no
C	8	14	22	14	22	0	yes
F	9	22	31	22	31	0	yes
H	10	22	32	29	39	7	no
G	12	30	42	30	42	0	yes
I	8	30	38	34	42	4	no