07-Str-B2 · May 2016

Question 1 of 6: Scheduling — critical path, total floats, and the effect of delaying E

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

Notes on this paper

Paper format: National Exams, May 2016 — 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 answered are marked. All six are worked below so that the paper can be used for revision whichever five a candidate chooses.

Reference texts: Hegazy, T., Computer-Based Construction Project Management (Prentice Hall) — precedence networks with lags, total and free float, project overhead versus general overhead, and the bar-chart/S-curve control method behind Questions 1, 3 and 6; Hendrickson, C. & Au, T., Project Management for Construction (2nd ed., Carnegie Mellon) — Chapters 5 (cost estimation), 10 (scheduling) and 12 (cost control, monitoring and accounting), the source of the earned-value quantities used in Question 6; Halpin, D.W. & Senior, B.A., Construction Management (4th ed., Wiley) — competitive bidding, unbalanced bids, indirect-cost structure and construction safety; Sullivan, W.G., Wicks, E.M. & Koelling, C.P., Engineering Economy (17th ed., Pearson) — Chapters 5 and 6, present-worth analysis and the repeatability assumption for alternatives with unequal lives, used in Question 4; Peurifoy, R.L. & Oberlender, G.D., Estimating Construction Costs (6th ed., McGraw-Hill) — job overhead versus general overhead; Canadian Construction Documents Committee, CCDC 2 — Stipulated Price Contract (2020) and CCDC 23 — A Guide to Calling Bids and Awarding Contracts — bid-call practice, bid security and award criteria for Question 2; Ron Engineering (M.J.B. Enterprises line of cases) as summarised in Goldsmith, I. & Heintzman, T.G., Goldsmith on Canadian Building Contracts (5th ed., Thomson Reuters) — the Contract A/Contract B doctrine that governs a Canadian public bid call; WorkSafeBC, Occupational Health and Safety Regulation (Parts 4, 8, 11, 13, 18, 19 and 20) and the BC Workers Compensation Act, together with CSA Z259 (fall protection), CSA Z94.4 (respirators) and CSA W117.2 (welding safety) — the Canadian rule set behind Question 5.

Question 1 (network). Every activity letter and duration is printed inside its box and reads cleanly. The link routing was traced at high magnification: Start feeds A, D and G; a riser from the right edge of D feeds B; the horizontal link D → E carries the only labelled lag on the sheet, FS 6; a riser from the right edge of G feeds E; G also feeds H, H feeds I, A feeds B, B feeds C, E feeds F; and C, F and I terminate at End. Every unlabelled arrow is an ordinary finish-to-start link with zero lag, which is the only reading consistent with the drawing.

Question 6 (bar chart). Every percentage label falls on a week boundary, so the printed figures are cumulative percent complete at each week end. Planned: A 20/60/100 in weeks 1–3; B 10/80 in weeks 1–2, finishing in week 3; C 20/70 in weeks 3–4, finishing in week 5. Actual: A 10/50/90 in weeks 1–3, its bar closing in week 4; B 70 at week 2, its bar closing in week 3; C 50 at week 3, its bar closing exactly on the week-4 gridline. A bar that closes is read as 100 % complete from that week end onward, which is the standard convention and the only reading that lets part (c) be answered at all.

Question 1: Scheduling — critical path, total floats, and the effect of delaying E (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 nine-activity precedence (activity-on-node) network with durations in working days and one labelled lag.

Activity	Duration (days)	Predecessors (relationship)
A	10	Start
B	12	A (FS 0), D (FS 0)
C	9	B (FS 0)
D	5	Start
E	7	D (FS 6), G (FS 0)
F	6	E (FS 0)
G	3	Start
H	4	G (FS 0)
I	6	H (FS 0)

Find. The critical path and project duration, the total float of every activity, and what happens to the project if activity E is delayed by one day.

[Figure not reproduced: Figure 1.1 — The precedence network as printed, redrawn with the forward-pass results and total floats above each box. Durations are in working days; the only lag on the sheet is the six-day finish-to-start lag on D → E. The critical path is highlighted. See the official exam paper.]

Approach. Run a forward pass from time zero to get early start and early finish for every activity, take the project duration as the largest early finish among the activities that feed End, then run a backward pass from that duration to get late start and late finish, and read total float as the difference between the late and early start of each activity.

Set the forward-pass rule, including the lag. Using the "day 0 = start of work" convention, an activity may start no earlier than the latest of its predecessors' constrained finishes: $$ ES_j=\max_{i\to j}\left(EF_i+\text{Lag}_{ij}\right),\qquad EF_j=ES_j+D_j $$ where $ES$ and $EF$ are early start and early finish and $D$ is duration. Only one link carries a lag: $\text{Lag}_{DE}=6$ days, so E cannot start until six days after D finishes.
Forward pass along the three chains. The three activities fed by Start begin at day 0, so $EF_A=0+10=10$, $EF_D=0+5=5$ and $EF_G=0+3=3$. B is fed by both A and D, and A governs: $$ ES_B=\max(EF_A,\;EF_D)=\max(10,\;5)=10,\qquad EF_B=10+12=22 $$ so $ES_C=22$ and $EF_C=22+9=31$. On the middle chain the lag governs rather than the longer of the two predecessors: $$ ES_E=\max\left(EF_D+6,\;EF_G\right)=\max(5+6,\;3)=11,\qquad EF_E=11+7=18 $$ and therefore $ES_F=18$, $EF_F=18+6=24$. On the bottom chain $ES_H=3$, $EF_H=7$, $ES_I=7$ and $EF_I=7+6=13$.
Read the project duration off the three terminal activities. C, F and I are the only activities with no successor, so the project cannot end before the last of them: $$ T=\max\left(EF_C,\;EF_F,\;EF_I\right)=\max(31,\;24,\;13) $$ $$ \boxed{T = 31\ \text{working days}} $$
Backward pass from day 31. Each terminal activity is given $LF=T=31$, and every other activity takes the tightest requirement its successors impose: $$ LF_i=\min_{i\to j}\left(LS_j-\text{Lag}_{ij}\right),\qquad LS_i=LF_i-D_i $$ Working right to left: $LS_C=31-9=22$, $LS_F=31-6=25$, $LS_I=31-6=25$; then $LF_B=LS_C=22$ so $LS_B=10$, $LF_E=LS_F=25$ so $LS_E=18$, and $LF_H=LS_I=25$ so $LS_H=21$. Activity A is fed only into B, so $LF_A=10$ and $LS_A=0$.
Handle the two activities with a branching successor set. D feeds B directly and E across the six-day lag, so both constraints must be honoured and the smaller wins: $$ LF_D=\min\left(LS_B,\;LS_E-6\right)=\min(10,\;18-6)=\min(10,\;12)=10 $$ giving $LS_D=10-5=5$. Similarly G feeds E and H: $$ LF_G=\min\left(LS_E,\;LS_H\right)=\min(18,\;21)=18,\qquad LS_G=18-3=15 $$ Note that for D it is the B link, not the lagged E link, that binds — the six-day lag is generous enough that it never controls D.
Compute total float and identify the critical path. Total float is the delay an activity can absorb without pushing the project end date: $$ TF_i=LS_i-ES_i=LF_i-EF_i $$ Applying this to each activity gives $TF_A=TF_B=TF_C=0$, $TF_D=5-0=5$, $TF_E=18-11=7$, $TF_F=25-18=7$, $TF_G=15-0=15$, $TF_H=21-3=18$ and $TF_I=25-7=18$. The zero-float chain is continuous from Start to End: $$ \boxed{\text{Critical path: Start}\to A\to B\to C\to\text{End},\quad 10+12+9=31\ \text{days}} $$
Assemble the full schedule table. Free float — the delay an activity can absorb without disturbing the early start of any successor — is worth carrying alongside total float, because it is the float a superintendent may consume unilaterally: $$ FF_i=\min_{i\to j}\left(ES_j-\text{Lag}_{ij}\right)-EF_i $$ For D this is $\min(ES_B,\;ES_E-6)-EF_D=\min(10,\;5)-5=0$, so D's five days of total float are shared with B and cannot be spent freely; for I it is $31-13=18$ days, all of it I's own.
Answer the delay question directly. Activity E carries $TF_E=7$ days. Delaying E by one day moves its start from day 11 to day 12 and its finish from day 18 to day 19, which pushes F to start on day 19 and finish on day 25. Since $LF_F=31$, the project end is untouched: $$ T_{\text{delayed}}=\max(31,\;25,\;13)=31\ \text{days} $$ $$ \boxed{\text{No effect on the project duration; E's remaining total float falls from 7 days to 6 days}} $$ The one caveat is that E's free float is zero, so the one-day slip is passed straight through to F's early start — the delay is absorbed by the schedule, not by E alone.

Final Results — Question 1
Activity	Duration	ES	EF	LS	LF	Total float	Free float
A	10	0	10	0	10	0 (critical)	0
B	12	10	22	10	22	0 (critical)	0
C	9	22	31	22	31	0 (critical)	0
D	5	0	5	5	10	5	0
E	7	11	18	18	25	7	0
F	6	18	24	25	31	7	7
G	3	0	3	15	18	15	0
H	4	3	7	21	25	18	0
I	6	7	13	25	31	18	18
Project duration						31 working days
Critical path						Start → A → B → C → End
Effect of delaying E by one day						None on completion; E's float 7 → 6 days

Topic. Critical-path method on a precedence (activity-on-node) network containing a finish-to-start lag.

Key relations. Forward pass $ES_j=\max(EF_i+\text{Lag}_{ij})$, $EF_j=ES_j+D_j$; backward pass $LF_i=\min(LS_j-\text{Lag}_{ij})$, $LS_i=LF_i-D_i$; total float $TF=LS-ES=LF-EF$; free float $FF=\min(ES_j-\text{Lag}_{ij})-EF_i$; the critical path is the continuous chain of zero-total-float activities from Start to End.

Why this works. The forward pass propagates the earliest each activity can happen given only its predecessors, and the backward pass propagates the latest each activity may happen without moving a fixed completion date. The gap between the two answers is exactly the schedule slack, because the two passes differ only in which end of the network is pinned. A lag is nothing more than an extra constant added to a link, so it enters the forward pass as $+\text{Lag}$ and the backward pass as $-\text{Lag}$; treating it symmetrically is what keeps the two passes consistent. When the gap closes to zero along a whole chain, no rearrangement can shorten the project without shortening one of those activities — that chain is the critical path, and here it is the top row, whose 31 days beat the middle chain's 24 and the bottom chain's 13.

Common pitfall. The two classic errors on this network are both about D. The first is applying the FS 6 lag in the forward pass but forgetting it in the backward pass, so that the E link is taken to constrain D to $LS_E=18$ instead of $LS_E-6=12$. On this particular network the slip is masked, because the B link ($LS_B=10$) binds either way and $LF_D$ stays at 10; on a network where the lagged link governed, it would overstate D's float by the full six days. The second is assuming the lagged link must be the binding one because it looks like the harsher constraint; in fact $LS_B=10$ is tighter than $LS_E-6=12$, so B controls D. A third, subtler slip is answering the delay question with "E is not critical, so nothing happens" without checking free float — E's free float is zero, so F really does move, and a candidate who is tracking activity start dates rather than just the end date needs to say so.

Source. Hegazy, Computer-Based Construction Project Management, Chapters 5–6 (precedence networks, lags, and float definitions); Hendrickson & Au, Project Management for Construction, Chapter 10 (fundamental scheduling procedures).

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