24-Bld-A3 Construction Engineering · Undated paper
Question 1 of 7: CPM Network, Critical Path and the Effect of Delaying Activity F
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations — 07-Bld-A3, May 2019 — Construction Engineering. Closed book; candidates may use one of the two approved calculators (Casio or Sharp). The paper prints seven 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 record any interpretive assumptions with their answers. All seven questions are worked below, because the set is intended as a study resource rather than as a single exam sitting.
Reference texts: Hendrickson, C. & Au, T., Project Management for Construction (2nd ed., Carnegie Mellon) — precedence networks with SS/FS lags, cash-flow financing, contract types; Halpin, D.W. & Senior, B.A., Construction Management (4th ed., Wiley) — CPM/LOB scheduling, formwork & equipment production, bonding and cash flow; Canadian Construction Documents Committee, CCDC 2 — Stipulated Price Contract (2020) — contract clauses, addenda, change orders, holdback; Canadian Foundation Engineering Manual (CFEM) & WorkSafeBC Occupational Health and Safety Regulation, Part 20 — excavation support and shoring.
Question 1: CPM Network, Critical Path and the Effect of Delaying Activity F (20 marks)
Given. An activity-on-node network of nine activities. A, B and C are the three start activities. A drives D through a Start-to-Start lag (SS = 2); D drives H (finish-to-start, no lag). B drives both E and F (finish-to-start, no lag). C drives G through a Finish-to-Start lag (FS = 3); F and G both drive I (finish-to-start, no lag). E and I are the network's other two end activities besides H.
Activity durations (days)
Activity
A
B
C
D
E
F
G
H
I
Duration
5
4
5
5
12
6
4
8
3
[Figure not reproduced: Fig. Q1-1 — activity-on-node network (redrawn from the source diagram): three start chains A→D→H, B→E and B→F→I, C→G→I, with the SS = 2 and FS = 3 lagged links shown. See the official exam paper.]
Find. The early/late schedule for every activity, the critical path and project duration, the Late Bar Chart, and the schedule impact of a 5-day delay to activity F.
Approach. Run a forward pass respecting the two lag types (for an SS link, ES(succ) = ES(pred) + lag; for an FS link, ES(succ) = EF(pred) + lag), then a backward pass using the symmetric late-time rules, read off total float = LS − ES, and re-run the forward pass with F's duration increased by 5 days to measure the delay's effect.
Forward pass — early start / early finish.
A, B and C have no predecessors, so $ES_A = ES_B = ES_C = 0$ and $EF = ES + duration$: $EF_A = 5$, $EF_B = 4$, $EF_C = 5$.
For D (SS = 2 from A): $ES_D = ES_A + 2 = 0 + 2 = 2$, $EF_D = 2 + 5 = 7$.
For E and F (plain FS from B): $ES_E = EF_B = 4$, $EF_E = 4 + 12 = 16$; $ES_F = EF_B = 4$, $EF_F = 4 + 6 = 10$.
For G (FS = 3 from C): $ES_G = EF_C + 3 = 5 + 3 = 8$, $EF_G = 8 + 4 = 12$.
For H (FS from D): $ES_H = EF_D = 7$, $EF_H = 7 + 8 = 15$.
For I (FS from both F and G, so the later of the two governs): $ES_I = \max(EF_F, EF_G) = \max(10, 12) = 12$, $EF_I = 12 + 3 = 15$.
The network has three "ends" — E, H and I — so $\boxed{\text{Project duration} = \max(EF_E, EF_H, EF_I) = \max(16, 15, 15) = 16 \text{ days}}$.
Backward pass — late finish / late start.
Working back from a 16-day project finish: $LF_E = 16 \Rightarrow LS_E = 4$; $LF_H = 16 \Rightarrow LS_H = 8$; $LF_I = 16 \Rightarrow LS_I = 13$.
$LF_G = LS_I = 13 \Rightarrow LS_G = 9$. $LF_F = LS_I = 13 \Rightarrow LS_F = 7$.
$LF_C = LS_G - 3 = 9 - 3 = 6 \Rightarrow LS_C = 1$ (the FS = 3 lag is subtracted going backward, mirroring how it was added going forward).
$LF_D = LS_H = 8 \Rightarrow LS_D = 3$.
$LF_B = \min(LS_E, LS_F) = \min(4, 7) = 4 \Rightarrow LS_B = 0$.
$LS_A$ is fixed by the SS link, not by $LF_A$: $LS_A = LS_D - 2 = 3 - 2 = 1$, so $LF_A = LS_A + 5 = 6$.
Total float and the critical path.
$Float = LS - ES$ for every activity, tabulated below. Only B and F… only B and E come out at zero float ($Float_B = 0-0=0$, $Float_E = 4-4=0$), so $\boxed{\text{Critical path} = B \to E}$, the 4 + 12 = 16-day chain that sets the whole project's duration. F carries $Float_F = 7-4=3$ days of total float even though it feeds the same converging node (I) as the critical-adjacent G — the float is "used up" by the fact that G, not F, is the later arrival at I under the as-planned durations.
Activity
ES
EF
LS
LF
Total float
Critical?
A
0
5
1
6
1
No
B
0
4
0
4
0
Yes
C
0
5
1
6
1
No
D
2
7
3
8
1
No
E
4
16
4
16
0
Yes
F
4
10
7
13
3
No
G
8
12
9
13
1
No
H
7
15
8
16
1
No
I
12
15
13
16
1
No
Project duration = 16 days; Critical path = B → E
Fig. Q1-2 — Late Bar Chart: every bar spans LS→LF for its activity. B and E (gold) are the zero-float, critical chain; the other seven bars carry the total float listed in the results table.
In the figure above, each bar spans its LS→LF window: e.g. F runs LS=7→LF=13, a 6-day span, matching the table.
Effect of delaying activity F by 5 days.
F's own total float is only 3 days ($Float_F = 3$), so a 5-day delay overruns that float by $5-3=2$ days. Re-running the forward pass with $duration_F = 6+5=11$: $EF_F = 4+11=15$. This now exceeds G's arrival at I ($EF_G=12$), so F becomes the governing predecessor at the converging node: $ES_I = \max(15,12)=15$, $EF_I = 15+3=18$. The B→E chain is untouched, so the new project duration is $\max(16,15,18)=18$ days.
$$\boxed{\text{Delaying F by 5 days extends the project from 16 to 18 days} - \text{a net 2-day delay} - \text{and the critical path shifts from B}\to\text{E to B}\to\text{F}\to\text{I} \ (4+11+3=18)}$$
E is still on time (float on the old critical path is now 2 days, since the new 18-day duration exceeds E's 16-day finish), while G picks up 3 days of float it did not have before, because I no longer waits on it.
