Question 6 of 7: Construction Project — CPM Network and an Accident-Driven Duration Change

Question 6: Construction Project — CPM Network and an Accident-Driven Duration Change (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. Ten activities with precedence and durations (days), late completion penalty $\$5,000$/day.

Find. (a) The project network and critical path; (b) earliest/latest start times (and slack) of every activity; (c) the effect on finish date if D becomes 15 days because of the accident investigation, plus recovery strategies.

Approach. Run a forward pass (earliest start/finish) then a backward pass (latest start/finish) through the precedence network to get every activity's slack; the critical path is the chain of zero-slack activities, and its total duration is the project length. Then re-run the forward pass with D's new 15-day duration to see whether the critical path itself relocates.

1. Forward pass (earliest times). Starting A at day 0 and working through the precedence chain ($ES_i=\max$ of predecessors' $EF$; $EF_i=ES_i+d_i$): $$EF_A=15,\ EF_B=27,\ EF_C=21,\ EF_D=20,\ EF_E=\max(27,21)+3=30,\ EF_G=21+8=29,$$ $$EF_H=20+9=29,\ EF_F=\max(30,29)+8=38,\ EF_J=\max(29,29)+14=43,\ EF_I=38+7=45.$$ The project duration is the later of the two end activities: $\boxed{T=\max(EF_I,EF_J)=\max(45,43)=45\ \text{days}}$.
2. Backward pass and slack (part b). Setting $LF_I=LF_J=45$ (project length) and working backward ($LF_i=\min$ of successors' $LS$; $LS_i=LF_i-d_i$; slack $=LS_i-ES_i$) gives every activity's latest start and slack, tabulated below. Zero-slack activities are critical.
3. Critical path (part a). The chain of zero-slack activities is $\boxed{A\to B\to E\to F\to I}$, with $15+12+3+8+7=45$ days, matching the project duration — this is the path shown in red in Figure 1. All other activities (C, D, G, H, J) carry positive slack (1–2 days) and are not critical — note D itself has only 2 days of slack.
4. Effect of D → 15 days, accident investigation (part c). D is not on the original critical path (2 days of slack), so the first 2 days of its 10-day increase (5→15) are absorbed harmlessly, but the remaining 8 days push its successor chain past the old critical path. Re-running the forward pass with $d_D=15$: $EF_D=15+15=30\Rightarrow EF_H=30+9=39\Rightarrow EF_J=\max(29,39)+14=53$, while the F/I side is unaffected ($EF_I=45$ still, since D does not feed F). The new project duration is $$\boxed{T'=\max(45,53)=53\ \text{days}},$$ a delay of $53-45=\boxed{8\ \text{days}}$ versus the original plan (D's duration increases from 5 to 15 days, a 10-day increase, of which 2 days of slack absorb part, giving a net delay of $10-2=8$ days, confirmed by the recomputation above). The critical path shifts entirely to $A\to D\to H\to J$ (new duration $15+15+9+14=53$ days). At $\$5{,}000$/day, an unmitigated 8-day delay costs $\boxed{8\times\$5{,}000=\$40{,}000}$ in late-completion penalties.

Recovery strategies (part c, continued). Because the delay is driven by an accident investigation at the subcontractor (a regulatory/safety hold, not simply a resourcing shortfall like a strike), the available strategies are shaped by what can and cannot be rushed: (1) Compress the new critical path downstream of D (A–D–H–J). The investigation itself typically cannot be shortened by throwing money at it (it has to run its course for safety and liability reasons), but once D is released, H and J can be crashed — authorize overtime or a second crew on H and J to buy back days at a cost that should be compared against the $\$5,000$/day penalty (any crashing move that costs less than $\$5,000$ to save a day is worth it). (2) Fast-track by overlapping H into D's tail end where safe. If the investigation only restricts specific work areas or activities (common in practice), identify which portions of H do not depend on the still-restricted scope and begin them as soon as permitted, rather than waiting for D's full 15 days to elapse; this must be coordinated with whoever is running the investigation to avoid compromising it. (3) Reduce exposure on the non-critical branch as a hedge. Rerunning the backward pass with $T'=53$ gives C and G 9 days of slack each and B, E, F and I 8 days each, so crews can safely be borrowed from those activities for the D–H–J recovery. The limit is the old path A–B–E–F–I (45 days): once H and J have been crashed by a combined 8 days, both paths are critical, and any further crashing has to shorten both paths.