Question 7 of 8: Construction Project — CPM Network and an Accident-Driven Duration Change

Question 7: 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, critical path, and earliest/latest start times (and slack) of every activity; (b) whether to hire the $\$1{,}500$/day-more sub-contractor who finishes D in 6 days instead of 12, once the accident investigation strikes.

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. Then re-run the forward pass with D's new duration (12 days for the disruption, 6 days for the alternate sub-contractor) to see whether the critical path relocates, and compare the net cost of the faster sub-contractor against the penalty it avoids.

1. Forward pass (earliest times). Starting A at day 0 and working through the precedence chain ($ES_i=\max$ of predecessors' $EF$; $EF_i=ES_i+d_i$): $$EF_A=15,\ EF_B=29,\ EF_C=21,\ EF_D=20,\ EF_E=\max(29,21)+3=32,\ EF_G=21+8=29,$$ $$EF_H=20+9=29,\ EF_F=\max(32,29)+8=40,\ EF_J=\max(29,29)+12=41,\ EF_I=40+7=47.$$ The project duration is the later of the two end activities: $\boxed{T=\max(EF_I,EF_J)=\max(47,41)=47\ \text{days}}$.
2. Backward pass, slack, and critical path (part a). Setting $LF_I=LF_J=47$ 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). The chain of zero-slack activities is $\boxed{A\to B\to E\to F\to I}$, with $15+14+3+8+7=47$ days, matching the project duration — the path shown in red in Figure 1. D itself carries 6 days of slack (its path A–D–H–J is 6 days shorter than the critical path).
3. Effect of D → 12 days, accident investigation (part b, first check). D is not on the critical path and carries 6 days of slack, but its duration increase (5→12, i.e. $+7$ days) exceeds that slack by 1 day. Re-running the forward pass with $d_D=12$: $EF_D=15+12=27\Rightarrow EF_H=27+9=36\Rightarrow EF_J=\max(29,36)+12=48$, while the F/I side ($EF_I=47$) is unaffected. The new project duration is $$\boxed{T'=\max(47,48)=48\ \text{days}},$$ a delay of $48-47=\boxed{1\ \text{day}}$ (D's 7-day increase minus its 6 days of slack). The critical path shifts to $A\to D\to H\to J$ (new length $15+12+9+12=48$ days). At $\$5{,}000$/day, keeping the original D subcontractor and simply absorbing the 1-day delay costs $\boxed{1\times\$5{,}000=\$5{,}000}$ in late-completion penalty.
4. New sub-contractor for D, $\$1{,}500$/day more, 6-day duration (part b, decision). With $d_D=6$ instead of 12, re-run the forward pass: $EF_D=15+6=21\Rightarrow EF_H=21+9=30\Rightarrow EF_J=\max(29,30)+12=42$, while $EF_I=47$ is unaffected (still the larger of the two). The new project duration is $T''=\max(47,42)=47$ days — exactly the original, undisrupted 47-day schedule, i.e. zero delay and zero penalty. The new sub-contractor costs $\$1{,}500$/day more, over D's 6-day duration, i.e. an extra $6\times\$1{,}500=\$9{,}000$. $$\text{Option 1 (keep original subcontractor): cost}=\boxed{\$5{,}000}\ \text{(1-day penalty, no extra fee)}$$ $$\text{Option 2 (hire new subcontractor): cost}=\boxed{\$9{,}000}\ \text{(no penalty, but the full \$1{,}500/day premium)}$$ Since $\$5{,}000<\$9{,}000$, the new sub-contractor is not worth hiring here: $\boxed{\text{No, do not hire them}}$ — absorbing the 1-day delay penalty saves $\$9{,}000-\$5{,}000=\$4{,}000$ versus paying the faster contractor's premium.