Construction Project — CPM Network and an Accident-Driven Duration Change: 23-Ind-A4 Production Management, December 2016, Q6

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.

Activity	Precedes	Duration (days)
A	B, C, D	15
B	E	12
C	E, G	6
D	H	5
E	F	3
F	I	8
G	F, J	8
H	J	9
I	END	7
J	END	14

Find. (a) The project network, critical path, and earliest/latest start times (and slack) of every activity; (b) the effect on finish date if D becomes 15 days because of the accident investigation, plus recovery strategies; (c) whether to hire a $\$1{,}000$/day-more sub-contractor who finishes D in 8 days instead of 15.

Figure 1 — Activity-on-node project network (original durations). Red boxes/arrows mark the original critical path A→B→E→F→I (45 days); part (b) shows this path shifting to A→D→H→J once D lengthens to 15 days.

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 15-day duration to see whether the critical path relocates, and for part (c) compare the net cost of crashing D against the penalty it avoids.

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: $\boxed{T=\max(EF_I,EF_J)=\max(45,43)=45\ \text{days}}$.
Backward pass, slack, and critical path (part a). Setting $LF_I=LF_J=45$ 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 $15+12+3+8+7=45$ days, matching the project duration — the path shown in red in Figure 1. All other activities carry 1–2 days of positive slack; D itself has only 2 days.
Effect of D → 15 days, accident investigation (part b). 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. The new project duration is $$\boxed{T'=\max(45,53)=53\ \text{days}},$$ a delay of $53-45=\boxed{8\ \text{days}}$ (D's 10-day increase minus its 2 days of slack). 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. Because the delay is caused by an accident investigation (a regulatory/safety hold, not a resourcing shortfall like a strike), it generally cannot be crashed with money once underway; realistic recovery strategies are to crash D's successors H and J once D is released (any crashing cost under $\$5,000$/day is worth it), or to fast-track by starting any portion of H not restricted by the investigation as soon as permitted.
New sub-contractor for D, $\$1{,}000$/day more, 8-day duration (part c). With $d_D=8$ instead of 15, re-run the forward pass: $EF_D=15+8=23\Rightarrow EF_H=23+9=32\Rightarrow EF_J=\max(29,32)+14=46$, while $EF_I=45$ is unaffected. The new project duration is $T''=\max(45,46)=46$ days — a delay of only $46-45=1$ day versus the original 45-day plan, instead of 8 days with the standard contractor. The penalty avoided is $(8-1)\times\$5{,}000=\boxed{\$35{,}000}$ in late-completion fees. The new sub-contractor costs $\$1{,}000$/day more over D's 8-day duration, i.e. an extra $8\times\$1{,}000=\boxed{\$8{,}000}$. Since the penalty saved ($\$35{,}000$) far exceeds the extra contractor cost ($\$8{,}000$), the net benefit is $\$35{,}000-\$8{,}000=\boxed{\$27{,}000}$, so $$\boxed{\text{Yes, hire the new sub-contractor.}}$$

Activity	ES	EF	LS	LF	Slack
A	0	15	0	15	0 (critical)
B	15	27	15	27	0 (critical)
C	15	21	16	22	1
D	15	20	17	22	2
E	27	30	27	30	0 (critical)
G	21	29	22	30	1
H	20	29	22	31	2
F	30	38	30	38	0 (critical)
J	29	43	31	45	2
I	38	45	38	45	0 (critical)
Project duration: original / D→15d (b) / D→8d new sub (c)					45 d / 53 d (+8 d, \$40,000) / 46 d (+1 d, net \$27,000 saved)

Concept & Theory

Topic. Critical Path Method (CPM) scheduling: forward/backward pass, slack, schedule recovery, and a crashing decision compared against the late-completion penalty it avoids.

Key relations. Forward pass $ES_i=\max_{p\in\text{pred}(i)}EF_p$, $EF_i=ES_i+d_i$; backward pass $LF_i=\min_{s\in\text{succ}(i)}LS_s$, $LS_i=LF_i-d_i$; total slack $=LS_i-ES_i$; crashing decision rule: hire if (penalty avoided) $>$ (extra contractor cost).

Why this works. The project duration is set by the longest path through the network, so only activities on that (possibly relocated) critical path affect the finish date; part (c) is a direct benefit-cost comparison because both quantities — the penalty saved and the extra contractor cost — scale linearly with days, once the new critical path (A–D–H–J even at $d_D=8$, since $46>45$) is confirmed by re-running the forward pass rather than assumed.

Common pitfall. Comparing the new sub-contractor's day-rate premium against the wrong baseline (the original 5-day D, instead of the 15-day disrupted D) or against the wrong day count (D's own duration, rather than confirming which path stays critical afterward); assuming every schedule-recovery strategy is equally available regardless of the delay's cause — a safety/regulatory hold generally cannot be crashed once already running, only shortened by replacing the resource before it starts, as in part (c).

Source. Hillier & Lieberman, Introduction to Operations Research, Ch. 10 (project management, CPM/PERT, crashing).

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)