Construction Project — CPM Network and an Accident-Driven Duration Change: 23-Ind-A4 Production Management, December 2018, Q7

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.

Activity	Precedes	Duration (days)
A	B, C, D	15
B	E	14
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	12

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 the disrupted 15.

Figure 1 — Activity-on-node project network (original durations). Red boxes/arrows mark the critical path A→B→E→F→I (47 days); part (b) checks whether this path relocates 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 duration (15 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.

Forward pass (earliest times, part a). Starting A at day 0 and working through the precedence chain ($ES_i=\max$ of predecessors' $EF$; $EF_i=ES_i+d_i$): $$EF_A=15,\ EF_B=29,\ EF_C=21,\ EF_D=20,\ EF_E=\max(29,21)+3=32,\ EF_G=21+8=29,$$ $$EF_H=20+9=29,\ EF_F=\max(32,29)+8=40,\ EF_J=\max(29,29)+12=41,\ EF_I=40+7=\boxed{47}.$$ Project duration $=\max(EF_I,EF_J)=\max(47,41)=\boxed{47\ \text{days}}$.
Backward pass and slack (part a). Working back from day 47 ($LF_i=\min$ of successors' $LS$; $LS_i=LF_i-d_i$) gives the slack table below; the critical path is every activity with zero slack: $$\boxed{\text{Critical path: A}\rightarrow\text{B}\rightarrow\text{E}\rightarrow\text{F}\rightarrow\text{I}\ (47\ \text{days})}.$$
Disrupted schedule, D→15 d (part b). Re-running the forward pass with $d_D=15$: $EF_D=15+15=30$, $EF_H=30+9=39$, $EF_J=\max(29,39)+12=51$; $EF_F,EF_I$ are unchanged (they depend on E/G, not D) at 40/47. New project duration $=\max(47,51)=\boxed{51\ \text{days}}$ — the critical path relocates to A–D–H–J, and the project is $51-47=4$ days late: $$\text{Penalty if not hired}=4\times\$5{,}000=\boxed{\$20{,}000}.$$
Alternate sub-contractor, D→6 d (part b). Re-running once more with $d_D=6$: $EF_D=15+6=21$, $EF_H=21+9=30$, $EF_J=\max(29,30)+12=42$; project duration $=\max(47,42)=\boxed{47\ \text{days}}$ — back to the original 47-day critical path (A–B–E–F–I), i.e. exactly on time, zero penalty. Extra cost of hiring $=\$1{,}500\text{/day}\times6\ \text{days}=\boxed{\$9{,}000}$. Net benefit of hiring: $$\$20{,}000\ (\text{penalty avoided})-\$9{,}000\ (\text{extra cost})=\boxed{\$11{,}000\ \text{net savings}}\implies\boxed{\text{hire the alternate sub-contractor}}.$$

Activity	ES	EF	LS	LF	Slack
A	0	15	0	15	0 (critical)
B	15	29	15	29	0 (critical)
C	15	21	18	24	3
D	15	20	21	26	6
E	29	32	29	32	0 (critical)
G	21	29	24	32	3
H	20	29	26	35	6
F	32	40	32	40	0 (critical)
J	29	41	35	47	6
I	40	47	40	47	0 (critical)
Project duration: original / D→15d (b) / D→6d new sub (b)					47 d / 51 d (+4 d, $20,000) / 47 d (net +$11,000 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; the part-(b) decision 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 once $d_D=15$) 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); and 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 (b).

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

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)