Question 7 of 8: Construction Project — CPM Network and an Accident-Driven Duration Change
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Technical Examinations — May 2018 — 17-Ind-A4 Production Management. Three-hour, closed-book exam; Casio or Sharp approved calculators only. Format: eight questions, each worth 20 marks (sub-part weights 10/10 as tabulated on the front-page marking scheme); candidates do two questions from Section A and three from Section B, and only the first five questions appearing in the answer book are marked. All eight are solved below for completeness. The paper asks for point-form answers wherever possible; the solutions below use full working for clarity.
Reference texts: Nahmias & Olsen, Production and Operations Analysis (7th ed., Waveland/McGraw-Hill) — forecasting, inventory (EOQ/EPQ) and aggregate planning; Sipper & Bulfin, Production: Planning, Control, and Integration — production scheduling, JIT/kanban and shop-floor implementation gaps; Hillier & Lieberman, Introduction to Operations Research (11th ed.) — LP formulation and project scheduling (CPM/PERT); Pinedo, Scheduling: Theory, Algorithms, and Systems (5th ed.) — parallel-machine scheduling and days-off workforce scheduling; Hopp & Spearman, Factory Physics (3rd ed.) — variability, buffering, and production scheduling; Liker, The Toyota Way, and Shingo, A Revolution in Manufacturing: The SMED System — 5S, Five Whys, SMED and lean root-cause analysis.
Question 7: Construction Project — CPM Network and an Accident-Driven Duration Change (20 marks)
The activity-precedence topology (A→B,C,D; B→E; C→E,G; D→H; E→F; F→I; G→F,J; H→J; I,J→END) and every base duration (A=15, B=14, C=6, D=5, E=3, F=8, G=8, H=9, I=7, J=12 days) are as printed in the question. Part (b)’s disruption is that D lengthens to 15 days, and the schedule-recovery analysis follows from that duration.
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)