Project Scheduling and Analysis: 24-MMP-B8 Rock Slope Engineering, May 2018, Q3

Question 3: Project Scheduling and Analysis (30 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.

New ore zone development — task list and dependencies
#	Task description	Duration (months)	Depends on task #
1	Drive ramp to base of new zone	11	none
2	Develop/equip new raises for hoisting and ventilation	16	1
3	Develop new u/g exploration drilling gallery	2	1
4	Complete new u/g exploration drilling program	18	3
5	Develop ore body model and mining schedule	2	4
6	Reconfigure mine ventilation system for new zone	4	2
7	Expand u/g diesel powered equipment fleet	4	6
8	Develop upper mining level for new zone	8	7
9	Develop lower mining level for new zone	10	7
10	Develop slot raises for initial stope blocks	2	8, 9
11	Drill open stoping blastholes for initial 2 stopes	1	10, 5
12	First production from stopes in new zone	0 (milestone)	11, 2

Find. (a) a Gantt chart of the schedule; (b) the critical path and the shortest possible project duration.

Approach. Build the activity-on-node network from the given dependencies, run a forward pass to get each task's earliest start/finish (ES/EF), then a backward pass from the project finish to get each task's latest start/finish (LS/LF); the critical path is the chain of tasks with zero total float (LS−ES=0), and the Gantt chart plots every task at its earliest start with float shown for non-critical tasks.

a) Gantt chart

Fig. 3.1 — Gantt chart, each task plotted at its earliest start (ES); dashed tails show total float for non-critical tasks.

b) Critical Path Method

Forward pass (ES/EF). $ES$ of a task is the largest $EF$ among its predecessors ($ES=0$ for Task 1). Task 1: ES 0, EF 11. Task 2 (dep. 1): ES 11, EF 27. Task 3 (dep. 1): ES 11, EF 13. Task 4 (dep. 3): ES 13, EF 31. Task 5 (dep. 4): ES 31, EF 33. Task 6 (dep. 2): ES 27, EF 31. Task 7 (dep. 6): ES 31, EF 35. Task 8 (dep. 7): ES 35, EF 43. Task 9 (dep. 7): ES 35, EF 45. Task 10 (dep. 8, 9): $ES=\max(43,45)=45$, EF 47. Task 11 (dep. 10, 5): $ES=\max(47,33)=47$, EF 48. Task 12 (dep. 11, 2): $$\boxed{ES_{12}=\max(48,27)=48\ \text{months} = \text{project duration}}$$
Backward pass (LS/LF). Starting from $LF_{12}=48$ and working back, $LS$ of a task is the smallest $LS$ among its successors minus its own duration. Task 12: LF 48, LS 48. Task 11 (feeds 12): LF 48, LS 47. Task 2 (feeds 12 and 6): $LF=\min(48,27)=27$, LS 11. Task 10 (feeds 11): LF 47, LS 45. Task 9 (feeds 10): LF 45, LS 35. Task 8 (feeds 10): LF 45, LS 37. Task 7 (feeds 8, 9): $LF=\min(37,35)=35$, LS 31. Task 6 (feeds 7): LF 31, LS 27. Task 5 (feeds 11): LF 47, LS 45. Task 4 (feeds 5): LF 45, LS 27. Task 3 (feeds 4): LF 27, LS 25. Task 1 (feeds 2, 3): $$LF_1=\min(11,25)=11,\qquad LS_1=11-11=0$$
Total float and the critical path. Float $=LS-ES$ for every task: Task 1 float 0; Task 2 float 0; Task 3 float 14; Task 4 float 14; Task 5 float 14; Task 6 float 0; Task 7 float 0; Task 8 float 2; Task 9 float 0; Task 10 float 0; Task 11 float 0; Task 12 float 0. The only zero-float chain is Tasks 1→2→6→7→9→10→11→12: $$\boxed{\text{Critical path: } 1\rightarrow2\rightarrow6\rightarrow7\rightarrow9\rightarrow10\rightarrow11\rightarrow12,\ \ 11+16+4+4+10+2+1=48\ \text{months}}$$

Fig. 3.2 — Activity-on-node network. Each box: task # (top), ES/EF (upper corners, months), LS/LF (lower corners, months). Red = critical path (zero float).

The result again rewards checking the arithmetic rather than the task count: Task 9 (lower-level development, 10 months) is critical while the parallel Task 8 (upper-level development, 8 months, same start) is not — both start together at month 35 once the diesel fleet expansion (Task 7) is done, but Task 9 finishes 2 months later and is the one Task 10 (slot raises) must actually wait for, leaving Task 8 with 2 months of float. Separately, the exploration-drilling branch (Tasks 3→4→5) has a generous 14 months of float even though it includes the single longest individual activity in the network (18 months for Task 4) — it simply starts and can finish well before the raise-development/ventilation/fleet chain that actually paces the project.

Question 3 — final results
Item	Result
(a) Gantt chart	see Fig. 3.1
(b) Critical path	1 → 2 → 6 → 7 → 9 → 10 → 11 → 12
Shortest project duration	48 months
Non-critical tasks (total float)	3 (14 mo), 4 (14 mo), 5 (14 mo), 8 (2 mo)

Topic. Gantt charting and Critical Path Method (CPM) scheduling for a mine-development project network.

Key relations. Forward pass: $ES_j=\max_i(EF_i)$ over predecessors $i$, $EF_j=ES_j+d_j$. Backward pass: $LF_i=\min_j(LS_j)$ over successors $j$, $LS_i=LF_i-d_i$. Total float $=LS-ES=LF-EF$; the critical path is every task with float $=0$.

Why this works. A milestone or a task that depends on two or more converging chains (here, Task 10 needs BOTH the upper-level, Task 8, and lower-level, Task 9, development finished) can only start once the SLOWEST of those chains finishes; the forward pass captures that by always taking the maximum EF among predecessors, and the backward pass then propagates back exactly how much slack every other chain was left with once the slowest chain is fixed as the schedule driver.

Common pitfall. Assuming two parallel tasks that start together and look similar in scope (Tasks 8 and 9) must both be critical, or must both have equal float — only the LONGER of two converging parallel tasks is critical; and reading a Gantt chart's bar lengths as a proxy for criticality instead of computing float explicitly.

Source. Project Management Institute, PMBOK Guide (Critical Path Method, Gantt charts); Hustrulid, Kuchta & Martin, Open Pit Mine Planning and Design (mine development scheduling).

Question 3 of 4: Project Scheduling and Analysis

Question 3: Project Scheduling and Analysis (30 marks)

a) Gantt chart

b) Critical Path Method