Project Scheduling — New Ore Zone Development: 24-MMP-B8 Rock Slope Engineering, May 2013, Q2

Question 3: Project Scheduling — New Ore Zone Development (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.

New Ore Zone Development — task list and dependencies
#	Task description	Duration (months)	Depends on task #
1	Drive new ramp Phase A	8	none
2	Develop new shaft for hoisting and ventilation	36	1
3	Drive new ramp Phase B	12	1
4	Develop new u/g exploration drilling gallery	2	1
5	Complete u/g exploration drilling program, ore body model and mining schedule	18	4
6	Reconfigure mine ventilation system for new zone	4	3
7	Expand u/g diesel powered equipment fleet	4	6
8	Develop upper mining level for new zone	12	7
9	Develop lower mining level for new zone	14	3
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) the sequence of tasks forming the critical path; (b) the shortest project duration; (c) every non-critical task and why it isn't critical (its total float).

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 every other task's float is how many months it can slip without delaying the milestone (Task 12).

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 8. Task 2 (dep. 1): ES 8, EF 44. Task 3 (dep. 1): ES 8, EF 20. Task 4 (dep. 1): ES 8, EF 10. Task 5 (dep. 4): ES 10, EF 28. Task 6 (dep. 3): ES 20, EF 24. Task 7 (dep. 6): ES 24, EF 28. Task 8 (dep. 7): ES 28, EF 40. Task 9 (dep. 3): ES 20, EF 34. Task 10 (dep. 8, 9): $ES=\max(40,34)=40$, EF 42. Task 11 (dep. 10, 5): $ES=\max(42,28)=42$, EF 43. Task 12 (dep. 11, 2): $$\boxed{ES_{12}=\max(43,44)=44\ \text{months} = \text{project duration}}$$
Backward pass (LS/LF). Starting from $LF_{12}=44$ and working back, $LS$ of a task is the smallest $LS$ among its successors minus its own duration. Task 12: LF 44, LS 44. Task 11 (feeds 12): LF 44, LS 43. Task 2 (feeds 12): LF 44, LS 8. Task 10 (feeds 11): LF 43, LS 41. Task 8 (feeds 10): LF 41, LS 29. Task 9 (feeds 10): LF 41, LS 27. Task 7 (feeds 8): LF 29, LS 25. Task 6 (feeds 7): LF 25, LS 21. Task 5 (feeds 11): LF 43, LS 25. Task 3 (feeds 6 and 9): $LF=\min(21,27)=21$, LS 9. Task 4 (feeds 5): LF 25, LS 23. Task 1 (feeds 2, 3, 4): $$LF_1=\min(8,9,23)=8,\qquad LS_1=8-8=0$$
Total float and the critical path. Float $=LS-ES$ for every task: Task 1 float 0; Task 2 float 0; Task 3 float 1; Task 4 float 15; Task 5 float 15; Task 6 float 1; Task 7 float 1; Task 8 float 1; Task 9 float 7; Task 10 float 1; Task 11 float 1; Task 12 float 0. The only zero-float chain is Tasks 1 → 2 → 12: $$\boxed{\text{Critical path: Task 1 (Ramp Phase A)}\rightarrow\text{Task 2 (Shaft development)}\rightarrow\text{Task 12 (First production)},\ \ 8+36=44\ \text{months}}$$

Fig. 3.1 — Activity-on-node network. Each box: task # (top), ES/EF (upper corners, months), LS/LF (lower corners, months), duration below. Red = critical path (zero float).

The result is a little counter-intuitive: the longest-looking chain of individual activities is the exploration/drilling → slot-raise → blasthole path (Tasks 1→3→9→10→11→12, or 1→4→5→11→12), but neither is critical — the single, standalone 36-month shaft-sinking activity (Task 2), which depends on nothing but the initial ramp, is the true bottleneck because nothing else can start production until both the shaft AND the full underground development chain are ready, and the shaft alone already takes as long as the whole development chain. The tightest non-critical branch (Tasks 3→6→7→8→10→11, ventilation/equipment/upper-level) has only 1 month of float and needs close monitoring even though it is not, strictly, critical.

Question 3 — final results
Item	Result
(a) Critical path	Task 1 → Task 2 → Task 12 (Ramp A → Shaft development → First production)
(b) Shortest project duration	44 months
(c) Non-critical tasks (total float)	3 (1 mo), 4 (15 mo), 5 (15 mo), 6 (1 mo), 7 (1 mo), 8 (1 mo), 9 (7 mo), 10 (1 mo), 11 (1 mo)

Topic. 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 that depends on two or more converging chains (here, Task 12 needs BOTH the shaft, Task 2, and the full stope-development chain via Task 11) 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 the chain with the most individual tasks, or the largest sum of "obviously important" activities, must be critical — here a single 36-month activity outweighs an entire 5-task development chain, and reading the network by task count rather than by ES/EF arithmetic would misidentify the critical path.

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

Question 2 of 4: Project Scheduling — New Ore Zone Development

Question 3: Project Scheduling — New Ore Zone Development (20 marks)