HSC Mathematics Standard 2 critical path analysis (2026 guide)

Critical path analysis (CPA) is the most heavily examined Networks topic in HSC Mathematics Standard 2. Every paper since the 2017 syllabus reform has had a CPA question, typically worth $5$-$8$ marks in Section II.

It is also one of the most predictable topics in the entire paper. The procedure is mechanical: build the activity network from the precedence table, forward scan for earliest times, backward scan for latest times, compute float, identify the critical path. The only real skill is being neat with the diagram and the table of values, so that markers can follow your work.

This guide covers everything you need: building the network, the scanning algorithms, computing float, identifying the critical path, and Australian-context worked examples.

What is critical path analysis

Critical path analysis is a project management technique used to:

• Compute the minimum possible duration of a project given the dependencies between its activities.
• Identify which activities are critical (cannot be delayed without delaying the whole project).
• Identify which activities have slack and can be delayed by some amount without affecting the project end date.

It is used in construction, software engineering, event planning, manufacturing rollouts, and any project with multiple interdependent tasks.

The activity network

NESA uses the activity-on-edge (AOE) convention:

• Vertices (nodes) represent events. An event is a state of completion, such as "Activity A complete and Activity B can begin".
• Edges (arrows) represent activities. Each activity is labelled with its name and its duration.
• The graph is directed: edges have arrows showing which event comes before which.
• The graph is acyclic: there are no cycles (a cycle would mean activities circularly depend on each other, which is impossible).

Building the network from a precedence table

A precedence table lists each activity, its duration, and its immediate predecessors. To build the network:

1. Identify activities with no predecessors. These start at the project-start node.
2. For each activity, find the event where all its predecessors have just finished. This is the start node for the activity.
3. Draw the activity as an arrow from its start node to a new (or shared) end node.
4. Continue until all activities are drawn. The project-end node is where the activities with no successors terminate.

You may need dummy activities (zero-duration edges) to enforce precedence without introducing false connections. Standard 2 questions usually avoid these.

Forward scanning (earliest times)

Label each event with the earliest time at which it can occur. This is the EST (earliest start time) of any activity that starts at that event.

The algorithm:

1. Start event: EST = $0$.
2. For each other event, EST is the maximum over all activities arriving at that event of (EST of activity's start event + activity's duration).

Take the maximum because all predecessors must finish before the event can occur.

After the forward scan completes, the EST of the project-end event equals the minimum possible project duration.

Backward scanning (latest times)

Label each event with the latest time it can occur without delaying the project. This is the LFT (latest finish time) of any activity that ends at that event.

The algorithm:

1. End event: LFT = minimum project duration (from forward scan).
2. For each other event, LFT is the minimum over all activities leaving that event of (LFT of activity's end event $-$ activity's duration).

Take the minimum because the event must complete in time for the earliest required successor.

Per-activity calculations

For an activity from event $i$ to event $j$ with duration $t$:

• EST (earliest start time) = $E_i$
• EFT (earliest finish time) = $E_i + t$
• LFT (latest finish time) = $L_j$
• LST (latest start time) = $L_j - t$
• Float = LST $-$ EST = LFT $-$ EFT = IMATH_15

The critical path

The critical path consists of all activities with float = $0$. These activities cannot be delayed without delaying the project. The critical path is the longest path through the network.

If two or more paths tie at the longest length, all of them are critical.

Worked example: kitchen renovation

A project to renovate an Australian kitchen has these activities:

Activity	Duration (days)	Predecessors
IMATH_17 - Demolition	IMATH_18	None
IMATH_19 - Plumbing rough-in	IMATH_20	IMATH_21
IMATH_22 - Electrical rough-in	IMATH_23	IMATH_24
IMATH_25 - Floor laying	IMATH_26	IMATH_27
IMATH_28 - Cabinet installation	IMATH_29	IMATH_30 , $C$, IMATH_32
IMATH_33 - Tiling and painting	IMATH_34	IMATH_35
IMATH_36 - Final inspection	IMATH_37	IMATH_38

Build the network

Events:

• Event $1$: project start.
• Event $2$: after $A$ (so $B$, $C$, $D$ can begin).
• Event $3$: after $B$, $C$, $D$ (so $E$ can begin).
• Event $4$: after $E$ (so $F$ can begin).
• Event $5$: after $F$ (so $G$ can begin).
• Event $6$: after $G$ (project end).

Forward scan

• Event $1$: EST = $0$.
• Event $2$: EST = $0 + 2 = 2$.
• Event $3$: EST = $\max(2 + 3, 2 + 2, 2 + 4) = \max(5, 4, 6) = 6$ (via $D$).
• Event $4$: EST = $6 + 5 = 11$.
• Event $5$: EST = $11 + 3 = 14$.
• Event $6$: EST = $14 + 1 = 15$.

Minimum project duration: **$15$ days**.

Backward scan

• Event $6$: LFT = $15$.
• Event $5$: LFT = $15 - 1 = 14$.
• Event $4$: LFT = $14 - 3 = 11$.
• Event $3$: LFT = $11 - 5 = 6$.
• Event $2$: LFT = $\min(6 - 3, 6 - 2, 6 - 4) = \min(3, 4, 2) = 2$ (via $D$).
• Event $1$: LFT = $2 - 2 = 0$.

Per-activity table

Activity	Duration	EST	EFT	LST	LFT	Float	Critical?
IMATH_85	IMATH_86	IMATH_87	IMATH_88	IMATH_89	IMATH_90	IMATH_91	Yes
IMATH_92	IMATH_93	IMATH_94	IMATH_95	IMATH_96	IMATH_97	IMATH_98	No
IMATH_99	IMATH_100	IMATH_101	IMATH_102	IMATH_103	IMATH_104	IMATH_105	No
IMATH_106	IMATH_107	IMATH_108	IMATH_109	IMATH_110	IMATH_111	IMATH_112	Yes
IMATH_113	IMATH_114	IMATH_115	IMATH_116	IMATH_117	IMATH_118	IMATH_119	Yes
IMATH_120	IMATH_121	IMATH_122	IMATH_123	IMATH_124	IMATH_125	IMATH_126	Yes
IMATH_127	IMATH_128	IMATH_129	IMATH_130	IMATH_131	IMATH_132	IMATH_133	Yes

Critical path: **$A$-$D$-$E$-$F$-$G$**. Minimum duration: $15$ days.

Plumbing ($B$) has $1$ day of float; electrical ($C$) has $2$ days of float. Both can be delayed by their float amount without affecting the project end date.

Worked example: Sydney apartment construction

A simplified apartment build:

Activity	Duration (weeks)	Predecessors
IMATH_144 - Foundation	IMATH_145	None
IMATH_146 - Structural frame	IMATH_147	IMATH_148
IMATH_149 - External walls	IMATH_150	IMATH_151
IMATH_152 - Roofing	IMATH_153	IMATH_154
IMATH_155 - Plumbing	IMATH_156	IMATH_157 , IMATH_158
IMATH_159 - Internal finishes	IMATH_160	IMATH_161

Forward scan: Event $1$: $0$. Event $2$: $4$. Event $3$ (after $B$): $10$. Event $4$ (after $C$ and $D$): $\max(10 + 3, 10 + 4) = 14$ (via $D$). Event $5$ (after $E$): $19$. Event $6$ (after $F$): $26$.

Backward scan: Event $6$: $26$. Event $5$: $19$. Event $4$: $14$. Event $3$: $\min(14 - 3, 14 - 4) = 10$ (via $D$). Event $2$: $4$. Event $1$: $0$.

Float:

Activity	Float	Critical?
IMATH_193	IMATH_194	Yes
IMATH_195	IMATH_196	Yes
IMATH_197	IMATH_198	No
IMATH_199	IMATH_200	Yes
IMATH_201	IMATH_202	Yes
IMATH_203	IMATH_204	Yes

Critical path: $A$-$B$-$D$-$E$-$F$. Minimum duration: $26$ weeks. External walls ($C$) has $1$ week of slack.

Exam strategy

Layout

The HSC marker wants a clean layout. Standard approach:

1. Draw the activity network with all activities and durations labelled.
2. Label each event with EST and LFT (a two-number tag at each node).
3. Compute float in a table.
4. State the critical path and the minimum duration explicitly.

Common loss-of-marks

• Skipping the network diagram and just listing paths. Markers want the network.
• Wrong max/min direction in the scans. Forward scan uses MAX over predecessors; backward scan uses MIN over successors.
• Mislabelling events with EST as LFT or vice versa.
• Forgetting that float = LFT $-$ EST $-$ duration. Easy to forget the duration subtraction.
• Stating the shortest path as critical. The critical path is the LONGEST.

Speed tactics

CPA questions usually have a fixed format. Once you have practised the scanning routine, it should take $5$-$8$ minutes for a $6$-$8$ mark question. Drill the routine in Term 2 of Year 12, before HSC trials, so it is automatic.

What CPA does not cover

CPA assumes activity durations are known exactly. In real projects, activities take variable amounts of time, and project management uses tools like PERT (Programme Evaluation and Review Technique) to handle uncertainty. NESA Standard 2 does not test PERT.

CPA also does not handle resource constraints (e.g. two activities cannot use the same crane at the same time). Real project management software handles these; HSC Standard 2 does not.