What is the critical path?

The critical path is the longest path through the project network from start to end. Its length is the minimum project duration. Activities on the critical path are critical: any delay to them delays the whole project. Activities not on the critical path have float (slack) and can be delayed without affecting the project end date.

Float (or slack) is the amount of time an activity can be delayed without delaying the project end date. Activities on the critical path have zero float. Float is calculated as $\text{Float} = \text{LST} - \text{EST} = \text{LFT} - \text{EFT}$, where LST is latest start, EST is earliest start, LFT is latest finish, EFT is earliest finish.

What are forward and backward scanning?

Forward scanning works through the project network from start to finish, computing the earliest time each event can occur. Backward scanning works from finish back to start, computing the latest time each event can occur without delaying the project. Together they give you all the information needed to identify the critical path and calculate float for every activity.

How do I draw an activity network from a precedence table?

Start with the project-start node. Find all activities with no predecessors and draw them as edges from the start node. For each activity, find the node where all its predecessors finish; this is its start node. Continue until all activities are drawn. The project-end node is where the final activities terminate.

Why is the longest path critical, not the shortest?

The longest path determines the minimum project duration because every activity on it must happen in sequence. You cannot do them in parallel because of the precedence constraints. The shortest path is irrelevant for the project duration; it just means some sub-sequence can finish quickly, but the project waits for the longest sequence.

§-Study guide

NSWMaths Standard 2

HSC Mathematics Standard 2 critical path analysis (2026 guide)

Q: What is critical path analysis?

Critical path analysis (CPA) is a project management technique to find the minimum duration of a project and identify which activities cannot be delayed without delaying the whole project. In HSC Maths Standard 2 it is examined as part of MS-N3 in the Networks topic, usually worth $5$-$8$ marks in Section II of the paper.

A complete 2026 guide to critical path analysis in HSC Mathematics Standard 2. Building activity networks from precedence tables, the forward and backward scanning algorithm, finding the critical path and float, and worked Australian construction examples.

Generated by Claude Opus 4.818 min readNESA Mathematics Standard Stage 6 Syllabus (2017), Year 12 Networks: MS-N3 Critical Path AnalysisUpdated 2026-07-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

What is critical path analysis
The activity network
Forward scanning (earliest times)
Backward scanning (latest times)
Per-activity calculations
The critical path
Dummy activities
Exam strategy
What CPA does not cover
Check your knowledge

Figure 1. Activity-on-edge network for a 7-activity kitchen renovation. The heavier accent edges form the critical path A $\to$ D $\to$ E $\to$ F $\to$ G with total duration $2 + 4 + 5 + 3 + 1 = 15$ days. The three trades B, C and D run in parallel between events 2 and 3; B and C finish with time to spare, so they carry float. Each activity has a box at its tail showing its earliest start time (EST, left) and latest start time (LST, right); a critical activity has EST $=$ LST.

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:

Identify activities with no predecessors. These start at the project-start node.
For each activity, find the event where all its predecessors have just finished. This is the start node for the activity.
Draw the activity as an arrow from its start node to a new (or shared) end node.
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:

Start event: EST = $0$ .
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:

End event: LFT = minimum project duration (from forward scan).
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 = $L_j - E_i - t$

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.

Kitchen renovation: full forward and backward scan

A project to renovate an Australian kitchen has these activities:

Activity	Duration (days)	Predecessors
$A$ - Demolition	$2$	None
$B$ - Plumbing rough-in	$3$	$A$
$C$ - Electrical rough-in	$2$	$A$
$D$ - Floor laying	$4$	$A$
$E$ - Cabinet installation	$5$	$B$ , $C$ , $D$
$F$ - Tiling and painting	$3$	$E$
$G$ - Final inspection	$1$	$F$

Step 1. Identify the events from the precedence table.

Each event is a moment when one set of activities has just finished and the next set can begin. Reading down the predecessor column gives six events: project start, after $A$ , after the parallel trio $B$ , $C$ and $D$ , after $E$ , after $F$ , and project end.

Event $1$ : project start.
Event $2$ : after $A$ (so $B$ , $C$ , $D$ can begin).
Event $3$ : after $B$ , $C$ and $D$ (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).

Step 2. Forward scan, event by event, taking the maximum at every join.

At each event we take the maximum over all incoming activities. The maximum is the earliest moment every predecessor has finished, which is what an event represents.

Event $1$ : EST $= 0$ .
Event $2$ : only $A$ arrives, so EST $= 0 + 2 = 2$ .
Event $3$ : three activities arrive, so EST $= \max(2 + 3,\ 2 + 2,\ 2 + 4) = \max(5,\ 4,\ 6) = 6$ , with the maximum coming from $D$ .
Event $4$ : only $E$ arrives, so EST $= 6 + 5 = 11$ .
Event $5$ : only $F$ arrives, so EST $= 11 + 3 = 14$ .
Event $6$ : only $G$ arrives, so EST $= 14 + 1 = 15$ .

Minimum project duration is therefore $15$ days.

Step 3. Backward scan, event by event, taking the minimum at every fork.

The backward scan starts at the project end and subtracts activity durations. Where one event has multiple activities leaving it we take the minimum, because each successor must start in time for the project to finish on schedule.

Event $6$ : LFT $= 15$ .
Event $5$ : LFT $= 15 - 1 = 14$ .
Event $4$ : LFT $= 14 - 3 = 11$ .
Event $3$ : LFT $= 11 - 5 = 6$ .
Event $2$ : three activities leave, so LFT $= \min(6 - 3,\ 6 - 2,\ 6 - 4) = \min(3,\ 4,\ 2) = 2$ , with the minimum coming from $D$ .
Event $1$ : LFT $= 2 - 2 = 0$ .

Step 4. Build the per-activity table, including float.

Float for each activity equals $L_j - E_i - t$ , where $i$ is the start event, $j$ is the end event, and $t$ is the duration.

Activity	Duration	EST	EFT	LST	LFT	Float	Critical?
$A$	$2$	$0$	$2$	$0$	$2$	$0$	Yes
$B$	$3$	$2$	$5$	$3$	$6$	$1$	No
$C$	$2$	$2$	$4$	$4$	$6$	$2$	No
$D$	$4$	$2$	$6$	$2$	$6$	$0$	Yes
$E$	$5$	$6$	$11$	$6$	$11$	$0$	Yes
$F$	$3$	$11$	$14$	$11$	$14$	$0$	Yes
$G$	$1$	$14$	$15$	$14$	$15$	$0$	Yes

Step 5. State the critical path and slack explicitly.

The critical activities (zero float) form the longest path through the network.

Answer. Critical path $A \to D \to E \to F \to G$ , minimum duration $15$ days. Plumbing $B$ has $1$ day of slack and electrical $C$ has $2$ days of slack, so both can be delayed by their float amount without pushing the project end date.

Sydney apartment construction (six-activity build)

A simplified Sydney apartment build:

Activity	Duration (weeks)	Predecessors
$A$ - Foundation	$4$	None
$B$ - Structural frame	$6$	$A$
$C$ - External walls	$3$	$B$
$D$ - Roofing	$4$	$B$
$E$ - Plumbing	$5$	$C$ , $D$
$F$ - Internal finishes	$7$	$E$

Step 1. Map the events from the precedence table.

After the foundation and frame we get a parallel branch where roofing and external walls both depend on the frame but neither depends on the other. They rejoin at the start of plumbing.

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

Step 2. Forward scan.

Take the maximum at the join where $C$ and $D$ both arrive at event $4$ , because event $4$ cannot occur until both finish.

Event $1$ : EST $= 0$ .
Event $2$ : EST $= 0 + 4 = 4$ .
Event $3$ : EST $= 4 + 6 = 10$ .
Event $4$ : EST $= \max(10 + 3,\ 10 + 4) = \max(13,\ 14) = 14$ via $D$ .
Event $5$ : EST $= 14 + 5 = 19$ .
Event $6$ : EST $= 19 + 7 = 26$ .

Minimum project duration is $26$ weeks.

Step 3. Backward scan.

Take the minimum at the fork where $C$ and $D$ both leave event $3$ , because both must start in time for event $4$ to be reached by its LFT.

Event $6$ : LFT $= 26$ .
Event $5$ : LFT $= 26 - 7 = 19$ .
Event $4$ : LFT $= 19 - 5 = 14$ .
Event $3$ : LFT $= \min(14 - 3,\ 14 - 4) = \min(11,\ 10) = 10$ via $D$ .
Event $2$ : LFT $= 10 - 6 = 4$ .
Event $1$ : LFT $= 4 - 4 = 0$ .

Step 4. Compute float per activity.

Activity	Float ( $L_j - E_i - t$ )	Critical?
$A$	$4 - 0 - 4 = 0$	Yes
$B$	$10 - 4 - 6 = 0$	Yes
$C$	$14 - 10 - 3 = 1$	No
$D$	$14 - 10 - 4 = 0$	Yes
$E$	$19 - 14 - 5 = 0$	Yes
$F$	$26 - 19 - 7 = 0$	Yes

Answer. Critical path $A \to B \to D \to E \to F$ , minimum duration $26$ weeks. External walls $C$ has $1$ week of slack and can drift by up to $1$ week without delaying handover.

Slack calculation for a single activity on a partial network

A school musical has the following partial precedence table:

Activity	Duration (days)	Predecessors
$P$ - Casting	$2$	None
$Q$ - Set build	$7$	$P$
$R$ - Costume making	$4$	$P$
$S$ - Rehearsals	$5$	$Q$ , $R$
$T$ - Final dress run	$1$	$S$

Find the float of costume making $R$ and explain what it means in plain language.

Step 1. Establish the events and run a forward scan to find EST values.

We need the EST of the event where $R$ starts and the EST of the event where it finishes, so we run the full forward scan.

Event $1$ (start): EST $= 0$ .
Event $2$ (after $P$ ): EST $= 0 + 2 = 2$ .
Event $3$ (after $Q$ and $R$ , before $S$ ): EST $= \max(2 + 7,\ 2 + 4) = \max(9,\ 6) = 9$ via $Q$ .
Event $4$ (after $S$ ): EST $= 9 + 5 = 14$ .
Event $5$ (project end): EST $= 14 + 1 = 15$ .

Step 2. Run a backward scan to find LFT values.

Minimum project duration is $15$ days, so LFT at event $5$ is $15$ .

Event $5$ : LFT $= 15$ .
Event $4$ : LFT $= 15 - 1 = 14$ .
Event $3$ : LFT $= 14 - 5 = 9$ .
Event $2$ : LFT $= \min(9 - 7,\ 9 - 4) = \min(2,\ 5) = 2$ via $Q$ .
Event $1$ : LFT $= 2 - 2 = 0$ .

Step 3. Apply the float formula to activity $R$ .

Costume making $R$ runs from event $2$ ( $E_2 = 2$ ) to event $3$ ( $L_3 = 9$ ) with duration $t = 4$ . The float formula is $\text{Float} = L_j - E_i - t$ .

\text{Float}(R) = 9 - 2 - 4 = 3 \text{ days}

Step 4. Translate the answer into project language.

A float of $3$ days means costume making can be delayed by up to $3$ days from its earliest start (or its finish can slip by $3$ days) without delaying the dress run on day $15$ . If the costumes slip a fourth day they push back rehearsals and therefore the final project finish.

Answer. Float of $R$ is $3$ days. The costume team can absorb up to a $3$ -day delay without pushing the show date.

Reading a given activity network and identifying the critical path

A given activity network has the following labelled events (the diagram is described in the question, you are not asked to draw it).

Activity	Edge (events)	Duration (hours)
$A$	$1 \to 2$	$5$
$B$	$1 \to 3$	$4$
$C$	$2 \to 4$	$6$
$D$	$3 \to 4$	$8$
$E$	$4 \to 5$	$3$

Find the critical path and the minimum project duration.

Step 1. Identify all paths from event $1$ to the project end ( $5$ ).

Reading the edges, only one path passes through event $2$ and one passes through event $3$ . They rejoin at event $4$ , then a single activity $E$ leads to event $5$ .

Path 1: $A \to C \to E$ via events $1 \to 2 \to 4 \to 5$ .
Path 2: $B \to D \to E$ via events $1 \to 3 \to 4 \to 5$ .

Step 2. Compute the length of each path.

The critical path is the longest from start to end. Add the durations along each path.

Length of Path 1: $5 + 6 + 3 = 14$ hours.
Length of Path 2: $4 + 8 + 3 = 15$ hours.

Step 3. Compare the two lengths and identify the critical path.

Path 2 is longer, so it is the critical path. Every activity on it has zero float.

Step 4. Confirm by computing float on Path 1.

Take $C$ for example. EST at event $2$ is $5$ (from $A$ ), and LFT at event $4$ is $15 - 3 = 12$ (event $5$ less duration of $E$ ). So float of $C$ is $12 - 5 - 6 = 1$ hour. Path 1 is not critical, which agrees with our path-comparison answer.

Answer. Critical path $B \to D \to E$ ( $1 \to 3 \to 4 \to 5$ ) with minimum project duration $15$ hours.

Dummy activities

Most Standard 2 networks can be drawn with one activity per edge, but sometimes two activities share some of their predecessors and not others, and a plain activity-on-edge diagram cannot capture that. The fix is a dummy activity: a zero-duration edge, drawn dashed, that carries a dependency from one event to another without representing any real work.

Take this project, where activity $D$ must wait for both $A$ and $B$ , but activity $C$ waits for $A$ only:

Activity	Duration	Predecessors
$A$	$3$	None
$B$	$2$	None
$C$	$5$	$A$
$D$	$4$	$A$ , $B$
$E$	$6$	$C$
$F$	$3$	$D$
$G$	$4$	$E$ , $F$

If you simply ran $A$ and $B$ into one shared event and started both $C$ and $D$ from it, you would wrongly force $C$ to wait for $B$ as well. Instead, give $A$ and $B$ separate end events, then add a dummy from the end of $A$ across to the end of $B$ . Now $C$ leaves the end of $A$ (so it needs $A$ only), while $D$ leaves the end of $B$ , which the dummy has tied back to $A$ as well (so $D$ needs both).

Figure 2. The dashed arrow is the dummy: zero duration, so it adds a dependency without adding time. It lets $D$ depend on both $A$ and $B$ while $C$ still depends on $A$ alone.

The scan runs exactly as before; the dummy passes the earliest and latest times across with no duration added. Forward and backward scanning produce the per-activity boxes shown, giving the critical path $A \to C \to E \to G$ with a minimum duration of $18$ . Activity $B$ has $5$ units of float, and $D$ and $F$ each have $4$ .

When you need a dummy. Reach for one whenever two activities have overlapping but different predecessor lists, or whenever two activities would otherwise share the same start and end event (every activity needs its own unique pair of events). If neither situation applies, you do not need a dummy.

Exam strategy

Layout

The HSC marker wants a clean layout. Standard approach:

Draw the activity network with all activities and durations labelled.
Label each event with EST and LFT (a two-number tag at each node).
Compute float in a table.
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.

Check your knowledge

Try these before checking the worked solutions. They progress from quick recall to a full network walkthrough.

For an activity with EST $= 4$ , LFT $= 12$ and duration $3$ , what is the float?
The forward scan of a network gives the project-end event an EST of $22$ days. What is the LFT of the project-end event?
Three activities arrive at event $7$ : $X$ from event $4$ (duration $5$ , EST of event $4$ is $6$ ), $Y$ from event $5$ (duration $3$ , EST of event $5$ is $9$ ), and $Z$ from event $6$ (duration $4$ , EST of event $6$ is $8$ ). What is the EST of event $7$ ?
Two activities leave event $3$ : $M$ (duration $6$ , ends at event $5$ with LFT $14$ ) and $N$ (duration $2$ , ends at event $4$ with LFT $9$ ). What is the LFT of event $3$ ?
Activity $K$ has EST $= 5$ , EFT $= 9$ , LST $= 7$ and LFT $= 11$ . Is $K$ on the critical path? Justify briefly.
A small project has activities $A$ (duration $3$ , no predecessors), $B$ (duration $5$ , predecessor $A$ ), $C$ (duration $4$ , predecessor $A$ ) and $D$ (duration $2$ , predecessors $B$ and $C$ ). Find the minimum project duration and the critical path.
For the project in question $6$ , find the float of activity $C$ and state in one sentence what it means.
A network has two paths from start to finish: path $P$ with durations $4, 3, 6, 2$ and path $Q$ with durations $5, 4, 7$ . Which is critical, and what is the minimum project duration?

Solutions to check-your-knowledge questions

These eight questions cover float calculations, EST and LFT rules, forward and backward scanning, and identifying the critical path. Work through each in order.

Step 1: Q1. Calculate float given LFT, EST, and duration

Float measures how long an activity can slip before it delays the whole project. The formula is Float $= \text{LFT} - \text{EST} - \text{duration}$ .

\text{Float} = 12 - 4 - 3 = 5

The activity can be delayed by $5$ time units without pushing the project finish.

Step 2: Q2. Find the LFT of the project-end event

The backward scan always sets the LFT of the project-end event equal to the minimum project duration found by the forward scan. This ensures the project can finish on time.

\text{LFT of project-end event} = 22 \text{ days}

Step 3: Q3. Compute EST(7) with three incoming activities

An event cannot occur until all its predecessors have finished. The EST is therefore the maximum over all incoming "EST of predecessor plus duration" values.

\text{EST}(7) = \max(6 + 5,\ 9 + 3,\ 8 + 4) = \max(11,\ 12,\ 12) = 12

Step 4: Q4. Compute LFT(3) with two outgoing activities

An event must be early enough for all its successors to start in time. The LFT is the minimum over all outgoing "LFT of successor minus duration" values.

\text{LFT}(3) = \min(14 - 6,\ 9 - 2) = \min(8,\ 7) = 7

Step 5: Q5. Determine whether activity $K$ is on the critical path

Float = LST $-$ EST. If float is zero the activity is critical; if positive it has slack.

\text{Float of } K = 7 - 5 = 2

Float is $2$ , which is non-zero, so $K$ is NOT on the critical path and has $2$ time units of slack.

Step 6: Q6. Forward scan for a four-activity project

Identify the events: event $1$ (start), event $2$ (after $A$ , duration 3), event $3$ (after $B$ duration 5 and $C$ duration 4), event $4$ (project end after $D$ , duration 2). Propagate the earliest times from left to right, taking the maximum when multiple paths arrive.

\text{EST}(1) = 0, \quad \text{EST}(2) = 0 + 3 = 3

\text{EST}(3) = \max(3 + 5,\ 3 + 4) = \max(8,\ 7) = 8 \text{ (via } B\text{)}

\text{EST}(4) = 8 + 2 = 10

Minimum project duration is $10$ days. The longest path is $A \to B \to D$ , so the critical path is $A \to B \to D$ .

Step 7: Q7. Backward scan and float of activity $C$

Start from the project-end event (LFT $= 10$ ) and propagate rightward to left, taking the minimum when multiple paths leave. Then compute float as $\text{LFT of end event} - \text{EST of start event} - \text{duration}$ .

\text{LFT}(4) = 10, \quad \text{LFT}(3) = 10 - 2 = 8

\text{LFT}(2) = \min(8 - 5,\ 8 - 4) = \min(3,\ 4) = 3 \text{ (via } B\text{)}

\text{Float of } C = 8 - 3 - 4 = 1 \text{ day}

Activity $C$ can be delayed by up to $1$ day without delaying the project.

Step 8: Q8. Identify the critical path by comparing total path lengths

The critical path is the longest path from start to finish, because every activity on it must run in sequence and cannot be overlapped.

Length of path $P$ : $4 + 3 + 6 + 2 = 15$ time units.

Length of path $Q$ : $5 + 4 + 7 = 16$ time units.

Final answer: path $Q$ is longer, so $Q$ is the critical path and the minimum project duration is $16$ time units.

networks
critical-path
project-scheduling
forward-scanning
backward-scanning
hsc-maths-standard-2
year-12
2026