HSC Mathematics Standard 2 critical path analysis (2026 guide)

A project to renovate an Australian kitchen has these activities:

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.

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.

A simplified Sydney apartment build:

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.

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.

A school musical has the following partial precedence table:

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 $\text{Float} = L_j - E_i - t$.

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.

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

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.

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 $= \text{LFT} - \text{EST} - \text{duration}$.

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.

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.

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.

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.

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.

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 $\text{LFT of end event} - \text{EST of start event} - \text{duration}$.

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.