Critical path analysis: Precedence tables and minimum project duration for HSC Maths Standard 2

Syllabus dot point

“Construct an activity network from a precedence table, identify the critical path and find the minimum project duration”

15 min readUpdated 20 Jun 2026

Quick answer

The critical path is the longest path through the project network from start to end; its length equals the minimum project duration; activities on the critical path are critical, and any delay to them delays the whole project.

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What this dot point is asking
The answer
Exam-style questions
Practice questions

What this dot point is asking

NESA gives you a precedence table for a project. A precedence table is a list of the activities, how long each one takes, and which other activities must finish before each one can start. From that table you build the activity network, find the critical path, and state the minimum project duration. This is the standard format for project management questions.

The answer

What is a project network

A project network is a diagram of a project drawn with dots joined by arrows. The dots, called vertices, are events (points where some stage is complete). The arrows, called edges, are activities (tasks to be done). Each activity has a duration.

The two main conventions:

Activity-on-edge (AOE) also called activity-on-arrow. Activities are edges with durations as weights. Vertices are events (e.g. "Activity A complete").
Activity-on-node (AON). Activities are nodes with weights; edges represent the precedence ordering only.

NESA uses AOE convention. Each activity is a labelled edge with its duration as the weight.

Precedence table

A precedence table lists:

Each activity.
Its duration.
Its immediate predecessors (activities that must finish first).

From this table, you build the network.

Building the network

Identify activities with no predecessors. These start at the project-start node.
For each activity, find the node where all its predecessors finish; this is the activity's start node.
Draw the activity as an edge from its start node to a new (or shared) end node.
The project end node is where all the final activities (those that are no predecessor for anything) terminate.

Dummy activities

Some precedence tables cannot be drawn with real activities alone. A dummy activity is a zero-duration edge (drawn dashed) that carries a precedence relationship without representing any real work. Two situations force one:

Overlapping but unequal predecessors. Suppose $E$ requires both $C$ and $D$ , while $F$ requires only $D$ . Let $C$ and $D$ finish at separate events, and start $F$ from the end of $D$ . Now add a dummy from the end of $D$ to the end of $C$ . That event now marks both $C$ and $D$ as complete, so $E$ can start there and waits for both. $F$ still depends on $D$ alone.
Two activities sharing the same start and end events. Drawing them as two edges between the same pair of nodes is unclear, because you should be able to name an activity by its start and end events. A dummy gives one of them its own separate event.

A dummy has duration $0$ , so it never adds to a path length, but it still enforces order. Standard 2 keeps dummy use light, but you are expected to recognise when one is needed and to draw it dashed; the ExamExplained critical path analysis guide works a full dummy example end to end.

Build the network, stage by stage

Turning a precedence table into a network is the part students find hardest. Below, the network shown at the top of this page is built one stage at a time, with the reasoning spelled out at each step.

Stage 1, place the activities with no predecessors. Read the precedence table and find the activities that depend on nothing. Here that is $A$ (duration $2$ ), so $A$ leaves the project start node.

Stage 2, add the activities that are now unblocked. $B$ and $C$ both list only $A$ as a predecessor, so once $A$ finishes they run in parallel. The next activity $D$ needs both of them, so $B$ and $C$ meet at a single shared event rather than ending separately.

Stage 3, finish the chain to the end. $D$ depends on $B$ and $C$ , so it leaves the shared event; $E$ then follows $D$ to the finish. Every activity from the table is now on the diagram.

Stage 4, trace the paths and mark the critical one. List every route from start to finish and add the durations: $A$ - $B$ - $D$ - $E$ gives $2 + 3 + 5 + 2 = 12$ , and $A$ - $C$ - $D$ - $E$ gives $2 + 4 + 5 + 2 = 13$ . The longer route, $13$ , is the critical path and the minimum project duration; $B$ carries $1$ unit of slack.

Paths through the network

A path is a sequence of activities from the project start to the project end. The length (duration) of a path is the sum of its activity durations.

The critical path

The critical path is the longest path through the network. Its length is the minimum possible project duration. No matter how you schedule the activities, the project cannot finish faster than the critical-path length, because that run of activities must happen in order.

Activities on the critical path are critical activities. Any delay to a critical activity delays the whole project.

Multiple critical paths

If two or more paths tie at the longest length, every one of them is critical, and every activity lying on any of them is a critical activity. Ties are common in exams and easy to miss if you stop at the first long path you find.

For example, take a project whose paths from start to finish work out to:

$A$ - $C$ - $F$ - $G$ : $4 + 3 + 5 + 2 = 14$
$A$ - $C$ - $E$ - $G$ : $4 + 3 + 6 + 2 = 15$
$B$ - $D$ - $E$ - $G$ : $3 + 4 + 6 + 2 = 15$
$B$ - $D$ - $F$ - $G$ : $3 + 4 + 5 + 2 = 14$

Two paths tie at $15$ , so both $A$ - $C$ - $E$ - $G$ and $B$ - $D$ - $E$ - $G$ are critical and the minimum duration is $15$ . Activities $E$ and $G$ lie on both, so they are doubly critical: a delay to either lengthens two paths at once.

Float: how the critical path relates to slack

Every non-critical activity has float (slack): the amount it can be delayed without pushing out the project end date. Float is computed from the forward and backward scan (the next dot point), but the headline result belongs here because it is what decides whether an activity is critical.

Key fact

An activity is critical exactly when its float is zero. Float is the gap between the latest and earliest times an activity can start, $\text{Float} = \text{LST} - \text{EST}$ . Critical activities have no slack, so they pin the project duration: the critical path is precisely the chain of zero-float activities from start to finish.

Reducing the project duration

A common follow-up asks how to finish sooner. The rule falls straight out of "the critical path sets the duration":

Shortening a non-critical activity does nothing to the end date; you only eat into its float.
Shortening a critical activity shortens the project, but only until a different path becomes the longest.

In the kitchen project below, the critical path runs through floor laying $D$ at $15$ days. The next-longest path runs through plumbing $B$ at $14$ days. Cut $D$ by one day and the project drops to $14$ days, but now $B$ 's path also measures $14$ , so there are two critical paths. Cut $D$ by a second day and the project stays at $14$ days, because $B$ 's path is now the new hold-up. So on its own $D$ can usefully be reduced by only one day. Any further saving needs the plumbing path shortened too. That "how far before another path takes over" margin is exactly the float of the next-longest path, and it is a favourite extension question.

Reading a network in the exam

Questions run in both directions, so rehearse both:

Table to network. Given the precedence table, draw the activity-on-edge network, then list and sum every path.
Network to answer. Given a drawn network, read the immediate predecessors off it (the activities arriving at the event where an activity starts), find the critical path by inspection or scanning, and state the duration with units. If asked for predecessors, give immediate predecessors only, not the whole chain leading up to an activity.

Why this matters

In real project management:

Critical activities need the most attention because they directly affect the deadline.
Non-critical activities have slack (covered in the "forward and backward scanning" dot point) which lets them be delayed without affecting the project end date.
Speeding up a critical activity shortens the project; speeding up a non-critical activity does not (until it becomes critical).

Worked example

Critical path for a kitchen renovation

A project to renovate a 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	$1$	$F$

Build the activity network from the precedence table: Demolition ( $A$ ) has no predecessors, so it starts the project. Once $A$ is complete, the three trades ( $B$ , $C$ , $D$ ) can all begin independently and run in parallel. Cabinet installation ( $E$ ) cannot start until every one of those three trades is done, so it waits for all of $B$ , $C$ and $D$ . Tiling ( $F$ ) and then the final inspection ( $G$ ) follow in sequence.

List all paths from start to finish and sum their durations: Every path must pass through $A$ , then $E$ , $F$ , $G$ . The only variable is which of the three parallel activities connects $A$ to $E$ .

$A$ - $B$ - $E$ - $F$ - $G$ : $2 + 3 + 5 + 3 + 1 = 14$ days.
$A$ - $C$ - $E$ - $F$ - $G$ : $2 + 2 + 5 + 3 + 1 = 13$ days.
$A$ - $D$ - $E$ - $F$ - $G$ : $2 + 4 + 5 + 3 + 1 = 15$ days.

Identify the critical path as the longest: The path via floor laying is longest at $15$ days. No matter how efficiently the other trades work, the project cannot finish faster than $15$ days because $A$ , $D$ , $E$ , $F$ and $G$ must happen in strict sequence.

So the critical path is $A$ - $D$ - $E$ - $F$ - $G$ , with a minimum project duration of $15$ days.

If floor laying ( $D$ ) takes a day longer, the whole project extends to $16$ days. If plumbing ( $B$ ) takes a day longer, its path becomes $15$ days, matching the critical path, so the project is unaffected (as long as $B$ does not exceed $D$ by more than its slack).

Read off the float for the parallel trades: The critical chain $A$ - $D$ - $E$ - $F$ - $G$ has zero float. For the other two trades, compare the path through each against the critical $15$ days:

Plumbing path $A$ - $B$ - $E$ - $F$ - $G$ is $14$ days, so $B$ has $15 - 14 = 1$ day of float.
Electrical path $A$ - $C$ - $E$ - $F$ - $G$ is $13$ days, so $C$ has $15 - 13 = 2$ days of float.

Plumbing can therefore start up to $1$ day late, and electrical up to $2$ days late, without delaying handover. The forward and backward scan in the next dot point reproduces these floats activity by activity.

Critical path for a Sydney apartment build

A simplified construction project:

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

Trace the parallel branches: After the foundation ( $A$ ) and frame ( $B$ ), two trades run in parallel: external walls ( $C$ , three weeks) and roofing ( $D$ , four weeks). Plumbing ( $E$ ) cannot begin until both are done, so it waits for the slower of the two. Finishes ( $F$ ) come last.

Sum both paths:

$A$ - $B$ - $C$ - $E$ - $F$ : $4 + 6 + 3 + 5 + 7 = 25$ weeks.
$A$ - $B$ - $D$ - $E$ - $F$ : $4 + 6 + 4 + 5 + 7 = 26$ weeks.

Identify the critical path: The roofing path is one week longer, making it critical.

So the critical path is $A$ - $B$ - $D$ - $E$ - $F$ at $26$ weeks, with a minimum project duration of $26$ weeks.

A one-week delay to the roof pushes the project to $27$ weeks. A one-week delay to external walls ( $C$ ) makes its path $26$ weeks, matching the critical path, so the project duration is unchanged.

Common traps

Forgetting precedence: An activity cannot start until all its immediate predecessors have finished. Check the table carefully.
Confusing shortest with critical: The critical path is the LONGEST, not the shortest. The shortest path would let you finish too fast and violate precedence.
Missing a path: Check every path from start to end. The critical one might not be obvious.
Skipping the network diagram: Markers want to see the network drawn before the paths are listed. A clean diagram earns easy marks.
Mis-summing path durations: Add the durations of all activities on the path, including the first and last.

Exam technique

Draw the network first and label every edge with its activity letter and duration; markers award method marks for a correct diagram even when the arithmetic slips. State the critical path as a sequence of activities, give the minimum duration with units, and if two paths tie, name both. A question asking which activities "can be delayed" is a float question: name the non-critical activities and how many days each can slip.

Note

Picture a big job, like renovating a kitchen, that is made up of lots of smaller tasks. Some tasks can happen at the same time, but others have to wait for an earlier task to finish first. If you trace the longest chain of must-wait tasks from start to end, that chain decides how soon the whole job can possibly be done. Speeding up a task that is not on that longest chain saves you nothing, because you would still be waiting on the chain. So the skill here is spotting which tasks are the real hold-ups and which ones have spare time to play with.

Exam-style questions

Questions in the style of NESA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

2022 HSC-style5 marks

A project has activities

A

G

. The table gives each activity's duration in days and its immediate predecessor(s). Construct the activity network, identify the critical path, and state the minimum project duration. | Activity |

A

B

C

D

E

F

G

| |---|---|---|---|---|---|---|---| | Duration |

4

2

3

4

6

5

2

A

B

C, D

C

E, F

Show worked answer →

Build the network. $A$ and $B$ leave the start node (no predecessors). $C$ follows $A$ , $D$ follows $B$ , and $F$ follows $C$ . $E$ needs both $C$ and $D$ , so those two meet at one node before $E$ . $G$ needs both $E$ and $F$ , which meet before the finish node.

List every start-to-finish path and its length, in days.

$A$ - $C$ - $E$ - $G$ : $4 + 3 + 6 + 2 = 15$
$A$ - $C$ - $F$ - $G$ : $4 + 3 + 5 + 2 = 14$
$B$ - $D$ - $E$ - $G$ : $2 + 4 + 6 + 2 = 14$

Identify the critical path. The longest path is $A$ - $C$ - $E$ - $G$ at $15$ days, so that is the critical path and the minimum project duration is $15$ days. Activities $B$ , $D$ and $F$ lie off it and so carry float.

Markers reward a correct network, the paths listed with totals, the longest named as the critical path, and the minimum duration of $15$ days with units.

2023 HSC-style4 marks

A kitchen renovation has five activities

A

E

, with durations in days and immediate predecessors as shown. Find the critical path and the minimum time to complete the renovation. | Activity |

A

B

C

D

E

| |---|---|---|---|---|---| | Duration |

2

4

3

2

3

| | Predecessors | none |

A

A

B

C, D

Show worked answer →

Build the network. $A$ leaves the start node. $B$ and $C$ both follow $A$ . $D$ follows $B$ . $E$ needs both $C$ and $D$ , so they meet at one node before $E$ , which runs to the finish.

List every start-to-finish path and its length, in days.

$A$ - $B$ - $D$ - $E$ : $2 + 4 + 2 + 3 = 11$
$A$ - $C$ - $E$ : $2 + 3 + 3 = 8$

Identify the critical path. The longest path is $A$ - $B$ - $D$ - $E$ at $11$ days, so that is the critical path and the minimum time is $11$ days. Activity $C$ is off the critical path; its float is $11 - 8 = 3$ days.

Markers reward a correct network, both paths summed, the longer named as the critical path, and the minimum time of $11$ days.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

A small landscaping job has three activities that must run strictly one after another. Activity

A

(clearing) takes

3

days and has no predecessor; activity

B

(digging) takes

5

days and needs

A

; activity

C

(planting) takes

2

days and needs

B

. Draw the activity network, then state the only path and the minimum project duration.

Show worked solution →

Build the network. With each activity needing the one before it, the network is a single chain: start, then $A$ , then $B$ , then $C$ , then finish. There are no parallel branches.

List the path and sum the durations. Only one route runs from start to finish:

A\text{-}B\text{-}C = 3 + 5 + 2 = 10 \text{ days}

State the answer. The minimum project duration is $10$ days, and since every activity lies on the single path, all of $A$ , $B$ and $C$ are critical. Check: a strict chain has no slack anywhere, so a duration equal to the sum of all three is exactly what we expect.

foundation2 marks

Two activities can start immediately:

A

(order tiles) takes

4

days and

B

(order timber) takes

6

days, neither with a predecessor. Activity

C

(delivery sign-off) takes

3

days and needs both

A

and

B

finished first. Find the critical path and the minimum project duration.

Show worked solution →

Build the network. $A$ and $B$ both leave the project start and run in parallel; because $C$ needs both, they meet at a shared event where $C$ begins, then $C$ runs to the finish.

List the paths and sum them. Two routes reach the finish:

A\text{-}C = 4 + 3 = 7 \qquad B\text{-}C = 6 + 3 = 9

Pick the longest as critical. $C$ cannot start until the slower of $A$ and $B$ is done, namely $B$ at $6$ days, so the critical path is $B\text{-}C$ at $9$ days. The minimum project duration is $9$ days. Check: $A$ finishes at day $4$ but waits until day $6$ for $B$ , giving $A$ a slack of $9 - 7 = 2$ days, which is consistent.

foundation3 marks

A project has activities

A

B

C

D

with this table. Activity

A

: duration

2

, no predecessor. Activity

B

: duration

4

, predecessor

A

. Activity

C

: duration

3

, predecessor

A

. Activity

D

: duration

5

, predecessors

B

and

C

. (a) State the minimum project duration. (b) State how many days activity

C

can be delayed without delaying the project.

Show worked solution →

Build the network. $A$ starts the project; $B$ and $C$ both follow $A$ and run in parallel; $D$ needs both, so it leaves the shared event where $B$ and $C$ meet, then runs to the finish.

Part (a): list the paths and take the longest. The two routes are:

A\text{-}B\text{-}D = 2 + 4 + 5 = 11 \qquad A\text{-}C\text{-}D = 2 + 3 + 5 = 10

The longer is $A\text{-}B\text{-}D$ , so the minimum project duration is $11$ days.

Part (b): float on the shorter branch. $B$ and $C$ both feed the same event, which the critical path reaches at day $6$ . $C$ finishes as early as $2 + 3 = 5$ , so it can slip by $6 - 5 = 1$ day. Activity $C$ can be delayed $1$ day. Check: $C$ 's path is $1$ day shorter than the critical path, matching the float of $1$ .

foundation3 marks

A project has these activities.

A

: duration

3

, no predecessor.

B

: duration

2

, predecessor

A

C

: duration

5

, predecessor

A

D

: duration

4

, predecessors

B

and

C

E

: duration

2

, predecessor

D

. List every path from start to finish with its length, then state the critical path and the minimum project duration.

Show worked solution →

Build the network. $A$ starts; $B$ and $C$ run in parallel after $A$ and meet at a shared event; $D$ leaves that event; $E$ follows $D$ to the finish.

List all paths and sum each. The only choice is whether $B$ or $C$ links $A$ to $D$ :

A\text{-}C\text{-}D\text{-}E = 3 + 5 + 4 + 2 = 14

A\text{-}B\text{-}D\text{-}E = 3 + 2 + 4 + 2 = 11

Identify the critical path. The longest is $A\text{-}C\text{-}D\text{-}E$ at $14$ , so that is the critical path and the minimum project duration is $14$ days. Check: $B$ 's path is $11$ , so $B$ carries $14 - 11 = 3$ days of slack while every activity on $A\text{-}C\text{-}D\text{-}E$ has none.

core3 marks

A project has activities

A

E

A

: duration

5

, no predecessor.

B

: duration

2

, predecessor

A

C

: duration

6

, predecessor

A

D

: duration

4

, predecessors

B

and

C

E

: duration

3

, predecessor

D

. Using

\text{Float} = \text{LST} - \text{EST}

, find the float of activity

B

Show worked solution →

Forward pass for the earliest start of $B$ . $A$ starts at $0$ and finishes at $5$ , so $B$ can start as early as $\text{EST}_B = 5$ .

Find when $D$ must start, then work back to $B$ 's latest start. $D$ needs both $B$ and $C$ . The earliest $D$ can start is the later of $B$ finishing ( $5 + 2 = 7$ ) and $C$ finishing ( $5 + 6 = 11$ ), so $\text{EST}_D = 11$ . Since the critical chain $A\text{-}C\text{-}D\text{-}E$ has no slack, $D$ 's latest start is also $11$ , so $B$ must finish by $11$ , giving $\text{LST}_B = 11 - 2 = 9$ .

Apply the float formula.

\text{Float}_B = \text{LST}_B - \text{EST}_B = 9 - 5 = 4 \text{ days}

Activity $B$ has $4$ days of float. Check: the path through $B$ is $5 + 2 + 4 + 3 = 14$ against the critical $5 + 6 + 4 + 3 = 18$ , and $18 - 14 = 4$ matches.

core3 marks

Two activities can start at once.

A

: duration

4

, no predecessor.

B

: duration

6

, no predecessor.

C

: duration

5

, predecessor

A

D

: duration

3

, predecessor

B

E

: duration

4

, predecessors

C

and

D

. Find the critical path (or paths) and the minimum project duration.

Show worked solution →

Build the network. $A$ and $B$ both leave the start. $C$ follows $A$ ; $D$ follows $B$ . $E$ needs both $C$ and $D$ , so the two branches meet at the event where $E$ begins, then $E$ runs to the finish.

List the paths and sum them. Two routes reach the finish:

A\text{-}C\text{-}E = 4 + 5 + 4 = 13 \qquad B\text{-}D\text{-}E = 6 + 3 + 4 = 13

Identify the critical path. Both routes measure $13$ , so they tie: the network has two critical paths, $A\text{-}C\text{-}E$ and $B\text{-}D\text{-}E$ , and the minimum project duration is $13$ weeks. Check: $E$ lies on both, and the two branches feeding it ( $A\text{-}C = 9$ and $B\text{-}D = 9$ ) are equal, so no activity has slack, consistent with every path being critical.

core4 marks

A project has activities

A

F

A

: duration

3

, no predecessor.

B

: duration

4

, predecessor

A

C

: duration

2

, predecessor

A

D

: duration

5

, predecessor

B

E

: duration

6

, predecessor

C

F

: duration

2

, predecessors

D

and

E

. Carry out a forward scan to find the earliest start time (EST) of every activity, and hence state the minimum project duration.

Show worked solution →

Set the start. The forward scan gives each event its earliest time. $A$ has no predecessor, so $\text{EST}_A = 0$ .

Scan forward, taking the largest incoming time at each step.

$\text{EST}_B = 0 + 3 = 3$ and $\text{EST}_C = 0 + 3 = 3$ (both follow $A$ ).
$\text{EST}_D = 3 + 4 = 7$ (follows $B$ ).
$\text{EST}_E = 3 + 2 = 5$ (follows $C$ ).
$F$ needs both $D$ and $E$ : take the larger of $7 + 5 = 12$ and $5 + 6 = 11$ , so $\text{EST}_F = 12$ .

Read off the duration. The project finishes at $\text{EST}_F + 2 = 12 + 2 = 14$ , so the minimum project duration is $14$ days. Check: the longest path $A\text{-}B\text{-}D\text{-}F = 3 + 4 + 5 + 2 = 14$ agrees.

core4 marks

A project has activities

A

F

A

: duration

2

, no predecessor.

B

: duration

5

, predecessor

A

C

: duration

3

, predecessor

A

D

: duration

4

, predecessor

B

E

: duration

2

, predecessor

C

F

: duration

3

, predecessors

D

and

E

. Find the float of every activity and hence list the critical activities.

Show worked solution →

Forward scan (EST): $\text{EST}_A = 0$ ; $\text{EST}_B = 2$ , $\text{EST}_C = 2$ ; $\text{EST}_D = 7$ , $\text{EST}_E = 5$ ; $F$ takes the larger of $7 + 4 = 11$ and $5 + 2 = 7$ , so $\text{EST}_F = 11$ . Duration $= 11 + 3 = 14$ .
Backward scan (LST), starting from $\text{LFT} = 14$: $\text{LST}_F = 14 - 3 = 11$ . Then $\text{LST}_D = 11 - 4 = 7$ and $\text{LST}_E = 11 - 2 = 9$ . Back through $B$ : $\text{LST}_B = 7 - 5 = 2$ ; through $C$ : $\text{LST}_C = 9 - 3 = 6$ . Finally $\text{LST}_A = 0$ .
Float $= \text{LST} - \text{EST}$ for each: $A: 0$ , $B: 0$ , $C: 6 - 2 = 4$ , $D: 0$ , $E: 9 - 5 = 4$ , $F: 0$ .
List the critical activities (zero float): They are $A$ , $B$ , $D$ , $F$ . Check: the path $A\text{-}B\text{-}D\text{-}F = 2 + 5 + 4 + 3 = 14$ equals the duration, confirming the critical chain.

exam5 marks

A road upgrade has activities

A

G

(durations in weeks).

A

4

, no predecessor.

B

3

, predecessor

A

C

6

, predecessor

A

D

5

, predecessor

B

E

2

, predecessor

C

F

4

, predecessors

D

and

E

G

3

, predecessor

F

. Carry out a forward and backward scan, then state the critical path (or paths) and the minimum project duration.

Show worked solution →

Forward scan (EST): $\text{EST}_A = 0$ ; $\text{EST}_B = 4$ , $\text{EST}_C = 4$ ; $\text{EST}_D = 4 + 3 = 7$ , $\text{EST}_E = 4 + 6 = 10$ ; $F$ takes the larger of $7 + 5 = 12$ and $10 + 2 = 12$ , so $\text{EST}_F = 12$ ; $\text{EST}_G = 12 + 4 = 16$ . Duration $= 16 + 3 = 19$ weeks.
Backward scan (LST) from $\text{LFT} = 19$: $\text{LST}_G = 16$ ; $\text{LST}_F = 12$ ; $\text{LST}_D = 12 - 5 = 7$ , $\text{LST}_E = 12 - 2 = 10$ ; $\text{LST}_B = 7 - 3 = 4$ , $\text{LST}_C = 10 - 6 = 4$ ; $\text{LST}_A = 0$ .
Compare to find float: Every activity has $\text{LST} = \text{EST}$ , so every float is $0$ .
State the critical path(s): Both branches are tight, so two paths tie: $A\text{-}B\text{-}D\text{-}F\text{-}G$ and $A\text{-}C\text{-}E\text{-}F\text{-}G$ , each $4 + 3 + 5 + 4 + 3 = 19$ and $4 + 6 + 2 + 4 + 3 = 19$ . Minimum duration is $19$ weeks. Check: both sums equal $19$ .

exam5 marks

A project has activities

A

G

A

: duration

4

, no predecessor.

B

: duration

3

, no predecessor.

C

: duration

3

, predecessor

A

D

: duration

4

, predecessor

B

E

: duration

6

, predecessors

C

and

D

F

: duration

2

, predecessor

E

G

: duration

1

, predecessor

F

. (a) Find the minimum project duration. (b) Name every critical path. (c) Name any activity that lies on more than one critical path and explain why a delay to it is especially serious.

Show worked solution →

Build and forward-scan: $A$ and $B$ both start the project; $C$ follows $A$ , $D$ follows $B$ ; $E$ needs both $C$ and $D$ ; then $F$ then $G$ . $\text{EST}_C = 4$ , $\text{EST}_D = 3$ ; $E$ takes the larger of $4 + 3 = 7$ and $3 + 4 = 7$ , so $\text{EST}_E = 7$ ; $\text{EST}_F = 13$ , $\text{EST}_G = 15$ .
Part (a): duration: Project finishes at $15 + 1 = 16$ , so the minimum project duration is $16$ days.
Part (b): critical paths: The two branches into $E$ are equal ( $A\text{-}C = 7$ , $B\text{-}D = 7$ ), so both routes tie at the longest:

A\text{-}C\text{-}E\text{-}F\text{-}G = 4 + 3 + 6 + 2 + 1 = 16

B\text{-}D\text{-}E\text{-}F\text{-}G = 3 + 4 + 6 + 2 + 1 = 16

Both are critical.

Part (c): shared activities. $E$ , $F$ and $G$ lie on both critical paths. A delay to any of them lengthens both critical paths at once, so it pushes out the finish no matter which branch is faster. Check: both path sums equal the $16$ found in part (a).

exam6 marks

A renovation has these activities (days).

A

2

, no predecessor.

B

6

, predecessor

A

C

3

, predecessor

A

D

7

, predecessor

A

E

5

, predecessors

B

C

D

F

3

, predecessor

E

. (a) Find the critical path and minimum project duration. (b) The builder can shorten activity

D

. By how many days can

D

be reduced before the project duration stops improving, and what is the shortest duration

D

alone can achieve?

Show worked solution →

Part (a): build and sum the paths. $A$ starts; $B$ , $C$ , $D$ run in parallel after $A$ ; $E$ needs all three; $F$ follows $E$ . The three routes are:

A\text{-}D\text{-}E\text{-}F = 2 + 7 + 5 + 3 = 17

A\text{-}B\text{-}E\text{-}F = 2 + 6 + 5 + 3 = 16

A\text{-}C\text{-}E\text{-}F = 2 + 3 + 5 + 3 = 13

The longest is $A\text{-}D\text{-}E\text{-}F$ , so the critical path is $A\text{-}D\text{-}E\text{-}F$ and the minimum project duration is $17$ days.

Part (b): how far $D$ helps. Cutting $D$ shortens the critical path until a different path takes over. The next-longest is $A\text{-}B\text{-}E\text{-}F$ at $16$ days, which is $17 - 16 = 1$ day below the critical path. So reducing $D$ by $1$ day drops the project to $16$ days; any further cut to $D$ leaves the project at $16$ because $A\text{-}B\text{-}E\text{-}F$ is now the bottleneck. $D$ can usefully be reduced by $1$ day, giving a shortest duration of $16$ days from $D$ alone. Check: with $D$ cut to $6$ , paths $A\text{-}D\text{-}E\text{-}F$ and $A\text{-}B\text{-}E\text{-}F$ both equal $16$ .

exam6 marks

A project has activities

A

H

A

: duration

3

, no predecessor.

B

: duration

5

, predecessor

A

C

: duration

2

, predecessor

A

D

: duration

4

, predecessor

B

E

: duration

6

, predecessor

C

F

: duration

3

, predecessors

D

and

E

G

: duration

2

, predecessor

F

H

: duration

4

, predecessor

G

. (a) Find the critical path and the minimum project duration. (b) State which activities have float and how many days each carries.

Show worked solution →

Forward scan (EST): $\text{EST}_A = 0$ ; $\text{EST}_B = 3$ , $\text{EST}_C = 3$ ; $\text{EST}_D = 3 + 5 = 8$ , $\text{EST}_E = 3 + 2 = 5$ ; $F$ takes the larger of $8 + 4 = 12$ and $5 + 6 = 11$ , so $\text{EST}_F = 12$ ; $\text{EST}_G = 15$ , $\text{EST}_H = 17$ . Duration $= 17 + 4 = 21$ days.
Part (a): critical path: The longest route is $A\text{-}B\text{-}D\text{-}F\text{-}G\text{-}H = 3 + 5 + 4 + 3 + 2 + 4 = 21$ , so that is the critical path and the minimum project duration is $21$ days.
Backward scan (LST) for the off-path activities: From $\text{LFT} = 21$ : $F$ must start by $12$ , so $\text{LST}_E = 12 - 6 = 6$ and $\text{LST}_C = 6 - 2 = 4$ .
Part (b): floats: $\text{Float}_C = \text{LST}_C - \text{EST}_C = 4 - 3 = 1$ and $\text{Float}_E = 6 - 5 = 1$ . Only $C$ and $E$ carry float, each $1$ day; every other activity has $0$ . Check: the side path $A\text{-}C\text{-}E\text{-}F\dots = 20$ is $1$ short of $21$ , matching a float of $1$ .

Practise this

Sources & how we know this

Mathematics Standard Stage 6 Syllabus (2017) — NESA
Mathematics Standard HSC exam papers — NESA