QLDGeneral MathematicsSyllabus dot point

Topic 3: Networks and decision mathematics - how do we schedule a project and find the shortest time to complete it?

Represent a project as an activity network with durations and dependencies, perform forward and backward scanning to find earliest and latest start times, identify the critical path and minimum completion time, and calculate float for non-critical activities

A focused answer to the QCE General Mathematics Unit 4 dot point on critical path analysis. Covers building an activity network, forward and backward scanning for earliest and latest start times, finding the critical path and minimum completion time, and calculating float for non-critical activities, with arithmetic-verified worked examples for IA3 and the external assessment.

Generated by Claude Opus 4.76 min answerUpdated 2026-05-29

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

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What this dot point is asking

QCAA wants you to plan a project made of activities that depend on one another, and find the shortest possible time to finish it. You draw an activity network with durations, scan forwards to find the earliest each activity can start, scan backwards to find the latest it can start without delaying the project, identify the critical path of activities that cannot slip, and compute the float (slack) on the rest. This is a distinct sub-topic of Unit 4 Topic 3 and one of the highest-value extended-response items in IA3 and the external assessment.

The answer

The activity network

A project is broken into activities, each with a duration and a list of activities that must finish before it can start (its predecessors). These are drawn as a directed network where edges or nodes carry the activity durations and the arrows show the dependencies. The network has a single start and a single finish.

Forward scanning: earliest start times

Forward scanning finds the earliest start time (EST) of each activity, working from the start to the finish. The first activities have EST $0$ . Any later activity can only start once all its predecessors have finished, so its EST is the largest of (predecessor EST plus predecessor duration) over all predecessors. The earliest finish time of the whole project is the largest finishing time at the end, which is the minimum completion time.

Backward scanning: latest start times

Backward scanning finds the latest start time (LST) of each activity, working from the finish back to the start, using the minimum completion time as the project deadline. An activity's latest finish is the smallest LST of the activities that follow it, and its LST is that latest finish minus its own duration. The first activities should end with LST $0$ if the arithmetic is correct.

The critical path

The critical path is the chain of activities from start to finish that have no spare time: for each, the EST equals the LST. These activities determine the project length, so any delay to a critical activity delays the whole project. The total duration along the critical path equals the minimum completion time.

Float

Float (or slack) is the spare time on a non-critical activity, the amount it can be delayed without pushing back the project finish.

\text{float} = \text{LST} - \text{EST}.

Critical activities have zero float. Float tells a manager where there is flexibility to reallocate resources.

Worked example

Finding completion time and float

A project has activities A (duration $3$ ), B (duration $5$ ) and C (duration $2$ ). A and B both start at the beginning; C can only start after A finishes. The project ends when B and C are both done.

Step 1: Forward scan (EST and earliest finish).

\text{A: EST } 0, \text{ finish } 0 + 3 = 3.

\text{B: EST } 0, \text{ finish } 0 + 5 = 5.

\text{C: EST } 3 \text{ (after A), finish } 3 + 2 = 5.

Step 2: Minimum completion time. The project finishes when both end activities are done: the larger of B's finish ( $5$ ) and C's finish ( $5$ ).

\max(5, 5) = 5.

Step 3: Backward scan (LST from deadline $5$ ).

\text{B: LST } 5 - 5 = 0.

\text{C: LST } 5 - 2 = 3, \text{ so A must finish by } 3.

\text{A: LST } 3 - 3 = 0.

Step 4: Float for each activity, float $=$ LST $-$ EST.

\text{A: } 0 - 0 = 0, \quad \text{B: } 0 - 0 = 0, \quad \text{C: } 3 - 3 = 0.

Step 5: Critical path. Every activity here has zero float, so both chains A-C and B are critical and the minimum completion time is $5$ . Verify: A then C is $3 + 2 = 5$ , and B alone is $5$ , matching.

Common traps

Taking the smallest predecessor finish in the forward scan: An activity waits for all its predecessors, so its EST is the largest predecessor finishing time, not the smallest.
Taking the largest successor LST in the backward scan: Working backwards you take the smallest of the following activities' latest starts, because the activity must not delay the earliest of them.
Calling the longest-duration activity the critical path: The critical path is the longest chain through the network, not the single longest activity. A short activity can be critical if it lies on that chain.
Forgetting that float equals LST minus EST: Subtracting durations or finishing times instead of start times gives the wrong slack.
Missing a parallel finishing branch: The completion time is the largest earliest-finish among all activities with no successors. Ignoring a parallel branch can understate the project length.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 QCAA5 marksThe activity table for a project is shown. A (prereq none, 2 days); B (none, 4); C (A, 3); D (B, 6); E (D, 3); F (C and E, 4); G (D, 8); H (F and G, 4). a) Use the activity table to construct a network diagram, including earliest and latest starting times. [3 marks] b) Determine the critical path. [1 mark] c) Determine the shortest completion time for the project. [1 mark]

Show worked answer →

a) Network and scanning (3 marks). Draw each activity as an edge respecting the prerequisites (1 mark): A and B start from the source; C follows A; D follows B; E follows D; G follows D; F follows both C and E; H follows both F and G. Label each activity with its letter and duration (1 mark). Forward scan to get earliest start times and backward scan to get latest start times (1 mark).

b) Critical path (1 mark). Forward scanning gives the longest chains: B - D - G - H = 4 + 6 + 8 + 4 = 22, while B - D - E - F - H = 4 + 6 + 3 + 4 + 4 = 21. The longest is BDGH, so the critical path is B, D, G, H.

c) Shortest completion time (1 mark). The minimum completion time equals the length of the critical path, 22 days.

The critical path is the longest path through the network; its length is the shortest possible completion time, and every activity on it has zero float.

2023 QCAA5 marksJed is preparing and serving a meal. Tasks with duration (minutes) and prerequisite: A Assemble equipment 2 (none); B Boil rice 20 (A); C Prepare curry ingredients 6 (A); D Make naan bread 8 (A); E Simmer curry 40 (C); F Fry naan bread 4 (D); G Serve meal 2 (B, E, F). a) Construct a network diagram showing the sequence of tasks, labelling all tasks and durations, and use forward and backward scanning to show the earliest and latest starting times. [3 marks] b) Determine the critical activities and minimum completion time. [2 marks]

Show worked answer →

a) Network and scanning (3 marks). All of B, C and D follow A; E follows C; F follows D; G follows B, E and F (1 mark for the correct structure). Label each task with its letter and duration (1 mark). Forward scan for earliest start times and backward scan for latest start times (1 mark).

b) Critical activities and minimum time (2 marks). Compare the path lengths: A - B - G = 2 + 20 + 2 = 24, A - C - E - G = 2 + 6 + 40 + 2 = 50, A - D - F - G = 2 + 8 + 4 + 2 = 16. The longest is A - C - E - G (1 mark for the critical activities A, C, E, G), so the minimum completion time is 50 minutes (1 mark).

The 40-minute simmer dominates, so the curry path is critical; the rice and naan tasks have float and can slip without delaying the meal, as long as they finish before serving.