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How do we schedule dependent tasks to finish a project in the shortest time?

Construct an activity network, compute earliest and latest starting times and float, and identify the critical path and minimum completion time.

How to build an activity network from a precedence table, run forward and backward passes for earliest and latest start times, compute float, and identify the critical path and minimum project duration.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

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

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What this dot point is asking
Building the activity network
The forward pass (earliest start times)
The backward pass (latest start times)
Float and the critical path
Using the analysis

What this dot point is asking

You must construct the network, compute earliest and latest start times and float, and identify the critical path and minimum duration.

Building the activity network

A precedence table lists each activity, its duration, and its immediate predecessors (the activities that must finish before it can start). From it you draw a directed network where activities follow their predecessors.

The forward pass (earliest start times)

Work left to right. The first activities start at time $0$ . Each later activity starts when all its predecessors have finished, so at a merge you take the maximum of the predecessors' finish times.

The minimum completion time of the whole project is the largest earliest finish time across all activities.

The backward pass (latest start times)

Work right to left from the project completion time. Each activity's latest finish is the minimum of the latest start times of the activities that follow it; its latest start is that finish minus its duration. Activities with no successors have a latest finish equal to the project completion time.

Float and the critical path

For each activity, $\text{float} = \text{LST} - \text{EST}$ . Activities with zero float cannot be delayed at all; they form the critical path, the longest chain through the network that determines the minimum project duration.

Worked example

Finding the critical path

A project has activities $A$ ( $4$ days, no predecessor), $B$ ( $3$ , no predecessor), $C$ ( $5$ , after $A$ ), $D$ ( $2$ , after $B$ ), $E$ ( $3$ , after $C$ and $D$ ).

Step 1: Run the forward pass to find the earliest start time of each activity.

Working left to right, each activity begins as soon as all its predecessors have finished. Activities with no predecessors start at time $0$ . At any merge point, we take the maximum of the predecessors' finish times, because the next activity must wait for the slowest one. This gives: $A$ starts at $0$ and finishes at $4$ ; $B$ starts at $0$ and finishes at $3$ ; $C$ starts at $4$ (after $A$ ) and finishes at $9$ ; $D$ starts at $3$ (after $B$ ) and finishes at $5$ . Activity $E$ waits for both $C$ and $D$ , so it starts at $\max(9, 5) = 9$ and finishes at $12$ .

Step 2: Read off the minimum completion time.

The project cannot end until every activity has finished, so the minimum completion time equals the largest earliest finish time across all activities. Here that is $12$ days, which is the earliest finish of activity $E$ .

Step 3: Run the backward pass to find the latest start time and float of each activity.

Working right to left from the project deadline of $12$ days, each activity's latest finish equals the minimum of the latest start times of all its successors. Subtracting each duration gives the latest start time (LST). Activity $E$ : LST $= 12 - 3 = 9$ , float $= 9 - 9 = 0$ . Activity $C$ : LST $= 9 - 5 = 4$ , float $= 4 - 4 = 0$ . Activity $A$ : LST $= 4 - 4 = 0$ , float $= 0 - 0 = 0$ . Activity $D$ : LST $= 9 - 2 = 7$ , EST $= 3$ , float $= 7 - 3 = 4$ . Activity $B$ : $D$ must start by $7$ , so $B$ must finish by $7$ , giving LST $= 7 - 3 = 4$ and float $= 4 - 0 = 4$ .

Step 4: Identify the critical path as the chain of zero-float activities.

Any activity with zero float cannot be delayed without pushing back the whole project. The zero-float activities are $A$ , $C$ , and $E$ , so the critical path is $A \to C \to E$ with a total length of $4 + 5 + 3 = 12$ days. Activities $B$ and $D$ each have $4$ days of float and can absorb minor delays without affecting the project finish.

Using the analysis

The critical path tells a project manager which activities must be watched: any delay to a critical activity delays the whole project, while non-critical activities can absorb delays up to their float. Reducing the project duration ("crashing") requires shortening activities on the critical path, not others.

Exam-style practice questions

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

WACE 20228 marksA project has activities with durations (days) and immediate predecessors:

A

(

3

, none),

B

(

5

, none),

C

(

4

A

D

(

6

B

E

(

2

C

and

D

). (a) Find the earliest start time of each activity. (b) Find the minimum completion time. (c) Identify the critical path.

Show worked answer →

Run a forward pass, taking the latest finishing predecessor at each merge.

(a) EST: $A = 0$ , $B = 0$ , $C = 0 + 3 = 3$ (after $A$ ), $D = 0 + 5 = 5$ (after $B$ ), $E$ starts after both $C$ and $D$ finish: $C$ finishes at $3 + 4 = 7$ , $D$ finishes at $5 + 6 = 11$ , so $E$ starts at $11$ . (4 marks)

(b) Minimum completion time $= E$ start $+$ duration $= 11 + 2 = 13$ days. (2 marks)

(c) The critical path is the longest path: $B \to D \to E$ , lengths $5 + 6 + 2 = 13$ days, matching the completion time. (2 marks)

Markers reward taking the maximum predecessor finish at the merge, the $13$ -day duration, and the critical path through $B$ , $D$ , $E$ .

WACE 20245 marksFor an activity with earliest start time

6

, latest start time

9

and duration

4

, (a) calculate the float and state its meaning. (b) Explain what it means for an activity to lie on the critical path.

Show worked answer →

Float is the slack between earliest and latest start.

(a) Float $=$ latest start $-$ earliest start $= 9 - 6 = 3$ days. It means this activity can be delayed by up to $3$ days without delaying the whole project. (3 marks)

(b) An activity on the critical path has zero float: any delay to it delays the entire project, because it lies on the longest path determining the minimum completion time. (2 marks)

Markers reward float as latest minus earliest start with its delay meaning, and zero float on the critical path.