NSWMaths Standard 2Syllabus dot point

How do forward and backward scanning give the earliest and latest start times of each activity, and how is float calculated?

Perform forward and backward scanning to find earliest start, latest start, earliest finish, latest finish times and float for each activity

A focused answer to the HSC Maths Standard 2 dot point on forward and backward scanning. Computing earliest start, latest start, earliest finish, latest finish and float for each activity in a project network, with worked Australian construction and renovation examples.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to use forward and backward scanning to find the earliest and latest start and finish times for each activity in a project network, compute float (slack), and confirm the critical path as the activities with zero float.

The answer

Forward scanning (earliest times)

Label each event (node) with the earliest time it can occur, computed by working forward from the project start.

Start event: EST $= 0$ .
Any other event: EST equals the maximum over all incoming activities of (EST of the activity's start event + the activity's duration).

If an event has multiple predecessors, take the maximum because all predecessors must finish before the event occurs.

The forward scan reaches the end event with the minimum project duration.

Backward scanning (latest times)

Label each event with the latest time it can occur without delaying the project, computed by working backward from the project end.

End event: LFT = minimum project duration (from the forward scan).
Any other event: LFT equals the minimum over all outgoing activities of (LFT of the activity's end event $-$ activity duration).

If an event has multiple successors, take the minimum because the event must complete in time for the earliest required successor.

Time computations for each activity

For activity from event $i$ to event $j$ with duration $t$ :

EST (earliest start time) = $E_i$ (the EST of the activity's start event).
EFT (earliest finish time) = $E_i + t$ .
LFT (latest finish time) = $L_j$ (the LFT of the activity's end event).
LST (latest start time) = $L_j - t$ .

Float

The float of an activity is the maximum delay possible without extending the project:

\text{Float} = LST - EST = LFT - EFT = L_j - E_i - t.

An activity with float $0$ is critical: any delay to it delays the project. An activity with positive float has slack; it can be delayed by up to its float without affecting the project end date.

Critical path

The critical path consists of activities with zero float. It is also the longest path through the network (the same path found in the previous dot point's analysis).

For activities not on the critical path, there is some flexibility in scheduling. This is useful in real project management for planning resource allocation, contingency, and parallel work.

Compact notation

The HSC often uses a four-quadrant notation per event:

EST	LFT
Event label	(sometimes float of the event)

Or two stacked numbers: EST on top, LFT below.

Always label clearly so the marker can follow.

Worked example

Worked through end-to-end

Project with these activities (kitchen renovation continued from previous dot point):

Activity	Duration	Predecessors
IMATH_0	IMATH_1	None
IMATH_2	IMATH_3	IMATH_4
IMATH_5	IMATH_6	IMATH_7
IMATH_8	IMATH_9	IMATH_10
IMATH_11	IMATH_12	IMATH_13 , $C$ , IMATH_15
IMATH_16	IMATH_17	IMATH_18
IMATH_19	IMATH_20	IMATH_21

Events: $1$ (project start), $2$ (after $A$ ), $3$ (after $B$ , $C$ , $D$ ready for $E$ ), $4$ (after $E$ ), $5$ (after $F$ ), $6$ (project end after $G$ ).

Forward scan:

Event $1$ : EST $= 0$ .
Event $2$ : EST $= 0 + 2 = 2$ (only one predecessor activity, $A$ ).
Event $3$ : EST $= \max(2 + 3, 2 + 2, 2 + 4) = \max(5, 4, 6) = 6$ (via $D$ ).
Event $4$ : EST $= 6 + 5 = 11$ .
Event $5$ : EST $= 11 + 3 = 14$ .
Event $6$ : EST $= 14 + 1 = 15$ . Project duration: $15$ days.

Backward scan:

Event $6$ : LFT $= 15$ .
Event $5$ : LFT $= 15 - 1 = 14$ .
Event $4$ : LFT $= 14 - 3 = 11$ .
Event $3$ : LFT $= 11 - 5 = 6$ .
Event $2$ : LFT $= \min(6 - 3, 6 - 2, 6 - 4) = \min(3, 4, 2) = 2$ (via $D$ ).
Event $1$ : LFT $= 2 - 2 = 0$ .

Float for each activity:

IMATH_64 : float $= 2 - 0 - 2 = 0$ . Critical.
IMATH_66 : float $= 6 - 2 - 3 = 1$ .
IMATH_68 : float $= 6 - 2 - 2 = 2$ .
IMATH_70 : float $= 6 - 2 - 4 = 0$ . Critical.
IMATH_72 : float $= 11 - 6 - 5 = 0$ . Critical.
IMATH_74 : float $= 14 - 11 - 3 = 0$ . Critical.
IMATH_76 : float $= 15 - 14 - 1 = 0$ . Critical.

Critical path: $A$ - $D$ - $E$ - $F$ - $G$ (zero-float activities), total duration $15$ . Same answer as listing paths and picking the longest.

Activity $B$ can be delayed by $1$ day, $C$ by $2$ days, without affecting the project end date.

Australian construction (extension to previous example)

Activity	Duration (weeks)	Predecessors
IMATH_88 - Foundation	IMATH_89	None
IMATH_90 - Frame	IMATH_91	IMATH_92
IMATH_93 - External walls	IMATH_94	IMATH_95
IMATH_96 - Roofing	IMATH_97	IMATH_98
IMATH_99 - Plumbing	IMATH_100	IMATH_101 , IMATH_102
IMATH_103 - Finishes	IMATH_104	IMATH_105

Forward: Event $1$ EST $0$ ; Event $2$ (after $A$ ) EST $4$ ; Event $3$ (after $B$ ) EST $10$ ; Event $4$ (after $C$ and $D$ ) EST $\max(10 + 3, 10 + 4) = 14$ ; Event $5$ (after $E$ ) EST $19$ ; Event $6$ (after $F$ ) EST $26$ .

Backward: LFT $6 = 26$ ; LFT $5 = 19$ ; LFT $4 = 14$ ; LFT $3 = \min(14 - 3, 14 - 4) = 10$ ; LFT $2 = 4$ ; LFT $1 = 0$ .

Float:

IMATH_130 : $4 - 0 - 4 = 0$ . Critical.
IMATH_132 : $10 - 4 - 6 = 0$ . Critical.
IMATH_134 : $14 - 10 - 3 = 1$ . Slack of $1$ week.
IMATH_137 : $14 - 10 - 4 = 0$ . Critical.
IMATH_139 : $19 - 14 - 5 = 0$ . Critical.
IMATH_141 : $26 - 19 - 7 = 0$ . Critical.

Critical path: $A$ - $B$ - $D$ - $E$ - $F$ . The external-wall installation ( $C$ ) has $1$ week of slack.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q295 marksA project network has activities with the following durations and precedences. For each activity, find the earliest start time (EST), latest start time (LST), and float. Identify the critical path.

Show worked answer →

Forward scan: for each node, set EST = max of (EST of predecessor + activity duration) over all predecessors. Start node has EST $= 0$ .

Backward scan: for each node, set LFT (latest finish time) = min of (LFT of successor $-$ successor activity duration) over all successors. End node has LFT = total project duration (from the forward scan).

For each activity:

EST = EST of its start node.
LFT = LFT of its end node.
EFT (earliest finish time) = EST + duration.
LST (latest start time) = LFT $-$ duration.
Float = LST $-$ EST = LFT $-$ EFT.

Critical path consists of activities with float = $0$ .

Markers reward the forward scan, the backward scan, the float calculation for each activity, and identification of the critical path as the set of zero-float activities.

2023 HSC Q284 marksUse forward and backward scanning on the given project network to find the float of each activity. State which activities can be delayed without affecting the project duration.

Show worked answer →

Forward scan gives EST for each event (node). Backward scan gives LFT.

For each activity (edge from event $i$ to event $j$ with duration $t$ ):

EST = $E_i$ , EFT = $E_i + t$ .
LFT = $L_j$ , LST = $L_j - t$ .
Float = $L_j - E_i - t$ .

Activities with float $> 0$ can be delayed by up to their float without affecting the project duration. Activities with float $= 0$ are critical; any delay to them pushes the project end date.

List each activity with its float. Activities with positive float can be delayed.

Markers reward complete float table, identification of activities with positive float, and the maximum delay (= float) for each.