Quick questions on Forward and backward scanning and activity float for HSC Maths Standard 2

Each activity in the diagram above carries a box at its tail (where it starts): the left cell is its EST (earliest start time) and the right cell is its LST (latest start time). Fill every left cell first with a forward pass. Then fill every right cell with a backward pass. Then read the float straight off each box as right minus left.

The diagram at the top of this page shows the finished scan. Below is the same network assembled one stage at a time, so you can see exactly where every number comes from. Each stage below adds one more layer and spells out the working behind it.

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

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

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

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

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).

An activity cannot begin until all of its predecessors have finished, so its EST is the largest "predecessor EST plus that predecessor's duration". Work left to right.

Now work right to left from the finish. An activity's latest finish is the earliest latest-start among the activities that follow it; subtract its own duration to get its own LST.

Float $=$ LFT $-$ EST $-$ duration. Easy to forget the duration.