The forward pass

What the forward pass finds

• The forward pass works left to right through the network to find, for every activity:
  • its earliest start time (EST) — the soonest it can begin,
  • its earliest finish time (EFT) — the soonest it can be complete.
• These go in the top row of each node box (EST top-left, EFT top-right).
• The pass also gives the minimum project completion time — the EFT of the final activity.

Earliest start and earliest finish

• For each activity:

$\text{EFT} = \text{EST} + \text{duration}$

• EST — earliest start time; the soonest all predecessors are done.
• duration — the fixed length of the activity.
• EFT — earliest finish time; add the duration to the earliest start.

The forward pass rule

• Start activities (no predecessor) have $\text{EST} = 0$.
• For any other activity, its EST is the earliest finish time of its predecessor:

$\text{EST of an activity} = \text{EFT of its predecessor}$

• Then apply $\text{EFT} = \text{EST} + \text{duration}$ and carry the EFT forward along each arrow.

Merging paths — take the largest

• At a merge point an activity has several predecessors and must wait for all of them.
• Its EST is the largest (latest) of the incoming earliest finish times:

$\text{EST at a merge} = \max(\text{EFTs of all predecessors})$

• Taking the smallest by mistake would start the activity before some predecessor is finished — the classic forward-pass error.

The completed forward pass

• Reading the top rows of the hall network: A finishes at 3, B at 7, C at 2. D merges B (7) and C (2), so $\text{EST}_D = \max(7, 2) = 7$ and $\text{EFT}_D = 12$. E finishes at 15, F at 14. G merges E (15) and F (14), so $\text{EST}_G = 15$ and $\text{EFT}_G = 19$.