Initial Residual Network Gf for initial f (0 assignment)
Forward residual capacities equal the original capacities; no backward edges yet.
| Iteration | Augmenting Path | Bottleneck Δ | Flow Value |
|---|
| 1 | s→a→c→t | 6 | 6 |
| 2 | s→b→a→t | 3 | 9 |
| 3 | s→b→c→t | 3 | 12 |
| 4 | s→b→d→t | 2 | 14 |
| 5 | s→a→t | 2 | 16 |
| 6 | s→d→t | 5 | 21 |
Interactive Walkthrough
Iteration 1 of 6
Iteration 1
- Path: s -> a -> c -> t
- Bottleneck capacity: min(9, 6, 9) = 6
- Flow value after augmentation: 6
We can now notice that all the edges going into vertex t has capacity 0, so we know for sure that no more simple path exists from s to t.