Given a feasible network flow , an original directed edge has forward residual capacity and a reverse residual arc of capacity . A source–sink graph path in this residual network permits an augmentation by the minimum residual capacity along it: increase forward flows and decrease the corresponding reversed flows. Capacities and flow conservation are preserved, and the flow value increases by that amount.
If no residual source–sink graph path exists, let be the vertices reachable from the source. The sink is outside . Every original arc from to its complement is saturated, and every original arc back into has zero flow, since a positive reverse residual arc would make its tail reachable. Conservation therefore gives
The flow equals this cut capacity, so the max-flow min-cut theorem proves it is maximum.
Termination needs a capacity or path-selection assumption. For integer capacities, starting from zero, every residual capacity and augmentation is integral; each augmentation increases the value by at least one, and the value is bounded by the total source-outgoing capacity. The algorithm therefore terminates. Rational capacities can be scaled to integers. With arbitrary real capacities, unrestricted graph path choices do not guarantee finite termination. A valid uniformly terminating implementation is the Edmonds–Karp algorithm, choosing a shortest residual graph path each time.
Here is why that choice works independently of numerical capacity sizes. Residual distances from the source cannot decrease after augmenting along a shortest graph path. Every newly created residual arc is the reverse of a graph path arc; if it could cause the first distance decrease, its previous distance relation would contradict that decrease. When an arc is saturated on a shortest graph path, . Before the same direction can be saturated again, it must have been restored by a reverse augmentation on a shortest graph path, at which time . Each arc is consequently critical only times. There are arcs and each augmentation saturates at least one, giving augmentations, each found by breadth-first search in time. Thus the shortest-path Ford–Fulkerson implementation terminates and finds a maximum flow in arithmetic operations.
For sports elimination by maximum flow, first give the target team all its remaining wins. This cannot hurt its chance to finish strictly ahead: changing a result to favour it also removes a rival's win. Let be its resulting win count. If any rival already has at least wins, strict victory is impossible. Otherwise give each unordered pair of rivals one game node. Its incoming capacity is , and its two outgoing arcs go to the participating team nodes. Their capacities may be replaced by the finite number
since no arc can need more than the total number of remaining rival games. Team has sink capacity .
An integral flow of value saturates all game-node incoming arcs. Its two outgoing amounts allocate exactly wins to the two teams; the team-sink caps keep each rival strictly below . Conversely any successful completion of the league specifies such a flow. Integral capacities and augmentation give an integral maximum flow, so
after the initial negative-capacity rejection. Each actual game is counted once: making independent nodes for both ordered pairs would double-count it.
The network has nodes and arcs. If each input integer has at most bits, its derived capacities have bits. The shortest-path algorithm above uses arithmetic operations, each on integers of polynomial bit length. Therefore the decision problem belongs to P, even with part of the input. If is fixed, even the basic integral-augmentation bound is polynomial in . If varies, that particular bound is only pseudopolynomial; it is the shortest-path implementation that establishes polynomial bit complexity.