Max-flow min-cut theorem (${\max }_f|f|={\min }_{S:s\in S,t\notin S}c(S,V\setminus S)$)

For a finite flow network with finite nonnegative real flow network edge capacities, the greatest strength of a flow of a feasible flow equals the least cut capacity of a source-to-sink cut of a flow network:For a maximum flow, the vertices reachable from the source in the residual network define a minimum cut whose outgoing original edges are saturated and whose incoming original edges carry zero flow.^[ref]

Why is a cut of a flow network an upper bound? Sum flow conservation on its source side $S$. Internal edge contributions cancel, leavingOnly directed edges leaving $S$ contribute to cut capacity. Directed edges entering $S$ are subtracted in net flow, so adding their capacities to the upper bound would lose the intended bottleneck. This inequality is weak duality; the theorem says the bound is attained, an instance of strong duality. Its beauty is that a routing plan and a separating boundary independently witness each other's optimality.

In the running flow network, the two internal vertices give four possible source sides. For $S=\{s\}$, the cut capacity is $c_{sa}+c_{sb}=3+2=5$. For $S=\{s,a\}$ it is $c_{sb}+c_{ab}+c_{at}=2+1+2=5$. For $S=\{s,b\}$ it is $c_{sa}+c_{bt}=3+3=6$; the incoming edge $a\to b$ contributes nothing to that cut capacity. Finally, $S=\{s,a,b\}$ gives $c_{at}+c_{bt}=2+3=5$. Thus the minimum cut has cut capacity $5$, matching the explicitly computed maximum flow.

Ford and Fulkerson's 1956 paper opens with a railway-capacity question attributed there to T. Harris and proves a minimal-cut theorem as Theorem 1. Its original formulation uses graph paths carrying nonnegative amounts; the residual reachability certificate for maximum flow below gives the finite directed flow network proof used here.^[ref]

In a finite flow network, a maximum flow exists because the feasible flow set contains the zero flow and is a compact set: the capacity constraints bound every coordinate, and they and flow conservation define a closed set. The strength of a flow is a continuous function, so the extreme value theorem gives an attained maximum. An augmenting path in the residual network would permit positive augmentation, contradicting maximality. The source-reachable vertices in the residual network therefore define a cut of a flow network whose outgoing original edges are saturated and incoming original edges carry zero flow. Flow conservation makes the strength of a flow equal to this cut capacity, proving the max-flow min-cut theorem.

Existence and strong duality for finite real flow network edge capacities do not imply finite termination for every choice of augmenting paths in the Ford-Fulkerson algorithm. With irrational capacities, some choices can continue forever. With integer capacities, Integrality of the Ford-Fulkerson algorithm instead forces finite termination; rational capacities reduce to that case by a common denominator. Choosing shortest augmenting paths as in the Edmonds–Karp algorithm avoids dependence on capacity magnitudes.^[ref]

If every directed edge capacity increases by $\epsilon$, a cut of a flow network with $m(S)$ outgoing edges gains $\epsilon m(S)$. By the max-flow min-cut theorem, the new maximum flow value isFor a finite flow network, this is a nondecreasing concave piecewise-linear function, and its active cut can change as $\epsilon$ varies. Two unit-capacity edge-disjoint two-edge source-to-sink paths gain $2\epsilon$, refuting an invariably exact $\epsilon$ increase. If there is a directed source-to-sink path, every cut has at least one outgoing edge and the increase is at least $\epsilon$; without a path it can be zero.

A flow network with integer capacities has a maximum flow whose value on every edge is an integer. The augmenting-path algorithm proves this because every augmentation preserves integrality.

In a finite bipartite graph, the maximum size of a matching equals the minimum size of a vertex cover. It follows by representing matching as an integral flow and vertex covers as finite-capacity cuts.

For a feasible flow $f$, each original directed edge $u\to v$ supplies a forward residual network edge of capacity $c_{uv}-f_{uv}$ and a backward one $v\to u$ of capacity $f_{uv}$; zero-capacity residual edges may be omitted. The forward edge permits an increase, while the backward edge permits cancellation of existing flow. In particular, backward residual capacity is the amount already sent, rather than unused original capacity. If both original orientations exist, keep their residual contributions separate, or add capacities for contributions with the same direction. An augmenting path uses these permitted changes while preserving flow conservation and the capacity constraints.

To see why cancellation matters, give all five directed edges $s\to a$, $s\to b$, $a\to b$, $a\to t$, $b\to t$ unit flow network edge capacities. Starting from zero flow, an augmenting path $s\to a\to b\to t$ carrying $1$ yieldsThe remaining unused original edges are $s\to b$ and $a\to t$, which do not form a source-to-sink graph path. However, the backward residual edge $b\to a$ has capacity $f_{ab}=1$. Consequently the residual network contains the augmenting path $s\to b\to a\to t$, with residual capacities $1,1,1$.

Augment by $1$: increase $f_{sb}$ and $f_{at}$ from $0$ to $1$, and decrease $f_{ab}$ from $1$ to $0$. The result isNow flow conservation is $1=0+1$ at $a$ and $1+0=1$ at $b$. The final flow sends one unit along each of $s\to a\to t$ and $s\to b\to t$. The cut of a flow network $S=\{s\}$ has cut capacity $1+1=2$, proving this is a maximum flow. Traversing $b\to a$ in the residual network revised the earlier routing decision; it did not send actual traffic along a nonexistent original edge.

For arc capacities $(c_{12},c_{13},c_{24},c_{34},c_{35},c_{46},c_{56})=(3,3,2,2,3,3,2)$, the feasible flow $(1,3,1,2,1,3,1)$ has value four. The displayed residual path has bottleneck one and traverses $4\to 3$ opposite to the original arc $3\to 4$. Augmentation cancels one unit there and gives $(2,3,2,1,2,3,2)$ of value five. The sink cut has capacity five, proving optimality. Reverse residual arcs are what permit rerouting instead of only adding flow along unused forward arcs.

For an undirected capacitated network, write the net edge flow as $y_{ij}=-y_{ji}$, and the directed residual capacity as $c_{ij}-y_{ij}$. On every source–sink cut, residual capacity is the original cut capacity minus the current net flow value $f$. Taking the minimum over cuts and applying the max-flow min-cut theorem to each network gives the displayed remaining-flow identity.

An augmenting path is a source-to-sink graph path $P$ of positive capacities in the residual network. Its bottleneck is $\Delta ={\min }_{e\in P}{c}_f(e)>0$. Add $\Delta$ to original flow on forward residual edges and subtract $\Delta$ on backward residual edges. Each internal vertex still satisfies flow conservation, and the bottleneck guarantees every capacity constraint. The resulting feasible flow has strength of a flow $|f|+\Delta$. In the unit-capacity residual network example, $\Delta =1$, and the subtraction on $a\to b$ is precisely what allows the value to rise from $1$ to $2$.

An augmenting path maximizing the possible increase in flow, namely its smallest residual network edge capacity. If the remaining maximum flow is $F>0$ and the original undirected network has $m$ edges, a path has bottleneck at least $F/m$. Otherwise the vertices reachable through arcs of capacity at least $F/m$ define a cut of capacity strictly less than $F$, contradicting the max-flow min-cut theorem. Only one directed arc per original crossing edge contributes to that cut.

Starting from zero flow, a widest augmenting path increases the current flow by at least $F_k/m$, where $F_k$ is its remaining optimality gap. Therefore $F_{k+1}\le (1-1/m)F_k$. Integer capacities keep this gap integer, so a gap below one is zero. The method takes $O(m\log f^{\ast })$ augmentations for $f^{\ast }\ge 2$, or uniformly $O(m\log (1+f^{\ast }))$; zero maximum flow requires no augmentation.

The Ford-Fulkerson algorithm repeatedly augments a feasible flow along an augmenting path. When no such path remains, the source-reachable vertices in the residual network define a cut of capacity equal to the flow value, proving optimality by the max-flow min-cut theorem.

Starting from the zero flow with integer capacities, every residual capacity and every augmenting bottleneck is an integer. Each augmentation therefore preserves integer edge flows and raises the flow value by at least one. Since the value is bounded by the total capacity leaving the source, the algorithm terminates after finitely many augmentations with an integral maximum flow.

If one edge leaving the source has capacity $x$, every cut capacity is affine in $x$ with coefficient zero or one. In the 2024 Cambridge Part IB example, cuts of capacities $x+5$ and $14$ are both sharp, giving