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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 $2$.

A flow network is a directed graph $G=(V,E)$ with distinct source and sink vertices $s,t$ and a nonnegative flow network edge capacity $c_{uv}$ on each directed edge $(u,v)$. In the finite setting here, every flow network edge capacity is a finite real number. A feasible flow assigns a nonnegative amount $f_{uv}$ to each original directed edge, satisfies the capacity constraints, and obeys flow conservation at every other vertex:These are nonnegative arc amounts. The antisymmetric convention in flow instead records net amounts ${\theta }_{uv}=f_{uv}-f_{vu}$, taking a missing arc amount as zero. The two conventions express the same flow conservation; keeping them distinct makes backward residual network edges easier to interpret.

Think of vertices as routers, directed edges as communication links, flow network edge capacities as available bandwidth, and flow as a steady transmission rate. An intermediate router can distribute incoming traffic among outgoing links, but flow conservation prevents it from creating traffic. The maximum flow problem asks how much total traffic can pass from $s$ to $t$, allowing several graph paths to share the work. The same flow network language models transportation and abstract resource allocation.^[ref]

Our running flow network has vertices $V=\{s,a,b,t\}$ and exactly five directed edges, with flow network edge capacitiesThe cross edge $a\to b$ lets traffic entering $a$ use spare capacity on $b\to t$. With this edge removed, the two separate graph paths can carry only $\min (3,2)+\min (2,3)=4$; including it will let the maximum flow reach $5$. Thus a link's usefulness depends on the rest of the flow network, rather than just its own flow network edge capacity.

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]

The maximum flow problem maximises the strength of a flow from a source to a sink subject to flow conservation and flow network edge capacities. For the nonnegative arc amounts used in flow network, its objective is net outflow from the source:The equality follows by summing flow conservation over all internal vertices. Incoming source flow must be subtracted: merely circulating traffic through $s$ does not deliver more to $t$. The max-flow min-cut theorem identifies the optimum with the least cut capacity of a cut of a flow network. A feasible flow and a cut of a flow network of equal value certify optimality without requiring a particular choice of augmenting paths.

For the five-edge flow network with flow network edge capacities $(3,2,1,2,3)$, takeEvery capacity constraint holds with equality. Flow conservation at $a$ is $3=1+2$, and at $b$ it is $2+1=3$. The strength of a flow is $|f|=3+2=5$, also $2+3=5$ entering $t$. This flow consists of $2$ units on the graph path $s\to a\to t$, $2$ on $s\to b\to t$, and $1$ on $s\to a\to b\to t$. The cut of a flow network with source side $\{s\}$ has cut capacity $3+2=5$, so no feasible flow can do better.

Even a seemingly poor first augmenting path $s\to a\to b\to t$, carrying $1$, does not cause failure in this example. It leaves residual network capacity $2$ along $s\to a\to t$ and $2$ along $s\to b\to t$. These two subsequent augmenting paths yield values $1$, $3$, and $5$, and produce exactly the flow above. Backward residual network edges become essential in the unit-capacity example under residual network, where this same first choice blocks every unused forward route.

Discrete assignment is another reason to study maximum flow. Put unit-capacity edges from $s$ to job vertices, from each job to eligible worker vertices, and from workers to $t$. An integral flow selects a matching in a graph: each job and worker participates at most once, and the strength of a flow counts assigned jobs. The integral max-flow theorem therefore turns maximum cardinality assignment into maximum flow. Minimizing assignment costs is the related assignment problem, modeled by minimum-cost flow.^[ref]

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.

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]