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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 $2$.

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 finite directed flow network with source $s$, sink $t$ and nonnegative finite flow network edge capacities $c_e$, the max-flow min-cut theorem statesHere feasible arc flows satisfy $0\le f_e\le c_e$ and flow conservation at every vertex except $s,t$. Their strength of a flow is the net outflow at $s$. A cut of a flow network has cut capacity equal to the sum of capacities on arcs directed from $S$ to $S^c$.

For any feasible flow and any such cut of a flow network, summing flow conservation over $S$ cancels internal arcs and givesThus every flow value is bounded by every cut capacity.

A maximum flow exists: the feasible arc-flow vectors form a nonempty closed subset of the finite-dimensional compact box ${\prod }_e[0,c_e]$, and flow value is continuous. Choose one, $f^{\ast }$. Its residual network has a forward arc of capacity $c_e-f_e^{\ast }$ and a reverse arc of capacity $f_e^{\ast }$ for each original arc. If a residual $s$–$t$ augmenting path existed, its smallest positive residual capacity would permit increasing forward flows and decreasing reverse flows along the path. This preserves all capacity bounds and internal flow conservation while strictly increasing $|f^{\ast }|$, a contradiction.

Let $S$ be the vertices reachable from $s$ through positive residual arcs. Then $t\notin S$. Every original arc leaving $S$ is saturated, since otherwise its endpoint would also be reachable. Every original arc entering $S$ carries zero flow, since a positive flow would create a residual reverse arc leaving $S$. ThereforeTogether with the universal upper bound, this proves both optimality claims and the theorem. This existence argument works for real capacities; it does not assume that arbitrary choices in the Ford-Fulkerson algorithm terminate. For integer capacities, positive augmentations are integral and increase the flow value by at least one, so the algorithm does terminate.

A finite flow network is a directed graph with distinguished source $s$, sink $t$, and nonnegative finite flow network edge capacities $c_{ij}$ on its directed edges. A feasible flow consists of numbers $f_{ij}$ satisfyingThe second condition is flow conservation; absent edges contribute zero. The strength of a flow is its net outflow from the source,which equals net inflow to the sink by summing flow conservation over the other vertices. The maximum flow problem isNo assumption that a capacity-saturating flow at the source is feasible downstream is made. On a finite network a maximum exists, because the constraints define a nonempty compact set of edge flows and the objective is continuous.

The Ford-Fulkerson algorithm starts with zero flow. For each original edge $i\to j$, the residual network has a forward edge with residual capacity $c_{ij}-f_{ij}$ and a reverse edge with residual capacity $f_{ij}$. Distinguish these labelled residual edges if the original graph has antiparallel edges. While there is a source-to-sink augmenting path with positive residual capacities, increase the flow by the minimum residual capacity along it, adding on forward edges and subtracting on reverse edges. This preserves capacity bounds and flow conservation and increases the strength of a flow by that positive bottleneck.

A cut of a flow network is a partition $(S,V\setminus S)$ with $s\in S$, $t\notin S$, of capacity $c(S)={\sum }_{i\in S,j\notin S}{c}_{ij}$. Summing flow conservation gives $|f|=f(S,V\setminus S)-f(V\setminus S,S)\le c(S)$. The max-flow min-cut theorem statesWhen no augmenting path remains, take $S$ to be the vertices reachable from $s$ in the residual network. Every original outgoing edge of this cut of a flow network is saturated and every incoming edge has zero flow; otherwise a positive residual edge would leave $S$. Thus $|f|=c(S)$, certifying optimality and the theorem. For general finite real capacities, one can also apply this argument to an existing maximum: it cannot admit an augmenting path.

With integer capacities, each augmentation increases the value by at least one, so the Ford-Fulkerson algorithm terminates. Rational capacities can be scaled to integers. Arbitrary path choices with irrational capacities need not terminate; the existence theorem does not justify claiming finite termination in that generality. The given network has integer capacities, so this caveat does not obstruct its computation.

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]

A unit flow from $a$ to $b$ is an antisymmetric flow with net outward strength of a flow one at $a$, net inward strength one at $b$, and zero divergence of a flow elsewhere. For a unit flow to infinity, only the finite source has nonzero divergence.