Flow network
Example

In Figure 1 you see a flow network with source labeled s, sink t, and four additional nodes. The flow and capacity is denoted f / c {\displaystyle f/c}. Notice how the network upholds the capacity constraint and flow conservation constraint. The total amount of flow from s to t is 5, which can be easily seen from the fact that the total outgoing flow from s is 5, which is also the incoming flow to t. By the skew symmetry constraint, from c to a is -2 because the flow from a to c is 2.

In Figure 2 you see the residual network for the same given flow. Notice how there is positive residual capacity on some edges where the original capacity is zero in Figure 1, for example for the edge (d, c) {\displaystyle (d,c)}. This network is not at maximum flow. There is available capacity along the paths (s, a, c, t) {\displaystyle (s,a,c,t)}, (s, a, b, d, t) {\displaystyle (s,a,b,d,t)} and (s, a, b, d, c, t) {\displaystyle (s,a,b,d,c,t)}, which are then the augmenting paths.

The bottleneck of the (s, a, c, t) {\displaystyle (s,a,c,t)} path is equal to min (c (s, a) − f (s, a), c (a, c) − f (a, c), c (c, t) − f (c, t)) {\displaystyle \min(c(s,a)-f(s,a),c(a,c)-f(a,c),c(c,t)-f(c,t))} = min (c f (s, a), c f (a, c), c f (c, t)) {\displaystyle =\min(c_{f}(s,a),c_{f}(a,c),c_{f}(c,t))} = min (5 − 3, 3 − 2, 2 − 1) {\displaystyle =\min(5-3,3-2,2-1)} = min (2, 1, 1) = 1 {\displaystyle =\min(2,1,1)=1}.
length_constraints:number_sentences[8]