EMIS 8374 The Maximum Flow Problem: Residual Flows and Networks Updated 3 March 2008

The Residual Network Given a flow x, the residual capacity rij of arc (i, j) is the maximum additional flow that can be sent from i to j using arcs (i, j) and (j, i) rij = uij – xij + xji The residual network is G(x) = (N, A) with the capacity of arc (i, j) = rij

The Residual Network rij = (uij – xij) + xji (uij – xij) = unused capacity on (i, j) xji = flow from j to i that can be reduced to increase the net flow from i to j

Residual Capacity Example xij = 8, uij = 10 i j xji = 2, uji = 5 Net flow from i to j = 8 – 2 = 6 Net flow from j to i = 2 – 8 = -6 rij = (10 – 8) + 2= 4 i j rji = (5 – 2) + 8 = 11

Residual Network Example: Feasible Flow x (2,2) 2 4 (4,5) (2,4) 1 (2,4) 6 t s (5,6) (7,7) 3 5 (5,5) (xij, uij > 0) i j

Residual Network for flow x 2 2 4 2 1 2 4 2 1 6 t s 2 1 7 3 5 5 5 rij i j

Residual Capacity of an s-t Cut Consider an s-t cut [S, T] An arc (i, j) with i in S and j in T is called a forward arc An arc (i, j) with i in T and j in S is called a backwards arc Residual capacity r[S, T] = sum of the residual capacities of the forward arcs in the cut.

Residual Capacity of Example Cut 1: S = {1}, T = {2, 3, 4, 5, 6} 7 3 5 5 5 r[S, T] = 1 + 1 = 2

Residual Capacity of Example Cut 2: S = {1, 3, 5}, T = {2, 4, 6} 7 3 5 5 5 r[S, T] = 1

Residual Capacity of Example Cut 3: S = {1, 2, 3, 5}, T = {4, 6} 2 4 2 1 6 t s 2 1 7 3 5 5 5 r[S, T] = 0