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

1. EMIS 8374 The Maximum Flow Problem: Residual Flows and NetworksUpdated 3 March 2008

2. The Residual Network • Given a flow x, the residual capacity rijof 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

3. 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

4. xij = 8, uij = 10 i j xji = 2, uji = 5 Residual Capacity Example 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

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

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

7. 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.

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

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

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