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

2. 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

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. 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

5. 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

6. Residual Network for flow x2 2 4 2 1 2 4 2 1 6 t s 2 1 7 3 5 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}7 3 5 5 5
r[S, T] = = 2

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

10. 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