This presentation is the property of its rightful owner.
1 / 10

# EMIS 8374 The Maximum Flow Problem: Residual Flows and Networks Updated 3 March 2008 PowerPoint PPT Presentation

EMIS 8374 The Maximum Flow Problem: Residual Flows and Networks Updated 3 March 2008. The Residual Network. Given a flow x , the residual capacity r ij of arc ( i , j ) is the maximum additional flow that can be sent from i to j using arcs ( i , j ) and ( j , i )

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

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

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

### 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

### 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

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)

3

5

(2,2)

2

4

(2,4)

(4,5)

1

6

(2,4)

t

s

(5,6)

(7,7)

(xij, uij > 0)

i

j

2

2

4

2

1

2

2

4

1

6

t

s

2

1

7

5

3

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}

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

2

2

4

2

1

2

2

4

1

6

t

s

2

1

7

5

3

5

5

r[S, T] = 1

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