Emis 8374 the maximum flow problem residual flows and networks updated 3 march 2008
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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 )

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

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Emis 8374 the maximum flow problem residual flows and networks updated 3 march 2008

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


The residual network

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 network1

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

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


Residual network example feasible flow x

(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


Residual network for flow x

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


Residual capacity of an s t cut

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

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


Residual capacity of example cut 2 s 1 3 5 t 2 4 6

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


Residual capacity of example cut 3 s 1 2 3 5 t 4 6

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


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