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

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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 NetworksUpdated 3 March 2008

- 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

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

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

- 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.

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