Maximum flow problems
This presentation is the property of its rightful owner.
Sponsored Links
1 / 66

Maximum flow problems PowerPoint PPT Presentation


  • 186 Views
  • Uploaded on
  • Presentation posted in: General

Maximum flow problems. Maximum flow problems. Introduction of: network, max-flow problem capacity, flow Ford-Fulkerson method pseudo code, residual networks, augmenting paths cuts of networks max-flow min-cut theorem example of an execution analysis, running time,

Download Presentation

Maximum flow problems

An Image/Link below is provided (as is) to download presentation

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 - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Maximum flow problems

Maximum flow problems


Maximum flow problems1

Maximum flow problems

  • Introduction of:

  • network, max-flow problem

  • capacity, flow

  • Ford-Fulkerson method

  • pseudo code, residual networks, augmenting paths

  • cuts of networks

  • max-flow min-cut theorem

  • example of an execution

  • analysis, running time,

  • variations of the max-flow problem


Introduction flow network

Introduction – flow network

  • Practical examples of a flow network

  • - liquids flowing through pipes

  • parts through assembly lines

  • current through electrical network

  • information through communication network

  • goods transported on the road…


Introduction flow network1

Introduction – flow network

Representation

Example: oil pipeline

Flow network: directed graph G=(V,E)

v3

v3

v1

v1

S

S

t

t

source

v2

v4

v2

v4

sink


Introduction max flow problem

Introduction – max-flow problem

Representation

Example: oil pipeline

Flow network: directed graph G=(V,E)

v3

v3

v1

v1

S

S

t

t

source

v2

v4

v2

v4

sink

Informal definition of the max-flow problem:

What is the greatest rate at which material can be shipped from the source to the sink without violating any capacity contraints?


Introduction capacity

3

v3

v1

6

8

3

S

t

3

6

8

v2

v4

6

c(u,v)=12

6

c(u,v)=6

u

u

v

v

Introduction - capacity

Representation

Example: oil pipeline

Flow network: directed graph G=(V,E)

v3

v1

S

t

v2

v4

12

Big pipe

Small pipe


Introduction capacity1

Introduction - capacity

Representation

Example: oil pipeline

Flow network: directed graph G=(V,E)

3

v3

v3

v1

v1

6

8

3

S

S

t

t

3

6

8

v2

v4

v2

v4

6

If (u,v)  E  c(u,v) = 0

6

v2

v4

0

0

v4

v3

0


Introduction flow

3

v3

v1

6

8

3

S

t

3

6

8

v2

v4

6

f(u,v)=6

f(u,v)=6

u

u

v

v

Introduction – flow

Representation

Example: oil pipeline

Flow network: directed graph G=(V,E)

v3

v1

S

t

v2

v4

6/12

Flow below capacity

6/6

Maximum flow


Introduction flow1

3

v3

v1

6

8

3

S

t

3

6

8

v2

v4

6

Introduction – flow

Representation

Example: oil pipeline

Flow network: directed graph G=(V,E)

v3

v1

S

t

v2

v4


Introduction flow2

3

v3

v1

6

8

3

S

t

3

6/6

6/8

v2

v4

6/6

Introduction – flow

Representation

Example: oil pipeline

Flow network: directed graph G=(V,E)

v3

v1

S

t

v2

v4


Introduction flow3

3

v3

v1

6

8

3

S

t

3

6/6

6/8

v2

v4

6/6

Introduction – flow

Representation

Example: oil pipeline

Flow network: directed graph G=(V,E)

v3

v1

S

t

v2

v4


Introduction flow4

3/3

v3

v1

3/6

3/8

3

S

t

3

6/6

6/8

v2

v4

6/6

Introduction – flow

Representation

Example: oil pipeline

Flow network: directed graph G=(V,E)

v3

v1

S

t

v2

v4


Introduction flow5

3/3

v3

v1

3/6

3/8

3

S

t

3

6/6

6/8

v2

v4

6/6

Introduction – flow

Representation

Example: oil pipeline

Flow network: directed graph G=(V,E)

v3

v1

S

t

v2

v4


Introduction flow6

3/3

v3

v1

3/6

5/8

2/3

S

t

3

6/6

8/8

v2

v4

6/6

Introduction – flow

Representation

Example: oil pipeline

Flow network: directed graph G=(V,E)

v3

v1

S

t

v2

v4


Introduction cancellation

3/3

v3

v1

3/6

5/8

2/3

S

t

3

6/6

8/8

v2

v4

6/6

Introduction – cancellation

Representation

Example: oil pipeline

Flow network: directed graph G=(V,E)

v3

v1

S

t

v2

v4


Introduction cancellation1

3/3

v3

v1

4/6

6/8

3/3

S

t

1/3

6/6

8/8

v2

v4

5/6

Introduction – cancellation

Representation

Example: oil pipeline

Flow network: directed graph G=(V,E)

v3

v1

S

t

v2

v4

u

u

u

u

8/10

3/4

5/10

4

10

4

8/10

4

v

v

v

v


Flow properties

Flow properties


Flow properties1

Flow properties

Flow network G = (V,E)

12/12

v3

v1

Note:

by skew symmetry

f (v3,v1) = - 12

11/16

15/20

10

1/4

7/7

S

t

4/9

4/4

8/13

v2

v4

11/14


Net flow and value of a flow

Net flow and value of a flow

u

u

f(u,v) = 5

f(v,u) = -5

8/10

3/4

5/10

4

v

v

3/3

v3

v1

4/6

6/8

3/3

S

t

1/3

6/6

8/8

v2

v4

5/6


The max flow problem

The max-flow problem

Informal definition of the max-flow problem:

What is the greatest rate at which material can be shipped from the source to the sink without violating any capacity contraints?

Formal definition of the max-flow problem:

The max-flow problem is to find a valid flow for a given weighted directed graph G, that has the maximum value over all valid flows.


The ford fulkerson method a way how to find the max flow

The Ford-Fulkerson methoda way how to find the max-flow

This method contains 3 important ideas:

  1) residual networks

    2) augmenting paths

    3) cuts of flow networks


Ford fulkerson pseudo code

Ford-Fulkerson – pseudo code

1 initialize flow f to 0

2 while there exits an augmenting path p

3do augment flow f along p

4 return f


Ford fulkerson residual networks

5

11

5

8

cf(u,v) = c(u,v) – f(u,v)

Ford Fulkerson – residual networks

The residual network Gf of a given flow network G with a valid flow f consists of the same vertices v  V as in G which are linked with residual edges (u,v)  Ef that can admit more strictly positive net flow.

The residual capacity cf represents the weight of each edge Ef and is the amount of additional net flow f(u,v) before exceeding the capacity c(u,v)

Flow network G = (V,E)

residual network Gf = (V,Ef)

12/12

v3

v3

v1

v1

11/16

15/20

10

1/4

7/7

S

t

S

t

4/9

4/4

8/13

v2

v4

v2

v4

11/14


Ford fulkerson residual networks1

cf(u,v) = c(u,v) – f(u,v)

Ford Fulkerson – residual networks

The residual network Gf of a given flow network G with a valid flow f consists of the same vertices v  V as in G which are linked with residual edges (u,v)  Ef that can admit more strictly positive net flow.

The residual capacity cf represents the weight of each edge Ef and is the amount of additional net flow f(u,v) before exceeding the capacity c(u,v)

Flow network G = (V,E)

residual network Gf = (V,Ef)

12

12/12

v3

v3

v1

v1

5

5

11/16

15/20

4

11

10

1/4

11

7/7

7

15

3

S

t

S

t

4/9

5

5

4/4

4

3

8/13

8

v2

v4

v2

v4

11/14

11


Ford fulkerson augmenting paths

Ford Fulkerson – augmenting paths

Definition: An augmenting path p is a simple (free of any cycle) path from s to t in the residual network Gf

Residual capacity of p

cf(p) = min{cf (u,v): (u,v) is on p}

Flow network G = (V,E)

residual network Gf = (V,Ef)

12

12/12

v3

v3

v1

v1

5

5

11/16

15/20

4

11

10

1/4

11

7/7

7

15

3

S

t

S

t

4/9

5

5

4/4

4

3

8/13

8

v2

v4

v2

v4

11/14

11


Ford fulkerson augmenting paths1

Ford Fulkerson – augmenting paths

Definition: An augmenting path p is a simple (free of any cycle) path from s to t in the residual network Gf

Residual capacity of p

cf(p) = min{cf (u,v): (u,v) is on p}

Flow network G = (V,E)

residual network Gf = (V,Ef)

12

12/12

v3

v3

v1

v1

5

5

11/16

15/20

4

11

10

1/4

11

7/7

7

15

3

S

t

S

t

4/9

5

5

4/4

4

3

8/13

8

v2

v4

v2

v4

11/14

11

Augmenting path


Ford fulkerson augmenting paths2

Ford Fulkerson – augmenting paths

We define a flow: fp: V x V  R such as:

cf(p) if (u,v) is on p

fp(u,v) = - cf(p) if (v,u) is on p

0 otherwise

Flow network G = (V,E)

residual network Gf = (V,Ef)

12

12/12

v3

v3

v1

v1

5

5

11/16

15/20

4

11

10

1/4

11

7/7

7

15

3

S

t

S

t

4/9

5

5

4/4

4

3

8/13

8

v2

v4

v2

v4

11/14

11


Ford fulkerson augmenting paths3

cf(p) if (u,v) is on p

fp(u,v) = - cf(p) if (v,u) is on p

0 otherwise

Ford Fulkerson – augmenting paths

We define a flow: fp: V x V  R such as:

Flow network G = (V,E)

residual network Gf = (V,Ef)

12

12/12

v3

v3

v1

v1

5

4/5

11/16

15/20

4/4

11

10

1/4

11

7/7

7

-4/15

3

S

t

S

t

4/9

-4/5

4/5

4/4

4

3

8/13

-4/8

v2

v4

v2

v4

11/14

11

Our virtual flow fp along the augmenting path p in Gf


Ford fulkerson augmenting the flow

cf(p) if (u,v) is on p

fp(u,v) = - cf(p) if (v,u) is on p

0 otherwise

Ford Fulkerson – augmenting the flow

We define a flow: fp: V x V  R such as:

Flow network G = (V,E)

residual network Gf = (V,Ef)

12

12/12

v3

v3

v1

v1

5

4/5

11/16

19/20

4/4

11

10

1/4

11

7/7

7

15

3

S

t

S

t

0/9

5

4/5

4/4

4

3

12/13

8

v2

v4

v2

v4

11/14

11

New flow:

f´: V x V  R : f´=f + fp

Our virtual flow fp along the augmenting path p in Gf


Ford fulkerson new valid flow proof of capacity constraint

cf(p) if (u,v) is on p

fp(u,v) = - cf(p) if (v,u) is on p

0 otherwise

Ford Fulkerson – new valid flowproof of capacity constraint

Lemma:

f´ : V x V  R : f´ = f + fp in G

cf(p) = min{cf (u,v): (u,v) is on p}

Capacity constraint:

For all u,v  V, we require f(u,v) < c(u,v)

cf(u,v) = c(u,v) – f(u,v)

Proof:

fp (u ,v) < cf (u ,v) = c (u ,v) – f (u ,v)

 (f + fp) (u ,v) = f (u ,v) + fp (u ,v) < c (u ,v)


Ford fulkerson new valid flow proof of skew symmetry

Ford Fulkerson – new valid flowproof of Skew symmetry

Lemma:

f´ : V x V  R : f´ = f + fp in G

Skew symmetry:

For all u,v  V, we require f(u,v) = - f(v,u)

Proof:

(f + fp)(u ,v) = f (u ,v) + fp (u ,v) = - f (v ,u) – fp (v ,u)

= - (f (v ,u) + fp (v ,u)) = - (f + fp) (v ,u)


Ford fulkerson new valid flow proof of flow conservation

Ford Fulkerson – new valid flowproof of flow conservation

Lemma:

f´ : V x V  R : f´ = f + fp in G

Flow conservation:

For all u  V \ {s,t} :  f(u,v) = 0

v  V

Proof:

u  V – {s ,t}  (f + fp) (u ,v) =  (f(u ,v) + fp (u ,v))

=  f (u ,v) +  fp (u ,v) = 0 + 0 = 0

v  V

v  V

v  V

v  V


Ford fulkerson new valid flow

Ford Fulkerson – new valid flow

Lemma:

| (f + fp) | = | f | + | fp |

Value of a Flow f:

 Def:

|f| =  f(s,v)

v  V

Proof:

| (f + fp) | =  (f + fp) (s ,v) =  (f (s ,v) + fp (s ,v))

=  f (s ,v) +  fp (s ,v) = | f | + | fp |

v  V

v  V

v  V

v  V


Ford fulkerson new valid flow1

Ford Fulkerson – new valid flow

Lemma:

f´ : V x V  R : f´ = f + fp in G

| (f + fp) | = | f | + | fp | > | f |

Lemma shows:

if an augmenting path can be found then the above flow augmentation will result in a flow improvement.

Question: If we cannot find any more an augmenting path is our flow then maximum?

Idea: The flow in G is maximum  the residual Gf contains no augmenting path.


Ford fulkerson cuts of flow networks

Practical example

In the example:

S = {s,v1,v2), T = {v3,v4,t}

Net flow f(S ,T) = f(v1,v3) + f(v2,v4) + f(v2,v3)

= 12 + 11 + (-0) = 23

Capacity c(S,T) = c(v1,v3) + c(v2,v4)

= 12 + 14 = 26

12/12

v3

v1

11/16

19/20

10

1/4

7/7

S

t

0/9

4/4

12/13

v2

v4

11/14

S

T

Implicit summation notation: f (S, T) =  f (u, v)

u  S

v  T

Ford Fulkerson – cuts of flow networks

New notion: cut (S,T) of a flow network

A cut (S,T) of a flow network G=(V,E) is a partiton of V into S and T = V \ S such that s  S and t  T.


Ford fulkerson cuts of flow networks1

Ford Fulkerson – cuts of flow networks

Lemma:

the value of a flow in a network is the net flow across any cut of the network

f (S ,T) = | f |

12/12

v3

v1

11/16

19/20

10

1/4

7/7

S

t

0/9

4/4

12/13

v2

v4

11/14

proof


Ford fulkerson cuts of flow networks2

Ford Fulkerson – cuts of flow networks

Assumption:

The value of any flow f in a flow network G is bounded from above by the capacity of any cut of G

Lemma:

| f | <c (S, T)

12/12

| f | = f (S, T)

=  f (u, v)

< c (u, v)

= c (S, T)

v3

v1

11/16

19/20

10

1/4

7/7

S

t

0/9

u S

v T

4/4

u S

v T

12/13

v2

v4

11/14


F fulkerson max flow min cut theorem

F. Fulkerson: Max-flow min-cut theorem

  • If f is a flow in a flow network G = (V,E) with source s and sink t, then the following conditions are equivalent:

  • f is a maximum flow in G.

  • The residual network Gf contains no augmenting paths.

  • | f | = c (S, T) for some cut (S, T) of G.

  • proof:

  •  (2):

  • We assume for the sake of contradiction that f is a maximum flow in G but that there still exists an augmenting path p in Gf.

  • Then as we know from above, we can augment the flow in G according to the formula: f´= f + fp. That would create a flow f´that is strictly greater than the former flow f which is in contradiction to our assumption that f is a maximum flow.


F fulkerson max flow min cut theorem1

residual network Gf

3/3

3

v3

v3

v1

v1

4/6

2

6/8

3/3

2

3

4

6

1

S

S

t

t

1/3

2

6/6

6

5

8/8

8

v2

v4

v2

v4

5/6

1

F. Fulkerson: Max-flow min-cut theorem

  • If f is a flow in a flow network G = (V,E) with source s and sink t, then the following conditions are equivalent:

  • f is a maximum flow in G.

  • The residual network Gf contains no augmenting paths.

  • | f | = c (S, T) for some cut (S, T) of G.

proof:

(2)  (3):

original flow network G


F fulkerson max flow min cut theorem2

F. Fulkerson: Max-flow min-cut theorem

  • If f is a flow in a flow network G = (V,E) with source s and sink t, then the following conditions are equivalent:

  • f is a maximum flow in G.

  • The residual network Gf contains no augmenting paths.

  • | f | = c (S, T) for some cut (S, T) of G.

proof:

(2)  (3): Define

S = {v V |  path p from s to v in Gf }

T = V \ S (note t  S according to (2))

residual network Gf

3

v3

v1

2

2

3

4

6

1

S

t

2

6

5

8

v2

v4

1


F fulkerson max flow min cut theorem3

3/3

v3

v1

4/6

6/8

3/3

S

t

1/3

6/6

8/8

v2

v4

5/6

F. Fulkerson: Max-flow min-cut theorem

  • If f is a flow in a flow network G = (V,E) with source s and sink t, then the following conditions are equivalent:

  • f is a maximum flow in G.

  • The residual network Gf contains no augmenting paths.

  • | f | = c (S, T) for some cut (S, T) of G.

  • proof:

  • (2)  (3): Define

  • S = {v V |  path p from s to v in Gf }

  • T = V \ S (note t  S according to (2))

  • for  u  S, v  T: f (u, v) = c (u, v) (otherwise (u, v)  Ef and v  S)

  • | f | = f (S, T) = c (S, T)

original network G

1


F fulkerson max flow min cut theorem4

F. Fulkerson: Max-flow min-cut theorem

  • If f is a fow in a flow network G = (V,E) with source s and sink t, then the following conditions are equivalent:

  • f is a maximum flow in G.

  • The residual network Gf contains no augmenting paths.

  • | f | = c (S, T) for some cut (S, T) of G.

proof:

(3)  (1): as proofed before | f | = f (S, T) < c (S, T)

the statement of (3) : | f | = c (S, T) implies that f is a maximum flow


The basic ford fulkerson algorithm

The basic Ford Fulkerson algorithm


The basic ford fulkerson algorithm example of an execution

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

16

20

10

4

7

S

t

9

4

13

v2

v4

14


The basic ford fulkerson algorithm example of an execution1

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

0/12

v3

v1

0/16

0/20

0/10

0/4

7

S

t

0/9

0/4

0/13

v2

v4

0/14


The basic ford fulkerson algorithm example of an execution2

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

16

20

10

4

7

S

t

9

4

13

v2

v4

14


The basic ford fulkerson algorithm example of an execution3

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

16

20

10

4

7

S

t

9

4

13

v2

v4

14

temporary variable:

cf (p) = 12


The basic ford fulkerson algorithm example of an execution4

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

16

20

10

4

7

S

t

9

4

13

v2

v4

14

new flow network G

12/12

v3

v1

12/16

temporary variable:

cf (p) = 12

12/20

10

4

7

S

t

9

4

13

v2

v4

14


The basic ford fulkerson algorithm example of an execution5

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

4

8

12

12

10

4

7

S

t

9

4

13

v2

v4

14

new flow network G

12/12

v3

v1

12/16

12/20

10

4

7

S

t

9

4

13

v2

v4

14


The basic ford fulkerson algorithm example of an execution6

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

4

8

12

12

10

4

7

S

t

9

4

13

v2

v4

14

new flow network G

12/12

v3

v1

12/16

12/20

10

4

7

S

t

9

4

13

v2

v4

14


The basic ford fulkerson algorithm example of an execution7

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

4

8

12

12

10

4

7

S

t

9

4

13

v2

v4

14

new flow network G

12/12

v3

v1

12/16

12/20

10

4

7

S

t

9

4

13

v2

v4

14


The basic ford fulkerson algorithm example of an execution8

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

4

8

12

12

10

4

7

S

t

9

4

13

v2

v4

14

new flow network G

12/12

v3

v1

12/16

12/20

10

4

7

S

t

9

temporary variable:

cf (p) = 4

4

13

v2

v4

14


The basic ford fulkerson algorithm example of an execution9

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

4

8

12

12

10

4

7

S

t

9

4

13

v2

v4

14

new flow network G

12/12

v3

v1

12/16

temporary variable:

cf (p) = 4

12/20

10

4

7

S

t

9

4/4

4/13

v2

v4

4/14


The basic ford fulkerson algorithm example of an execution10

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

4

8

12

12

10

4

7

S

t

9

4

4

4

9

v2

v4

10

new flow network G

12/12

v3

v1

12/16

12/20

10

4

7

S

t

9

4/4

4/13

v2

v4

4/14


The basic ford fulkerson algorithm example of an execution11

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

4

8

12

12

10

4

7

S

t

9

4

4

4

9

v2

v4

10

new flow network G

12/12

v3

v1

12/16

12/20

10

4

7

S

t

9

4/4

4/13

v2

v4

4/14


The basic ford fulkerson algorithm example of an execution12

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

4

8

12

12

10

4

7

S

t

9

4

4

4

9

v2

v4

10

new flow network G

12/12

v3

v1

12/16

12/20

10

4

7

S

t

9

4/4

4/13

v2

v4

4/14


The basic ford fulkerson algorithm example of an execution13

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

4

8

12

12

10

4

7

S

t

9

4

4

4

9

v2

v4

10

new flow network G

12/12

v3

v1

12/16

temporary variable:

cf (p) = 7

12/20

10

4

7

S

t

9

4/4

4/13

v2

v4

4/14


The basic ford fulkerson algorithm example of an execution14

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

4

8

12

12

10

4

7

S

t

9

4

4

4

9

v2

v4

10

new flow network G

12/12

v3

v1

12/16

temporary variable:

cf (p) = 7

19/20

10

4

7/7

S

t

9

4/4

11/13

v2

v4

11/14


The basic ford fulkerson algorithm example of an execution15

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

4

1

12

19

10

4

7

S

t

9

11

4

11

2

v2

v4

3

new flow network G

12/12

v3

v1

12/16

19/20

10

4

7/7

S

t

9

4/4

11/13

v2

v4

11/14


The basic ford fulkerson algorithm example of an execution16

The basic Ford Fulkerson algorithmexample of an execution

(residual) network Gf

12

v3

v1

4

1

12

19

10

4

7

S

t

9

11

4

11

2

v2

v4

3

new flow network G

12/12

v3

v1

12/16

19/20

Finally we have:

| f | = f (s, V) = 23

10

4

7/7

S

t

9

4/4

11/13

v2

v4

11/14


Analysis of the ford fulkerson algorithm running time

Analysis of the Ford Fulkerson algorithmRunning time

The running time depends on how the augmenting path p in line 4 is determined.


Analysis of the ford fulkerson algorithm running time arbitrary choice of p

Analysis of the Ford Fulkerson algorithmRunning time (arbitrary choice of p)

O(|E|)

O(|E|)

O(|E| |fmax|)

O(|E|)

running time: O ( |E| |fmax| )

with fmax as maximum flow

(1) The augmenting path is chosen arbitrarily and all capacities are integers


Analysis of the ford fulkerson algorithm running time arbitrary choice of p1

Analysis of the Ford Fulkerson algorithmRunning time (arbitrary choice of p)

Consequencies of an arbitrarily choice:

Example if |f*| is large:

v1

1,000,000

1,000,000

s

1

t

1,000,000

1,000,000

v2

running time: O ( |E| |fmax| )

with fmax as maximum flow

(1) The augmenting path is chosen arbitrarily and all capacities are integers


Analysis of the ford fulkerson algorithm running time arbitrary choice of p2

Analysis of the Ford Fulkerson algorithmRunning time (arbitrary choice of p)

Consequencies of an arbitrarily choice:

Example if |f*| is large:

residual network Gf

v1

v1

1 / 1,000,000

1,000,000

999,999

1,000,000

1

1 / 1

s

t

s

1

t

1

1,000,000

1,000,000

1 / 1,000,000

999,999

v2

v2

running time: O ( |E| |fmax| )

with fmax as maximum flow

(1) The augmenting path is chosen arbitrarily and all capacities are integers


Analysis of the ford fulkerson algorithm running time arbitrary choice of p3

Analysis of the Ford Fulkerson algorithmRunning time (arbitrary choice of p)

Consequencies of an arbitrarily choice:

Example if |f*| is large:

residual network Gf

v1

v1

1/ 1,000,000

1 / 1,000,000

999,999

999,999

1

1

1

s

t

s

1

t

1

1

999,999

1 / 1,000,000

1/ 1,000,000

999,999

v2

v2

running time: O ( |E| |fmax| )

with fmax as maximum flow

(1) The augmenting path is chosen arbitrarily and all capacities are integers


Analysis of the ford fulkerson algorithm running time with edmonds karp algorithm

Analysis of the Ford Fulkerson algorithmRunning time with Edmonds-Karp algorithm

  • Informal idea of the proof:

  • for all vertices v  V\{s,t}:

  • the shortest path distance f(s,v) in Gf increases monotonically with each flow augmentation

  • .f(s,v) < f´(s,v)

4

v3

u1

6

6

2

2

S

t

4

4

u2

v4

4

f(s,u2) = 1

running time: O ( |V| |E|² )

(2) Edmonds-Karp algorithm

augmenting path is found by breath-first search and has to be a shortest path from p to t


  • Login