Download presentation

Presentation is loading. Please wait.

Published byGuadalupe Modrell Modified over 2 years ago

1
Max Flow Problem Given network N=(V,A), two nodes s,t of V, and capacities on the arcs: uij is the capacity on arc (i,j). Find non-negative flow fij for each arc (i,j) such that for each node, except s and t, the total incoming flow is equal to the total outgoing flow (flow conservation), such that fij is at most the capacity uij on arc (i,j); The total amount of flow going out of s (which is equal to the total amount of flow coming into t ) is maximized

2
**the black numbers next to an arc is its capacity**

1 3 2 4 3 2 3 1 t 1 1 s 2 4 2 4 4 the black numbers next to an arc is its capacity

3
**Set costs all other arcs at 0 The minimum cost flow circulation (Af=0) **

1 3 2 4 3 2 3 1 t 1 1 s 2 4 2 4 4 Cts= -1 Set costs all other arcs at 0 The minimum cost flow circulation (Af=0) maximises the s-t flow

4
**the black numbers next to an arc is its capacity **

1 3 2 4 3 2 3 1 t 1 1 s 2 4 2 4 4 the black numbers next to an arc is its capacity Push flow over the path s-1-4-t The bottlenecks on this path are the edges {1,4} and {4,t}. So we can send flow 2 along this path

5
**the black number next to an arc is its capacity **

1 3 2 4 2 3 2 2 3 1 t 1 1 s 2 2 4 2 4 4 the black number next to an arc is its capacity the green number next to an arc is the flow on it We can push another extra flow of 1 on the path s-1-3-t. The bottleneck is now {s,1} with remaining capacity 1.

6
**the black number next to an arc is its capacity **

1 3 1 2 1 4 3 3 2 2 3 1 t 1 1 s 2 2 4 2 4 4 the black number next to an arc is its capacity the green number next to an arc is the flow on it Since {4,t} is on its capacity, the only path remaining through which we can send extra flow is s t. Here {4,3} is the bottleneck and thus we can send extra flow of 1

7
**the black number next to an arc is its capacity **

1 3 1 2 2 4 3 3 2 2 3 1 t 1 1 1 s 2 2 1 4 2 1 4 4 the black number next to an arc is its capacity the green number next to an arc is the flow on it There are no s-t paths left on which we can send extra flow. So is this the maximal flow?

8
**the residual graph blue arcs (i,j) are forward arcs (fij<uij)**

1 3 3 3 2 2 3 3 1 1 t 1 1 1 1 s 2 2 2 4 blue arcs (i,j) are forward arcs (fij<uij) green arcs (j,i) are backward arcs (fij>0) the blue number is the residual capacity of a blue arc the green number is the capacity of a green arc the black number is the original capacity of the arc

9
**the residual graph red arcs form an s-t-path in the residual graph**

1 3 1 1 2 2 3 2 3 1 t 1 1 s 2 1 3 2 1 3 4 red arcs form an s-t-path in the residual graph and therefore a flow-augmenting path in the original network

10
**the residual graph red arcs form an augmenting path**

1 3 1 1 2 2 3 2 3 1 t 1 1 s 2 1 3 2 1 3 4 red arcs form an augmenting path Augment the flow by the minimum capacity of a red arc, i.e, 1

11
**the residual graph Augment the flow by the minimum capacity of a**

1 3 1 1 2 2 3 2 3 1 t 1 1 s 2 1 3 2 1 3 4 Augment the flow by the minimum capacity of a red arc, i.e, 1: - increase the flow by 1 on all arcs corresponding to forward red arcs - decrease the flow by 1 on all arcs corresponding to backward red arcs

12
**the residual graph Augment the flow by the minimum capacity of a**

1 3 2 3 1 3 3 1 1 1 t 1 1 s 2 2 2 2 2 2 4 Augment the flow by the minimum capacity of a red arc, i.e, 1: - increase the flow by 1 on all arcs corresponding to forward red arcs - decrease the flow by 1 on all arcs corresponding to backward red arcs Construct the new residual graph

13
**An s-t cut is defined by a set S of the nodes with **

1 3 2 4 2 2 3 1 t 1 1 s 2 4 2 4 4 An s-t cut is defined by a set S of the nodes with s in S and t not in S.

14
**An s-t cut is defined by a set S of the nodes with **

1 3 2 4 2 2 3 1 t 1 1 s 2 4 2 4 4 S={s,1,2} An s-t cut is defined by a set S of the nodes with s in S and t not in S.

15
**An s-t cut is defined by a set S of the nodes with **

1 3 2 4 2 2 3 1 t 1 1 s 2 4 2 4 4 S={s,1,2} An s-t cut is defined by a set S of the nodes with s in S and t not in S Size of cut S is the sum of the capacities on the arcs from S to N\S.

16
**An s-t cut is defined by a set S of the nodes with **

1 3 2 4 2 2 3 1 t 1 1 s 2 4 2 4 4 C(S)=u13+u14+u24= 2+2+4=8 An s-t cut is defined by a set S of the nodes with s in S and t not in S Size of cut S is the sum of the capacities on the arcs from S to N\S.

17
S1 1 3 2 4 2 2 3 1 t 1 1 s 2 4 2 4 4 C(S1)=us1+u24= 2+4=6

18
**S2 S1 C(S1)=us1+u24= 2+4=6 C(S2)=u13+u43+u4t= 2+1+2=5 t s 1 3 2 4 2 2**

19
**S2 C(S2)=u13+u43+u4t= 2+1+2=5 Max Flow ≤ Min Cut t s 1 3 2 4 2 2 3 1 1**

20
**S2 C(S2)=u13+u43+u4t= 2+1+2=5 Max Flow ≤ Min Cut fs1+fs2 ≤ Min Cut ≤ 5**

21
**= the residual graph S2 fs1+fs2=2+3=5 C(S2)=u13+u43+u4t= 2+1+2=5**

3 1 3 3 1 1 1 t 1 1 s 2 2 2 2 2 2 4 = fs1+fs2=2+3=5 C(S2)=u13+u43+u4t= 2+1+2=5 Max Flow = Min Cut

22
**= the residual graph S2 fs1+fs2=2+3=5 C(S2)=u13+u43+u4t= 2+1+2=5**

3 1 3 3 1 1 1 t 1 1 s 2 2 2 2 2 2 4 = fs1+fs2=2+3=5 C(S2)=u13+u43+u4t= 2+1+2=5 Max Flow = Min Cut Theorem: Max Flow = Min Cut

23
**= Proof sketch of Max Flow=Min Cut S2 fs1+fs2=2+3=5**

3 1 3 3 1 1 1 t 1 1 s 2 2 2 2 2 2 4 = fs1+fs2=2+3=5 C(S2)=u13+u43+u4t= 2+1+2=5 There is no s-t-path in the residual graph!!! S2={s,1,2,4} the set of nodes reachable from S

24
**= Proof sketch of Max Flow=Min Cut S2 fs1+fs2=2+3=5**

2 2 4 3 2 2 3 0 2 1 1 0 t 1 1 1 0 s 2 2 4 3 2 4 2 4 = fs1+fs2=2+3=5 C(S2)=u13+u43+u4t= 2+1+2=5 from S2={s,1,2,4} to {3,t} f13 = u13 = 2 f43 = u43 = 1 f4t = u4t = 2

25
**= Proof sketch of Max Flow=Min Cut S2 fs1+fs2=2+3=5**

2 2 4 3 2 2 3 0 2 1 1 0 t 1 1 1 0 s 2 2 4 3 2 4 2 4 = fs1+fs2=2+3=5 C(S2)=u13+u43+u4t= 2+1+2=5 from S2={s,1,2,4} to {3,t} f13 = u13 = 2 f43 = u43 = 1 f4t = u4t = 2 Full capacity of cut is used

26
**= Proof sketch of Max Flow=Min Cut S2 fs1+fs2=2+3=5**

2 2 4 3 2 2 3 0 2 1 1 0 t 1 1 1 0 s 2 2 4 3 2 4 2 4 = fs1+fs2=2+3=5 C(S2)=u13+u43+u4t= 2+1+2=5 from S2={s,1,2,4} to {3,t} f13 = u13 = 2 f43 = u43 = 1 f4t = u4t = 2 Full capacity of cut is used from {3,t} to {S2={s,1,2,4} f32 = 0 f34 = 0

27
**= Proof sketch of Max Flow=Min Cut S2 fs1+fs2=2+3=5**

2 2 4 3 2 2 3 0 2 1 1 0 t 1 1 1 0 s 2 2 4 3 2 4 2 4 = fs1+fs2=2+3=5 C(S2)=u13+u43+u4t= 2+1+2=5 from S2={s,1,2,4} to {3,t} f13 = u13 = 2 f43 = u43 = 1 f4t = u4t = 2 Full capacity of cut is used from {3,t} to {S2={s,1,2,4} f32 = 0 f34 = 0 nothing flows back across the cut

28
**= Proof sketch of Max Flow=Min Cut S2 fs1+fs2=2+3=5**

2 2 4 3 2 2 3 0 2 1 1 0 t 1 1 1 0 s 2 2 4 3 2 4 2 4 = fs1+fs2=2+3=5 C(S2)=u13+u43+u4t= 2+1+2=5 from S2={s,1,2,4} to {3,t} f13 = u13 = 2 f43 = u43 = 1 f4t = u4t = 2 from {3,t} to {S2={s,1,2,4} f32 = 0 f34 = 0 Capacity of the cut is equal to the flow from s to t

Similar presentations

Presentation is loading. Please wait....

OK

Outline LP formulation of minimal cost flow problem

Outline LP formulation of minimal cost flow problem

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on historical monuments in delhi Ppt on industrial revolution in america Ppt on history of cricket for class 9 Ppt on solar power plant in india Ppt on world environment day logo Ppt on statistics and probability problems Ppt on earth day 2013 Ppt on review of related literature example Ppt on world environment day 2013 Ppt on art and craft movement era