{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

09_Max_Flows_2

# 09_Max_Flows_2 - 15.082 and 6.855J March 6 2003 Maximum...

This preview shows pages 1–14. Sign up to view the full content.

1 15.082 and 6.855J March 6, 2003 Maximum Flows 2

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 Network Reliability Communication Network What is the maximum number of arc disjoint paths from s to t? How can we determine this number? Theorem . Let G = (N,A) be a directed graph. Then the maximum number of arc-disjoint paths from s to t is equal to the minimum number of arcs upon whose deletion there is no directed s-t path. t s
3 There are 3 arc-disjoint s-t paths s t 1 2 3 4 5 6 7 8 9 10 11 12

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
4 Deleting 3 arcs disconnects s and t t 1 2 5 6 7 9 10 11 12 s 3 4 8 Let S = {s, 3, 4, 8}. The only arcs from S to T = N\S are the 3 deleted arcs.
5 Node disjoint paths Two s-t paths P and P' f are said to be node- disjoint if the only nodes in common to P and P' are s and t). How can one determine the maximum number of node disjoint s-t paths? Answer: node splitting Theorem. Let G = (N,A) be a network with no arc from s to t. The maximum number of node- disjoint paths from s to t equals the minimum number of nodes whose removal from G disconnects all paths from nodes s to node t.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
6 There are 2 node disjoint s-t paths. s t 1 2 3 4 5 6 7 8 9 10 11 12
7 Deleting 5 and 6 disconnects t from s? t 5 6 7 9 10 11 12 s 1 2 3 4 8 Let S = {s, 1, 2, 3, 4, 8} Let T = {7, 9, 10, 11, 12, t} There is no arc directed from S to T.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
8 Matchings An undirected network G = (N, A) is bipartite if N can be partitioned into N 1 and N 2 so that for every arc (i,j) either i N 1 and j N 2 . A matching in N is a set of arcs no two of which are incident to a common node. Matching Problem : Find a matching of maximum cardinality 1 2 3 4 5 6 7 8 9 10
9 Transformation to a Max Flow Problem 1 2 3 4 5 6 7 8 9 10 s t Each arc (s, i) has a capacity of 1. Each arc (j, t) has a capacity of 1. Replace original arcs by pairs, and put infinite capacity on original arcs.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
10 Find a max flow 1 2 3 4 5 6 7 8 9 10 s t The maximum s-t flow is 4. The max matching has cardinality 4.
11 Determine the minimum cut 1 2 3 4 5 6 7 8 9 10 s t S = {s, 1, 3, 4, 6, 8}. T = {2, 5, 7, 9, 10, t}. There is no arc from {1, 3, 4} to {7, 9, 10} or from {6, 8} to {2, 5}. Any such arc would have an infinite capacity.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
12 Interpret the minimum cut 1 2 3 4 5 6 7 8 9 10 s t Look at the original nodes incident to the minimum cut. Every original arc is incident to one of them.
13 Matching Duality 1 1 2 3 4 5 6 7 8 9 10 The maximum cardinality of a matching is the minimum number of nodes that “covers” all of the arcs.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 37

09_Max_Flows_2 - 15.082 and 6.855J March 6 2003 Maximum...

This preview shows document pages 1 - 14. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online