Successive shortest paths: Difference between revisions

Revision as of 10:40, 23 October 2014

Definition:

For a node [math]\displaystyle{ v\in V }[/math], let [math]\displaystyle{ \Delta f(v):=\sum_{w:(v,w)\in A}f(v,w)-\sum_{w:(w,v)\in A}f(w,v) }[/math].
The imbalance of a node [math]\displaystyle{ v\in V }[/math] is defined as [math]\displaystyle{ I_f(v):=\Delta f(v)-b(v) }[/math].
The imbalance of a node [math]\displaystyle{ v\in V }[/math] is underestimating if [math]\displaystyle{ 0\leq \Delta f(v)\leq b(v) }[/math] or [math]\displaystyle{ 0\geq\Delta f(v)\geq b(v) }[/math].
The total imbalance of [math]\displaystyle{ f }[/math] is the defined as [math]\displaystyle{ \sum_{v\in V}|I_f(v)| }[/math].

Invariant:

The capacity constraints are fulfilled, that is, [math]\displaystyle{ 0\leq f(a)\leq u(a) }[/math] for all [math]\displaystyle{ a\in A }[/math].
There is no negative cycle in the residual network of [math]\displaystyle{ f }[/math].
The imbalance of every node is underestimating.

Variant: The total imbalance strictly decreases.

Break condition: The imbalances of all nodes are zero.

Abstract view: Start with the zero flow.

Proof: Obvious.

@@ Line 5: / Line 5: @@
 # The '''imbalance''' of a node <math>v\in V</math> is defined as <math>I_f(v):=\Delta f(v)-b(v)</math>.
 # The imbalance of a node <math>v\in V</math> is '''underestimating''' if <math>0\leq \Delta f(v)\leq b(v)</math> or <math>0\geq\Delta f(v)\geq b(v)</math>.
-# The '''total imbalance''' of <math>f</math> is the defined as <math>\sum_{v\in V}|I_f(v)</math>.
+# The '''total imbalance''' of <math>f</math> is the defined as <math>\sum_{v\in V}|I_f(v)|</math>.
 '''Invariant:'''