Answer :
Answer:
See explanation and answer below.
Step-by-step explanation:
The tranformation
For this case we need to construct G' dividing making a division for each vertex v of G into 3 edges that on this case are [tex] v_1, v_2 and l(v)[/tex].
We assume that the edges from the begin are the incoming edges of [tex] v_1[/tex] and all the outgoing edges from v are outgoing edges from [tex]v_2[/tex]
We need to construct [tex] G' = (V', E')[/tex] with capacity function a' and we need to satisfy the follwoing:
For every [tex] v \in V[/tex] we create 2 vertices [tex] v_1, v_2 \in V'[/tex]
Now we can add a new edge asscoiated to [tex] v_1, v_2 \in E' [/tex] with the condition [tex] a' (v_1,v_2) = l(v)[/tex]
Now for each edges [tex] (u,v)\in E[/tex] we can create the following edge [tex]( u_r, v_1) \in E'[/tex] and the capacity is given by: [tex] a' (u_r, v_1) = a (u,v)[/tex]
And for this case we can see this:
[tex] |V'| = 2|V|, |E'|= |E| +|V|[/tex]
Now we assume that x is the flow who belongs to G respect vertex capabilities. We can create a flow function x' who belongs to G' with the following steps:
For every edge [tex](u,v) \in G[/tex] we can assume that [tex]x' (u_r ,v_1) = x(u,v)[/tex]
Then for each vertex [tex] u \in V -t[/tex] and we can define [tex] x\(u_1,u_r) = \sum_{v \in V} x(u,v)[/tex] and [tex] x' (t_1,t_2) = \sum_{v \in V} x(v,t)[/tex]
And after see that the capacity constraint on this case would be satisfied since for every edge in G' on the form [tex] (u_r, u_1)[/tex] we have a corresponding edge in G because:
[tex] u \in V -(s,t) [/tex] we have that:
[tex] x' (u_1, u_r) = \sum_{v \in V} x(u,v) \leq l(u) = a' (u_1, u_r)[/tex]
[tex] x' (t_1,t_2) = \sum_{v \in V} x(v,t) \leq (t) = a' (t_1,t_2)[/tex]
And with this we have the maximization problem solved.
We assume that we have K vertices using the max scale algorithm.