Mathematics, 10.01.2020 03:31, parisw8

Suppose that, in addition to edge capacities, a flow network has vertex capacities. that is each vertex has a limit l./ on how much flow can pass though . show how to transform a flow network g d .v; e/ with vertex capacities into an equivalent flow network g0 d .v 0 ; e0 / without vertex capacities, such that a maximum flow in g0 has the same value as a maximum flow in g. how many vertices and edges does g0 have?

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answered: aa010002

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 v_1, v_2 and l(v) .

We assume that the edges from the begin are the incoming edges of v_1 and all the outgoing edges from v are outgoing edges from v_2

We need to construct G' = (V', E') with capacity function a' and we need to satisfy the follwoing:

For every $v \in V$ we create 2 vertices $v_1, v_2 \in V'$

Now we can add a new edge asscoiated to $v_1, v_2 \in E'$ with the condition a' (v_1,v_2) = l(v)

Now for each edges $(u,v)\in E$ we can create the following edge $( u_r, v_1) \in E'$ and the capacity is given by: a' (u_r, v_1) = a (u,v)

And for this case we can see this:

|V'| = 2|V|, |E'|= |E| +|V|

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 $(u,v) \in G$ we can assume that x' (u_r ,v_1) = x(u,v)

Then for each vertex $u \in V -t$ and we can define $x\(u_1,u_r) = \sum_{v \in V} x(u,v)$ and $x' (t_1,t_2) = \sum_{v \in V} x(v,t)$

And after see that the capacity constraint on this case would be satisfied since for every edge in G' on the form (u_r, u_1) we have a corresponding edge in G because:

$u \in V -(s,t)$ we have that:

$x' (u_1, u_r) = \sum_{v \in V} x(u,v) \leq l(u) = a' (u_1, u_r)$

$x' (t_1,t_2) = \sum_{v \in V} x(v,t) \leq (t) = a' (t_1,t_2)$

And with this we have the maximization problem solved.

We assume that we have K vertices using the max scale algorithm.

answered: Guest

This is a robot. try again lol. maybe you attached the wrong image.
Can some find the area and perimeter for this triangle !

answered: Guest

Answer: question no. 1 -a. yes, the sides are in the ratio 2 : 5.question no. 2 -b. 59°question no. 3 -c. angles x and y are remote interior angles.question no. 4 -d. 90°, 45° and 45°question no. 5 -b. mn and qrstep-by-step explanation: question no. 1 -rs : xu = 3 : 7.5 = 2 : 5rt : x? = 6.4 : 16 = 2 : 5st : u? = 6 : 15 = 2 : 5as the ratios of all sides are equivalent, the triangles must be congruent and the sides must be proportional.therefore, the answer is: a. yes, the sides are in the ratio 2 : 5.question no. 2 -in the second diagram, angle b is the only letter not shown. therefore, the number must be equivalent to angle b.if we substitute the value of angle b as being 22° in the first diagram, we can now use the angle sum of a triangle to obtain the value of angle a.angle sum of a triangle = 180°therefore: let angle a = aa + 99 + 22 = 180a = 180 - 99 - 22a = 59°therefore, the answer is: b. 59°question no. 3 -angles x and y don't share the vertex of the triangle with the exterior angle w. remote interior angles don't share a vertex of a triangle with an exterior angle. this means that angles x and y are remote interior angles.therefore, the answer is: c. angles x and y are remote interior angles.question no. 4 -the sum of all interior angles of a triangle must be equivalent to 180°.90°, 45° and 45° are the only set of angle measures that are equivalent to 180°.therefore, the answer is: d. 90°, 45° and 45°question no. 5 -the only pair of sides stated in the options which are corresponding are mn and qr.therefore, the answer is: b. mn and qr.

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Suppose that, in addition to edge capacities, a flow network has vertex capacities. that is each ver...