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VladimirAG [237]
3 years ago
9

Suppose that a flow network G=(V,E)G = (V, E)G=(V,E) violates the assumption that the network contains a path s⇝v⇝ts \leadsto v

\leadsto ts⇝v⇝t for all vertices v∈Vv \in Vv∈V. Let uuu be a vertex for which there is no path s⇝u⇝ts \leadsto u \leadsto ts⇝u⇝t. Show that there must exist a maximum flow fff in GGG such that f(u,v)=f(v,u)=0f(u, v) = f(v, u) = 0f(u,v)=f(v,u)=0 for all vertices v∈Vv \in Vv∈V.
a. True
b. False
Computers and Technology
1 answer:
Alex17521 [72]3 years ago
4 0

Answer:

Hi your question is poorly written attached below is the complete question

answer : TRUE ( a )

Explanation:

The statement about a flow network is true because when there is a vertex in the flow network that inhibits the source from reaching the sink, the vertex can be successfully removed without altering the maximum flow in the flow network.

as given in the question ;  if f(u,v) = f(v,u) =0   we can remove the vertex.

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<h2>Answer:</h2>

The correct options are

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<h2>Explanation:</h2>

There are two types of hypervisors:

Type 1 hypervisor are those hypervisors that run directly on the system hardware such as a “bare metal” embedded hypervisor,

Similarly

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