This statement is true or false about minimum spanning tree?

Question

If average weight of a minimum spanning tree is Aavg.Then minimum spanning tree will have weight almost (n-1)Aavg, where n is no of vertices in the graph. It is true or false?why?

Answer 1

Let a disconnected graph. In disconnected graph, minimumcostspanningtree is not possible. Given statement is wrong.

Answer 2

Yes True. If there are n vertices, then minimum number of edges in MST is (n-1) , As MST must be connected with minimum weight edges

If avg weight of 1 edge is A_avg

then, for (n-1) edges weight (n-1)A_avg

Comments

A_avg is average Weight of whole minimum spanning tree and not of only one edge. So I didn't get ur answer. Plz explain

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@shrestha Aavg is average Weight of whole minimum spanning tree and not of only one edge. So I didn't get ur answer. Plz explain

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we can have more than one minimum spanning tree for a given connected graph(when weights of edges are not unique) but minimum weight in each tree is always same

which is given as Aavg is this question.

and when all the weights are same then avg wt. of MST =(n-1) * weight of an edge

but here Aavg is not wt. of an edge it is weight of whole spanning tree.

hence this statement should be false.