103s11chap7solutions

103s11chap7solutions - Math 103, Spring 2011, Solutions to...

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Math 103, Spring 2011, Solutions to Chapter 7 Homework Assignment Problem 12 : (a) There are many spanning trees in the given network, hence many possible correct answers to the question. Here is one spanning tree, though there are many more. The key point is that we must draw a graph which contains all the vertices of the original graph, and just enough of its edges to be a tree, i.e. connected and without circuits. Note that if we drew in one more edge we would create a circuit, whereas if we removed one edge we would no longer have a network (i.e. our graph would no longer be connected); in either case we would no longer have a tree. (b) To find the redundancy of the given network, compute M - (N - 1 ) , where M equals the number of edges in the graph, and N equals the number of vertices. In this case M = 24 and N = 16, so the redundancy equals 24 - (16 - 1) = 9 . Conceptually, this means that 9 is just the right number of edges to remove from the original network to produce a spanning tree. Problem 14
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This note was uploaded on 01/20/2012 for the course MATH 103 taught by Professor Berkowitz during the Spring '07 term at Rutgers.

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103s11chap7solutions - Math 103, Spring 2011, Solutions to...

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