Unit6-2011-01-25_235730_network2

Unit6-2011-01-25_235730_network2 - #1) Brantley College You...

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#1) Brantley College You are looking for a minimal span solution because you have to visit all nodes (buildings) in the network. Draw a map of the buildings and the distances between them. It should look something like this: Beginning at node 1, proceed through the network, choosing the shortest segment leaving each node, provided that it does not lead to a node that has already been visited. The shortest path to an unvisited node leaving Node 1 goes to Node 2, with a length of 3. The shortest path to an unvisited node leaving Node 2 goes to Node 4, with a length of 2. The shortest path to an unvisited node leaving Node 4 goes to Node 6, with a length of 3. Up to this point the shortest path is 1 2 4 6, with a total length of 800 feet. There are two paths leaving Node 6 with equal lengths, one going to Node 3 and one going to Node 5. Each of these paths has a length of 4. The two paths are now: 1 2 4 6 3, with a total length of 1200 feet 1 2 4 6 5, with a total length of 1200 feet For the first of these two paths, the only remaining unvisited node is Node 5, with a length of 4, making the final path 1 2 4 6 3 5, with a total length of 1600 feet.
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For the second of these two paths, the only remaining unvisited node is Node 3, with a length of 4, making the final path 1 2 4 6 5 3, with a total length of 1600 feet. If you list all of the distances between the buildings in order, you will see that each of these paths uses the five shortest segments, so there is no possibility of another
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This note was uploaded on 09/22/2011 for the course ACCOUNTING MT482 taught by Professor Das during the Spring '11 term at Kaplan University.

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Unit6-2011-01-25_235730_network2 - #1) Brantley College You...

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