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HW10Sol

# HW10Sol - Fall 2009 IEOR 160 Industrial Engineering...

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Fall 2009 IEOR 160 Industrial Engineering & Operations Research November 29, 2009 Page 1 of 2 HOMEWORK 10 SOLUTIONS 1. Following Kruskal’s algorithm we first include the edge (1,2). Then we add in order the edges (9,12), (10,11), (5,6), (8,11), (11,12), (4,5), (6,9) and (1,4). At this moment there is a path from node 1 to node 12. The path is 1-4-5-8-11-12 and the maximum altitude is 5. 2. As long as the arc (i,j) does not disconnect the graph G , we can find the MST for G 0 by adding an arc with minimum weight that does not create a cycle. Section 8.2 1. Since the graph is acyclic we can find the shortest path as follows: l (1) = 0. l (2) = l (1) + c 12 = 0 + 7 = 7 and b (2) = 1. l (3) = min { 0 + 12 , 7 + 7 } = 12 and b (3) = 1. l (4) = min { 0 + 21 , 7 + 12 , 12 + 7 } = 19 and b (4) = 2. l (5) = min { 0 + 31 , 7 + 21 , 12 + 12 , 19 + 7 } = 24 and b (5) = 3. l (6) = min { 0 + 44 , 7 + 31 , 12 + 21 , 19 + 12 , 24 + 7 } = 31 and b (6) = 4. Thus 1-2-4-6 is a shortest path (of length 31) from node 1 to node 6. 2. Even though this is also an acyclic graph, we will use Dijkstra’s algorithm to find the shortest path. We begin by permanently labeling node 1 and assigning temporary labels to other nodes: [0* 2 8 ∞ ∞ ]. Then we give node 2 a permanent label: [0* 2* 8

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