HW10_Sol - E160 Operations Research I Fall 2009 Homework...

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E160 Operations Research I Fall 2009 Homework #10 Solution 1. The min-max path from node 1 to node 12 is 1->4->5->8->11->12. It is found through finding the minimum spanning tree of the graph. 2. Soution Let T denote the minimum spanning tree of the graph G. Let N1:={ k| k and i are connected in T\{(i,j)}, N2:={ k| k and j are connected in T\{(i,j)} Since G’ is connected. There must be an edge having one end node in N1 and the other end node in N2. Choose the edge (k*,l*) in G’ such that w(k*,l*)=min{w(k,l)≤w(k,l)| k N1, l N2} and k* N1, l* N2. Chapter 8 Section 8.2 1. First label node 1 with a permanent label: [0* 7 12 21 31 44] Now node 2 receives a permanent label [0* 7* 12 21 31 44]. Node Temporary Label (* denotes next assigned permanent label) 3 min{12,7+7} = 12* 4 min{21,7+12} = 19 5 min{31,7+21} = 28 6 min{44,7+31} = 38 Now labels are [0* 7* 12* 19 28 38] Node Temporary Label (* denotes next assigned permanent label) 4 min{19,12+7} = 19* 5 min{28,12+12} = 24 6 min{38,12+21} = 33 Now labels are [0* 7* 12* 19* 24 33] Node Temporary Label (* denotes next assigned permanent label)
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This note was uploaded on 02/17/2010 for the course IEOR 160 taught by Professor Hochbaum during the Fall '07 term at University of California, Berkeley.

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HW10_Sol - E160 Operations Research I Fall 2009 Homework...

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