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

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E160 Operations Research I Spring 2009 Homework #10 Solution 1. The min-max path from node 1 to node 12 is 1->4->5->8->11->12. 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) 5 min{24,19+7} = 24* 6 min{33, 19+12} = 31 Now labels are [0* 7* 12* 19* 24* 31] Node Temporary Label (* denotes next assigned permanent label) 6 min{31,24+7} = 31 Now labels are [0* 7* 12* 19* 24* 31*] 31 -24 = c 56 , 24 - 12 = c 35 , 12 - 0 = c 13 . Thus 1-3-6 is the shortest path (of length 31) from node 1 to node 6. 2.
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This note was uploaded on 09/06/2009 for the course IEOR 160 taught by Professor Hochbaum during the Spring '07 term at University of California, Berkeley.

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

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