hw12solutions

# hw12solutions - IEOR 162 Linear Programming Spring 2007...

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IEOR 162 Linear Programming Spring 2007 Homework 12 Solutions 8.2.1 Find the shortest path from 1 to 6 for the graph with V = { 1 , 2 , 3 , 4 , 5 , 6 } using Dijkstra’s Algorithm. Iteration 1 : S = , d ( V ) = (0 , , , , , ) , p ( V ) = ( , , , , , ) argmin { d ( j ) : j V \ S } = 1 (1 , 2) : d (2) = > d (1) + c 12 = 7 d (2) = 7 , p (2) = 1 (1 , 3) : d (3) = > d (1) + c 13 = 12 d (3) = 12 , p (3) = 1 (1 , 4) : d (4) = > d (1) + c 14 = 21 d (4) = 21 , p (4) = 1 (1 , 5) : d (5) = > d (1) + c 15 = 31 d (5) = 31 , p (5) = 1 (1 , 6) : d (6) = > d (1) + c 16 = 44 d (6) = 44 , p (6) = 1 Iteration 2 : S = { 1 } , d ( V ) = (0 , 7 , 12 , 21 , 31 , 44) , p ( V ) = ( , 1 , 1 , 1 , 1 , 1) argmin { d ( j ) : j V \ S } = 2 (2 , 3) : d (3) = 12 6 > d (2) + c 23 = 14 (2 , 4) : d (4) = 21 > d (2) + c 24 = 19 d (4) = 19 , p (4) = 2 (2 , 5) : d (5) = 31 > d (2) + c 25 = 28 d (5) = 28 , p (5) = 2 (2 , 6) : d (6) = 44 > d (2) + c 26 = 38 d (6) = 38 , p (6) = 2 Iteration 3 : S = { 1 , 2 } , d ( V ) = (0 , 7 , 12 , 19 , 28 , 38) , p ( V ) = ( , 1 , 1 , 2 , 2 , 2) argmin { d ( j ) : j V \ S } = 3 (3 , 4) : d (4) = 19 6 > d (3) +

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## This homework help was uploaded on 04/02/2008 for the course IEOR 162 taught by Professor Zhang during the Spring '07 term at University of California, Berkeley.

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hw12solutions - IEOR 162 Linear Programming Spring 2007...

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