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Unformatted text preview: C&O 350 Linear Optimization – Winter 2009 – Solutions to Assignment 9 Question # Max. marks Part marks 1 No Credit 2 No Credit 3 No Credit Total No Credit 1. Exercise 11.4.3 from the course notes. Solution : (a) The second equation of the tableau is x 2 + 1 2 x 3 1 2 x 4 = 3 2 . We split each of the coefficients for x 3 and x 4 and the righthandside value into their integer part and fractional part. x 2 + 1 2 x 3 + ( 1 + 1 2 ) x 4 = (1 + 1 2 ) . We further reorganize this equation to get: 1 2 x 3 + 1 2 x 4 1 2 = 1 x 2 + x 4 . Since we want all the variables to be integers, 1 x 2 + x 4 must be integer. Thus, 1 2 x 3 + 1 2 x 4 1 2 must be integer as well. We also need all the variables to be nonnegative, thus 1 2 x 3 + 1 2 x 4 ≥ 0, which implies 1 2 x 3 + 1 2 x 4 1 2 ≥  1 2 . Since 1 2 x 3 + 1 2 x 4 1 2 is an integer, if it is larger than 1 2 , it must be greater than or equal to 0. Thus the resulting Gomory cut is: 1 2 x 3 + 1 2 x 4 ≥ 1 2 . In the basic feasible solution for B = { 1 , 2 } , x 3 and x 4 are 0, thus the current solution violates this Gomory cut. The constraint 1 2 x 3 + 1 2 x 4 1 2 ≥  1 2 is satisfied by all nonnegative solutions but we also know that the lefthandside must be integer (from above). So integer solutions that satisfyalso know that the lefthandside must be integer (from above)....
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This note was uploaded on 02/05/2010 for the course CO 350 taught by Professor S.furino,b.guenin during the Spring '07 term at Waterloo.
 Spring '07
 S.Furino,B.Guenin

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