asst9sol - C&O 350 Linear Optimization – Winter 2009 –...

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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 right-hand-side 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 left-hand-side must be integer (from above). So integer solutions that satisfy 1 2 x 3 + 1 2 x 4 - 1 2 ≥ - 1 2 will satisfy 1 2
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