class_examples_9

# class_examples_9 - 1 First multiply each constraint with-1...

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1. First, multiply each constraint with -1 in order to get a positive 1 coefficient for x3 and x4. With this we can use x3 and x4 as basic variables. Initial table can be constructed as follows. All reduced costs are postivie, therefore dual feasible. However, XB values are negatif, therefore primal infeasible. We perform dual simplex algorithm. Choose the row with most negative column0 column 0 column1 column2 column3 column4 z x1 x2 x3 x4 row0 0 1 1 0 0 row1 -2 -1 -2 1 0 row2 -1 -1 0 0 1 Row1 is selected to be the pivot row. Look at the negative entries.Calculate ratio Row0/Row1 for each. Take the minimum as a pivot column. Which is column2 In column 2, we want 0,1,0. To achieve this, perform row operations. column 0 column1 column2 column3 column4 z x1 x2 x3 x4 row0 -1 0.5 0 0.5 0 row1 1 0.5 1 -0.5 -0 row2 -1 -1 0 0 1 Row2 is pivot row. Only negative entry is column 1. We need to enter x1 to basis and leave x4 from basis. column 0

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## This note was uploaded on 01/09/2012 for the course IE 521 taught by Professor Zelihaakça during the Fall '11 term at Fatih Üniversitesi.

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class_examples_9 - 1 First multiply each constraint with-1...

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