phase1ex3

phase1ex3 - 0 1 0 3 2 0 1 ] A=A0; disp('x_3 enters basis...

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disp('Example 3 Phase I with remaining artificial variable x_5 ') disp('optimal problem (A) data with only x_5 an artificial variable') A0=[ 1 0 0 0 -2 0 0 0 0 1 4 3 0 2 0 0 0 -2 -2 1 0
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Unformatted text preview: 0 1 0 3 2 0 1 ] A=A0; disp('x_3 enters basis and x_5 leaves' ) pause A=piv(A,3,4) disp('Not optimal tableau - but optimal value 0 - so B={2,3,1}')...
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This note was uploaded on 06/11/2011 for the course C 350 taught by Professor Wolkowicz during the Fall '97 term at Waterloo.

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