# cut - A=piv(A,5,6) A2=A; disp('temporary pivot for plot')...

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clf clear all A0=[ 1 -2 -1 0 0 0 0 0 1 1 1 0 0 5 0 -1 1 0 1 0 0 0 6 2 0 0 1 21 ] hold on axis square axis equal x=0:1:6; y=0*x; plot(x,y,'k-') y=0:1:6; x=0*y; plot(x,y,'k-') pause(.5) x=0:1:5; y=5-x; plot(x,y,'b-') pause(.5) y=0+x; plot(x,y,'g-') pause(.5) x=1.8:.2:3.7; y=(21-6*x)/2; plot(x,y,'r-') pause(.5) plot(11/4,9/4,'ko') disp('now display the lattice points') pause for i=0:1:4, for j=0:1:3, plot(i,j,'kx') end end disp('now start to pivot') pause A=A0; A=piv(A,2,2) A=piv(A,3,3) A=piv(A,4,5) %%%%%%%%%%%%%%%%%%%%%%%%%%% disp('add cut from x_4 row ') pause A1=A; A=[ A(:,1:6) zeros(4,1) A(:,7) 0 0 0 0 0 -.5 1 -.5] disp('apply dual simplex - s_1 leaves and x_5 enters')

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Unformatted text preview: A=piv(A,5,6) A2=A; disp('temporary pivot for plot') pause A=piv(A,2,4) pause A=piv(A,3,7) disp('plot the cut') pause x=1.2:.1:3.4; y=(10-3*x); plot(x,y,'k-') A=A2 %%%%%%%%%%%%%%%%%%%%%%%%%%% disp('now add cut from x_1 row ') pause A3=A; A=[ A(:,1:7) zeros(5,1) A(:,8) 0 0 0 -.5 0 0 -.5 1 -.5] disp('apply dual simplex - s_1 leaves and x_3 enters') A=piv(A,6,4) plot(3,1,'mo') disp('we have an optimal ILP tableau') A4=A; disp('temporary pivot for plot') pause A=piv(A,2,7) pause A=piv(A,3,8) disp('plot the cut') pause x=1.2:.1:3.4; y=(7-2*x); plot(x,y,'m-') A=A4 hold off...
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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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cut - A=piv(A,5,6) A2=A; disp('temporary pivot for plot')...

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