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
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}')
