In Q1, in the first inequality we have x+y >=3
First of all I set y as y=y1-y2, and added the constraint y1,y2 >= 0, so that now I have x+y1-y2 >= 3
Next I changed it to the standard input, so that now I have: -x-y1+y2 <= -3
Then I changed it to the slack format: t1 = -3+x+y1-y2 >= 0, with x=y1=y2=0
The problem is that now I have t1=-3 which is not >=0 …
This also applies the second equation, where i get t2 = -12+3x+2y1-2y2 >= 0
How should we handle this situation?
Date: 16 Jan 2015 14:51
Number of posts: 4
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0,0 is indeed infeasible.
You should find a feasible solution first!
In the example you gave, there is only one bi < 0, and in our case there are two.
so should we define two new "x0" variables?
It seems as the definition of x0(single new argument) will satisfy both bi < 0.
We add x0 to all constraints and choose to pivot (init simplex step 5) with the smallest bi, hence the found x0 will also cover for any larger bi's in our way.