- #1
mertcan
- 340
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<Moderator's note: Continued from a technical forum and thus no template. Re-opening has been approved by moderator.>
Hi, my question is related to simplex algorithm anticycling rule called Bland's rule. While I was working with the proof in the link https://www.math.ubc.ca/~anstee/math340/340blandsrule.pdf I got stuck with the very important part and my question is so simple by the way. Why the variable $$xs=q$$ can be everything??
I believe $$xs$$ only be zero because among degenerate iterations entering variable and leaving variable must be 0...
Also if you look at this link you can examine a similar proof (of my previous link) in the link http://profs.sci.univr.it/~rrizzi/classes/LP/homeworks/Bland/Bland.html. And there is sentence that "it must admit any solution which is admitted from
. In particular, it must be satisfied by the following choice of values": $$xs=y$$
Why must it admit any solution thus why must the choice of values be $$xs=y$$??
Again I consider that it must be only 0 can not be any choice of value..
Hi, my question is related to simplex algorithm anticycling rule called Bland's rule. While I was working with the proof in the link https://www.math.ubc.ca/~anstee/math340/340blandsrule.pdf I got stuck with the very important part and my question is so simple by the way. Why the variable $$xs=q$$ can be everything??
I believe $$xs$$ only be zero because among degenerate iterations entering variable and leaving variable must be 0...
Also if you look at this link you can examine a similar proof (of my previous link) in the link http://profs.sci.univr.it/~rrizzi/classes/LP/homeworks/Bland/Bland.html. And there is sentence that "it must admit any solution which is admitted from
Why must it admit any solution thus why must the choice of values be $$xs=y$$??
Again I consider that it must be only 0 can not be any choice of value..