Linear programming, affine scaling

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I'm having some trouble understanding the intuition behind the theory of affine scaling(algorithm used in linear programming)



$D_k$ is a diagonal matrix with $x_k$ on the diagonal
the vector of dual variables:
$w^k = (AD^2_kA^T)^-1AD^2_kc$



we then compute the vector of reduced costs
$r^k = c-A^Tw^k$



then the theory states (this part I don't quite understand)
that if $r^k > 0$ and $1^TD_kr^k < epsilon$



that the current solution $x_k$ is optimal. why is that?



I am looking at this: https://en.wikipedia.org/wiki/Affine_scaling



thank you for any help







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    why us a fake profile photo? topbeautyful.blogspot.com/2015/02/…
    – LinAlg
    Aug 2 at 15:36










  • @LinAlg: Gets more attention, presumably. Opportunity for mansplaining.
    – copper.hat
    Aug 2 at 16:46















up vote
0
down vote

favorite












I'm having some trouble understanding the intuition behind the theory of affine scaling(algorithm used in linear programming)



$D_k$ is a diagonal matrix with $x_k$ on the diagonal
the vector of dual variables:
$w^k = (AD^2_kA^T)^-1AD^2_kc$



we then compute the vector of reduced costs
$r^k = c-A^Tw^k$



then the theory states (this part I don't quite understand)
that if $r^k > 0$ and $1^TD_kr^k < epsilon$



that the current solution $x_k$ is optimal. why is that?



I am looking at this: https://en.wikipedia.org/wiki/Affine_scaling



thank you for any help







share|cite|improve this question















  • 1




    why us a fake profile photo? topbeautyful.blogspot.com/2015/02/…
    – LinAlg
    Aug 2 at 15:36










  • @LinAlg: Gets more attention, presumably. Opportunity for mansplaining.
    – copper.hat
    Aug 2 at 16:46













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'm having some trouble understanding the intuition behind the theory of affine scaling(algorithm used in linear programming)



$D_k$ is a diagonal matrix with $x_k$ on the diagonal
the vector of dual variables:
$w^k = (AD^2_kA^T)^-1AD^2_kc$



we then compute the vector of reduced costs
$r^k = c-A^Tw^k$



then the theory states (this part I don't quite understand)
that if $r^k > 0$ and $1^TD_kr^k < epsilon$



that the current solution $x_k$ is optimal. why is that?



I am looking at this: https://en.wikipedia.org/wiki/Affine_scaling



thank you for any help







share|cite|improve this question











I'm having some trouble understanding the intuition behind the theory of affine scaling(algorithm used in linear programming)



$D_k$ is a diagonal matrix with $x_k$ on the diagonal
the vector of dual variables:
$w^k = (AD^2_kA^T)^-1AD^2_kc$



we then compute the vector of reduced costs
$r^k = c-A^Tw^k$



then the theory states (this part I don't quite understand)
that if $r^k > 0$ and $1^TD_kr^k < epsilon$



that the current solution $x_k$ is optimal. why is that?



I am looking at this: https://en.wikipedia.org/wiki/Affine_scaling



thank you for any help









share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 2 at 15:01







user581239














  • 1




    why us a fake profile photo? topbeautyful.blogspot.com/2015/02/…
    – LinAlg
    Aug 2 at 15:36










  • @LinAlg: Gets more attention, presumably. Opportunity for mansplaining.
    – copper.hat
    Aug 2 at 16:46













  • 1




    why us a fake profile photo? topbeautyful.blogspot.com/2015/02/…
    – LinAlg
    Aug 2 at 15:36










  • @LinAlg: Gets more attention, presumably. Opportunity for mansplaining.
    – copper.hat
    Aug 2 at 16:46








1




1




why us a fake profile photo? topbeautyful.blogspot.com/2015/02/…
– LinAlg
Aug 2 at 15:36




why us a fake profile photo? topbeautyful.blogspot.com/2015/02/…
– LinAlg
Aug 2 at 15:36












@LinAlg: Gets more attention, presumably. Opportunity for mansplaining.
– copper.hat
Aug 2 at 16:46





@LinAlg: Gets more attention, presumably. Opportunity for mansplaining.
– copper.hat
Aug 2 at 16:46
















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