Derive LCP from KKT conditions of a QP
$begingroup$
I'm working through this tutorial on LCPs and interior point methods. In it, the authors claim that the following quadratic program
$$
begin{aligned}
min quad& frac{1}{2}u^TQu - c^Tu\
text{subject to} quad& Auleq b, 0 leq u
end{aligned}
$$
has the following KKT conditions
$$
begin{align}
y &= Mx+q,; x^Ty = 0, ; 0 leq x, ; 0 leq y
end{align}
$$
where
$$
begin{align}
x &= begin{bmatrix}u\vend{bmatrix}\
M &= begin{bmatrix}Q & A^T\-A & 0end{bmatrix}\
q &= begin{bmatrix}-c\bend{bmatrix}.
end{align}
$$
(Note, I already corrected $q$ to have $-c$.) The authors claim the KKT conditions are equivalent to a linear complementarity problem.
I am trying to derive the above LCP from the KKT conditions as I know them. I can write the KKT conditions down as
$$
begin{align}
0&leq u\
0&leq Au-b\
0&leq v\
0&leq lambda\
0&=v^T(Au-b)\
0&=lambda^Tu\
0&=Qu-c+A^Tv+lambda
end{align}
$$
where $v$ and $lambda$ are Lagrange multipliers. I believe the source of my confusion is understanding which relation ($=,subset,supset,capneqemptyset$) should go in the middle of
$$
SOL(QP) quad?quad SOL(LCP)
$$
This is confusion from a number of directions. First, if $lambda = 0$ and the LCP is feasible, then I can argue that $(u,v) in text{SOL(QP)}$ $Rightarrow$ $(u,v) in text{SOL(LCP)}$. However, what if $lambda neq 0$? Second, the complimentarity constraint in the LCP evaluates to
$$
x^Ty = u^T(Qu -c + A^Tv) -v^T(Au-b) = 0
$$
which isn't the same as requiring the equality constraints in the KKT conditions because $u^T(Qu -c + A^Tv) = v^T(Au-b)$ might happen.
I noticed similar issues also show up in Cotte's book, Lemma 3.1.1. Again, the Lagrange multipliers on the inequality constraint $xgeq0$ are ignored.
quadratic-programming karush-kuhn-tucker
asked Nov 28 '18 at 6:34
ehuangehuang
18217
$endgroup$
add a comment |
$begingroup$
I'm working through this tutorial on LCPs and interior point methods. In it, the authors claim that the following quadratic program
$$
begin{aligned}
min quad& frac{1}{2}u^TQu - c^Tu\
text{subject to} quad& Auleq b, 0 leq u
end{aligned}
$$
has the following KKT conditions
$$
begin{align}
y &= Mx+q,; x^Ty = 0, ; 0 leq x, ; 0 leq y
end{align}
$$
where
$$
begin{align}
x &= begin{bmatrix}u\vend{bmatrix}\
M &= begin{bmatrix}Q & A^T\-A & 0end{bmatrix}\
q &= begin{bmatrix}-c\bend{bmatrix}.
end{align}
$$
(Note, I already corrected $q$ to have $-c$.) The authors claim the KKT conditions are equivalent to a linear complementarity problem.
I am trying to derive the above LCP from the KKT conditions as I know them. I can write the KKT conditions down as
$$
begin{align}
0&leq u\
0&leq Au-b\
0&leq v\
0&leq lambda\
0&=v^T(Au-b)\
0&=lambda^Tu\
0&=Qu-c+A^Tv+lambda
end{align}
$$
where $v$ and $lambda$ are Lagrange multipliers. I believe the source of my confusion is understanding which relation ($=,subset,supset,capneqemptyset$) should go in the middle of
$$
SOL(QP) quad?quad SOL(LCP)
$$
This is confusion from a number of directions. First, if $lambda = 0$ and the LCP is feasible, then I can argue that $(u,v) in text{SOL(QP)}$ $Rightarrow$ $(u,v) in text{SOL(LCP)}$. However, what if $lambda neq 0$? Second, the complimentarity constraint in the LCP evaluates to
$$
x^Ty = u^T(Qu -c + A^Tv) -v^T(Au-b) = 0
$$
which isn't the same as requiring the equality constraints in the KKT conditions because $u^T(Qu -c + A^Tv) = v^T(Au-b)$ might happen.
I noticed similar issues also show up in Cotte's book, Lemma 3.1.1. Again, the Lagrange multipliers on the inequality constraint $xgeq0$ are ignored.
quadratic-programming karush-kuhn-tucker
asked Nov 28 '18 at 6:34
ehuangehuang
18217
$endgroup$
add a comment |
$begingroup$
I'm working through this tutorial on LCPs and interior point methods. In it, the authors claim that the following quadratic program
$$
begin{aligned}
min quad& frac{1}{2}u^TQu - c^Tu\
text{subject to} quad& Auleq b, 0 leq u
end{aligned}
$$
has the following KKT conditions
$$
begin{align}
y &= Mx+q,; x^Ty = 0, ; 0 leq x, ; 0 leq y
end{align}
$$
where
$$
begin{align}
x &= begin{bmatrix}u\vend{bmatrix}\
M &= begin{bmatrix}Q & A^T\-A & 0end{bmatrix}\
q &= begin{bmatrix}-c\bend{bmatrix}.
end{align}
$$
(Note, I already corrected $q$ to have $-c$.) The authors claim the KKT conditions are equivalent to a linear complementarity problem.
I am trying to derive the above LCP from the KKT conditions as I know them. I can write the KKT conditions down as
$$
begin{align}
0&leq u\
0&leq Au-b\
0&leq v\
0&leq lambda\
0&=v^T(Au-b)\
0&=lambda^Tu\
0&=Qu-c+A^Tv+lambda
end{align}
$$
where $v$ and $lambda$ are Lagrange multipliers. I believe the source of my confusion is understanding which relation ($=,subset,supset,capneqemptyset$) should go in the middle of
$$
SOL(QP) quad?quad SOL(LCP)
$$
This is confusion from a number of directions. First, if $lambda = 0$ and the LCP is feasible, then I can argue that $(u,v) in text{SOL(QP)}$ $Rightarrow$ $(u,v) in text{SOL(LCP)}$. However, what if $lambda neq 0$? Second, the complimentarity constraint in the LCP evaluates to
$$
x^Ty = u^T(Qu -c + A^Tv) -v^T(Au-b) = 0
$$
which isn't the same as requiring the equality constraints in the KKT conditions because $u^T(Qu -c + A^Tv) = v^T(Au-b)$ might happen.
I noticed similar issues also show up in Cotte's book, Lemma 3.1.1. Again, the Lagrange multipliers on the inequality constraint $xgeq0$ are ignored.
quadratic-programming karush-kuhn-tucker
asked Nov 28 '18 at 6:34
ehuangehuang
18217
$endgroup$
I'm working through this tutorial on LCPs and interior point methods. In it, the authors claim that the following quadratic program
$$
begin{aligned}
min quad& frac{1}{2}u^TQu - c^Tu\
text{subject to} quad& Auleq b, 0 leq u
end{aligned}
$$
has the following KKT conditions
$$
begin{align}
y &= Mx+q,; x^Ty = 0, ; 0 leq x, ; 0 leq y
end{align}
$$
where
$$
begin{align}
x &= begin{bmatrix}u\vend{bmatrix}\
M &= begin{bmatrix}Q & A^T\-A & 0end{bmatrix}\
q &= begin{bmatrix}-c\bend{bmatrix}.
end{align}
$$
(Note, I already corrected $q$ to have $-c$.) The authors claim the KKT conditions are equivalent to a linear complementarity problem.
I am trying to derive the above LCP from the KKT conditions as I know them. I can write the KKT conditions down as
$$
begin{align}
0&leq u\
0&leq Au-b\
0&leq v\
0&leq lambda\
0&=v^T(Au-b)\
0&=lambda^Tu\
0&=Qu-c+A^Tv+lambda
end{align}
$$
where $v$ and $lambda$ are Lagrange multipliers. I believe the source of my confusion is understanding which relation ($=,subset,supset,capneqemptyset$) should go in the middle of
$$
SOL(QP) quad?quad SOL(LCP)
$$
This is confusion from a number of directions. First, if $lambda = 0$ and the LCP is feasible, then I can argue that $(u,v) in text{SOL(QP)}$ $Rightarrow$ $(u,v) in text{SOL(LCP)}$. However, what if $lambda neq 0$? Second, the complimentarity constraint in the LCP evaluates to
$$
x^Ty = u^T(Qu -c + A^Tv) -v^T(Au-b) = 0
$$
which isn't the same as requiring the equality constraints in the KKT conditions because $u^T(Qu -c + A^Tv) = v^T(Au-b)$ might happen.
I noticed similar issues also show up in Cotte's book, Lemma 3.1.1. Again, the Lagrange multipliers on the inequality constraint $xgeq0$ are ignored.
quadratic-programming karush-kuhn-tucker
quadratic-programming karush-kuhn-tucker
asked Nov 28 '18 at 6:34
ehuangehuang
18217
asked Nov 28 '18 at 6:34
ehuangehuang
18217
asked Nov 28 '18 at 6:34
ehuangehuang
18217
asked Nov 28 '18 at 6:34
ehuangehuang
18217
asked Nov 28 '18 at 6:34
ehuangehuang
18217
18217
add a comment |
add a comment |
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