How can I solve $inf_{x in X} x^T(Ax + b) + sum_{i=1}^n x_i log(x_i)$ in a closed form depending on...
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I'm trying to solve the minimization problem
$$inf_{x in X} x^T(Ax + b) + sum_{i=1}^n x_i log(x_i)$$
where $b in mathbb{R}^n$, $X subset mathbb{R}^n$ is a convex set. And $A$ is a symmetric and positive definite matrix. The optimality conditions give me something like
$$-b in 2Ax + vec{1} + L(x) + N_X(x)$$
where $L(x)$ represents the vector in $mathbb{R}^n$ obtained by taking the elementwise logarithm of $x$; i.e., $[L(x)]_i=log x_i$.
The problem is how do I take that cone out of the inclusion. If there wasn't that log term it would become a projection on the set $X$. Is there a trick to make this as a projection on $X$ of other vector or something?
I'd appreciate any help or any paper/book reference where there's some information about this kind of problem
optimization convex-analysis convex-optimization
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add a comment |
$begingroup$
I'm trying to solve the minimization problem
$$inf_{x in X} x^T(Ax + b) + sum_{i=1}^n x_i log(x_i)$$
where $b in mathbb{R}^n$, $X subset mathbb{R}^n$ is a convex set. And $A$ is a symmetric and positive definite matrix. The optimality conditions give me something like
$$-b in 2Ax + vec{1} + L(x) + N_X(x)$$
where $L(x)$ represents the vector in $mathbb{R}^n$ obtained by taking the elementwise logarithm of $x$; i.e., $[L(x)]_i=log x_i$.
The problem is how do I take that cone out of the inclusion. If there wasn't that log term it would become a projection on the set $X$. Is there a trick to make this as a projection on $X$ of other vector or something?
I'd appreciate any help or any paper/book reference where there's some information about this kind of problem
optimization convex-analysis convex-optimization
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1
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There is no closed form. You have to solve this numerically, even if $Xequivmathbb{R}^n$.
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– Michael Grant
Mar 23 '15 at 19:38
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I see. In this case do you know where can I find info on numerically solving this?
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– karlabos
Mar 23 '15 at 22:46
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I would recommend looking into CVX (incidentally, it is partially authored by the previous commenter): cvxr.com/cvx.
$endgroup$
– ChrKroer
Mar 24 '15 at 18:49
add a comment |
$begingroup$
I'm trying to solve the minimization problem
$$inf_{x in X} x^T(Ax + b) + sum_{i=1}^n x_i log(x_i)$$
where $b in mathbb{R}^n$, $X subset mathbb{R}^n$ is a convex set. And $A$ is a symmetric and positive definite matrix. The optimality conditions give me something like
$$-b in 2Ax + vec{1} + L(x) + N_X(x)$$
where $L(x)$ represents the vector in $mathbb{R}^n$ obtained by taking the elementwise logarithm of $x$; i.e., $[L(x)]_i=log x_i$.
The problem is how do I take that cone out of the inclusion. If there wasn't that log term it would become a projection on the set $X$. Is there a trick to make this as a projection on $X$ of other vector or something?
I'd appreciate any help or any paper/book reference where there's some information about this kind of problem
optimization convex-analysis convex-optimization
$endgroup$
I'm trying to solve the minimization problem
$$inf_{x in X} x^T(Ax + b) + sum_{i=1}^n x_i log(x_i)$$
where $b in mathbb{R}^n$, $X subset mathbb{R}^n$ is a convex set. And $A$ is a symmetric and positive definite matrix. The optimality conditions give me something like
$$-b in 2Ax + vec{1} + L(x) + N_X(x)$$
where $L(x)$ represents the vector in $mathbb{R}^n$ obtained by taking the elementwise logarithm of $x$; i.e., $[L(x)]_i=log x_i$.
The problem is how do I take that cone out of the inclusion. If there wasn't that log term it would become a projection on the set $X$. Is there a trick to make this as a projection on $X$ of other vector or something?
I'd appreciate any help or any paper/book reference where there's some information about this kind of problem
optimization convex-analysis convex-optimization
optimization convex-analysis convex-optimization
edited Mar 23 '15 at 19:40
Michael Grant
15.2k12045
15.2k12045
asked Mar 23 '15 at 18:58
karlaboskarlabos
401414
401414
1
$begingroup$
There is no closed form. You have to solve this numerically, even if $Xequivmathbb{R}^n$.
$endgroup$
– Michael Grant
Mar 23 '15 at 19:38
$begingroup$
I see. In this case do you know where can I find info on numerically solving this?
$endgroup$
– karlabos
Mar 23 '15 at 22:46
$begingroup$
I would recommend looking into CVX (incidentally, it is partially authored by the previous commenter): cvxr.com/cvx.
$endgroup$
– ChrKroer
Mar 24 '15 at 18:49
add a comment |
1
$begingroup$
There is no closed form. You have to solve this numerically, even if $Xequivmathbb{R}^n$.
$endgroup$
– Michael Grant
Mar 23 '15 at 19:38
$begingroup$
I see. In this case do you know where can I find info on numerically solving this?
$endgroup$
– karlabos
Mar 23 '15 at 22:46
$begingroup$
I would recommend looking into CVX (incidentally, it is partially authored by the previous commenter): cvxr.com/cvx.
$endgroup$
– ChrKroer
Mar 24 '15 at 18:49
1
1
$begingroup$
There is no closed form. You have to solve this numerically, even if $Xequivmathbb{R}^n$.
$endgroup$
– Michael Grant
Mar 23 '15 at 19:38
$begingroup$
There is no closed form. You have to solve this numerically, even if $Xequivmathbb{R}^n$.
$endgroup$
– Michael Grant
Mar 23 '15 at 19:38
$begingroup$
I see. In this case do you know where can I find info on numerically solving this?
$endgroup$
– karlabos
Mar 23 '15 at 22:46
$begingroup$
I see. In this case do you know where can I find info on numerically solving this?
$endgroup$
– karlabos
Mar 23 '15 at 22:46
$begingroup$
I would recommend looking into CVX (incidentally, it is partially authored by the previous commenter): cvxr.com/cvx.
$endgroup$
– ChrKroer
Mar 24 '15 at 18:49
$begingroup$
I would recommend looking into CVX (incidentally, it is partially authored by the previous commenter): cvxr.com/cvx.
$endgroup$
– ChrKroer
Mar 24 '15 at 18:49
add a comment |
1 Answer
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oldest
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Starting from your inclusion
$$-b in 2Ax + vec{1} + L(x) + N_X(x),$$
we can add $x$ on both sides and re-arrange to get:
$$x-2Ax-vec{1}-L(x)-b in (I+N_X)(x),$$
where $I$ denotes the identity operator $xmapsto x$.
We can now invert to get
$$x = P_Xbig(x-2Ax-vec{1}-L(x)-bbig).$$
This reformulation takes the normal cone operator out of
your problem and converts it to a fixed point problem
that you might try to approach with fixed-point theoretic
algorithms.
$endgroup$
add a comment |
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$begingroup$
Starting from your inclusion
$$-b in 2Ax + vec{1} + L(x) + N_X(x),$$
we can add $x$ on both sides and re-arrange to get:
$$x-2Ax-vec{1}-L(x)-b in (I+N_X)(x),$$
where $I$ denotes the identity operator $xmapsto x$.
We can now invert to get
$$x = P_Xbig(x-2Ax-vec{1}-L(x)-bbig).$$
This reformulation takes the normal cone operator out of
your problem and converts it to a fixed point problem
that you might try to approach with fixed-point theoretic
algorithms.
$endgroup$
add a comment |
$begingroup$
Starting from your inclusion
$$-b in 2Ax + vec{1} + L(x) + N_X(x),$$
we can add $x$ on both sides and re-arrange to get:
$$x-2Ax-vec{1}-L(x)-b in (I+N_X)(x),$$
where $I$ denotes the identity operator $xmapsto x$.
We can now invert to get
$$x = P_Xbig(x-2Ax-vec{1}-L(x)-bbig).$$
This reformulation takes the normal cone operator out of
your problem and converts it to a fixed point problem
that you might try to approach with fixed-point theoretic
algorithms.
$endgroup$
add a comment |
$begingroup$
Starting from your inclusion
$$-b in 2Ax + vec{1} + L(x) + N_X(x),$$
we can add $x$ on both sides and re-arrange to get:
$$x-2Ax-vec{1}-L(x)-b in (I+N_X)(x),$$
where $I$ denotes the identity operator $xmapsto x$.
We can now invert to get
$$x = P_Xbig(x-2Ax-vec{1}-L(x)-bbig).$$
This reformulation takes the normal cone operator out of
your problem and converts it to a fixed point problem
that you might try to approach with fixed-point theoretic
algorithms.
$endgroup$
Starting from your inclusion
$$-b in 2Ax + vec{1} + L(x) + N_X(x),$$
we can add $x$ on both sides and re-arrange to get:
$$x-2Ax-vec{1}-L(x)-b in (I+N_X)(x),$$
where $I$ denotes the identity operator $xmapsto x$.
We can now invert to get
$$x = P_Xbig(x-2Ax-vec{1}-L(x)-bbig).$$
This reformulation takes the normal cone operator out of
your problem and converts it to a fixed point problem
that you might try to approach with fixed-point theoretic
algorithms.
answered Jan 31 at 22:03
max_zornmax_zorn
3,44061429
3,44061429
add a comment |
add a comment |
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1
$begingroup$
There is no closed form. You have to solve this numerically, even if $Xequivmathbb{R}^n$.
$endgroup$
– Michael Grant
Mar 23 '15 at 19:38
$begingroup$
I see. In this case do you know where can I find info on numerically solving this?
$endgroup$
– karlabos
Mar 23 '15 at 22:46
$begingroup$
I would recommend looking into CVX (incidentally, it is partially authored by the previous commenter): cvxr.com/cvx.
$endgroup$
– ChrKroer
Mar 24 '15 at 18:49