Transform SDP Optimization Constraints
A problem of the form was hinted at on an upcoming assignment:
$minimize$ $tr(P)$
$s.t.$
$A^TP+PA = -Q$,
$P=P^T>0$
It was hinted that the desired solution require the problem be transformed into a form similar to the one shown below and solved via KKT optimality conditions. Where the transformed constraints will be four equalities or inequalities.
$minimize$ $(C^TX)$
$s.t.$
$AX = b$
My question is how are these four transformed constraints found or derived?
---update---
Perhaps I am over complicating this. If $A$, $Q$ and $P$ are $2x2$, that would be four linear equalities of the form $Ax=b$. For $P=begin{bmatrix}p_1&p_3\p_3&p_2end{bmatrix}$, would it be valid specify the problem as $minimize(p_1+p_2)$, since we want to minimize $trace(P)$? Then taking the 1st equality piecewise:
$(1)$ $2A_{11}p_1+2A_{21}p_3=-Q_{11}$
$(2)$ $A_{12}p_1+A_{21}p_2+(A_{11}p_1+A_{22})p_3=0$
$(3)$ $2A_{22}p_2+2A_{12}p_3=-Q_{22}$
And to define P as PSD:
$(4)$ $p_1 geq 0$
$(5)$ $p_1p_2-p_3^2 geq 0$
Then apply the KKT conditions to $min(p_1+p_2)$ and constraints 1-5, treating $p_1, p_2$ & $p_3$ as the variables for which to be solved?
If that is the case, why did the instructor hint that the problem should be transformed to $minimize(C^TX)$?
convex-optimization
asked Nov 28 '18 at 2:22
JDD
11
|
show 1 more comment
A problem of the form was hinted at on an upcoming assignment:
$minimize$ $tr(P)$
$s.t.$
$A^TP+PA = -Q$,
$P=P^T>0$
It was hinted that the desired solution require the problem be transformed into a form similar to the one shown below and solved via KKT optimality conditions. Where the transformed constraints will be four equalities or inequalities.
$minimize$ $(C^TX)$
$s.t.$
$AX = b$
My question is how are these four transformed constraints found or derived?
---update---
Perhaps I am over complicating this. If $A$, $Q$ and $P$ are $2x2$, that would be four linear equalities of the form $Ax=b$. For $P=begin{bmatrix}p_1&p_3\p_3&p_2end{bmatrix}$, would it be valid specify the problem as $minimize(p_1+p_2)$, since we want to minimize $trace(P)$? Then taking the 1st equality piecewise:
$(1)$ $2A_{11}p_1+2A_{21}p_3=-Q_{11}$
$(2)$ $A_{12}p_1+A_{21}p_2+(A_{11}p_1+A_{22})p_3=0$
$(3)$ $2A_{22}p_2+2A_{12}p_3=-Q_{22}$
And to define P as PSD:
$(4)$ $p_1 geq 0$
$(5)$ $p_1p_2-p_3^2 geq 0$
Then apply the KKT conditions to $min(p_1+p_2)$ and constraints 1-5, treating $p_1, p_2$ & $p_3$ as the variables for which to be solved?
If that is the case, why did the instructor hint that the problem should be transformed to $minimize(C^TX)$?
convex-optimization
asked Nov 28 '18 at 2:22
JDD
11
why do you call the constraints of the original problem quadratic?
– LinAlg
Nov 28 '18 at 2:48
Are they not? Isn't the Lyapunov equality usually associated with the form $x^TPx$?
– JDD
Nov 28 '18 at 2:50
That's irrelevant to the immediate task at hand. The point is your problem in $P$ already has a linear objective, a set of linear equations in $P$, and a semidefinite constraint.
– Michael Grant
Nov 28 '18 at 2:56
$C^{T}X$ is a matrix rather than a scalar- are you actually meant to be minimizing $mbox{tr}(C^{T}X)$?
– Brian Borchers
Nov 28 '18 at 5:01
Michael Grant, I think I understand. Are you saying that everything is already correct for applying KKT, similar to what I added to the original question?
– JDD
Nov 28 '18 at 5:03
|
show 1 more comment
A problem of the form was hinted at on an upcoming assignment:
$minimize$ $tr(P)$
$s.t.$
$A^TP+PA = -Q$,
$P=P^T>0$
It was hinted that the desired solution require the problem be transformed into a form similar to the one shown below and solved via KKT optimality conditions. Where the transformed constraints will be four equalities or inequalities.
$minimize$ $(C^TX)$
$s.t.$
$AX = b$
My question is how are these four transformed constraints found or derived?
---update---
Perhaps I am over complicating this. If $A$, $Q$ and $P$ are $2x2$, that would be four linear equalities of the form $Ax=b$. For $P=begin{bmatrix}p_1&p_3\p_3&p_2end{bmatrix}$, would it be valid specify the problem as $minimize(p_1+p_2)$, since we want to minimize $trace(P)$? Then taking the 1st equality piecewise:
$(1)$ $2A_{11}p_1+2A_{21}p_3=-Q_{11}$
$(2)$ $A_{12}p_1+A_{21}p_2+(A_{11}p_1+A_{22})p_3=0$
$(3)$ $2A_{22}p_2+2A_{12}p_3=-Q_{22}$
And to define P as PSD:
$(4)$ $p_1 geq 0$
$(5)$ $p_1p_2-p_3^2 geq 0$
Then apply the KKT conditions to $min(p_1+p_2)$ and constraints 1-5, treating $p_1, p_2$ & $p_3$ as the variables for which to be solved?
If that is the case, why did the instructor hint that the problem should be transformed to $minimize(C^TX)$?
convex-optimization
asked Nov 28 '18 at 2:22
JDD
11
A problem of the form was hinted at on an upcoming assignment:
$minimize$ $tr(P)$
$s.t.$
$A^TP+PA = -Q$,
$P=P^T>0$
It was hinted that the desired solution require the problem be transformed into a form similar to the one shown below and solved via KKT optimality conditions. Where the transformed constraints will be four equalities or inequalities.
$minimize$ $(C^TX)$
$s.t.$
$AX = b$
My question is how are these four transformed constraints found or derived?
---update---
Perhaps I am over complicating this. If $A$, $Q$ and $P$ are $2x2$, that would be four linear equalities of the form $Ax=b$. For $P=begin{bmatrix}p_1&p_3\p_3&p_2end{bmatrix}$, would it be valid specify the problem as $minimize(p_1+p_2)$, since we want to minimize $trace(P)$? Then taking the 1st equality piecewise:
$(1)$ $2A_{11}p_1+2A_{21}p_3=-Q_{11}$
$(2)$ $A_{12}p_1+A_{21}p_2+(A_{11}p_1+A_{22})p_3=0$
$(3)$ $2A_{22}p_2+2A_{12}p_3=-Q_{22}$
And to define P as PSD:
$(4)$ $p_1 geq 0$
$(5)$ $p_1p_2-p_3^2 geq 0$
Then apply the KKT conditions to $min(p_1+p_2)$ and constraints 1-5, treating $p_1, p_2$ & $p_3$ as the variables for which to be solved?
If that is the case, why did the instructor hint that the problem should be transformed to $minimize(C^TX)$?
convex-optimization
convex-optimization
asked Nov 28 '18 at 2:22
JDD
11
asked Nov 28 '18 at 2:22
JDD
11
edited Nov 28 '18 at 5:32
asked Nov 28 '18 at 2:22
JDD
11
asked Nov 28 '18 at 2:22
JDD
11
asked Nov 28 '18 at 2:22
JDD
11
11
why do you call the constraints of the original problem quadratic?
– LinAlg
Nov 28 '18 at 2:48
Are they not? Isn't the Lyapunov equality usually associated with the form $x^TPx$?
– JDD
Nov 28 '18 at 2:50
That's irrelevant to the immediate task at hand. The point is your problem in $P$ already has a linear objective, a set of linear equations in $P$, and a semidefinite constraint.
– Michael Grant
Nov 28 '18 at 2:56
$C^{T}X$ is a matrix rather than a scalar- are you actually meant to be minimizing $mbox{tr}(C^{T}X)$?
– Brian Borchers
Nov 28 '18 at 5:01
Michael Grant, I think I understand. Are you saying that everything is already correct for applying KKT, similar to what I added to the original question?
– JDD
Nov 28 '18 at 5:03
|
show 1 more comment
why do you call the constraints of the original problem quadratic?
– LinAlg
Nov 28 '18 at 2:48
Are they not? Isn't the Lyapunov equality usually associated with the form $x^TPx$?
– JDD
Nov 28 '18 at 2:50
That's irrelevant to the immediate task at hand. The point is your problem in $P$ already has a linear objective, a set of linear equations in $P$, and a semidefinite constraint.
– Michael Grant
Nov 28 '18 at 2:56
$C^{T}X$ is a matrix rather than a scalar- are you actually meant to be minimizing $mbox{tr}(C^{T}X)$?
– Brian Borchers
Nov 28 '18 at 5:01
Michael Grant, I think I understand. Are you saying that everything is already correct for applying KKT, similar to what I added to the original question?
– JDD
Nov 28 '18 at 5:03
why do you call the constraints of the original problem quadratic?
– LinAlg
Nov 28 '18 at 2:48
why do you call the constraints of the original problem quadratic?
– LinAlg
Nov 28 '18 at 2:48
Are they not? Isn't the Lyapunov equality usually associated with the form $x^TPx$?
– JDD
Nov 28 '18 at 2:50
Are they not? Isn't the Lyapunov equality usually associated with the form $x^TPx$?
– JDD
Nov 28 '18 at 2:50
That's irrelevant to the immediate task at hand. The point is your problem in $P$ already has a linear objective, a set of linear equations in $P$, and a semidefinite constraint.
– Michael Grant
Nov 28 '18 at 2:56
That's irrelevant to the immediate task at hand. The point is your problem in $P$ already has a linear objective, a set of linear equations in $P$, and a semidefinite constraint.
– Michael Grant
Nov 28 '18 at 2:56
$C^{T}X$ is a matrix rather than a scalar- are you actually meant to be minimizing $mbox{tr}(C^{T}X)$?
– Brian Borchers
Nov 28 '18 at 5:01
$C^{T}X$ is a matrix rather than a scalar- are you actually meant to be minimizing $mbox{tr}(C^{T}X)$?
– Brian Borchers
Nov 28 '18 at 5:01
Michael Grant, I think I understand. Are you saying that everything is already correct for applying KKT, similar to what I added to the original question?
– JDD
Nov 28 '18 at 5:03
Michael Grant, I think I understand. Are you saying that everything is already correct for applying KKT, similar to what I added to the original question?
– JDD
Nov 28 '18 at 5:03
|
show 1 more comment
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why do you call the constraints of the original problem quadratic?
– LinAlg
Nov 28 '18 at 2:48
Are they not? Isn't the Lyapunov equality usually associated with the form $x^TPx$?
– JDD
Nov 28 '18 at 2:50
That's irrelevant to the immediate task at hand. The point is your problem in $P$ already has a linear objective, a set of linear equations in $P$, and a semidefinite constraint.
– Michael Grant
Nov 28 '18 at 2:56
$C^{T}X$ is a matrix rather than a scalar- are you actually meant to be minimizing $mbox{tr}(C^{T}X)$?
– Brian Borchers
Nov 28 '18 at 5:01
Michael Grant, I think I understand. Are you saying that everything is already correct for applying KKT, similar to what I added to the original question?
– JDD
Nov 28 '18 at 5:03