Lagrange dual of quadratic optimization problem with quadratic equality constraintsQuadratic optimisation with quadratic equality constraintsHow can we constrain lagrange multipliers in svm dual by adding constraints in primal problem?Minimization problem with infinitely many variables and linear constraintsFinding the Lagrange DualMinimizing quadratic objective subject to a quadratic equality constraintAm I stating the theorem about the dual optimization problem correctly?How to solve this non-convex optimization problem?Minimize $ mboxtr ( X^T A X ) + lambda mboxtr ( X^T B ) $ subject to $ X^T X = I $ - Linear Matrix Function with Norm Equality ConstraintCan I get a closed form of this optimization problem?Data sets for linear inequality and equality constrained quadratic optimization.
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Lagrange dual of quadratic optimization problem with quadratic equality constraints
Quadratic optimisation with quadratic equality constraintsHow can we constrain lagrange multipliers in svm dual by adding constraints in primal problem?Minimization problem with infinitely many variables and linear constraintsFinding the Lagrange DualMinimizing quadratic objective subject to a quadratic equality constraintAm I stating the theorem about the dual optimization problem correctly?How to solve this non-convex optimization problem?Minimize $ mboxtr ( X^T A X ) + lambda mboxtr ( X^T B ) $ subject to $ X^T X = I $ - Linear Matrix Function with Norm Equality ConstraintCan I get a closed form of this optimization problem?Data sets for linear inequality and equality constrained quadratic optimization.
$begingroup$
What is the Lagrange dual of the following optimization problem in $w in mathbbR^2$?
$$beginarrayll textminimize & w^T Q , w\ textsubject to & w_1^2 = 1\ & w_2^2 = 1endarray$$
optimization duality-theorems non-convex-optimization qcqp
edited Mar 20 at 17:38
Rodrigo de Azevedo
13.1k41960
asked Mar 20 at 17:22
user7080user7080
174
$endgroup$
add a comment |
$begingroup$
What is the Lagrange dual of the following optimization problem in $w in mathbbR^2$?
$$beginarrayll textminimize & w^T Q , w\ textsubject to & w_1^2 = 1\ & w_2^2 = 1endarray$$
optimization duality-theorems non-convex-optimization qcqp
edited Mar 20 at 17:38
Rodrigo de Azevedo
13.1k41960
asked Mar 20 at 17:22
user7080user7080
174
$endgroup$
add a comment |
$begingroup$
What is the Lagrange dual of the following optimization problem in $w in mathbbR^2$?
$$beginarrayll textminimize & w^T Q , w\ textsubject to & w_1^2 = 1\ & w_2^2 = 1endarray$$
optimization duality-theorems non-convex-optimization qcqp
edited Mar 20 at 17:38
Rodrigo de Azevedo
13.1k41960
asked Mar 20 at 17:22
user7080user7080
174
$endgroup$
What is the Lagrange dual of the following optimization problem in $w in mathbbR^2$?
$$beginarrayll textminimize & w^T Q , w\ textsubject to & w_1^2 = 1\ & w_2^2 = 1endarray$$
optimization duality-theorems non-convex-optimization qcqp
optimization duality-theorems non-convex-optimization qcqp
edited Mar 20 at 17:38
Rodrigo de Azevedo
13.1k41960
asked Mar 20 at 17:22
user7080user7080
174
edited Mar 20 at 17:38
Rodrigo de Azevedo
13.1k41960
asked Mar 20 at 17:22
user7080user7080
174
edited Mar 20 at 17:38
Rodrigo de Azevedo
13.1k41960
edited Mar 20 at 17:38
Rodrigo de Azevedo
13.1k41960
edited Mar 20 at 17:38
Rodrigo de Azevedo
13.1k41960
13.1k41960
asked Mar 20 at 17:22
user7080user7080
174
asked Mar 20 at 17:22
user7080user7080
174
asked Mar 20 at 17:22
user7080user7080
174
174
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The standard representation of the constraints is$$omega_1^2=1iff omega^Tbeginbmatrix1&0\0&0endbmatrixomega=1\omega_2^2=1iff omega^Tbeginbmatrix0&0\0&1endbmatrixomega=1$$therefore
$$L(omega,lambda,nu)=omega^TQomega+omega^Tcdotleft(nu_1beginbmatrix1&0\0&0endbmatrix+nu_2beginbmatrix0&0\0&1endbmatrixright)cdotomega\=omega^Tcdotleft(Q+nu_1beginbmatrix1&0\0&0endbmatrix+nu_2beginbmatrix0&0\0&1endbmatrixright)cdotomega$$and the dual function would be:$$g(lambda,nu)=min_omegaL(omega,lambda,nu)$$which yields to the following dual problem:$$max_lambda,nu g(lambda,nu)\s.t.\lambdage 0$$
answered Mar 20 at 17:30
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
$begingroup$
Thank you. But this is only the Lagrange equation, right? How do I get the dual from this?
$endgroup$
– user7080
Mar 20 at 17:35
$begingroup$
Your welcome. The dual is only an optimization over $omega$. I will add it to my answer...
$endgroup$
– Mostafa Ayaz
Mar 20 at 17:38
$begingroup$
Thank you, but what is $lambda$ here?
$endgroup$
– user7080
Mar 20 at 17:46
$begingroup$
Actually, $lambda$ stands for a most general form as a factor of inequality constraints which don't exist here. In our case $$g(lambda,nu)=g(nu)$$
$endgroup$
– Mostafa Ayaz
Mar 20 at 17:49
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The standard representation of the constraints is$$omega_1^2=1iff omega^Tbeginbmatrix1&0\0&0endbmatrixomega=1\omega_2^2=1iff omega^Tbeginbmatrix0&0\0&1endbmatrixomega=1$$therefore
$$L(omega,lambda,nu)=omega^TQomega+omega^Tcdotleft(nu_1beginbmatrix1&0\0&0endbmatrix+nu_2beginbmatrix0&0\0&1endbmatrixright)cdotomega\=omega^Tcdotleft(Q+nu_1beginbmatrix1&0\0&0endbmatrix+nu_2beginbmatrix0&0\0&1endbmatrixright)cdotomega$$and the dual function would be:$$g(lambda,nu)=min_omegaL(omega,lambda,nu)$$which yields to the following dual problem:$$max_lambda,nu g(lambda,nu)\s.t.\lambdage 0$$
answered Mar 20 at 17:30
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
$begingroup$
Thank you. But this is only the Lagrange equation, right? How do I get the dual from this?
$endgroup$
– user7080
Mar 20 at 17:35
$begingroup$
Your welcome. The dual is only an optimization over $omega$. I will add it to my answer...
$endgroup$
– Mostafa Ayaz
Mar 20 at 17:38
$begingroup$
Thank you, but what is $lambda$ here?
$endgroup$
– user7080
Mar 20 at 17:46
$begingroup$
Actually, $lambda$ stands for a most general form as a factor of inequality constraints which don't exist here. In our case $$g(lambda,nu)=g(nu)$$
$endgroup$
– Mostafa Ayaz
Mar 20 at 17:49
add a comment |
$begingroup$
The standard representation of the constraints is$$omega_1^2=1iff omega^Tbeginbmatrix1&0\0&0endbmatrixomega=1\omega_2^2=1iff omega^Tbeginbmatrix0&0\0&1endbmatrixomega=1$$therefore
$$L(omega,lambda,nu)=omega^TQomega+omega^Tcdotleft(nu_1beginbmatrix1&0\0&0endbmatrix+nu_2beginbmatrix0&0\0&1endbmatrixright)cdotomega\=omega^Tcdotleft(Q+nu_1beginbmatrix1&0\0&0endbmatrix+nu_2beginbmatrix0&0\0&1endbmatrixright)cdotomega$$and the dual function would be:$$g(lambda,nu)=min_omegaL(omega,lambda,nu)$$which yields to the following dual problem:$$max_lambda,nu g(lambda,nu)\s.t.\lambdage 0$$
answered Mar 20 at 17:30
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
$begingroup$
Thank you. But this is only the Lagrange equation, right? How do I get the dual from this?
$endgroup$
– user7080
Mar 20 at 17:35
$begingroup$
Your welcome. The dual is only an optimization over $omega$. I will add it to my answer...
$endgroup$
– Mostafa Ayaz
Mar 20 at 17:38
$begingroup$
Thank you, but what is $lambda$ here?
$endgroup$
– user7080
Mar 20 at 17:46
$begingroup$
Actually, $lambda$ stands for a most general form as a factor of inequality constraints which don't exist here. In our case $$g(lambda,nu)=g(nu)$$
$endgroup$
– Mostafa Ayaz
Mar 20 at 17:49
add a comment |
$begingroup$
The standard representation of the constraints is$$omega_1^2=1iff omega^Tbeginbmatrix1&0\0&0endbmatrixomega=1\omega_2^2=1iff omega^Tbeginbmatrix0&0\0&1endbmatrixomega=1$$therefore
$$L(omega,lambda,nu)=omega^TQomega+omega^Tcdotleft(nu_1beginbmatrix1&0\0&0endbmatrix+nu_2beginbmatrix0&0\0&1endbmatrixright)cdotomega\=omega^Tcdotleft(Q+nu_1beginbmatrix1&0\0&0endbmatrix+nu_2beginbmatrix0&0\0&1endbmatrixright)cdotomega$$and the dual function would be:$$g(lambda,nu)=min_omegaL(omega,lambda,nu)$$which yields to the following dual problem:$$max_lambda,nu g(lambda,nu)\s.t.\lambdage 0$$
answered Mar 20 at 17:30
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
The standard representation of the constraints is$$omega_1^2=1iff omega^Tbeginbmatrix1&0\0&0endbmatrixomega=1\omega_2^2=1iff omega^Tbeginbmatrix0&0\0&1endbmatrixomega=1$$therefore
$$L(omega,lambda,nu)=omega^TQomega+omega^Tcdotleft(nu_1beginbmatrix1&0\0&0endbmatrix+nu_2beginbmatrix0&0\0&1endbmatrixright)cdotomega\=omega^Tcdotleft(Q+nu_1beginbmatrix1&0\0&0endbmatrix+nu_2beginbmatrix0&0\0&1endbmatrixright)cdotomega$$and the dual function would be:$$g(lambda,nu)=min_omegaL(omega,lambda,nu)$$which yields to the following dual problem:$$max_lambda,nu g(lambda,nu)\s.t.\lambdage 0$$
answered Mar 20 at 17:30
Mostafa AyazMostafa Ayaz
18.2k31040
edited Mar 20 at 17:40
answered Mar 20 at 17:30
Mostafa AyazMostafa Ayaz
18.2k31040
answered Mar 20 at 17:30
Mostafa AyazMostafa Ayaz
18.2k31040
answered Mar 20 at 17:30
Mostafa AyazMostafa Ayaz
18.2k31040
18.2k31040
$begingroup$
Thank you. But this is only the Lagrange equation, right? How do I get the dual from this?
$endgroup$
– user7080
Mar 20 at 17:35
$begingroup$
Your welcome. The dual is only an optimization over $omega$. I will add it to my answer...
$endgroup$
– Mostafa Ayaz
Mar 20 at 17:38
$begingroup$
Thank you, but what is $lambda$ here?
$endgroup$
– user7080
Mar 20 at 17:46
$begingroup$
Actually, $lambda$ stands for a most general form as a factor of inequality constraints which don't exist here. In our case $$g(lambda,nu)=g(nu)$$
$endgroup$
– Mostafa Ayaz
Mar 20 at 17:49
add a comment |
$begingroup$
Thank you. But this is only the Lagrange equation, right? How do I get the dual from this?
$endgroup$
– user7080
Mar 20 at 17:35
$begingroup$
Your welcome. The dual is only an optimization over $omega$. I will add it to my answer...
$endgroup$
– Mostafa Ayaz
Mar 20 at 17:38
$begingroup$
Thank you, but what is $lambda$ here?
$endgroup$
– user7080
Mar 20 at 17:46
$begingroup$
Actually, $lambda$ stands for a most general form as a factor of inequality constraints which don't exist here. In our case $$g(lambda,nu)=g(nu)$$
$endgroup$
– Mostafa Ayaz
Mar 20 at 17:49
$begingroup$
Thank you. But this is only the Lagrange equation, right? How do I get the dual from this?
$endgroup$
– user7080
Mar 20 at 17:35
$begingroup$
Thank you. But this is only the Lagrange equation, right? How do I get the dual from this?
$endgroup$
– user7080
Mar 20 at 17:35
$begingroup$
Your welcome. The dual is only an optimization over $omega$. I will add it to my answer...
$endgroup$
– Mostafa Ayaz
Mar 20 at 17:38
$begingroup$
Your welcome. The dual is only an optimization over $omega$. I will add it to my answer...
$endgroup$
– Mostafa Ayaz
Mar 20 at 17:38
$begingroup$
Thank you, but what is $lambda$ here?
$endgroup$
– user7080
Mar 20 at 17:46
$begingroup$
Thank you, but what is $lambda$ here?
$endgroup$
– user7080
Mar 20 at 17:46
$begingroup$
Actually, $lambda$ stands for a most general form as a factor of inequality constraints which don't exist here. In our case $$g(lambda,nu)=g(nu)$$
$endgroup$
– Mostafa Ayaz
Mar 20 at 17:49
$begingroup$
Actually, $lambda$ stands for a most general form as a factor of inequality constraints which don't exist here. In our case $$g(lambda,nu)=g(nu)$$
$endgroup$
– Mostafa Ayaz
Mar 20 at 17:49
add a comment |
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