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“Convert” quadratic constraint to quadratic objective


Linear programming with one quadratic equality constraintHow to linearize a quadratic objective function with linear constraints?Minimizing quadratic objective function on the unit $ell_1$ sphereLasso with non-linear objectiveLinear objective function with quadratic constraintsIs it possible to delete a constraint after replacing it in another constraint?Optimization problem with quadratic objective and a bilinear constraintOptimization of linear objective with non-convex quadratic constraintMinimizing quadratic objective subject to a quadratic equality constraintRank-dependent constraints in linear programming













1












$begingroup$


I have a large sparse quadratic optimization problem with a single quadratic constraint:



$$beginarrayll textmaximize & c'x\ textsubject to & l leq Ax leq u\ & x'Qx + b'x leq u_qendarray$$



where



  • $x = (x_1, x_2, dots, x_n)'$.


  • $A$ is a $k times n$ matrix of linear constraint coefficients.


  • $l$ and $u$ are $k times 1$ vectors of lower and upper bounds, respectively.


  • $Q$ is a $n times n$ positive definite matrix.


  • $b$ is a $n times 1$ vector.


  • $u_q$ is a scalar.


The solver I am using can handle quadratic objective function but only linear constraints. I was wondering if there is a way to reformulate the problem as:



$$beginarrayll textminimize & hat c'x + x'hat Qx\ textsubject to & hat l leq hat Ax leq hat u\endarray$$










share|cite|improve this question











$endgroup$











  • $begingroup$
    If you actually have the inequality constraint with $l_q$, I do not believe so. But if you do not, then it might be possible.
    $endgroup$
    – Alex Shtof
    Mar 22 at 17:20










  • $begingroup$
    Since $Q$ is positive definite, try to convert $x'Qx + b'x$ to something like $(x-c)' M (x - c)$. Then make $y := x - c$ and work with $y$.
    $endgroup$
    – Rodrigo de Azevedo
    Mar 23 at 5:05










  • $begingroup$
    This is a non-convex problem because the constraint $ l_q leq x'Qx + b'x $ is not convex. I think it should be reformulatable as a Mixed-Integer QP (MIQP).
    $endgroup$
    – Mark L. Stone
    Mar 23 at 19:37










  • $begingroup$
    Apologies, there is no $l_q$ bound actually. Edited the problem setup now.
    $endgroup$
    – user10203644
    Mar 25 at 9:51















1












$begingroup$


I have a large sparse quadratic optimization problem with a single quadratic constraint:



$$beginarrayll textmaximize & c'x\ textsubject to & l leq Ax leq u\ & x'Qx + b'x leq u_qendarray$$



where



  • $x = (x_1, x_2, dots, x_n)'$.


  • $A$ is a $k times n$ matrix of linear constraint coefficients.


  • $l$ and $u$ are $k times 1$ vectors of lower and upper bounds, respectively.


  • $Q$ is a $n times n$ positive definite matrix.


  • $b$ is a $n times 1$ vector.


  • $u_q$ is a scalar.


The solver I am using can handle quadratic objective function but only linear constraints. I was wondering if there is a way to reformulate the problem as:



$$beginarrayll textminimize & hat c'x + x'hat Qx\ textsubject to & hat l leq hat Ax leq hat u\endarray$$










share|cite|improve this question











$endgroup$











  • $begingroup$
    If you actually have the inequality constraint with $l_q$, I do not believe so. But if you do not, then it might be possible.
    $endgroup$
    – Alex Shtof
    Mar 22 at 17:20










  • $begingroup$
    Since $Q$ is positive definite, try to convert $x'Qx + b'x$ to something like $(x-c)' M (x - c)$. Then make $y := x - c$ and work with $y$.
    $endgroup$
    – Rodrigo de Azevedo
    Mar 23 at 5:05










  • $begingroup$
    This is a non-convex problem because the constraint $ l_q leq x'Qx + b'x $ is not convex. I think it should be reformulatable as a Mixed-Integer QP (MIQP).
    $endgroup$
    – Mark L. Stone
    Mar 23 at 19:37










  • $begingroup$
    Apologies, there is no $l_q$ bound actually. Edited the problem setup now.
    $endgroup$
    – user10203644
    Mar 25 at 9:51













1












1








1





$begingroup$


I have a large sparse quadratic optimization problem with a single quadratic constraint:



$$beginarrayll textmaximize & c'x\ textsubject to & l leq Ax leq u\ & x'Qx + b'x leq u_qendarray$$



where



  • $x = (x_1, x_2, dots, x_n)'$.


  • $A$ is a $k times n$ matrix of linear constraint coefficients.


  • $l$ and $u$ are $k times 1$ vectors of lower and upper bounds, respectively.


  • $Q$ is a $n times n$ positive definite matrix.


  • $b$ is a $n times 1$ vector.


  • $u_q$ is a scalar.


The solver I am using can handle quadratic objective function but only linear constraints. I was wondering if there is a way to reformulate the problem as:



$$beginarrayll textminimize & hat c'x + x'hat Qx\ textsubject to & hat l leq hat Ax leq hat u\endarray$$










share|cite|improve this question











$endgroup$




I have a large sparse quadratic optimization problem with a single quadratic constraint:



$$beginarrayll textmaximize & c'x\ textsubject to & l leq Ax leq u\ & x'Qx + b'x leq u_qendarray$$



where



  • $x = (x_1, x_2, dots, x_n)'$.


  • $A$ is a $k times n$ matrix of linear constraint coefficients.


  • $l$ and $u$ are $k times 1$ vectors of lower and upper bounds, respectively.


  • $Q$ is a $n times n$ positive definite matrix.


  • $b$ is a $n times 1$ vector.


  • $u_q$ is a scalar.


The solver I am using can handle quadratic objective function but only linear constraints. I was wondering if there is a way to reformulate the problem as:



$$beginarrayll textminimize & hat c'x + x'hat Qx\ textsubject to & hat l leq hat Ax leq hat u\endarray$$







optimization convex-analysis convex-optimization constraints qclp






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 25 at 9:58







user10203644

















asked Mar 22 at 14:43









user10203644user10203644

83




83











  • $begingroup$
    If you actually have the inequality constraint with $l_q$, I do not believe so. But if you do not, then it might be possible.
    $endgroup$
    – Alex Shtof
    Mar 22 at 17:20










  • $begingroup$
    Since $Q$ is positive definite, try to convert $x'Qx + b'x$ to something like $(x-c)' M (x - c)$. Then make $y := x - c$ and work with $y$.
    $endgroup$
    – Rodrigo de Azevedo
    Mar 23 at 5:05










  • $begingroup$
    This is a non-convex problem because the constraint $ l_q leq x'Qx + b'x $ is not convex. I think it should be reformulatable as a Mixed-Integer QP (MIQP).
    $endgroup$
    – Mark L. Stone
    Mar 23 at 19:37










  • $begingroup$
    Apologies, there is no $l_q$ bound actually. Edited the problem setup now.
    $endgroup$
    – user10203644
    Mar 25 at 9:51
















  • $begingroup$
    If you actually have the inequality constraint with $l_q$, I do not believe so. But if you do not, then it might be possible.
    $endgroup$
    – Alex Shtof
    Mar 22 at 17:20










  • $begingroup$
    Since $Q$ is positive definite, try to convert $x'Qx + b'x$ to something like $(x-c)' M (x - c)$. Then make $y := x - c$ and work with $y$.
    $endgroup$
    – Rodrigo de Azevedo
    Mar 23 at 5:05










  • $begingroup$
    This is a non-convex problem because the constraint $ l_q leq x'Qx + b'x $ is not convex. I think it should be reformulatable as a Mixed-Integer QP (MIQP).
    $endgroup$
    – Mark L. Stone
    Mar 23 at 19:37










  • $begingroup$
    Apologies, there is no $l_q$ bound actually. Edited the problem setup now.
    $endgroup$
    – user10203644
    Mar 25 at 9:51















$begingroup$
If you actually have the inequality constraint with $l_q$, I do not believe so. But if you do not, then it might be possible.
$endgroup$
– Alex Shtof
Mar 22 at 17:20




$begingroup$
If you actually have the inequality constraint with $l_q$, I do not believe so. But if you do not, then it might be possible.
$endgroup$
– Alex Shtof
Mar 22 at 17:20












$begingroup$
Since $Q$ is positive definite, try to convert $x'Qx + b'x$ to something like $(x-c)' M (x - c)$. Then make $y := x - c$ and work with $y$.
$endgroup$
– Rodrigo de Azevedo
Mar 23 at 5:05




$begingroup$
Since $Q$ is positive definite, try to convert $x'Qx + b'x$ to something like $(x-c)' M (x - c)$. Then make $y := x - c$ and work with $y$.
$endgroup$
– Rodrigo de Azevedo
Mar 23 at 5:05












$begingroup$
This is a non-convex problem because the constraint $ l_q leq x'Qx + b'x $ is not convex. I think it should be reformulatable as a Mixed-Integer QP (MIQP).
$endgroup$
– Mark L. Stone
Mar 23 at 19:37




$begingroup$
This is a non-convex problem because the constraint $ l_q leq x'Qx + b'x $ is not convex. I think it should be reformulatable as a Mixed-Integer QP (MIQP).
$endgroup$
– Mark L. Stone
Mar 23 at 19:37












$begingroup$
Apologies, there is no $l_q$ bound actually. Edited the problem setup now.
$endgroup$
– user10203644
Mar 25 at 9:51




$begingroup$
Apologies, there is no $l_q$ bound actually. Edited the problem setup now.
$endgroup$
– user10203644
Mar 25 at 9:51










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