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Henrik Alsing Friberg
Henrik Alsing Friberg
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MOSEK hasMOSEK offers a whitepaper on quadratic constraints (https://docs.mosek.com/whitepapers/qmodel.pdfwhitepaper on quadratic constraints) in which it is concluded that if $Q$ has a suitable factor model, then it is often beneficial to represent your model in a way that exploits this factor model. In your case you have $Q=H^T H$ and although $H$ has no particular properties of interest (besides full rank with high probability), I suggest you try to reformulate $$ \frac{1}{2} x^T Q x + q^T x + q_0 \leq 0 $$

as $$ \frac{1}{2} y^T y + q^T x + q_0 \leq 0,\quad y = Hx, $$

or perhaps even better as $$ \frac{1}{2} y^T y + p^T y + q_0 \leq 0,\quad y = Hx, $$

where $p$ is the solution to $H^T p = q$, such that $p^T y = p^T H x = q^T x$ by definition. In your case, the potential benefit comes from having the quadratic coefficient matrix (the only nonlinear part) becoming an identity matrix which is numerically stable. There is no eigenvalue distribution to worry about.

MOSEK has a whitepaper on quadratic constraints (https://docs.mosek.com/whitepapers/qmodel.pdf) in which it is concluded that if $Q$ has a suitable factor model, then it is often beneficial to represent your model in a way that exploits this factor model. In your case you have $Q=H^T H$ and although $H$ has no particular properties of interest (besides full rank with high probability), I suggest you try to reformulate $$ \frac{1}{2} x^T Q x + q^T x + q_0 \leq 0 $$

as $$ \frac{1}{2} y^T y + q^T x + q_0 \leq 0,\quad y = Hx, $$

or even better as $$ \frac{1}{2} y^T y + p^T y + q_0 \leq 0,\quad y = Hx, $$

where $p$ is the solution to $H^T p = q$ such that $p^T y = p^T H x = q^T x$ by definition. In your case, the potential benefit comes from having the quadratic coefficient matrix (the only nonlinear part) becoming an identity matrix which is numerically stable. There is no eigenvalue distribution to worry about.

MOSEK offers a whitepaper on quadratic constraints in which it is concluded that if $Q$ has a suitable factor model, then it is often beneficial to represent your model in a way that exploits this factor model. In your case you have $Q=H^T H$ and although $H$ has no particular properties of interest (besides full rank with high probability), I suggest you try to reformulate $$ \frac{1}{2} x^T Q x + q^T x + q_0 \leq 0 $$

as $$ \frac{1}{2} y^T y + q^T x + q_0 \leq 0,\quad y = Hx, $$

or perhaps even better as $$ \frac{1}{2} y^T y + p^T y + q_0 \leq 0,\quad y = Hx, $$

where $p$ is the solution to $H^T p = q$, such that $p^T y = p^T H x = q^T x$ by definition. In your case, the potential benefit comes from having the quadratic coefficient matrix (the only nonlinear part) becoming an identity matrix which is numerically stable. There is no eigenvalue distribution to worry about.

Source Link
Henrik Alsing Friberg
Henrik Alsing Friberg
  • 2k
  • 8
  • 16

MOSEK has a whitepaper on quadratic constraints (https://docs.mosek.com/whitepapers/qmodel.pdf) in which it is concluded that if $Q$ has a suitable factor model, then it is often beneficial to represent your model in a way that exploits this factor model. In your case you have $Q=H^T H$ and although $H$ has no particular properties of interest (besides full rank with high probability), I suggest you try to reformulate $$ \frac{1}{2} x^T Q x + q^T x + q_0 \leq 0 $$

as $$ \frac{1}{2} y^T y + q^T x + q_0 \leq 0,\quad y = Hx, $$

or even better as $$ \frac{1}{2} y^T y + p^T y + q_0 \leq 0,\quad y = Hx, $$

where $p$ is the solution to $H^T p = q$ such that $p^T y = p^T H x = q^T x$ by definition. In your case, the potential benefit comes from having the quadratic coefficient matrix (the only nonlinear part) becoming an identity matrix which is numerically stable. There is no eigenvalue distribution to worry about.