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I have seen that QP can be rewritten as a SOCP from several resources online. Why is this the case? More precisely, using relaxation, QP can be written as $$ \min_{x,t} c^Tx + t \quad \text{ subject to } Ax = b, \quad Dx \le d, \quad \frac{1}{2}x^TQx \le t. $$ Resources I have encountered said $$ \frac{1}{2}x^TQx \le t \Longleftrightarrow \left|\left|\left(\frac{1}{\sqrt{2}}Q^{\frac{1}{2}}x, \frac{1}{2}(1-t)\right)\right|\right|_2 \le \frac{1}{2}(1+t) $$ I understand how these two constraints are the same, but why is the right-hand side of the form of the second-order norm cone? Clearly, $t$ is in both the left-hand side and the right-hand side, so I am not sure why this can be treated as the second-order norm-cone.

If there is an alternative way to see why QP is an SOCP, I would appreciate it if one could elaborate it.

Rodrigo de Azevedo
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user808843
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    Let $S$ be the second-order cone (the ice cream cone). Your reformulated constraint can be expressed as $M \begin{bmatrix} x \ t \end{bmatrix} + b \in S$, for a particular choice of the matrix $M$ and vector $b$. – littleO Jul 27 '20 at 05:01
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    Not every QP, only convex ($Q$ must be PSD). The new constraint is of the form "some scalar affine expression is greater than the norm of some vector affine expression" which is exactly (one way of) writing a SOC. See also https://docs.mosek.com/modeling-cookbook/cqo.html#convex-quadratic-sets and surrounding sections. – Michal Adamaszek Jul 27 '20 at 06:01
  • Is that a QP or a QCQP? – Rodrigo de Azevedo May 07 '23 at 10:33
  • https://math.stackexchange.com/questions/1932532/converting-from-qp-to-socp – RobPratt Jul 09 '24 at 01:49

1 Answers1

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Since $a^2 + b^2 \le c^2$, we obtain $\left \|\begin{bmatrix}a \\ b\end{bmatrix} \right\|_2^2 \le c^2$. For $1/2 \|Q^{1/2}x\|_2^2 \le t$, we rewrite $t = c^2 - b^2$: $t = ((1 + t)^2 - (1 - t)^2)/4$.

We immediately arrive: $a = Q^{1/2}x/\sqrt{2}$   $b = (1 - t)/2$   $c = (1 +t)/2$.

Specially, we notice that all of $a,b,c$ are linear functions of $(x,t)$. As we know that if $S$ is a convex set and $f$ is a linear function, then both $f(S)$ and $f^{-1}(S)$ are convex. So the set of $(x,t)$ is convex.

One obviously wrong approach is as follows:

$1/2 \|Q^{1/2}x\|_2^2 \le t$ $\rightarrow$ $\|Q^{1/2}x/\sqrt{2}\|_2 \le \sqrt{t}$.

We can not ensure that the set of $(x,t)$ is convex because the transformation $\sqrt{t}$ is not a linear function.

xbjin
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