On the implementation and strengthening of intersection cuts for QCQPs

Abstract

The generation of strong linear inequalities for QCQPs has been recently tackled by a number of authors using the intersection cut paradigm—a highly studied tool in integer programming whose flexibility has triggered these renewed efforts in non-linear settings. In this work, we consider intersection cuts using the recently proposed construction of maximal quadratic-free sets. Using these sets, we derive closed-form formulas to compute intersection cuts which allow for quick cut-computations by simply plugging-in parameters associated to an arbitrary quadratic inequality being violated by a vertex of an LP relaxation. Additionally, we implement a cut-strengthening procedure that dates back to Glover and evaluate these techniques with extensive computational experiments.

Explicit coefficient computations

In this section we present the computations of the coefficients in the quadratic equations presented in Sect. 4.1. These coefficients are presented in Tables 1, 2.

1.1 Case 1: ${\bar{b}}_{I_0} = 0$ and $\kappa = 0$

In this case, we would like to find the coefficients of (4.1), which reads

$$\begin{aligned} \sqrt{A_r t^2 + B_r t + C_r} - (D_r t + E_r) = 0, \end{aligned}$$

when we are using it to represent equation (4.2), which reads

$$\begin{aligned} \Vert y({\bar{s}}+ t r)\Vert -\frac{x({\bar{s}})^{\mathsf {T}}}{\Vert x({\bar{s}})\Vert } x({\bar{s}}+ t r) = 0. \end{aligned}$$

Recall that,

$$\begin{aligned} x(s)&= \left( \sqrt{\theta _i} v_i^{\mathsf {T}}\left( s+ \frac{b}{2\theta _i}\right) \right) _{i \in I_+} \\ y(s)&= \left( \sqrt{-\theta _i} v_i^{\mathsf {T}}\left( s+ \frac{b}{2\theta _i}\right) \right) _{i \in I_-}. \end{aligned}$$

Thus,

$$\begin{aligned} x({\bar{s}}+ tr)&= x({\bar{s}}) + t \left( \sqrt{\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_+} \\ y({\bar{s}}+ tr)&= y({\bar{s}}) + t \left( \sqrt{-\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_-} \end{aligned}$$

With this, we process the first component of (4.2):

$$\begin{aligned} \Vert y({\bar{s}}+ t r)\Vert ^2&= \left\| y({\bar{s}}) + t \left( \sqrt{-\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_-}\right\| ^2 \\&= \Vert y({\bar{s}})\Vert ^2 + 2 t y({\bar{s}})^{\mathsf {T}}\left( \sqrt{-\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_-} + t^2 \left\| \left( \sqrt{-\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_-}\right\| ^2. \end{aligned}$$

On the other hand,

$$\begin{aligned} x({\bar{s}})^{\mathsf {T}}x({\bar{s}}+ t r)&= x({\bar{s}})^{\mathsf {T}}\left( x({\bar{s}}) + t \left( \sqrt{\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_+} \right) \\&= t x({\bar{s}})^{\mathsf {T}}\left( \sqrt{\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_+} + \Vert x({\bar{s}})\Vert ^2, \end{aligned}$$

from where we obtain

$$\begin{aligned} \frac{x({\bar{s}})^{\mathsf {T}}}{\Vert x({\bar{s}})\Vert } x({\bar{s}}+ t r) = t \frac{x({\bar{s}})^{\mathsf {T}}}{\Vert x({\bar{s}})\Vert } \left( \sqrt{\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_+} + \Vert x({\bar{s}})\Vert . \end{aligned}$$

Altogether, we see that

$$\begin{aligned} A_r&= \left\| \left( \sqrt{-\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_-}\right\| ^2&= -\sum _{i \in I_-} \theta _i (v_i^{\mathsf {T}}r)^2, \\ B_r&= 2 y({\bar{s}})^{\mathsf {T}}\left( \sqrt{-\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_-}&= - 2\sum _{i \in I_-} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) (v_i^{\mathsf {T}}r), \\ C_r&= \Vert y({\bar{s}})\Vert ^2&= - \sum _{i \in I_-} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) ^2, \\ D_r&= \frac{x({\bar{s}})^{\mathsf {T}}}{\Vert x({\bar{s}})\Vert } \left( \sqrt{\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_+}&= \frac{1}{E} \sum _{i \in I_+} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) (v_i^{\mathsf {T}}r), \\ E_r&= \Vert x({\bar{s}})\Vert&= \sqrt{ \sum \nolimits _{i \in I_+} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) ^2}. \end{aligned}$$

1.2 Case 2: ${\bar{b}}_{I_0} = 0$ and $\kappa > 0$

As in the previous case, we would like to find the coefficients of (4.1), which reads

$$\begin{aligned} \sqrt{A_r t^2 + B_r t + C_r} - (D_r t + E_r) = 0, \end{aligned}$$

when we are using it to represent equation (4.3), which reads

$$\begin{aligned} \Vert y({\bar{s}}+ t r)\Vert - \frac{(x({\bar{s}}), \sqrt{\kappa })}{\Vert (x({\bar{s}}), \sqrt{\kappa })\Vert }^{\mathsf {T}}(x({\bar{s}}+ t r), \sqrt{\kappa }) = 0. \end{aligned}$$

It is not hard to see that the $A_r, B_r,$ and $C_r$ coefficients are the same as in the previous section. For the rest, we compute

$$\begin{aligned} (x({\bar{s}}), \sqrt{\kappa })^{\mathsf {T}}(x({\bar{s}}+ t r), \sqrt{\kappa })&= x({\bar{s}})^{\mathsf {T}}x({\bar{s}}+ t r) + \kappa \\&= t x({\bar{s}})^{\mathsf {T}}\left( \sqrt{\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_+} + \Vert x({\bar{s}})\Vert ^2 + \kappa \end{aligned}$$

thus,

$$\begin{aligned} \frac{(x({\bar{s}}), \sqrt{\kappa })}{\Vert (x({\bar{s}}), \sqrt{\kappa })\Vert }^{\mathsf {T}}(x({\bar{s}}+ t r), \sqrt{\kappa }) = t \frac{x({\bar{s}})^{\mathsf {T}}\left( \sqrt{\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_+}}{\Vert (x({\bar{s}}), \sqrt{\kappa })\Vert } + \Vert (x({\bar{s}}), \sqrt{\kappa })\Vert . \end{aligned}$$

Collecting all terms we obtain that in this case

$$\begin{aligned} A_r&= -\sum _{i \in I_-} \theta _i (v_i^{\mathsf {T}}r)^2, \\ B_r&= - 2\sum _{i \in I_-} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) (v_i^{\mathsf {T}}r), \\ C_r&= - \sum _{i \in I_-} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) ^2, \\ D_r&= \frac{1}{E} \sum _{i \in I_+} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) (v_i^{\mathsf {T}}r), \\ E_r&= \Vert (x({\bar{s}}), \sqrt{\kappa })\Vert = \sqrt{ \kappa + \sum _{i \in I_+} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) ^2}. \end{aligned}$$

1.3 Case 3: ${\bar{b}}_{I_0} = 0$ and $\kappa < 0$

We would like to find the coefficients of (4.1), which reads

$$\begin{aligned} \sqrt{A_r t^2 + B_r t + C_r} - (D_r t + E_r) = 0, \end{aligned}$$

when we are using it to represent equation (4.4), which reads

$$\begin{aligned} \Vert (y({\bar{s}}+ t r), \sqrt{-\kappa })\Vert - \frac{x({\bar{s}})^{\mathsf {T}}}{\Vert x({\bar{s}})\Vert } x({\bar{s}}+ t r) = 0. \end{aligned}$$

This case is almost identical as the previous case. Indeed, only the expressions defining the C and E coefficients change. As

$$\begin{aligned} \Vert (y({\bar{s}}+ t r), \sqrt{-\kappa })\Vert ^2 = \Vert y({\bar{s}}+ t r)\Vert ^2 -\kappa , \end{aligned}$$

we have that

$$\begin{aligned} A_r&= -\sum _{i \in I_-} \theta _i (v_i^{\mathsf {T}}r)^2, \\ B_r&= - 2\sum _{i \in I_-} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) (v_i^{\mathsf {T}}r), \\ C_r&= -\kappa - \sum _{i \in I_-} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) ^2, \\ D_r&= \frac{1}{E} \sum _{i \in I_+} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) (v_i^{\mathsf {T}}r), \\ E_r&= \sqrt{ \sum _{i \in I_+} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) ^2}. \end{aligned}$$

1.4 Case 4: ${\bar{b}}_{I_0} \ne 0$

1.4.1 Root defined by (4.8)

As before, we would like to compute the coefficients of

$$\begin{aligned} \sqrt{A_r t^2 + B_r t + C_r} - (D_r t + E_r) = 0, \end{aligned}$$

when we are using it to represent equation (4.8), which reads

$$\begin{aligned} \Vert {{\hat{y}}}({\bar{s}}+ tr)\Vert - \lambda ^{\mathsf {T}}{{\hat{x}}}({\bar{s}}+ tr) = 0. \end{aligned}$$

In this case we have

$$\begin{aligned} {{\hat{x}}}_{p_{+}+ 1}(s) = \frac{1}{2\sqrt{1+\kappa ^2}} \left( w(s) + \kappa + \sqrt{1+\kappa ^2}\right) , \\ {{\hat{y}}}_{p_{-}+ 1}(s) = \frac{1}{2\sqrt{1+\kappa ^2}} \left( w(s) + \kappa - \sqrt{1+\kappa ^2}\right) . \end{aligned}$$

Thus,

$$\begin{aligned} {{\hat{x}}}_{p_{+}+ 1}({\bar{s}}+ tr)&= t \left( \frac{w(r)}{2\sqrt{1+\kappa ^2}} \right) + {{\hat{x}}}_{p_{+}+ 1}({\bar{s}}), \\ {{\hat{y}}}_{p_{-}+ 1}({\bar{s}}+ tr)&= t \left( \frac{w(r)}{2\sqrt{1+\kappa ^2}} \right) + {{\hat{y}}}_{p_{-}+ 1}({\bar{s}}). \end{aligned}$$

Then,

$$\begin{aligned}&\Vert ({{\hat{y}}}({\bar{s}}+ t r)\Vert ^2 \\&\quad = \frac{1}{\sqrt{1+\kappa ^2}} \Vert y({\bar{s}}+ t r)\Vert ^2 + \left( t \left( \frac{w(r)}{2\sqrt{1+\kappa ^2}} \right) + {{\hat{y}}}_{p_{-}+1}({\bar{s}})\right) ^2 \\&\quad = \frac{1}{\sqrt{1+\kappa ^2}} \Vert y({\bar{s}}+ t r)\Vert ^2 + t^2 \left( \frac{w(r)^2}{4(1+\kappa ^2)} \right) + 2t \left( \frac{w(r)}{2\sqrt{1+\kappa ^2}} \right) {{\hat{y}}}_{p_{-}+1}({\bar{s}}) + {{\hat{y}}}_{p_{-}+1}({\bar{s}})^2\\ \end{aligned}$$

From here we obtain

$$\begin{aligned} A_r&= \frac{w(r)^2}{4(1+\kappa ^2)} - \frac{1}{\sqrt{1+\kappa ^2}} \sum _{i \in I_-} \theta _i (v_i^{\mathsf {T}}r)^2, \\ B_r&= 2 \left( \frac{w(r)}{2\sqrt{1+\kappa ^2}} \right) {{\hat{y}}}_{p_{-}+1}({\bar{s}}) - \frac{2}{\sqrt{1+\kappa ^2}} \sum _{i \in I_-} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) (v_i^{\mathsf {T}}r), \\ C_r&= {{\hat{y}}}_{p_{-}+1}({\bar{s}})^2 - \frac{1}{\sqrt{1+\kappa ^2}} \sum _{i \in I_-} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) ^2. \end{aligned}$$

For the other coefficients we compute:

$$\begin{aligned} {{\hat{x}}}({\bar{s}})^{\mathsf {T}}{{\hat{x}}}({\bar{s}}+ t r) =&\frac{1}{\sqrt{1+\kappa ^2}} x({\bar{s}})^{\mathsf {T}}x({\bar{s}}+ tr) + {{\hat{x}}}_{p_{+}+1}({\bar{s}}) {{\hat{x}}}_{p_{+}+1}({\bar{s}}+ tr) \\ =&\frac{1}{\sqrt{1+\kappa ^2}} x({\bar{s}})^{\mathsf {T}}x({\bar{s}}+ tr) + {{\hat{x}}}_{p_{+}+1}({\bar{s}}) \left( t \left( \frac{w(r)}{2\sqrt{1+\kappa ^2}} \right) + {{\hat{x}}}_{p_{+}+ 1}({\bar{s}})\right) \\ =&\frac{1}{\sqrt{1+\kappa ^2}} \left( t x({\bar{s}})^{\mathsf {T}}\left( \sqrt{\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_+} + \Vert x({\bar{s}})\Vert ^2 \right) \\&+ {{\hat{x}}}_{p_{+}+1}({\bar{s}}) \left( t \left( \frac{w(r)}{2\sqrt{1+\kappa ^2}} \right) + {{\hat{x}}}_{p_{+}+ 1}({\bar{s}})\right) \\ =&\frac{t}{\sqrt{1+\kappa ^2}} \left( x({\bar{s}})^{\mathsf {T}}\left( \sqrt{\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_+} + {{\hat{x}}}_{p_{+}+1}({\bar{s}}) \frac{w(r)}{2} \right) + \Vert {\hat{x}}({\bar{s}})\Vert ^2. \end{aligned}$$

so

$$\begin{aligned} \lambda ^{\mathsf {T}}{{\hat{x}}}({\bar{s}}+ t r) = \frac{t}{\sqrt{1+\kappa ^2} \Vert {\hat{x}}({\bar{s}}) \Vert }\left( x({\bar{s}})^{\mathsf {T}}\left( \sqrt{\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_+} + {{\hat{x}}}_{p_{+}+1}({\bar{s}}) \frac{w(r)}{2}\right) + \Vert {\hat{x}}({\bar{s}})\Vert . \end{aligned}$$

Thus,

$$\begin{aligned} D_r&= \frac{1}{E\sqrt{1+\kappa ^2}} \left( \sum _{i \in I_+} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) (v_i^{\mathsf {T}}r) + {{\hat{x}}}_{p_{+}+1}({\bar{s}}) \frac{w(r)}{2} \right) \\&= \frac{1}{E\sqrt{1+\kappa ^2}} \left( \sum _{i \in I_+} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) (v_i^{\mathsf {T}}r) + \frac{(w({\bar{s}}) + \kappa + \sqrt{1 + \kappa ^2})w(r)}{4\sqrt{1+\kappa ^2}} \right) ,\\ E_r&= \frac{1}{\root 4 \of {1+\kappa ^2}} \sqrt{ \frac{(w({\bar{s}}) + \kappa + \sqrt{1+\kappa ^2})^2}{4\sqrt{1+\kappa ^2}} + \sum _{i \in I_+} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) ^2}. \end{aligned}$$

1.4.2 Root defined by (4.9)

In this case, we look at the equation

$$\begin{aligned}\frac{\Vert x({\bar{s}})\Vert }{\sqrt{1+\kappa ^2}} \Vert y({\bar{s}}+ tr)\Vert + {{\hat{x}}}_{p_{+}+1}({\bar{s}}) {{\hat{y}}}_{p_{-}+ 1}({\bar{s}}+ tr) - {{\hat{x}}}({\bar{s}})^{\mathsf {T}}{{\hat{x}}}({\bar{s}}+ tr) = 0,\end{aligned}$$

which we rewrite as

$$\begin{aligned} \frac{\Vert x({\bar{s}})\Vert }{\sqrt{1+\kappa ^2}} \Vert y({\bar{s}}+ tr)\Vert - ({{\hat{x}}}({\bar{s}})^{\mathsf {T}}{{\hat{x}}}({\bar{s}}+ tr) - {{\hat{x}}}_{p_{+}+1}({\bar{s}}){{\hat{y}}}_{p_{-}+ 1}({\bar{s}}+ tr)) = 0. \end{aligned}$$

which is also an equation of the form

$$\begin{aligned}\sqrt{A_r t^2 + B_r t + C_r} - (D_r t + E_r) = 0.\end{aligned}$$

From the term involving $\Vert y({\bar{s}}+ tr)\Vert$ we see that

$$\begin{aligned} A_r&= -\frac{\Vert x({\bar{s}})\Vert ^2}{1+\kappa ^2}\sum _{i \in I_-} \theta _i (v_i^{\mathsf {T}}r)^2, \\ B_r&= - 2\frac{\Vert x({\bar{s}})\Vert ^2}{1+\kappa ^2}\sum _{i \in I_-} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) (v_i^{\mathsf {T}}r), \\ C_r&= -\frac{\Vert x({\bar{s}})\Vert ^2}{1+\kappa ^2}\sum _{i \in I_-} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) ^2. \end{aligned}$$

To find the remaining coefficients, we compute the term

$$\begin{aligned}&{{\hat{x}}}({\bar{s}})^{\mathsf {T}}{{\hat{x}}}({\bar{s}}+ tr) - {{\hat{x}}}_{p_{+}+1}({\bar{s}}){{\hat{y}}}_{p_{-}+ 1}({\bar{s}}+ tr) \\&= \frac{1}{\sqrt{1+\kappa ^2}} x({\bar{s}})^{\mathsf {T}}x({\bar{s}}+ tr) + {{\hat{x}}}_{p_{+}+1}({\bar{s}})({{\hat{x}}}_{p_{+}+1}({\bar{s}}+ tr) - {{\hat{y}}}_{p_{-}+ 1}({\bar{s}}+ tr)) \\&= \frac{1}{\sqrt{1+\kappa ^2}} x({\bar{s}})^{\mathsf {T}}x({\bar{s}}+ tr) + {{\hat{x}}}_{p_{+}+1}({\bar{s}}) \\&= \frac{1}{\sqrt{1+\kappa ^2}} \left( t x({\bar{s}})^{\mathsf {T}}\left( \sqrt{\theta _i} v_i^{\mathsf {T}}r\right) _{i \in I_+} + \Vert x({\bar{s}})\Vert ^2 \right) + {{\hat{x}}}_{p_{+}+1}({\bar{s}}). \end{aligned}$$

Thus,

$$\begin{aligned} D_r&= \frac{1}{\sqrt{1+\kappa ^2}} \sum _{i \in I_+} \theta _i \left( v_i^{\mathsf {T}}({\bar{s}}+ \frac{b}{2\theta _i})\right) (v_i^{\mathsf {T}}r), \\ E_r&= \frac{1}{\sqrt{1+\kappa ^2}} \left( \Vert x({\bar{s}})\Vert ^2 + \frac{w({\bar{s}}) + \kappa + \sqrt{1 + \kappa ^2}}{2}\right) . \end{aligned}$$