I have a quadratic equality constraint as below. \(x\) is a binary variable, \(z,y,w,v\) are real variables, the rest are constants. \(z = yx  Ayx  Byx  Cyx + Dwx + Evx + Fx + G(1x)\) I try to refer to link text to solve the quadratic constraint, but since there are 4 real variables, I need some idea how to linearize the quadratic constraint... Would appreciate any suggestion or new method. Well, I did something like this... probably introducing three new quadratic equality constraints will be neat? \( \begin{align} z \leq & \overline{y} x A\overline{y} x B\overline{y} x C\overline{y} x + D\overline{w} x + E\overline{v} x + Fx + G(1x)\\ z \geq & \underline{y}x A\underline{y} x B\underline{y} x C\underline{y} x + D\underline{w} x + E\underline{v} x + Fx + G(1x)\\ z \leq & [y\underline{y}(1x)] [A(y\underline{y}(1x))] [B(y\underline{y}(1x))] [C(y\underline{y}(1x))]\\ & + [D(w\underline{w}(1x))] + [E(v\underline{v}(1x))] + Fx + G(1x)\\ z \geq & [y\overline{y}(1x)] [A(y\overline{y}(1x))] [B(y\overline{y}(1x))] [C(y\overline{y}(1x))]\\ & + [D(w\overline{w}(1x))] + [E(v\overline{v}(1x))] + Fx + G(1x) \end{align} \) 
Every quadratic term of the constraint at hand consists of exactly one binary and one real(valued) variable. As the blog post points out:
Hence, you should be able to apply the provided reformulation "asis". answered 15 Dec '13, 05:22 fbahr ♦ PS: Start the reformulation by replacing \(yx\), \(wx\), and \(vx\) with real variables \(p,~q,~r\) respectively – which will make your equality constraint linear ...at the cost of introducing three new quadratic equality constraints, though. However, these three new constraints are actually good news, since they have the form \(p=yx,~q=wx,~r=vx\) – and that's where you can plug in the "template" discussed in @Paul Rubin's blog post.
(15 Dec '13, 06:12)
fbahr ♦
