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I have a simple QP as below:

$\min L(x,y) = (x-5.1)^2+y^2$

such that

$(x-3)^2+y^2\geq1$

$(x-5.3)^2+y^2\geq1$

$(x-7)^2+y^2\geq1$

Intuitively, I think the optimal solution of the problem is $x^*=4.3,y^*=0$.

I want the closest point x in a plane to (5.1,0), while it must be far enough from (3,0), (5.3,0), and (7,0).

I think I can use Lagrangian relaxation here, but I could not find a good starting point.

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1 Answer 1

up vote 4 down vote accepted

First of all just delete y. Modify the constraints and use dummy variables to get rid of functional inequalities

$$g_1(x)+s_1=-(x-3.0)^2+s_1=-1$$

$$g_2(x)+s_2=-(x-5.3)^2+s_2=-1$$

$$g_3(x)+s_3=-(x-7.0)^2+s_3=-1$$

$$s_1,s_2,s_3\ge 0$$

Construct your Lagrangian

$$Z=L(x)+\lambda_1(r_1-g_1(x)-s_1)+\lambda_2(r_2-g_2(x)-s_2)+\lambda_3(r_3-g_3(x)-s_3)$$

where $r_1=r_2=r_3=-1$. The first order conditions for regular variables are

$$\frac{\partial Z}{\partial x}= \frac{\partial L}{\partial x}-\lambda_1\frac{\partial g_1}{\partial x}-\lambda_2\frac{\partial g_2}{\partial x}-\lambda_3\frac{\partial g_3}{\partial x}=0$$

$$\frac{\partial Z}{\partial \lambda_1}= r_1-g_1-s_1=0$$

$$\frac{\partial Z}{\partial \lambda_2}= r_2-g_2-s_2=0$$

$$\frac{\partial Z}{\partial \lambda_3}= r_3-g_3-s_3=0$$

For non-negative dummy variables the Kuhn-Tucker conditions yield (I skip derivation since can be found anywhere)

$$s_1\cdot \frac{\partial Z}{\partial s_1}= -s_1\lambda_1=0$$

$$s_2\cdot \frac{\partial Z}{\partial s_2}= -s_2\lambda_2=0$$

$$s_3\cdot \frac{\partial Z}{\partial s_3}= -s_3\lambda_3=0$$

If you solve the system (I used NSolve of Mathematica) there are 7 sets of solution from which just 4 sets satisfy the conditions $s_1,s_2,s_3\ge0$. By checking the Lagrangian function the optimal set is found to be $$x=4.3\quad \lambda_1=0\quad \lambda_2=-0.8\quad \lambda_3=0\quad s_1=0.69\quad s_2=0\quad s_3=6.29$$

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