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I am trying to find Lagrange Multipliers in QP problem:

$$\begin{aligned}\min_x~~&\frac{1}{2}\begin{bmatrix}x_1 & x_2\end{bmatrix}\begin{bmatrix}1 & 1\\1 & 2\end{bmatrix}\begin{bmatrix}x1\\x2\end{bmatrix} + \begin{bmatrix}8\\9\end{bmatrix}^T\begin{bmatrix}x1\\x2\end{bmatrix} \\ \mathrm{s.t}~~ &x_1 + x_2 = 0 \\ &2x_1 + x_2 = 0 \\ &3x_1 + 2x_2 = 0 \end{aligned} $$

Lagrangian of this problem is

$$L(x,\lambda) = \frac{1}{2}x_1^2 +x_1x_2 + x_2^2 + 8x_1 + 9x_2 + \lambda_1(x_1 + x_2) + \lambda_2(2x_1 + x_2) + \lambda_3(3x_1+2x_2)$$

$\nabla_xL(x,\lambda)$ set to zero gives (for feasible solution $x_1=0,~x_2=0)$

$$\begin{aligned}\lambda_1 + 2\lambda_2 + 3\lambda_3 &= -8\\ \lambda_1 + \lambda_2 + 2\lambda_3 &= -9\end{aligned}$$

And I ended up with two equations with three unknowns, although solving above QP problem with quadprog in MATLAB gives:

$$\lambda_1 = 0,~~\lambda_2 =11 ,~~\lambda_3 = -10$$

Am I missing some equation or how can I figure out that the first multiplier is zero?

  • 0
    Are you sure about your problem. The only admissible is $x_{1}=x_{2}=0$. About the multipliers, you have 2 variables and 3 constraints, why do you expect the three multipliers to be pinned down uniquelly? – 2017-01-24
  • 0
    In general it's possible for mutiple sets of Lagrange multipliers to be dual optimal. – 2017-01-24

1 Answers 1

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The third constraint is a consequence ( a sum) of the other two, throw it away, and be happy.