Convex optimization lagrange multiplier nash bargaining problem

65 Views Asked by At

I have an optimization problem that looks like this:

let $\mathbf{X} = \begin{bmatrix} x_{0,0} & x_{0,1} \ ... & x_{0,L}\\ x_{1,0} & x_{1,1} \ ... & x_{1,L}\\ \vdots \\ x_{K,0} & x_{K,1} \ ... & x_{K,L}\\ \end{bmatrix}$

$\max_{\mathbf{X}} \sum_{L}\ln\phi_{L}$ where $\phi_{L} = \sum_{k} q_{k,L}x_{k, L}$,

$\sum_{L} x_{k,L} = N_{k}$

and $d_{L}\leq\phi_{L}$

$x_{k,L}$ being free variables, $d_{L}, q_{k,L}, N_{k}$ being constants.

I've tried solving with the LaGrange multiplier method, the lagrangian then became:

$\mathcal{L}(\mathbf{X}) = \sum_{L}\ln\phi_{L} + \sum_{k}\lambda_{k}\left(N_{k} - \sum_{L} x_{k,l}\right) + \sum_{L}m_{L}(\phi_{L}-d_{L}) $,

deriving and equalling to zero:

$\partial_{x_{k,L}}\mathcal{L} = 0 = \frac{q_{k,L}}{\phi_{L}} -\lambda_{k} + m_{L}q_{k,L} $

isolating $\phi_{L}$:

$\phi_{L} = \frac{q_{k,L}}{\lambda_{k} - q_{k,L} m_{L}}$

the KKT conditions being:

$m_{L}(d_{L} - \phi_{L}) = 0$ and

The problems arise when we perceive that the $\phi_{L}$ depends only on the variable L, but the RHS depends on k and L. What am i doing wrong here?