# Karush-Kuhn-Tucker conditions

A typical constrained optimisation problem is as follows. \begin{aligned} \min_{x\in\mathbb{R}^n}&f(x) \\ s.t.~h_i(x) &= 0 \\ g_j(x) &\leq 0 \end{aligned}

## Karush-Kuhn-Tucker conditions

If the negative of the gradient (of $f$) has any component along an equality constraint $h(x)=0$, then we can take small steps along this surface to reduce $f(x)$.

Since $\nabla h(x)$, the gradient of the equality constraint is always perpendicular to the constraint surface $h(x)=0$, at optimum, $-\nabla f(x)$ should be either parallel or anti-parallel to $\nabla h(x)$ $$-\nabla f(x) = \mu \nabla h(x)$$ A similar argument can be made for inequality constraints. These form KKT conditions. So at an optimum point $x^\ast$ we have, \begin{aligned} h_i(x^\ast)&=0 & g_j(x^\ast) &\leq 0 \\ \lambda_j g_j(x^\ast) &= 0 & \lambda_j &\geq 0 \end{aligned} $$\nabla f(x^\ast) +\sum_{i} \mu_i\nabla h(x^\ast) + \sum_j \lambda_j \nabla g_j(x^\ast)= 0$$ These are the KKT conditions for constrained optimisation.

## Lagrange multipliers

The method of Lagrange multipliers relies on KKT conditions. For a constrained optimisation problem, we introduce a Lagrange function \begin{aligned} \mathcal{L}(x,\mu,\lambda) = f(x) + {\mu}^{T} h(x) + {\lambda}^{T} g(x) \end{aligned} Stationary points are points where derivative of the function is zero.

The sationary points of Lagrange function satisfies all of the KKT conditions. Hence we can solve for $\nabla_x\mathcal{L} =0$ to find the optimum point of $f(x)$. $\nabla_\mu \mathcal{L}=0$ and $\nabla_\lambda \mathcal{L} =0$ gives us the constraints.

tags: optimisation