Why do we need Complementary Slackness Condition for Karush-Kuhn-Tucker Conditions

Question

Complementary slackness condition (CSC) state that $\lambda_j[g_j(x) − c_j] = 0 \hspace{5pt} \text{for} \hspace{5pt} j = 1, ..., m.$ Therefore, every constraint either needs to be an equality constraint ($\lambda_j=0$) or we need to have an active constraint ($g_j(x) − c_j = 0$). What if we have a passive constraint? Why would that prevent us from finding the optimum point? In that case, couldn't we just ignore that constraint and find the optimum without it because it doesn't affect our optimal point? If yes, wouldn't that make CSC irrelevant?

Answer 1

Solving a non-linear programming (with inequality constraints) is about trial and error. You don't know a priori if a constraint is active. You consider all the possible cases satisfying your constraints and see if they optimize your objective function (maximize the utility function, for instance) .

Also $\lambda_j =0$ is not an equality constraint, it is just the multiplier attached to your constraint. You have a single distinct multiplier per each constraint.$\lambda_j =0$ if your constraint is NOT active. On the other hand, $\lambda_j >0$ if your constraint is active

The simplest example you can consider is:

Say you want to max. $U=xy$ , $s.t.$

$i)x+y\le 100$

$ii) x,y \ge0$

You build up the Lagrangian:

$L(x,y,\lambda)=xy+ \lambda(100-x-y)$

KT conditions:

i) $\frac{\partial L}{\partial x}=y-\lambda \le 0$ ; $x\ge 0$

Note that one between the partial derivative and $x$ must be zero. This is called complementary slackness and can be summarized as $x \frac{\partial L}{\partial x}=0$

ii) $\frac{\partial L}{\partial y}=x-\lambda \le 0$ ; $y\ge 0$,

which can be summarized as $y \frac{\partial L}{\partial y}=0$

iii) $\frac{\partial L}{\partial \lambda}=100-x-y \ge 0$ ; $\lambda \ge 0$,

which can be summerized as $\lambda \frac{\partial L}{\partial \lambda}=0$, and this is exactly the CSC you wrote in your question.

As I said, you don't know a priori which constraint is binding. You must consider all the possible cases satisfying your constraint and see if these maximize the utility function. In our example, it does not make sense to assume that $x$ or $y$ are zero, because o.w. $U(x,y)=0$. It does make sense to assume that $\frac{\partial L}{\partial x}=\frac{\partial L}{\partial y}=0$, from complementary slackness. Thus, since $\frac{\partial L}{\partial x}=\frac{\partial L}{\partial y}$, then, you get $y -\lambda = x - \lambda$, which means $x=y$, and since your wealth is $100$, you realize that $x=y=50$

Answer 2

Opportunity given, there can be cases where we have both $\lambda_j=0$ and $g_j(x^*) = c_j$. This happens when the unconstrained optimum $x^*_u$ equals the constrained one. In such a case, while the constraint is satisfied with equality at the optimum, it is not really binding, in the sense that it does not affect the solution.