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When I am dealing with a constrained maximization, once I face constraints with inequality sign, I have to understand which one are binding and which others are slack.

If I find that a constraint is binding, I can easily substitute it in the objective function, but what if I find that a constraint is slack (it holds with inequality)? Should I keep it in the maximization problem or can I get rid of it?

I am asking it since my professor gives us two methods in order to understand if a constraint is slack: one is to compute the Lagrangian, using slackness conditions; the other is trying to understand if the constraint is implicitly satisfied, given the other constraints. In this last case, it seems I can get rid of implicitly satisfied constraints since they seem redundant to me.

I'll give you this problem as an example: there two types of person $\theta_L, \theta_H$ with $\theta_H>\theta_L$. To make the problem interesting, we suppose $\psi > \beta$.

\begin{gather*} max_{c_H,c_L,q_H,q_L}\ \psi(c_L - v(\frac{q_L}{\theta_L})) + (1-\psi)(c_H - v(\frac{q_H}{\theta_H})) \end{gather*} \begin{gather*} \beta(q_L-c_L)+(1-\beta)(q_H-c_H) \geq 0\ (BC) \end{gather*} \begin{gather*} c_L - v(\frac{q_L}{\theta_L})\geq 0 \ (IR_L) \end{gather*} \begin{gather*} c_H - v(\frac{q_H}{\theta_H}) \geq 0 \ (IR_H) \end{gather*}

In perfect information at the optimum, I know from the solutions that both $IR_H$ and $BC$ are binding, whereas $IR_L$ is slack.

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  • $\begingroup$ I think one option is to transform the inequality constraint s.t. it holds with equality and then to proceed using methods of equality constrained optimization. $\endgroup$
    – 123
    Commented May 10, 2016 at 17:17
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    $\begingroup$ Yes. That's the whole point. Some constraints are binding, others are not. A priori you don;t know which is which. Once you know which is which, you can 'solve the model', as you say by forgetting the non-binding ones and us gin the binding ones as equalities, because,hm, they are binding. $\endgroup$
    – Fix.B.
    Commented May 11, 2016 at 17:55
  • $\begingroup$ OK , so if I were able to prove a constraint to be slack, I can remove it from the Lagrangian since its multiplier would be equal to 0 ? $\endgroup$
    – PostDocing
    Commented May 12, 2016 at 13:18

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