<p>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.</p>

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

<p>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.</p>

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

<p>\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*}</p>

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

CC BY-SA 3.0
2016-05-10T17:17:22.273Z
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.
123
CC BY-SA 3.0
2016-05-11T17:55:30.757Z
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.
Fix.B.
CC BY-SA 3.0
2016-05-12T13:18:27.603Z
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 ?
PostDocing
<microeconomics><adverse-selection>
Binding and slack constraints in adverse selection problem
PostDocing
# Binding and slack constraints in adverse selection problem

(Asked by: username_0 on 2016-05-10)

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.

**Tags:** microeconomics, adverse-selection

> **Comments:**
> **username_1**: 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.  
> **username_2**: 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.  
> **username_0**: 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 ?  

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## 0 answers

