# Meaning of Lagrange multiplier always being 0?

I'm optimizing a family of objectives over $x_1,\ldots,x_n$ with a single constraint that $x$'s add up to $1$, and when using method of Lagrange multipliers, the single multiplier always ends up $0$ for members of this family... does this tell us anything interesting about these objectives?

Edit: the answer was -- it means the constraint it inactive, ie, removing the constraint doesn't change the answer of optimization problem

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It seems that the constraints are always satisfied. Can you post the formulation of your problem? –  Tom Stephens Aug 10 '10 at 2:46
One more thing... are you solving this numerically? Your algorithm may be choosing $q$'s satisfying the conservation constraint that $\sum q_i = 1$. –  Tom Stephens Aug 10 '10 at 2:59

## 1 Answer

One of two things that I can think of are happening here. The global minimum (or max, whatever your goal is here) of your function may actually lie within the simplex defined by $\sum_{i=1}^{4} q_{i} = 1$. In this case your algorithm is finding the correct solution and the constraint is trivially satisfied. (Consider minimizing $x^2+y^2$ constrained to the line $x = 0$, so the Lagrange formulation is $L(x,y,\nu) = x^2 + y^2 + \nu x$. Solving $\nabla L=0$ demands that $\nu=0$. )

Alternatively, your algorithm could be sampling only points that already satisfy the constraint - so you are only checking those points (intentionally or otherwise). Then the algorithm is hopefully finding a local minimum (or max) along this constraint and your multiplier will always be zero.

I hope this helps you identify where a problem may be in your model - or clears up a conceptual issue and allows you to move forward with the solution you have. As far as interpreting the meaning of Lagrange multipliers, look here.

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OK, good point, for that objective it seems the former holds -- maximum is inside the simplex. –  Yaroslav Bulatov Aug 10 '10 at 4:48