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Currently I am reading a paper and the author has an optimization problem

$$\max_w\frac{w^2\alpha}{w^2\beta+v}$$

Then he substitutes $w^2$ with $x$ and defines an objective function using a Lagrangian multiplier $\lambda$ as follows

$$g(x)=\frac{x\alpha}{x\beta+v}-x\lambda$$

then taking the first two derivatives concludes that $g$ is convex and

$$x=\frac{\sqrt{\frac{av}{\lambda}}-v}{\beta}$$

is the optimum solution.

My question is: why do we introduce $-x\lambda$? or why dont we introduce something else such as $x\lambda$ in the function $g$?

Thank you very much.

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up vote 2 down vote accepted

Note the $w^2$ term in your function, so when he changes the variable to $x=w^2$, he implicitly assumes that $x>=0$, so actually he is considering the problem,

\begin{align} \max_{x}~\frac{x\alpha}{x\beta+v}\\ subject~to~x\geq0 \end{align} Note that he (most probably) substitutes with $x=w^{2}$, than the substitution you have showed. I will explain in detail. Let us consider a general case, consider you are dealing with the problem, \begin{align} \max_{x}~f_{o}(x) \\ subject~to~f_{c}(x)\geq0 \end{align} where $f_{o}(x)$ is the function you want to maximize and $f_c(x)$ is the constraint function. Now the turn of the lagrangian function comes. It is defined as \begin{align} g(x,\lambda)=f_{o}(x)-\lambda f_{c}(x) \end{align} Here $\lambda\geq 0$ is the lagrangian multiplier. The lagrangian function is like you move your constraint to your objective function with a penalty to it, the penalty being $\lambda$. Note that at the optimal solution point, your lagrangian function will also be equal to your objective function value at that point. But at other feasible points, the term $\lambda f_{c}(x)$ always adds a penalty by decreasing the objective function term. Now see this with your case.

EDIT--- In a pure mathematical sense, if the point $x^{*}$ is your maximizer, then there always exists a $\lambda^{*}$, such that $(x^{*},\lambda^{*})$ is a stationary point (a point where partial derivatives are zero) for $g(x,\lambda)$. So the idea is to search among the stationary points of the lagrangian multiplier. Note that all points are not optimal solutions, hence lagrangian multiplier technique gives a necessary condition. In the case of convex problems, this also becomes sufficient.

In an intuitive way, solving the lagrangian problem instead of the orginal problem makes sure that you don't forget the constraints while in the hurry to maximize the objective function. Note that the (positive) term $\lambda f_c (x)$ always resist your attempt to increase $g(x,\lambda)$ by acting as a negative term. Hope this makes it clear to you.

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Oh sorry. yes yes you are right. Mistake because of copy paste. I am correcting. –  Seyhmus Güngören Nov 6 '12 at 18:18
    
Ok this exactly describes my the point in the paper. So I guess if it is maximization then $-\lambda x$ if minimisation $\lambda x$ bu t still didnt get why. –  Seyhmus Güngören Nov 6 '12 at 18:39
    
@SeyhmusGüngören see the edit –  dineshdileep Nov 7 '12 at 2:55
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