B Constrained Optimization with the KKT Approach

[SilverSoldier]

I'm reading the book Deep Learning by Ian Goodfellow, Yoshua Bengio, and Aaron Courville, and currently reading this chapter on numerical methods--specifically, the section on constrained optimization.

The book states the following.

Suppose we wish to minimize a function $f\left(\textbf{x}\right)$ over some set $$S=\left\{\textbf{x}\,|\,\forall\,i,g^{(i)}\left(\textbf{x}\right)=0,\,\forall\,j,h^{(j)}\left(\textbf{x}\right)\leq0\right\},$$ where $g^{(i)}$ and $h^{(j)}$ are functions called the equality constraints and inequality constraints respectively; i.e., we wish to find $\displaystyle\min_{\textbf{x}\in S}f\left(\textbf{x}\right)$.

To do this, we start by defining a function $L\left(\textbf{x}, \boldsymbol{\lambda}, \boldsymbol{\alpha}\right)$ called the generalized Lagrangian as follows; $$L\left(\textbf{x}, \boldsymbol{\lambda}, \boldsymbol{\alpha}\right)=f\left(\textbf{x}\right)+\sum_i\lambda_i\cdot g^{(i)}\left(\textbf{x}\right)+\sum_j\alpha_j\cdot h^{(j)}\left(\textbf{x}\right).$$ Then, $$\min_{\textbf{x}\in S}f\left(\textbf{x}\right)=\min_{\textbf{x}}\max_{\boldsymbol{\lambda}}\max_{\boldsymbol{\alpha}, \boldsymbol{\alpha}\geq0}L\left(\textbf{x}, \boldsymbol{\lambda}, \boldsymbol{\alpha}\right).$$ The way I understand this is that we start by picking some random values for $\textbf{x}$ and $\boldsymbol{\lambda}$, and then find the best non-negative $\boldsymbol{\alpha}$ for which $L\left(\textbf{x}, \boldsymbol{\lambda}, \boldsymbol{\alpha}\right)$ is maximum.

Say the $\textbf{x}$ we picked randomly happened to lie in $S$. Then we should end up at the result that $\boldsymbol{\alpha}=\textbf{0}$.
It remains now to maximize $L\left(\textbf{x}, \boldsymbol{\lambda}, \textbf{0}\right)$ over $\boldsymbol{\lambda}$, and the minimize the resulting function over $\textbf{x}$. Again, because the $\textbf{x}$ we have picked randomly happens to lie in $S$, $L\left(\textbf{x}, \boldsymbol{\lambda}, \textbf{0}\right)$ is already maximum for any $\boldsymbol{\lambda}$, because $\forall\,i, g^{(i)}\left(\textbf{x}\right)=0$.

Therefore, in such a case where the $\textbf{x}$ we pick randomly just so happens to lie in $S$, we see that $$\max_{\boldsymbol{\lambda}}\max_{\boldsymbol{\alpha}, \boldsymbol{\alpha}\geq0}L\left(\textbf{x}, \boldsymbol{\lambda}, \boldsymbol{\alpha}\right)=f\left(\textbf{x}\right).$$ What I do not understand is how minimizing $\displaystyle\max_{\boldsymbol{\lambda}}\max_{\boldsymbol{\alpha}, \boldsymbol{\alpha}\geq0}L\left(\textbf{x}, \boldsymbol{\lambda}, \boldsymbol{\alpha}\right)=f\left(\textbf{x}\right)$ over $\textbf{x}$ now will keep $\textbf{x}$ within $S$, and find a minimum in this constrained region.

If $$\max_{\boldsymbol{\lambda}}\max_{\boldsymbol{\alpha}, \boldsymbol{\alpha}\geq0}L\left(\textbf{x}, \boldsymbol{\lambda}, \boldsymbol{\alpha}\right)=f\left(\textbf{x}\right),$$ mustn't we have $$\min_{\textbf{x}}\max_{\boldsymbol{\lambda}}\max_{\boldsymbol{\alpha}, \boldsymbol{\alpha}\geq0}L\left(\textbf{x}, \boldsymbol{\lambda}, \boldsymbol{\alpha}\right)=\min_{\textbf{x}}f\left(\textbf{x}\right),$$ which is different from $\displaystyle\min_{\textbf{x}\in S}f\left(\textbf{x}\right)$?

 

[FactChecker]

It remains now to ..... minimize the resulting function over $\textbf{x}$. .....

mustn't we have $$\min_{\textbf{x}}\max_{\boldsymbol{\lambda}}\max_{\boldsymbol{\alpha}, \boldsymbol{\alpha}\geq0}L\left(\textbf{x}, \boldsymbol{\lambda}, \boldsymbol{\alpha}\right)=\min_{\textbf{x}}f\left(\textbf{x}\right),$$ which is different from $\displaystyle\min_{\textbf{x}\in S}f\left(\textbf{x}\right)$?

Yes. You already said that you need to minimize over $\textbf{x}$. If $\textbf{x}$ was picked randomly, then you would have to repeat this process enough to feel confident that you have found a minimum over $\textbf{x}$, or do a methodical search for the minimum.

 

[SilverSoldier]

Yes. You already said that you need to minimize over $\textbf{x}$. If $\textbf{x}$ was picked randomly, then you would have to repeat this process enough to feel confident that you have found a minimum over $\textbf{x}$, or do a methodical search for the minimum.

But it is necessary to find the minimum over values in $S$ only. How does an unconstrained minimization of $f$ over all values of $\textbf{x}$ result in its minimum over values in $S$?

 

[FactChecker]

But it is necessary to find the minimum over values in $S$ only. How does an unconstrained minimization of $f$ over all values of $\textbf{x}$ result in its minimum over values in $S$?

Oh, I am sorry. I missed the point. By maximizing over $\lambda_x$ and $\alpha_x \ge 0$ for any given $x$, you are finding the multipliers that provide the best penalty for violating the constraints of $g$ and $h$. So that forces the minimization over $x$ to stay within $S$.

UPDATE: After thinking more about this, I do not feel I am knowledgeable enough to give authoritative help here. I will leave this to others.

 

[Office_Shredder]

I think fact checker has it. If you pick a value of x for which $g^i(x)>0$ you can pick $\lambda_i>>>0$ and $L$ is very positive. Similarly if $g^i(x) <0$ you can pick $\lambda_i<<<<0$. So $g^i(x)=0$ when you minimize $L$.

But for the inequality constraints you only get $h^i(x)\leq0$ since otherwise we make $\alpha^i>>>>0$

 

[SilverSoldier]

I'm sorry I still don't understand it and I am quite confused. Shouldn't we have $\displaystyle\min_{\textbf{x}\in S}f\left(\textbf{x}\right)=\max_{\boldsymbol{\lambda}}\max_{\boldsymbol{\alpha}, \boldsymbol{\alpha}\geq0}\min_{\textbf{x}}L\left(\textbf{x}, \boldsymbol{\lambda}, \boldsymbol{\alpha}\right)$ or $\displaystyle\min_{\textbf{x}\in S}f\left(\textbf{x}\right)=\max_{\boldsymbol{\alpha}, \boldsymbol{\alpha}\geq0}\max_{\boldsymbol{\lambda}}\min_{\textbf{x}}L\left(\textbf{x}, \boldsymbol{\lambda}, \boldsymbol{\alpha}\right)$ instead?

Say we wanted to find the minimum of $\dfrac{x^2+y^2}{5}$ subject to the constraints $x^2+y-2\leq0$ and $10x-y-22=0$. The actual minimum, which is $1.6$ given these constraints, is achieved at $x=2$ and $y=-2$. To find this minimum using the KKT approach, we first define the generalized Lagrangian, as follows; $$\dfrac{x^2+y^2}{5}+\lambda\left(10x-y-22\right)+\alpha\left(x^2+y-2\right).$$ Say we start by picking $\left(1.8, -4\right)$ randomly for $\left(x,y\right)$. Notice $\left(1.8, -4\right)$ satisfies both constraints and so $\displaystyle\max_{\boldsymbol{\lambda}}\max_{\boldsymbol{\alpha}, \boldsymbol{\alpha}\geq0}L\left(\textbf{x}, \boldsymbol{\lambda}, \boldsymbol{\alpha}\right)=f\left(1.8, -4\right)=3.85$. What does it now mean to minimize this? Does this mean we now vary $x$ and $y$ until we look for values that minimize $\dfrac{x^2+y^2}{5}$?

Can you also please describe how this approach may be implemented in code?

 

[Office_Shredder]

I think you're thinking about the ordering wrong. The first thing is min over x. So you can imagine having some function $K(x)$ that e are trying to minimize. That function is max over $\lambda$, max over $\alpha$, $L(x,\lambda, \alpha)$. So the ordering in the OP says for each $x$, pick the worst choice of $\lambda$ and $\alpha$ to compute $K(x)$. Then find the $x$ that makes this as small as possible.

 

[FactChecker]

I'm sorry I still don't understand it and I am quite confused. Shouldn't we have $\displaystyle\min_{\textbf{x}\in S}f\left(\textbf{x}\right)=\max_{\boldsymbol{\lambda}}\max_{\boldsymbol{\alpha}, \boldsymbol{\alpha}\geq0}\min_{\textbf{x}}L\left(\textbf{x}, \boldsymbol{\lambda}, \boldsymbol{\alpha}\right)$ or $\displaystyle\min_{\textbf{x}\in S}f\left(\textbf{x}\right)=\max_{\boldsymbol{\alpha}, \boldsymbol{\alpha}\geq0}\max_{\boldsymbol{\lambda}}\min_{\textbf{x}}L\left(\textbf{x}, \boldsymbol{\lambda}, \boldsymbol{\alpha}\right)$ instead?

If the first thing you do is minimize wrt $x$ then you are already too late to try to keep $x$ within $S$. The first thing you must do is to maximize wrt $\lambda$ and $\alpha$ to force $x$ to stay within $S$.