Optimization: How to find the dual problem?

[Master1022]

Homework Statement

Given the primal problem below: (i) is it convex?, (ii) find the dual problem

Relevant Equations

Lagrangian, hessian

Hi,

I am working on the following optimization problem, and am confused how to apply the Lagrangian in the following scenario:

Question:
Let us look at the following problem
\min_{x \in \mathbb{R}_{+} ^{n}} \sum_{j=1}^{m} x_j log(x_j)
\text{subject to} A \vec{x} \leq b \rightarrow A\vec{x} - b \leq \vec{0}
\sum_{j=1}^{n} x_j = 1

(i) Is this problem convex?
(ii) Find the dual problem

Attempt:
(i) For the convexity, I argued that it is convex and used the hessian argument to show that (although I don't know if I applied it correctly):
\frac{\partial ^2 (obj)}{\partial x_k ^2} = \frac{1}{x_k} \geq 0 \forall x_k \geq 0
\frac{\partial ^2 (obj)}{\partial x_k \partial x_j} = 0
Thus the second order convexity condition is satisfied and the problem is convex.

(ii) This is where I get confused as I am not sure how to deal with the equality constraint. We can define the Lagrangian to be:
L(\vec{x}, \vec{\lambda}, \vec{\nu}) = \sum_{j=1}^{m} x_j log(x_j) + \vec{\lambda}^{T} \left( A\vec{x} - b \right) + \text{something with inequality constraint}

The first thing that comes to mind is: -1 + \sum_{j=1}^{n} \nu_j x_j = 0. However, this ignores the fact that the one is also present in the expression. Other options I have thought about include:
(a) Introducing extra slack variables
(b) Re-formatting the equation

But I am not sure how to make any progress.

 

[BvU]

Hi,
(i) I agree
(ii) I don't even understand the inequality constraint. Is $\vec A$ an unknown vector of length $n$ and $b$ an unknown number $\in \mathbb{R}^n$ ?
And is there any limit on $m$ ? Or is $m=n$ ?

You make it look like $A \vec{x} - b =0$ is the the equality constraint. Isn't it the inequality constraint ?

And isn't $\displaystyle \sum_{j=1}^{n} x_j - 1 = 0$ the equality constraint ?

Anyway,

 

[Master1022]

Thanks for the reply!

Hi,
(i) I agree

Great

(ii) I don't even understand the inequality constraint. Is $\vec A$ an unknown vector of length $n$ and $b$ an unknown number $\in \mathbb{R}^n$ ?

$A$ is a matrix. $\vec{x}$ is a vector of dimension $n$. $\vec{b}$ is a vector whose dimension we are not told.

And is there any limit on $m$ ? Or is $m=n$ ?

Apologies, that was a typo and sum on $x_i log(x_i)$ should be from $j = 1$ to $m$

You make it look like $A \vec{x} - b =0$ is the the equality constraint. Isn't it the inequality constraint ?

Once again, another typo on my part. Yes, it is the inequality constraint

And isn't $\displaystyle \sum_{j=1}^{n} x_j - 1 = 0$ the equality constraint ?

Yes

I hope that makes more sense now.

 

[BvU]

$A$ is a matrix. $\vec{x}$ is a vector of dimension $n$. $\vec{b}$ is a vector whose dimension we are not told.

So how are we to interpret ${\sf A}\vec x -\vec b \le 0$ ? As an unknown number of equations $({\sf A}\vec x)_k -\vec b_k \le 0$ for $0\le k\le n$ ?

Re limit on $m$ :

Apologies, that was a typo and sum on $x_i log(x_i)$ should be from $j = 1$ to $m$

no typo, therefore: in post #1 you had $\displaystyle \min_{x \in \mathbb{R}_{+} ^{n}} \sum_{j=1}^{m} x_j\,\log x_j\$ . But I suppose we can take $1\le m\le n$

So the Lagrange problem with only the equality constraint would be something like $$
\sum_{j=1}^m(1+\log x_j) - \lambda _j = 0$$ and $\lambda_k =0$ for $k=m+1\;... \; n$, plus the constraint $\displaystyle\sum_{j = 1}^n x_i = 1$ ?

$\$

 

[Master1022]

Thanks for taking the time to respond.

So how are we to interpret ${\sf A}\vec x -\vec b \le 0$ ? As an unknown number of equations $({\sf A}\vec x)_k -\vec b_k \le 0$ for $0\le k\le n$ ?

Yes I suppose so. Why does the number of equations matter in terms of setting up the dual problem when we know it is a vector and we are just having an inner product between it and some vector $\nu$?

Re limit on $m$ :
no typo, therefore: in post #1 you had $\displaystyle \min_{x \in \mathbb{R}_{+} ^{n}} \sum_{j=1}^{m} x_j\,\log x_j\$ . But I suppose we can take $1\le m\le n$

Apologies, somehow I mis-typed it again (even after double checking the problem!) - it should be $n$ not $m$ in the limit.

So the Lagrange problem with only the equality constraint would be something like $$
\sum_{j=1}^m(1+\log x_j) - \lambda _j = 0$$ and $\lambda_k =0$ for $k=m+1\;... \; n$, plus the constraint $\displaystyle\sum_{j = 1}^n x_i = 1$ ?

$\$

Sorry, could you perhaps explain a bit further about how you got to this for the equality constraint?

Many thanks for taking the time to help me.

 

[BvU]

how you got to this for the equality constraint?

I have learned (a long long time ago ) that an extremum of $f$ under the condition $g=0$ can be found by requiring$$\nabla f -\lambda \nabla g = 0\ .$$
Here we have (after some back and forh ) $\ f = \displaystyle \sum_{j=1}^{n} x_j\,\log x_j\$ and $\ g= \displaystyle\left [\sum_{j = 1}^n x_i \right ]\ - 1\$ , so $\ \ {\partial f\over \partial x_j} = 1+ \log x_j \$ and $\ \ {\partial g\over \partial x_j} =1$

And that means the Lagrange equations are $$1+\log x_j - \lambda = 0 \qquad j = 1\ldots n\tag 1$$ plus the constraint $\displaystyle\sum_{j = 1}^n x_i = 1$ .

If I try this out for $n=3$ I get a very reasonable result ​

When I try to revert to the complete problem: In the notation here we have found $\nu$ (the role played by the $\lambda$ above. There is only one component because there is only one constraint).

So it's possible to write down the Lagrangian, more or less as you posted in #1, but now with the 'something with the equality constraint' properly filled in. And the dual Lagrangian function.
Which may well be all that is required for this exercise: you don't have to solve the thing

$\$

 

[Master1022]

Thanks @BvU for taking the time to write this out! I am going to take some time to thoroughly read and (try to) understand everything in the post before getting back to you with any follow-up questions.