Convex Optimization: Dual Function Definition

In summary: Why are we concerned with ## \lambda \leq -1 ## when we have restricted ## \lambda \geq 0 ##?- What about the other part of what I said, about the maximal value being 5?
  • #1
Master1022
611
117
Homework Statement
Given the following problem: find the dual and the dual solution.
Relevant Equations
Lagrange dual
Hi,

I was working through the following problem and I am getting confused with the solution's definition of the dual.

Problem:
Given the optimization problem:
minimize ## x^2 + 1 ##
s.t. ## (x - 2) (x - 4) \leq 0 ##

Attempt:
I can define the Lagrangian as:
[tex] L(x, \lambda) = (x^2 + 1) + \lambda (x^2 - 6x + 8) [/tex]
where ## \lambda \geq 0 ## (this is where the question arises in a later part of the question)

Now we can define the primal problem as: ## min_{x} \left( max_{\lambda} \left[ L(x, \lambda) \right] \right) ##
and thus the dual problem is: ## max_{\lambda} \left( min_{x} \left[ L(x, \lambda) \right] \right) ##

if we follow the algebra through, we see that the minimum of ## L(x, \lambda) ## occurs at ## x_{min} = \frac{3 \lambda}{\lambda + 1} ## and therefore the dual function ## g(\lambda) ## is defined as :
[tex] g(\lambda) = L(x_{min}, \lambda) = \frac{1 + 9\lambda - \lambda^2}{1 + \lambda} [/tex]

The answer now specifies that the dual function as:
[tex] g(\lambda) = \begin{cases} \frac{1 + 9\lambda - \lambda^2}{1 + \lambda} & \text{if } \lambda > -1 \\ -\infty & \text{if } x \leq - \infty \end{cases} [/tex]

Question: Why are we concerned with ## \lambda \leq -1 ## when we have restricted ## \lambda \geq 0 ##?
- Is this because we are dealing with ## g(\lambda) ##, which is inside the maximum function and thus the restriction isn't applied yet?

Thanks in advance.

[EDIT]: Also, I don't fully understand where the ## -\infty## comes from. We can substitute values ## \lambda < -1 ## and get finite numbers, so am slightly confused...
 
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  • #2
Master1022 said:
Homework Statement:: Given the following problem: find the dual and the dual solution.
Relevant Equations:: Lagrange dual

Hi,

I was working through the following problem and I am getting confused with the solution's definition of the dual.

Problem:
Given the optimization problem:
minimize ## x^2 + 1 ##
s.t. ## (x - 2) (x - 4) \leq 0 ##

Attempt:
I can define the Lagrangian as:
[tex] L(x, \lambda) = (x^2 + 1) + \lambda (x^2 - 6x + 8) [/tex]
where ## \lambda \geq 0 ## (this is where the question arises in a later part of the question)

Now we can define the primal problem as: ## min_{x} \left( max_{\lambda} \left[ L(x, \lambda) \right] \right) ##
and thus the dual problem is: ## max_{\lambda} \left( min_{x} \left[ L(x, \lambda) \right] \right) ##

if we follow the algebra through, we see that the minimum of ## L(x, \lambda) ## occurs at ## x_{min} = \frac{3 \lambda}{\lambda + 1} ## and therefore the dual function ## g(\lambda) ## is defined as :
[tex] g(\lambda) = L(x_{min}, \lambda) = \frac{1 + 9\lambda - \lambda^2}{1 + \lambda} [/tex]

The answer now specifies that the dual function as:
[tex] g(\lambda) = \begin{cases} \frac{1 + 9\lambda - \lambda^2}{1 + \lambda} & \text{if } \lambda > -1 \\ -\infty & \text{if } x \leq - \infty \end{cases} [/tex]

Question: Why are we concerned with ## \lambda \leq -1 ## when we have restricted ## \lambda \geq 0 ##?
- Is this because we are dealing with ## g(\lambda) ##, which is inside the maximum function and thus the restriction isn't applied yet?

Thanks in advance.

[EDIT]: Also, I don't fully understand where the ## -\infty## comes from. We can substitute values ## \lambda < -1 ## and get finite numbers, so am slightly confused...
Quick sanity check. ##(x - 2)(x - 4) \le 0 \Leftrightarrow 2 \le x \le 4##, so the minimal value of ##x^2 + 1## is attained when x = 2, thus ##x^2 + 1 \ge 5## is the minimum value on the interval I listed.

[tex] g(\lambda) = \begin{cases} \frac{1 + 9\lambda - \lambda^2}{1 + \lambda} & \text{if } \lambda > -1 \\ -\infty & \text{if } x \leq - \infty \end{cases} [/tex]
Part of this makes no sense to me. How can ##x \le -\infty##? Was this a typo, but should say ##\lambda \le -1##?
 
  • #3
Mark44 said:
Part of this makes no sense to me. How can ##x \le -\infty##? Was this a typo, but should say ##\lambda \le -1##?
Yes, apologies. This was a typo. It should read:

[tex] g(\lambda) = \begin{cases} \frac{1 + 9\lambda - \lambda^2}{1 + \lambda} & \text{if } \lambda > -1 \\ -\infty & \text{if } \lambda \leq - 1 \end{cases} [/tex]
 
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  • #4
What about the other part of what I said, about the maximal value being 5? If my understanding is correct, any work you do should agree with that.
 
  • #5
Thanks for the response @Mark44 !
Mark44 said:
What about the other part of what I said, about the maximal value being 5? If my understanding is correct, any work you do should agree with that.
Yeh that makes sense - I understood that part of the problem. We can differentiate that expression for ## g(\lambda) ## and then the optimal lambda (which is 2) leads us to dual problem optimal of 5. The strong duality makes sense because Slater's condition holds.

The parts that confused me were:
1. The definition of ## g(\lambda) ## : I was unsure why we needed to define it over all ## \lambda ## when we are optimizing over ## \lambda \geq 0 ##. The only reason I have is that the maximization (and hence restriction on ## \lambda ##) is applied after defining ## g (\lambda)## and therefore we need to define it over the whole range
2. Where did the ##- \infty## come from in the case of ## \lambda \leq -1 ##. I can understand for the case of -1, but not for numbers less than that.
 

FAQ: Convex Optimization: Dual Function Definition

1. What is the dual function in convex optimization?

The dual function in convex optimization is a lower bound on the optimal value of the primal problem. It is defined as the infimum of the Lagrangian function over all feasible solutions of the primal problem.

2. How is the dual function related to the primal problem?

The dual function is closely related to the primal problem as it provides a lower bound on its optimal value. This allows for the dual problem to be used as an approximation or relaxation of the primal problem.

3. What is the purpose of the dual function in convex optimization?

The dual function serves as a tool for solving convex optimization problems by providing a lower bound on the optimal value of the primal problem. It also allows for the use of duality theory to analyze and solve convex optimization problems.

4. How is the dual function calculated?

The dual function is calculated by taking the infimum of the Lagrangian function over all feasible solutions of the primal problem. This involves solving the primal problem and then optimizing the Lagrangian function with respect to the dual variables.

5. Can the dual function be used to solve non-convex optimization problems?

No, the dual function can only be used to solve convex optimization problems. This is because convex optimization problems have a unique optimal solution and the dual function provides a lower bound on this optimal value. For non-convex problems, there may be multiple optimal solutions and the dual function may not provide a tight bound on the optimal value.

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