- #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...
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...