[Ipopt] Ipopt for Convex Problems
Andreas Waechter
awaechter.iems at gmail.com
Mon May 12 09:52:17 EDT 2014
Bailin,
The problem with you constraint (1) is that its gradient will be zero if
all of the variables are zero at the optimal solution, and in that case
there will be no KKT multipliers for the optimal solution (the linear
independence constraint qualification fails) and Ipopt, which is looking
for those multipliers, will probably fail, or at least perform poorly.
This is why there are specialized methods for these second-order cone
constraints.
Andreas Waechter
Associate Professor
Department of Industrial Engineering and Management Sciences
McCormick School of Engineering
Northwestern University
Evanston, IL 60208
USA
http://users.iems.northwestern.edu/~andreasw/
On 05/11/2014 06:51 AM, Bailin Deng wrote:
> Hi all,
>
> A few years back, on the mailing list there was a post (http://list.coin-or.org/pipermail/ipopt/2009-November/001733.html) saying that Ipopt can find the global optimal solution for the following problem if f(x), d(x) are convex and c(x) is linear:
>
> min f(x)
> s.t. c(x) = 0
> d(x) <= 0
>
>
> Now if d(x) is not convex, but d(x) <=0 defines a convex feasible set, this is still a convex optimization problem. Can Ipopt still find the global minimum? One example is the following constraints:
>
> x_1^2 + x_2^2 - x_3^2 <=0, (1)
> x_3 >=0. (2)
>
> This defines a quadratic cone, which is convex. But the function x_1^2 + x_2^2 - x_3^2 itself is not convex.
>
>
> Interestingly, we can also write these constraints as
>
> \sqrt{ x_1^2 + x_2^2 } - x_3 <=0.
>
> In this case, function \sqrt{ x_1^2 + x_2^2 } - x_3 is convex, but it is not differentiable at points where x_1 = x_2 = 0. So to use it in Ipopt, we would wnat to write the constraints in the form of Equations (1) and (2). But then it is unclear to me whether Ipopt will still find the global minimum. Thanks a lot for your help!
>
>
> Best,
> Bailin
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