[Ipopt] Question concerning presence of indefinite terms in function

[Stefan Vigerske]

Hi,

I think the only thing to add is that Ipopt may relax bounds a little, 
so setting the lower bound to 0.0 will probably not be sufficient. See 
also option bounds_relax_factor:
https://www.coin-or.org/Ipopt/documentation/node44.html#SECTION000114010000000000000

Stefan


On 11/30/2017 02:42 PM, Daniel Feenberg wrote:
> 
> 
> On Tue, 28 Nov 2017, Maxime Boulay wrote:
> 
>>
>> Hello,
>>
>>
>>
>> I am using ipopt to solve an objective function which contains an 
>> expression of the form
>> x*log(x/(a+x)). This results in a derivative of the form log(x/(a+x)) 
>> and a second derivative of the
>> form a/(x*(x+a)) when considering other terms in the function. Since 
>> this will lead to problems
>> whenever x = 0 or x < 0, I am not sure about the best way to treat 
>> those cases. I have tried equating
>> the expression to 0 whenever those cases are met but this seems wrong 
>> since the expression doesn’t
>> tend to 0 when approaching the critical values. This also seems to 
>> slow down the time taken to find a
>> solution whenever the initial values given or the solution include an 
>> x=0.
> 
> I hesitate to respond, since I am a rank beginner at IPOPT, but no one 
> else has, so I will risk embarrassment. The Fortran example program in 
> the package shows how to set limits on the x values (see X_L and X_U), 
> and in my limited experience, these are respected. Am I missing something?
> 
>> Given the fact that many of these may pop out since the function 
>> includes a summation, what would be
>> the most efficient and correct way to deal with this?
> 
> "pop out"?
> 
> Daniel Feenberg
> 
>>
>>
>>
>> Thank you.
>>
>>
>>
>>
>>
> 
> 
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