Hi,
thanks for your reply.
There are only a few variables that have negative bounds.
But I treat them in a similar fashion. For example:
U("Mco2b_Transp"): R("Mco2b_Transp") + 10000 y("Mco2b_Transp") <= 10000
L("Mco2b_Transp"): R("Mco2b_Transp") - 10000 y("Mco2b_Transp") >= -10000
-10000 <= R("Mco2b_Transp") <= 10000
You can take a look at the model on
http://nikosonnenschein.de/modelMaximalCombi.lp
I am sorry for the attachment in my first email.
Best,
Nikolaus Sonnenschein
On Tue, Jun 10, 2008 at 8:06 PM, Xie Zhengwei <[EMAIL PROTECTED]> wrote:
> How do you deal with double non-zero bounds?
>
> On Tue, Jun 10, 2008 at 3:28 AM, Nikolaus Sonnenschein <
> [EMAIL PROTECTED]> wrote:
>
>> Hi.
>> I have a problem with a mip and I am struggling now for quite a long time
>> to find a solution for it.
>>
>> A short description of what I am doing:
>> 1. I solve a lp in which I maximize the value of the structural variable
>> z.
>> 2. Then I reformulate the lp into a mip in which the bounds of z are fixed
>> near the lp solution
>> and I utilize additional binary variables to switch between zero and
>> default bounds for all structural variables.
>> The new objective function is just the sum over all binary variables.
>>
>> I want to find the largest combination of structural variables that can be
>> removed from the system/constrained to zero
>> without a decrease in z.
>>
>> e.g. usage of of the binary variable y("R_PMDPHT") to switch between two
>> behaviors.
>>
>> U("R_PMDPHT"): R("R_PMDPHT") + 10000 y("R_PMDPHT") <= 10000
>> L("R_PMDPHT"): R("R_PMDPHT") >= 0
>> 0 <= R("R_PMDPHT") <= 10000
>>
>> If y("R_PMDPHT") becomes 1 than hopefully 0 <= R("R_PMDPHT") <= 0 .
>> But I get this strange solution when I solve the mip with
>> "glpsol --cpxlp --exact modelMaximalCombi.lp".
>>
>> From the solution output:
>> ...
>> 479 U("R_PMDPHT") 10000 10000
>> 791 L("R_PMDPHT") 0.001 0
>> ...
>> 229 y("R_PMDPHT") * 1 0 1
>> ...
>> 380 R("R_PMDPHT") 0.001 0 10000
>> ...
>>
>> So R("R_PMDPHT") has a value greater zero although it should not.
>> Yeah, I now the bounds are really big and surely mip rounding issues come
>> into play.
>> (I refer here to the "[Help-glpk] MIP rounding" email from Jean-Sebastien
>> Roy. Date: Sun, 6 Jul 2003 22:55:23 +0200).
>>
>> But even when I solve the problem with lp relaxation like this --
>> "glpsol --cpxlp --exact --nomip modelMaximalCombi.lp" --
>> I get the following result:
>> ...
>> 479 U("R_PMDPHT") NU 10000 10000
>> 0.0001
>> 791 L("R_PMDPHT") B 0.001 0
>>
>> ...
>> 229 y("R_PMDPHT") B 1 0
>> 1
>> ...
>> 380 R("R_PMDPHT") B 0.001 0 10000
>> ...
>>
>> y("R_PMDPHT") is exactly 1 and so R("R_PMDPHT") should be zero.
>> Is this normal? Am I overseeing something?
>> Could it be that the normal boundary statement
>> 0 <= R("R_PMDPHT") <= 10000
>> plays a role?
>>
>> I attached the mip problem in cplex format.
>>
>> Best,
>>
>> Nikolaus Sonnenschein
>>
>>
>>
>> _______________________________________________
>> Help-glpk mailing list
>> [email protected]
>> http://lists.gnu.org/mailman/listinfo/help-glpk
>>
>>
>
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