Originally posted by: osarood

Hello

I am new to optimization and solvers. How can I run the blending example found at:
ibm/ILOG/CPLEX_Studio124/opl/examples/opl/blending

What command do I need to give. Or is there any documentation that I can read?
Thanks
Osman
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Originally posted by: SystemAdmin

The documentation is available here: http://pic.dhe.ibm.com/infocenter/cosinfoc/v12r4/index.jsp (as well as in your installation directory CPLEX_Studio124/doc

You can execute the sample with the executable oplrun:

oplrun -p ibm/ILOG/CPLEX_Studio124/opl/examples/opl/blending

or on Linux and Windows, you can import the example in the IDE and solve it there.

Hope this helps
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Originally posted by: osarood

Thanks a lot for the reply

I am trying put a simple indicator constraint in the blending problem. I add a boolean array

dvar boolean xxMetals;

and put the following indicator constraint:
ct3:
forall( j in Metals) xx[j]==1 => m[j]>0;

It gives the following error:

• ERRORGENERATE_022 at 57:28-47 /opt/ibm/ILOG/CPLEX_Studio124/opl/examples/opl/blending/blending.mod: Function operator>(dvar float,int) not available in context CPLEX.
  1. OPL exception: Impossible to load model.

Can you point out the problem? Thanks
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Originally posted by: SystemAdmin

The ">" is not supported in the constraint declaration.
You have to have a ">=".
Therefore, you should write your constraint as

ct3:
forall( j in Metals) xx[j]==1 => m[j]>=0.000001;

Hope this wil help.
Regards
Norbert
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Originally posted by: osarood

Thanks

It still doesnt produce the correct results.

I declared xx as boolean:
dvar boolean xx Metals ;

and am using the exact same constraint mentioned:
ct3:
forall( j in Metals) xx[j]==1 => m[j]>=0.000001;
but I am getting xx as all zeros whereas it should be all 1's
m = http://3.55 24.85 42.6;
xx = 0 0 0;
Is there something I am missing?

Thanks
Osman
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Originally posted by: osarood

sorry the values for m and x are

m = 3.55 24.85 42.6;
xx = 0 0 0;
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Originally posted by: SystemAdmin

Where is the problem?
Your constraints just says that when x=1 the m should be positive.
In your solution all x are 0, thus no restriction on the m.

Regards
Norbert
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