Setting a a given precision (of decimal places) for a variable

Answered

Jonathan Ayala Marcelo

June 29, 2023 14:24

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Hi!

I have a positive variable (Iq) in my model that need to have a precision of 8 decimal places. So, I though than doing 100*I >=0 could help (since the constraint tolerance is 1e-6), but the results doesn't change (sometimes I get Iq=-1.5e-7 for example). After some tests I realized that the constraint 100*I>=0 is discarded by the presolve phase, but in constrant other scaled constraints (e.g.100*(Iq-V)>=0 ) are preserved. Actually, I think that doing 100*I >=0 is not a good practice, so I changed Iq by Iq2/100 in the model and that works well. Finally, I would like to ask when the presolve phase can affect the custom scaling of the model, and if there are better ways to model decimal precision of variables.

Thanks 

 

Related to

• Scaling

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  Maliheh Aramon

  • Gurobi Staff

  July 01, 2023 01:05

  Hi, 

  Gurobi has a parameter named ScaleFlag that automatically scales the model. Automatic scaling implicitly changes feasibility tolerance values.

  • The internal substitution of \(x\) with \(0.1x^{\prime}\) implicitly multiplies the feasibility tolerance for bounds of \(x^{\prime}\) by 10.
  • The internal substitution of \(x\) with \(10x^{\prime}\) implicitly divides the feasibility tolerance for bounds of \(x^{\prime}\) by 0.1.

  By setting the ScaleFlag parameter to 0, you can turn the automatic scaling off. 

  • If the smallest meaningful value in your model is in the order of 1e-8, the rule of thumb is to set the feasibility tolerance to a value which is at least one order of magnitude smaller, i.e., 1e-9. 
  • Another idea is for the user to substitute \(x\) by \(0.001x^{\prime}\) where the default feasibility tolerance of 1e-6 for variable \(x^{\prime}\) means a feasibility tolerance of 1e-9 for variable \(x\).

  Best regards,

  Maliheh

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