How to deal with the restriction of not being able to use strictly greater/smaller constraints?

Answered

Yi Ting Lin

• November 19, 2023 22:45

Hi,

I tried to use Big-M for conditional constraints.

The length of Part 1,2,3,4, can be denoted as Q1,Q2,Q3,Q4.

The accumulated length after part 1,2,3,4 join can be denoted as AQ1, AQ2, AQ3, AQ4 respectively.

The order for these parts is also part of the optimized variable. 

I have a threshold TH, I want to know which part starts to be longer than the threshold. The part that meets: AQ-Q < TH < AQ is the part I am looking for. (Binary4=1)

(Q1,Q2,Q3,Q4 are all variables)

Four binary variables Binary1, Binary2, Binary3, and Binary4 are used to formulate this.

However, when AQ= TH, it would lead to the problem of having two parts that meet the constraints.

Assume the order of the part is 1-->2-->3-->4

AQ2=AQ1+AQ2

AQ3=AQ1+AQ2+AQ3

If AQ2=TH, then AQ3-Q3 would also equal TH.

In this case, Binary4[2] and Binary4[3] all equal 1.

How do I modify the equation to let Binbary4[2]=0?

I tried to add another Binary5 to identify the situation when AQ=TH, however, in my case, the two values are not exactly the same (i.e., AQ2=10.818181818181806, TH=10.81818181818)

For i in range(1,5):
 m.addConstr((AQ[i]-Q[i])>=(TH+eps-(1-Binary1[i])*M))
 m.addConstr((AQ[i]-Q[i])<=(TH+eps+(Binary1[i]*M)))
 m.addConstr((AQ[i]+eps-(1-Binary2[i])*M)<=TH)
 m.addConstr((AQ[i]+eps+(Binary2[i]*M))>=TH)

 m.addConstr(Binary3[i]==Binary1[i]+Binary2[i])
 m.addConstr((Binary3[i]==1)>>(Binary4[i]==0))
 m.addConstr((Binary3[i]==0)>>(Binary4[i]==1))

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• 

  Marika Karbstein

  Gurobi Staff

  • November 22, 2023 08:45

  I am not sure if I understand your setting. But isn't it enough to change the sign of eps, i.e.,

   m.addConstr((AQ[i]-Q[i])>=(TH-eps-(1-Binary1[i])*M))
   m.addConstr((AQ[i]-Q[i])<=(TH-eps+(Binary1[i]*M)))

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• 

  Yi Ting Lin

  • November 22, 2023 16:53

  

  

  

  Hi Marika,

  I tried to change the sign of eps, but it would lead to both Binary4[2] and Binary4[3] =0.
  I want to have Binary4[2]=0 but Binary4[3]=1

  Is there any other way that might be able to solve this issue?
  
  Thanks for helping

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• 

  Marika Karbstein

  Gurobi Staff

  • November 22, 2023 18:00

  I assume there is only one part for which AQ-Q <= TH < AQ holds. You need to have B1=1 if AQ-Q <= TH and B2=1 if TH < AQ. Then with Model.addGenConstrAnd() you can get B4=1 if both B1 and B2 are one. This should then only hold for 3 in your example. For 2 TH < AQ is not fulfilled.

  For i in range(1,5):
   m.addConstr((AQ[i]-Q[i])<=(TH + tol + (1-Binary1[i])*M))
   m.addConstr((AQ[i]-Q[i])>=(TH + tol - (Binary1[i]*M)))
   m.addConstr(TH +eps <= (AQ[i] + (1-Binary2[i])*M))
   m.addConstr(TH +eps >= (AQ[i] - (Binary2[i]*M)))
  
   m.addGenConstrAnd(Binary4[i], [Binary1[i],Binary2[i]])

  tol is a small tolerance, e.g. FeasibilityTol. But for the < check, eps needs to be a bigger value.

   

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• 

  Yi Ting Lin

  • November 27, 2023 23:29

  

  

  

  Hi Marika,

  Thanks for the reply. I'll check out how to use Model.addGenConstrAnd().

  Thanks again for your help.

   

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