Create variables in a loop

Answered

Ana Guasque Ortega

February 01, 2023 11:24

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Good morning, 

I want to create a variable whose size depends on different parameters. I tried something like (I is a list like [0,1,2,3], H is an integer and T is a list like [5,10,15, 50]:

for i in I:
    for j in I:
        for ni in range(int(H/T[i])):
            for nj in range(int(H/T[j])):
                y[i,j,ni,nj] = m.addVar(vtype=GRB.BINARY, name="y%s" %str([i,j,ni,nj]))

I get the error: NameError: name 'y' is not defined. How could I solve the problem? I followed this post.

Thanks in adavance.

Related to

• Python

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• 

  David Torres Sanchez

  • Gurobi Staff

  February 01, 2023 12:55

  Hi,

  You can do this using list comprehension and model.addVars:

  y = model.addVars(
      [
          (i, j, ni, nj)
          for i in range(I)
          for j in range(I)
          for ni in range(int(H / T[i]))
          for nj in range(int(H / T[j]))
      ],
      vtype=GRB.BINARY,
      name="y",
  )

  Cheers, 
  David

  2
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  Ana Guasque Ortega

  February 02, 2023 13:53

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  Many thanks, David! I didn't know this possibility.

  An another question related to indicator constraints (if I should write the question at any other thread, please tell me)

  I read that it is not possible to compare optimization variables. What I want is:

  if x != y then z=0 (x,y and z are binary variables) and if x==y then z can be 0 or 1. I read in this link that I need an auxiliary binary variable and then use indicator constraints. The point is that following the example, I think it doesn't work for me:

  if x!=y -> b=1

  if x==y -> b=0

  Equations:

  x!=y+eps-M(1-b)  (equation 1)

  x==y+Mb  (equation 2)
  
  When x=0 and y=1, equation 2 is not accomplished:

  0==1+10*b -> b cannot b 1.
  
  I think that the point is != and ==, because in the example they work with > and <. Could you please help me or send me another post, where this problem is solved? I have not found it. 

  Thanks again.!

  0
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  David Torres Sanchez

  • Gurobi Staff

  February 03, 2023 15:51 Edited

  Hi Ana,

  Thanks for the new question, I can try and answer it here.

  • Since \(\texttt{x}\) and \(\texttt{y}\) are binary, you can model the \(\texttt{!=}\) relationship as \(\texttt{x = 1 - y}\).
  • You can add an extra variable to check for this relationship

  I think this will model what you want:

  m = gp.Model()
  
  # Add variables
  x = m.addVar(vtype=gp.GRB.BINARY, name="x")
  y = m.addVar(vtype=gp.GRB.BINARY, name="y")
  z = m.addVar(vtype=gp.GRB.BINARY, name="z")
  aux = m.addVar(vtype=gp.GRB.BINARY, name="aux")
  
  # If aux = 1 then x != y, otherwise they are equal.
  m.addConstr((aux == 1) >> (x == 1 - y))
  m.addConstr((aux == 0) >> (x == y))
  
  # If aux = 1, then z = 0, otherwise, z can be either 0 or 1.
  m.addConstr((aux == 1) >> (z == 0))
  m.addConstr((aux == 0) >> (z <= 1))

  Cheers, 
  David

  2
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  Ana Guasque Ortega

  February 06, 2023 12:25

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  Thanks again for your great response :)

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