Performance issues with large variables

Answered

Ritesh Bhalerao

August 14, 2023 03:57

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I have a distance matrix D of size 2000×2000. Mvars A and B of same size as above A is continous and B is Binary. Forming a part of objective function is the expression D*A*B representing that Dij*Aij is selected or not for all i and j in 2000×2000

But formulation of this expression takes too much time and I have many more such expressions and adding them all to model as objective function taked another big time.

How do I optimize it

Related to

• Matrix API

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  Simranjit Kaur

  • Gurobi Staff

  August 14, 2023 05:00

  Hi Ritesh,

  Can you please share the code you are using to build the objective function of your model and the Gurobi version you are using?

  Best regards,
  Simran

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  Ritesh Bhalerao

  August 14, 2023 05:16

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  #Distance.shape = (2000,2000) type = np.matrix
  B = m.addMVar((2000,2000), vtype= GRB.CONTINUOUS , lb = 0 , name = B_name )
  X = m.addMVar((2000,2000), vtype= GRB.BINARY)
  exp1 = (Distance * B * X).sum() 
  
  P = m.addMVar((2000,2000), vtype= GRB.CONTINUOUS , lb = 0 , name=P_name)
  exp2 = (Distance * P * Z).sum()
  
  exp3 = gp.quicksum(900 - gp.quicksum(B[i][j] for i in range(Di)) for j in range(Dj))
  
  exp4 = gp.quicksum(1800 - gp.quicksum(P[i][j] for i in range(Di)) for j in range(Dj))
  
  m.setObjective(exp1 + exp2 + exp3 + exp4 , GRB.MINIMIZE)

  Gurobi Optimizer version 10.0.2 build v10.0.2rc0 (win64)
  
  

   

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  Simranjit Kaur

  • Gurobi Staff

  August 15, 2023 07:17

  Hi Ritesh,

  Thanks for sharing your code.

  For exp1 and exp2, we suggest flattening the MVars to 1D arrays and using the matrix multiplication overloaded @ to compute your expression. For example,

  exp1 = B.reshape(-1) @ sp.sparse.diags(Distance.reshape(-1)) @ X.reshape(-1)

  With this, the building of exp1 and exp2 will improve significantly.

  For exp3 and exp4, we suggest avoiding the use of MVars to add one variable at a time in a term-based fashion, instead, you can do the following to speed up the building time of these expressions:

  exp3 = 900*Dj - B[:Di,:Dj].sum()

  Best regards,
  Simran

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