Quadratic Programming （QP）with gurobipy，why the answer is wrong?

回答済み

Haury Miao

2022年03月04日 16:06

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import gurobipy as gp
from gurobipy import *
import numpy as np

a = np.array([-1,-1,-1,-1])
b = np.array([1, 1, 1, 1])
c = np.array([1, 1, 1, 1])

model = gp.Model()
x = model.addMVar(4, vtype=GRB.CONTINUOUS)

model.addConstr(x >= a)
model.addConstr(x <= b)
model.setObjective(x@x+x@c, GRB.MINIMIZE)

model.write('model_4.mps')

model.optimize()

# Output
print('obj=', model.objVal)
for v in model.getVars():
    print(v.varName, ':', v.x)

My code is above, I want to solve a simple QP here, I think the optimal solution should be x =-1/2c=[-0.5,-0.5,-0.5,-0.5], but the solver returns the following result to me, I just want to know where is the wrong? Can any one help me? Thanks sooooo much!

Academic license - for non-commercial use only - expires 2023-02-19
Gurobi Optimizer version 9.5.1 build v9.5.1rc2 (win64)
Thread count: 6 physical cores, 12 logical processors, using up to 12 threads
Optimize a model with 8 rows, 4 columns and 8 nonzeros
Model fingerprint: 0x8fab40bf
Model has 4 quadratic objective terms
Coefficient statistics:
  Matrix range     [1e+00, 1e+00]
  Objective range  [1e+00, 1e+00]
  QObjective range [2e+00, 2e+00]
  Bounds range     [0e+00, 0e+00]
  RHS range        [1e+00, 1e+00]
Presolve removed 8 rows and 4 columns
Presolve time: 0.00s
Presolve: All rows and columns removed

Barrier solved model in 0 iterations and 0.00 seconds (0.00 work units)
Optimal objective 0.00000000e+00
obj= 0.0
C0 : 0.0
C1 : 0.0
C2 : 0.0
C3 : 0.0

Process finished with exit code 0

 

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• 正式なコメント

  

  Simranjit Kaur

  • Gurobi Staff

  2025年11月16日 23:42

  This post is more than three years old. Some information may not be up to date. For current information, please check the Gurobi Documentation or Knowledge Base. If you need more help, please create a new post in the community forum, or try Gurobot, our chatbot interface offering instant, expert-level support.
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  Haury Miao

  2022年03月04日 17:23

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  So it is because I forget to add the LB and UB in the definition of the decision variable x?

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  Eli Towle

  • Gurobi Staff

  2022年03月04日 18:40 編集済み

  By default, variables added to the model with Model.addMVar(), Model.addVar(), and Model.addVars() have a lower bound of \( 0 \) and no upper bound.

  If you specify your lower and upper bounds on \( x \) by adding \( \texttt{lb=a} \) and \( \texttt{ub=b} \) to your call to Model.addMVar(), you obtain the expected optimal solution:

  obj= -1.0
  C0 : -0.5
  C1 : -0.5
  C2 : -0.5
  C3 : -0.5

  0

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