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

Answered

Haury Miao

March 04, 2022 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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• Official comment

  

  Simranjit Kaur

  • Gurobi Staff

  November 16, 2025 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

  March 04, 2022 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

  March 04, 2022 18:40 Edited

  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

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