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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  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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