Pyomo 找到列表值的最小总和

Question

我想要一个索引二进制变量，所以 pyomo 优化它以最小化列表的总和，同时至少选择 2 个元素。当我删除（imo 冗余）model.q 时，我收到：

ValueError: No variables appear in the Pyomo model constraints or objective. This is not supported by the NL file interface

pyomo 给我的解决方案 model.q 包含 q=0，这违反了约束 c1。

5 Declarations: i x q y objective
q 0.0
y[0] 1
y[1] 1
y[2] 1

from pyomo.environ import *


# create a model instance
model = ConcreteModel()

#Parameters
model.i = RangeSet(0, 2)

model.x = Param(model.i, initialize=[5,1,2])

#Variables
model.q = Var(domain=Binary, initialize=1)

model.y = Var(model.i, domain=Binary)

#Constraints
model.c1 = model.Constraint(expr=model.q == 1)
model.c2 = model.Constraint(expr=sum(model.y[i] for i in model.i) >= 2)

#Objective function
model.objective = Objective(expr = sum(model.x[i]*model.y[i]*model.q for i in model.i), sense=minimize)

# compute a solution
results = SolverFactory('mindtpy').solve(model, mip_solver='glpk', nlp_solver='ipopt', tee=True)
model.pprint()

Answer 1

欢迎来到本站。

您有几个错误导致了问题。

当你构造你的参数时，你需要传入一个字典，这样pyomo就可以将集合中的项目与值相关联。您不能传入列表并假设事情按顺序发生......该集合可以有任何顺序等。
您在制作约束 C2 时出现了一个可怕的拼写错误。请参阅我在代码注释中的注释
您的变量 q 完全没有必要。并且，通过将 q 乘以 y，您通过乘以变量来解决问题 non-linear。

有点固定：

from pyomo.environ import *


# create a model instance
model = ConcreteModel()

#Parameters
model.i = RangeSet(0, 2)

values = {0:5, 1:1, 2:2}

model.x = Param(model.i, initialize=values)

#Variables
#model.q = Var(domain=Binary, initialize=1)

model.y = Var(model.i, domain=Binary)

#Constraints
#model.c1 = model.Constraint(expr=model.q == 1)

# NOTE:  you mistakenly had "model.Constraint" which is a sneaky & bad typo!!
model.c2 = Constraint(expr=sum(model.y[i] for i in model.i) >= 2)

#Objective function
model.objective = Objective(expr = sum(model.x[i]*model.y[i] for i in model.i), sense=minimize)

# compute a solution
results = SolverFactory('glpk').solve(model) #, mip_solver='glpk', nlp_solver='ipopt', tee=True)
print(results)
model.display()
model.pprint()

产生（有点长，但我认为它会帮助您查看所有 3 个项目...

Problem: 
- Name: unknown
  Lower bound: 3.0
  Upper bound: 3.0
  Number of objectives: 1
  Number of constraints: 2
  Number of variables: 4
  Number of nonzeros: 4
  Sense: minimize
Solver: 
- Status: ok
  Termination condition: optimal
  Statistics: 
    Branch and bound: 
      Number of bounded subproblems: 1
      Number of created subproblems: 1
  Error rc: 0
  Time: 0.006919145584106445
Solution: 
- number of solutions: 0
  number of solutions displayed: 0

Model unknown

  Variables:
    y : Size=3, Index=i
        Key : Lower : Value : Upper : Fixed : Stale : Domain
          0 :     0 :   0.0 :     1 : False : False : Binary
          1 :     0 :   1.0 :     1 : False : False : Binary
          2 :     0 :   1.0 :     1 : False : False : Binary

  Objectives:
    objective : Size=1, Index=None, Active=True
        Key  : Active : Value
        None :   True :   3.0

  Constraints:
    c2 : Size=1
        Key  : Lower : Body : Upper
        None :   2.0 :  2.0 :  None
1 RangeSet Declarations
    i : Dimen=1, Size=3, Bounds=(0, 2)
        Key  : Finite : Members
        None :   True :   [0:2]

1 Param Declarations
    x : Size=3, Index=i, Domain=Any, Default=None, Mutable=False
        Key : Value
          0 :     5
          1 :     1
          2 :     2

1 Var Declarations
    y : Size=3, Index=i
        Key : Lower : Value : Upper : Fixed : Stale : Domain
          0 :     0 :   0.0 :     1 : False : False : Binary
          1 :     0 :   1.0 :     1 : False : False : Binary
          2 :     0 :   1.0 :     1 : False : False : Binary

1 Objective Declarations
    objective : Size=1, Index=None, Active=True
        Key  : Active : Sense    : Expression
        None :   True : minimize : 5*y[0] + y[1] + 2*y[2]

1 Constraint Declarations
    c2 : Size=1, Index=None, Active=True
        Key  : Lower : Body               : Upper : Active
        None :   2.0 : y[0] + y[1] + y[2] :  +Inf :   True

5 Declarations: i x y c2 objective
[Finished in 564ms]

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