我正在努力解决成本优化问题中的“关联”变量。总的来说,我有四个变量:varX1
、varX2
、varY1
和varY2
,它们以以下方式配对:
if 0 <= varX1 < 1 then varY1 = 0
if 1 <= varX1 <= 6 then varY1 = 1
及
if 0 <= varX2 < 1 then varY2 = 0
if 1 <= varX2 <= 5 then varY2 = 1
我尝试使用pyomos分段建模varX1
和varY1as well as
varX2and
varY2`的关系,尝试从https://github.com/Pyomo/pyomo/blob/main/examples/pyomo/piecewise/step.py重新创建示例
完整型号代码为:
import pyomo.environ as po
costsX1 = 4
costsX2 = 6
costsY1 = 2
costsY2 = 1
a1 = 4
a2 = 3
model = po.ConcreteModel()
model.VarX1 = po.Var(bounds=(0,6))
model.VarX2 = po.Var(bounds=(0,5))
model.VarY1 = po.Var(within=po.Binary)
model.VarY2 = po.Var(within=po.Binary)
model.cons1 = po.Constraint(expr=model.VarX1+model.VarX2==5)
model.cons2 = po.Constraint(expr=a1*model.VarY1+ a2*model.VarY2>=3)
model.obj = po.Objective(expr=costsX1*model.VarX1+costsY1*a1*model.VarY1+costsX2*model.VarX2+costsY2*a2*model.VarY2,
sense=po.minimize)
DOMAIN_PTS_X1 = [0, 1, 1, 6]
RANGE_PTS_Y1 = [0, 0, 1, 1]
DOMAIN_PTS_X2 = [0, 1, 1, 5]
RANGE_PTS_Y2 = [0, 0, 1, 1]
model.piece1 = po.Piecewise(model.VarX1, model.VarY1,
pw_pts=DOMAIN_PTS_X1,
pw_constr_type='LB',
f_rule=RANGE_PTS_Y1,
pw_repn='INC')
model.piece2 = po.Piecewise(model.VarX2, model.VarY2,
pw_pts=DOMAIN_PTS_X2,
pw_constr_type='LB',
f_rule=RANGE_PTS_Y2,
pw_repn='INC')
opt = po.SolverFactory('cbc')
result_obj = opt.solve(model, tee=True)
model.pprint()
我得到以下结果
4 Var Declarations
VarX1 : Size=1, Index=None
Key : Lower : Value : Upper : Fixed : Stale : Domain
None : 0 : 5.0 : 6 : False : False : Reals
VarX2 : Size=1, Index=None
Key : Lower : Value : Upper : Fixed : Stale : Domain
None : 0 : 0.0 : 5 : False : False : Reals
VarY1 : Size=1, Index=None
Key : Lower : Value : Upper : Fixed : Stale : Domain
None : 0 : 0.0 : 1 : False : False : Binary
VarY2 : Size=1, Index=None
Key : Lower : Value : Upper : Fixed : Stale : Domain
None : 0 : 1.0 : 1 : False : False : Binary
正如我们所看到的,不存在varX1
和varY1
的配对
有人能帮我吗
在AirSquid的帮助下,我找到了解决问题的简单方法。我放弃了使用pyomos分段函数,并为模型引入了额外的约束。以下模型可以根据需要工作
import pyomo.environ as po
costsX1 = 4
costsX2 = 6
costsY1 = 2
costsY2 = 1
a1 = 4
a2 = 3
model = po.ConcreteModel()
model.VarX1 = po.Var(bounds=(0,6))
model.VarX2 = po.Var(bounds=(0,5))
model.VarY1 = po.Var(within=po.Binary)
model.VarY2 = po.Var(within=po.Binary)
model.cons1 = po.Constraint(expr=model.VarX1+model.VarX2==5)
model.cons2 = po.Constraint(expr=a1*model.VarY1+ a2*model.VarY2>=3)
model.con3 = po.Constraint(expr=model.VarY1 <= model.VarX1)
model.con4 = po.Constraint(expr=model.VarY1 >= model.VarX1/6)
model.con5 = po.Constraint(expr=model.VarY2 <= model.VarX2)
model.con6 = po.Constraint(expr=model.VarY2 >= model.VarX2/5)
model.obj = po.Objective(expr=costsX1*model.VarX1+costsY1*a1*model.VarY1+costsX2*model.VarX2+costsY2*a2*model.VarY2,
sense=po.minimize)
opt = po.SolverFactory('cbc')
result_obj = opt.solve(model, tee=True)
model.pprint()
换个角度思考这个问题。考虑^ {CD1}}是一个“指示器”变量。在这种情况下,它表示x在哪个范围内,或者更准确地说,它表示x的上下限。所以,现在的任务是用二进制数做一点代数,使之工作
让我们考虑一下低
适用于
y in {0, 1}
的两个值和高。。。如果y=0,我们想要1,如果y=1,我们想要6。。。一点代数:
相关问题 更多 >
编程相关推荐