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我在成本优化问题中与“链接”变量作斗争。总的来说，我有四个变量：varX1、和varX2，它们以下列方式配对：varY1varY2

if 0 <= varX1 < 1 then varY1 = 0
if 1 <= varX1 <= 6 then varY1 = 1

和

if 0 <= varX2 < 1 then varY2 = 0 
if 1 <= varX2 <= 5 then varY2 = 1

我尝试使用 pyomos 分段尝试从https://github.com/Pyomo/pyomo/blob/main/examples/pyomo/piecewise/step.py重新创建示例，对 varY1 varX2 varY2`varX1和 varY1的关系进行建模。as well asand

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

可以看出，没有varX1和的配对varY1。

任何人都可以帮助我吗？

解决方案：

在 AirSquid 的帮助下，我能够为我的问题找到一个简单的解决方案。我放弃了使用 pyomos 分段函数，并为模型引入了额外的约束。以下模型按需要工作

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()
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1 回答 1

0

换个角度思考这个问题。认为y是一个“指标”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 的上限和下限。所以，现在的任务是用二进制数做一些代数来完成这项工作......

让我们想想低...

x >= 1 * y

适用于y in {0, 1}

高......如果y = 0，我们想要1，如果y = 1，我们想要6......一个小代数：

x <= 1 + 5 * y
于 2021-06-17T16:27:32.640 回答