Python中使用Pyomo库的强对偶约束

Question

我正在实现一个逆优化问题，该问题不使用KKT条件，而是应用强对偶定理(原最优目标和对偶最优目标相等)。为了这个目的，需要公式化原始问题和对偶问题。我有4台发电机，每个发电机有8个能量块(b)和固定需求。这是每小时一次的市场清算。主要问题是一个简单的市场清算，使用库Pyomo:如下

model = ConcreteModel()
model.g1=Var(b, within=NonNegativeReals)
model.g2=Var(b, within=NonNegativeReals)
model.g3=Var(b, within=NonNegativeReals)
model.g4=Var(b, within=NonNegativeReals)
model.obj = Objective(expr=
(sum(g1price[i]*model.g1[i] for i in b)+
sum(g2price[i]*model.g2[i] for i in b)+
sum(g3price[i]*model.g3[i] for i in b)+
sum(g4price[i]*model.g4[i] for i in b)))
model.con_power_balance=Constraint(
expr=
(sum(model.g1[i] for i in b)+
sum(model.g2[i] for i in b)+
sum(model.g3[i] for i in b)+
sum(model.g4[i] for i in b)- demand) == 0)

model.con_g1max=ConstraintList()
for i in b:
model.con_g1max.add(model.g1[i] <= gsize[i])
model.con_g2max=ConstraintList()
for i in b:
model.con_g2max.add(model.g2[i] <= gsize[i])
model.con_g3max=ConstraintList()
for i in b:
model.con_g3max.add(model.g3[i]< = gsize[i])
model.con_g4max=ConstraintList()
for i in b:
model.con_g4max.add(model.g4[i] <= gsize[i])

假设前面的方程被命名为(1a-1f)。如果没有说明，则设置为默认值，以最小化目标函数。由于原始问题是一个最小化问题，它的对偶问题必须是一个最大化问题。但它等价于max[F(x)]==min[-F(x)]，这就是我所应用的。这是对偶问题(方程2a-2e)：

model = ConcreteModel()
model.mu_g1max=Var(b, within=NonNegativeReals)
model.mu_g2max=Var(b, within=NonNegativeReals)
model.mu_g3max=Var(b, within=NonNegativeReals)
model.mu_g4max=Var(b, within=NonNegativeReals)
model.mu_g1min=Var(b, within=NonNegativeReals)
model.mu_g2min=Var(b, within=NonNegativeReals)
model.mu_g3min=Var(b, within=NonNegativeReals)
model.mu_g4min=Var(b, within=NonNegativeReals)
model.lambda=Var(b, within=NonNegativeReals)
model.obj = Objective(expr=
(sum(gsize[i]*model.mu_g1max[i] for i in b)+
sum(gsize[i]*model.mu_g2max[i] for i in b)+
sum(gsize[i]*model.mu_g3max[i] for i in b)+
sum(gsize[i]*model.mu_g4max[i] for i in b))
model.con_g1_dual=ConstraintList()
for i in b:
model.con_g1_dual.add(model.lambda[i]+model.mu_g1min[i]-model.mu_g1max <= gsize[i])
model.con_g2_dual=ConstraintList()
for i in b:
model.con_g2_dual.add(model.lambda[i]+model.mu_g2min[i]-model.mu_g2max <= gsize[i])
model.con_g3_dual=ConstraintList()
for i in b:
model.con_g3_dual.add(model.lambda[i]+model.mu_g3min[i]-model.mu_g3max <= gsize[i])
model.con_g4_dual=ConstraintList()
for i in b:
model.con_g4_dual.add(model.lambda[i]+model.mu_g4min[i]-model.mu_g4max <= gsize[i])

一旦提出了原始问题和对偶问题，我必须解决的真正问题如下：

model = ConcreteModel()
model.lambda=Var(b, within=NonNegativeReals)
model.obj = Objective(expr=
np.absolute(sum(model.lambda[i]-lambda_ini[i] for i in b*4)))

在这里，我必须把之前写的对偶问题(2b-2e)的约束作为约束，并且应用强对偶约束，它将是：

min(目标函数原问题)=max(目标函数对偶问题)

我的问题是：最后一个约束是如何编写的

Answer 1

好吧，我真的不理解你的问题，我还不能发表评论，所以我发布了一个答案，希望它能帮助你。

首先，在pyomo中，就像@Qi Chen提到的那样，您可以访问解算器本身生成的对偶值。为此，您需要在您的模型中添加以下内容：

model.dual = Suffix(direction=Suffix.IMPORT)

然后，你可以在其他地方调用你的模型的实例，你可以通过获得模型内部的约束的对偶值

dual_value = inst.dual[inst.constraint_name]

例如：

dual_con_g1max = inst.dual[inst.con_g1max]

但是，由于con_g1max是Constraint_List，因此需要将其索引提供给对偶获取算法。类似：

inst.dual[inst.con_g1max[1]]