如何通过 Gurobi 变量索引字典 - 线性规划

Question

在我的优化问题中，我有一个包含数据（给定）的字典，并且对应于每个键有一个决策变量。我想添加一个约束，我需要使用决策变量作为字典的索引。

import gurobipy as gp
from gurobipy import GRB

model = gp.Model()

dic = {1: 0.5, 2: 1.45, 3: 3, 4: 0.3, 5: 2, 6: 2.8, 7: 1}

# domain of x is defined by the indices of the above dictionary
x = model.addVar(vtype = GRB.INTEGER, lb = 1, ub = 7)

# now, I want to put a constraint on the following lines
# model.addConstr(dic[x] >= 2.1)

我是 Gurobi 的新手，任何帮助将不胜感激。

Answer 1

您不能像您所做的那样使用决策变量在 Gurobi 中对字典进行索引。为了达到预期的目的，一种方法是引入二元变量。请参阅下面的代码片段：

import gurobipy as gp
from gurobipy import GRB

model = gp.Model()

dic = {1: 0.5, 2: 1.45, 3: 3, 4: 0.3, 5: 2, 6: 2.8, 7: 1}

len_dic = len(dic)
bool_varb = model.addVars(range(1, len_dic + 1), vtype=GRB.BINARY)
# assuming we want to select only one unique index, basis some constraints
model.addConstr(bool_varb.sum() == 1)

# x = model.addVar(vtype = GRB.INTEGER)
# model.addConstr(dic[x] >= 2.1)
# use below to express above
model.addConstr(bool_varb.prod(dic) >= 2.1)

# just added for illustration
# below expression in the constraint would resolve to 'x'
model.addConstr(gp.quicksum(i * j for i, j in zip(bool_varb.values(), dic.keys())) >= 4)

model.optimize()

正如预期的那样，

x

（使用您的术语）解析为

6

dic[x]

（使用您的术语）解决为

2.8