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我在对指数方程求和并将其用作目标函数时遇到问题。

我还尝试将指数方程写成约束，因为我认为这可能是解决此问题的另一种方法，但这对我也不起作用。

对此的任何帮助将不胜感激。

import pandas as pd
from pyscipopt import Model, quicksum, multidict, exp

num_fac_to_open = 1
order_to_open = []
opened_fac = []
closed_fac = [0, 1, 2]
S = [0, 1, 2]
R = [10, 11, 12]
distance_dict = {(0, 10): 300.8, (1, 10): 150.6, (2, 10): 1567.8, (0, 11): 1241.0, (1, 11): 2012.1, (2, 11): 789.2, (0, 12): 563.2, (1, 12): 1798.3, (2, 12): 946.3}
population_dict = {10:54, 11:46, 12:22}

# n is the desired number of facilities to open
n = len(opened_fac) + num_fac_to_open
# create a model
model = Model()

z, y= {}, {}
for s in S:
    # x_i is binary, 1 if service facility i is opened, 0 otherwise
    z[s] = model.addVar(vtype="B")
    for r in R:
        # y_i,j is binary, 1 if service facility i is assigned to residential area j, 0 otherwise
        y[s, r] = model.addVar(vtype="B")

for r in R:
    want_list = (distance_dict[s, r]*y[s, r] for s in S)
    want_list_quicksum = quicksum(want_list)
    exp_power = population_dict[r]*want_list_quicksum
    w = exp(exp_power)

model.setObjective(quicksum([w]), 'minimize')
model.optimize()

此代码发生的错误是：

Traceback (most recent call last):
  File "stack_overflow_code.py", line 38, in <module>
    model.setObjective(quicksum([w]), 'minimize')
  File "src/pyscipopt/scip.pyx", line 1246, in pyscipopt.scip.Model.setObjective
AssertionError: given coefficients are neither Expr or number but SumExpr

我的理解是目标函数的格式应该是（这是我打印 exp_power 得到的结果）：

Expr({Term(): 0.0, Term(x4): 12390.400000000001, Term(x8): 39562.6, Term(x12): 20818.6})

但是，一旦添加指数项 (w)，格式变为：

exp(sum(0.0,prod(12390.400000000001,x4),prod(39562.6,x8),prod(20818.6,x12)))

此外，当添加 quicksum[w] 时，格式变为：

sum(0.0,exp(sum(0.0,prod(12390.400000000001,x4),prod(39562.6,x8),prod(20818.6,x12))))

目的是最小化 \sum_{r=1}^N W_r

其中 \W_r = exp(population_dict[r]*sum_{s∈S} d_r,s * y_r,s) ∀r ∈ R

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1 回答 1

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quicksum接受一个可迭代对象，但distance_dict[s, r]*y[s, r] for s in S没有像您预期的那样返回一个可迭代对象。

尝试将该行简化为多行，并确保在将其输入quicksum.

为了在 Python 中使用列表推导，我们必须将语句括在方括号中。[distance_dict[s, r]*y[s, r] for s in S].

于 2021-01-11T04:55:13.300 回答