作为预算约束的决策变量的累计和(纸浆)。

问题描述 投票:0回答:1

我的程序是 不依不饶.

这是我的全部代码。

    import sys
!{sys.executable} -m pip install pulp
import pulp
import pandas as pd

dates = [d.strftime('%Y-%m-%d') for d in pd.date_range("2020-01-01","2020-12-31")]


days = range(367)
preise = range(367)
capacity = 100000

#Decision variable
volumes = pulp.LpVariable.dicts("volumes", days,
                                 lowBound=-2400, 
                                 upBound=1500, 
                                 cat=pulp.LpContinuous)

#Initialize Problem
storage = pulp.LpProblem("Storage Valuation", pulp.LpMaximize)

#Objective Function
storage += pulp.lpSum([volumes[(i)] * preise[(i)] 
                       for i in days]), "objective function"

for j in days:
    storage += sum(volumes) <= capacity, "capacity constraint {}".format(j)


storage.solve()
print("Status : ", pulp.LpStatus[storage.status])
print("Result : ", pulp.value(storage.objective))
data = []
for v in storage.variables():
    data.append({'wert': v.varValue})
result = pd.DataFrame (data)


result.insert(1, "Storage level", result['wert'].cumsum(), True)
result.insert(2, "Date", dates, True)

with pd.option_context('display.max_rows', None, 'display.max_columns', None):  
    print(result)

我的决策变量的累积总和 在每天都不允许超过一定的容量。

对于约束条件,我也试过类似的东西。

    for j in days:
    storage += pulp.lpSum([volumes[i][j] for i in days) <= capacity, "Capacity Limit {}".format(j)

我相信lpSum语句有问题。也许我的程序需要一个类似计数变量的东西。如果是这样,我该如何实现?

谢谢你!我的程序没有采取我的约束条件。

python optimization constraints linear-programming pulp
1个回答
0
投票

下面的修改给出了一个结果(虽然我不太确定这是否是你需要的)。

#days = range(367)
#preise = range(367)
days = range(366)
preise = range(366)

改变约束条件(限制累计总和):

#for j in days:
#    storage += sum(volumes) <= capacity, "capacity constraint {}".format(j)    
for j in days:
    storage += pulp.lpSum([volumes[(i)] for i in range(j+1)]) <= capacity

不要在字典上循环,变量的顺序不能保证。相反。

#for v in storage.variables():
#    data.append({'wert': v.varValue})
for j in days:
    data.append({'wert': volumes[(j)].varValue}) 
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