在使用纸浆的线性编程优化中添加约束

Question

以下代码为我提供了去度假的最佳场所，以保持费用低：

from pulp import *
import numpy as np
import pandas as pd
import re 
#write a scaper before hand
data = pd.read_csv('clymb_adventures.csv')
problem_name = 'GoingOnVacation'
aval_vacation_days = 10
def optimize_vacation_schedule(aval_vacation_days):
# create the LP object, set up as a minimization problem --> since we 
want to minimize the costs 
prob = pulp.LpProblem(problem_name, pulp.LpMinimize)

#create decision variables
decision_variables = []
for rownum, row in data.iterrows():
    variable = str('x' + str(rownum))
    variable = pulp.LpVariable(str(variable), lowBound = 0, upBound = 1, cat= 'Integer') #make variables binary
    decision_variables.append(variable)
print ("Total number of decision_variables: " + str(len(decision_variables)))
#create objective Function -minimize the costs for the trip
total_cost = ""
for rownum, row in data.iterrows():
    for i, schedule in enumerate(decision_variables):
        if rownum == i:
            formula = row['cost']*schedule
            total_cost += formula
prob += total_cost
print ("Optimization function: " + str(total_cost)) 

#create constrains - total vacation days should be no more than 14
total_vacation_days = ""
for rownum, row in data.iterrows():
    for i, schedule in enumerate(decision_variables):
        if rownum == i:
            formula = row['duration']*schedule
            total_vacation_days += formula
prob += (total_vacation_days == aval_vacation_days)

#now run optimization
optimization_result = prob.solve()
assert optimization_result == pulp.LpStatusOptimal
prob.writeLP(problem_name + ".lp" )
print("Status:", LpStatus[prob.status])
print("Optimal Solution to the problem: ", value(prob.objective))
print ("Individual decision_variables: ")
for v in prob.variables():
    print(v.name, "=", v.varValue)
if __name__ == "__main__":
    optimize_vacation_schedule(aval_vacation_days)

示例数据集：

destination duration    cost    description                 location
0   Baja          7      899    Hike Bike                [31223,23123]
1   Nepal         11     899    culture of the Himalayas [91223,28123]
2   Spain         8      568    Sport climb              [66223,23123]
3   Yosemite      3      150    Guided hiking            [0223,23123]
4   Utah          6      156    Hike.                    [35523,23123]
5   Okla          1      136    Hike.                    [25523,23123]

我在数据集中添加了一个额外的字段"位置"。

我要实现的是，如果求解器将三个3个位置作为最佳解决方案，则必须确保使用位置坐标之间的两个连续建议的引用之间的最大曼哈顿距离不超过3000？p>示例：如果索尔弗建议优胜美地，犹他州和俄克拉荷马州，然后在建议它们之前检查从优胜美地到犹他州的距离低于3000犹他州到俄克拉荷马州的犹他州低于3000。

这也使其成为路由问题。

因此，如何添加一个约束，该约束使用位置坐标保持两个连续的建议城市之间的距离。请帮助

谢谢！！！

Answer 1

如果要添加条件x（i，j）== 1作为约束，那么您将创建第二组决策变量。让关键是元组（I，J），并且值是cat ='binarle'的lpvaria。然后，您必须设置一些其他约束。

注意：我假设X是键，其中键是位置，而值是决策变量。我不确定为什么在这里使用列表。匹配您的结构需要进行一些修改。

import itertools
locations = ['Baja', 'Nepal', 'Spain', ...]
x = LpVariable.dicts('x', destinations, cat='Binary'
prob = LpProblem('Vacation', pulp.LpMinimize)
# non-duplicative cominations
destination_pairs = itertools.combinations(locations, 2)
# new decision variables
# i is destination1, j is destination2
y = {(i, j): LpVariable('y', cat='Binary') for i, j in destination_pairs}
# new constraints
# if x[i] or x[j] is 0, then y[(i,j)] has to be zero
# if y[(i, j)] is 1, then x[i] and x[j] both have to be one
for i, j in destination_pairs:
    prob += y[(i, j)] <= x[i]
    prob += y[(i, j)] <= x[j]
    prob += y[(i,j] >= x[i] + x[j] - 1
# this ensures that a combination is chosen
prob += lpSum(y.values()) >= 1