我需要解决类似于背包问题的优化问题。我在这篇文章中详细介绍了优化问题：
knapsack optimization with dynamic variables
实际上，我实际上需要使用python而不是OPL，因此为了使用cplex优化框架，我已经安装了docplex和clpex软件包。
所以这是我想使用docplex转换为python的OPL代码
{string} categories=...;
{string} groups[categories]=...;
{string} allGroups=union (c in categories) groups[c];
{string} products[allGroups]=...;
{string} allProducts=union (g in allGroups) products[g];
float prices[allProducts]=...;
int Uc[categories]=...;
float Ug[allGroups]=...;
float budget=...;
dvar boolean z[allProducts]; // product out or in ?
dexpr int xg[g in allGroups]=(1<=sum(p in products[g]) z[p]);
dexpr int xc[c in categories]=(1<=sum(g in groups[c]) xg[g]);
maximize
sum(c in categories) Uc[c]*xc[c]+
sum(c in categories) sum(g in groups[c]) Uc[c]*Ug[g]*xg[g];
subject to
{
ctBudget:
sum(p in allProducts) z[p]*prices[p]<=budget;
}
{string} solution={p | p in allProducts : z[p]==1};
execute
{
writeln("solution = ",solution);
}
这是我的第一次代码尝试：
from collections import namedtuple
from docplex.mp.model import Model
# --------------------------------------------------------------------
# Initialize the problem data
# --------------------------------------------------------------------
Categories_groups = {"Carbs": ["Meat","Milk"],"Protein":["Pasta","Bread"], "Fat": ["Oil","Butter"]}
Groups_Products = {"1":["Product11","Product12"], "2": ["Product21","Product22","Product23"], "3":["Product31","Product32"],"4":["Product41","Product42"], "5":["Product51"],"6":["Product61","Product62"]}
Products_Prices ={"Product11":1,"Product12":4,"Product21":1,"Product22":3,"Product23":2,"Product31":4,"Product32":2,"Product41":1,"Product42":3,"Product51":1,"Product61":2,"Product62":1}
Uc=[1,1,0];
Ug=[0.8,0.2,0.1,1,0.01,0.6];
budget=3;
def build_diet_model(**kwargs):
allcategories = Categories_groups.keys()
allgroups = Groups_Products.keys()
prices=Products_Prices.values()
# Model
mdl = Model(name='summary', **kwargs)
for g, products in Groups_Products.items():
xg = mdl.sum(z[p] for p in products)# this line is not correct as I dont know how to add the condition like in the OPL code, and I was unable to model the variable z and add it as decision variable to the model.
mdl.add_constraint(mdl.sum(Products_Prices[p] * z[p] for p in Products_Prices.keys() <= budget)
mdl.maximize(mdl.sum(Uc[c] * xc[c] for c in Categories_groups.keys()) +
model.sum(xg[g] * Uc[c] * Ug[g] for c, groups in Categories_groups.items() for g in groups))
mdl.solve()
if __name__ == '__main__':
build_diet_model()
我实际上不知道如何像OPL代码中那样正确建模变量xg，xc和z？
关于如何正确建模的任何想法。
先感谢您
编辑：这是@HuguesJuille建议后的编辑，我已经清理了代码，现在可以正常工作。
from docplex.mp.model import Model
from docplex.util.environment import get_environment
# ----------------------------------------------------------------------------
# Initialize the problem data
# ----------------------------------------------------------------------------
Categories_groups = {"Carbs": ["Meat","Milk"],"Protein":["Pasta","Bread"], "Fat": ["Oil","Butter"]}
Groups_Products = {"Meat":["Product11","Product12"], "Milk": ["Product21","Product22","Product23"], "Pasta": ["Product31","Product32"],
"Bread":["Product41","Product42"], "Oil":["Product51"],"Butter":["Product61","Product62"]}
Products_Prices ={"Product11":1,"Product12":4, "Product21":1,"Product22":3,"Product23":2,"Product31":4,"Product32":2,
"Product41":1,"Product42":3, "Product51": 1,"Product61":2,"Product62":1}
Uc={"Carbs": 1,"Protein":1, "Fat": 0 }
Ug = {"Meat": 0.8, "Milk": 0.2, "Pasta": 0.1, "Bread": 1, "Oil": 0.01, "Butter": 0.6}
budget=3;
def build_userbasket_model(**kwargs):
allcategories = Categories_groups.keys()
allgroups = Groups_Products.keys()
allproducts = Products_Prices.keys()
# Model
mdl = Model(name='userbasket', **kwargs)
z = mdl.binary_var_dict(allproducts, name='z([%s])')
xg = {g: 1 <= mdl.sum(z[p] for p in Groups_Products[g]) for g in allgroups}
xc = {c: 1 <= mdl.sum(xg[g] for g in Categories_groups[c]) for c in allcategories}
mdl.add_constraint(mdl.sum(Products_Prices[p] * z[p] for p in allproducts) <= budget)
mdl.maximize(mdl.sum(Uc[c] * xc[c] for c in allcategories) + mdl.sum(
xg[g] * Uc[c] * Ug[g] for c in allcategories for g in Categories_groups[c]))
mdl.solve()
return mdl
if __name__ == '__main__':
"""DOcplexcloud credentials can be specified with url and api_key in the code block below.
Alternatively, Context.make_default_context() searches the PYTHONPATH for
the following files:
* cplex_config.py
* cplex_config_<hostname>.py
* docloud_config.py (must only contain context.solver.docloud configuration)
These files contain the credentials and other properties. For example,
something similar to::
context.solver.docloud.url = "https://docloud.service.com/job_manager/rest/v1"
context.solver.docloud.key = "example api_key"
"""
url = None
key = None
mdl = build_userbasket_model()
# will use IBM Decision Optimization on cloud.
if not mdl.solve(url=url, key=key):
print("*** Problem has no solution")
else:
mdl.float_precision = 3
print("* model solved as function:")
mdl.print_solution()
# Save the CPLEX solution as "solution.json" program output
with get_environment().get_output_stream("solution.json") as fp:
mdl.solution.export(fp, "json")
我希望这将帮助像我这样的初学者遇到同样的问题。
最佳答案
如果我正确理解了您的数据模型（我不确定您的示例中的数据是否一致（Categories_groups和Groups_Products的“ groups”值集合不同）。），则决策变量和表达式的定义将看起来像这样：
z = mdl.binary_var_dict(allProducts, name='z([%s])')
xg = {g: 1 <= mdl.sum(z[p] for p in Groups_Products[g]) for g in allgroups}
xc = {c: 1 <= mdl.sum(xg[g] for g in Categories_groups[c]) for c in allcategories}
在此，将“ z”决策变量定义为字典。然后可以轻松对其进行索引。
您还可以在此处找到有关编写docplex模型的文档：https://rawgit.com/IBMDecisionOptimization/docplex-doc/master/docs/mp/creating_model.html
请注意，如果您需要构建处理大型数据集的模型，则使用pandas可能更有效地定义复杂切片。