我正在解决类似于pulp example的混合问题
我有这种约束来确保生产的数量是所需的数量
prob += lpSum([KG[i] * deposit_vars[i] for i in deposit]) == 64, "KGRequirement"
但是我还需要为最小值（不等于零）添加另一个约束，这是因为例如，我服用0.002KG一种成分并不方便，我必须服用0或至少2kg，因此有效的情况是例如0、2、2.3、6、3.23。
我试图通过这种方式：
for i in deposit:
prob += (KG[i] * deposit_vars[i] == 0) or (TM[i] * deposit_vars[i] >= 30)
但这是行不通的，只会使问题变得不可行。
编辑
这是我当前的代码：
import pulp
from pulp import *
import pandas as pd
food = ["f1","f2","f3","f4"]
KG = [10,20,50,80]
Protein = [18,12,16,18]
Grass = [13,14,13,16]
price_per_kg = [15,11,10,22]
## protein,carbohydrates,kg
df = pd.DataFrame({"tkid":food,"KG":KG,"Protein":Protein,"Grass":Grass,"value":price_per_kg})
deposit = df["tkid"].values.tolist()
factor_volumen = 1
costs = dict((k,v) for k,v in zip(df["tkid"],df["value"]))
Protein = dict((k,v) for k,v in zip(df["tkid"],df["Protein"]))
Grass = dict((k,v) for k,v in zip(df["tkid"],df["Grass"]))
KG = dict((k,v) for k,v in zip(df["tkid"],df["KG"]))
prob = LpProblem("The Whiskas Problem", LpMinimize)
deposit_vars = LpVariable.dicts("Ingr",deposit,0)
prob += lpSum([costs[i]*deposit_vars[i] for i in deposit]), "Total Cost of Ingredients per can"
#prob += lpSum([deposit_vars[i] for i in deposit]) == 1.0, "PercentagesSum"
prob += lpSum([Protein[i] *KG[i] * deposit_vars[i] for i in deposit]) >= 17.2*14, "ProteinRequirement"
prob += lpSum([Grass[i] *KG[i] * deposit_vars[i] for i in deposit]) >= 12.8*14, "FatRequirement"
prob += lpSum([KG[i] * deposit_vars[i] for i in deposit]) == 14, "KGRequirement"
prob += lpSum([KG[i] * deposit_vars[i] for i in deposit]) <= 80, "KGRequirement1"
prob.writeLP("WhiskasModel.lp")
prob.solve()
# The status of the solution is printed to the screen
print ("Status:", LpStatus[prob.status])
# Each of the variables is printed with it's resolved optimum value
for v in prob.variables():
print (v.name, "=", v.varValue)
# The optimised objective function value is printed to the screen
print ("Total Cost of Ingredients per can = ", value(prob.objective))
我要添加的新禁忌素在这一部分：
prob += lpSum([KG[i] * deposit_vars[i] for i in deposit]) <= 80, "KGRequirement1"
我希望乘积KG [i] * deposit_vars [i]为0或介于a和b之间
最佳答案
在传统的线性规划公式中，所有变量，目标函数和约束都必须是连续的。您要问的是如何使此变量成为离散变量，即它只能接受值a，b，...，而不能接受两者之间的任何值。当您具有连续变量和离散变量的组合时，这称为混合整数问题（MIP）。 See PuLP documentation that reflects this explanation.我建议您仔细阅读“整数”中提到的混合问题；他们分散在页面上。根据PuLP的文档，它可以通过调用外部MIP求解器来解决MIP问题，其中一些已经包括在内。
如果没有最低限度的工作示例，那么解释如何实现这一点将有些棘手。一种方法是将变量指定为整数，并将其作为dict的值。保留默认的求解器COIN-OR's CBC solver求解器，然后将求解MIP。同时，这里有一些可供您继续使用的资源：
https://www.toptal.com/algorithms/mixed-integer-programming#example-problem-scheduling
注意它如何使用CBC求解器（这是默认的求解器）来解决此问题
http://yetanothermathprogrammingconsultant.blogspot.com/2018/08/scheduling-easy-mip.html
关于如何设置整数变量并调用CBC求解器的更明确的示例