如何在CPLEXpython中添加和删除约束？

Question

我在python中使用了新的Cplex。我已经知道如何正确地编写数学优化问题的代码，并创建一个函数来友好地显示解决方案（F1）。目前，我想在不创建新模型的情况下修改模型的某些功能。我的想法是求解一个模型P，然后是模型P1（改变决策变量域），P2（放松一些集合o约束），等等。我想用我的函数F1得到这些模型的解

我的代码如下：

import time
import numpy as np
import cplex
from cplex import Cplex
from cplex.exceptions import CplexError
import sys
from openpyxl import Workbook
import xlrd

def GAP_RL(I,J,data):
    #TIEMPO INICIAL
    inicio=time.process_time() #------------------------------- INICIO DEL CONTEO DE TIEMPO
    #CONJUNTOS
    maquinas=range(I)  #CONJUNTO DE TRABAJOS
    trabajos=range(J)  #CONJUNTO DE MÁQUINAS
    #print("MÁQUINAS={} | TRABAJOS= {} |\n".format(list(maquinas),list(trabajos)))
    print("\n")
    #PARÁMETROS
    wb = Workbook()
    ws = wb.active
    book = xlrd.open_workbook(data+'.xlsx')           
    sheet = book.sheet_by_name("c")
    c_a=[[int(sheet.cell_value(r,c)) for c in range(sheet.ncols)] for r in range(sheet.nrows)]
    c_a=np.matrix(c_a)  
    print("COSTO DE ASIGNACIÓN")      
    print("")         
    print(c_a) # ------------------------------------------------ COSTO DE ASIGNACIÓN
    print("\n")
    sheet = book.sheet_by_name("a")
    a=[[int(sheet.cell_value(r,c)) for c in range(sheet.ncols)] for r in range(sheet.nrows)]
    a=np.matrix(a)  
    print("UTILIZACIÓN")      
    print("")         
    print(a) #---------------------------------------------------- REQUERIMIENTOS   
    print("\n")
    sheet = book.sheet_by_name("b")
    b=[[int(sheet.cell_value(r,c)) for c in range(sheet.ncols)] for r in range(sheet.nrows)]
    b=np.matrix(b)  
    print("DISPONIBILIDAD")      
    print("")         
    print(b) #---------------------------------------------------- CAPACIDAD MÁXIMA DE PROCESO
    print("\n")

    Model=cplex.Cplex()                                                  #CREACIÓN DEL MODELO.

    Model.parameters.mip.tolerances.integrality.set(0)                   #ESPECIFICA LA CANTIDAD POR LA CUAL UNA 
                                                                         #VARIABLE ENTERA PUEDE SER DIFERENTE DE UN 
                                                                         #ENTERO Y AÚN ASÍ CONSIDERASE FACTIBLE
    Model.objective.set_sense(Model.objective.sense.minimize)            #SENTIDO DE OPTIMIZACIÓN.

    #Model.parameters.timelimit.set(float(7200))                         #TIEMPO MÁXIMO DE EJECUCIÓN [SEGUNDOS].

    #Model.parameters.mip.tolerances.mipgap.set(float(0.1))              #GAP DE TÉRMINO.

    #VARIABLES DE DECISIÓN
    x_vars=np.array([["x("+str(i)+","+str(j)+")" for j in trabajos] for i in maquinas])
    x_varnames = x_vars.flatten()
    x_vartypes='B'*I*J
    x_varlb = [0.0]*len(x_varnames)
    x_varub = [1.0]*len(x_varnames)
    x_varobj = []
    for i in maquinas:
        for j in trabajos:
            x_varobj.append(float(c_a[i,j]))
    Model.variables.add(obj = x_varobj, lb = x_varlb, ub = x_varub, types = x_vartypes, names = x_varnames)
    #RESTRICCIONES

    #PRIMER CONJUNTO DE RESTRICCIONES: CADA TRABAJO ES ASIGNADO A UNA ÚNICA MÁQUINA.
    for j in trabajos:
        row1=[]
        val1=[]
        for i in maquinas:
            row1.append(x_vars[i,j])
            val1.append(float(1.0))
        Model.linear_constraints.add(lin_expr = [cplex.SparsePair(ind = row1, val= val1)], senses = 'E', rhs = [float(1.0)])

    #SEGUNDO CONJUNTO DE RESTRICCIONES: LAS ASIGNACIONES DE TRABAJOS CONSIDERAN LA CAPACIDAD MÁXIMA DE PROCESAMIENTO DE LAS MÁQUINAS
    for i in maquinas:
        row2=[]
        val2=[]
        for j in trabajos:
            row2.append(x_vars[i,j])
            val2.append(float(a[i,j]))
        Model.linear_constraints.add(lin_expr = [cplex.SparsePair(ind = row2, val= val2)], senses = 'L', rhs = [float(b[i])])
    #RESOLVER MODELO Y EXPANDIR
    solution=Model.solve() 
    Model.write('modelo_GAP.lp') 
    #MOSTRAR SOLUCION
    def show_solution():
        print("--------------------------------------------------------------------------------------------------------------")
        print("\nVALOR FUNCIÓN OBJETIVO - COSTO TOTAL DE ASIGNACIÓN = {}".format(Model.solution.get_objective_value()))
        print("--------------------------------------------------------------------------------------------------------------")

        for i in maquinas:
            for j in trabajos:
                if(Model.solution.get_values("x("+str(i)+","+str(j)+")")!=0.0):
                    print("x("+str(i+1)+","+str(j+1)+")"+" = "+str(Model.solution.get_values("x("+str(i)+","+str(j)+")")))
        fin=time.process_time()
        tiempo=float(fin)-float(inicio)
        print("")
        print("TIEMPO DE CÓMPUTO [s]= ", float(tiempo))
    show_solution()
GAP_RL(I=2,J=7,data='data')

数据文件如下所示：

矩阵c_a如下所示：

矩阵a如下所示：

矩阵b如下所示：

因此，我想知道如何在以这种方式编写的模型中进行这些更改。提前谢谢

Answer 1

以下是一些让你朝正确方向前进的提示：

1. add方法（例如，Cplex.variables.add，Cplex.linear_constraints.add）返回一个迭代器，其中包含添加到模型中的索引。您可以使用它来记住要修改的变量类或约束类的索引。例如：

   varind = list(Model.variables.add(obj = x_varobj, lb = x_varlb, ub = x_varub, types = x_vartypes, names = x_varnames))
   

2. 或者，如果为变量/约束命名，则可以按名称查询或修改它们。不过，这可能会导致性能下降，因此请阅读用户手册中的this部分

3. 可以使用Cplex.variables.set_lower_bounds和Cplex.variables.set_upper_bounds更改变量的上下限

4. 可以使用Cplex.linear_constraints.delete删除线性约束

5. 最后，您可以通过将现有模型传递给Cplex构造函数来创建模型的副本（也称为克隆）。例如：

   Model1 = cplex.Cplex(Model)
   

Answer 2

使用docplex python API可以很容易地做到这一点

from docplex.mp.model import Model

# original model

mdl = Model(name='buses')
nbbus40 = mdl.integer_var(name='nbBus40')
nbbus30 = mdl.integer_var(name='nbBus30')
mdl.add_constraint(nbbus40*40 + nbbus30*30 >= 300, 'kids')
mdl.minimize(nbbus40*500 + nbbus30*400)

mdl.solve()



for v in mdl.iter_integer_vars():
    print(v," = ",v.solution_value)

#now remove the constraint

print()
print("now 0 kids instead of 300")    

mdl.get_constraint_by_name("kids").rhs=0;
mdl.solve()

for v in mdl.iter_integer_vars():
    print(v," = ",v.solution_value)

# and now let 's adapt to Covid19 1 seat out of 2

print("now 40 seats buses take 20 children and same ratio for 30 seats buses") 

mdl.add_constraint(nbbus40*20 + nbbus30*15 >= 300, 'kidsCov19')

mdl.solve()

for v in mdl.iter_integer_vars():
    print(v," = ",v.solution_value)

给

nbBus40  =  6.0
nbBus30  =  2.0

now 0 kids instead of 300
nbBus40  =  0
nbBus30  =  0
now 40 seats buses take 20 children and same ratio for 30 seats buses
nbBus40  =  15.0
nbBus30  =  0

我使用了zoo and bus example