如何在存在GEKKO未知变量的情况下对numpy数组进行数学运算

Question

我试图通过Gekko-python包解决以下整数程序，但在目标函数部分遇到了问题。非常感谢你的帮助。

from gekko import GEKKO
import numpy
import pandas as pd
d=numpy.zeros((3,2))
d[:,0]=numpy.array([10,30,50])
w1=numpy.array([0,1.2,2])
w2=numpy.array([0,2,0])
w3=numpy.array([0.1,0,0.2])

def obj_fun(x1, x2, x3):
d[:,1]=d[:,0]-w1[:,0]*x1-w1[:,0]*x2-w1[:,0]*x3
obf=d.min(axis=1).sum()
return(obf)
m = GEKKO(remote=False)
m.options.SOLVER=1
# Initialize variables
x1 = m.Var(value=1,lb=1,ub=3,integer=True)
x2 = m.Var(value=1,lb=1,ub=4,integer=True)
x3 = m.Var(value=1,lb=1,ub=5,integer=True)
obf= m.Var()
m.Obj(obj_fun(x1, x2, x3))
m.solve(disp=False) # Solve
print('Results')
print('x1: ' + str(x1.value))
print('x2: ' + str(x2.value))
print('x3: ' + str(x3.value))
print('Objective: ' + str(m.options.objfcnval))

Answer 1

尝试使用列表理解：

c = [d[i]-w1[0]*x1-w2[0]*x2-w3[0]*x3 for i in range(n)]
minc = [m.min3(c[i],d[i]) for i in range(n)]

或定义目标函数的循环：

n = 3; c = [None]*n; minc = [None]*n
def obj_fun(x1, x2, x3):
for i in range(n):
c[i] = d[i]-w1[0]*x1-w2[0]*x2-w3[0]*x3
minc[i] = m.min3(c[i],d[i])
obf=m.sum(minc)
return(obf)

以下是完整的脚本和列表理解：

from gekko import GEKKO
import numpy
import pandas as pd
d=numpy.array([10,30,50])
w1=numpy.array([0,1.2,2])
w2=numpy.array([0,2,0])
w3=numpy.array([0.1,0,0.2])
n = 3
def obj_fun(x1, x2, x3):
c = [d[i]-w1[0]*x1-w2[0]*x2-w3[0]*x3 for i in range(n)]
minc = [m.min3(c[i],d[i]) for i in range(n)]
obf=m.sum(minc)
return(obf)
m = GEKKO(remote=False)
m.options.SOLVER=1
# Initialize variables
x1 = m.Var(value=1,lb=1,ub=3,integer=True)
x2 = m.Var(value=1,lb=1,ub=4,integer=True)
x3 = m.Var(value=1,lb=1,ub=5,integer=True)
obf= m.Var()
m.Obj(obj_fun(x1, x2, x3))
m.solve(disp=True) # Solve
print('Results')
print('x1: ' + str(x1.value))
print('x2: ' + str(x2.value))
print('x3: ' + str(x3.value))
print('Objective: ' + str(m.options.objfcnval))

它产生了解决方案：

Results
x1: [1.0]
x2: [1.0]
x3: [5.0]
Objective: 88.5