如何求解带约束的多个多元方程组

Question

我正在尝试使用 3 个方程组解决混合问题，并且我有 3 个目标要达到，或者尝试使它们中的三个尽可能接近的值：方程是：

def sat (c,s,a,f):
  return (100*c)/(2.8*s+1.18*a+0.65*f) #For this I need sat = 98.5

def ms (s,a,f):
  return s/(a+f)            #For this I need ms = 2.5
 
def ma (a,f):
  return (a/f)              #For this I need ms = 1.3


#The total mix ratio:
r1+r2+r3+r4+r5+r6 = 1

material_1:
c = 51.29
s = 4.16
a = 0.97
f = 0.38

material_2:
c = 51.42
s = 4.16
a = 0.95
f = 0.37

material_3:

c = 6.88
s = 63.36
a = 13.58
f = 3.06

material_4:
c = 32.05
s = 1.94
a = 0.0
f = 0.0

material_5:
c = 4.56
s = 21.43
a = 3.82
f = 52.28

material_6:
c = 0.19
s = 7.45
a = 4.58
f = 0.42

#The aproximate values I am trying to find are around: 

0.300 <= r1 <= 0.370 
0.300 <= r2 <= 0.370 
0.070 <= r3 <= 0.130 
0.005 <= r4 <= 0.015 
0.010 <= r5 <= 0.030 
0.110 <= r6 <= 0.130 

那么如何计算每个比率“r”的值，以便将壁橱值与 3 个方程的目标相匹配？ 我查看了一些优化器，但由于我是新手，我仍然无法理解如何设置问题、方程和约束。

Answer 1

我想我做到了，当然代码很糟糕，但我会尽力让它看起来更好。 我添加了组件的成本，所以我可以给出一个“最小化”的功能，当然这是因为我知道近似的材料比例，所以它可以引导求解器。

我将发布它的代码：

c1 = 51.42
c2 = 51.42
c3 = 6.88
c5 = 32.05
c6 = 4.56
c7 = 0.19

s1 = 4.16
s2 = 4.16
s3 = 63.36
s5 = 1.94
s6 = 21.43
s7 = 7.45

a1 = 0.97
a2 = 0.95
a3 = 13.58
a5 = 0.0
a6 = 3.82
a7 = 4.58

f1 = 0.38
f2 = 0.37
f3 = 3.06
f5 = 0.0
f6 = 52.28
f7 = 0.42

r7 = 0.125

r1 = cp.Variable()
r2 = cp.Variable()
r3 = cp.Variable()
r5 = cp.Variable()
r6 = cp.Variable()

#Costos
caliza = 10
arcilla = 20
hierro = 170
yeso = 80


objective = cp.Minimize(r1*caliza+r2*caliza+r3*arcilla+r5*yeso+r6*hierro)
constraints = [
               r1-r2 == 0,
               r1>= 0.20,
               r1<= 0.40,
               r3<=0.14,
               r3>=0.06,
               r5>=0.001,
               r5<=0.008,
               r6>=0.01,
               r6<=0.03,
               2.5*((r1*a1+r2*a2+r3*a3+r5*a5+r6*a6+r7*a7)+(f1*r1+f2*r2+f3*r3+f5*r5+f6*r6+f7*r7))-(r1*s1+r2*s2+r3*s3+r5*s5+r6*s6+r7*s7)==0,
               (98.5*(2.8*(r1*s1+r2*s2+r3*s3+r5*s5+r6*s6+r7*s7)+1.18*(r1*a1+r2*a2+r3*a3+r5*a5+r6*a6+r7*a7)+0.65*(f1*r1+f2*r2+f3*r3+f5*r5+f6*r6+f7*r7))-100*(r1*c1+r2*c2+r3*c3+r5*c5+r6*c6+r7*c7)) == 0,
               #1.3*(f1*r1+f2*r2+f3*r3+f5*r5+f6*r6+f7*r7)-(r1*a1+r2*a2+r3*a3+r5*a5+r6*a6+r7*a7) == 0,
               r1+r2+r3+r5+r6+r7 == 1]
problem = cp.Problem(objective,constraints)
problem.solve()
print(r1.value,r2.value,r3.value,r5.value,r6.value)
print(problem.status)

这给了我结果：

0.3644382497863931 0.3644382497863931 0.12287226775076901 0.0009999999955268117 0.022251232680917873
optimal

无论如何，获得可行结果的唯一方法是只考虑三个约束函数中的 2 个，因为组件无法达到其中的 3 个，这表明我需要在尝试达到 3 个约束之前检查材料组件（其中是SAT，MA和MS）。 现在我将尝试使用 pandas 使代码更好，这样我就可以通过某种 for 循环获取材料组件，并将其用于比率。 非常感谢您的帮助👍。

Answer 2

因此，这是一个简单/琐碎的示例，用于显示评论中提到的最小化误差平方的意图......而不是使用约束将值固定到确切的结果，我们让求解器找到最小化的最佳结果误差的平方，其中误差 = 值 - 目标。 我认为我在下面写的内容相当清楚。 CVXPY喜欢在线性代数领域工作，我确信这可以转换为向量/矩阵格式，但其概念是消除约束，让求解器找出最佳组合。 显然，如果存在硬约束，则需要添加这些约束，但请注意，我刚刚用您的 3 个目标中的 2 个（带有一些微不足道的数学）做了一个示例，并将其移至目标中。

您的 3 个不能同时满足的约束的问题可能是这样的转换的候选者......

import cvxpy as cp

r1 = cp.Variable()
r2 = cp.Variable()

ma = 2.5
ms = 3.4

delta_1 = (r1 + r2 - ma)**2      # diff from r1 + r2 and ma
delta_2 = (3*r1 + 2*r2 - ms)**2  # diff from 3r1 + 2r2 and ms

prob = cp.Problem(cp.Minimize(delta_1 + delta_2))

prob.solve()
print(prob.value)
print(r1.value, r2.value)

输出

9.860761315262648e-31
-1.6000000000000014 4.100000000000002

Answer 3

好的，这就是我所做的并且工作正常：

#I call the values from a pandas DF:
    c1 = df.at[0, 'MAX']
    c2 = df.at[4, 'MAX']
    c3 = df.at[8, 'MAX']
    c5 = df.at[12, 'MAX']
    c6 = df.at[16, 'MAX']
    c7 = df.at[20, 'MAX']
    s1 = df.at[1, 'MAX']
    s2 = df.at[5, 'MAX']
    s3 = df.at[9, 'MAX']
    s5 = df.at[13, 'MAX']
    s6 = df.at[17, 'MAX']
    s7 = df.at[21, 'MAX']
    a1 = df.at[2, 'MAX']
    a2 = df.at[6, 'MAX']
    a3 = df.at[10, 'MAX']
    a5 = df.at[14, 'MAX']
    a6 = df.at[18, 'MAX']
    a7 = df.at[22, 'MAX']
    f1 = df.at[3, 'MAX']
    f2 = df.at[7, 'MAX']
    f3 = df.at[11, 'MAX']
    f5 = df.at[15, 'MAX']
    f6 = df.at[19, 'MAX']
    f7 = df.at[23, 'MAX']


r1 = cp.Variable()
r2 = cp.Variable()
r3 = cp.Variable()
r5 = cp.Variable()
r6 = cp.Variable()
r7 = 12.5

#Objectives
ma = 1.3
ms = 2.50
lsf = 98.5


delta1 =(ms*((r1*a1+r2*a2+r3*a3+r5*a5+r6*a6+r7*a7)+(f1*r1+f2*r2+f3*r3+f5*r5+f6*r6+f7*r7))-(r1*s1+r2*s2+r3*s3+r5*s5+r6*s6+r7*s7))**2
delta2 =(ma*(f1*r1+f2*r2+f3*r3+f5*r5+f6*r6+f7*r7)-(r1*a1+r2*a2+r3*a3+r5*a5+r6*a6+r7*a7))**2
delta3 =((lsf*(2.8*(r1*s1+r2*s2+r3*s3+r5*s5+r6*s6+r7*s7)+1.18*(r1*a1+r2*a2+r3*a3+r5*a5+r6*a6+r7*a7)+0.65*(f1*r1+f2*r2+f3*r3+f5*r5+f6*r6+f7*r7))-100*(r1*c1+r2*c2+r3*c3+r5*c5+r6*c6+r7*c7)))**2

objective = cp.Minimize(delta1+delta2+delta3)

constraints = [r1-r2 == 0, #I added this to make r1=r2.
               r1>= 0.20,
               r3>=0,      #I added these to make it non negative.
               r5>=0,
               r5<=0.008,
               r6>=0,
               r1+r2+r3+r5+r6+r7 == 1]
problem = cp.Problem(objective,constraints)
problem.solve()
print(r1.value,r2.value,r3.value,r5.value,r6.value)
print(problem.status)

再次感谢你们的帮助。

也许您知道如何改进获取变量值的代码，也许有使用 for 循环来获取值而不是直接从 DF 中获取值的示例，DF 看起来像这样：

    DATO    MAX
0   c1  51.95000    
1   s1  3.07000 
2   a1  0.83000 
3   f1  0.31000 
4   c2  52.26000    
5   s2  2.82000 
6   a2  0.75000
...