Question

为了计算电线,我在计算同时连接的电路时遇到了从最小功能中作出不正确反应的问题。例如:

“parallel

Creating a matrix of connections and fill in the resistance matrix: Equation of the form: A*x = V ,where function sum(F * x**3) -> minimum

import numpy as np
from scipy.optimize import minimize, LinearConstraint

def CreateArrays(A,V):
    A = np.array(A) #matrix of connections
    V = np.array(V)
    buf = [0.1 for i in range(len(A[0]))]
    x0 = np.array(buf)
    return A,V,x0

def start(A,V,F):
    def obj_func(x):
        return np.sum(F * x**3)
    A,V,x0 = CreateArrays(A,V)
    lincon = LinearConstraint(A, lb=V, ub=V)
    
    res = minimize(obj_func, x0, method= trust-constr , constraints=lincon)
    return res.x
#Matrix of connections, vector V, vector F
I = start([[1, 0, 1], [0, -1, -1], [-1, 1, 0]], 
      [100, -100, 0],[1e-4, 1e-4, 8e-4])
print(I)

The result is [33.36666667 33.36666667 66.63333333] but it s a mistake; the true result is [66.63333333, 66.63333333, 33.36666667], where value of function is minimum. The higher the resistance, the lower the electric currents.

我发现这一法典最低:

f = [1e-4,1e-4,8e-4]
s=0

save = 0.5
x = [0.5,0.5,99.5]
for i in range(len(x)):
    s = s + f[i] * x[i]**3
mini = s                     #start minimum
print(mini)
j = 0.5
while j < 100:
    x = [j, j, 100-j]  
    s=0
    for i in range(len(x)):
        s = s + f[i] * x[i]**3 
    if s < mini:
        mini = s
        save = x  
    j =j +  0.5
print(mini,save)

Answer 1

问题似乎在这里:

lincon = LinearConstraint(A, lb=V, ub=V)

这一制约因素似乎出现在SLSQP、COBYLA和信托公司。 SLSQP和COBYLA要么抱怨单体制约矩阵,要么(在信任的情况下)在找到解决办法后不取得进展。

我发现有两点帮助。

首先是使用更好的先头guesX0,首先采用服从这些限制的东西。

def CreateArrays(A,V):
    A = np.array(A) #matrix of connections
    V = np.array(V)
    x0 = np.linalg.lstsq(A, V, rcond=None)[0]
    return A,V,x0

第二件事是给一条线性限制室,让它要么被一丝不.带走。最初,线性限制允许在0.001上有所变化,并且从那里减少。

def start(A,V,F):
    def obj_func(x):
        ret = np.sum(F * x**3)
        return ret
    A,V,x0 = CreateArrays(A,V)
    for eps in [1e-3, 1e-6, 1e-9, 0]:
        lincon = LinearConstraint(A, lb=V - eps, ub=V + eps)
        res = minimize(obj_func, x0, method= trust-constr , constraints=lincon)
        x0 = res.x
    return res.x

友情链接