python - scipy.optimize.minimize 函数确定多个变量

Question

我想找出最佳参数，例如 bo、ho、t1 和 t2，以使误差函数 (f) 最小(即为零或接近于零)。完整代码如下所示，

import numpy
import scipy.optimize as optimize

#OPTIMISATION USING SCIPY
def Obj_func(x):
    bo,ho,t1,t2=x
    f=-321226.4817 + (10400000*bo*ho**3 - 10400000*(bo - 2*t2)*(ho - 2*t1)**3)/(125 + (10400000*bo*ho**3 - 10400000*(bo - 2*t2)*(ho - 2*t1)**3)/(1563920*t1*(bo - 2*t2)))
    return f

initial_guess=[2,2,0.2,0.2]
cons = ({'type': 'eq', 'fun': lambda bo,ho,t1,t2: (bo*ho - (bo - 2*t2)*(ho - 2*t1)-7.55)})
bnds = ((0, None), (0, None),(0, None), (0, None)) #all four variables are positive and greater than zero
#Always t1 and t2 should always be lesser than bo and ho
#res=optimize.minimize(Obj_func, method='SLSQP',initial_guess, bounds=bnds,constraints=cons)
res=optimize.minimize(Obj_func,initial_guess, bounds=bnds,constraints=cons)
print ("Result",res)

我在这里遇到了一些问题，Q1。为 cons 创建的 lambda 函数给我一个错误，指出“TypeError: () missing 3 required positional arguments: 'ho', 't1', and 't2'”。为什么会这样？即使我正确地遵循了语法，我还是遗漏了一些东西。Q2。我需要最小化目标函数，即这里将误差函数 (f) 最小化为零或接近零。我已经提到了 optimize.minimize() 而没有提到优化方法的类型。可以吗？它会给我正在寻找的正确答案吗？

Answer 1

Q1(语法):您的约束必须采用单个向量参数:

import numpy
import scipy.optimize as optimize

#OPTIMISATION USING SCIPY
def Obj_func(x):
    bo,ho,t1,t2=x
    f=-321226.4817 + (10400000*bo*ho**3 - 10400000*(bo - 2*t2)*(ho - 2*t1)**3)/(125 + (10400000*bo*ho**3 - 10400000*(bo - 2*t2)*(ho - 2*t1)**3)/(1563920*t1*(bo - 2*t2)))
    return f

initial_guess=[2,2,0.2,0.2]
cons = ({'type': 'eq', 'fun': lambda bhtt: (bhtt[0]*bhtt[1] - (bhtt[0] - 2*bhtt[3])*(bhtt[1] - 2*bhtt[2])-7.55)})
bnds = ((0, None), (0, None),(0, None), (0, None)) #all four variables are positive and greater than zero
#Always t1 and t2 should always be lesser than bo and ho
#res=optimize.minimize(Obj_func, method='SLSQP',initial_guess, bounds=bnds,constraints=cons)
res=optimize.minimize(Obj_func,initial_guess, bounds=bnds,constraints=cons)
print ("Result",res)

输出:

Result      fun: -15467.04696553346
     jac: array([  165915.125     ,   147480.57421875,  1243506.8828125 ,
        -130354.05859375])
 message: 'Positive directional derivative for linesearch'
    nfev: 6
     nit: 5
    njev: 1
  status: 8
 success: False
       x: array([ 2. ,  2. ,  0.2,  0.2])

Q2(数学):求解者似乎不喜欢提出的问题。通过消除等式约束和一个变量，我设法让它吐出一个解决方案。默认求解器仍然不起作用，但 COBYLA 起作用了:

import numpy
import scipy.optimize as optimize
from operator import itemgetter

#OPTIMISATION USING SCIPY
def Obj_func_orig(x):
    bo,ho,t1,t2=x
    f=-321226.4817 + (10400000*bo*ho**3 - 10400000*(bo - 2*t2)*(ho - 2*t1)**3)/(125 + (10400000*bo*ho**3 - 10400000*(bo - 2*t2)*(ho - 2*t1)**3)/(1563920*t1*(bo - 2*t2)))
    return f

def Obj_func(x):
    bh, h, t = x
    btht = bh - 7.55
    D = bh*h*h - btht*(h-t)*(h-t)
    f = -321226.4817 + D / (125/10400000 + D*(h-t)/(781960*t*btht))
    return f

def cons(x):
    bh, h, t = x
    btht = bh - 7.55
    return (btht * h - bh * (h-t)) * (t-h)

def ge(i):
    return {'type': 'ineq', 'fun': itemgetter(i)}

initial_guess=[1,1,1]
initial_guess=[64,8,0.5]
cons = ({'type': 'ineq', 'fun': cons}, {'type': 'ineq', 'fun': Obj_func}, ge(0), ge(1), ge(2))
res=optimize.minimize(lambda x: Obj_func(x)**2,initial_guess,constraints=cons,method='COBYLA')
print ("Result",res)
bh, h, t = res.x
bo, ho, t1, t2 = bh / h, h, t/2, (bh / h - (bh - 7.55) / (h-t))/2
print('bo, ho, t1, t2, f', bo, ho, t1, t2, Obj_func_orig([bo,ho,t1,t2]))

用 OP 的初始猜测回答:

# Result      fun: 285138.38958324661
#    maxcv: 0.0
#  message: 'Optimization terminated successfully.'
#     nfev: 47
#   status: 1
#  success: True
#        x: array([  6.40732657e+01,   8.29209901e+00,   6.03312595e-02])
# bo, ho, t1, t2, f 7.72702612177 8.29209901408 0.0301656297535 0.430273240999 533.983510591

请注意，这些显然是取决于初始猜测的局部最小值。例如，如果我对转换后的变量使用 1,1,1:

# Result      fun: 5829.8452437661899
#    maxcv: 0.0
#  message: 'Optimization terminated successfully.'
#     nfev: 48
#   status: 1
#  success: True
#        x: array([ 4.39911932,  0.93395108,  0.89739524])
# bo, ho, t1, t2, f 4.71022456896 0.933951079923 0.44869761948 45.4519279253 76.3534232616