Question

我想找出最佳参数，如bo，ho，t1和t2，使得误差函数（f）最小（即零或接近零）。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)

我在这里面临一些问题， Q1。为cons创建的lambda函数给出了一个错误，指出“TypeError :(）缺少3个必需的位置参数：'ho'，'t1'和't2'”。为什么会这样？即使我正确地遵循了语法，我还缺少一些东西。 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。我已经提到过optimize.minimize（）而没有提到优化方法的类型。好吗？它会给我正确的答案吗？

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）：解算器似乎并不像它所提出的那样喜欢这个问题。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。默认解算器仍然无效，但COBYLA执行了操作：

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]))

回答OP的初步猜测：

# 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，如果我使用1,1,1作为转换变量：

# 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

Scipy.optimize.minimize函数确定多个变量

1 个答案: