用Python优化函数的多个输出变量

Question

我目前正在研究一种算法来确定风力涡轮机支撑结构的成本。我正在编写的算法需要优化初始输入支撑结构的重量，这样应力水平不会超过，而是接近所用材料性能的破坏准则。另一项要求是，结构的固有频率必须在两个值之间有界。为了优化结构，可以修改4个变量。

我是否可以使用Scipy.Optimize库中的一个函数来优化这个结构的重量，使用几个设计参数，但要考虑到支撑结构中的固有频率和最大应力值？

我正在优化的函数如下所示：

def func(self, x):
    self.properties.D_mp = x[0]                    # Set a new diameter for the monopile
    self.properties.Dtrat_tower = x[1]             # Set a new thickness ratio for the tower
    self.properties.Dtrat_tp = x[2]                # Set a new thickness ratio for the transition piece  
    self.properties.Dtrat_mud = x[3]               # Set a new thickness ratio for the mudline region of the monopile

    self.UpdateAll()                               # Update the support structure based on the changes in variables above

    eig = self.GetEigenFrequency()                 # Get the natural frequency
    maxUtil = self.GetMaximumUtilisationFactor()   # Get the maximum utilisation ratio on the structure (more than 1 means stress is higher than maximum allowed)

    # Natural frequency of 0.25 and utilisation ratio of 1 are ideal
    # Create some penalty...
    penalty = (100000 * abs((eig - 0.25)))
    penalty +=  (100000 * abs(maxUtil - 1)) 

    return self.GetTotalMass() + penalty

提前感谢！

Answer 1

通过将频率和应力约束折叠为对整体适应度的惩罚，这可能是最简单的方法。

LOW_COST = 10.
MID_COST = 150.
HIGH_COST = 400.

def weight(a, b, c, d):
    return "calculated weight of structure"

def frequency(a, b, c, d):
    return "calculated resonant frequency"

def freq_penalty(freq):
    # Example linear piecewise penalty function -
    #   increasing cost for frequencies below 205 or above 395
    if freq < 205:
        return MID_COST * (205 - freq)
    elif freq < 395:
        return 0.
    else:
        return MID_COST * (freq - 395)

def stress_fraction(a, b, c, d):
    return "calculated stress / failure criteria"

def stress_penalty(stress_frac):
    # Example linear piecewise penalty function -
    #   low extra cost for stress fraction below 0.85,
    #   high extra cost for stress fraction over 0.98
    if stress_frac < 0.85:
        return LOW_COST * (0.85 - stress_frac)
    elif stress_frac < 0.98:
        return 0.
    else:
        return HIGH_COST * (stress_frac - 0.98)

def overall_fitness(parameter_vector):
    a, b, c, d = parameter_vector
    return (
        # D'oh! it took me a while to get this right -
        # we want _minimum_ weight and _minimum_ penalty
        # to get _maximum_ fitness.
       -weight(a, b, c, d)
      - freq_penalty(frequency(a, b, c, d))
      - stress_penalty(stress_fraction(a, b, c, d)
    )

..。当然，你会希望找到更合适的点球函数，并发挥相对权重，但这应该会给你一个大致的想法。然后你就可以把它最大化

from scipy.optimize import fmin

initial_guess = [29., 45., 8., 0.06]
result = fmin(lambda x: -overall_fitness(x), initial_guess, maxfun=100000, full_output=True)

(使用lambda让fmin (一个最小化器)为overall_fitness找到最大值)。

或者，fmin允许在每次迭代之后应用回调函数；如果您知道如何适当地调整a、b、c、d，则可以使用此函数对频率施加硬限制，比如

def callback(x):
    a, b, c, d = x     # unpack parameter vector
    freq = frequency(a, b, c, d)
    if freq < 205:
        # apply appropriate correction to put frequency back in bounds
        return [a, b, c + 0.2, d]
    elif freq < 395:
        return x
    else:
        return [a, b, c - 0.2, d]

Answer 2

您可以使用最少的spicy.optimize函数。

在我的例子中，它是用两个变量来拟合指数函数：

def func_exp(p, x, z):                                     
# exponential function with multiple parameters  
    a, b, c, d, t, t2 = p[0], p[1], p[2], p[3], p[4], p[5]
    return a*np.exp(b + pow(x,c)*t + pow(z,d)*t2)

但是要使用最少的an函数，您需要创建一个错误函数，这个函数是您要优化的。

def err(p, x,z, y):                                         
# error function compare the previous to the estimate to
    return func_exp(p, x,z) - y                            
# minimise the residuals 

使用它：

p0=[1e4,-1e-3,1,1,-1e-2, -1e-6]                           
# First parameters 
pfit_fin, pcov, infodict, errmsg, success = leastsq(err, p0, args=(X,Y,Z), full_output=1, epsfcn=0.000001)

因此，返回最佳参数，以生成结果：

Y_2= func_exp(pfit_fin, X,Y)        

我希望这能帮到你

克里斯。