作为多启动优化的一部分,我正在运行差分进化(DE),我将其输出作为初始值输入到带有SLSQP的scipy最小化(我需要约束)
我正在测试Ackley函数的程序。即使在DE返回最佳值(零)的情况下,scipy最小化也会偏离最佳初始值,并返回高于最佳值的值
您知道如何使scipy最小化返回最佳值吗?我注意到它有助于指定scipy最小化的容差,但它确实完全解决了这个问题。缩放目标函数会让事情变得更糟。COBYLA解算器不存在此问题
以下是优化步骤:
# Set up
x0min = -20
x0max = 20
xdim = 4
fun = ackley
bounds = [(x0min,x0max)] * xdim
tol = 1e-12
# Get a DE solution
result = differential_evolution(fun, bounds,
maxiter=10000,
tol=tol,
workers = 1,
init='latinhypercube')
# Initialize at DE output
x0 = result.x
# Estimate the model
r = minimize(fun, x0, method='SLSQP', tol=1e-18)
在我的例子中,这会产生
result.fun = -4.440892098500626e-16
r.fun = 1.0008238682246429e-09
result.x = array([0., 0., 0., 0.])
r.x = array([-1.77227927e-10, -1.77062108e-10, 4.33179228e-10, -2.73031830e-12])
以下是Ackley函数的实现:
def ackley(x):
# Computes the value of Ackley benchmark function.
# ACKLEY accepts a matrix of size (dim,N) and returns a vetor
# FVALS of size (N,)
# Parameters
# ----------
# x : 1-D array size (dim,) or a 2-D array size (dim,N)
# Each row of the matrix represents one dimension.
# Columns have therefore the interpretation of different points at which
# the function is evaluated. N is number of points to be evaluated.
# Returns
# -------
# fvals : a scalar if x is a 1-D array or
# a 1-D array size (N,) if x is a 2-D array size (dim,N)
# in which each row contains the function value for each column of X.
n = x.shape[0]
ninverse = 1 / n
sum1 = np.sum(x**2, axis=0)
sum2 = np.sum(np.cos(2 * np.pi * x), axis=0)
fvals = (20 + np.exp(1) - (20 * np.exp(-0.2 * np.sqrt( ninverse * sum1)))
- np.exp( ninverse * sum2))
return fvals
在SLSQP选项中减少“用于雅可比矩阵数值近似的步长”为我解决了这个问题
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