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LogoutMultimodal Optimization With Local Optima
The Ackley function is an example of an objective function that has a single global optima and multiple local optima in which a local search might get stuck.
As such, a global optimization technique is required. It is a two-dimensional objective function that has a global optima at [0,0], which evaluates to 0.0.
The example below implements the Ackley and creates a three-dimensional surface plot showing the global optima and multiple local optima.
ackley multimodal function
from numpy import arange
from numpy import exp
from numpy import sqrt
from numpy import cos
from numpy import e
from numpy import pi
from numpy import meshgrid
from matplotlib import pyplot
from mpl_toolkits.mplot3d import Axes3D
objective function
def objective(x, y):
return -20.0 * exp(-0.2 * sqrt(0.5 * (x2 + y2))) - exp(0.5 * (cos(2 * pi * x) + cos(2 * pi * y))) + e + 20
define range for input
r_min, r_max = -5.0, 5.0
sample input range uniformly at 0.1 increments
xaxis = arange(r_min, r_max, 0.1)
yaxis = arange(r_min, r_max, 0.1)
create a mesh from the axis
x, y = meshgrid(xaxis, yaxis)
compute targets
results = objective(x, y)
create a surface plot with the jet color scheme
figure = pyplot.figure()
axis = figure.gca(projection=‘3d’)
axis.plot_surface(x, y, results, cmap=‘jet’)
show the plot
pyplot.show()
ackley multimodal function
from numpy import arange
from numpy import exp
from numpy import sqrt
from numpy import cos
from numpy import e
from numpy import pi
from numpy import meshgrid
from matplotlib import pyplot
from mpl_toolkits.mplot3d import Axes3D
objective function
def objective(x, y):
return -20.0 * exp(-0.2 * sqrt(0.5 * (x2 + y2))) - exp(0.5 * (cos(2 * pi * x) + cos(2 * pi * y))) + e + 20
define range for input
r_min, r_max = -5.0, 5.0
sample input range uniformly at 0.1 increments
xaxis = arange(r_min, r_max, 0.1)
yaxis = arange(r_min, r_max, 0.1)
create a mesh from the axis
x, y = meshgrid(xaxis, yaxis)
compute targets
results = objective(x, y)
create a surface plot with the jet color scheme
figure = pyplot.figure()
axis = figure.gca(projection=‘3d’)
axis.plot_surface(x, y, results, cmap=‘jet’)
show the plot
pyplot.show()
Running the example creates the surface plot of the Ackley function showing the vast number of local optima.
3D Surface Plot of the Ackley Multimodal Function
3D Surface Plot of the Ackley Multimodal Function
We can apply the basin hopping algorithm to the Ackley objective function.
In this case, we will start the search using a random point drawn from the input domain between -5 and 5.
…
define the starting point as a random sample from the domain
pt = r_min + rand(2) * (r_max - r_min)
…
define the starting point as a random sample from the domain
pt = r_min + rand(2) * (r_max - r_min)
We will use a step size of 0.5, 200 iterations, and the default local search algorithm. This configuration was chosen after a little trial and error.
…
perform the basin hopping search
result = basinhopping(objective, pt, stepsize=0.5, niter=200)
…
perform the basin hopping search
result = basinhopping(objective, pt, stepsize=0.5, niter=200)
After the search is complete, it will report the status of the search and the number of iterations performed as well as the best result found with its evaluation.
…
summarize the result
print(‘Status : %s’ % result[‘message’])
print(‘Total Evaluations: %d’ % result[‘nfev’])
evaluate solution
solution = result[‘x’]
evaluation = objective(solution)
print(‘Solution: f(%s) = %.5f’ % (solution, evaluation))
…
summarize the result
print(‘Status : %s’ % result[‘message’])
print(‘Total Evaluations: %d’ % result[‘nfev’])
evaluate solution
solution = result[‘x’]
evaluation = objective(solution)
print(‘Solution: f(%s) = %.5f’ % (solution, evaluation))
Tying this together, the complete example of applying basin hopping to the Ackley objective function is listed below.
basin hopping global optimization for the ackley multimodal objective function
from scipy.optimize import basinhopping
from numpy.random import rand
from numpy import exp
from numpy import sqrt
from numpy import cos
from numpy import e
from numpy import pi
objective function
def objective(v):
x, y = v
return -20.0 * exp(-0.2 * sqrt(0.5 * (x2 + y2))) - exp(0.5 * (cos(2 * pi * x) + cos(2 * pi * y))) + e + 20
define range for input
r_min, r_max = -5.0, 5.0
define the starting point as a random sample from the domain
pt = r_min + rand(2) * (r_max - r_min)
perform the basin hopping search
result = basinhopping(objective, pt, stepsize=0.5, niter=200)
summarize the result
print(‘Status : %s’ % result[‘message’])
print(‘Total Evaluations: %d’ % result[‘nfev’])
evaluate solution
solution = result[‘x’]
evaluation = objective(solution)
print(‘Solution: f(%s) = %.5f’ % (solution, evaluation))
basin hopping global optimization for the ackley multimodal objective function
from scipy.optimize import basinhopping
from numpy.random import rand
from numpy import exp
from numpy import sqrt
from numpy import cos
from numpy import e
from numpy import pi
objective function
def objective(v):
x, y = v
return -20.0 * exp(-0.2 * sqrt(0.5 * (x2 + y2))) - exp(0.5 * (cos(2 * pi * x) + cos(2 * pi * y))) + e + 20
define range for input
r_min, r_max = -5.0, 5.0
define the starting point as a random sample from the domain
pt = r_min + rand(2) * (r_max - r_min)
perform the basin hopping search
result = basinhopping(objective, pt, stepsize=0.5, niter=200)
summarize the result
print(‘Status : %s’ % result[‘message’])
print(‘Total Evaluations: %d’ % result[‘nfev’])
evaluate solution
solution = result[‘x’]
evaluation = objective(solution)
print(‘Solution: f(%s) = %.5f’ % (solution, evaluation))
Running the example executes the optimization, then reports the results.
Note: Your results may vary given the stochastic nature of the algorithm or evaluation procedure, or differences in numerical precision. Consider running the example a few times and compare the average outcome.
In this case, we can see that the algorithm located the optima with inputs very close to zero and an objective function evaluation that is practically zero.
We can see that 200 iterations of the algorithm resulted in 86,020 function evaluations.
Status: [‘requested number of basinhopping iterations completed successfully’]
Total Evaluations: 86020
Solution: f([ 5.29778873e-10 -2.29022817e-10]) = 0.00000
Status: [‘requested number of basinhopping iterations completed successfully’]
Total Evaluations: 86020
Solution: f([ 5.29778873e-10 -2.29022817e-10]) = 0.00000