Simulated annealing (dual_annealing) in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of dual_annealing on a simple quadratic function

What is the output of the following code snippet using scipy.optimize.dual_annealing on a quadratic function?

SciPy

from scipy.optimize import dual_annealing

def func(x):
    return (x[0] - 3)**2 + (x[1] + 1)**2

bounds = [(-5, 5), (-5, 5)]
result = dual_annealing(func, bounds)
print((round(result.x[0], 1), round(result.x[1], 1)))

A (-3.0, 1.0)

B (0.0, 0.0)

C (3.0, -1.0)

D (-5.0, -5.0)

💡 Hint

The function has its minimum where both squared terms are zero.

The function is minimized when x[0] = 3 and x[1] = -1, so the optimizer should find values close to these.

❓ data_output

intermediate

Number of iterations in dual_annealing result

After running dual_annealing on a function, what is the value of result.nit representing the number of iterations?

SciPy

from scipy.optimize import dual_annealing
import numpy as np

def func(x):
    return np.sin(x[0]) + np.cos(x[1])

bounds = [(0, 10), (0, 10)]
result = dual_annealing(func, bounds)
print(result.nit)

A A float representing the function minimum value

B An integer greater than zero representing iterations performed

C A tuple with the best solution coordinates

D A boolean indicating success or failure

💡 Hint

Check the documentation for result.nit attribute.

result.nit is the number of iterations the algorithm ran before stopping, always an integer >= 0.

🔧 Debug

advanced

Identify the error in dual_annealing usage

What error will this code raise when running dual_annealing with incorrect bounds?

SciPy

from scipy.optimize import dual_annealing

def func(x):
    return x[0]**2

bounds = [(0, 5), (1, 0)]  # Note the second bound is reversed
result = dual_annealing(func, bounds)

A No error, runs successfully

B TypeError: func() missing 1 required positional argument

C RuntimeError: Maximum number of iterations exceeded

D ValueError: bounds must be in increasing order

💡 Hint

Bounds must be specified with lower bound less than upper bound.

The second bound (1, 0) is invalid because the lower bound is greater than the upper bound, causing a ValueError.

🧠 Conceptual

advanced

Why use dual_annealing over other optimizers?

Which reason best explains why dual_annealing is chosen for optimization problems?

A It combines stochastic and local search to escape local minima

B It requires no bounds and works on infinite domains

C It only works for linear functions but is very fast

D It is guaranteed to find the global minimum for any function

💡 Hint

Think about how simulated annealing helps avoid local traps.

dual_annealing uses a combination of stochastic sampling and local search to escape local minima and find better solutions.

🚀 Application

expert

Interpreting dual_annealing output for a noisy function

Given this noisy function and dual_annealing output, what is the best interpretation of the result's fun value?

SciPy

from scipy.optimize import dual_annealing
import numpy as np

def noisy_func(x):
    noise = np.random.normal(0, 0.1)
    return (x[0] - 2)**2 + noise

bounds = [(0, 4)]
result = dual_annealing(noisy_func, bounds)
print(round(result.fun, 2))

A An approximate minimum value affected by noise

B The maximum value found during optimization

C The exact minimum value of the function without noise

D A constant zero because noise averages out

💡 Hint

Consider how noise affects function evaluations during optimization.

Because the function includes random noise, the fun value is an approximation and may vary slightly each run.