SciPy Optimization

Optimization means finding the best value: the lowest cost, the highest profit, or the point where an equation equals zero.

What scipy.optimize Can Do

Finding a Minimum of One Variable

A shop's cost to make x hundred items is cost(x) = x² - 8x + 20 thousand rupees. Which x gives the lowest cost?

from scipy.optimize import minimize_scalar

def cost(x):
    return x ** 2 - 8 * x + 20

result = minimize_scalar(cost)
print("Best x:", round(result.x, 3))
print("Lowest cost:", round(result.fun, 3))

Best x: 4.0
Lowest cost: 4.0

The result object holds the answer: x is where the minimum is and fun is the value of the function there.

Finding a Maximum

SciPy only minimises. To find a maximum, minimise the negative of your function.

Profit from selling at price p is profit(p) = -2p² + 400p. Which price gives the most profit?

from scipy.optimize import minimize_scalar

def profit(p):
    return -2 * p ** 2 + 400 * p

result = minimize_scalar(lambda p: -profit(p))
print("Best price:", round(result.x, 2))
print("Max profit:", round(profit(result.x), 2))

Best price: 100.0
Max profit: 20000.0

Several Variables With minimize

minimize() needs a starting guess x0. Here we find the lowest point of f(x, y) = (x - 1)² + (y + 2)²:

import numpy as np
from scipy.optimize import minimize

def f(v):
    x, y = v
    return (x - 1) ** 2 + (y + 2) ** 2

result = minimize(f, x0=[0, 0])
print(np.round(result.x, 3))
print(result.success)

[ 1. -2.]
True

Finding Roots

A root is where a function equals zero. Solve x³ - 2x - 5 = 0:

from scipy.optimize import root_scalar

result = root_scalar(lambda x: x ** 3 - 2 * x - 5, bracket=[2, 3])
print(round(result.root, 4))

2.0946

bracket=[2, 3] tells SciPy the root lies between 2 and 3. The function must have different signs at the two ends.

Fitting a Curve to Data

curve_fit() finds the parameters of a model that best match your data. A shop's monthly customers grow roughly in a straight line. Let us fit y = a·x + b:

import numpy as np
from scipy.optimize import curve_fit

months = np.array([1, 2, 3, 4, 5, 6])
customers = np.array([110, 132, 149, 171, 188, 212])

def line(x, a, b):
    return a * x + b

params, _ = curve_fit(line, months, customers)
a, b = params
print(f"a = {a:.2f}, b = {b:.2f}")
print("Month 9 forecast:", round(line(9, a, b)))

a = 20.00, b = 90.33
Month 9 forecast: 270

The shop gains about 20 customers each month, and the model predicts about 270 customers in month 9.

Exercise