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Compute fixedpoint of f(x, y)= [x^2 + y^3; xy - 0.5]

Compute fixedpoint of \(f(x, y)= [x^2 + y^3; xy - 0.5]\)

Randall Romero Aguilar, PhD

This demo is based on the original Matlab demo accompanying the Computational Economics and Finance 2001 textbook by Mario Miranda and Paul Fackler.

Original (Matlab) CompEcon file: demslv04.m

Running this file requires the Python version of CompEcon. This can be installed with pip by running

!pip install compecon --upgrade

Last updated: 2022-Sept-04

About

Compute fixedpoint of

(2)\[\begin{equation} f(x, y)= \begin{bmatrix}x^2 + y^3 \\ xy - 0.5 \end{bmatrix} \end{equation}\]

using Newton, Broyden, and function iteration methods.

Initial values generated randomly. Some algorithms may fail to converge, depending on the initial value.

True fixedpoint is \(x = -0.09\), \(y=-0.46\).

import numpy as np
import pandas as pd
from compecon import NLP, tic, toc
np.random.seed(12)

Set up the problem

def g(z):
    x, y = z
    return [x **2 + y ** 3, x * y - 0.5]

problem_as_fixpoint = NLP(g, maxit=1500)

Equivalent Rootfinding Formulation

def f(z):
    x, y = z
    fval = [x - x ** 2 - y ** 3,
            y - x * y + 0.5]
    fjac = [[1 - 2 * x, -3 * y **2],
            [-y, 1 - x]]

    return fval, fjac

problem_as_zero = NLP(f, maxit=1500)

Randomly generate starting point

xinit = np.random.randn(2)

Compute fixed-point using Newton method

t0 = tic()
z1 = problem_as_zero.newton(xinit)
t1 = 100 * toc(t0)
n1 = problem_as_zero.fnorm

Compute fixed-point using Broyden method

t0 = tic()
z2 = problem_as_zero.broyden(xinit)
t2 = 100 * toc(t0)
n2 = problem_as_zero.fnorm

Compute fixed-point using function iteration

t0 = tic()
z3 = problem_as_fixpoint.fixpoint(xinit)
t3 = 100 * toc(t0)
n3 = np.linalg.norm(problem_as_fixpoint.fx - z3)

print('Hundredths of seconds required to compute fixed-point of ')
print('\n\t\tg(x1,x2)=[x1^2+x2^3; x1*x2-0.5]')
print('\nusing Newton, Broyden, and function iteration methods, starting at')
print('\n\t\tx1 = {:4.2f}  x2 = {:4.2f}\n\n'.format(*xinit))

pd.DataFrame({
    'Time': [t1, t2, t3],
    'Norm of f': [n1, n2, n3],
    'x1': [z1[0], z2[0], z3[0]],
    'x2': [z1[1], z2[1], z3[1]]},
    index=['Newton', 'Broyden', 'Function']
)

Hundredths of seconds required to compute fixed-point of 

		g(x1,x2)=[x1^2+x2^3; x1*x2-0.5]

using Newton, Broyden, and function iteration methods, starting at

		x1 = 0.47  x2 = -0.68
Time Norm of f x1 x2
Newton 0.100231 1.665335e-15 -0.088909 -0.459175
Broyden 0.700378 8.015544e-10 -0.088909 -0.459175
Function 0.000000 8.421821e-09 -0.088909 -0.459175

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