Compute root of Rosencrantz function
Contents
Compute root of Rosencrantz function¶
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: demslv02.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 root of
(1)¶\[\begin{equation}
f(x_1,x_2)= \begin{bmatrix}200x_1(x_2-x_1^2) + 1-x_1 \\ 100(x_1^2-x_2)\end{bmatrix}
\end{equation}\]
using Newton and Broyden methods. Initial values generated randomly. True root is \(x_1=1, \quad x_2=1\).
import numpy as np
import pandas as pd
from compecon import NLP, tic, toc
Set up the problem¶
def f(x):
x1, x2 = x
fval = [200 * x1 * (x2 - x1 ** 2) + 1 - x1, 100 * (x1 ** 2 - x2)]
fjac = [[200 * (x2 - x1 ** 2) - 400 * x1 ** 2 - 1, 200 * x1],
[200 * x1, -100]]
return fval, fjac
problem = NLP(f)
Randomly generate starting point¶
problem.x0 = np.random.randn(2)
Compute root using Newton method¶
t0 = tic()
x1 = problem.newton()
t1 = 100 * toc(t0)
n1 = problem.fnorm
Compute root using Broyden method¶
t0 = tic()
x2 = problem.broyden()
t2 = 100 * toc(t0)
n2 = problem.fnorm
Print results¶
print('Hundreds of seconds required to compute root of Rosencrantz function')
print('f(x1,x2)= [200*x1*(x2-x1^2)+1-x1;100*(x1^2-x2)] via Newton and Broyden')
print('methods, starting at x1 = {:4.2f} x2 = {:4.2f}'.format(*problem.x0))
pd.DataFrame({
'Time': [t1, t2],
'Norm of f': [n1, n2],
'x1': [x1[0], x2[0]],
'x2': [x1[1], x2[1]]},
index=['Newton', 'Broyden']
)
Hundreds of seconds required to compute root of Rosencrantz function
f(x1,x2)= [200*x1*(x2-x1^2)+1-x1;100*(x1^2-x2)] via Newton and Broyden
methods, starting at x1 = -1.66 x2 = -0.22
| Time | Norm of f | x1 | x2 | |
|---|---|---|---|---|
| Newton | 4.298520 | 2.581214 | -1.427721 | 2.037849 |
| Broyden | 7.571697 | 3.257524 | -1.132487 | 1.277559 |