Jacobi method in python

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In numerical linear algebra, the Jacobi method is an iterative algorithm for determining the solutions of a strictly diagonally dominant system of linear equations. This program implements Jacobi Method in python programming language. Email me if you have any questions about this code.

motisoltani/Jacobin-Method-Python

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README.md

In numerical linear algebra, the Jacobi method is an iterative algorithm for determining the solutions of a strictly diagonally dominant system of linear equations. This program implements Jacobi Method in python programming language. Email me if you have any questions about this code.

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In numerical linear algebra, the Jacobi method is an iterative algorithm for determining the solutions of a strictly diagonally dominant system of linear equations. This program implements Jacobi Method in python programming language. Email me if you have any questions about this code.

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Jacobi method in python

Jacobi Method in Python and NumPy

Jacobi Method in Python and NumPy

This article will discuss the Jacobi Method in Python. We’ve already looked at some other numerical linear algebra implementations in Python, including three separate matrix decomposition methods: LU Decomposition, Cholesky Decomposition and QR Decomposition. The Jacobi method is a matrix iterative method used to solve the equation $Ax=b$ for a known square matrix $A$ of size $n\times n$ and known vector $b$ or length $n$.

Jacobi’s method is used extensively in finite difference method (FDM) calculations, which are a key part of the quantitative finance landscape. The Black-Scholes PDE can be formulated in such a way that it can be solved by a finite difference technique. The Jacobi method is one way of solving the resulting matrix equation that arises from the FDM.

The algorithm for the Jacobi method is relatively straightforward. We begin with the following matrix equation:

$A$ is split into the sum of two separate matrices, $D$ and $R$, such that $A=D+R$. $D_ = A_$, but $D_=0$, for $i\neq j$. $R$ is essentially the opposite. $R_ = 0$, but $R_ = A_$ for $i \neq j$. The solution to the equation, i.e. the value of $x$, is given by the following iterative equation:

We will make use of the NumPy library to speed up the calculation of the Jacobi method. NumPy is significantly more efficient than writing an implementation in pure Python. The iterative nature of the Jacobi method means that any increases in speed within each iteration can have a large impact on the overall calculation.

Note that this implementation uses a predetermined number of steps when converging upon the correct solution. Depending upon your needs in production, you may wish to use a residual tolerance method. For that you will need to take a look at the spectral radius.

Here is the implementation via NumPy:

from pprint import pprint from numpy import array, zeros, diag, diagflat, dot def jacobi(A,b,N=25,x=None): """Solves the equation Ax=b via the Jacobi iterative method.""" # Create an initial guess if needed if x is None: x = zeros(len(A[0])) # Create a vector of the diagonal elements of A # and subtract them from A D = diag(A) R = A - diagflat(D) # Iterate for N times for i in range(N): x = (b - dot(R,x)) / D return x A = array([[2.0,1.0],[5.0,7.0]]) b = array([11.0,13.0]) guess = array([1.0,1.0]) sol = jacobi(A,b,N=25,x=guess) print "A:" pprint(A) print "b:" pprint(b) print "x:" pprint(sol) 

The output from the NumPy implementation is given below:

A: array([[ 2., 1.], [ 5., 7.]]) b: array([ 11., 13.]) x: array([ 7.11110202, -3.22220342]) 

We can see that after 25 iterations, the output is as given by the Wikipedia article on the Jacobi Method, up to the difference in significant figures.

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Jacobi Method in Python and Numpy

ln this, we will discuss the Jacobi Method implementation in Python. The Jacobi method (or the Jacobi iterative method is an algorithm for determining solutions for a system of linear equations that diagonally dominant.

Jocobi Method

Jacobi method is a matrix iterative method used to solve the linear equation Ax = b of a known square matrix of magnitude n * n and vector b or length n.

Jacobi’s method is widely used in boundary calculations (FDM), which is an important part of the financial world. It can be done in such a way that it is solved by finite difference technique. Jacobi method is one of the ways to solve the resulting matrix equation that arises from FDM.

Jocobi Method with Numpy

We will use the NumPy library to speed up the calculation of the Jacobi method. NumPy works much better than writing implementations in pure Python.

The iterative nature of the Jacobi method means that any increase in speed within each iteration can have a significant impact on the overall calculation.

Now take a look at how to implement Jocobi Method in Python Programming using Numpy library.

Example 1: Implement of the following

A = [[ 4. -2. 1.] , b [1.0, 2.0, 3.0] , x [1.0, 1.0, 1.0]
[ 1. -3. 2.]
[-1. 2. 6.]]

import numpy as np
from scipy.linalg import solve
def jacobi(A,b,x,n): 
D = np.diag(A) 
R = A-np.diagflat(D) 
for i in range(n): 
x = (b-np.dot(R,x))/D 
return x
A = np.array([[4,-2.0,1.0],[1.0,-3.0,2.0],[-1.0,2.0,6.0]])
b = [1.0,2.0,3.0]x = [1.0,1.0,1.0]n = 25
x = jacobi(A,b,x,n)
print(solve(A,b))

A [[ 4. -2. 1.] [ 1. -3. 2.] [-1. 2. 6.]] b [1.0, 2.0, 3.0] x [1.0, 1.0, 1.0]

[-0.04109589 -0.28767123 0.5890411 ] 

Example 2: Implement of the following

A= [[2 1] , b= [11, 13] , x= [1.0, 1.0]
[5 7]]

import numpy as np
from scipy.linalg import solve
def jacobi(A,b,x,n): 
D = np.diag(A) 
R = A-np.diagflat(D) 
for i in range(n): 
x = (b-np.dot(R,x))/D 
return x
A = np.array([[2,1],[5,7]])
print(';A=',A)
b = [11,13]print('b=',b)
x = [1.0,1.0]print('x=',x)
n = 25
x = jacobi(A,b,x,n)
print(solve(A,b))

A= [[2 1] [5 7]] b= [11, 13] x= [1.0, 1.0] [ 7.11111111 -3.22222222] 

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Python Program for Jacobi Iteration Method with Output

In Jacobi method, we first arrange given system of linear equations in diagonally dominant form. For example, if system of linear equations are:

3x + 20y - z = -18 2x - 3y + 20z = 25 20x + y - 2z = 17

Then they will be arranged like this:

20x + y - 2z = 17 3x + 20y -z = -18 2x - 3y + 20z = 25

After arranging equations in diagonally dominant form, we form equations for x, y, & z like this:

x = (17-y+2z)/20 y = (-18-3x+z)/20 z = (25-2x+3y)/20

These equations are defined later in Jacobi python program using lambda expression.

Python Source Code: Jacobi Method

 # Defining equations to be solved # in diagonally dominant form f1 = lambda x,y,z: (17-y+2*z)/20 f2 = lambda x,y,z: (-18-3*x+z)/20 f3 = lambda x,y,z: (25-2*x+3*y)/20 # Initial setup x0 = 0 y0 = 0 z0 = 0 count = 1 # Reading tolerable error e = float(input('Enter tolerable error: ')) # Implementation of Jacobi Iteration print('\nCount\tx\ty\tz\n') condition = True while condition: x1 = f1(x0,y0,z0) y1 = f2(x0,y0,z0) z1 = f3(x0,y0,z0) print('%d\t%0.4f\t%0.4f\t%0.4f\n' %(count, x1,y1,z1)) e1 = abs(x0-x1); e2 = abs(y0-y1); e3 = abs(z0-z1); count += 1 x0 = x1 y0 = y1 z0 = z1 condition = e1>e and e2>e and e3>e print('\nSolution: x=%0.3f, y=%0.3f and z = %0.3f\n'% (x1,y1,z1)) 

Python Program Output: Jacobi Method

Enter tolerable error: 0.00001 Count x y z 1 0.8500 -0.9000 1.2500 2 1.0200 -0.9650 1.0300 3 1.0012 -1.0015 1.0032 4 1.0004 -1.0000 0.9996 5 1.0000 -1.0001 1.0000 6 1.0000 -1.0000 1.0000 7 1.0000 -1.0000 1.0000 Solution: x=1.000, y=-1.000 and z = 1.000

Recommended Readings

• Algorithm for Bisection Method
• Pseudocode for Bisection Method
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• C++ Program for Bisection Method
• MATLAB Program for Bisection Method
• Python Program for Bisection Method
• Bisection Method Advantages
• Bisection Method Disadvantages
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• Convergence of Bisection Method
• Bisection Method Online Calculator

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