A) Use Jacobi or Gauss-Seidel iteration and perform three
iterations by hand.
B) Use Jacobi or Gauss-Siedel iteration for ten iterations with
a MAT-LAB function.
* A= [10 -2 1;-2 10 -2;-2 -5 10] , B=[9;12;18]
Find the solution to the given linear systems by Jacobi and
Gauss Seidel iteration methods.
2x + 5y =
16
20x + y – 2z =
17
5x – y +2z = 12
3x + y =
11
3x +20y – z =
-18 3x
+8y -2z = -25
2x – 3y +20z = 25
x
+ y +4z = 6
2. Solve the equation Ax = b by using the LU decomposition
method given the following...
The Gauss-Seidel method as an iterative technique often refers
to an improved version of the Jacobi method, since the Gauss-Seidel
method generally achieves a faster convergence. Describe the
difference between the Gauss-Seidel and Jacobi methods.
QUESTION:
USING MATLAB, Carry out three iterations of the
Gauss-Seidel method, starting from the initial vector Use
the , and norm to calculate the
residual error after each iteration, until all errors are below
0.0001.
[Use 5 decimal place accuracy in you
calculations]
How do the Gauss-Seidel, Newton-Raphson, and
Fast-Decoupled-Newton-Raphson iteration methods differ from task
layout, iteration terminations, and
from the point of convergence?
What does the Gauss-Seidel method acceleration factor means, how
does it affect the calculation?
Use the Gauss-Seidel method (a) without relaxation and (b) with
relaxation (l 5 0.95) to solve the following system to a tolerance
of es 5 5%. If necessary, rearrange the equations to achieve
convergence. 23x1 1 x2 1 12x3 5 50 6x1 2 x2 2 x3 5 3 6x1 1 9x2 1 x3
5 40
Use ten iterations of the appropriate MATLAB function, with
x^(0)=[0,...,0]', to solve Ax=b (approximately).
A)use Jacobi iteration.
B) use Gauss-siedel iteration.
1) make sure to use SOR with w=1.25, w=1.5, w=1.75,w=1.9, and
optimal value if given.
* A=[1,-2,0,0;-2,5,-1,0;0,-1,2,-0.5;0,0,-0.5,1.25]] ,
B=[-3;5;2;3.5]. , (optimal w is 1.5431.)
% Solves Ax = b by Gauss-Seidel method with relaxation.
% USAGE: [x,numIter,omega] =
gaussSeidel(func,x,maxIter,epsilon)
% INPUT:
% func = handle of function that returns improved x using
% x = starting solution vector
% maxIter = allowable number of iterations (default is 500)
% epsilon = error tolerance (default is 1.0e-9)
% OUTPUT:
% x = solution vector
% numIter = number of iterations carried out
% omega = computed relaxation factor
if nargin < 4; epsilon = 1.0e-9;...
Use ten iterations of the appropriate MATLAB function, with
x^(0)=[0,...,0]', to solve Ax=b (approximately).
B) use Gauss-siedel iteration.
C)use SOR with w=1.25, w=1.5, w=1.75,w=1.9, and optimal value if
given.
* A=[4,8,0,0;8,18,2,0;0,2,5,1.5;0,0,1.5,1.75] ,
B=[8;18;0.5;-1.75]. , (optimal w is 1.634.)