Numerical Methods and Techniques - Using Gauss-Seidel Method

Gauss-Seidel Method

One of the most useful methods which can be used to solve a large number of simultaneous equations. Although, the Gauss-Seidel method has the risks of not always converging to a solution and when it does converse, very slowly. However, this technique will usually converge to a solution when the magnitude of a coefficient of a different unknown in every equation of the set, is sufficiently dominant with respect to the magnitudes of the other coefficients in the equation. It is challenging to define the specific minimum margin by means of which such a coefficient has to dominate the other coefficients to make certain convergence for some combination of coefficient values when convergence exists. However, when the absolute value of the dominant coefficient for a distinct unknown in every equation is larger than the sum of the absolute values of the different coefficients in that equation is unknown as a "diagonal system".

A "diagonal system" is enough to make sure convergence however is no longer necessary.

Assignment 1: Using the Gauss-Seidel Method

Problem 4: Using the Gauss-Seidel iteration, determine the unknown variables. Verify the error up to five significant figures.

Given:

Solution:

Arrange the equation into a diagonal system:

Rewrite the equation to solve for the value

Equation 1:

x1 = 2.70 + 2.2x1 - 1.5x3

2.7

Equation 2:

x2 = 2.35 - 1.7x1 + 1.5x3

2.3

Equation 3:

x3 = 0.90 + 1.1x1 + 1.6x3

1.9

Note:

1. Use Equation 1 to solve the value of x1 for the new given value x2 & x3.

2. Use Equation 2 to solve the value of x2 for the new given value x1 & x3.

3. Use Equation 3 to solve the value of x3 for the new given value x1 & x2.

Initial Value of:

X1 = 0, X2 = 0, X3 = 0

1st Iteration:

X1 = 2.70 + 2.2 (0) - 1.5 (0) = 1

2.7

X2 = 2.35 - 1.7 (1) + 1.5 (0) = 0.28261

2.3

X3 = 0.90 + 1.1 (1) + 1.6 (0.28261) = 1.29062

1.9

2nd Iteration:

X1 = 2.70 + 2.2 (0.28261) - 1.5 (1.29062) = 0.51326

2.7

X2 = 2.35 - 1.7 (0.51326) + 1.5 (1.29062) = 1.48408

2.3

X3 = 0.90 + 1.1 (0.51326) + 1.6 (1.48408) = 2.02059

1.9

3rd Iteration:

X1 = 2.70 + 2.2 (1.48408) - 1.5 (2.02059) = 1.08670

2.7

X2 = 2.35 - 1.7 (1.08670) + 1.5 (2.02059) = 1.53630

2.3

X3 = 0.90 + 1.1 (1.08670) + 1.6 (1.53630) = 2.39655

1.9

4th Iteration:

X1 = 2.70 + 2.2 (1.53630) - 1.5 (2.39655) = 0.92038

2.7

X2 = 2.35 - 1.7 (0.92038) + 1.5 (2.39655) = 1.90443

2.3

X3 = 0.90 + 1.1 (0.92038) + 1.6 (1.90443) = 2.61027

1.9

5th Iteration:

X1 = 2.70 + 2.2 (1.90443) - 1.5 (2.61027) = 2.61027

2.7

X2 = 2.35 - 1.7 (1.10161) + 1.5 (2.61027) = 1.90986

2.3

X3 = 0.90 + 1.1 (1.1061) + 1.6 (1.90986) = 2.71977

1.9

6th Iteration:

X1 = 2.70 + 2.2 (1.90986) - 1.5 (2.71977) = 1.04520

2.7

X2 = 2.35 - 1.7 (1.04520) + 1.5 (2.71977) = 2.02296

2.3

X3 = 0.90 + 1.1 (1.04520) + 1.6 (2.02296) = 2.78235

1.9

7th Iteration:

X1 = 2.70 + 2.2 (2.02296) - 1.5 (2.78235) = 1.10259

2.7

X2 = 2.35 - 1.7 (1.10259) + 1.5 (2.78235) = 2.02136

2.3

X3 = 0.90 + 1.1 (1.10259) + 1.6 (2.02136) = 2.81422

1.9

Iteration Table:

For | εa | ≤ | εs |

X1: | εs | = 0.50 x 102-n = 0.0005%

X2: | εs | = 0.50 x 102-n = 0.0005%

X3: | εs | = 0.50 x 102-n = 0.0005%

Therefore the value of the unknown variables are:

X1 = 1.10071

X2 = 2.06896

X3 = 2.85323