Chapter 11, Problem 13P

Use the Gauss-Seidel method(a) without relaxation and (b) with relaxation $\left(\lambda =1.2\right)$ to solve the following system to a tolerance of ${\epsilon }_s=5\%$. If necessary, rearrange the equations to achieve convergence.

$\begin{array}{c}2x_1-6x_2-x_3=-38\\ -3x_1-x_2+7x_3=-34\\ -8x_1+x_2-2x_3=-20\end{array}$

To determine

To calculate: The solution of following system of equation with Gauss-Seidel method without relaxation to a tolerance of ${\epsilon }_s=5\%$. If necessary, rearrange the equations to achieve convergence.

$\begin{array}{c}2x_1-6x_2-x_3=-38\\ -3x_1-x_2+7x_3=-34\\ -8x_1+x_2-2x_3=-20\end{array}$

Answer to Problem 13P

Solution:

Using Gauss-Seidel method three iterations are performed to get the values $x_1=4.0046,x_2=7.9916\text{and}{x}_3=-1.9991$.

Explanation of Solution

Given:

A system of equation:

$\begin{array}{c}2x_1-6x_2-x_3=-38\\ -3x_1-x_2+7x_3=-34\\ -8x_1+x_2-2x_3=-20\end{array}$

With ${\epsilon }_s=5\%$.

Formula used:

(1) The values of $x_1,x_2\text{and}{x}_3$ are given by the formula:

$x_1=\frac{b_1-a_{12}{x}_2-a_{13}{x}_3}{a_{11}}$, $x_2=\frac{b_2-a_{21}{x}_1-a_{23}{x}_3}{a_{22}}$, $x_3=\frac{b_3-a_{31}{x}_1-a_{32}{x}_2}{a_{33}}$

(2) Convergence can be checked using the criterion

${\epsilon }_{a,i}=\left|\frac{x_i{}^j-x_i{}^{j-1}}{x_i{}^j}\right|100\mathbf{\text{\%<}}{\epsilon }_s$

For all i, where j and j- 1 are the present and previous iterations.

Calculation:

Consider the system of equation:

$\begin{array}{c}2x_1-6x_2-x_3=-38\\ -3x_1-x_2+7x_3=-34\\ -8x_1+x_2-2x_3=-20\end{array}$

The sufficient condition for convergence is:

The diagonal coefficient in each of the equations must be larger than the sum of the absolute values of the other coefficients in the equation. The systems where this condition holds are called diagonally dominant.

Thus, the equations should first be arranged so that they are diagonally dominant.

$\begin{array}{c}-8x_1+x_2-2x_3=-20\\ 2x_1-6x_2-x_3=-38\\ -3x_1-x_2+7x_3=-34\end{array}$

It can be written in the form:

$\begin{array}{ccc}-8& 1& -2\\ 2& -6& -1\\ -3& -1& 7\end{array}\left\{\begin{array}{l}{x}_1\\ x_2\\ x_3\end{array}\right\}=\left\{\begin{array}{l}-20\\ -38\\ -34\end{array}\right\}$

Where,

$\left[A\right]=\begin{array}{ccc}-8& 1& -2\\ 2& -6& -1\\ -3& -1& 7\end{array}$

$\left\{X\right\}=\left\{\begin{array}{l}{x}_1\\ x_2\\ x_3\end{array}\right\}$

$\left\{B\right\}=\left\{\begin{array}{l}-20\\ -38\\ -34\end{array}\right\}$

First, solve each of the equations for its unknown on the diagonal

$x_1=\frac{-20-\left(1\right)x_2-\left(-2\right)x_3}{-8}$.. .. .. (1)

$x_2=\frac{-38-\left(2\right)x_1-\left(-1\right)x_3}{-6}$.. .. .. (2)

$x_3=\frac{-34-\left(-3\right)x_1-\left(-1\right)x_2}{7}$.. .. .. (3)

For initial guess, assume $x_2=0$ and $x_3=0$.

Thus, equation (1) becomes,

$\begin{array}{c}{x}_1=\frac{-20-x_2+2x_3}{-8}\\ =\frac{-20-0+2\left(0\right)}{-8}\\ =2.5\end{array}$

This value, along with the assumed value of $x_3=0$, can be substituted in equation (2) to calculate,

$\begin{array}{c}{x}_2=\frac{-38-2x_1+x_3}{-6}\\ =\frac{-38-2\times \left(2.5\right)+0}{-6}\\ =7.1667\end{array}$

Now, substitute the calculated values of $x_1=2.5$ and $x_2=7.1667$ into equation (3) to calculate,

$\begin{array}{c}{x}_3=\frac{-34+3x_1+x_2}{7}\\ =\frac{-34+3\times \left(2.5\right)+7.1667}{7}\\ =-2.7619\end{array}$

For the second iteration, the same process is repeated with $x_1=2.5,x_2=7.\text{1667 and }{x}_3=-2.7619$ to compute,

$\begin{array}{c}{x}_1=\frac{-20-x_2+2x_3}{-8}\\ =\frac{-20-\left(7.1667\right)+2\left(-2.7619\right)}{-8}\\ =4.0863\end{array}$

The value of $x_1=4.0863$, along with the calculated value of $x_3=-2.7619$, can be used to get,

$\begin{array}{c}{x}_2=\frac{-38-2x_1+x_3}{-6}\\ =\frac{-38-2\times \left(4.0863\right)+\left(-2.7619\right)}{-6}\\ =8.1557\end{array}$

Now, substitute the calculated values of $x_1=4.0863$ and $x_2=8.1557$ to calculate,

$\begin{array}{c}{x}_3=\frac{-34+3x_1+x_2}{7}\\ =\frac{-34+3\times \left(4.0863\right)+\left(8.1557\right)}{7}\\ =-1.9407\end{array}$

Now, the error can be computed as:

$\begin{array}{c}{\epsilon }_{a,1}=\left|\frac{4.08631-2.5}{4.08631}\right|\times 100\%\\ =38.82\%\end{array}$

$\begin{array}{c}{\epsilon }_{a,2}=\left|\frac{8.1557-7.1667}{8.1557}\right|\times 100\%\\ =12.13\%\end{array}$

$\begin{array}{c}{\epsilon }_{a,3}=\left|\frac{-1.9407-\left(-2.7619\right)}{-1.9407}\right|\times 100\%\\ =42.31\%\end{array}$

For the third iteration, the same process is repeated with $x_1=4.0863,x_2=8.155\text{7 }\text{and }{x}_3=-1.9407$ to compute,

$\begin{array}{c}{x}_1=\frac{-20-x_2+2x_3}{-8}\\ =\frac{-20-\left(8.1557\right)+2\left(-1.9407\right)}{-8}\\ =4.0046\end{array}$

The value of $x_1=4.0046$, along with the calculated value of $x_3=-1.9407$, can be used to get,

$\begin{array}{c}{x}_2=\frac{-38-2x_1+x_3}{-6}\\ =\frac{-38-2\times \left(4.0046\right)+\left(-1.9407\right)}{-6}\\ =7.9916\end{array}$

Now, substitute the calculated values of $x_1=4.0046$ and $x_2=7.9916$ to calculate,

$\begin{array}{c}{x}_3=\frac{-34+3x_1+x_2}{7}\\ =\frac{-34+3\times \left(4.0046\right)+\left(7.9916\right)}{7}\\ =-1.9992\end{array}$

Now, the error can be computed as:

$\begin{array}{c}{\epsilon }_{a,1}=\left|\frac{4.0046-4.08631}{4.0046}\right|\times 100\%\\ =2.0404\%\end{array}$

$\begin{array}{c}{\epsilon }_{a,2}=\left|\frac{7.9916-8.1557}{7.9916}\right|\times 100\%\\ =2.0534\%\end{array}$

$\begin{array}{c}{\epsilon }_{a,3}=\left|\frac{-1.9992-\left(-1.9407\right)}{-1.9992}\right|\times 100\%\\ =2.92\%\end{array}$

Thus, after three iterations the maximum error is 2.92% which is less than ${\epsilon }_s=5\%$ and the values are $x_1=4.0046,x_2=7.9916\text{and}{x}_3=-1.9991$.

To determine

To calculate: The solution of following system of equation with Gauss-Seidel method with relaxation $\lambda =1.2$ to a tolerance of ${\epsilon }_s=5\%$. If necessary, rearrange the equations to achieve convergence.

$\begin{array}{c}-3x_1+x_2+12x_3=50\\ 6x_1-x_2-x_3=3\\ 6x_1+9x_2+x_3=40\end{array}$

Answer to Problem 13P

Solution:

Using Gauss-Seidel method six iterations are performed with relaxation $\lambda =1.2$ to get the values $x_1=4.0122,x_2=8.0272\text{and }{x}_3=-1.9790$.

Explanation of Solution

Given:

A system of equation:

$\begin{array}{c}-3x_1+x_2+12x_3=50\\ 6x_1-x_2-x_3=3\\ 6x_1+9x_2+x_3=40\end{array}$

With relaxation $\lambda =1.2$ and ${\epsilon }_s=5\%$.

Formula used:

(1) The values of $x_1,x_2\text{and}{x}_3$ are given by the formula:

$x_1=\frac{b_1-a_{12}{x}_2-a_{13}{x}_3}{a_{11}}$, $x_2=\frac{b_2-a_{21}{x}_1-a_{23}{x}_3}{a_{22}}$, $x_3=\frac{b_3-a_{31}{x}_1-a_{32}{x}_2}{a_{33}}$

(2) Relaxation

$x_i{}^{\text{new}}=\lambda x_i{}^{\text{new}}+\left(1-\lambda \right)x_i{}^{\text{old}}$

Where $\lambda$ is a weighting factor that is assigned a value between 0 and 2.

(3) Convergence can be checked using the criterion

${\epsilon }_{a,i}=\left|\frac{x_i{}^j-x_i{}^{j-1}}{x_i{}^j}\right|100\mathbf{\text{\%<}}{\epsilon }_s$

For all i, where j and j- 1 are the present and previous iterations.

Calculation:

Consider the system of equation:

$\begin{array}{c}2x_1-6x_2-x_3=-38\\ -3x_1-x_2+7x_3=-34\\ -8x_1+x_2-2x_3=-20\end{array}$

The sufficient condition for convergence is:

The diagonal coefficient in each of the equations must be larger than the sum of the absolute values of the other coefficients in the equation. The systems where this condition holds are called diagonally dominant.

Thus, the equations should first be arranged so that they are diagonally dominant.

$\begin{array}{c}-8x_1+x_2-2x_3=-20\\ 2x_1-6x_2-x_3=-38\\ -3x_1-x_2+7x_3=-34\end{array}$

It can be written in the form:

$\begin{array}{ccc}-8& 1& -2\\ 2& -6& -1\\ -3& -1& 7\end{array}\left\{\begin{array}{l}{x}_1\\ x_2\\ x_3\end{array}\right\}=\left\{\begin{array}{l}-20\\ -38\\ -34\end{array}\right\}$

Where,

$\left[A\right]=\begin{array}{ccc}-8& 1& -2\\ 2& -6& -1\\ -3& -1& 7\end{array}$

$\left\{X\right\}=\left\{\begin{array}{l}{x}_1\\ x_2\\ x_3\end{array}\right\}$

$\left\{B\right\}=\left\{\begin{array}{l}-20\\ -38\\ -34\end{array}\right\}$

First, solve each of the equations for its unknown on the diagonal

$x_1=\frac{-20-\left(1\right)x_2-\left(-2\right)x_3}{-8}$.. .. .. (1)

$x_2=\frac{-38-\left(2\right)x_1-\left(-1\right)x_3}{-6}$.. .. .. (2)

$x_3=\frac{-34-\left(-3\right)x_1-\left(-1\right)x_2}{7}$.. .. .. (3)

For initial guess, assume $x_1=0$, $x_2=0$ and $x_3=0$.

Thus, equation (1) becomes,

$\begin{array}{c}{x}_1=\frac{-20-x_2+2x_3}{-8}\\ =\frac{-20-0+2\left(0\right)}{-8}\\ =2.5\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_1=\lambda \left(2.5\right)+\left(1-\lambda \right)\left(0\right)\\ =1.2\times \left(2.5\right)-0.2\times \left(0\right)\\ =3\end{array}$

This value, along with the assumed value of $x_3=0$, can be substituted in equation (2) to calculate,

$\begin{array}{c}{x}_2=\frac{-38-2x_1+x_3}{-6}\\ =\frac{-38-2\times \left(3\right)+0}{-6}\\ =7.3333\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_2=\lambda \left(7.3333\right)+\left(1-\lambda \right)\left(0\right)\\ =1.2\times \left(7.3333\right)-0.2\times \left(0\right)\\ =8.8\end{array}$

Now, substitute the calculated values of $x_1=3$ and $x_2=8.8$ into equation (3) to calculate,

$\begin{array}{c}{x}_3=\frac{-34+3x_1+x_2}{7}\\ =\frac{-34+3\times \left(3\right)+\left(8.8\right)}{7}\\ =-2.3142\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_3=\lambda \left(-2.3142\right)+\left(1-\lambda \right)\left(0\right)\\ =1.2\times \left(-2.3142\right)-0.2\times \left(0\right)\\ =-2.7771\end{array}$

For the second iteration, the same process is repeated with $x_1=3,x_2=8.8\text{ and }{x}_3=-2.7771$ to compute,

$\begin{array}{c}{x}_1=\frac{-20-x_2+2x_3}{-8}\\ =\frac{-20-\left(8.8\right)+2\left(-2.7771\right)}{-8}\\ =4.2942\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_1=\lambda \left(4.2942\right)+\left(1-\lambda \right)\left(3\right)\\ =1.2\times \left(4.2942\right)-0.2\times \left(3\right)\\ =4.5531\end{array}$

The value of $x_1=4.5531$, along with the calculated value of $x_3=-2.7771$, can be used to get,

$\begin{array}{c}{x}_2=\frac{-38-2x_1+x_3}{-6}\\ =\frac{-38-2\times \left(4.5531\right)+\left(-2.7771\right)}{-6}\\ =8.3139\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_2=\lambda \left(8.3139\right)+\left(1-\lambda \right)\left(8.8\right)\\ =1.2\times \left(8.3139\right)-0.2\times \left(8.8\right)\\ =8.2166\end{array}$

Now, substitute the calculated values of $x_1=4.5531$ and $x_2=8.2166$ to calculate,

$\begin{array}{c}{x}_3=\frac{-34+3x_1+x_2}{7}\\ =\frac{-34+3\times \left(4.5531\right)+\left(8.2166\right)}{7}\\ =-1.7319\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_3=\lambda \left(-1.7319\right)+\left(1-\lambda \right)\left(-2.7771\right)\\ =1.2\times \left(-1.7319\right)-0.2\times \left(-2.7771\right)\\ =-1.52294\end{array}$

Now, the error can be computed as:

$\begin{array}{c}{\epsilon }_{a,1}=\left|\frac{4.5531-3}{4.5531}\right|\times 100\%\\ =34.11\%\end{array}$

$\begin{array}{c}{\epsilon }_{a,2}=\left|\frac{8.2166-8.8}{8.2166}\right|\times 100\%\\ =7.1\%\end{array}$

$\begin{array}{c}{\epsilon }_{a,3}=\left|\frac{-1.5229-\left(-2.7771\right)}{-1.5229}\right|\times 100\%\\ =82.35\%\end{array}$

For the third iteration, the same process is repeated with $x_1=4.5531,x_2=8.2166\text{ and }{x}_3=-1.5229$ to compute,

$\begin{array}{c}{x}_1=\frac{-20-x_2+2x_3}{-8}\\ =\frac{-20-\left(8.2166\right)+2\left(-1.5229\right)}{-8}\\ =3.9078\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_1=\lambda \left(3.9078\right)+\left(1-\lambda \right)\left(4.5531\right)\\ =1.2\times \left(3.9078\right)-0.2\times \left(4.5531\right)\\ =3.7787\end{array}$

The value of $x_1=3.7787$, along with the calculated value of $x_3=-1.5229$, can be used to get,

$\begin{array}{c}{x}_2=\frac{-38-2x_1+x_3}{-6}\\ =\frac{-38-2\times \left(3.7787\right)+\left(-1.5229\right)}{-6}\\ =7.8467\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_2=\lambda \left(7.8467\right)+\left(1-\lambda \right)\left(8.2166\right)\\ =1.2\times \left(7.8467\right)-0.2\times \left(8.2166\right)\\ =7.7727\end{array}$

Now, substitute the calculated values of $x_1=3.7787$ and $x_2=7.7727$ to calculate,

$\begin{array}{c}{x}_3=\frac{-34+3x_1+x_2}{7}\\ =\frac{-34+3\times \left(3.7787\right)+\left(7.7727\right)}{7}\\ =-2.1273\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_3=\lambda \left(-2.1273\right)+\left(1-\lambda \right)\left(-1.5229\right)\\ =1.2\times \left(-2.1273\right)-0.2\times \left(-1.5229\right)\\ =-2.2481\end{array}$

Now, the error can be computed as:

$\begin{array}{c}{\epsilon }_{a,1}=\left|\frac{3.7787-4.5531}{3.7787}\right|\times 100\%\\ =20.49\%\end{array}$

$\begin{array}{c}{\epsilon }_{a,2}=\left|\frac{7.7727-8.2166}{7.7727}\right|\times 100\%\\ =5.71\%\end{array}$

$\begin{array}{c}{\epsilon }_{a,3}=\left|\frac{-2.2481-\left(-1.5229\right)}{-2.2481}\right|\times 100\%\\ =32.25\%\end{array}$

The table showing values calculated for further iterations is given below

$\begin{array}{cccccc}i& \mathrm{var}& \text{value}& \text{relaxation}& \epsilon \left(\%\right)& \max \epsilon \\ 1& \begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}& \begin{array}{c}2.5\\ 7.3333\\ -2.3142\end{array}& \begin{array}{c}3\\ 8.8\\ -2.777\end{array}& \begin{array}{c}100\\ 100\\ 100\end{array}& \\ 2& \begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}& \begin{array}{c}4.2942\\ 8.3139\\ -1.7319\end{array}& \begin{array}{c}4.553\\ 8.2166\\ -1.5229\end{array}& \begin{array}{c}34.11\\ 7.10\\ 82.35\end{array}& 82.35\\ 3& \begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}& \begin{array}{c}3.9078\\ 7.8467\\ -2.1272\end{array}& \begin{array}{c}3.7787\\ 7.7727\\ -2.2481\end{array}& \begin{array}{c}20.49\\ 5.71\\ 32.24\end{array}& 32.25\\ 4& \begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}& \begin{array}{c}4.0336\\ 8.0695\\ -1.9453\end{array}& \begin{array}{c}4.0846\\ 8.1289\\ -1.8842\end{array}& \begin{array}{c}7.50\\ 4.40\\ 19.30\end{array}& 19.30\\ 5& \begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}& \begin{array}{c}9.9873\\ 7.9700\\ -2.0225\end{array}& \begin{array}{c}3.9678\\ 7.9383\\ -2.0500\end{array}& \begin{array}{c}2.94\\ 2.40\\ 8.07\end{array}& 8.07\\ 6& \begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}& \begin{array}{c}4.0048\\ 8.0124\\ -1.9908\end{array}& \begin{array}{c}4.0122\\ 8.0272\\ -1.9790\end{array}& \begin{array}{c}1.11\\ 1.11\\ 3.60\end{array}& 3.60\end{array}$

Therefore, after six iterations, the maximum error is 3.60% which is less than 5%. Thus, the results are $x_1=4.0122,x_2=8.0272\text{and }{x}_3=-1.9790$.