Chapter 11, Problem 12P

Use the Gauss-Seidel method (a) without relaxation and (b) with relaxation $\left(\lambda =0.95\right)$ to solve the following system to a tolerance of ${\epsilon }_s=5\%$. If necessary, rearrange the equations to achieve convergence.

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

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.

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

Answer to Problem 12P

Solution:

Using Gauss-Seidel method four iterations are performed to get the values $x_1=1.696789,x_2=2.829356\text{and}{x}_3=4.355084$.

Explanation of Solution

Given Information:

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

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:

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

The sufficient condition for convergence is:

If the diagonal coefficient of each equations is greater than the sum of the absolute values of the other coefficients present in the equation. The systems where this condition holds are called diagonally dominant.

Thus, the firstarrange the equation so that theywill be diagonally dominant.

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

It can be written in the form:

$\begin{array}{ccc}6& -1& -1\\ 6& 9& 1\\ -3& 1& 12\end{array}\left\{\begin{array}{l}{x}_1\\ x_2\\ x_3\end{array}\right\}=\left\{\begin{array}{l}3\\ 40\\ 50\end{array}\right\}$

Where,

$\left[A\right]=\begin{array}{ccc}6& -1& -1\\ 6& 9& 1\\ -3& 1& 12\end{array}$

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

$\left\{B\right\}=\left\{\begin{array}{l}3\\ 40\\ 50\end{array}\right\}$

First, solve the equations for its unknown and find the value of $x_1,x_2,\text{ and }{x}_3$.

$x_1=\frac{3-\left(-1\right)x_2-\left(-1\right)x_3}{6}$.. .. .. (1)

$x_2=\frac{40-\left(6\right)x_1-\left(1\right)x_3}{9}$.. .. .. (2)

$x_3=\frac{50-\left(-3\right)x_1-\left(1\right)x_2}{12}$.. .. .. (3)

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

Thus, equation (1) becomes,

$\begin{array}{c}{x}_1=\frac{3-\left(-1\right)x_2-\left(-1\right)x_3}{6}\\ =\frac{3-\left(-1\right)\left(0\right)-\left(-1\right)\left(0\right)}{6}\\ =0.5\end{array}$

Substitute the value of $x_1$ and $x_3=0$ in equation (2) to calculate $x_2$.

$\begin{array}{c}{x}_2=\frac{40-\left(6\right)x_1-\left(1\right)x_3}{9}\\ =\frac{40-\left(6\right)\left(0.5\right)-\left(1\right)\left(0\right)}{9}\\ =4.11111\end{array}$

Now, substitute the calculated values of $x_1=0.5$ and $x_2=4.11111$ into equation (3) to calculate,

$\begin{array}{c}{x}_3=\frac{50-\left(-3\right)x_1-\left(1\right)x_2}{12}\\ =\frac{50-\left(-3\right)\left(0.5\right)-\left(1\right)\left(4.11111\right)}{12}\\ =3.949074\end{array}$

For the second iteration, the same process is repeated with $x_2=4.11111$ and $x_3=3.949074$ to compute,

$\begin{array}{c}{x}_1=\frac{3-\left(-1\right)x_2-\left(-1\right)x_3}{6}\\ =\frac{3-\left(-1\right)\left(4.11111\right)-\left(-1\right)\left(3.949074\right)}{6}\\ =1.843364\end{array}$

The value of $x_1=1.843364$, along with the calculated value of $x_3=3.949074$, can be used to get,

$\begin{array}{c}{x}_2=\frac{40-\left(6\right)x_1-\left(1\right)x_3}{9}\\ =\frac{40-\left(6\right)\left(1.843364\right)-\left(1\right)\left(3.949074\right)}{9}\\ =2.776749\end{array}$

Now, substitute the calculated values of $x_1=1.843364$ and $x_2=2.776749$ to calculate,

$\begin{array}{c}{x}_3=\frac{50-\left(-3\right)x_1-\left(1\right)x_2}{12}\\ =\frac{50-\left(-3\right)\left(1.843364\right)-\left(1\right)\left(2.776749\right)}{12}\\ =4.396112\end{array}$

Now, the error can be computed as:

$\begin{array}{c}{\epsilon }_{a,1}=\left|\frac{1.843364-0.5}{1.843364}\right|\times 100\mathbf{\text{\%}}\\ \mathbf{\text{=0}}\mathbf{\text{.7288}}\times 100\mathbf{\text{\%}}\\ =72.88\mathbf{\text{\%}}\end{array}$

$\begin{array}{c}{\epsilon }_{a,2}=\left|\frac{2.776749-4.11111}{2.776749}\right|\times 100\mathbf{\text{\%}}\\ \mathbf{\text{=0}}\mathbf{\text{.4805}}\times 100\mathbf{\text{\%}}\\ =48.05\mathbf{\text{\%}}\end{array}$

$\begin{array}{c}{\epsilon }_{a,3}=\left|\frac{4.396112-3.949074}{4.396112}\right|\times 100\mathbf{\text{\%}}\\ \mathbf{\text{=0}}\mathbf{\text{.1017}}\times 100\mathbf{\text{\%}}\\ =10.17\mathbf{\text{\%}}\end{array}$

For the third iteration, the same process is repeated with $x_2=2.776749$ and $x_3=4.396112$ to compute,

$\begin{array}{c}{x}_1=\frac{3-\left(-1\right)x_2-\left(-1\right)x_3}{6}\\ =\frac{3-\left(-1\right)\left(2.776749\right)-\left(-1\right)\left(4.396112\right)}{6}\\ =1.695477\end{array}$

The value of $x_1=1.695477$, along with the calculated value of $x_3=4.396112$, can be used to get,

$\begin{array}{c}{x}_2=\frac{40-\left(6\right)x_1-\left(1\right)x_3}{9}\\ =\frac{40-\left(6\right)\left(1.695477\right)-\left(1\right)\left(4.396112\right)}{9}\\ =2.82567\end{array}$

Now, substitute the calculated values of $x_1=1.695477$ and $x_2=2.82567$ to calculate,

$\begin{array}{c}{x}_3=\frac{50-\left(-3\right)x_1-\left(1\right)x_2}{12}\\ =\frac{50-\left(-3\right)\left(1.695477\right)-\left(1\right)\left(2.82567\right)}{12}\\ =4.355063\end{array}$

Now, the error can be computed as:

$\begin{array}{c}{\epsilon }_{a,1}=\left|\frac{1.695477-1.843364}{1.695477}\right|\times 100\mathbf{\text{\%}}\\ \mathbf{\text{=0}}\mathbf{\text{.0872}}\times 100\mathbf{\text{\%}}\\ =8.72\mathbf{\text{\%}}\end{array}$

$\begin{array}{c}{\epsilon }_{a,2}=\left|\frac{2.82567-2.776749}{2.82567}\right|\times 100\mathbf{\text{\%}}\\ \mathbf{\text{=0}}\mathbf{\text{.0173}}\times 100\mathbf{\text{\%}}\\ =1.73\mathbf{\text{\%}}\end{array}$

$\begin{array}{c}{\epsilon }_{a,3}=\left|\frac{4.355063-4.396112}{4.355063}\right|\times 100\mathbf{\text{\%}}\\ \mathbf{\text{=0}}\mathbf{\text{.0094}}\times 100\mathbf{\text{\%}}\\ =0.94\mathbf{\text{\%}}\end{array}$

For the fourth iteration, the same process is repeated with $x_2=2.82567$ and $x_3=4.355063$ to compute,

$\begin{array}{c}{x}_1=\frac{3-\left(-1\right)x_2-\left(-1\right)x_3}{6}\\ =\frac{3-\left(-1\right)\left(2.82567\right)-\left(-1\right)\left(4.355063\right)}{6}\\ =1.696789\end{array}$

The value of $x_1=1.696789$, along with the calculated value of $x_3=4.355063$, can be used to get,

$\begin{array}{c}{x}_2=\frac{40-\left(6\right)x_1-\left(1\right)x_3}{9}\\ =\frac{40-\left(6\right)\left(1.696789\right)-\left(1\right)\left(4.355063\right)}{9}\\ =2.829356\end{array}$

Now, substitute the calculated values of $x_1=1.696789$ and $x_2=2.829356$ to calculate,

$\begin{array}{c}{x}_3=\frac{50-\left(-3\right)x_1-\left(1\right)x_2}{12}\\ =\frac{50-\left(-3\right)\left(1.696789\right)-\left(1\right)\left(2.829356\right)}{12}\\ =4.355084\end{array}$

Now, the error can be computed as:

$\begin{array}{c}{\epsilon }_{a,1}=\left|\frac{1.696789-1.695477}{1.696789}\right|\times 100\mathbf{\text{\%}}\\ \mathbf{\text{=0}}\mathbf{\text{.0008}}\times 100\mathbf{\text{\%}}\\ =0.08\mathbf{\text{\%}}\end{array}$

$\begin{array}{c}{\epsilon }_{a,2}=\left|\frac{2.829356-2.82567}{2.829356}\right|\times 100\mathbf{\text{\%}}\\ \mathbf{\text{=0}}\mathbf{\text{.0013}}\times 100\mathbf{\text{\%}}\\ =00.13\mathbf{\text{\%}}\end{array}$

$\begin{array}{c}{\epsilon }_{a,3}=\left|\frac{4.355084-4.355063}{4.355084}\right|\times 100\mathbf{\text{\%}}\\ \mathbf{\text{=0}}\mathbf{\text{.00004}}\times 100\mathbf{\text{\%}}\\ =0.004\mathbf{\text{\%}}\end{array}$

Thus, after four iterations the maximum error is 0.13% which is less than ${\epsilon }_s=5\%$ and the values are $x_1=1.696789,x_2=2.829356\text{and}{x}_3=4.355084$.

To determine

To calculate: The solution of following system of equation with Gauss-Seidel method withrelaxation $\lambda =0.95$ 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 12P

Solution:

Using Gauss-Seidel method four iterations are performed with relaxation $\lambda =0.95$ to get the values $x_1=1.70969,x_2=2.83003\text{and}{x}_3=4.35641$.

Explanation of Solution

Given Information:

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 $\lambda =0.95$ 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:

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

If the diagonal coefficient of each equations is greater than the sum of the absolute values of the other coefficients present in the equation. The systems where this condition holds are called diagonally dominant.

Thus, the first arrange the equation so that they will be diagonally dominant.

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

It can be written in the form:

$\begin{array}{ccc}6& -1& -1\\ 6& 9& 1\\ -3& 1& 12\end{array}\left\{\begin{array}{l}{x}_1\\ x_2\\ x_3\end{array}\right\}=\left\{\begin{array}{l}3\\ 40\\ 50\end{array}\right\}$

Where,

$\left[A\right]=\begin{array}{ccc}6& -1& -1\\ 6& 9& 1\\ -3& 1& 12\end{array}$

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

$\left\{B\right\}=\left\{\begin{array}{l}3\\ 40\\ 50\end{array}\right\}$

First, solve the equations for its unknown and find the value of $x_1,x_2,\text{ and }{x}_3$.

$x_1=\frac{3-\left(-1\right)x_2-\left(-1\right)x_3}{6}$.. .. .. (1)

$x_2=\frac{40-\left(6\right)x_1-\left(1\right)x_3}{9}$.. .. .. (2)

$x_3=\frac{50-\left(-3\right)x_1-\left(1\right)x_2}{12}$.. .. .. (3)

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

Thus, equation (1) becomes,

$\begin{array}{c}{x}_1=\frac{3-\left(-1\right)x_2-\left(-1\right)x_3}{6}\\ =\frac{3-\left(-1\right)\left(0\right)-\left(-1\right)\left(0\right)}{6}\\ =0.5\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_1=\lambda \left(0.5\right)+\left(1-\lambda \right)\left(0\right)\\ =0.95\left(0.5\right)+\left(1-0.95\right)\left(0\right)\\ =0.95\left(0.5\right)+\left(0.05\right)\left(0\right)\\ =0.475\end{array}$

Substitute the value of $x_1$ and $x_3=0$ in equation (2) to calculate $x_2$.

$\begin{array}{c}{x}_2=\frac{40-\left(6\right)x_1-\left(1\right)x_3}{9}\\ =\frac{40-\left(6\right)\left(0.475\right)-\left(1\right)\left(0\right)}{9}\\ =4.12778\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_2=\lambda \left(4.12778\right)+\left(1-\lambda \right)\left(0\right)\\ =0.95\left(4.12778\right)+\left(1-0.95\right)\left(0\right)\\ =0.95\left(4.12778\right)+\left(0.05\right)\left(0\right)\\ =3.92139\end{array}$

Now, substitute the calculated values of $x_1=0.475$ and $x_2=3.92139$ into equation (3) to calculate,

$\begin{array}{c}{x}_3=\frac{50-\left(-3\right)x_1-\left(1\right)x_2}{12}\\ =\frac{50-\left(-3\right)\left(0.475\right)-\left(1\right)\left(3.92139\right)}{12}\\ =3.95863\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_3=\lambda \left(3.95863\right)+\left(1-\lambda \right)\left(0\right)\\ =0.95\left(3.95863\right)+\left(1-0.95\right)\left(0\right)\\ =0.95\left(3.95863\right)+\left(0.05\right)\left(0\right)\\ =3.76070\end{array}$

For the second iteration, the same process is repeated with $x_2=3.92139$ and $x_3=3.76070$ to compute,

$\begin{array}{c}{x}_1=\frac{3-\left(-1\right)x_2-\left(-1\right)x_3}{6}\\ =\frac{3-\left(-1\right)\left(3.92139\right)-\left(-1\right)\left(3.76070\right)}{6}\\ =1.78035\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_1=\lambda \left(1.78035\right)+\left(1-\lambda \right)\left(0.475\right)\\ =0.95\left(1.78035\right)+\left(1-0.95\right)\left(.475\right)\\ =0.95\left(1.78035\right)+\left(0.05\right)\left(0.475\right)\\ =1.71508\end{array}$

The value of $x_1=1.71508$, along with the calculated value of $x_3=3.76070$, can be used to get,

$\begin{array}{c}{x}_2=\frac{40-\left(6\right)x_1-\left(1\right)x_3}{9}\\ =\frac{40-\left(6\right)\left(1.71508\right)-\left(1\right)\left(3.76070\right)}{9}\\ =2.88320\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_2=\lambda \left(2.88320\right)+\left(1-\lambda \right)\left(3.92139\right)\\ =0.95\left(2.88320\right)+\left(1-0.95\right)\left(3.92139\right)\\ =0.95\left(2.88320\right)+\left(0.05\right)\left(3.92139\right)\\ =2.93511\end{array}$

Now, substitute the calculated values of $x_1=1.71508$ and $x_2=2.93511$ to calculate,

$\begin{array}{c}{x}_3=\frac{50-\left(-3\right)x_1-\left(1\right)x_2}{12}\\ =\frac{50-\left(-3\right)\left(1.71508\right)-\left(1\right)\left(2.93511\right)}{12}\\ =4.35084\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_3=\lambda \left(4.35084\right)+\left(1-\lambda \right)\left(3.76070\right)\\ =0.95\left(4.35084\right)+\left(1-0.95\right)\left(3.76070\right)\\ =0.95\left(4.35084\right)+\left(0.05\right)\left(3.76070\right)\\ =4.32134\end{array}$

Now, the error can be computed as:

$\begin{array}{c}{\epsilon }_{a,1}=\left|\frac{1.71508-0.475}{1.71508}\right|\times 100\mathbf{\text{\%}}\\ \mathbf{\text{=0}}\mathbf{\text{.7230}}\times 100\mathbf{\text{\%}}\\ =72.3\mathbf{\text{\%}}\end{array}$

$\begin{array}{c}{\epsilon }_{a,2}=\left|\frac{2.93511-3.92139}{2.93511}\right|\times 100\mathbf{\text{\%}}\\ \mathbf{\text{=0}}\mathbf{\text{.3360}}\times 100\mathbf{\text{\%}}\\ =33.6\mathbf{\text{\%}}\end{array}$

$\begin{array}{c}{\epsilon }_{a,3}=\left|\frac{4.32134-3.76070}{4.32134}\right|\times 100\mathbf{\text{\%}}\\ \mathbf{\text{=0}}\mathbf{\text{.1297}}\times 100\mathbf{\text{\%}}\\ =12.97\mathbf{\text{\%}}\end{array}$

For the third iteration, the same process is repeated with $x_2=2.93511$ and $x_3=4.32134$ to compute,

$\begin{array}{c}{x}_1=\frac{3-\left(-1\right)x_2-\left(-1\right)x_3}{6}\\ =\frac{3-\left(-1\right)\left(2.93511\right)-\left(-1\right)\left(4.32134\right)}{6}\\ =1.70941\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_1=\lambda \left(1.70941\right)+\left(1-\lambda \right)\left(1.71508\right)\\ =0.95\left(1.70941\right)+\left(1-0.95\right)\left(1.71508\right)\\ =0.95\left(1.70941\right)+\left(0.05\right)\left(1.71508\right)\\ =1.70969\end{array}$

The value of $x_1=1.70969$, along with the calculated value of $x_3=4.32134$, can be used to get,

$\begin{array}{c}{x}_2=\frac{40-\left(6\right)x_1-\left(1\right)x_3}{9}\\ =\frac{40-\left(6\right)\left(1.70969\right)-\left(1\right)\left(4.32134\right)}{9}\\ =2.82450\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_2=\lambda \left(2.82450\right)+\left(1-\lambda \right)\left(2.93511\right)\\ =0.95\left(2.82450\right)+\left(1-0.95\right)\left(2.93511\right)\\ =0.95\left(2.82450\right)+\left(0.05\right)\left(2.93511\right)\\ =2.83003\end{array}$

Now, substitute the calculated values of $x_1=1.70969$ and $x_2=2.83003$ to calculate,

$\begin{array}{c}{x}_3=\frac{50-\left(-3\right)x_1-\left(1\right)x_2}{12}\\ =\frac{50-\left(-3\right)\left(1.70969\right)-\left(1\right)\left(2.83003\right)}{12}\\ =4.35825\end{array}$

Relaxation yields:

$\begin{array}{c}{x}_3=\lambda \left(4.35825\right)+\left(1-\lambda \right)\left(4.32134\right)\\ =0.95\left(4.35825\right)+\left(1-0.95\right)\left(4.32134\right)\\ =0.95\left(4.35825\right)+\left(0.05\right)\left(4.32134\right)\\ =4.35641\end{array}$

Now, the error can be computed as:

$\begin{array}{c}{\epsilon }_{a,1}=\left|\frac{1.70969-1.71508}{1.70969}\right|\times 100\mathbf{\text{\%}}\\ \mathbf{\text{=0}}\mathbf{\text{.0031}}\times 100\mathbf{\text{\%}}\\ =0.31\mathbf{\text{\%}}\end{array}$

Also,

$\begin{array}{c}{\epsilon }_{a,2}=\left|\frac{2.83003-2.93511}{2.83003}\right|\times 100\mathbf{\text{\%}}\\ \mathbf{\text{=0}}\mathbf{\text{.0371}}\times 100\mathbf{\text{\%}}\\ =3.71\mathbf{\text{\%}}\end{array}$

$\begin{array}{c}{\epsilon }_{a,3}=\left|\frac{4.35641-4.32134}{4.35641}\right|\times 100\mathbf{\text{\%}}\\ \mathbf{\text{=0}}\mathbf{\text{.0080}}\times 100\mathbf{\text{\%}}\\ =0.80\mathbf{\text{\%}}\end{array}$

Thus, after four iterations the maximum error is 3.71% which is less than ${\epsilon }_s=5\%$ and the values are $x_1=1.70969,x_2=2.83003\text{and}{x}_3=4.35641$.