Difference between revisions of "Scilab/C4/Linear-equations-Iterative-Methods/English-timed"
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|'''Time''' | |'''Time''' | ||
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|'''Narration''' | |'''Narration''' | ||
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| 00:01 | | 00:01 | ||
− | |Dear Friends, | + | |Dear Friends, Welcome to the Spoken Tutorial on '''Solving System of Linear Equations using Iterative Methods'''. |
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|00:18 | |00:18 | ||
− | |Develop '''Scilab code''' to solve '''linear equations''' | + | |Develop '''Scilab code''' to solve '''linear equations'''. |
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|00:25 | |00:25 | ||
|'''Ubuntu 12.04''' as the operating system | |'''Ubuntu 12.04''' as the operating system | ||
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| 00:28 | | 00:28 | ||
− | |and '''Scilab 5.3.3''' version | + | |and '''Scilab 5.3.3''' version. |
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| 00:33 | | 00:33 | ||
− | | Before | + | | Before practicing this tutorial, a learner should have basic knowledge of |
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|00:38 | |00:38 | ||
− | |'''Scilab''' | + | |'''Scilab''', and '''solving linear equations'''. |
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|00:56 | |00:56 | ||
− | |Given a '''system of linear equations, with n equations and n unknowns''' | + | |Given a '''system of linear equations, with n equations and n unknowns''', |
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|01:02 | |01:02 | ||
− | | | + | | we rewrite the equations such that ''' x of i k plus one is equal to b i minus summation of a i j x j k from j equal to one to n divided by a i i''' where ''' i''' is from '''one to n'''. |
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|01:24 | |01:24 | ||
− | |We assume values for each '''x of i''' | + | |We assume values for each '''x of i'''. |
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| 01:39 | | 01:39 | ||
− | |Let us solve this example using '''Jacobi Method''' | + | |Let us solve this example using '''Jacobi Method'''. |
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| 01:48 | | 01:48 | ||
− | || We use '''format''' method to specify the format of the displayed answers on the '''Scilab console. ''' | + | || We use '''format''' method to specify the format of the displayed answers on the '''Scilab console.''' |
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| And '''twenty''' specifies the number of digits to be displayed. | | And '''twenty''' specifies the number of digits to be displayed. | ||
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|02:06 | |02:06 | ||
|Then we use '''input''' function to get the values for | |Then we use '''input''' function to get the values for | ||
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|02:14 | |02:14 | ||
|'''initial values matrix,''' | |'''initial values matrix,''' | ||
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− | | | + | | 02:17 |
|'''maximum number of iteration and''' | |'''maximum number of iteration and''' | ||
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− | | 02 | + | | 02:19 |
− | ||'''convergence tolerance''' | + | ||'''convergence tolerance'''. |
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− | |02 | + | |02:22 |
||Then we use '''size''' function to check if '''A matrix''' is a '''square matrix.''' | ||Then we use '''size''' function to check if '''A matrix''' is a '''square matrix.''' | ||
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− | |02 | + | |02:29 |
− | + | | If it isn't, we use '''error''' function to display an error. | |
− | | If it isn't, we use '''error | + | |
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− | |02 | + | |02:34 |
| We then check if '''matrix A''' is '''diagonally dominant.''' | | We then check if '''matrix A''' is '''diagonally dominant.''' | ||
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− | | 02 | + | | 02:40 |
|| The first half calculates the sum of each row of the '''matrix.''' | || The first half calculates the sum of each row of the '''matrix.''' | ||
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− | | 02 | + | | 02:45 |
| Then it checks if twice the product of the '''diagonal element''' is greater than the sum of the elements of that row. | | Then it checks if twice the product of the '''diagonal element''' is greater than the sum of the elements of that row. | ||
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− | |02 | + | |02:54 |
− | | If it isn't, an error is displayed using ''' error | + | | If it isn't, an error is displayed using '''error''' function. |
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− | |03 | + | |03:01 |
| Then we define the function '''Jacobi Iteration''' with input arguments | | Then we define the function '''Jacobi Iteration''' with input arguments | ||
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− | | 03 | + | | 03:07 |
− | | '''A, b , x zero, ''' | + | | '''A, b , x zero,''' |
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− | | 03 | + | | 03:09 |
− | |'''maximum iteration and tolerance level | + | |'''maximum iteration''' and '''tolerance level'''. |
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− | | 03 | + | | 03:14 |
− | |Here '''x zero''' is the '''initial values matrix. ''' | + | |Here '''x zero''' is the '''initial values matrix.''' |
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− | | 03 | + | | 03:19 |
|We check if the size of '''A matrix''' and '''initial values matrix''' are compatible with each other. | |We check if the size of '''A matrix''' and '''initial values matrix''' are compatible with each other. | ||
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− | |03 | + | |03:28 |
| We calculate the value for '''x k p one''' and then check if the '''relative error''' is lesser than '''tolerance level.''' | | We calculate the value for '''x k p one''' and then check if the '''relative error''' is lesser than '''tolerance level.''' | ||
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− | | 03 | + | | 03:38 |
− | | If it is lesser than '''tolerance level''', we break the iteration and the solution is returned. | + | | If it is lesser than '''tolerance level''', we '''break''' the iteration and the solution is returned. |
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− | | 03 | + | | 03:45 |
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+ | |Finally we '''end''' the function. | ||
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− | | 03 | + | | 03:48 |
|| Let us save and execute the function. | || Let us save and execute the function. | ||
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− | |03 | + | |03:51 |
− | ||Switch to '''Scilab console. ''' | + | ||Switch to '''Scilab console.''' |
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− | | 03 | + | | 03:54 |
| Let us enter the values at each prompt. | | Let us enter the values at each prompt. | ||
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− | | 03 | + | | 03:57 |
| The '''coefficient matrix A is open square bracket two space one semi colon five space seven close square bracket ''' | | The '''coefficient matrix A is open square bracket two space one semi colon five space seven close square bracket ''' | ||
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− | |04 | + | |04:08 |
| Press '''Enter. ''' | | Press '''Enter. ''' | ||
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− | | 04 | + | | 04:10 |
| Then we type '''open square bracket eleven semicolon thirteen close square bracket''' | | Then we type '''open square bracket eleven semicolon thirteen close square bracket''' | ||
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− | |04 | + | |04:17 |
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+ | ||Press '''Enter.''' | ||
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− | |04 | + | |04:20 |
|The '''initial values matrix is open square bracket one semi colon one close square bracket''' | |The '''initial values matrix is open square bracket one semi colon one close square bracket''' | ||
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− | | 04 | + | | 04:28 |
| Press '''Enter.''' | | Press '''Enter.''' | ||
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− | | 04 | + | | 04:30 |
|The '''maximum number of iterations''' is twenty five. | |The '''maximum number of iterations''' is twenty five. | ||
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− | | 04 | + | | 04:34 |
| Press '''Enter. ''' | | Press '''Enter. ''' | ||
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− | | 04 | + | | 04:36 |
| Let the '''convergence tolerance level be zero point zero zero zero zero one ''' | | Let the '''convergence tolerance level be zero point zero zero zero zero one ''' | ||
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− | | 04 | + | | 04:44 |
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+ | ||Press '''Enter.''' | ||
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− | | 04 | + | | 04:46 |
||We call the function by typing | ||We call the function by typing | ||
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− | | 04 | + | | 04:48 |
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+ | ||'''Jacobi Iteration open parenthesis A comma b comma x zero comma M a x I t e r comma t o l close parenthesis''' | ||
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− | | 05 | + | | 05:04 |
|Press '''Enter. ''' | |Press '''Enter. ''' | ||
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− | | 05 | + | | 05:06 |
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+ | |The values for '''x one''' and '''x two''' are shown on the '''console.''' | ||
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− | |05 | + | |05:11 |
|The number of iterations are also shown. | |The number of iterations are also shown. | ||
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− | |05 | + | |05:14 |
| Let us now study '''Gauss Seidel method. ''' | | Let us now study '''Gauss Seidel method. ''' | ||
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− | | 05 | + | | 05:19 |
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+ | | 'Given a '''system of linear equations''' with '''n equations''' and ''' n unknowns ''' | ||
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− | |05 | + | |05:26 |
− | + | ||
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+ | ||we rewrite the equations for each unknown | ||
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− | | 05 | + | | 05:29 |
| by subtracting the other variables and their coefficients from the corresponding right hand side element. | | by subtracting the other variables and their coefficients from the corresponding right hand side element. | ||
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− | | 05 | + | | 05:37 |
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+ | | Then we divide this by the '''coefficient a i i of the''' unknown variable' for that variable.''' | ||
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− | | 05 | + | | 05:45 |
|This is done for every given equation. | |This is done for every given equation. | ||
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− | | 05 | + | | 05:49 |
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+ | |In '''Jacobi method,''' for the computation of '''x of i k plus one,''' every element of '''x of i k''' is used except '''x of i k plus one '''. | ||
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− | | 06 | + | | 06:03 |
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− | + | ||
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+ | | In '''Gauss Seidel method,''' we over write the value of '''x of i k''' with '''x of i k plus one'''. | ||
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− | | 06 | + | | 06:12 |
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− | + | ||
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+ | |Let us solve this example using '''Gauss Seidel Method'''. | ||
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− | | 06 | + | | 06:17 |
− | | Let us look at the code for '''Gauss Seidel Method''' | + | | Let us look at the code for '''Gauss Seidel Method'''. |
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− | | 06 | + | | 06:21 |
− | |The first line specifies the '''format''' of the displayed answer on the '''console''' using '''format | + | |The first line specifies the '''format''' of the displayed answer on the '''console''' using '''format''' function. |
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− | | 06 | + | | 06:29 |
− | | Then we use '''input | + | | Then we use '''input''' function to get the values of |
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− | | 06 | + | | 06:32 |
| '''coefficient matrix, ''' | | '''coefficient matrix, ''' | ||
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− | | 06 | + | | 06:34 |
| '''right hand side matrix,''' | | '''right hand side matrix,''' | ||
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− | | 06 | + | | 06:36 |
| '''initial values of the variables matrix, ''' | | '''initial values of the variables matrix, ''' | ||
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− | | 06 | + | | 06:38 |
| '''maximum number of iterations''' and | | '''maximum number of iterations''' and | ||
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− | | 06 | + | | 06:40 |
− | | '''tolerance level''' | + | | '''tolerance level'''. |
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− | | 06 | + | | 06:43 |
− | | Then we define the function '''Gauss Seidel''' with ''' | + | | Then we define the function '''Gauss Seidel''' with input arguments '''A comma b comma x zero comma max iterations''' and '''tolerance level''' and output argument '''solution'''. |
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− | | 06 | + | | 06:58 |
− | | We check if '''matrix A is square''' and the sizes of '''initial vector and matrix A''' are compatible using '''size and length | + | | We check if '''matrix A is square''' and the sizes of '''initial vector and matrix A''' are compatible using '''size''' and '''length''' function. |
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− | | 07 | + | | 07:10 |
|Then we start the iterations. | |Then we start the iterations. | ||
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− | | 07 | + | | 07:13 |
|We equate the '''initial values vector x zero to x k. ''' | |We equate the '''initial values vector x zero to x k. ''' | ||
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− | | 07 | + | | 07:19 |
|We create a '''matrix of zeros''' with the same size of ''' x k''' and call it '''x k p one.''' | |We create a '''matrix of zeros''' with the same size of ''' x k''' and call it '''x k p one.''' | ||
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− | | 07 | + | | 07:28 |
|We solve for each equation to get the value of the '''unknown variable''' for that equation using '''x k p one. ''' | |We solve for each equation to get the value of the '''unknown variable''' for that equation using '''x k p one. ''' | ||
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− | | 07 | + | | 07:38 |
|At each iteration, the value of '''x k p one''' gets updated. | |At each iteration, the value of '''x k p one''' gets updated. | ||
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− | | 07 | + | | 07:44 |
|Also, we check if '''relative error''' is lesser than specified '''tolerance level.''' | |Also, we check if '''relative error''' is lesser than specified '''tolerance level.''' | ||
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− | | 07 | + | | 07:50 |
− | |If it is, we break the iteration. | + | |If it is, we '''break''' the iteration. |
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− | | 07 | + | | 07:54 |
|Then equate '''x k p one''' to the ''' variable solution.''' | |Then equate '''x k p one''' to the ''' variable solution.''' | ||
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− | | 07 | + | | 07:59 |
− | |Finally, we end the function. | + | |Finally, we '''end''' the function. |
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− | | 08 | + | | 08:02 |
|Let us save and execute the function. | |Let us save and execute the function. | ||
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− | | 08 | + | | 08:06 |
− | |Switch to '''Scilab console''' | + | |Switch to '''Scilab console'''. |
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− | | 08 | + | | 08:09 |
− | |For the first prompt, we type ''' matrix A.''' | + | |For the first prompt, we type '''matrix A.''' |
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− | | 08 | + | | 08:12 |
|Type '''open square bracket two space one semi colon five space seven close square bracket''' | |Type '''open square bracket two space one semi colon five space seven close square bracket''' | ||
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− | | 08 | + | | 08:21 |
− | |Press '''Enter''' | + | |Press '''Enter'''. For the next prompt, |
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− | | 08 | + | | 08:24 |
− | + | ||
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|type '''open square bracket eleven semi colon thirteen close square bracket''' | |type '''open square bracket eleven semi colon thirteen close square bracket''' | ||
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− | | 08 | + | | 08:31 |
|Press '''Enter. ''' | |Press '''Enter. ''' | ||
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− | | 08 | + | | 08:33 |
|We provide the values of '''initial value vector''' by typing | |We provide the values of '''initial value vector''' by typing | ||
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− | | 08 | + | | 08:38 |
− | |'''open square bracket one semicolon one close square bracket''' | + | |'''open square bracket one semicolon one close square bracket''' . |
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− | | 08 | + | | 08:43 |
− | |Press '''Enter. ''' | + | |Press '''Enter.''' |
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− | | 08 | + | | 08:45 |
− | |Then we specify the ''' maximum number of iterations''' to be twenty five | + | |Then we specify the ''' maximum number of iterations''' to be twenty five. |
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− | | 08 | + | | 08:50 |
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+ | |Press '''Enter.''' | ||
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− | | 08 | + | | 08:52 |
− | |Let us define '''tolerance level'' to be zero point zero zero zero zero one | + | |Let us define '''tolerance level'' to be zero point zero zero zero zero one. |
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− | | 08 | + | | 08:58 |
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+ | |Press '''Enter'''. | ||
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− | | 09 | + | | 09:01 |
|Finally we call the function by typing | |Finally we call the function by typing | ||
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− | | 09 | + | | 09:04 |
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+ | |'''G a u s s S e i d e l open parenthesis A comma b comma x zero comma M a x I t e r comma t o l close parenthesis''' | ||
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− | | 09 | + | | 09:24 |
|Press '''Enter'''. | |Press '''Enter'''. | ||
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− | | 09 | + | | 09:26 |
|The values of '''x one''' and '''x two''' are displayed. | |The values of '''x one''' and '''x two''' are displayed. | ||
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− | | 09 | + | | 09:30 |
|The number of iterations to solve the same problem are lesser than '''Jacobi method.''' | |The number of iterations to solve the same problem are lesser than '''Jacobi method.''' | ||
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− | | 09 | + | | 09:37 |
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+ | |Solve this problem on your own using '''Jacobi''' and '''Gauss Seidel methods'''. | ||
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− | | 09 | + | | 09:43 |
|In this tutorial, we have learnt to: | |In this tutorial, we have learnt to: | ||
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− | | 09 | + | | 09:47 |
− | |Develop '''Scilab code''' for solving system of linear equations | + | |Develop '''Scilab code''' for solving system of linear equations. |
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− | | 09 | + | | 09:52 |
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+ | |Find the value of the '''unknown variables''' of a system of '''linear equations'''. | ||
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− | |09 | + | |09:58 |
− | | Watch the video available at the following link | + | | Watch the video available at the following link. |
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− | | 10 | + | | 10:01 |
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+ | | It summarizes the Spoken Tutorial project. | ||
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− | |10 | + | |10:04 |
− | ||If you do not have good bandwidth, you can download and watch it | + | ||If you do not have good bandwidth, you can download and watch it. |
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− | |10 | + | |10:09 |
||The spoken tutorial project Team | ||The spoken tutorial project Team | ||
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− | |10 | + | |10:11 |
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+ | ||conducts workshops using spoken tutorials, | ||
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− | |10 | + | |10:15 |
− | + | ||
− | + | ||
+ | ||gives certificates to those who pass an online test. | ||
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− | |10 | + | |10:18 |
− | + | ||
− | + | ||
+ | ||For more details, please write to contact@spoken-tutorial.org . | ||
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− | |10 | + | |10:25 |
− | + | ||
− | + | ||
− | + | ||
+ | |Spoken Tutorial Project is a part of the Talk to a Teacher project. | ||
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− | | 10 | + | | 10:30 |
| It is supported by the National Mission on Eduction through ICT, MHRD, Government of India. | | It is supported by the National Mission on Eduction through ICT, MHRD, Government of India. | ||
|- | |- | ||
− | | 10 | + | | 10:37 |
− | |More information on this mission is available at http://spoken-tutorial.org/NMEICT-Intro | + | |More information on this mission is available at http://spoken-tutorial.org/NMEICT-Intro. |
|- | |- | ||
− | | 10 | + | | 10:49 |
− | |This is Ashwini Patil signing off. | + | |This is Ashwini Patil. signing off. |
|- | |- | ||
− | |10 | + | |10:51 |
| Thank you for joining. | | Thank you for joining. |
Latest revision as of 23:54, 3 January 2018
Time | Narration |
00:01 | Dear Friends, Welcome to the Spoken Tutorial on Solving System of Linear Equations using Iterative Methods. |
00:10 | At the end of this tutorial, you will learn how to: |
00:14 | Solve system of linear equations using iterative methods |
00:18 | Develop Scilab code to solve linear equations. |
00:22 | To record this tutorial, I am using |
00:25 | Ubuntu 12.04 as the operating system |
00:28 | and Scilab 5.3.3 version. |
00:33 | Before practicing this tutorial, a learner should have basic knowledge of |
00:38 | Scilab, and solving linear equations. |
00:42 | For Scilab, please refer to the relevant tutorials available on the Spoken Tutorial website. |
00:50 | The first iterative method we will be studying is Jacobi method. |
00:56 | Given a system of linear equations, with n equations and n unknowns, |
01:02 | we rewrite the equations such that x of i k plus one is equal to b i minus summation of a i j x j k from j equal to one to n divided by a i i where i is from one to n. |
01:24 | We assume values for each x of i. |
01:27 | Then we substitute the values in the equations obtained in the previous step. |
01:34 | We continue the iteration until the solution converges. |
01:39 | Let us solve this example using Jacobi Method. |
01:44 | Let us look at the code for Jacobi Method. |
01:48 | We use format method to specify the format of the displayed answers on the Scilab console. |
01:56 | Here e denotes the answer should be in scientific notation. |
02:01 | And twenty specifies the number of digits to be displayed. |
02:06 | Then we use input function to get the values for |
02:10 | the matrices coefficient matrix, |
02:12 | right hand side matrix, |
02:14 | initial values matrix, |
02:17 | maximum number of iteration and |
02:19 | convergence tolerance. |
02:22 | Then we use size function to check if A matrix is a square matrix. |
02:29 | If it isn't, we use error function to display an error. |
02:34 | We then check if matrix A is diagonally dominant. |
02:40 | The first half calculates the sum of each row of the matrix. |
02:45 | Then it checks if twice the product of the diagonal element is greater than the sum of the elements of that row. |
02:54 | If it isn't, an error is displayed using error function. |
03:01 | Then we define the function Jacobi Iteration with input arguments |
03:07 | A, b , x zero, |
03:09 | maximum iteration and tolerance level. |
03:14 | Here x zero is the initial values matrix. |
03:19 | We check if the size of A matrix and initial values matrix are compatible with each other. |
03:28 | We calculate the value for x k p one and then check if the relative error is lesser than tolerance level. |
03:38 | If it is lesser than tolerance level, we break the iteration and the solution is returned. |
03:45 | Finally we end the function. |
03:48 | Let us save and execute the function. |
03:51 | Switch to Scilab console. |
03:54 | Let us enter the values at each prompt. |
03:57 | The coefficient matrix A is open square bracket two space one semi colon five space seven close square bracket |
04:08 | Press Enter. |
04:10 | Then we type open square bracket eleven semicolon thirteen close square bracket |
04:17 | Press Enter. |
04:20 | The initial values matrix is open square bracket one semi colon one close square bracket |
04:28 | Press Enter. |
04:30 | The maximum number of iterations is twenty five. |
04:34 | Press Enter. |
04:36 | Let the convergence tolerance level be zero point zero zero zero zero one |
04:44 | Press Enter. |
04:46 | We call the function by typing |
04:48 | Jacobi Iteration open parenthesis A comma b comma x zero comma M a x I t e r comma t o l close parenthesis |
05:04 | Press Enter. |
05:06 | The values for x one and x two are shown on the console. |
05:11 | The number of iterations are also shown. |
05:14 | Let us now study Gauss Seidel method. |
05:19 | 'Given a system of linear equations with n equations and n unknowns |
05:26 | we rewrite the equations for each unknown |
05:29 | by subtracting the other variables and their coefficients from the corresponding right hand side element. |
05:37 | Then we divide this by the coefficient a i i of the unknown variable' for that variable. |
05:45 | This is done for every given equation. |
05:49 | In Jacobi method, for the computation of x of i k plus one, every element of x of i k is used except x of i k plus one . |
06:03 | In Gauss Seidel method, we over write the value of x of i k with x of i k plus one. |
06:12 | Let us solve this example using Gauss Seidel Method. |
06:17 | Let us look at the code for Gauss Seidel Method. |
06:21 | The first line specifies the format of the displayed answer on the console using format function. |
06:29 | Then we use input function to get the values of |
06:32 | coefficient matrix, |
06:34 | right hand side matrix, |
06:36 | initial values of the variables matrix, |
06:38 | maximum number of iterations and |
06:40 | tolerance level. |
06:43 | Then we define the function Gauss Seidel with input arguments A comma b comma x zero comma max iterations and tolerance level and output argument solution. |
06:58 | We check if matrix A is square and the sizes of initial vector and matrix A are compatible using size and length function. |
07:10 | Then we start the iterations. |
07:13 | We equate the initial values vector x zero to x k. |
07:19 | We create a matrix of zeros with the same size of x k and call it x k p one. |
07:28 | We solve for each equation to get the value of the unknown variable for that equation using x k p one. |
07:38 | At each iteration, the value of x k p one gets updated. |
07:44 | Also, we check if relative error is lesser than specified tolerance level. |
07:50 | If it is, we break the iteration. |
07:54 | Then equate x k p one to the variable solution. |
07:59 | Finally, we end the function. |
08:02 | Let us save and execute the function. |
08:06 | Switch to Scilab console. |
08:09 | For the first prompt, we type matrix A. |
08:12 | Type open square bracket two space one semi colon five space seven close square bracket |
08:21 | Press Enter. For the next prompt, |
08:24 | type open square bracket eleven semi colon thirteen close square bracket |
08:31 | Press Enter. |
08:33 | We provide the values of initial value vector by typing |
08:38 | open square bracket one semicolon one close square bracket . |
08:43 | Press Enter. |
08:45 | Then we specify the maximum number of iterations to be twenty five. |
08:50 | Press Enter. |
08:52 | Let us define 'tolerance level to be zero point zero zero zero zero one. |
08:58 | Press Enter. |
09:01 | Finally we call the function by typing |
09:04 | G a u s s S e i d e l open parenthesis A comma b comma x zero comma M a x I t e r comma t o l close parenthesis |
09:24 | Press Enter. |
09:26 | The values of x one and x two are displayed. |
09:30 | The number of iterations to solve the same problem are lesser than Jacobi method. |
09:37 | Solve this problem on your own using Jacobi and Gauss Seidel methods. |
09:43 | In this tutorial, we have learnt to: |
09:47 | Develop Scilab code for solving system of linear equations. |
09:52 | Find the value of the unknown variables of a system of linear equations. |
09:58 | Watch the video available at the following link. |
10:01 | It summarizes the Spoken Tutorial project. |
10:04 | If you do not have good bandwidth, you can download and watch it. |
10:09 | The spoken tutorial project Team |
10:11 | conducts workshops using spoken tutorials, |
10:15 | gives certificates to those who pass an online test. |
10:18 | For more details, please write to contact@spoken-tutorial.org . |
10:25 | Spoken Tutorial Project is a part of the Talk to a Teacher project. |
10:30 | It is supported by the National Mission on Eduction through ICT, MHRD, Government of India. |
10:37 | More information on this mission is available at http://spoken-tutorial.org/NMEICT-Intro. |
10:49 | This is Ashwini Patil. signing off. |
10:51 | Thank you for joining. |