Difference between revisions of "Scilab/C4/Linear-equations-Iterative-Methods/English"

• Solve system of linear equations using iterative methods
• Develop Scilab code to solve linear equations

• Ubuntu 12.04 as the operating system
• and Scilab 5.3.3 version

• Scilab
• and Solving Linear Equations

For Scilab, please refer to the relevant tutorials available on the Spoken Tutorial website.

• Given a system of linear equations, with n equations and n unknowns
• 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
• We assume values for each x of i
• Then we substitute the values in the equations obtained in the previous step.
• We continue the iteration until the solution converges.

format('e',20)

Here e denotes the answer should be in scientific notation.

And twenty specifies the number of digits to be displayed.

A=input("Enter the coeffiecient matrix of nxn: ")

b=input("Enter the right-hand side matrix nx1: ")

x0=input("Initial approximation nx1: ")

MaxIter=input("Maximum no. of iterations: ")

tol=input("Enter the convergence tolerance :")//stop if norm change in x < tol

• the matrices coefficient matrix,
• right hand side matrix,
• initial values matrix,
• maximum number of iteration and
• convergence tolerance

[m, n] = size( A )

If it isn't, we use error function to display an error.

for i=1:1:m

sum=0;

for j=1:1:m

sum=sum+abs(A(i,j));

end

if 2*abs(A(i,i))<sum then

error("the matrix is not diagonally dominant")

end

end

• We then check if matrix A is diagonally dominant.
• The first half calculates the sum of each row of the matrix.
• Then it checks if twice the product of the diagonal element is greater than the sum of the elements of that row.
• If it isn't, an error is displayed using error function.

function [ solution ] = JacobiIteration( A, b, x0, MaxIter, tol )

A, b , x zero,

maximum iteration and tolerance level.

Here x zero is the initial values matrix.

if ( length(x0) ~= n )

error( "Sizes of input matrix and input vector do not match" )

end

xk = x0

for k = 1 : 1 : MaxIter

for i = 1 : n

xkp1( i ) = (b(i) - A(i, 1:i-1)*xk(1:i-1) - A(i, i+1:n)*xk(i+1:n)) / A(i,i) //Computes the jacobian updates

end

RelativeError = norm( xkp1 - xk, 'inf' ) / norm( xkp1, 'inf' )

printf( "iter = %d, Relative Error = %g\n", k, RelativeError )

xk = xkp1

if ( RelativeError < tol )

break

end

end

solution = xkp1

endfunction

• We calculate the value for x k p one and then check if the relative error is lesser than tolerance level.
• If it is lesser than tolerance level, we break the iteration and the solution is returned.
• Finally we end the function.

Enter the coeffiecient matrix of nxn: [2 1;5 7]

The coefficient matrix A is open square bracket two space one semi colon five space seven close square bracket

Press Enter.

Enter the right-hand side matrix nx1: [11; 13]

Then we type open square bracket eleven semicolon thirteen

Press Enter.

Initial approximation nx1: [1;1]

Press Enter.

Maximum no. of iterations: 25

Press Enter.

Enter the convergence tolerance :0.00001

Press Enter.

JacobiIteration( A, b, x0, MaxIter, tol )

Jacobi Iteration open paranthesis A comma b comma x zero comma M a x I t e r comma t o l close paranthesis

Press Enter.

The values for x one and x two are shown on the console.

The number of iterations are also shown.

• Given a system of linear equations, with n equations and n unknowns
• We rewrite the equations for each unknown
  • by subtracting the other variables and their coefficients
  • from the corresponding right hand side element.
• Then we divide this by the coefficient a i i of the unknown variable' for that variable.

This is done for every given equation.

In Gauss Seidel method, we over write the value of x of i k with x of i k plus one

format('e',20)

A=input("Enter the coeffiecient matrix of nxn: ")

b=input("Enter the right-hand side matrix nx1: ")

x0=input("Initial approximation nx1: ")

MaxIter=input("Maximum no. of iterations: ")

tol=input("Enter the convergence tolerance :")//stop if norm change in x < tol

• coefficient matrix,
• right hand side matrix,
• initial values of the variables matrix,
• maximum number of iterations and
• tolerance level

function [ solution ] = GaussSeidel( A, b, x0, MaxIter, tol )

// Check that the matrix is square

[m, n] = size( A )

if ( m ~= n )

error( "The input matrix is not square" )

end

// Check that the initial vector is of the same size

if ( length(x0) ~= n )

error( "Sizes of input matrix and input vector do not match" )

end

xk = x0

xkp1 = zeros( xk )

• We equate the initial values vector x zero to x k.
• We create a matrix of zeros with the same size of x k and call it x k p one.

for k = 1 : 1 : MaxIter

// Computes the Gauss-Seidel update

for i = 1 : n

xkp1( i ) = (b(i) - A(i, 1:i-1)*xkp1(1:i-1) - A(i,i+1:n)*xk(i+1:n)) / A(i,i)

// x^{k+1}

end

// Applies the relaxation

RelativeError = norm( xkp1 - xk, 'inf' ) / norm( xkp1, 'inf' )

printf( "iter = %d, Relative Error = %g\n", k, RelativeError )

xk = xkp1

if ( RelativeError < tol )

break

end

end

solution = xkp1

endfunction

• At each iteration, the value of x k p one gets updated.
• Also, we check if relative error is lesser than specified tolerance level.
• If it is, we break the iteration.
• Then equate x k p one to the variable solution.
• Finally, we end the function.

For first prompt

[2 1;5 7]

Type open square bracket two space one semi colon five space seven close square bracket

Press Enter.

[11; 13]

type left square bracket eleven semi colon thirteen right square bracket

Press Enter.

[1;1]

open square bracket one semicolon one close square bracket

Press Enter.

25

Press Enter.

0.00001

Press Enter.

GaussSeidel( A, b, x0, MaxIter, tol )

G a u s s S e i d e l open paranthesis A comma b comma x zero comma M a x I t e r comma t o l close paranthesis

Press Enter.

The values of x one and x two are displayed.

The number of iterations to solve the same problem are lesser than Jacobi method.

• Develop Scilab code for solving system of linear equations
• Find the value of the unknown variables of a system of linear equations

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