ITERATIVE METHODS Prepared by Dr L Benedict Michael

ITERATIVE METHODS Prepared by, Dr. L. Benedict Michael Raj Associate Professor Department of Mathematics, St. Joseph’s College(Autonomous), Trichy

Iterative Methods This method is also called as indirect method. There are two types of iterative methods. (1)Gauss Jacobi Method (2)Gauss Seidel Method The condition for convergence Gauss Jacobi & Gauss Seidel methods is given by the following rule. The process of iteration will converge if each equation of the system, the absolute value of the largest coefficient is greater than the sum of the absolute values of all the remaining coefficients.

Gauss Jacobi Method Problem solve the equations by Jacobi iteration method the system. 8 x-3 y+2 z=20 6 x+3 y+12 z=35 4 x+11 y-z=33 Solution consider the given system as 8 x-3 y+2 z=20 4 x+11 y-z=33 6 x+3 y+12 z=35 We write the equations in the form x=1/8(20+3 y-2 z) y=1/11(33 -4 x+z) z=1/12(35 -6 x-3 y)

We start from an approximation I. NO X=1/8(20+3 y-2 z) x =y =x Y=1/11(33 -4 x+z) Z=1/12(35 -6 x-3 y) 1. 2. 5 3 2. 9166 2. 8958 2. 3560 0. 9166 3. 1543 2. 0303 0. 8797 4. 3. 0419 1. 9329 0. 8319 5. 3. 0168 1. 9696 0. 9127 6. 3. 0104 1. 9859 0. 9158 7. 3. 0157 1. 9885 0. 9149 8. 3. 0169 1. 9865 0. 9116

I. NO X=1/8(20+3 y-2 z) Y=1/11(33 -4 x+z) Z=1/12(35 -6 x-3 y) 9. 3. 0170 1. 9857 0. 9115 10. 3. 0167 1. 9857 0. 9116 11. 3. 0167 1. 9858 0. 9118 12. 3. 0167 1. 9858 0. 9118 Therefore, from the 11 th &12 th iterations, the values x, y, z are same correct to four decimal places. Stopping at this stage, we get x=3. 0167 y=1. 9858 z=0. 9118

Gauss Seidel Method * This is modification of Gauss Jacobi method. * The current values of the unknowns at each stage of iteration are used in preceeding to the next stage of iteration, this method is more rapid in convergence than Gauss Jacobi method. Problem Solve the equations by Gauss Seidel method. 8 x-3 y+2 z=20 6 x+3 y+12 z=35 4 x+11 y-z=33

Solution consider the given system as 8 x-3 y+2 z=20 4 x+11 y-z=33 6 x+3 y+12 z=35 we write the equations in the form x=1/8(20+3 y-2 z) y=1/11(33 -4 x+z) z=1/12(35 -6 x-3 y) x=y=z=0

Y=1/11(33 -4 x+z) Z=1/12(35 -6 x-3 y) I. NO X=1/8(20+3 y-2 z) 1. 2. 5 2. 0909 1. 1439 2. 9981 2. 0137 0. 9141 3. 0266 1. 9825 0. 9077 4. 3. 0165 1. 9856 0. 9120 5. 3. 0166 1. 9859 0. 9118 6. 3. 0167 1. 9858 0. 9118 7. 3. 0167 1. 9858 0. 9118

since at the 6 th & 7 th iterations the values of x, y, z are the same. They are x=3. 0167 y=1. 9858 z=0. 9118

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