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MAT3005 - Applied Numerical Methods

Dr. B.S.R.V. Prasad Department of Mathematics School of Advanced Sciences VIT, Vellore

[email protected] [email protected]

1

Gauss-Seidel Iterative Method

Exercise Problems

Problem. Monthly employee salary in three departments of a company is given below. Assuming that the salary for a particular category is same in all the departments, calculate the salary of each category of employee by using Gauss-Seidel method.

Department Manager Asoc. Manager Asst. Manager Salary A 2 3 6 60K B 4 1 2 60K C 1 4 3 40K

3

Gauss-Seidel Iterative Method

Exercise Problems

Now, the Gauss-Seidel iterative scheme for the above system is given by:

x ( n +^1 )^ =

60 − y ( n )^ − 2 z ( n )

y ( n +^1 )^ =

40 − x ( n +^1 )^ − 3 z ( n )

z ( n +^1 )^ =

60 − 2 x ( n +^1 )^ − 3 y ( n +^1 )

Iteration #1: Now, starting with x (^0 )^ = y (^0 )^ = z (^0 )^ = 0 , we obtain

x (^1 )^ =

y (^1 )^ =

z (^1 )^ =

4

Gauss-Seidel Iterative Method

Exercise Problems

Iteration #2:

x (^2 )^ =

y (^2 )^ =

z (^2 )^ =

Iteration #3:

x (^3 )^ =

y (^3 )^ =

z (^3 )^ =

6

Gauss-Seidel Iterative Method

Exercise Problems

Problem. The following system of equations is designed to determine concentrations (the ci s in g/m^3 ) in a series of coupled reactors as a function of the amount of mass input to each reactor (the right-hand sides in g/day):

15 c 1 − 3 c 2 − c 3 = 3800 − 3 c 1 + 18 c 2 − 6 c 3 = 1200 − 4 c 1 − c 2 + 12 c 3 = 2350

Solve the problem with Gauss-Seidel method to  s = 4 %

7

Solving System of Linear Equations

LU Decomposition Method

I (^) We are interested to find the solution of the system AX = B by factorising the matrix A into the form LU , where L is a unit lower triangular matrix and U is upper triangular matrix. I (^) This decomposition is possible and also unique if all the principal minors of A are nonsingular, i.e., if

a 11 6 = 0 ,

a 11 a 12 a 21 a 22

∣ 6 =^0 ,

a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33

= 0 , etc.

9

LU Decomposition Method

Working Process

Let A = LU , (1)

where

L =

l 21 1 0 l 31 l 32 1

 (^) and U =

u 11 u 12 u 13 0 u 22 u 23 0 0 u 33

Hence, the (1) becomes LUX = B. (3)

10

LU Decomposition Method

Working Process

Now, setting UX = Y (4)

(where Y =

[

y 1 y 2 y 3

] T

), the equation (3) can be written as

LY = B , (5)

which is equivalent to the system

y 1 = b 1 , l 21 y 1 + y 2 = b 2 , l 31 y 1 + l 32 y 2 + y 3 = b 3.

and therefore, we can solve for y 1 , y 2 and y 3 by the forward substitution.

12

Scheme for computing the matrices L and U

From AX = B and A = LU , we have  

l 21 1 0 l 31 l 32 1

u 11 u 12 u 13 0 u 22 u 23 0 0 u 33

a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33

Multiplying the matrices on the left and equating the corresponding elements of both sides, we obtain

u 11 = a 11 , u 12 = a 12 , u 13 = a 13 l 21 u 11 = a 21 , l 31 u 11 = a 31 l 21 u 12 + u 22 = a 22 , l 21 u 13 + u 23 = a 23 l 31 u 12 + l 32 u 22 = a 32 l 31 u 13 + l 32 u 23 + u 33 = a 33

13

Scheme for computing the matrices L and U

Solving the above equations, we get

l 21 =

a 21 a 11

; l 31 =

a 31 a 11 u 22 = a 22 −

a 21 a 11

a 12 ; u 23 = a 23 −

a 21 a 11

a 13

l 32 =

a 32 − ( a 31 / a 11 ) a 12 u 22

and finally we compute

u 33 = a 33 − l 31 u 13 − l 32 u 23. (8)

15

Example

Problem

Solve the equations

2 x + 3 y + z = 9 x + 2 y + 3 z = 6 3 x + y + 2 z = 8

by LU decomposition.

16

Example

We have

A =

Let (^) 



l 21 1 0 l 31 l 32 1

u 11 u 12 u 13 0 u 22 u 23 0 0 u 33

So, u 11 = 2 , u 12 = 3 , u 13 = 1

and l 21 u 11 = 1 =⇒ l 21 =

l 31 u 11 = 3 =⇒ l 31 =

18

Example

Finally, l 32 and u 33 are obtained from

l 31 u 12 + l 32 u 22 = 1 and l 31 u 13 + l 32 u 23 + u 33 = 2.

Solving, we have l 32 = −7 and u 33 = 18.

So,

A =

19

Example

Hence, the system of equations AX = B can be written as  

x y z

or, as (^) 



y 1 y 2 y 3

where (^) 



y 1 y 2 y 3

x y z