Numerical analysis Example No. 3

7/27/2019 Numerical analysis Example No. 3

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Numerical Analysis Assignment NO.3 17-OCT-2010

DR.Ahmed Amir Bayoumy

Prepared By: Ahmed Shaban Mahmoud

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For the shown 11x11 banded symmetric matrix, use only the 2x11 matrixto decompose the matrix into upper and lower matrix using Choleski

technique:-

11x11 matrix

25 10 0 0 0 0 0 0 0 0 0

10 8 2 0 0 0 0 0 0 0 0

0 2 5 4 0 0 0 0 0 0 0

0 0 4 13 6 0 0 0 0 0 00 0 0 6 29 15 0 0 0 0 0

0 0 0 0 15 34 10 0 0 0 0

0 0 0 0 0 10 29 10 0 0 0

0 0 0 0 0 0 10 40 12 0 0

0 0 0 0 0 0 0 12 20 8 0

0 0 0 0 0 0 0 0 8 13 6

0 0 0 0 0 0 0 0 0 6 53

Banded matrix

25 10

8 2

5 4

13 6

29 15

34 10

29 10

40 12

20 8

13 6

53 0


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Solution

See attached excel sheet for detailed calculation of

below equations and matrices.

L(i,i) = sqrt(A(i,i) - L(i,1:i-1)*L(i,1:i-1)'); For diagonal numbers

L(i,j) = (A(i,j) - L(i,1:j-1)*L(j,1:j-1)')/L(j,j); For all other numbers

Lower Matrix

5 0 0 0 0 0 0 0 0 0 0

2 2 0 0 0 0 0 0 0 0 0

0 1 2 0 0 0 0 0 0 0 0

0 0 2 3 0 0 0 0 0 0 0

0 0 0 2 5 0 0 0 0 0 0

0 0 0 0 3 5 0 0 0 0 0

0 0 0 0 0 2 5 0 0 0 0

0 0 0 0 0 0 2 6 0 0 0

0 0 0 0 0 0 0 2 4 0 0

0 0 0 0 0 0 0 0 2 3 0

0 0 0 0 0 0 0 0 0 2 7

Upper Matrix

5 2 0 0 0 0 0 0 0 0 0

0 2 1 0 0 0 0 0 0 0 00 0 2 2 0 0 0 0 0 0 0

0 0 0 3 2 0 0 0 0 0 0

0 0 0 0 5 3 0 0 0 0 0

0 0 0 0 0 5 2 0 0 0 0

0 0 0 0 0 0 5 2 0 0 0

0 0 0 0 0 0 0 6 2 0 0

0 0 0 0 0 0 0 0 4 2 0

0 0 0 0 0 0 0 0 0 3 2

0 0 0 0 0 0 0 0 0 0 7


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The lower matrix can be re written in 11x2 matrix

5 2

2 1

2 2

3 2

5 3

5 2

5 2

6 2

4 2

3 2

7 0


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Algorithm :

for i = 1 : n

for j = 1: m %% m is number of columns in banded matrix

if i == 1 & j == 1

s = A(i,i); %% i=1, j=1 is special case

else

s = A (i , j) -

C ( y , i y + 1)'

*

C ( y , j + i - y);

end

if j > 1

C( i , j ) = s / C( i , 1 );

else

if s


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Using the above algorithm and apply in it step by step we found that value of S

for the banded matrix is as below

i NO. j = 1 j = 2

1 S = A ( 1 , 1 ) S = A ( 1 , 2 )

2 S = A ( 2 , 1 ) C ( 1 , 2 ) * C ( 1 , 2 ) S = A ( 2 , 2 )

3 S = A ( 3 , 1 ) C ( 2 , 2 ) * C ( 2 , 2 ) S = A ( 3 , 2 )

4 S = A ( 4 , 1 ) C ( 3 , 2 ) * C ( 3 , 2 ) S = A ( 4 , 2 )

5 S = A ( 5 , 1 ) C ( 4 , 2 ) * C ( 4 , 2 ) S = A ( 5 , 2 )

6 S = A ( 6 , 1 ) C ( 5 , 2 ) * C ( 5 , 2 ) S = A ( 6 , 2 )

7 S = A ( 7 , 1 ) C ( 6 , 2 ) * C ( 6 , 2 ) S = A ( 7 , 2 )

8 S = A ( 8 , 1 ) C ( 7 , 2 ) * C ( 7 , 2 ) S = A ( 8 , 2 )

9 S = A ( 9 , 1 ) C ( 8 , 2 ) * C ( 8 , 2 ) S = A ( 9 , 2 )

10 S = A ( 10 , 1 ) C ( 9 , 2 ) * C ( 9 , 2 ) S = A ( 10 , 2 )

11 S = A ( 11 , 1 ) C ( 10 , 2 ) * C ( 10 , 2 ) S = A ( 11 , 2 )


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Continue in the above algorithm we can reach to value of the banded lower

matrix in form of 11x2 as shown below

i NO. j = 1 j = 2

1 Sqrt ( 25 ) = 5 10 / 5 = 2

2 Sqrt ( 8 2 * 2 ) = 2 2 / 2 = 1

3 Sqrt ( 5 1 * 1 ) = 2 4 / 2 = 2

4 Sqrt ( 13 2 * 2 ) = 3 6 / 3 = 2

5 Sqrt ( 29 2 * 2 ) = 5 15 / 5 = 3

6 Sqrt ( 34 3 * 3 ) = 5 10 / 5 = 2

7 Sqrt ( 29 2 * 2 ) = 5 10 / 5 = 2

8 Sqrt ( 40 2 * 2 ) = 6 12 / 6 = 2

9 Sqrt ( 20 2 * 2 ) =4 8 / 4 = 2

10 Sqrt ( 13 2 * 2 ) =3 6 / 3 = 2

11 Sqrt ( 53 2 * 2 ) =7 0 / 7 = 0


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Conclusion:-

As following of the new algorithm for banded matrix reached the same

result for Choleski technique. So, the new algorithm may be right.

Numerical analysis Example No. 3

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