Solution of Linear Algebric Equations Using MATLAB

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    EEE 212: Numerical Technique Laboratory

    No. of Experiment : 5

    Name of Experiment : Solution of Simultaneous Linear Algebric

    Equations

    Date of Performance: 23-05-2012

    Date of Submission: 30-05-2012

    Tahmid Sami Rahman

    0906016

    Department: EEE

    Section: A

    Md. Ayaz Masud

    0906021

    Department: EEE

    Section: A

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    Problem

    Solve the boundary value problem for

    = 1.2510 where h=1

    Plot the solution using

    i) PINVii) Gauss Seidel Method (error

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    Graph

    PINV:

    Elapsed time is 0.047521 seconds.

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    Gauss Elimination with Pivoting:

    Elapsed time is 0.444923 seconds.

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    Gauss Seidel Method:

    Elapsed time is 3.965348 seconds.

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    Combined Graph:

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    Appendix

    Code for the Solution

    cl ear al l ; h=10 - 6; A=zer os( 100, 100) ; A( 1, 1)=1; A( 101, 101) =1; f or i =2: 100

    j =i ; A( i , j - 1) =1; A( i , j ) =- 2;

    A( i , j +1) =1;

    endx=0: h: 10 - 4; B=zeros( 1, 101) ; B( 1, 1)=5; B( 1, 101)=0;

    f or i =2: 101B( i ) =1. 25. *10 12. *exp( - 5. *10 5. *x( i ) ) . *( h 2) ; endt i ce=pi nv( A) *B' ; t oc

    di sp( e) ;

    % pl ot ( d, ' r ' ) ; % hol d on;

    pl ot ( e, ' b' ) ; hol d on;

    a=A; b=B;

    aj i =0;

    t i cs=si ze( a) ; f or i =1: s(1) - 1

    f or j =i +1: s( 1) aj i =a( j , i ) ;

    f or k=1: s( 2) a( j , k) =a( j , k) - ( aj i . / a( i , i ) ) . *a( i , k) ;

    end

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    b( j ) =b( j ) - ( aj i . / a( i , i ) ) . *b( i ) ; end

    end

    di sp( a) ; di sp( b) ;

    x=b;

    n=s( 2) ; sum=0;

    x( n) =b( n) . / a( n, n) ;

    f or i =n- 1: - 1: 1sum=b( i ) ;

    f or j =i +1: nsum=sum- a( i , j ) . *x(j ) ;

    endx( i ) =sum. / a( i , i ) ;

    end

    %di sp( x) ;

    pl ot ( x, ' o' ) ;hol d on;

    t oc%gauss sei del

    t i c

    x=b; h=10 - 6; i =1; f or h=0: 10 - 6: 10 - 4x(i ) =5. *exp( - 5000. *h) ; i =i +1;

    end

    xr ef =x;

    check=0;

    whi l e( 1) f or i =1: 101

    sum=b( i ) ; f or j =1: 101

    i f i ~=j sum=sum- ( a( i , j ) . *x( j ) ) ; end

    endx( i ) =sum. / a( i , i ) ;

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    end

    xsol =xref - x;

    f i ni sh=0;

    f or i =1: 101i f xsol ( i ) >10 - 9

    f i ni sh=1; end

    end

    i f( f i ni sh==0) break;

    end

    xref =x; % check=check+1; % i f check==1000% br eak; % endend

    pl ot ( xref , ' g' ) ;

    t oc