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Broyden Algorithm Non Linear System-Numerical Analysis-MATLAB Code, Exercises of Mathematical Methods for Numerical Analysis and Optimization

This is solution to one of problems in Numerical Analysis. This is matlab code. Its helpful to students of Computer Science, Electrical and Mechanical Engineering. This code also help to understand algorithm and logic behind the problem. This code includes: Broyden, Algorithm, Nonlinear, System, Function, Program, Equations, Tolerance, Iterations, Components, Jacobian, Ordering

Typology: Exercises

2011/2012

Uploaded on 07/31/2012

saripella
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% BROYDEN ALGORITHM 10.2
%
% To approximate the solution of the nonlinear system F(X) = 0
% given an initial approximation X.
%
% INPUT: Number n of equations and unknowns; initial
% approximation X = (X(1),...,X(n)); tolerance TOL;
% maximum number of iterations N.
%
% OUTPUT: Approximate solution X = (X(1),...,X(n)) or a message
% that the number of iterations was exceeded.
syms('OK', 'N', 'I', 'ZZ', 'J', 's', 'ss', 'TOL', 'NN', 'X');
syms('FLAG', 'NAME', 'OUP', 'A', 'V', 'B', 'I1', 'I2');
syms('C', 'K', 'SN', 'S', 'VV', 'Y', 'ZN', 'Z', 'P', 'U', 'KK');
TRUE = 1;
FALSE = 0;
fprintf(1,'This is the Broyden Method for Nonlinear Systems.\n');
fprintf(1,'The functions could be input or defined in code.\n');
fprintf(1,'This code assumes input of functions - see \n');
fprintf(1,'comments in code for alternate version.\n');
fprintf(1,'This program also uses M-files JAC.M and FN.M \n');
fprintf(1,'If the number of equations exceeds 7 then JAC.M\n');
fprintf(1,'and FN.M must be changed.\n');
OK = FALSE;
while OK == FALSE
fprintf(1,'Input the number n of equations.\n');
N = input(' ');
if N >= 2
OK = TRUE;
else
fprintf(1,'N must be an integer greater than 1.\n');
end;
end;
for I = 1 : N
fprintf(1,'Input the function F_(%d) in terms of y1...y%d \n',I,N);
s(I) = input(' ','s');
end;
% Define components of F as follows:
% s(1) = '3*y1-cos(y2*y3)-0.5';
% s(2) = 'y1^2-81*(y2+0.1)^2+sin(y3)+1.06';
% s(3) = 'exp(-y1*y2)+20*y3+(10*pi-3)/3';
for I = 1 : N
for J = 1 : N
fprintf(1,'Input the partial of F_(%d) with respect to x_%d \n',I,J);
fprintf(1,'in terms of y1, ..., y%d \n',N);
ss((I-1)*N+J) = input(' ','s');
end;
end;
% Define the entries of the Jacobian in row major ordering.
% ss(1) = '3';
% ss(2) = 'y3*sin(y2*y3)';
% ss(3) = 'y2*sin(y2*y3)';
% ss(4) = '2*y1';
% ss(5) = '-162*(y2+0.1)';
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% BROYDEN ALGORITHM 10.

% To approximate the solution of the nonlinear system F(X) = 0 % given an initial approximation X. % % INPUT: Number n of equations and unknowns; initial % approximation X = (X(1),...,X(n)); tolerance TOL; % maximum number of iterations N. % % OUTPUT: Approximate solution X = (X(1),...,X(n)) or a message % that the number of iterations was exceeded. syms('OK', 'N', 'I', 'ZZ', 'J', 's', 'ss', 'TOL', 'NN', 'X'); syms('FLAG', 'NAME', 'OUP', 'A', 'V', 'B', 'I1', 'I2'); syms('C', 'K', 'SN', 'S', 'VV', 'Y', 'ZN', 'Z', 'P', 'U', 'KK'); TRUE = 1; FALSE = 0; fprintf(1,'This is the Broyden Method for Nonlinear Systems.\n'); fprintf(1,'The functions could be input or defined in code.\n'); fprintf(1,'This code assumes input of functions - see \n'); fprintf(1,'comments in code for alternate version.\n'); fprintf(1,'This program also uses M-files JAC.M and FN.M \n'); fprintf(1,'If the number of equations exceeds 7 then JAC.M\n'); fprintf(1,'and FN.M must be changed.\n'); OK = FALSE; while OK == FALSE fprintf(1,'Input the number n of equations.\n'); N = input(' '); if N >= 2 OK = TRUE; else fprintf(1,'N must be an integer greater than 1.\n'); end; end; for I = 1 : N fprintf(1,'Input the function F_(%d) in terms of y1...y%d \n',I,N); s(I) = input(' ','s'); end; % Define components of F as follows: % s(1) = '3y1-cos(y2y3)-0.5'; % s(2) = 'y1^2-81(y2+0.1)^2+sin(y3)+1.06'; % s(3) = 'exp(-y1y2)+20y3+(10pi-3)/3'; for I = 1 : N for J = 1 : N fprintf(1,'Input the partial of F_(%d) with respect to x_%d \n',I,J); fprintf(1,'in terms of y1, ..., y%d \n',N); ss((I-1)N+J) = input(' ','s'); end; end; % Define the entries of the Jacobian in row major ordering. % ss(1) = '3'; % ss(2) = 'y3sin(y2y3)'; % ss(3) = 'y2sin(y2y3)'; % ss(4) = '2y1'; % ss(5) = '-162*(y2+0.1)';

% ss(6) = 'cos(y3)'; % ss(7) = '-y2exp(-y1y2)'; % ss(8) = '-y1exp(-y1y2)'; % ss(9) = '20'; OK = FALSE; while OK == FALSE fprintf(1,'Input tolerance\n'); TOL = input(' '); if TOL > 0 OK = TRUE; else fprintf(1,'Tolerance must be positive.\n'); end; end; OK = FALSE; while OK == FALSE fprintf(1,'Input the maximum number of iterations.\n'); NN = input(' '); if NN > 0 OK = TRUE; else fprintf(1,'Must be a positive integer.\n'); end; end; X = zeros(1,N); A = zeros(N,N); B = zeros(N,N); V = zeros(1,N); S = zeros(1,N); Y = zeros(1,N); U = zeros(1,N); Z = zeros(1,N); for I = 1 : N fprintf(1,'Input initial approximation X(%d).\n', I); X(I) = input(' '); end; if OK == TRUE fprintf(1,'Select output destination\n'); fprintf(1,'1. Screen\n'); fprintf(1,'2. Text file\n'); fprintf(1,'Enter 1 or 2\n'); FLAG = input(' '); if FLAG == 2 fprintf(1,'Input the file name in the form - drive:\name.ext\n'); fprintf(1,'for example A:\OUTPUT.DTA\n'); NAME = input(' ','s'); OUP = fopen(NAME,'wt'); else OUP = 1; end; fprintf(1,'Select amount of output\n'); fprintf(1,'1. Answer only\n'); fprintf(1,'2. All intermediate approximations\n'); fprintf(1,'Enter 1 or 2\n');

if OK == TRUE for J = 1 : N if J ~= I C = A(J,I)/A(I,I); for K = 1 : N A(J,K) = A(J,K)-CA(I,K); B(J,K) = B(J,K)-CB(I,K); end; end; end; end; I = I+1; end; if OK == TRUE for I = 1 : N C = A(I,I); for J = 1 : N A(I,J) = B(I,J)/C; end; end; else fprintf(1,'Jacobian has no inverse\n'); end; if OK == TRUE % STEP 3 K = 2; % Note: S = S(1) % Compute the product S = -Av and the L2 norm SN of S SN = 0; for I = 1 : N S(I) = 0; for J = 1 : N S(I) = S(I)-A(I,J)*V(J); end; SN = SN+S(I)^2; end; SN = sqrt(SN); for I = 1 : N X(I) = X(I)+S(I); end; if FLAG == 2 fprintf(OUP,' %d',K-1); for I = 1 : N fprintf(OUP,' %11.8f',X(I)); end; fprintf(OUP,'\n %12.6e\n',SN); end; % STEP 4 while K <= NN & OK == TRUE % STEP 5 for I = 1 : N VV = FN(I,N,X,s); Y(I) = VV-V(I); V(I) = VV;

end; % Note: V = F(X(K)) and Y = Y(K) % STEP 6 % Form Z = -Ay and norm ZN of Z. ZN = 0; for I = 1 : N Z(I) = 0; for J = 1 : N Z(I) = Z(I)-A(I,J)Y(J); end; ZN = ZN+Z(I)Z(I); end; ZN = sqrt(ZN); % Note = Z = -A(K-1)^(-1)Y(K) % STEP 7 P = 0; % P will be S(K)^TA(K)^(-1)Y(K) for I = 1 : N P = P-S(I)Z(I); end; % STEP 8 for I = 1 : N U(I) = 0; for J = 1 : N U(I) = U(I)+S(J)A(J,I); end; end; % STEP 9 for I = 1 : N for J = 1 : N A(I,J) = A(I,J)+(S(I)+Z(I))U(J)/P; end; end; % STEP 10 % Form S = -Av and norm SN of S. SN = 0; for I = 1 : N S(I) = 0; for J = 1 : N S(I) = S(I)-A(I,J)V(J); end; SN = SN+S(I)^2; end; SN = sqrt(SN); % Note = A = A(K)^(-1) and S = -A(K)^(-1)F(X(K)) % STEP 11 for I = 1 : N X(I) = X(I)+S(I);; end; % Note: X = X(K+1) KK = K+1; if FLAG == 2 fprintf(OUP, ' %2d', K); for I = 1 : N