Stop the iteration loop at desire Ea

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Siti Salmiah
Siti Salmiah 2020-11-9
回答： Mathieu NOE 2020-11-9
I need help with my fixed point MATLAB coding. My iteration seems does not stop at my desired Ea. I want the iteration to stop at Ea<5 but the iteration will always stop at the max number of iteration, N. I have try to use the "break else return" but than it not show any of the iteration. Below are my MATLAB code:
function FixedPoint
clear
clc
cinitial = input('\nGive the initial value of c : '); % Inputs initial value of x
iterations = 0;
N = 100; % max no. iterations.
fprintf('\n\nIteration c Ea(%)')
%-------------------- Runs Fixed Point algorithm --------------------------
while iterations < N
c = f(cinitial);
Ea = abs((c-cinitial)/c)*100; % Calculates error.
iterations = iterations + 1; % Calculates iteration no.
cinitial = c; % Replaces old value with new one.
if abs((c-cinitial)/c)*100 <= 5
fprintf('\n%5.0f%16.8f %f', iterations,c,Ea )
end
if iterations > 1000
Ea <= 5;
disp('No Convergence (Ignore answer below)') % Halts algorithm after 10000
end % iterations if convergence is
end % not achieved.
fprintf('\nSolution is %f \n', double(c))
end
%-------------------- Ends algorithm --------------------------------------
function F = f(c)
F = ((9.81 * 82).* (1-exp(-0.04878.*c)))/36; % defines function
end
%-------------------- End of program --------------------------------------

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回答（2 个）

Alan Stevens
Alan Stevens 2020-11-9
编辑：Alan Stevens 2020-11-9
Like so
%cinitial = input('\nGive the initial value of c : '); % Inputs initial value of x
cinitial = 1;
iterations = 0;
N = 100; % max no. iterations.
% This will need to be much larger if you tighten
% the convergence tolerance
Ea = 1;
fprintf('\n\nIteration c Ea(%)')
%-------------------- Runs Fixed Point algorithm --------------------------
while iterations < N && Ea > 10^-2 % You need an error test
c = f(cinitial);
Ea = abs((c-cinitial)/c)*100; % Calculates error.
iterations = iterations + 1; % Calculates iteration no.
if Ea <= 5
fprintf('\n%5.0f%16.8f %f', iterations,c,Ea )
end
cinitial = c; % Replaces old value with new one.
if iterations > N
Ea <= 5;
disp('No Convergence (Ignore answer below)') % Halts algorithm after N
end % iterations if convergence is
end % not achieved.
fprintf('\nSolution is %f \n', double(c))
%-------------------- Ends algorithm --------------------------------------
function F = f(c)
F = ((9.81 * 82).* (1-exp(-0.04878.*c)))/36; % defines function
end
%-------------------- End of program --------------------------------------
Mathieu NOE
Mathieu NOE 2020-11-9
hello
I believe it is working better now
I tested it even with much tighter error constraint (down to 0.1)
I'll let you check
all the best
clear
clc
c_initial = input('\nGive the initial value of c : '); % Inputs initial value of x
iterations = 0;
N = 1000; % max no. iterations.
fprintf('\n\nIteration c Ea(%)')
%-------------------- Runs Fixed Point algorithm --------------------------
c_old = c_initial;
while iterations < N
c = f(c_old);
Ea = abs((c-c_old)/c)*100; % Calculates error.
iterations = iterations + 1; % Calculates iteration no.
c_old = c; % Replaces old value with new one.
if Ea <= 0.1
fprintf('\n%5.0f%16.8f %f', iterations,c,Ea )
return
end
if iterations > 1000
Ea <= 5;
disp('No Convergence (Ignore answer below)') % Halts algorithm after 10000
end % iterations if convergence is
end % not achieved.
fprintf('\nSolution is %f \n', double(c))
% end
%-------------------- Ends algorithm --------------------------------------
function F = f(c)
F = ((9.81 * 82).* (1-exp(-0.04878.*c)))/36; % defines function
end

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