How to add tolerance to iteration with if statement

Question

Bethany Sinclair ，2022-10-27

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Direct link to this question

https://ww2.mathworks.cn/matlabcentral/answers/1837108-how-to-add-tolerance-to-iteration-with-if-statement

回答： Chunru ，2022-10-27

I have a Gauss-seidel iteration code, which works to 99 iterations with a relaxation factor. However, when I put an if statement in (no other changes), to say if the different between iteration a^x+1-a^x is less than a certain tolerance say 0.1 then stop the iteration sequence before the specified 99 iterations.

Any help to spot any issues would be appreciated!

a2(1,1)=0;
b2(1,1)=0;
c2(1,1)=0;
R2=0.3;
for i=1:99
    if a2(i+1,1)-a2(i,1)<0.1
       a2(i+1,1)=(1-R2)*a2(i,1)+R2*(4-b2(i,1)-c2(i,1));
       b2(i+1,1)=(1-R2)*b2(i,1)+R2*(2*a2(i+1,1)+4*c2(i,1)-33)/3;
       c2(i+1,1)=(1-R2)*c2(i,1)+R2*(3*a2(i+1,1)-2*b2(i+1,1)-2)/2;
    end
end
figure(5);
plot(a1);
hold on
plot (b1);
plot (c1);

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Answer 1

Chunru 2022-10-27

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1837108-how-to-add-tolerance-to-iteration-with-if-statement#answer_1085093

x(1,1)=0; %Initial guess, x=0

y(1,1)=0; %Initial guess, y=0

z(1,1)=0; %Initial guess, z=0

R=0.3; %relaxation factor

for i=1:99 %99 is the number of iterations

x(i+1,1)=(1-R)*x(i,1)+R*(4-y(i,1)-z(i,1)); %what multiplied with R is the "new" x value produced by the equation

y(i+1,1)=(1-R)*y(i,1)+R*(2*x(i,1)+4*z(i,1)-33)/3; %what multiplied with R is the "new" y value produced by the equation

z(i+1,1)=(1-R)*z(i,1)+R*(3*x(i,1)-2*y(i,1)-2)/2; %what multiplied with R is the "new" z value produced by the equation

if abs(x(i+1)-x(i)) < 0.1

break

end

plot(x);

hold on

plot(y);

plot(z);

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Answer 2

Chunru 2022-10-27

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1837108-how-to-add-tolerance-to-iteration-with-if-statement#answer_1085018

% if a2(i+1,1)-a2(i,1)<0.1
% change the above to:
if abs(a2(i+1,1)-a2(i,1))<0.1

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Bethany Sinclair 2022-10-27

Below is a code for iteration to find 3 values of a simulteneous equation to 99 iterations. I want to be able to stop the iteration algorithm early (before the 99 iterations are up) if the results fal within a certain tolerance. I.e - the result for A is within 0.1 of the previous iteration. The code below works, it just doesn't like it when I put the if statement in?

x(1,1)=0;  %Initial guess, x=0
y(1,1)=0;  %Initial guess, y=0
z(1,1)=0;  %Initial guess, z=0
R=0.3;     %relaxation factor
for i=1:99 %99 is the number of iterations
x(i+1,1)=(1-R)*x(i,1)+R*(4-y(i,1)-z(i,1)); %what multiplied with R is the "new" x value produced by the equation
y(i+1,1)=(1-R)*y(i,1)+R*(2*x(i,1)+4*z(i,1)-33)/3; %what multiplied with R is the "new" y value produced by the equation
z(i+1,1)=(1-R)*z(i,1)+R*(3*x(i,1)-2*y(i,1)-2)/2; %what multiplied with R is the "new" z value produced by the equation
end
plot(x);
hold on  
plot(y);
plot(z);