How can I reduce the execution time of my code?

Question

Surath Ghosh 2024-4-12

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2106006-how-can-i-reduce-the-execution-time-of-my-code

评论： Surath Ghosh 2024-4-18

I wrote this code. it is working for m=32 and I am getting output with in 2 minutes. But if I am putting m=64, it is running. I checked for 24 hours. I am not getting the output. Can you tell me how I can reduce the execution time of this code.

clc;
clear all;
close all;
a=1/3;
delta=1.3;
Gamma=0.6;
global m t matrix matrix1
m=32;
matrix = zeros(m, m);
matrix1=zeros(m,m);
for i=1:m
    matrix(1,i)= 1/sqrt(m);
    for l=1:m
        t(l)=(l-0.5)./m;
    end
end
for i = 1:m-1
    for j = 0:(i - 1)
        for k = 1:(2^j + 1)-1
            if i == 2^j + k - 1
                fprintf('For i = %d, j = %d, and k = %d\n', i, j, k)
                for l = 1:m % Changed variable name from j to l
                    t_val = t(l);
                    matrix1(0+1,l) = (t_val^a)/(gamma(1+a)*sqrt(m));
                    if (k - 1) / 2^j <= t_val
                        if t_val < (k - 0.5) / 2^j
                            matrix(i+1, l) = 1/sqrt(m) *(2^(j/2));
                            matrix1(i+1, l) = 2^(j/2) *(1/(gamma(1+a)*sqrt(m)) * (t_val - (k - 1) / 2^j)^a);
                        elseif t_val < k / 2^j
                            matrix(i+1, l) = 1/sqrt(m) * (-2^(j/2));
                            matrix1(i+1, l) = 2^(j/2) * (1/(gamma(1+a)*sqrt(m)) * (t_val - (k - 1) / 2^j)^a - 2 / (gamma(1+a)*sqrt(m)) * (t_val - (k - 0.5) / 2^j)^(a));
                            
                        elseif t_val < 1
                            matrix1(i+1, l) = 2^(j/2) * (1/(gamma(1+a)*sqrt(m))* (t_val - (k - 1) / 2^j)^(a) - (2 / (gamma(1+a)*sqrt(m)) * (t_val - (k - 0.5) / 2^j)^(a)) + (1 / (gamma(1+a)*sqrt(m)) * (t_val - k / 2^j)^(a)))
                            
                        else
                            matrix(i+1, l) = 0;
                            matrix1(i+1, l) = 0;
                        end
                    end
                end
            end
        end
    end
end
disp(matrix) % Display the resulting matrix
disp(matrix1) % Operational matrix
cstart=zeros(1,2*m);
G=fsolve(@nddt,cstart);
G
for i=1:m
    G1(i)=G(i);
end
for i=1:m
    u(i)=delta+G1*matrix1(:,i);
end
plot(t,u,'-b')
hold on
p=1;
for j=m+1:2*m
    G2(p)=G(j);
    p=p+1;
end
for i=1:m
    v(i)=Gamma+G2*matrix1(:,i);
end
plot(t,v,'-r')
function [F]=nddt(E)
delta = 1.3;
Gamma = 0.6;
global m t matrix matrix1
for l=1:m
    F1(l)=E(1:m)* matrix(:,l) - t(l)*E(1:m)* matrix1(:,1)- t(l)*delta +(E(1:m)* matrix1(:,1) + delta)*(E(m+1:2*m)* matrix1(:,1) + Gamma);
    F2(l)=E(m+1:2*m)* matrix(:,l) - (E(1:m) * matrix1(:,1) + delta)*(E(m+1:2*m)*matrix1(:,1)+ Gamma) + t(l)*(E(m+1:2*m) * matrix1(:,1) + Gamma);
end
F=[F1,F2];
end

5 个评论
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Walter Roberson 2024-4-17

在 MATLAB Online 中打开

for i = 1:m-1
    for j = 0:(i - 1)
        for k = 1:(2^j + 1)-1
            if i == 2^j + k - 1

That can be rewritten,

for i = 1:m-1
    for j = 0:(i - 1)
        k = i - 2^j + 1
        if k >= 1 & k <= 2^j

Surath Ghosh 2024-4-18

Thank you very much sir for your comments. But I am getting the output that are the matrix and matrix1 within 10 minutes. That 4 loops are used only for matrix and matrix1. Main problem is in the G=fsolve(@nddt,cstart), when I called the nddt function. For this comment, output is not coming. For m=64, basically using fsolve, I want to solve 64*2=128 equations. The code is stucked. I tried to run this code for 96 hours but I did not get the output.

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Answer 1

Dinesh 2024-4-12

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2106006-how-can-i-reduce-the-execution-time-of-my-code#answer_1440276

在 MATLAB Online 中打开

Hi Surath,

Here are some optimizations that I could identify in your code:

Precompute Reusable Values: Avoid repetitive computation of constants or expressions within loops.

codegammaFactor = 1 / (gamma(1 + a) * sqrt(m));

Vectorize Conditional Assignments: Use logical indexing to apply conditions across a vector or matrix.

codecondition = (k - 1) / 2^j <= t & t < k / 2^j;
matrix1(i, condition) = 2^(j/2) * gammaFactor .* (t(condition) - (k - 1) / 2^j).^a;

Optimize Matrix Operations: For operations applied to each row or column, use matrix multiplication or dot products.

codeG1 = G(1:m);
u = delta + G1 * matrix1;

Further Optimization Strategies:

Parallel Computing: For large-scale matrix operations or loops that can run independently, consider using MATLAB's parallel computing capabilities (parfor loops, distributed arrays).
Profiling and Algorithm Refinement: Utilize MATLAB's profiler to identify bottlenecks. In some cases, rethinking the algorithm or applying mathematical simplifications can lead to significant performance gains.

Reference Links:

https://www.mathworks.com/help/parallel-computing/getting-started-with-parallel-computing-toolbox.html

https://www.mathworks.com/help/matlab/matlab_prog/profiling-for-improving-performance.html

2 个评论
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Surath Ghosh 2024-4-12

Thank you very much for your comments. I am getting matrix and matrix1 within 10 minutes. So, it is not taking more time. But using the below code, I am not getting the output (checked for 24 hours). When I called the nddt function, I am not getting the output.

function [F]=nddt(E)

delta = 1.3;

Gamma = 0.6;

global m t matrix matrix1

for l=1:m

F1(l)=E(1:m)* matrix(:,l) - t(l)*E(1:m)* matrix1(:,1)- t(l)*delta +(E(1:m)* matrix1(:,1) + delta)*(E(m+1:2*m)* matrix1(:,1) + Gamma);

F2(l)=E(m+1:2*m)* matrix(:,l) - (E(1:m) * matrix1(:,1) + delta)*(E(m+1:2*m)*matrix1(:,1)+ Gamma) + t(l)*(E(m+1:2*m) * matrix1(:,1) + Gamma);

end

F=[F1,F2];

end

Torsten 2024-4-12

You could try "lsqnonlin" or "fmincon" instead of "fsolve" to solve your quadratic system of equations, but these are the only options you have.

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Answer 2

Walter Roberson 2024-4-17

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2106006-how-can-i-reduce-the-execution-time-of-my-code#answer_1443426

在 MATLAB Online 中打开

for i = 1:m-1
    for j = 0:(i - 1)
        for k = 1:(2^j + 1)-1
            if i == 2^j + k - 1
                fprintf('For i = %d, j = %d, and k = %d\n', i, j, k)
                for l = 1:m % Changed variable name from j to l

You have 4 nested loops, three of which effectively have an upper limit of m, and the fourth has a much higher upper limit. When you double your m you have to expect that it will take notably longer.

1 个评论
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Surath Ghosh 2024-4-18

Thank you very much sir for your comments. But I am getting the output that are the matrix and matrix1 within 10 minutes. That 4 loops are used only for matrix and matrix1. Main problem is in the G=fsolve(@nddt,cstart), when I called the nddt function. For this comment, output is not coming. For m=64, basically using fsolve, I want to solve 64*2=128 equations. The code is stucked. I tried to run this code for 96 hours but I did not get the output.

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