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How can I reduce the execution time of my code?

조회 수: 2 (최근 30일)
Surath Ghosh
Surath Ghosh 2024년 4월 12일
댓글: Surath Ghosh 2024년 4월 18일
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
  댓글 수: 5
Walter Roberson
Walter Roberson 2024년 4월 17일
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
Surath Ghosh 2024년 4월 18일
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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답변 (2개)

Dinesh
Dinesh 2024년 4월 12일
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:
  댓글 수: 2
Surath Ghosh
Surath Ghosh 2024년 4월 12일
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
Torsten 2024년 4월 12일
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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Walter Roberson
Walter Roberson 2024년 4월 17일
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.
  댓글 수: 1
Surath Ghosh
Surath Ghosh 2024년 4월 18일
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.

댓글을 달려면 로그인하십시오.

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