How can I reduce the execution time of my code?

2 Ansichten (letzte 30 Tage)
Surath Ghosh
Surath Ghosh am 12 Apr. 2024
Kommentiert: Surath Ghosh am 18 Apr. 2024
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 Kommentare
Walter Roberson
Walter Roberson am 17 Apr. 2024
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 am 18 Apr. 2024
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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Antworten (2)

Dinesh
Dinesh am 12 Apr. 2024
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:
  2 Kommentare
Surath Ghosh
Surath Ghosh am 12 Apr. 2024
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 am 12 Apr. 2024
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 am 17 Apr. 2024
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 Kommentar
Surath Ghosh
Surath Ghosh am 18 Apr. 2024
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.

Melden Sie sich an, um zu kommentieren.

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