Why is MATLAB running slow? 2013a vs 2011a

Hi everyone,

I am running an m-file in Matlab 2013a, which takes roughly 20sec to complete. However, when a friend runs the same code in 2011a it takes a fraction of a second to complete. What could be causing this in my 2013a installation?

Our PCs have the same components so that is not the culprit.

Just a note; the code I am testing is very simple, for the most part just for loops.

Thanks

 clc; clear all; close all;
 format compact

 % Define number of cells in x and y
nx = 20; 
ny = 20;
uw = 1;

 % Define cell length in x and y
dx = 1.0/nx; 
dy = 1.0/ny;

 % Relaxation Parametres
relax1 = 0.2; 
relax2 = 1.2;
relax3 = 0.2;

 % Reynolds Number
re = 100; 

 nu = 1/re;
 kappa = 1/re;
 rex = 2.0/dx/dx + 2.0/dy/dy;

 % Initialise the streamfunction and vorticity everywhere.
for i = 1:nx+1 

    for j = 1:ny+1

        psi(i,j) = 0.0; % Streamfunction
        w(i,j) = 0.0;
        x(i,j) = (i-1)*dx;
        y(i,j) = (j-1)*dy;
        T(i,j) = 0.0;

        if j == ny+1
            T(i,j) = 1.0;
        end

    end;

 end;

 for k = 1:20000 %main loop - iterate until convergence.

    adu = 0.0;
    adv = 0.0;
    psimin = 1000000.0;
    psimax = -1000000.0;

    for i = 2:nx %sweep through all internal points on the grid

        for j = 2:ny %updating until convergence.
        wimj0 = w(i-1,j); %for the vorticity equations these values
        wipj0 = w(i+1,j); %may have to be modified at the points
        wi0jm = w(i,j-1); %adjacent to the walls
        wi0jp = w(i,j+1);

            % Boundary Adjacent Conditions
            if(i == 2) 
                wimj0 = -2.0*psi(i,j)/dx/dx;
            end;

            if(i == nx)
                wipj0 = -2.0*psi(i,j)/dx/dx;
            end;

            if(j == 2)
                wi0jm = -2.0*psi(i,j)/dy/dy;
            end;

            if(j == ny)
                wi0jp = -2.0*(psi(i,j)+dy*uw)/dy/dy;
            end;

            %this is the update for w(i,j) from the FDE.

            wij = (nu*((wipj0+wimj0)/dx/dx + (wi0jp+wi0jm)/dy/dy)...
                + (((psi(i+1,j)-psi(i-1,j))*(wi0jp-wi0jm)-((psi(i,j+1)...
                -psi(i,j-1))*(wipj0-wimj0)))/(4*dx*dy)))/(nu*rex);
            dw = wij - w(i,j);

            %compute the difference between new and old value.
            w(i,j) = w(i,j) + relax1*dw;

            %update using relaxation
            %update equation for the streamfunction

            psiij = ((psi(i+1,j)+psi(i-1,j))/dx/dx + ...
                (psi(i,j+1)+psi(i,j-1))/dy/dy + w(i,j))/rex;
            dpsi = psiij - psi(i,j);
            psi(i,j) = psi(i,j) + relax2*dpsi;

            if(psi(i,j)>psimax) %determine maximum and minimum streamfunction
                psimax = psi(i,j);
            end;

            if(psi(i,j)<psimin)
                psimin = psi(i,j);
            end;

        ddu = abs(dpsi); %compute maximum change in solution over full sweep.
        ddv = abs(dw);

            if(ddu>adu)
            adu = ddu;
            end;

            if(ddv>adv)
            adv = ddv;
            end;

        end;

    end;
 % k
 % psimax
 % psimin
 % adu
 % adv

    % Exit loop if converged
    if(adu<1.0e-4 && adv<1.0e-4) 
        break;
    end;

 end;

 % Compute the temperature distribution
for l = 1:10000

    adT = 0.0;

    for i = 1:nx+1

        for j = 1:ny+1

            if j ~= 1 && j ~= ny+1

                if i == 1 % Using a forward difference scheme on the psi differential

                    Tij = (kappa*((T(i,j+1)+T(i,j-1))/dy/dy) + ...
                        (psi(i+1,j)-psi(i,j))*(T(i,j+1)-T(i,j-1))/(2*dx*dy))/(kappa*2/dy/dy);

                elseif i == nx+1 % Using a backward difference scheme on the psi differential

                    Tij = (kappa*((T(i,j+1)+T(i,j-1))/dy/dy) + ...
                        (psi(i,j)-psi(i-1,j))*(T(i,j+1)-T(i,j-1))/(2*dx*dy))/(kappa*2/dy/dy);

                else

                    Tij = (kappa*((T(i+1,j)+T(i-1,j))/dx/dx + ...
                        (T(i,j+1)+T(i,j-1))/dy/dy) + ...
                        (psi(i+1,j)-psi(i-1,j))*(T(i,j+1)-T(i,j-1))/(4*dx*dy)...
                        - (psi(i,j+1)-psi(i,j-1))*(T(i+1,j)-T(i-1,j))/(4*dx*dy))/(kappa*rex);

                end;

                dT = Tij - T(i,j);
                T(i,j) = T(i,j) + dT*relax3;
                ddT = abs(dT);

                if(ddT>adT)
                    adT = ddT;
                end;

            end;

        end;

    end;

    if adT < 1e-6
        break
    end
 % l    
 % adT   

 end;

 k
 psimin
 adu
 adv
 l
 adT

 % Plot solution.
 for i = 1:10
 v(i) = -(i-1)*0.01;
 end;

 for i = 11:20
 v(i) = (i-10)*psimax/10;
 end;

 figure(1)
 contourf(x,y,psi,v);
 xlabel('x')
 ylabel('y')
 title('Streamfunction')
 colorbar

 figure(2)
 contourf(x,y,T)
 xlabel('x')
 ylabel('y')
 title('Temperature Distribution (\circC)')
 colorbar

[Oleg Komarov]

Paste the code so we can test it. Without it's impossible to say.

[Oleg Komarov]

I confirm it takes ~20 sec on R2014a.

Answer 1

[Edric Ellis]

I tried your code on my machine in both R2011a and R2014a, and as posted it is indeed quite a bit slower in R2014a. You can get nearly all the speed back if you make your code a function rather than a script. See this page for more about functions if you're not already used to writing them.

[Andreas Goser]

I did my copy / paste know using Chrome instead of IE :-D Putting it into a function gives me a speed-up from 24 to 14 seconds on my machine

[Adam]

I guess this isn't really part of the question, but I am just interested - do you know why putting it into a function suddenly speeds it up by that much? Is it simply because of the reduced scope of its workspace? I rarely use scripts myself except for temporary stuff, but I have colleagues who do and performance is an important consideration in what they are doing.

I did try putting the current script I am working on into a function, but it is a fast script anyway and there was no statistically significant time difference.

[Edric Ellis]

I'm afraid I'm not an expert on the interpretation of the MATLAB language, but I believe basically you're on the right lines that there are a bunch of optimisations that can be applied inside a function because of things like the reduced variable scope.

[Dominic]

Thanks for your reply Edric,

I will try making it a function instead

Answer 2

[Andreas Goser]

When I copy / paste the code, I need to correct too many smaller issues as of unintended line breaks that I can be sure I run the code you are using. A couple of comments:

I understand you say your machines have "the same components". I can however share that the majority of such effects is happening not from release to relase, but when changing hardware. I suggest you do an R2011a/b installation on your machine in parallel to narrow down the issue. Even then, there can be issues with stuff like the math kernel library.

Second, I wonder if your question is really finding out the why or if you need to reduce runtime. There might be opportunities to get rid of loops.

[Andreas Goser]

Memory should not be an issue - all matrices involed appear to be small.

What I wonder if - as you do iterations - the code breaks out of the loop at different times?

[Dominic]

Thanks for your response Andreas,

I did a 2011a install and it did reduce the time by roughly x4, still not anywhere near as fast as it is for my friend. This speed will suffice for now and I will try to make it into a function instead, as suggested below.

The reason I would like an increase in the speed is for when I increase some of the parameters (re) and hence must increase the grid number (nx,ny) and in turn the matrix size, after increasing this it took roughly 2-3 hours to run on my pc and on my friend's roughly 15-20min. <-- This is why i dont think it is hardware related

Answer 3

[yonatan gerufi]

there are many reason this could happen.

any other programs are running?

try

memory

to see how much matlab use, and also do

clear all

perhaps it will help.

if it doesn't help you can use "profiler" to see your time for each line.

you can upload the code, so other people will compare there time to yours.

[Dominic]

Thanks, I have checked the memory, it uses about 1GB out of eight and whilst MATLAB is running there is about 4GB free. I also tried increasing the Java heap memory to no avail.

Using profiler, nothing in particular takes up a lot of time, total time ~20sec, and the rest of the lines take a fraction of a second.