Asked by Jakob Sørensen
on 27 Apr 2012

I tried making a very simple algorithm for testing the computers speed, like this

tic;

for r = 1:2000;

for c = 1:2000;

A(r,c) = r^2+sqrt(c);

end

end

toc

And it's obviously on purpose that I didn't preallocate A, since I want to stress the processor a bit. Anyway, i ran it on three different computers with follow results:

University desktop, with Ubuntu and Matlab 2010b. Time ~18 sek

Home desktop, with Windows and Matlab 2011 (i think). Time ~7 sek

Home laptop, with Ubuntu and Matlab 2011 or 2012 (dont remember) Time ~4 sek

The thing that confuses me a bit, is that my Laptop is that much faster than my home desktop, since both processor, graphics card and ram should be superior on the home desktop. Can anyone help me solve this mystery?

Answer by Jan
on 27 Apr 2012

Accepted Answer

Filling an array without pre-allocation does not stress the processor, but the memory. Therefore the processor is most of the time in sleep mode during your program runs. Matlab's collaboration with the memory manager of Window 7 is obviously better than with the manager of Ubuntu.

Matlab 2012a uses a new technique to reduce the drawbacks of a missing pre-allocation. It seems, like it allocates memory in larger chunks predictively, buit I do not know any details.

Therefore your measurements are not surprising. If you need a more detailed answer, post the exact description of the used operating systems and Matlab versions (version numbers and 32/64 bits). In addition post the amount of installed and available RAM as well as the type and speed of the RAM.

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Answer by Daniel Shub
on 27 Apr 2012

Why not just use bench?

doc bench

Your lack of preallocation is probably distorting your results for your university desktop since automatic array growth gets a lot faster in R2011a

As for your two home machines, again the version may matter since MATLAB is always improving the JIT.

Answer by Andreas Goser
on 27 Apr 2012

Other contributing factors could be

- 32/64 architecture of machine in combination of 32/64 MATLAB
- BLAS routines, e.g. the MKL library for Intel processors

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Answer by Jakob Sørensen
on 27 Apr 2012

Daniel Shub
on 27 Apr 2012

Jakob Sørensen
on 27 Apr 2012

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Answer by Jeremy Irons
on 13 Feb 2018

clear all;

theta=0.002;

lambda1=0.0005;

lambda2=0.0008;

lambda3=0.0012;

I0=1;

x=linspace(-0.2,0.2);

I1=I0*(cos((pi./2)+(2.*pi.*x.*tan(theta))/lambda1).^2);

I2=I0*(cos((pi./2)+(2.*pi.*x.*tan(theta))/lambda2).^2);

I3=I0*(cos((pi./2)+(2.*pi.*x.*tan(theta))/lambda3).^2);

plot(x,I1,x,I2,x,I3)

legend('lambda=0.0005','lambda=0.0008','lambda=0.0012');

Jeremy Irons
on 13 Feb 2018

%2

clear all;

D=4000;

lambda=0.05;

I0=1;

d=5;

a=4;

tx=-80:1:80;

ty=tx;

[X,Y]=meshgrid(tx,ty);

I1=I0.*((sin((pi.*d.*Y)./(lambda.*D)))./((pi*d*Y)/(lambda.*D))).^2*I0.*((sin((pi.*d.*X)./(lambda.*D)))./((pi*d*X)/(lambda.*D))).^2;

hold on;

subplot(2,1,1)

surf(X,Y,I1)

xlabel('x')

ylabel('y');

subplot(2,1,2)

contour(X,Y,I1,200)

xlabel('x')

ylabel('y');

Jeremy Irons
on 13 Feb 2018

%3

clear;

Na = 6.022*10^23;

kB = 1.38*10^-23;

u = 1.66*10^-27;

mH2 = 2*u;

mO2 = 32*u;

mN2 = 28*u;

v = 0:2000;

fvH2 = 4.*pi.*(mH2./(2.*pi.*kB.*300)).^(3./2).*v.^2.*exp(-(mH2.*v.^2)./(2.*kB.*300));

fvO2 = 4.*pi.*(mO2./(2.*pi.*kB.*300)).^(3./2).*v.^2.*exp(-(mO2.*v.^2)./(2.*kB.*300));

fvN2 = 4.*pi.*(mN2./(2.*pi.*kB.*300)).^(3./2).*v.^2.*exp(-(mN2.*v.^2)./(2.*kB.*300));

fvH2_2 = 4.*pi.*(mH2./(2.*pi.*kB.*70)).^(3./2).*v.^2.*exp(-(mH2.*v.^2)./(2.*kB.*70));

fvH2_3 = 4.*pi.*(mH2./(2.*pi.*kB.*500)).^(3./2).*v.^2.*exp(-(mH2.*v.^2)./(2.*kB.*500));

subplot(2,1,1);

hold on;

plot(v,fvH2)

plot(v,fvO2)

plot(v,fvN2)

xlabel('v [m/s]');

ylabel('f(v)');

legend('H2','O2','N2');

subplot(2,1,2);

hold on;

plot(v,fvH2)

plot(v,fvH2_2)

plot(v,fvH2_3)

xlabel('v [m/s]');

ylabel('f(v)');

legend('300K','70K','500K');

Andrew Jordan
on 13 Feb 2018

1

clear all;

theta = 0.002;

lambda = [0.0005, 0.0008, 0.0012];

I0 = 1;

x = linspace(-0.2, 0.2);

I1 = I0*(cos((pi./2)+(2.*pi.*x.*tan(theta))/lambda(1)).^2);

I2 = I0*(cos((pi./2)+(2.*pi.*x.*tan(theta))/lambda(2)).^2);

I3 = I0*(cos((pi./2)+(2.*pi.*x.*tan(theta))/lambda(3)).^2);

plot(x, I1, x, I2, x, I3)

xlabel('X');

ylabel('I');

legend('0.0005','0.0008','0.0012');

title('Rozklad natezenia');

for j = length(lambda)

d(j) = lambda(j)/sin(theta);

end

figure

plot(lambda, d)

title('d(lambda)')

xlabel('Dlugosc swiatla lambda')

ylabel('Odleglosc d')

2

clear all;

D = 4000;

lambda = 0.05;

I0 = 1;

d = 5;

a = 4;

tx = -75:1:75;

ty = -75:1:75;

[X,Y] = meshgrid(tx,ty);

I = I0.*((sin((pi.*d.*Y)./(lambda.*D)))./((pi*d*Y)/(lambda.*D))).^2*I0.*((sin((pi.*d.*X)./(lambda.*D)))./((pi*d*X)/(lambda.*D))).^2;

hold on;

subplot(1,2,1)

surf(X,Y,I)

title('Rozklad natezenia')

xlabel('x')

ylabel('y');

zlabel('I')

subplot(1,2,2)

contour(X,Y,I,200)

title('Kontur natezenia')

xlabel('x')

ylabel('y');

zlabel('I')

hold off;

3

clear;

Na = 6.022*10^23;

kB = 1.38*10^-23;

u = 1.66*10^-27;

mH = 2*u;

mO = 32*u;

mN = 28*u;

v = 0:5000;

fH = 4.*pi.*(mH./(2.*pi.*kB.*300)).^(3./2).*v.^2.*exp(-(mH.*v.^2)./(2.*kB.*300));

fO = 4.*pi.*(mO./(2.*pi.*kB.*300)).^(3./2).*v.^2.*exp(-(mO.*v.^2)./(2.*kB.*300));

fN = 4.*pi.*(mN./(2.*pi.*kB.*300)).^(3./2).*v.^2.*exp(-(mN.*v.^2)./(2.*kB.*300));

fK70 = 4.*pi.*(mH./(2.*pi.*kB.*70)).^(3./2).*v.^2.*exp(-(mH.*v.^2)./(2.*kB.*78));

fK500 = 4.*pi.*(mH./(2.*pi.*kB.*500)).^(3./2).*v.^2.*exp(-(mH.*v.^2)./(2.*kB.*500));

subplot(2,1,1);

hold on;

plot(v,fH)

plot(v,fO)

plot(v,fN)

title('Rozklad predkosci czastek gazu Maxwella temperaturze 300K');

xlabel('v [m/s]');

ylabel('f(v)');

legend('H2','O2','N2');

subplot(2,1,2);

hold on;

plot(v,fK70)

plot(v,fH)

plot(v,fK500)

title('Rozklad predkosci czastek gazu Maxwella dla H2');

xlabel('v [m/s]');

ylabel('f(v)');

legend('Temperatura = 78K', 'Temepratura = 300K', 'Temperatura = 500K')

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