How to reduce computation time

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Yassine Hmamouche on 23 May 2020
Commented: Walter Roberson on 25 May 2020
Hi,
The function "funct" takes up almost 0.06 seconds and it is the first responsible for a slow execution of an integral (more than 3 hours) when it is used in it with regards to y.
Any help please to reduce the computation time of the following function
function [out] = funct(mu,lmax,wmax,hmax,lambdab,hu,hb,y,x)
nx=length(x);
out=zeros(1,nx);
A=(mu./pi).*(wmax+lmax);
B=(mu.*wmax.*lmax)./4;
C1=(hb-(2.*hmax))./(2*hmax);
C2=(hu+hb-(2.*hmax))./(2*hmax);
for i=1:nx
S1=@(u,theta) u.*(1-exp(C1.*((A.*sqrt(u.^2+x(i).^2-(2.*x(i).*u.*cos(theta))))+B)));
S2=@(theta) (1-exp(C1.*((A.*sqrt(y.^2+x(i).^2-(2.*x(i).*y.*cos(theta))))+B)));
end
end

Show 1 older comment
Yassine Hmamouche on 24 May 2020
Thank you so much for the feedback.
All input data are positive real numbers such that C1 and C2 are negative numbers. Hereafter some typical values:
mu=1; lmax=4;wmax=6;hmax=30;lambdab=1; hu=40; hb=10;y=4;x=2;
Best,
Walter Roberson on 25 May 2020
On my system, I time as closer to 0.012 .
Yassine Hmamouche on 25 May 2020
Yes this is the self computing time of the function, but when used as the integrand of another integral it gives a huge total computing time (i.e., more than 4 hours).

Walter Roberson on 25 May 2020
By going back and replacing that 1000 limit with infinity, and adding the assumption that all values are non-negative, and that hb < 2*hmax, Maple is able to combine terms to remove one integral. The symbolic form is
-exp(lambdab * int( int( u * (exp(1/2 * (1/4 * lmax * wmax * pi + (lmax+wmax) * ...
(u^2 - 2 * cos(theta) * u * x + x^2)^(1/2)) * mu * (hb-2*hmax) / pi / hmax)-1), u, 0, y), ...
theta, 0, 2*pi)) * lambdab * y * int(-1 + exp(1/8 * ((4 * lmax + 4 * wmax) * ...
(x^2 - 2 * cos(theta) * x * y + y^2)^(1/2) + lmax * wmax * pi) * (hb-2*hmax) * mu / pi / hmax), ...
theta, 0, 2*pi)
This should be more efficient

Walter Roberson on 25 May 2020
You can see in the below that I input all 5 levels of integration, and that Maple believes that it can be reduced to 3 integrations.
Yassine Hmamouche on 25 May 2020
Dear Walter and isakson,
Thank you so much for your feedback and effort.
You have demonstrated in fact that we have a great symbolic simplification. The function f3 in
equals 1 all the time.
The processing time reduces then as
tic
for jj = 1 : 10000
out1 = f_Y0_X0_bis_bis(4,3,-4,-3,1,20,60,10,4,1); % Old function
end
toc
%%
tic
for jj = 1 : 10000
out2 = f_Y0_X0_bis_bis2(4,3,-4,-3,1,20,60,10,4,1);% Function after simplification
end
toc
disp(out2-out1)
%%%%%%%%%%%%%%%
Elapsed time is 55.496311 seconds.
Elapsed time is 39.585332 seconds.
Let's summarize here the lessons learned from these discussions:
• Use a recent version of Matlab and a new vanilla PC.
• Doing steps 0, 1 and 2 manually or through powerful symbolic tools like Maple and Mathematica before considering vectorial analysis through Matlab.
• Avoid computing unecessary tasks inside loops, e.g., constants, wide integral ranges.
Many thanks and best regards.
Walter Roberson on 25 May 2020
Also, it helps to know any constraints that might apply. Even just knowing that a particular variable is real-valued might be enough to permit a calculation to simplify.

per isakson on 24 May 2020
Edited: per isakson on 25 May 2020
On R2018b and a new vanilla PC I found a significant decrease in computation time by replacing quad2d() by integral2() (with the same signature, i.e. defaults are used). With your input data this replacement didn't affect the result, out. (Caveat, I do this because I'm curious, not because I know much about the topic.) There are two posts on Double Integration in MATLAB ( and here ) at Loren on the Art of MATLAB. I don't find backing for the difference in speed that I've seen, which makes me a bit nervous. What's the catch? There ought to exist a discussion on how to chose between the two functions, but I fail to find it.
Have you tried to simplify the expressions with the help of the Symbolic Toolbox? See https://se.mathworks.com/matlabcentral/answers/33399-problem-with-double-integral-dblquad-and-quad2d#answer_42126 it's a bit old but shows the idea. Maybe that toolbox sees something that I don't see.
The difference in speed between integral2(), introduced in R2012a, and quad2d(), introduced in R2009a, that I see is irking.
I've compared your function, funct(), to funct_int2(), in which I replaced quad2d() by integral2(). (And split the long statement into three to make the code and the result of profile() easier to read.)
>> cssm % first call to cssm after starting Matlab
Elapsed time is 0.364935 seconds.
Elapsed time is 0.140186 seconds.
0
>> cssm
Elapsed time is 0.104307 seconds.
Elapsed time is 0.021930 seconds.
0
>> cssm
Elapsed time is 0.088423 seconds.
Elapsed time is 0.016356 seconds.
0
...
...
>> cssm
Elapsed time is 0.091209 seconds.
Elapsed time is 0.017180 seconds.
0
where cssm.m is a script containing
%%
mu=1; lmax=4; wmax=6; hmax=30; lambdab=1; hu=40; hb=10; y=4; x=2;
%%
tic
for jj = 1 : 10
out1 = funct(mu,lmax,wmax,hmax,lambdab,hu,hb,y,x);
end
toc
%%
tic
for jj = 1 : 10
out2 = funct_int2(mu,lmax,wmax,hmax,lambdab,hu,hb,y,x);
end
toc
disp( out2-out1 )
and
function [out] = funct_int2(mu,lmax,wmax,hmax,lambdab,hu,hb,y,x)
nx=length(x);
out=zeros(1,nx);
A=(mu./pi).*(wmax+lmax);
B=(mu.*wmax.*lmax)./4;
C1=(hb-(2.*hmax))./(2*hmax);
C2=(hu+hb-(2.*hmax))./(2*hmax); %#ok<NASGU>
for ii=1:nx
S1=@(u,theta) u.*(1-exp(C1.*((A.*sqrt(u.^2+x(ii).^2-(2.*x(ii).*u.*cos(theta))))+B)));
S2=@(theta) (1-exp(C1.*((A.*sqrt(y.^2+x(ii).^2-(2.*x(ii).*y.*cos(theta))))+B)));
f1 = lambdab.*y.*integral(S2,0,2*pi,'ArrayValued',true);
f2 = exp(-lambdab.*integral2(S1,0,y,0,2*pi));
f3 = (1-exp(-lambdab.*integral2(S1,0,1000,0,2*pi)));
out(ii) = f1.*f2./f3;
end
end
• Fine print; don't blame me for mistakes.
• In this comparison both integral2() and quad2d() use the "tiled method"
• The fact that the two return exactly the same result makes me believe that deep inside the same calculation is made.

Yassine Hmamouche on 25 May 2020
@per isakson, I am using Matlab R2013a.
per isakson on 25 May 2020
@Yassine Hmamouche, Then you have integral2() and should be able to rum "my" funct_int2. See integral2, Numerically evaluate double integral. I'm curious to hear whether you can reproduce my result.
Yassine Hmamouche on 25 May 2020
I tried to use integral2() but it doesn't work.
R2013a has typically the following integral functions:
• For one fold integral: integral, quad, ...,etc.
• For two fold integral: quad2D.
• For three fold integral: integral3.

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