# MATLAB parfor is slower than for -- any help to speed up of my loop cacluations within KDL function?

4 views (last 30 days)
Mehdi on 11 Jan 2024
Commented: Torsten on 13 Jan 2024
II=9;
JJ=9;
M=11;
W=rand(II+1,JJ+1,3,M);
k_d=01e12;
syms x y
for i=1:II+1
for j=1:JJ+1
lgij(i,j) = legendreP(i-1, x)*legendreP(j-1, y);
end
end
wxy2 = sym('wxy2',[1 M]);
wxy3 = sym('wxy3',[1 M]);
wxy2(1:M) = sym('0');
wxy3(1:M) = sym('0');
for r=1:M
for i=1:II+1
for j=1:JJ+1
wxy2(r) = W(i, j, 2, r)*lgij(i,j) + wxy2(r);
wxy3(r) = W(i, j, 3, r)*lgij(i,j) + wxy3(r);
end
end
end
wxxyy2=simplify(wxy2);
wxxyy3=simplify(wxy3);
for r=1:M
for o=1:M
Wijklmo(r,o)= expand(wxxyy2(r)*wxxyy2(o) + wxxyy3(r)*wxxyy3(o)- wxxyy2(r)*wxxyy3(o) - wxxyy2(o)*wxxyy3(r));
end
end
dt = 3.8655e-08;
T=1e9*dt;
q = zeros(M, floor(taim/dt)+1);
RA=111;
s=1;
for tn=t__0:dt:T
p_hat(1:M,s) = KDL(wxxyy2, wxxyy3, M, q, RA, k_d, Wijklmo, s)
s=s+1
end
a
function kdl = KDL( wxxyy2, wxxyy3, M, q, RA, k_d, Wijklmo, s)
xx = sym('xx');
yy = sym('yy');
Wxy2 = sym('0');
Wxy3 = sym('0');
for r=1:M
Wxy2 = wxxyy2(r)*q(r, s) + Wxy2;
Wxy3 = wxxyy3(r)*q(r, s) + Wxy3;
end
w23 = expand(Wxy3-Wxy2);
kdl=zeros(M,1);
xspan = [-1 1];
yspan = [-1 1];
Gy0 = 0;
H(xx,yy) =(0.5*(1+tanh(k_d*RA*w23)));
g = matlabFunction(H,'Vars',[xx yy]);
if w23==0
kdl(:,1) =0;
else
tic
for i=1:M
%parfor i=1:M
ff(i,1)= vpa(Wijklmo(i,:)*q(1:M, s));
f = matlabFunction(ff(i,1),'Vars',[xx yy]);
D = @(xx,yy)f(xx,yy).*(g(xx ,yy)>0);
if w23==0
kdl(i,1) =0;
else
[~,G] = ode45(@(y,Gy)fun(y,Gy,D,xspan),yspan,Gy0,odeset('RelTol',1e-10,'AbsTol',1e-10));
kdl(i,1) = k_d*real(G(end));
end
end
toc
end
a
function dGydy = fun(y,Gy,g,xspan)
% Compute the x-integrals at y = y
Gx0 = 0;
[~,Gx] = ode45(@(x,~)g(x,y),xspan,Gx0,odeset('RelTol',1e-10,'AbsTol',1e-10));
dGydy = Gx(end);
end
Mehdi on 13 Jan 2024
I did this, but did not speed-up. The problem is that parfor is slower than for.
Torsten on 13 Jan 2024
Ok, then I think @Matt J is correct: the gain in speed is overcompensated by the overhead of parallel computing.

Matt J on 12 Jan 2024
Edited: Matt J on 12 Jan 2024
M=11 is a rather small number of iterations. If this is a realistic value, there may not be enough iterative work to make the overhead of parfor worthwhile. It might be better to move the parfor to the outer loop that calls KDL.
Tn=t__0:dt:T; %where are these values used???
parfor s=1:numel(Tn)
p_hat(:,s) = KDL(wxxyy2, wxxyy3, M, q, RA, k_d, Wijklmo, s);
end
A few other remarks,
(1) Things like this loop
for r=1:M
Wxy2 = wxxyy2(r)*q(r, s) + Wxy2;
Wxy3 = wxxyy3(r)*q(r, s) + Wxy3;
end
can be replaced with vectorized statements,
Wxy2 = wxxyy2*q(:, s);
Wxy3 = wxxyy3*q(:, s);
(2) Also, q(:, s) will probably evaluate faster than q(1:M,s).
##### 3 CommentsShow 1 older commentHide 1 older comment
Matt J on 12 Jan 2024
Edited: Matt J on 12 Jan 2024
You should pre-allocate p_hat
M=11;
N=5;
p_hat=zeros(M,N);
parfor s=1:N
p_hat(:,s) = KDL(s);
end
Starting parallel pool (parpool) using the 'Processes' profile ... Parallel pool using the 'Processes' profile is shutting down.
p_hat
p_hat = 11×5
1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5
function s=KDL(s)
end
Mehdi on 12 Jan 2024
Edited: Mehdi on 12 Jan 2024
this method does not work for me since in my original problem KDL(s+1)=f(KDL(s)).(because of complexities I have not brought this part of my code here.

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