How to replace a for loop with something faster

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Maël
Maël 2013-11-22
评论： Maël 2013-11-22
Hi,
I have to speed up a portion of my code as much as possible, since the associated computational time is way too high (especially because this part of the code belongs to a nested function, which is called in an optimization process). I 'd like to avoid using a for loop for my vector recursion since its pretty time consuming
I have heard about using the "filter" function or a Cmex file to accelerate the process in a similar discussion (<http://www.mathworks.com/matlabcentral/answers/28396-speed-up-recursive-loop)>, but I dont have a clue of how to apply them to my problem. Therefore, I would really appreciate any kind of help :)
phi=phi'; % the input has to be a row vector
% recursion for calculating A(t,T,Phi)=A_ and B(t,T,Phi)=B_ (see
% p.592 in Heston and Nandi article)
%A=zeros(phi.*r);
row=size(phi,1);
A=zeros(row,T-1);
B=zeros(row,T-1);
A(:,T)=0;
B(:,T)=0;
*for i=1:T-1
A(:,T-i)=A(:,T-i+1)+phi.*r+B(:,T-i+1).*w-.5*log(1-2*a.*B(:,T-i+1));
B(:,T-i)=phi.*(lam_+g_)-.5*g_^2+b.*B(:,T-i+1)+.5.*(phi-g_).^2./(1-2.*a.*B(:,T-i+...
1));
end*
A_=A(:,1)+phi.*r+B(:,1).*w-.5*log(1-2.*a.*B(:,1)); % A(t;T,phi)
B_=phi.*(lam_+g_)-.5*g_^2+b.*B(:,1)+.5*(phi-g_).^2./(1-2.*a.*B(:,1)); % B(t;T,phi)
Thanks a lot!!!
Notes: r w a b lam_ g_ S_0 Sig_ are scalars, phi is a vector that might contain complex numbers
Basically the backward recursion is as described in the for loop, from final conditions A(T,T)=B(T,T)=0
  2 个评论
Simon
Simon 2013-11-22
Hi!
Can you give example values for the parameters so that we can execute the code?
Maël
Maël 2013-11-22
Sure:
a= 1.32e-6; b= 0.589; lam_= -0.5; g_= 422.0950; w=5.02e-6 T= 60;S_0= 1132.60;Sig_=1.4416e-04;r=6.6048e-05;
phi= [-79.6315i; -93.2105i]
Thanks a lot for your help Simon!

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回答（2 个）

Andrei Bobrov
Andrei Bobrov 2013-11-22
Please try is it:
p1 = phi.*r;
p2 = phi.*(lam_+g_)-.5*g_^2;
p3 = .5.*(phi-g_).^2;
a2 = 2*a;
for ii = 1:T-1
k = T-ii+1;
p4 = 1-a2.*B(:,k);
A(:,k-1) = A(:,k) + p1 + B(:,k).*w - .5*log(p4);
B(:,k-1) = p2 + b.*B(:,k) + p3./p4;
end
Simon
Simon 2013-11-22
Hi!
If you don't need the results of every loop, you can write:
p1 = phi.*r;
p2 = phi.*(lam_+g_)-.5*g_^2;
p3 = .5.*(phi-g_).^2;
a2 = 2*a;
A=zeros(row,1);
B=zeros(row,1);
for ii = 1:T-1
p4 = 1-a2*B;
A = A + p1 + B*w - .5*log(p4);
B = p2 + b*B + p3./p4;
end
It saves some more time.
  5 个评论
显示 3更早的评论
Simon
Simon 2013-11-22
No, this will give you the independent calculation of A and B, but this is not possible since A depends on B.
Maël
Maël 2013-11-22
You're absolutely right, it missed that.. Thank you for your help!

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