Why is the execution time changing?

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MRC
MRC 2013-11-14
评论: the cyclist 2013-11-15
Hi, I have two versions of a code. Both versions are divided into two parts and the two versions differ for the first part, while the second part is identical in both. Here the time of execution reported by using tic toc: Version 1 -part 1: 0.03 sec -part 2: 0.19 sec
Version 2 -part 1: 0.05 sec -part 2: 0.48 sec
My question is: why the execution time of part 2 in the second version is higher, although the code of the second part is exactly the same in both versions? I'm probably doing a very stupid mistake but I can't see it. Here the code and attached the .mat files used:
clear all
r=30;
m=20;
n=6;
load data
load freq
load EC
load C
index=makeindex(n-1);
repindex=repmat(index,m,1);
param=[-0.7 0.6]';
j=1;
s=2;
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Version 1 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%Part 1 (Different for the two versions)
tic
upp=[freq(:,1:3,ones(1,n)) zeros(size(freq,1),r,n)];
low=[freq(:,1:3,ones(1,n)) zeros(size(freq,1),r,n)];
datasub=data(m*j-(m-1):m*j,:);
gamma=param(1);
delta= param(2);
piece1=datasub(:,2+n+1:2+n+n-1)*gamma;
others=zeros(2^(n-1),n-1);
for i=1:n-1
behavothers=index;
behavothers(:,i)=[];
others(:,i)=sum(behavothers,2);
end
repothers=repmat(others,m,1);
piece2=delta*repothers;
piece1=kron(piece1, ones(2^(n-1),1));
obsprof= piece1+piece2;
epsi=kron(datasub(:,2+n+n-1+n-1+(n-1)*s-(n-1)+1:2+n+n-1+n-1+(n-1)*s), ones(2^(n-1),1));
prof=obsprof+epsi;
prof=(prof>0)==repindex;
sumequil=sum(prof,2);
vectorequil=(sumequil==(n-1)*ones(2^(n-1)*m,1));
toc
%%Part 2 (Identical for both versions)
tic
cumsumvectorequil= cumsum(vectorequil);
sumvectorequil=cumsumvectorequil(2^(n-1):2^(n-1):m*2^(n-1));
a=zeros((m-1),1);
for h=1:(m-1)
a(h)=sumvectorequil(h+1)-sumvectorequil(h);
end
sumvectorequilsep=[sumvectorequil(1); a];
ec=zeros(sumvectorequil(m), n+n-1+2);
previous=0;
for h=1:m
ec(previous+1:previous+sumvectorequilsep(h),1:n+1)=ones(sumvectorequilsep(h),1)*datasub(h,2:2+n);
previous=previous+sumvectorequilsep(h);
end
previous=0;
for h=1:m
indexequilibrium=find(vectorequil(2^(n-1)*h-2^(n-1)+1:2^(n-1)*h));
ec(previous+1:previous+size(indexequilibrium),1+n+1:1+n+n-1)=index(indexequilibrium,:);
previous=previous+size(indexequilibrium,1);
end
for h=1:size(ec,1)
for i=1:size(EC,1)
if all(ec(h,2:1+n+n-1)==EC(i,(1:n+n-1)))
ec(h,size(ec,2))=EC(i,n+n-1+2);
end
end
end
tmpupp = zeros(1, size(freq,1));
for i=1:size(ec,1)
for w=1:size(freq,1)
if all([ec(i,1) ec(i,1+n+n-1+1)]==[freq(w,1) freq(w,3)])
tmpupp(w)=1;
end
end
end
upp(:, s+3, j) = tmpupp';
tmplow = zeros(1, size(freq,1));
for i=1:size(ec,1)
for w=1:size(freq,1)
if all([ec(i,1) ec(i,1+n+n-1+1)]==[freq(w,1) freq(w,3)]) && sumvectorequilsep(ec(i),1)==1
tmplow(w)=1;
end
end
end
low(:, s+3, j) = tmplow';
toc
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Version 2 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%Part 1 (Different for the two versions)
tic
upp=[freq(:,1:3,ones(1,n)) zeros(size(freq,1),r,n)];
low=[freq(:,1:3,ones(1,n)) zeros(size(freq,1),r,n)];
datasub=data(m*j-(m-1):m*j,:);
gamma=param(1);
delta= param(2);
obsprofgreat= datasub(:,2+n+1:2+n+n-1)*gamma+delta*(n-2)*ones(m,n-1);
obsproflow= datasub(:,2+n+1:2+n+n-1)*gamma+zeros(m,n-1);
equilgreat=ones(m,n-1);
equillow=zeros(m,n-1);
responsegreat=(obsprofgreat+datasub(:,2+n+n-1+n-1+(n-1)*s-(n-1)+1:2+n+n-1+n-1+(n-1)*s)>0);
while any(abs( all(responsegreat==equilgreat,2)-ones(size(responsegreat,1),1)))==true
equilgreat=responsegreat;
others=zeros(m,n-1);
for i=1:n-1
behavothers=equilgreat;
behavothers(:,i)=[];
others(:,i)=sum(behavothers,2);
end
responsegreat=(datasub(:,2+n+1:2+n+n-1)*gamma+delta*others+datasub(:,2+n+n-1+n-1+(n-1)*s-(n-1)+1:2+n+n-1+n-1+(n-1)*s)>0);
end
responselow=(obsproflow+datasub(:,2+n+n-1+n-1+(n-1)*s-(n-1)+1:2+n+n-1+n-1+(n-1)*s)>0);
while any(abs( all(responselow==equillow,2)-ones(size(responselow,1),1)))==true
equillow=responselow;
others=zeros(m,n-1);
for i=1:n-1
behavothers=equillow;
behavothers(:,i)=[];
others(:,i)=sum(behavothers,2);
end
responselow=(datasub(:,2+n+1:2+n+n-1)*gamma+delta*others+datasub(:,2+n+n-1+n-1+(n-1)*s-(n-1)+1:2+n+n-1+n-1+(n-1)*s)>0);
end
repgreat=kron(responsegreat, ones(2^(n-1),1));
replow=kron(responselow, ones(2^(n-1),1));
sumgreat=kron(sum(responsegreat,2),ones(2^(n-1),1));
sumlow=kron(sum(responselow,2),ones(2^(n-1),1));
sumindex=repmat(sum(index,2),m,1);
vectorequil=((sumindex<sumgreat & sumindex>sumlow) | ((all(repgreat==repindex,2) | all(replow==repindex,2))));
toc
%%Part 2 (Identical for both versions)
tic
cumsumvectorequil= cumsum(vectorequil);
sumvectorequil=cumsumvectorequil(2^(n-1):2^(n-1):m*2^(n-1));
a=zeros((m-1),1);
for h=1:(m-1)
a(h)=sumvectorequil(h+1)-sumvectorequil(h);
end
sumvectorequilsep=[sumvectorequil(1); a];
ec=zeros(sumvectorequil(m), n+n-1+2);
previous=0;
for h=1:m
ec(previous+1:previous+sumvectorequilsep(h),1:n+1)=ones(sumvectorequilsep(h),1)*datasub(h,2:2+n);
previous=previous+sumvectorequilsep(h);
end
previous=0;
for h=1:m
indexequilibrium=find(vectorequil(2^(n-1)*h-2^(n-1)+1:2^(n-1)*h));
ec(previous+1:previous+size(indexequilibrium),1+n+1:1+n+n-1)=index(indexequilibrium,:);
previous=previous+size(indexequilibrium,1);
end
for h=1:size(ec,1)
for i=1:size(EC,1)
if all(ec(h,2:1+n+n-1)==EC(i,(1:n+n-1)))
ec(h,size(ec,2))=EC(i,n+n-1+2);
end
end
end
tmpupp = zeros(1, size(freq,1));
for i=1:size(ec,1)
for w=1:size(freq,1)
if all([ec(i,1) ec(i,1+n+n-1+1)]==[freq(w,1) freq(w,3)])
tmpupp(w)=1;
end
end
end
upp(:, s+3, j) = tmpupp';
tmplow = zeros(1, size(freq,1));
for i=1:size(ec,1)
for w=1:size(freq,1)
if all([ec(i,1) ec(i,1+n+n-1+1)]==[freq(w,1) freq(w,3)]) && sumvectorequilsep(ec(i),1)==1
tmplow(w)=1;
end
end
end
low(:, s+3, j) = tmplow';
toc
  2 个评论
Matt J
Matt J 2013-11-14
If you show us the code, maybe we can say.
MRC
MRC 2013-11-14
编辑:MRC 2013-11-14
I have included the code in the question.

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回答(2 个)

the cyclist
the cyclist 2013-11-14
I recommend you analyze your code using the profiler.
>> doc profile
for details.
  2 个评论
MRC
MRC 2013-11-14
Thanks for your suggestion. However, the profile doesn't show anything different between part 2 of version 1 and part 2 of version 2.
the cyclist
the cyclist 2013-11-15
I'm confused. The profiler doesn't show any timing difference between these two sections of code, but tic-toc does?

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Matt J
Matt J 2013-11-14
If you aren't doing so already, put these routines in a function mfile (not in a script mfile). Then redo the timing tests.

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