Performance 如何在matlab中检查并行处理的性能？

为了检查matlab中的并行处理是如何工作的，我尝试了下面的代码并测量了执行时间。但是我发现并行处理代码比普通代码花费更多的时间，这是出乎意料的。我在哪里做错了吗

并行处理代码

function t = parl()
matlabpool('open',2);
tic;
A = 5:10000000;
parfor i = 1:length(A)
    A(i) = 3*A(i) + (A(i)/5);
    A(i) = 0.456*A(i) + (A(i)/45);
end
tic;
matlabpool('close');
t = toc;
end

>> parl Starting matlabpool using the 'local' profile ... connected to 2 workers. Sending a stop signal to all the workers ... stopped.

ans =

    3.3332


function t = parl()
tic;
A = 5:10000000;
for i = 1:length(A)
    A(i) = 3*A(i) + (A(i)/5);
    A(i) = 0.456*A(i) + (A(i)/45);
end
tic;
t = toc;
end

并行处理没有结果

function t = parl()
matlabpool('open',2);
tic;
A = 5:10000000;
parfor i = 1:length(A)
    A(i) = 3*A(i) + (A(i)/5);
    A(i) = 0.456*A(i) + (A(i)/45);
end
tic;
matlabpool('close');
t = toc;
end

>> parl Starting matlabpool using the 'local' profile ... connected to 2 workers. Sending a stop signal to all the workers ... stopped.

ans =

    3.3332


function t = parl()
tic;
A = 5:10000000;
for i = 1:length(A)
    A(i) = 3*A(i) + (A(i)/5);
    A(i) = 0.456*A(i) + (A(i)/45);
end
tic;
t = toc;
end

无并行处理代码的结果

>> parl

ans =

   2.8737e-05

看看（显然）执行串行版本代码的时间，它实际上是

0

。这很可疑，看看密码

tic;
t = toc;

嗯，这会启动秒表并立即停止。是的，大约需要0秒。看看并行代码

tic;
matlabpool('close');
t = toc;

啊，在这种情况下，代码乘以关闭工人池的执行次数。这需要相当多的工作，所需的时间，

3.33s

，是在Matlab中使用并行计算的部分开销

是的，我确实认为你做错了什么，你没有测量你（可能）认为你在测量的东西

tic

启动秒表并

toc

读取。您的代码启动秒表两次并读取一次，它可能只开始计时一次