How to use parpool worker again, after it times out?

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Nasser M. Abbasi
Nasser M. Abbasi 2021-6-25
编辑： Nasser M. Abbasi 2021-6-25
I am new to using Matlab's parpool. The only reason I am using it, is because I need to set a timeout on a call to int() in the symbolic toolbox, and this is the only way I found in Matlab to set a timeout on a function call. I am not using parpool to do any actual parallel processing. So that is why below I create only one worker.
I'll explain in words the problem, then give a MWE to reproduce it. I create a parpool using
parpool('local',1)
Then a function is called to integrate a problem, with a timeout of one minute using
fut = parfeval(@mupad_proc,0,the_integrand,the_variable);
ok = wait(fut, 'finished', 60); % Wait to finish
Then check if it timedout or not by checking the ok result. I noticed when the worker timeout, then each next call after that also timesout, when it should not.
I have thousands on integrals that I am processing this way, which means once one integral times out, then all the rest will also timeout because that worker is stuck/hanged.
What is the correct way to terminate the worker that times out and create a new worker in its place, so the next call do not hang?
Here is a MWE
function matlab_CAS()
%delete(gcp('nocreate'))
parpool('local',1) %create one worker
%%%%%%%%%%%%%%%
disp('trying integral which should not time out')
the_integrand = evalin(symengine,'sin(x)');
the_variable=evalin(symengine,'x');
fut = parfeval(@mupad_proc,0,the_integrand,the_variable);
ok = wait(fut, 'finished', 60); % Wait to finish
if ~ok
disp('Hanged. Terminated');
else
disp('completed on time');
end
%%%%%%%%%%%%%%%
disp('trying integral which will time out')
the_integrand = evalin(symengine,'(c + d*x)^3/sin(a + b*x)^3');
fut = parfeval(@mupad_proc,0,the_integrand,the_variable);
ok = wait(fut, 'finished', 60); % Wait to finish
if ~ok
disp('Hanged. Terminated');
else
disp('completed on time');
end
%%%%%%%%%%%%%%%
disp('trying first integral which should not time out, but it now hangs')
the_integrand = evalin(symengine,'sin(x)');
fut = parfeval(@mupad_proc,0,the_integrand,the_variable);
ok = wait(fut, 'finished', 60); % Wait to finish
if ~ok
disp('Hanged. Terminated');
else
disp('completed on time');
end
end
%this function is the one that the worker will run
function[]= mupad_proc(the_integrand,the_variable)
int(the_integrand,the_variable);
end
When I run the above, it prints on console the following
trying integral which should not time out
completed on time
trying integral which will time out
Hanged. Terminated
trying first integral which should not time out, but it now hangs
Hanged. Terminated
You see, how it times out on the 3rd call now, which it should not. This is because second call caused timeout. So I need to do some sort of clean up, to either restart the worker, or terminate it and start a new one?.
How does one handle such a problem in Matlab? Need to have symbolic toolbox to run the above. This is using Matlab 2021a on widnows 10.
UPDATE
I was trying things, and found that calling `cancel(...)` each time it times out seems to have solved the problem in the above test. I do not know now if this the correct way to handle this. I need to test this more on my main script. This is what it now looks like
fut = parfeval(@mupad_proc,0,the_integrand,the_variable);
ok = wait(fut, 'finished', 60); % Wait to finish
if ~ok
disp('Hanged. Terminated');
cancel(fut); %ADDED THIS
else
disp('completed on time');
end
Now, when I run the same function above, the output is
trying integral which should not time out
completed on time
trying integral which will time out
Hanged. Terminated
trying first integral which should not time out, now it no longer hangs. good.
completed on time
So now next call no longer hangs.
Is this the correct way to handle this issue?

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采纳的回答

Edric Ellis
Edric Ellis 2021-6-25
As @Walter Roberson says, if the wait call times out and you don't want the result, you must use cancel to stop execution. So your updated approach is correct. The timeout argument to wait simply allows you to do something else in between checking for completion - nothing happens if the timeout expires other than the wait call terminates.

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