How to run computations for n seconds in Julia?

Question

I'd like to run heavy computations in Julia for a fixed duration, for example 10 seconds. I tried this:

timer = Timer(10.0)
while isopen(timer)
    computation()
end

But this does not work, since the computations never let Julia's task scheduler take control. So I added yield() in the loop:

timer = Timer(10.0)
while isopen(timer)
    yield()
    computation()
end

But now there is significant overhead from calling yield(), especially when one call to computation() is short. I guess I could call yield() and isopen() only every 1000 iterations or so, but I would prefer a solution where I would not have to tweak the number of iterations every time I change the computations. Any ideas?

Do you expect each execution of computation to take approximately the same duration ? — High Performance Mark
– High Performance Mark, Commented Jul 11, 2020 at 12:26

Przemyslaw Szufel · Accepted Answer · 2020-07-15 21:41:37Z

1

This pattern below uses threads and on my laptop has a latency of around 35ms for each 1,000,000 calls which is more than acceptable for any job.

Tested on Julia 1.5 release candidate:

function should_stop(timeout=10)
    handle = Threads.Atomic{Bool}(false)
    mytask = Threads.@spawn begin
        sleep(timeout)
        Threads.atomic_or!(handle, true)
    end
    handle
end


function do_some_job_with_timeout()
    handle = should_stop(5)
    res = BigInt() # save results to some object
    mytask = Threads.@spawn begin
        for i in 1:10_000_000
            #TODO some complex computations here
            res += 1 # mutate the result object
            handle.value && break
        end
    end
    wait(mytask) # wait for the job to complete
    res
end

You can also used Distributed instead. The code below seems to have a much better latency - only about 1ms for each 1,000,000 timeout checks.

using Distributed
using SharedArrays
addprocs(1)
function get_termination_handle(timeout=5,workerid::Int=workers()[end])::SharedArray{Bool}
    handle = SharedArray{Bool}([false])
    proc = @spawnat workerid begin
        sleep(timeout)
        handle[1]=true
    end
    handle
end


function fun_within_timeout()
    res = 0
    h = get_termination_handle(0.1)
    for i = 1:100_000_000
        res += i % 2 == 0 ? 1 : 0
        h[1] && break
    end
    res
end

edited Jul 15, 2020 at 21:41

answered Jul 12, 2020 at 22:15

Przemyslaw Szufel

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8 Comments

MiniQuark Over a year ago

Thanks Przemyslaw Szufel, but unfortunately I cannot get this to work. If I run it as is, the computations are finished after 2 seconds on my computer. If I bump up the iteration to 1:10^9, the computations do not stop after 5 seconds. If I add yield() in the loop, then it works. I suspect that both sleep & loop tasks get scheduled in the same thread. Note that I did set JULIA_NUM_THREADS=16 before starting Julia. Also note that I tried running the job in the main thread rather than a second Thread.@spawn, but it made no difference.

Przemyslaw Szufel Over a year ago

What Julia version you have? I have been testing it in Julia 1.5 rc1. This code worked for me only when these are two separate threads (hence that is why I spawn twice). You could perhaps try to add another variable and store Threads.threadid() to see if they get spawned separately. As fat as I know there is no method to select threadid so if this does not help I will propose code using Distributed.

MiniQuark Over a year ago

I'm using 1.4.2. I'll give 1.5 rc1 a try.

Przemyslaw Szufel Over a year ago

I added the Distributed code. It also seems to have much lower latency.

MiniQuark Over a year ago

Thanks for the distributed version, it seems to work in Julia 1.4.2. Also, I can confirm that the 1st solution works for me when using 1.5rc1. However, the distributed version does not work for me in 1.5rc1, this is weird.

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