Julia Threads@ threads比单线程性能慢

编程问答 2022-06-13

问题描述

我试图用数值方法求解一维热方程：

enter image description here

我使用的是有限差分，在使用Julia中的@threads指令时遇到一些麻烦。特别是在下面，有两个相同代码的版本：第一个是单线程，而另一个使用@threads（除了@thread指令之外，它们是相同的）

function heatSecLoop(;T::Float64)

    println("start")
    L = 1
    ν = 0.5
    Δt = 1e-6
    Δx = 1e-3

    Nt = ceil(Int,T/Δt )
    Nx = ceil(Int,L/Δx + 2)
    u = zeros(Nx)    
    u[round(Int,Nx/2)] = 1
    
    println("starting loop")
    for t=1:Nt-1
        u_old = copy(u)
        for i=2:Nx-1
            u[i] = u_old[i] + ν * Δt/(Δx^2)*(u_old[i.-1]-2u_old[i] + u_old[i.+1])
        end

        if t % round(Int,Nt/10) == 0
            println("time = " * string(round(t*Δt,digits=4)) )
        end
    end
    println("done")
    return u
end

function heatParLoop(;T::Float64)

    println("start")
    L = 1
    ν = 0.5
    Δt = 1e-6
    Δx = 1e-3

    Nt = ceil(Int,Nx/2)] = 1
    
    println("starting loop")
    for t=1:Nt-1
        u_old = copy(u)
        Threads.@threads for i=2:Nx-1
            u[i] = u_old[i] + ν * Δt/(Δx^2)*(u_old[i.-1]-2u_old[i] + u_old[i.+1])
        end

        if t % round(Int,digits=4)) )
        end
    end
    println("done")
    return u
end

问题是顺序线程比多线程线程要快。这是时间（经过一轮编译）

julia> Threads.nthreads()
2

julia> @time heatParLoop(T=1.0)
start
starting loop
time = 0.1
time = 0.2
time = 0.3
time = 0.4
time = 0.5
time = 0.6
time = 0.7
time = 0.8
time = 0.9
done
  5.417182 seconds (12.14 M allocations: 9.125 GiB,6.59% gc time)

julia> @time heatSecLoop(T=1.0)
start
starting loop
time = 0.1
time = 0.2
time = 0.3
time = 0.4
time = 0.5
time = 0.6
time = 0.7
time = 0.8
time = 0.9
done
  3.892801 seconds (1.00 M allocations: 7.629 GiB,8.06% gc time)

当然，热方程只是一个更复杂的问题的例子。我还尝试将SharedArrays等其他库与distributed一起使用，但效果较差。

感谢您的帮助。

解决方法

这似乎仍然成立，可能是由于

  1. Threads.@threads 的开销
  2. 也许在较小程度上，Julia 中的垃圾收集是单线程的，而这里的原始版本会产生大量垃圾。

此外，基于链接的话语线程的建议，可能值得注意的是，现在有一个来自 LoopVectorization.jl 的 @avx（现在是 @turbo）宏的线程版本，它使用来自 Polyester.jl 的非常轻量级的线程，尽管线程开销仍然不小，但仍设法勉强维持了稍微好一点的性能：

function heatSecLoop(;T::Float64)

    println("start")
    L = 1
    ν = 0.5
    Δt = 1e-6
    Δx = 1e-3

    Nt = ceil(Int,T/Δt )
    Nx = ceil(Int,L/Δx + 2)
    u = zeros(Nx)    
    u[round(Int,Nx/2)] = 1
    u_old = similar(u)

    println("starting loop")
    for t=1:Nt-1
        u_old,u = u,u_old
        for i=2:Nx-1
            u[i] = u_old[i] + ν * Δt/(Δx^2)*(u_old[i.-1]-2u_old[i] + u_old[i.+1])
        end

        if t % round(Int,Nt/10) == 0
            println("time = " * string(round(t*Δt,digits=4)) )
        end
    end
    println("done")
    return u
end
function heatVecLoop(;T::Float64)
    println("start")
    L = 1
    ν = 0.5
    Δt = 1e-6
    Δx = 1e-3

    Nt = ceil(Int,L/Δx + 2)
    u = zeros(Nx)
    u[round(Int,Nx/2)] = 1
    u_old = similar(u)

    println("starting loop")
    for t=1:Nt-1
       u_old,u_old
       @tturbo for i=2:Nx-1
           u[i] = u_old[i] + ν * Δt/(Δx^2)*(u_old[i-1]-2u_old[i] + u_old[i+1])
       end

       if t % round(Int,Nt/10) == 0
           println("time = " * string(round(t*Δt,digits=4)) )
       end
    end
    println("done")
    return u
end

function heatTVecLoop(;T::Float64)
    println("start")
    L = 1
    ν = 0.5
    Δt = 1e-6
    Δx = 1e-3

    Nt = ceil(Int,digits=4)) )
       end
    end
    println("done")
    return u
end

julia> @time heatSecLoop(T=1.0)
start
starting loop
time = 0.1
time = 0.2
time = 0.3
time = 0.4
time = 0.5
time = 0.6
time = 0.7
time = 0.8
time = 0.9
done
  1.786011 seconds (114 allocations: 22.094 KiB)

julia> @time heatVecLoop(T=1.0)
start
starting loop
time = 0.1
time = 0.2
time = 0.3
time = 0.4
time = 0.5
time = 0.6
time = 0.7
time = 0.8
time = 0.9
done
  0.314305 seconds (114 allocations: 22.094 KiB)

julia> @time heatTVecLoop(T=1.0)
start
starting loop
time = 0.1
time = 0.2
time = 0.3
time = 0.4
time = 0.5
time = 0.6
time = 0.7
time = 0.8
time = 0.9
done
  0.300656 seconds (114 allocations: 22.094 KiB)

自从第一次提出这个问题以来，单线程 @turbo 向量化版本的性能似乎也有显着提高，多线程 @tturbo 版本的性能可能会继续提高问题大小。

juliamultithreadingparallel-processingpde