本篇博文转载于https://www.cnblogs.com/1024incn/tag/CUDA/,仅用于学习。
这部分主要介绍并行分析,涉及掌握nvprof的几个metric参数,具体的这些调节为什么会影响性能会在后续博文解释。
下面是我们的kernel函数sumMatrixOnGPUD:
__global__ void sumMatrixOnGPU2D(float *A, float *B, float *C, int NX, int NY) {
unsigned int ix = blockIdx.x * blockDim.x + threadIdx.x;
unsigned int iy = blockIdx.y * blockDim.y + threadIdx.y;
unsigned int idx = iy * NX + ix;
if (ix < NX && iy < NY) {
C[idx] = A[idx] + B[idx];
}
}
我们指定一个比较大的数据矩阵,包含16384个元素:
int nx = 1<<14; int ny = 1<<14;
下面的代码用来配置main函数的参数,也就是block的维度配置:
if (argc > 2) { dimx = atoi(argv[1]); dimy = atoi(argv[2]); } dim3 block(dimx, dimy); dim3 grid((nx + block.x - 1) / block.x, (ny + block.y - 1) / block.y);
编译:
$ nvcc -O3 -arch=sm_20 sumMatrix.cu -o sumMatrix
在做各项数据比较的时候需要有个基准,这里使用四个block配置的时间消耗作为基准观察,分别为(32,32)(32,16)(16,32)和(16,16),本文开始时有提到,第一个参数是x象限维度,第二个参数是y象限维度。
下面是几种配置的时间消耗输出结果:
$ ./sumMatrix 32 32
sumMatrixOnGPU2D <<< (512,512), (32,32) >>> elapsed 60 ms
$ ./sumMatrix 32 16
sumMatrixOnGPU2D <<< (512,1024), (32,16) >>> elapsed 38 ms
$ ./sumMatrix 16 32
sumMatrixOnGPU2D <<< (1024,512), (16,32) >>> elapsed 51 ms
$ ./sumMatrix 16 16
sumMatrixOnGPU2D <<< (1024,1024),(16,16) >>> elapsed 46 ms
比较这几个结果,不难发现,最慢的是第一个(32,32),最快的是第二个(32,16),这里可以猜测的到的是,拥有更多的block并行性更好。这个猜测可以使用nvprof 的achieved_occupancy这个metric参数来验证。该参数的定义公式在上一篇博文有介绍,实际上就是指每个SM在每个cycle能够达到的最大active warp数目占总warp的比例。下面是使用该参数后得到的结果:
$ nvprof --metrics achieved_occupancy ./sumMatrix 32 32
sumMatrixOnGPU2D <<<(512,512), (32,32)>>> Achieved Occupancy 0.501071
$ nvprof --metrics achieved_occupancy ./sumMatrix 32 16
sumMatrixOnGPU2D <<<(512,1024), (32,16)>>> Achieved Occupancy 0.736900
$ nvprof --metrics achieved_occupancy ./sumMatrix 16 32
sumMatrixOnGPU2D <<<(1024,512), (16,32)>>> Achieved Occupancy 0.766037
$ nvprof --metrics achieved_occupancy ./sumMatrix 16 16
sumMatrixOnGPU2D <<<(1024,1024),(16,16)>>> Achieved Occupancy 0.810691
从上面的输出可以得知两件事儿:
对于C[idx] = A[idx] + B[idx]来说共有三个memory操作:两个memory load和一个memory store。要查看这些操作的效率可以使用nvprof的两个metric参数,如果想要查看memory的throughput,则可使用gld_throughput:
$ nvprof --metrics gld_throughput./sumMatrix 32 32
sumMatrixOnGPU2D <<<(512,512), (32,32)>>> Global Load Throughput 35.908GB/s
$ nvprof --metrics gld_throughput./sumMatrix 32 16
sumMatrixOnGPU2D <<<(512,1024), (32,16)>>> Global Load Throughput 56.478GB/s
$ nvprof --metrics gld_throughput./sumMatrix 16 32
sumMatrixOnGPU2D <<<(1024,512), (16,32)>>> Global Load Throughput 85.195GB/s
$ nvprof --metrics gld_throughput./sumMatrix 16 16
sumMatrixOnGPU2D <<<(1024,1024),(16,16)>>> Global Load Throughput 94.708GB/s
不难看到,第四个拥有最高的load throughput,但是却比第二个慢(第二个也就是第四个的一半),所以,较高的load throughput也不一定就有较高的性能。之后讲到memory transaction时会具体分析这种现象的原因,简单说,就是高load throughput有可能是一种假象,如果需要的数据在memory中存储格式未对齐不连续,会导致许多额外的不必要的load操作,所以本文中的efficiency会这么低。
然后,我们可以使用nvprof的gld_efficiency来度量load efficiency,该metric参数是指我们确切需要的global load throughput与实际得到global load memory的比值。这个metric参数可以让我们知道,APP的load操作利用device memory bandwidth的程度:
$ nvprof --metrics gld_efficiency ./sumMatrix 32 32
sumMatrixOnGPU2D <<<(512,512), (32,32)>>> Global Memory Load Efficiency 100.00%
$ nvprof --metrics gld_efficiency ./sumMatrix 32 16
sumMatrixOnGPU2D <<<(512,1024), (32,16)>>> Global Memory Load Efficiency 100.00%
$ nvprof --metrics gld_efficiency ./sumMatrix 16 32
sumMatrixOnGPU2D <<<(1024,512), (16,32)>>> Global Memory Load Efficiency 49.96%
$ nvprof --metrics gld_efficiency ./sumMatrix 16 16
sumMatrixOnGPU2D <<<(1024,1024),(16,16)>>> Global Memory Load Efficiency 49.80%
从上述结果可知,最后两个的load efficiency只是前两个的一半。这也可以解释,为什么较高的throughput和较高的Occupancy却没有产生较好的性能。尽管最后两个的load操作数目要多不少(因为二者throughput较高),但是他们的load effecitiveness却低不少(由于efficiency较低)。
观察最后两个可以发现,他们block的x象限配置是warp的一半,由前文推测知,该象限应该保持为warp大小的整数倍。关于其具体原因将在后续博文详细解释。
我们现在可以得出一个结论就是blockDim.x应该是warp大小的整数倍。这样做是很容易就提升了load efficiency。现在,我们可能还有其他疑惑,比如:
现在,我们重新整一个基准数据出来,这两个问题可以从这个基准分析个大概:
$ ./sumMatrix 64 2
sumMatrixOnGPU2D <<<(256,8192), (64,2) >>> elapsed 0.033567 sec
$ ./sumMatrix 64 4
sumMatrixOnGPU2D <<<(256,4096), (64,4) >>> elapsed 0.034908 sec
$ ./sumMatrix 64 8
sumMatrixOnGPU2D <<<(256,2048), (64,8) >>> elapsed 0.036651 sec
$ ./sumMatrix 128 2
sumMatrixOnGPU2D <<<(128,8192), (128,2)>>> elapsed 0.032688 sec
$ ./sumMatrix 128 4
sumMatrixOnGPU2D <<<(128,4096), (128,4)>>> elapsed 0.034786 sec
$ ./sumMatrix 128 8
sumMatrixOnGPU2D <<<(128,2048), (128,8)>>> elapsed 0.046157 sec
$ ./sumMatrix 256 2
sumMatrixOnGPU2D <<<(64,8192), (256,2)>>> elapsed 0.032793 sec
$ ./sumMatrix 256 4
sumMatrixOnGPU2D <<<(64,4096), (256,4)>>> elapsed 0.038092 sec
$ ./sumMatrix 256 8
sumMatrixOnGPU2D <<<(64,2048), (256,8)>>> elapsed 0.000173 sec
Error: sumMatrix.cu:163, code:9, reason: invalid configuration argument
从上面数据,我们能够分析出来的是:
现在,我们又得猜测了,拥有block最少的应该会有一个最低的achieved Occupancy吧?而拥有最多block的应该会达到最高的achieved Occupancy吧?为了验证这些想法,我们再看一组数据:
$ nvprof --metrics achieved_occupancy ./sumMatrix 64 2
sumMatrixOnGPU2D <<<(256,8192), (64,2) >>> Achieved Occupancy 0.554556
$ nvprof --metrics achieved_occupancy ./sumMatrix 64 4
sumMatrixOnGPU2D <<<(256,4096), (64,4) >>> Achieved Occupancy 0.798622
$ nvprof --metrics achieved_occupancy ./sumMatrix 64 8
sumMatrixOnGPU2D <<<(256,2048), (64,8) >>> Achieved Occupancy 0.753532
$ nvprof --metrics achieved_occupancy ./sumMatrix 128 2
sumMatrixOnGPU2D <<<(128,8192), (128,2)>>> Achieved Occupancy 0.802598
$ nvprof --metrics achieved_occupancy ./sumMatrix 128 4
sumMatrixOnGPU2D <<<(128,4096), (128,4)>>> Achieved Occupancy 0.746367
$ nvprof --metrics achieved_occupancy ./sumMatrix 128 8
sumMatrixOnGPU2D <<<(128,2048), (128,8)>>> Achieved Occupancy 0.573449
$ nvprof --metrics achieved_occupancy ./sumMatrix 256 2
sumMatrixOnGPU2D <<<(64,8192), (256,2) >>> Achieved Occupancy 0.760901
$ nvprof --metrics achieved_occupancy ./sumMatrix 256 4
sumMatrixOnGPU2D <<<(64,4096), (256,4) >>> Achieved Occupancy 0.595197
看到了吧,(64,2)的achieved Occupancy竟然是最低的,尽管他有最多的block(高中做物理题也是这感觉),它达到了硬件对block数量的限制。
第四个(128,2)和第七个(256,2)拥有拥有差不多的achieved Occupancy。我们对这两个再做一个试验,再次增大,将blockDim.y设置为1,这也减少了block的大小:
$ ./sumMatrix 128 1
sumMatrixOnGPU2D <<<(128,16384),(128,1)>>> elapsed 0.032602 sec
$ ./sumMatrix 256 1
sumMatrixOnGPU2D <<<(64,16384), (256,1)>>> elapsed 0.030959 sec
这次配置产生了最佳的性能,特别是,(256,1)要比(128,1)要更好,,再次检查achieved Occupancy,load throughput和load efficiency:
$ nvprof --metrics achieved_occupancy ./sumMatrix 256 1
$ nvprof --metrics gld_throughput ./sumMatrix 256 1
$ nvprof --metrics gld_efficiency ./sumMatrix 256 1
输出:
Achieved Occupancy 0.808622
Global Load Throughput 69.762GB/s
Global Memory Load Efficiency 100.00%
现在可以看出,最佳配置既不是拥有最高achieved Occupancy也不是最高load throughput的。所以不存在唯一metric来优化计算性能,我么需要从众多metric中寻求一个平衡。
总结