Question

我尝试从下面的代码中达到每个 SM 的最佳效果。峰值介于25 GFlops（GTX275-GT200 Arch。）之间。此代码最多可提供8个GFlops

__global__ void new_ker(float *x)
{
  int index = threadIdx.x+blockIdx.x*blockDim.x;
  float a,b;
  a=0;
  b=x[index];
  //LOOP=10000000
  //No. of blocks = 1
  //Threads per block = 512 (I'm using GTX 275 - GT200 Arch.)
  #pragma unroll 2048
  for(int i=0;i<LOOP;i++){
       a=a*b+b;
  }  

  x[index] = a;

 }

我不想在代码中增加ILP。任何想法为什么它没有达到峰值？

int main(int argc,char **argv)
{

   //Initializations
   float *x;
   float *dx;
   cudaEvent_t new_start,new_stop;
   float elapsed;
   double gflops;
   x = 0;
   flag = 0;
   cudaMalloc((void **)&dx,sizeof(float)*THPB);

   //ILP=1  
   cudaEventCreate(&new_start);
   cudaEventCreate(&new_stop);
   printf("Kernel1:\n");
   cudaEventRecord(new_start, 0);
   new_ker<<<BLOCKS,THPB>>>(dx);
   cudaEventRecord(new_stop,0);
   cudaEventSynchronize(new_stop);
   cudaEventElapsedTime(&elapsed,new_start,new_stop);
   x = (float *)malloc(sizeof(float)*THPB);
   cudaMemcpy(x,dx,sizeof(float)*THPB,cudaMemcpyDeviceToHost);

   gflops = ((double)(BLOCKS)*(THPB)*LOOP/elapsed)/1000000;
   printf("\t%f",gflops);
   cudaEventDestroy(new_start);
   cudaEventDestroy(new_stop);
   return 0;
}

平台： CUDA 3.0 NVIDIA GeForce GTX275（GT200）

Answer 1

如果我使用正确的FLOP计算从您的代码中整理出一个完整的repro案例：

#include <stdio.h> 

#define LOOP (10000000)
#define BLOCKS (30)
#define THPB (512)

__global__ void new_ker(float *x)
{
  int index = threadIdx.x+blockIdx.x*blockDim.x;
  float a,b;
  a=0;
  b=x[index];
  #pragma unroll 2048
  for(int i=0;i<LOOP;i++){
       a=a*b+b;
  }  

  x[index] = a;
}

int main(int argc,char **argv)
{

   //Initializations
   float *x;
   float *dx;
   cudaEvent_t new_start,new_stop;
   float elapsed;
   double gflops;
   x = 0;
   cudaMalloc((void **)&dx,sizeof(float)*THPB);

   //ILP=1  
   cudaEventCreate(&new_start);
   cudaEventCreate(&new_stop);
   printf("Kernel1:\n");
   cudaEventRecord(new_start, 0);
   new_ker<<<BLOCKS,THPB>>>(dx);
   cudaEventRecord(new_stop,0);
   cudaEventSynchronize(new_stop);
   cudaEventElapsedTime(&elapsed,new_start,new_stop);
   x = (float *)malloc(sizeof(float)*THPB*BLOCKS);
   cudaMemcpy(x,dx,sizeof(float)*THPB*BLOCKS,cudaMemcpyDeviceToHost);

   gflops = 2.0e-6 * ((double)(LOOP)*double(THPB*BLOCKS)/(double)elapsed);
   printf("\t%f\n",gflops);
   cudaEventDestroy(new_start);
   cudaEventDestroy(new_stop);
   return 0;
}

我编译它并在64位Linux平台上使用CUDA 3.2在1.4GHz GTX275上运行它：

$ nvcc -arch=sm_13 -Xptxas="-v" -o perf perf.cu
ptxas info    : Compiling entry function '_Z7new_kerPf' for 'sm_13'
ptxas info    : Used 4 registers, 8+16 bytes smem, 8 bytes cmem[1]
$ ./perf 
Kernel1:
        671.806039

对于运行纯FMAD码（1.4 GHz * 2 FLOP * 8核/ MP * 30 MP）= 672 GFLOP / s的卡，我得到峰值FLOP / s的0.01％以内。

所以看起来代码实际上每个多处理器只有一个块达到峰值FLOP / s，但你只是没有正确计算FLOP / s数。

无法达到最佳表现

1 个答案: