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我是 Cuda 的新手。我试图以 Ricky 动量的形式求解具有初始条件的波动方程。代码的性能是 12 GFlops，虽然我的 GPU 性能是 3900。为什么代码对我来说如此无效，我该如何解决？

主文件

#include <iostream>
#include <cmath>
#include "step.cu"
#include <cuda.h>
#include "err.cu"
#include "err.h"
using namespace std;
int main(int argc, char const *argv[])
{
        if (argc <= 3)
        {
                perror("Error in argc: argc<=3 (wait h, tau, C) \n");
                exit(1);
        }

  char *eptr;
  errno = 0;

  long long int size,tmax;
  double tau,cour,h,C, cour2;

  h = std::strtod(argv[1], &eptr);
  tau = std::strtod(argv[2], &eptr);
  C = std::strtod(argv[3], &eptr);

  tmax = 2000;
  cour = C*tau/h;
  cour2 = cour* cour;
  size = 18*13*1024;

  double *nxt_layer=nullptr;
  double *layer_1=nullptr;
  double *layer_2=nullptr;
  double *rev_layer=nullptr;

  dim3 blockSize = dim3(1024);
  dim3 gridSize = dim3(size/blockSize.x);

  float time;
  cudaTimer timer;

  cudaError_t ret = cudaMallocManaged(&nxt_layer, sizeof(double) * size);

  if (ret != cudaSuccess)
  {
    std::cout << cudaGetErrorString(ret) << std::endl;
    return 1;
  }
  ret = cudaMallocManaged(&layer_1, sizeof(double) * size);

 if (ret != cudaSuccess)
 {
    std::cout << cudaGetErrorString(ret) << std::endl;
    return 1;
 }

 ret = cudaMallocManaged(&layer_2, sizeof(double) * size);

 if (ret != cudaSuccess)
 {
    std::cout << cudaGetErrorString(ret) << std::endl;
    return 1;
 }

  for (int i = 0; i < size; ++i)
  {
    layer_1[i] = exp(-(i*h-7)*(i*h-7)/2)*((i*h-7)*(i*h-7)-1);
  }
  for (int i = 1; i < size/2; ++i)
  {
    nxt_layer[i] = layer_1[i+1]+0.5*cour2*(layer_1[i+1]-2*layer_1[i]+layer_1[i-1]);
  }

  nxt_layer[0] = 0; nxt_layer[size-1] = 0;

  for (int i = size/2; i < size-1; ++i)
  {
    nxt_layer[i] = layer_1[i+1]+0.25*0.5*cour2*(layer_1[i+1]-2*layer_1[i]+layer_1[i-1]);
  }

  for (int i = 0; i < size-1; ++i)
  {
    layer_2[i] = layer_1[i];
    layer_1[i] = nxt_layer[i];
  }

  nxt_layer[0] = 0; nxt_layer[size-1] = 0;

  timer.start();
  for (double t = 0; t < tmax; t=t+tau)
  {
         step<<<gridSize, blockSize>>>(nxt_layer, layer_1, layer_2, cour2, size);
         if (CHECK_ERROR(cudaDeviceSynchronize()))
                throw(-1);
         nxt_layer[size-1]=0;
         nxt_layer[0]=0;
  }
  time = timer.stop();

  for (int i = 0; i < size; ++i)
  {
          cout<<i*h<<" "<<nxt_layer[i]<<endl;
  }

}

步骤.cu

inline __device__ double compute(double *layer_1_tmp, double layer_2_tmp, double cour2)
{
        return __fmaf_rd(cour2, layer_1_tmp[0]+layer_1_tmp[2], __fmaf_rd(2.0-2*cour2,layer_1_tmp[1],-layer_2_tmp));
}

__global__ void step(double *tmp_layer, double *layer_1, double *layer_2, double cour2, int Nx)
{
        int node = threadIdx.x + blockDim.x * blockIdx.x;

        if(node >= Nx-1 || node<=0) return;

        double layer_1_tmp[3];

        layer_1_tmp[0]=layer_1[node-1];
        layer_1_tmp[1]=layer_1[node];
        layer_1_tmp[2]=layer_1[node+1];

        double layer_2_tmp=layer_2[node];

        if(node<=Nx/2)
        {
              tmp_layer[node] = compute(layer_1_tmp, layer_2_tmp, 0.25*cour2);
        }
        else
        {
               tmp_layer[node] = compute(layer_1_tmp, layer_2_tmp, cour2);
        }

        layer_2[node]=layer_1[node];
        layer_1[node]=tmp_layer[node];
}

我将 GFlops 计算为

long long int perfomance = size*tmax/tau;
long long int perftime = 1000*perfomance/time;
double gflops =(8*perfomance/time)/1000000;

我将不胜感激您的任何意见和提示。

4

2 回答 2

-1

在内核中，每个工作项只做几次乘法和加法。与每个 cuda 线程的内核启动开销和每个 layer_1 元素的内存访问延迟相比，这可以忽略不计。这相当于在内核时间的微秒内测量几纳秒。尝试围绕 compute() 函数调用进行时钟测量。它至少会给出一些“每次计算周期”测量，并且您可以在计算调用期间找到总体性能。

clock_t c1 = clock();
compute();
clock_t c2 = clock();
timings[node] = c2-c1;

即使这也不是真正的性能测量，因为当多个计算调用一个接一个地进行时，它没有考虑流水线。您可以在第一个计算调用之后添加另一个计算调用，并由于流水线和延迟隐藏而获得更高的性能。

于 2022-01-09T16:03:23.093 回答
-2

许多（更多面向消费者或半专业的）显卡具有比双精度性能更好的单精度性能。GTX 970 的单精度性能是其双精度性能的 32 倍。

将使用的数据类型从 double 更改为 float。

于 2021-12-23T16:50:16.690 回答