Question

我正在测试台式机和服务器上的内存带宽。

Sklyake desktop 4 cores/8 hardware threads
Skylake server Xeon 8168 dual socket 48 cores (24 per socket) / 96 hardware threads

系统的峰值带宽是

Peak bandwidth desktop = 2-channels*8*2400 = 38.4 GB/s
Peak bandwidth server  = 6-channels*2-sockets*8*2666 = 255.94 GB/s

我正在使用自己的triad function from STREAM来测量带宽（稍后将提供完整代码）

void triad(double *a, double *b, double *c, double scalar, size_t n) {
  #pragma omp parallel for
  for(int i=0; i<n; i++) a[i] = b[i] + scalar*c[i];
}

这是我得到的结果

         Bandwidth (GB/s)
threads  Desktop  Server         
1             28      16
2(24)         29     146
4(48)         25     177
8(96)         24     189

对于1个线程，我不明白为什么台式机比服务器快得多。根据这个答案https://stackoverflow.com/a/18159503/2542702，SSE足以获得双通道系统的全部带宽。这就是我在桌面上观察到的。两个线程只会稍微有所帮助，而4和8个线程会产生较差的结果，但是在服务器上，单线程带宽要少得多。 这是为什么？

在服务器上，使用96个线程可获得最佳结果。我本来会以更少的线程饱和它。 为什么要使用这么多线程来饱和服务器上的带宽？结果中存在很大的误差，并且我不包括误差估计。我获得了几次跑步的最佳成绩。

代码

//gcc -O3 -march=native triad.c -fopenmp
//gcc -O3 -march=skylake-avx512 -mprefer-vector-width=512 triad.c -fopenmp
#include <stdio.h>
#include <omp.h>
#include <x86intrin.h>

void triad_init(double *a, double *b, double *c, double k, size_t n) {
  #pragma omp parallel for
  for(size_t i=0; i<n; i++) a[i] = k, b[i] = k, c[i] = k;
}

void triad(double *a, double *b, double *c, double scalar, size_t n) {
  #pragma omp parallel for
  for(size_t i=0; i<n; i++) a[i] = b[i] + scalar*c[i];
}

void triad_stream(double *a, double *b, double *c, double scalar, size_t n) {
#if defined ( __AVX512F__ ) || defined ( __AVX512__ )
  __m512d scalarv = _mm512_set1_pd(scalar);
  #pragma omp parallel for
  for(size_t i=0; i<n/8; i++) {
    __m512d bv = _mm512_load_pd(&b[8*i]), cv = _mm512_load_pd(&c[8*i]);
    _mm512_stream_pd(&a[8*i], _mm512_add_pd(bv, _mm512_mul_pd(scalarv, cv)));
  }        
#else
  __m256d scalarv = _mm256_set1_pd(scalar);
  #pragma omp parallel for
  for(size_t i=0; i<n/4; i++) {
    __m256d bv = _mm256_load_pd(&b[4*i]), cv = _mm256_load_pd(&c[4*i]);
    _mm256_stream_pd(&a[4*i], _mm256_add_pd(bv, _mm256_mul_pd(scalarv, cv)));
  }        
#endif
}

int main(void) {
  size_t n = 1LL << 31LL; 
  double *a = _mm_malloc(sizeof *a * n, 64), *b = _mm_malloc(sizeof *b * n, 64), *c = _mm_malloc(sizeof *c * n, 64);
  //double peak_bw = 2*8*2400*1E-3; // 2-channels*8-bits/byte*2400MHz
  double peak_bw = 2*6*8*2666*1E-3; // 2-sockets*6-channels*8-bits/byte*2666MHz
  double dtime, mem, bw;
  printf("peak bandwidth %.2f GB/s\n", peak_bw);

  triad_init(a, b, c, 3.14159, n);
  dtime = -omp_get_wtime();
  triad(a, b, c, 3.14159, n);  
  dtime += omp_get_wtime();
  mem = 4*sizeof(double)*n*1E-9, bw = mem/dtime;
  printf("triad:       %3.2f GB, %3.2f s, %8.2f GB/s, bw/peak_bw %8.2f %%\n", mem, dtime, bw, 100*bw/peak_bw);

  triad_init(a, b, c, 3.14159, n);
  dtime = -omp_get_wtime();
  triad_stream(a, b, c, 3.14159, n);  
  dtime += omp_get_wtime();
  mem = 3*sizeof(double)*n*1E-9, bw = mem/dtime;
  printf("triads:      %3.2f GB, %3.2f s, %8.2f GB/s, bw/peak_bw %8.2f %%\n", mem, dtime, bw, 100*bw/peak_bw);
}

许多通道x86系统的内存带宽

0 个答案: