多通道 x86系统的内存带宽

我在测试桌面和服务器的内存带宽。

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

我正在使用我自己的 来自 STREAM 的三角函数来测量带宽(稍后完整的代码)

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，我不明白为什么桌面比服务器快那么多。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。这就是我在桌面上看到的。两个线程只有轻微的帮助，4个和8个线程的结果更差，但在服务器上，单线程带宽要小得多。为什么会这样？

在服务器上，我使用96个线程得到了最好的结果。我还以为里面的线会少很多呢。我的结果有很大的误差范围，我不包括一个误差估计。我在几次比赛中取得了最好的成绩。

密码

//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);
}