I'm testing the memory bandwidth on a desktop and a server.
Sklyake desktop 4 cores/8 hardware threads
Skylake server Xeon 8168 dual socket 48 cores (24 per socket) / 96 hardware threads
The peak bandwidth of the system is
Peak bandwidth desktop = 2-channels*8*2400 = 38.4 GB/s
Peak bandwidth server = 6-channels*2-sockets*8*2666 = 255.94 GB/s
I'm using my own triad function from STREAM to measure the bandwidth (full code later)
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];
}
Here are results I get
Bandwidth (GB/s)
threads Desktop Server
1 28 16
2(24) 29 146
4(48) 25 177
8(96) 24 189
For 1 thread I don't understand why the desktop is so much faster than the server. According to this answer https://stackoverflow.com/a/18159503/2542702 SSE is sufficient to get the full bandwidth of a dual channel system. That's what I observe on the desktop. Two threads only helps slightly and 4 and 8 threads give a worse result But on the server the single threaded bandwidth is much less. Why is this?
On the server I get the best results using 96 threads. I would have thought it would be saturated with far fewer threads. Why are so many threads necessary to saturate the bandwidth on the server? There is a large margin of error in my results and I don't include an error estimate. I took the best result of several runs.
The code
//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);
}
