Workloads consisting of a large number of processes running the same program
with a very large shared data section may experience performance problems
when numa balancing attempts to migrate the shared cow pages. This manifests
itself with many processes or tasks in TASK_UNINTERRUPTIBLE state waiting
for the shared pages to be migrated.
This patch changes change_pte_range() to skip shared copy-on-write pages when
called from change_prot_numa().
The program listed below simulates the conditions with these results when
run with 288 processes on a 144 core/8 socket machine.
Average throughput Average throughput Average throughput
with numa_balancing=0 with numa_balancing=1 with numa_balancing=1
without the patch with the patch
--------------------- --------------------- ---------------------
2118782 2021534 2107979
Complex production environments show less variability and fewer poorly
performing outliers accompanied with a smaller number of processes waiting
on NUMA page migration. In some cases, %iowait drops from 16%-26% to 0.
// SPDX-License-Identifier: GPL-2.0
/*
* Copyright (c) 2017 Oracle and/or its affiliates. All rights reserved.
*/
#include <sys/time.h>
#include <stdio.h>
#include <wait.h>
#include <sys/mman.h>
int a[1000000] = {13};
int main(int argc, const char **argv)
{
int n = 0;
int i;
pid_t pid;
int stat;
int *count_array;
int cpu_count = 288;
long total = 0;
struct timeval t1, t2 = {(argc > 1 ? atoi(argv[1]) : 10), 0};
if (argc > 2)
cpu_count = atoi(argv[2]);
count_array = mmap(NULL, cpu_count * sizeof(int),
(PROT_READ|PROT_WRITE),
(MAP_SHARED|MAP_ANONYMOUS), 0, 0);
if (count_array == MAP_FAILED) {
perror("mmap:");
return 0;
}
for (i = 0; i < cpu_count; ++i) {
pid = fork();
if (pid <= 0)
break;
if ((i & 0xf) == 0)
usleep(2);
}
if (pid != 0) {
if (i == 0) {
perror("fork:");
return 0;
}
for (;;) {
pid = wait(&stat);
if (pid < 0)
break;
}
for (i = 0; i < cpu_count; ++i)
total += count_array[i];
printf("Total %ld\n", total);
munmap(count_array, cpu_count * sizeof(int));
return 0;
}
gettimeofday(&t1, 0);
timeradd(&t1, &t2, &t1);
while (timercmp(&t2, &t1, <)) {
int b = 0;
int j;
for (j = 0; j < 1000000; j++)
b += a[j];
gettimeofday(&t2, 0);
n++;
}
count_array[i] = n;
return 0;
}
mm/mprotect.c | 5 +++++
1 file changed, 5 insertions(+)
--
1.8.3.1
