On Wed, 25 Aug 2010 03:25:25 -0700 (PDT) David Rientjes <rientjes@xxxxxxxxxx> wrote: > > > 3) No reason to implement ABI breakage. > > old tuning parameter mean) > > oom-score = oom-base-score x 2^oom_adj > > Everybody knows this is useless beyond polarizing a task for kill or > making it immune. > > > new tuning parameter mean) > > oom-score = oom-base-score + oom_score_adj / (totalram + totalswap) > > This, on the other hand, has an actual unit (proportion of available > memory) that can be used to prioritize tasks amongst those competing for > the same set of shared resources and remains constant even when a task > changes cpuset, its memcg limit changes, etc. > > And your equation is wrong, it's > > ((rss + swap) / (available ram + swap)) + oom_score_adj > > which is completely different from what you think it is. > I'm now trying to write a userspace tool to calculate this, for me. Then, could you update documentation ? == 3.2 /proc/<pid>/oom_score - Display current oom-killer score ------------------------------------------------------------- This file can be used to check the current score used by the oom-killer is for any given <pid>. Use it together with /proc/<pid>/oom_adj to tune which process should be killed in an out-of-memory situation. == add a some documentation like: == (For system monitoring tool developpers, not for usual users.) oom_score calculation is implemnentation dependent and can be modified without any caution. But current logic is oom_score = ((proc's rss + proc's swap) / (available ram + swap)) + oom_score_adj proc's rss and swap can be obtained by /proc/<pid>/statm and available ram + swap is dependent on the situation. If the system is totaly under oom, available ram == /proc/meminfo's MemTotal available swap == in most case == /proc/meminfo's SwapTotal When you use memory cgroup, When swap is limited, avaliable ram + swap == memory cgroup's memsw limit. When swap is unlimited, avaliable ram + swap = memory cgroup's memory limit + SwapTotal Then, please be careful that oom_score's order among tasks depends on the situation. Assume 2 proceses A, B which has oom_score_adj of 300 and 0 And A uses 200M, B uses 1G of memory under 4G system Under the 4G system. A's socre = (200M *1000)/4G + 300 = 350 B's score = (1G * 1000)/4G = 250. In the memory cgroup, it has 2G of resource. A's score = (200M * 1000)/2G + 300 = 400 B's socre = (1G * 1000)/2G = 500 You shoudn't depend on /proc/<pid>/oom_score if you have to handle OOM under cgroups and cpuset. But the logic is simple. == If you don't want, I'll add text and a sample tool to cgroup/memory.txt. Thanks, -Kame -- To unsubscribe, send a message with 'unsubscribe linux-mm' in the body to majordomo@xxxxxxxxxx For more info on Linux MM, see: http://www.linux-mm.org/ . Don't email: <a href=mailto:"dont@xxxxxxxxx";> email@xxxxxxxxx </a>