The numbers are not easy to derive in a closed form (certainly mere protections ratios do not apply), therefore use a simulation to obtain expected numbers.
Signed-off-by: Michal Koutný mkoutny@suse.com --- MAINTAINERS | 1 + .../selftests/cgroup/memcg_protection.m | 89 +++++++++++++++++++ .../selftests/cgroup/test_memcontrol.c | 29 +++--- 3 files changed, 107 insertions(+), 12 deletions(-) create mode 100644 tools/testing/selftests/cgroup/memcg_protection.m
diff --git a/MAINTAINERS b/MAINTAINERS index 78c57046fa93..b28b6aeb8636 100644 --- a/MAINTAINERS +++ b/MAINTAINERS @@ -5029,6 +5029,7 @@ L: linux-mm@kvack.org S: Maintained F: mm/memcontrol.c F: mm/swap_cgroup.c +F: tools/testing/selftests/cgroup/memcg_protection.m F: tools/testing/selftests/cgroup/test_kmem.c F: tools/testing/selftests/cgroup/test_memcontrol.c
diff --git a/tools/testing/selftests/cgroup/memcg_protection.m b/tools/testing/selftests/cgroup/memcg_protection.m new file mode 100644 index 000000000000..051daa3477b6 --- /dev/null +++ b/tools/testing/selftests/cgroup/memcg_protection.m @@ -0,0 +1,89 @@ +% SPDX-License-Identifier: GPL-2.0 +% +% run as: octave-cli memcg_protection.m +% +% This script simulates reclaim protection behavior on a single level of memcg +% hierarchy to illustrate how overcommitted protection spreads among siblings +% (as it depends also on their current consumption). +% +% Simulation assumes siblings consumed the initial amount of memory (w/out +% reclaim) and then the reclaim starts, all memory is reclaimable, i.e. treated +% same. It simulates only non-low reclaim and assumes all memory.min = 0. +% +% Input configurations +% -------------------- +% E number parent effective protection +% n vector nominal protection of siblings set at the given level (memory.low) +% c vector current consumption -,,- (memory.current) + +% example from testcase (values in GB) +E = 50 / 1024; +n = [75 25 0 500 ] / 1024; +c = [50 50 50 0] / 1024; + +% Reclaim parameters +% ------------------ + +% Minimal reclaim amount (GB) +cluster = 32*4 / 2**20; + +% Reclaim coefficient (think as 0.5^sc->priority) +alpha = .1 + +% Simulation parameters +% --------------------- +epsilon = 1e-7; +timeout = 1000; + +% Simulation loop +% --------------- + +ch = []; +eh = []; +rh = []; + +for t = 1:timeout + % low_usage + u = min(c, n); + siblings = sum(u); + + % effective_protection() + protected = min(n, c); % start with nominal + e = protected * min(1, E / siblings); % normalize overcommit + + % recursive protection + unclaimed = max(0, E - siblings); + parent_overuse = sum(c) - siblings; + if (unclaimed > 0 && parent_overuse > 0) + overuse = max(0, c - protected); + e += unclaimed * (overuse / parent_overuse); + endif + + % get_scan_count() + r = alpha * c; % assume all memory is in a single LRU list + + % commit 1bc63fb1272b ("mm, memcg: make scan aggression always exclude protection") + sz = max(e, c); + r .*= (1 - (e+epsilon) ./ (sz+epsilon)); + + % uncomment to debug prints + % e, c, r + + % nothing to reclaim, reached equilibrium + if max(r) < epsilon + break; + endif + + % SWAP_CLUSTER_MAX roundup + r = max(r, (r > epsilon) .* cluster); + % XXX here I do parallel reclaim of all siblings + % in reality reclaim is serialized and each sibling recalculates own residual + c = max(c - r, 0); + + ch = [ch ; c]; + eh = [eh ; e]; + rh = [rh ; r]; +endfor + +t +c, e diff --git a/tools/testing/selftests/cgroup/test_memcontrol.c b/tools/testing/selftests/cgroup/test_memcontrol.c index 4924425639b0..dc2c7d6e3572 100644 --- a/tools/testing/selftests/cgroup/test_memcontrol.c +++ b/tools/testing/selftests/cgroup/test_memcontrol.c @@ -248,7 +248,7 @@ static int cg_test_proc_killed(const char *cgroup) /* * First, this test creates the following hierarchy: * A memory.min = 50M, memory.max = 200M - * A/B memory.min = 50M, memory.current = 50M + * A/B memory.min = 50M * A/B/C memory.min = 75M, memory.current = 50M * A/B/D memory.min = 25M, memory.current = 50M * A/B/E memory.min = 0, memory.current = 50M @@ -259,10 +259,13 @@ static int cg_test_proc_killed(const char *cgroup) * Then it creates A/G and creates a significant * memory pressure in it. * + * Then it checks actual memory usages and expects that: * A/B memory.current ~= 50M - * A/B/C memory.current ~= 33M - * A/B/D memory.current ~= 17M - * A/B/F memory.current ~= 0 + * A/B/C memory.current ~= 29M + * A/B/D memory.current ~= 21M + * A/B/E memory.current ~= 0 + * A/B/F memory.current = 0 + * (for origin of the numbers, see model in memcg_protection.m.) * * After that it tries to allocate more than there is * unprotected memory in A available, and checks @@ -365,10 +368,10 @@ static int test_memcg_min(const char *root) for (i = 0; i < ARRAY_SIZE(children); i++) c[i] = cg_read_long(children[i], "memory.current");
- if (!values_close(c[0], MB(33), 10)) + if (!values_close(c[0], MB(29), 10)) goto cleanup;
- if (!values_close(c[1], MB(17), 10)) + if (!values_close(c[1], MB(21), 10)) goto cleanup;
if (c[3] != 0) @@ -405,7 +408,7 @@ static int test_memcg_min(const char *root) /* * First, this test creates the following hierarchy: * A memory.low = 50M, memory.max = 200M - * A/B memory.low = 50M, memory.current = 50M + * A/B memory.low = 50M * A/B/C memory.low = 75M, memory.current = 50M * A/B/D memory.low = 25M, memory.current = 50M * A/B/E memory.low = 0, memory.current = 50M @@ -417,9 +420,11 @@ static int test_memcg_min(const char *root) * * Then it checks actual memory usages and expects that: * A/B memory.current ~= 50M - * A/B/ memory.current ~= 33M - * A/B/D memory.current ~= 17M - * A/B/F memory.current ~= 0 + * A/B/C memory.current ~= 29M + * A/B/D memory.current ~= 21M + * A/B/E memory.current ~= 0 + * A/B/F memory.current = 0 + * (for origin of the numbers, see model in memcg_protection.m.) * * After that it tries to allocate more than there is * unprotected memory in A available, @@ -512,10 +517,10 @@ static int test_memcg_low(const char *root) for (i = 0; i < ARRAY_SIZE(children); i++) c[i] = cg_read_long(children[i], "memory.current");
- if (!values_close(c[0], MB(33), 10)) + if (!values_close(c[0], MB(29), 10)) goto cleanup;
- if (!values_close(c[1], MB(17), 10)) + if (!values_close(c[1], MB(21), 10)) goto cleanup;
if (c[3] != 0)
On Wed, May 18, 2022 at 06:18:57PM +0200, Michal Koutny wrote:
The numbers are not easy to derive in a closed form (certainly mere protections ratios do not apply), therefore use a simulation to obtain expected numbers.
Signed-off-by: Michal Koutný mkoutny@suse.com
Acked-by: Roman Gushchin roman.gushchin@linux.dev
Thanks, Michal!
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