Messages in this thread Patch in this message | | | Date | Thu, 25 Oct 2012 14:16:19 +0200 | From | Peter Zijlstra <> | Subject | [PATCH 02/31] sched, numa, mm: Describe the NUMA scheduling problem formally |
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This is probably a first: formal description of a complex high-level computing problem, within the kernel source.
Signed-off-by: Peter Zijlstra <a.p.zijlstra@chello.nl> Cc: Linus Torvalds <torvalds@linux-foundation.org> Cc: Andrew Morton <akpm@linux-foundation.org> Cc: Peter Zijlstra <a.p.zijlstra@chello.nl> Cc: "H. Peter Anvin" <hpa@zytor.com> Cc: Mike Galbraith <efault@gmx.de> Rik van Riel <riel@redhat.com> [ Next step: generate the kernel source from such formal descriptions and retire to a tropical island! ] Signed-off-by: Ingo Molnar <mingo@kernel.org> --- Documentation/scheduler/numa-problem.txt | 230 +++++++++++++++++++++++++++++++ 1 file changed, 230 insertions(+) create mode 100644 Documentation/scheduler/numa-problem.txt
Index: tip/Documentation/scheduler/numa-problem.txt =================================================================== --- /dev/null +++ tip/Documentation/scheduler/numa-problem.txt @@ -0,0 +1,230 @@ + + +Effective NUMA scheduling problem statement, described formally: + + * minimize interconnect traffic + +For each task 't_i' we have memory, this memory can be spread over multiple +physical nodes, let us denote this as: 'p_i,k', the memory task 't_i' has on +node 'k' in [pages]. + +If a task shares memory with another task let us denote this as: +'s_i,k', the memory shared between tasks including 't_i' residing on node +'k'. + +Let 'M' be the distribution that governs all 'p' and 's', ie. the page placement. + +Similarly, lets define 'fp_i,k' and 'fs_i,k' resp. as the (average) usage +frequency over those memory regions [1/s] such that the product gives an +(average) bandwidth 'bp' and 'bs' in [pages/s]. + +(note: multiple tasks sharing memory naturally avoid duplicat accounting + because each task will have its own access frequency 'fs') + +(pjt: I think this frequency is more numerically consistent if you explicitly + restrict p/s above to be the working-set. (It also makes explicit the + requirement for <C0,M0> to change about a change in the working set.) + + Doing this does have the nice property that it lets you use your frequency + measurement as a weak-ordering for the benefit a task would receive when + we can't fit everything. + + e.g. task1 has working set 10mb, f=90% + task2 has working set 90mb, f=10% + + Both are using 9mb/s of bandwidth, but we'd expect a much larger benefit + from task1 being on the right node than task2. ) + +Let 'C' map every task 't_i' to a cpu 'c_i' and its corresponding node 'n_i': + + C: t_i -> {c_i, n_i} + +This gives us the total interconnect traffic between nodes 'k' and 'l', +'T_k,l', as: + + T_k,l = \Sum_i bp_i,l + bs_i,l + \Sum bp_j,k + bs_j,k where n_i == k, n_j == l + +And our goal is to obtain C0 and M0 such that: + + T_k,l(C0, M0) =< T_k,l(C, M) for all C, M where k != l + +(note: we could introduce 'nc(k,l)' as the cost function of accessing memory + on node 'l' from node 'k', this would be useful for bigger NUMA systems + + pjt: I agree nice to have, but intuition suggests diminishing returns on more + usual systems given factors like things like Haswell's enormous 35mb l3 + cache and QPI being able to do a direct fetch.) + +(note: do we need a limit on the total memory per node?) + + + * fairness + +For each task 't_i' we have a weight 'w_i' (related to nice), and each cpu +'c_n' has a compute capacity 'P_n', again, using our map 'C' we can formulate a +load 'L_n': + + L_n = 1/P_n * \Sum_i w_i for all c_i = n + +using that we can formulate a load difference between CPUs + + L_n,m = | L_n - L_m | + +Which allows us to state the fairness goal like: + + L_n,m(C0) =< L_n,m(C) for all C, n != m + +(pjt: It can also be usefully stated that, having converged at C0: + + | L_n(C0) - L_m(C0) | <= 4/3 * | G_n( U(t_i, t_j) ) - G_m( U(t_i, t_j) ) | + + Where G_n,m is the greedy partition of tasks between L_n and L_m. This is + the "worst" partition we should accept; but having it gives us a useful + bound on how much we can reasonably adjust L_n/L_m at a Pareto point to + favor T_n,m. ) + +Together they give us the complete multi-objective optimization problem: + + min_C,M [ L_n,m(C), T_k,l(C,M) ] + + + +Notes: + + - the memory bandwidth problem is very much an inter-process problem, in + particular there is no such concept as a process in the above problem. + + - the naive solution would completely prefer fairness over interconnect + traffic, the more complicated solution could pick another Pareto point using + an aggregate objective function such that we balance the loss of work + efficiency against the gain of running, we'd want to more or less suggest + there to be a fixed bound on the error from the Pareto line for any + such solution. + +References: + + http://en.wikipedia.org/wiki/Mathematical_optimization + http://en.wikipedia.org/wiki/Multi-objective_optimization + + +* warning, significant hand-waving ahead, improvements welcome * + + +Partial solutions / approximations: + + 1) have task node placement be a pure preference from the 'fairness' pov. + +This means we always prefer fairness over interconnect bandwidth. This reduces +the problem to: + + min_C,M [ T_k,l(C,M) ] + + 2a) migrate memory towards 'n_i' (the task's node). + +This creates memory movement such that 'p_i,k for k != n_i' becomes 0 -- +provided 'n_i' stays stable enough and there's sufficient memory (looks like +we might need memory limits for this). + +This does however not provide us with any 's_i' (shared) information. It does +however remove 'M' since it defines memory placement in terms of task +placement. + +XXX properties of this M vs a potential optimal + + 2b) migrate memory towards 'n_i' using 2 samples. + +This separates pages into those that will migrate and those that will not due +to the two samples not matching. We could consider the first to be of 'p_i' +(private) and the second to be of 's_i' (shared). + +This interpretation can be motivated by the previously observed property that +'p_i,k for k != n_i' should become 0 under sufficient memory, leaving only +'s_i' (shared). (here we loose the need for memory limits again, since it +becomes indistinguishable from shared). + +XXX include the statistical babble on double sampling somewhere near + +This reduces the problem further; we loose 'M' as per 2a, it further reduces +the 'T_k,l' (interconnect traffic) term to only include shared (since per the +above all private will be local): + + T_k,l = \Sum_i bs_i,l for every n_i = k, l != k + +[ more or less matches the state of sched/numa and describes its remaining + problems and assumptions. It should work well for tasks without significant + shared memory usage between tasks. ] + +Possible future directions: + +Motivated by the form of 'T_k,l', try and obtain each term of the sum, so we +can evaluate it; + + 3a) add per-task per node counters + +At fault time, count the number of pages the task faults on for each node. +This should give an approximation of 'p_i' for the local node and 's_i,k' for +all remote nodes. + +While these numbers provide pages per scan, and so have the unit [pages/s] they +don't count repeat access and thus aren't actually representable for our +bandwidth numberes. + + 3b) additional frequency term + +Additionally (or instead if it turns out we don't need the raw 'p' and 's' +numbers) we can approximate the repeat accesses by using the time since marking +the pages as indication of the access frequency. + +Let 'I' be the interval of marking pages and 'e' the elapsed time since the +last marking, then we could estimate the number of accesses 'a' as 'a = I / e'. +If we then increment the node counters using 'a' instead of 1 we might get +a better estimate of bandwidth terms. + + 3c) additional averaging; can be applied on top of either a/b. + +[ Rik argues that decaying averages on 3a might be sufficient for bandwidth since + the decaying avg includes the old accesses and therefore has a measure of repeat + accesses. + + Rik also argued that the sample frequency is too low to get accurate access + frequency measurements, I'm not entirely convinced, event at low sample + frequencies the avg elapsed time 'e' over multiple samples should still + give us a fair approximation of the avg access frequency 'a'. + + So doing both b&c has a fair chance of working and allowing us to distinguish + between important and less important memory accesses. + + Experimentation has shown no benefit from the added frequency term so far. ] + +This will give us 'bp_i' and 'bs_i,k' so that we can approximately compute +'T_k,l' Our optimization problem now reads: + + min_C [ \Sum_i bs_i,l for every n_i = k, l != k ] + +And includes only shared terms, this makes sense since all task private memory +will become local as per 2. + +This suggests that if there is significant shared memory, we should try and +move towards it. + + 4) move towards where 'most' memory is + +The simplest significance test is comparing the biggest shared 's_i,k' against +the private 'p_i'. If we have more shared than private, move towards it. + +This effectively makes us move towards where most our memory is and forms a +feed-back loop with 2. We migrate memory towards us and we migrate towards +where 'most' memory is. + +(Note: even if there were two tasks fully trashing the same shared memory, it + is very rare for there to be an 50/50 split in memory, lacking a perfect + split, the small will move towards the larger. In case of the perfect + split, we'll tie-break towards the lower node number.) + + 5) 'throttle' 4's node placement + +Since per 2b our 's_i,k' and 'p_i' require at least two scans to 'stabilize' +and show representative numbers, we should limit node-migration to not be +faster than this. + + n) poke holes in previous that require more stuff and describe it.
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