Re: [PATCH 04/19] sched, numa, mm: Describe the NUMA schedulingproblem formally

[abhishek agarwal]

as per 4) move towards where "most" memory. If we have a large shared
memory than private memnory. Why not we just move the process towrds
the memory.. instead of the memory moving towards the node. This will
i guess be less cumbersome, then moving all the shared memory

On Fri, Nov 16, 2012 at 9:55 PM, Ingo Molnar <mingo@xxxxxxxxxx> wrote:
> From: Peter Zijlstra <a.p.zijlstra@xxxxxxxxx>
>
> 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@xxxxxxxxx>
> Cc: Linus Torvalds <torvalds@xxxxxxxxxxxxxxxxxxxx>
> Cc: Andrew Morton <akpm@xxxxxxxxxxxxxxxxxxxx>
> Cc: Peter Zijlstra <a.p.zijlstra@xxxxxxxxx>
> Cc: "H. Peter Anvin" <hpa@xxxxxxxxx>
> Cc: Mike Galbraith <efault@xxxxxx>
> Rik van Riel <riel@xxxxxxxxxx>
> Link: http://lkml.kernel.org/n/tip-mmnlpupoetcatimvjEld16Pb@xxxxxxxxxxxxxx
> [ Next step: generate the kernel source from such formal descriptions and retire to a tropical island! ]
> Signed-off-by: Ingo Molnar <mingo@xxxxxxxxxx>
> ---
>  Documentation/scheduler/numa-problem.txt | 230 +++++++++++++++++++++++++++++++
>  1 file changed, 230 insertions(+)
>  create mode 100644 Documentation/scheduler/numa-problem.txt
>
> diff --git a/Documentation/scheduler/numa-problem.txt b/Documentation/scheduler/numa-problem.txt
> new file mode 100644
> index 0000000..a5d2fee
> --- /dev/null
> +++ b/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.
> --
> 1.7.11.7
>
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