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// SPDX-License-Identifier: GPL-2.0
// Copyright (C) 2016, Linaro Ltd - Daniel Lezcano <daniel.lezcano@linaro.org>
#include <linux/kernel.h>
#include <linux/percpu.h>
#include <linux/slab.h>
#include <linux/static_key.h>
#include <linux/interrupt.h>
#include <linux/idr.h>
#include <linux/irq.h>
#include <linux/math64.h>
#include <linux/log2.h>
#include <trace/events/irq.h>
#include "internals.h"
DEFINE_STATIC_KEY_FALSE(irq_timing_enabled);
DEFINE_PER_CPU(struct irq_timings, irq_timings);
static DEFINE_IDR(irqt_stats);
void irq_timings_enable(void)
{
static_branch_enable(&irq_timing_enabled);
}
void irq_timings_disable(void)
{
static_branch_disable(&irq_timing_enabled);
}
/*
* The main goal of this algorithm is to predict the next interrupt
* occurrence on the current CPU.
*
* Currently, the interrupt timings are stored in a circular array
* buffer every time there is an interrupt, as a tuple: the interrupt
* number and the associated timestamp when the event occurred <irq,
* timestamp>.
*
* For every interrupt occurring in a short period of time, we can
* measure the elapsed time between the occurrences for the same
* interrupt and we end up with a suite of intervals. The experience
* showed the interrupts are often coming following a periodic
* pattern.
*
* The objective of the algorithm is to find out this periodic pattern
* in a fastest way and use its period to predict the next irq event.
*
* When the next interrupt event is requested, we are in the situation
* where the interrupts are disabled and the circular buffer
* containing the timings is filled with the events which happened
* after the previous next-interrupt-event request.
*
* At this point, we read the circular buffer and we fill the irq
* related statistics structure. After this step, the circular array
* containing the timings is empty because all the values are
* dispatched in their corresponding buffers.
*
* Now for each interrupt, we can predict the next event by using the
* suffix array, log interval and exponential moving average
*
* 1. Suffix array
*
* Suffix array is an array of all the suffixes of a string. It is
* widely used as a data structure for compression, text search, ...
* For instance for the word 'banana', the suffixes will be: 'banana'
* 'anana' 'nana' 'ana' 'na' 'a'
*
* Usually, the suffix array is sorted but for our purpose it is
* not necessary and won't provide any improvement in the context of
* the solved problem where we clearly define the boundaries of the
* search by a max period and min period.
*
* The suffix array will build a suite of intervals of different
* length and will look for the repetition of each suite. If the suite
* is repeating then we have the period because it is the length of
* the suite whatever its position in the buffer.
*
* 2. Log interval
*
* We saw the irq timings allow to compute the interval of the
* occurrences for a specific interrupt. We can reasonibly assume the
* longer is the interval, the higher is the error for the next event
* and we can consider storing those interval values into an array
* where each slot in the array correspond to an interval at the power
* of 2 of the index. For example, index 12 will contain values
* between 2^11 and 2^12.
*
* At the end we have an array of values where at each index defines a
* [2^index - 1, 2 ^ index] interval values allowing to store a large
* number of values inside a small array.
*
* For example, if we have the value 1123, then we store it at
* ilog2(1123) = 10 index value.
*
* Storing those value at the specific index is done by computing an
* exponential moving average for this specific slot. For instance,
* for values 1800, 1123, 1453, ... fall under the same slot (10) and
* the exponential moving average is computed every time a new value
* is stored at this slot.
*
* 3. Exponential Moving Average
*
* The EMA is largely used to track a signal for stocks or as a low
* pass filter. The magic of the formula, is it is very simple and the
* reactivity of the average can be tuned with the factors called
* alpha.
*
* The higher the alphas are, the faster the average respond to the
* signal change. In our case, if a slot in the array is a big
* interval, we can have numbers with a big difference between
* them. The impact of those differences in the average computation
* can be tuned by changing the alpha value.
*
*
* -- The algorithm --
*
* We saw the different processing above, now let's see how they are
* used together.
*
* For each interrupt:
* For each interval:
* Compute the index = ilog2(interval)
* Compute a new_ema(buffer[index], interval)
* Store the index in a circular buffer
*
* Compute the suffix array of the indexes
*
* For each suffix:
* If the suffix is reverse-found 3 times
* Return suffix
*
* Return Not found
*
* However we can not have endless suffix array to be build, it won't
* make sense and it will add an extra overhead, so we can restrict
* this to a maximum suffix length of 5 and a minimum suffix length of
* 2. The experience showed 5 is the majority of the maximum pattern
* period found for different devices.
*
* The result is a pattern finding less than 1us for an interrupt.
*
* Example based on real values:
*
* Example 1 : MMC write/read interrupt interval:
*
* 223947, 1240, 1384, 1386, 1386,
* 217416, 1236, 1384, 1386, 1387,
* 214719, 1241, 1386, 1387, 1384,
* 213696, 1234, 1384, 1386, 1388,
* 219904, 1240, 1385, 1389, 1385,
* 212240, 1240, 1386, 1386, 1386,
* 214415, 1236, 1384, 1386, 1387,
* 214276, 1234, 1384, 1388, ?
*
* For each element, apply ilog2(value)
*
* 15, 8, 8, 8, 8,
* 15, 8, 8, 8, 8,
* 15, 8, 8, 8, 8,
* 15, 8, 8, 8, 8,
* 15, 8, 8, 8, 8,
* 15, 8, 8, 8, 8,
* 15, 8, 8, 8, 8,
* 15, 8, 8, 8, ?
*
* Max period of 5, we take the last (max_period * 3) 15 elements as
* we can be confident if the pattern repeats itself three times it is
* a repeating pattern.
*
* 8,
* 15, 8, 8, 8, 8,
* 15, 8, 8, 8, 8,
* 15, 8, 8, 8, ?
*
* Suffixes are:
*
* 1) 8, 15, 8, 8, 8 <- max period
* 2) 8, 15, 8, 8
* 3) 8, 15, 8
* 4) 8, 15 <- min period
*
* From there we search the repeating pattern for each suffix.
*
* buffer: 8, 15, 8, 8, 8, 8, 15, 8, 8, 8, 8, 15, 8, 8, 8
* | | | | | | | | | | | | | | |
* 8, 15, 8, 8, 8 | | | | | | | | | |
* 8, 15, 8, 8, 8 | | | | |
* 8, 15, 8, 8, 8
*
* When moving the suffix, we found exactly 3 matches.
*
* The first suffix with period 5 is repeating.
*
* The next event is (3 * max_period) % suffix_period
*
* In this example, the result 0, so the next event is suffix[0] => 8
*
* However, 8 is the index in the array of exponential moving average
* which was calculated on the fly when storing the values, so the
* interval is ema[8] = 1366
*
*
* Example 2:
*
* 4, 3, 5, 100,
* 3, 3, 5, 117,
* 4, 4, 5, 112,
* 4, 3, 4, 110,
* 3, 5, 3, 117,
* 4, 4, 5, 112,
* 4, 3, 4, 110,
* 3, 4, 5, 112,
* 4, 3, 4, 110
*
* ilog2
*
* 0, 0, 0, 4,
* 0, 0, 0, 4,
* 0, 0, 0, 4,
* 0, 0, 0, 4,
* 0, 0, 0, 4,
* 0, 0, 0, 4,
* 0, 0, 0, 4,
* 0, 0, 0, 4,
* 0, 0, 0, 4
*
* Max period 5:
* 0, 0, 4,
* 0, 0, 0, 4,
* 0, 0, 0, 4,
* 0, 0, 0, 4
*
* Suffixes:
*
* 1) 0, 0, 4, 0, 0
* 2) 0, 0, 4, 0
* 3) 0, 0, 4
* 4) 0, 0
*
* buffer: 0, 0, 4, 0, 0, 0, 4, 0, 0, 0, 4, 0, 0, 0, 4
* | | | | | | X
* 0, 0, 4, 0, 0, | X
* 0, 0
*
* buffer: 0, 0, 4, 0, 0, 0, 4, 0, 0, 0, 4, 0, 0, 0, 4
* | | | | | | | | | | | | | | |
* 0, 0, 4, 0, | | | | | | | | | | |
* 0, 0, 4, 0, | | | | | | |
* 0, 0, 4, 0, | | |
* 0 0 4
*
* Pattern is found 3 times, the remaining is 1 which results from
* (max_period * 3) % suffix_period. This value is the index in the
* suffix arrays. The suffix array for a period 4 has the value 4
* at index 1.
*/
#define EMA_ALPHA_VAL 64
#define EMA_ALPHA_SHIFT 7
#define PREDICTION_PERIOD_MIN 2
#define PREDICTION_PERIOD_MAX 5
#define PREDICTION_FACTOR 4
#define PREDICTION_MAX 10 /* 2 ^ PREDICTION_MAX useconds */
#define PREDICTION_BUFFER_SIZE 16 /* slots for EMAs, hardly more than 16 */
struct irqt_stat {
u64 last_ts;
u64 ema_time[PREDICTION_BUFFER_SIZE];
int timings[IRQ_TIMINGS_SIZE];
int circ_timings[IRQ_TIMINGS_SIZE];
int count;
};
/*
* Exponential moving average computation
*/
static u64 irq_timings_ema_new(u64 value, u64 ema_old)
{
s64 diff;
if (unlikely(!ema_old))
return value;
diff = (value - ema_old) * EMA_ALPHA_VAL;
/*
* We can use a s64 type variable to be added with the u64
* ema_old variable as this one will never have its topmost
* bit set, it will be always smaller than 2^63 nanosec
* interrupt interval (292 years).
*/
return ema_old + (diff >> EMA_ALPHA_SHIFT);
}
static int irq_timings_next_event_index(int *buffer, size_t len, int period_max)
{
int i;
/*
* The buffer contains the suite of intervals, in a ilog2
* basis, we are looking for a repetition. We point the
* beginning of the search three times the length of the
* period beginning at the end of the buffer. We do that for
* each suffix.
*/
for (i = period_max; i >= PREDICTION_PERIOD_MIN ; i--) {
int *begin = &buffer[len - (i * 3)];
int *ptr = begin;
/*
* We look if the suite with period 'i' repeat
* itself. If it is truncated at the end, as it
* repeats we can use the period to find out the next
* element.
*/
while (!memcmp(ptr, begin, i * sizeof(*ptr))) {
ptr += i;
if (ptr >= &buffer[len])
return begin[((i * 3) % i)];
}
}
return -1;
}
static u64 __irq_timings_next_event(struct irqt_stat *irqs, int irq, u64 now)
{
int index, i, period_max, count, start, min = INT_MAX;
if ((now - irqs->last_ts) >= NSEC_PER_SEC) {
irqs->count = irqs->last_ts = 0;
return U64_MAX;
}
/*
* As we want to find three times the repetition, we need a
* number of intervals greater or equal to three times the
* maximum period, otherwise we truncate the max period.
*/
period_max = irqs->count > (3 * PREDICTION_PERIOD_MAX) ?
PREDICTION_PERIOD_MAX : irqs->count / 3;
/*
* If we don't have enough irq timings for this prediction,
* just bail out.
*/
if (period_max <= PREDICTION_PERIOD_MIN)
return U64_MAX;
/*
* 'count' will depends if the circular buffer wrapped or not
*/
count = irqs->count < IRQ_TIMINGS_SIZE ?
irqs->count : IRQ_TIMINGS_SIZE;
start = irqs->count < IRQ_TIMINGS_SIZE ?
0 : (irqs->count & IRQ_TIMINGS_MASK);
/*
* Copy the content of the circular buffer into another buffer
* in order to linearize the buffer instead of dealing with
* wrapping indexes and shifted array which will be prone to
* error and extremelly difficult to debug.
*/
for (i = 0; i < count; i++) {
int index = (start + i) & IRQ_TIMINGS_MASK;
irqs->timings[i] = irqs->circ_timings[index];
min = min_t(int, irqs->timings[i], min);
}
index = irq_timings_next_event_index(irqs->timings, count, period_max);
if (index < 0)
return irqs->last_ts + irqs->ema_time[min];
return irqs->last_ts + irqs->ema_time[index];
}
static inline void irq_timings_store(int irq, struct irqt_stat *irqs, u64 ts)
{
u64 old_ts = irqs->last_ts;
u64 interval;
int index;
/*
* The timestamps are absolute time values, we need to compute
* the timing interval between two interrupts.
*/
irqs->last_ts = ts;
/*
* The interval type is u64 in order to deal with the same
* type in our computation, that prevent mindfuck issues with
* overflow, sign and division.
*/
interval = ts - old_ts;
/*
* The interrupt triggered more than one second apart, that
* ends the sequence as predictible for our purpose. In this
* case, assume we have the beginning of a sequence and the
* timestamp is the first value. As it is impossible to
* predict anything at this point, return.
*
* Note the first timestamp of the sequence will always fall
* in this test because the old_ts is zero. That is what we
* want as we need another timestamp to compute an interval.
*/
if (interval >= NSEC_PER_SEC) {
irqs->count = 0;
return;
}
/*
* Get the index in the ema table for this interrupt. The
* PREDICTION_FACTOR increase the interval size for the array
* of exponential average.
*/
|