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explain a bit

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This commit is contained in:
maxv 2017-07-08 15:15:43 +00:00
parent aff8e6cfa5
commit 5dc461da23
1 changed files with 82 additions and 65 deletions
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147
sys/kern/sched_4bsd.c
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@ -1,6 +1,6 @@
/* $NetBSD: sched_4bsd.c,v 1.30 2014/06/24 10:08:45 maxv Exp $ */ /* $NetBSD: sched_4bsd.c,v 1.31 2017/07/08 15:15:43 maxv Exp $ */
/*- /*
* Copyright (c) 1999, 2000, 2004, 2006, 2007, 2008 The NetBSD Foundation, Inc. * Copyright (c) 1999, 2000, 2004, 2006, 2007, 2008 The NetBSD Foundation, Inc.
* All rights reserved. * All rights reserved.
* *
@ -31,7 +31,7 @@
* POSSIBILITY OF SUCH DAMAGE. * POSSIBILITY OF SUCH DAMAGE.
*/ */
/*- /*
* Copyright (c) 1982, 1986, 1990, 1991, 1993 * Copyright (c) 1982, 1986, 1990, 1991, 1993
* The Regents of the University of California. All rights reserved. * The Regents of the University of California. All rights reserved.
* (c) UNIX System Laboratories, Inc. * (c) UNIX System Laboratories, Inc.
@ -68,7 +68,7 @@
*/ */
#include <sys/cdefs.h> #include <sys/cdefs.h>
__KERNEL_RCSID(0, "$NetBSD: sched_4bsd.c,v 1.30 2014/06/24 10:08:45 maxv Exp $"); __KERNEL_RCSID(0, "$NetBSD: sched_4bsd.c,v 1.31 2017/07/08 15:15:43 maxv Exp $");
#include "opt_ddb.h" #include "opt_ddb.h"
#include "opt_lockdebug.h" #include "opt_lockdebug.h"
@ -80,13 +80,10 @@ __KERNEL_RCSID(0, "$NetBSD: sched_4bsd.c,v 1.30 2014/06/24 10:08:45 maxv Exp $")
#include <sys/cpu.h> #include <sys/cpu.h>
#include <sys/proc.h> #include <sys/proc.h>
#include <sys/kernel.h> #include <sys/kernel.h>
#include <sys/signalvar.h>
#include <sys/resourcevar.h> #include <sys/resourcevar.h>
#include <sys/sched.h> #include <sys/sched.h>
#include <sys/sysctl.h> #include <sys/sysctl.h>
#include <sys/kauth.h>
#include <sys/lockdebug.h> #include <sys/lockdebug.h>
#include <sys/kmem.h>
#include <sys/intr.h> #include <sys/intr.h>
static void updatepri(struct lwp *); static void updatepri(struct lwp *);
@ -95,7 +92,7 @@ static void resetpriority(struct lwp *);
extern unsigned int sched_pstats_ticks; /* defined in kern_synch.c */ extern unsigned int sched_pstats_ticks; /* defined in kern_synch.c */
/* Number of hardclock ticks per sched_tick() */ /* Number of hardclock ticks per sched_tick() */
static int rrticks; static int rrticks __read_mostly;
/* /*
* Force switch among equal priority processes every 100ms. * Force switch among equal priority processes every 100ms.
@ -133,7 +130,7 @@ sched_tick(struct cpu_info *ci)
if (spc->spc_flags & SPCF_SHOULDYIELD) { if (spc->spc_flags & SPCF_SHOULDYIELD) {
/* /*
* Process is stuck in kernel somewhere, probably * Process is stuck in kernel somewhere, probably
* due to buggy or inefficient code. Force a * due to buggy or inefficient code. Force a
* kernel preemption. * kernel preemption.
*/ */
cpu_need_resched(ci, RESCHED_KPREEMPT); cpu_need_resched(ci, RESCHED_KPREEMPT);
@ -170,71 +167,90 @@ sched_tick(struct cpu_info *ci)
#define ESTCPULIM(e) min((e), ESTCPU_MAX) #define ESTCPULIM(e) min((e), ESTCPU_MAX)
/* /*
* Constants for digital decay and forget: * The main parameter used by this algorithm is 'l_estcpu'. It is an estimate
* 90% of (l_estcpu) usage in 5 * loadav time * of the recent CPU utilization of the thread.
* 95% of (l_pctcpu) usage in 60 seconds (load insensitive) *
* Note that, as ps(1) mentions, this can let percentages * l_estcpu is:
* total over 100% (I've seen 137.9% for 3 processes). * - increased each time the hardclock ticks and the thread is found to
* be executing, in sched_schedclock() called from hardclock()
* - decreased (filtered) on each sched tick, in sched_pstats_hook()
* If the lwp is sleeping for more than a second, we don't touch l_estcpu: it
* will be updated in sched_setrunnable() when the lwp wakes up, in burst mode
* (ie, we decrease it n times).
* *
* Note that hardclock updates l_estcpu and l_cpticks independently. * Note that hardclock updates l_estcpu and l_cpticks independently.
* *
* We wish to decay away 90% of l_estcpu in (5 * loadavg) seconds. * -----------------------------------------------------------------------------
* That is, the system wants to compute a value of decay such *
* that the following for loop: * Here we describe how l_estcpu is decreased.
* for (i = 0; i < (5 * loadavg); i++) *
* l_estcpu *= decay; * Constants for digital decay (filter):
* will compute * 90% of l_estcpu usage in (5 * loadavg) seconds
* l_estcpu *= 0.1; *
* for all values of loadavg: * We wish to decay away 90% of l_estcpu in (5 * loadavg) seconds. That is, we
* want to compute a value of decay such that the following loop:
* for (i = 0; i < (5 * loadavg); i++)
* l_estcpu *= decay;
* will result in
* l_estcpu *= 0.1;
* for all values of loadavg.
* *
* Mathematically this loop can be expressed by saying: * Mathematically this loop can be expressed by saying:
* decay ** (5 * loadavg) ~= .1 * decay ** (5 * loadavg) ~= .1
* *
* The system computes decay as: * And finally, the corresponding value of decay we're using is:
* decay = (2 * loadavg) / (2 * loadavg + 1) * decay = (2 * loadavg) / (2 * loadavg + 1)
* *
* We wish to prove that the system's computation of decay * -----------------------------------------------------------------------------
* will always fulfill the equation: *
* decay ** (5 * loadavg) ~= .1 * Now, let's prove that the value of decay stated above will always fulfill
* the equation:
* decay ** (5 * loadavg) ~= .1
* *
* If we compute b as: * If we compute b as:
* b = 2 * loadavg * b = 2 * loadavg
* then * then
* decay = b / (b + 1) * decay = b / (b + 1)
* *
* We now need to prove two things: * We now need to prove two things:
* 1) Given factor ** (5 * loadavg) ~= .1, prove factor == b/(b+1) * 1) Given [factor ** (5 * loadavg) =~ .1], prove [factor == b/(b+1)].
* 2) Given b/(b+1) ** power ~= .1, prove power == (5 * loadavg) * 2) Given [b/(b+1) ** power =~ .1], prove [power == (5 * loadavg)].
* *
* Facts: * Facts:
* For x close to zero, exp(x) =~ 1 + x, since * * For x real: exp(x) = 0! + x**1/1! + x**2/2! + ...
* exp(x) = 0! + x**1/1! + x**2/2! + ... . * Therefore, for x close to zero, exp(x) =~ 1 + x.
* therefore exp(-1/b) =~ 1 - (1/b) = (b-1)/b. * In turn, for b large enough, exp(-1/b) =~ 1 - (1/b) = (b-1)/b.
* For x close to zero, ln(1+x) =~ x, since *
* ln(1+x) = x - x**2/2 + x**3/3 - ... -1 < x < 1 * * For b large enough, (b-1)/b =~ b/(b+1).
* therefore ln(b/(b+1)) = ln(1 - 1/(b+1)) =~ -1/(b+1). *
* ln(.1) =~ -2.30 * * For x belonging to [-1;1[, ln(1-x) = - x - x**2/2 - x**3/3 - ...
* Therefore ln(b/(b+1)) = ln(1 - 1/(b+1)) =~ -1/(b+1).
*
* * ln(0.1) =~ -2.30
* *
* Proof of (1): * Proof of (1):
* Solve (factor)**(power) =~ .1 given power (5*loadav): * factor ** (5 * loadavg) =~ 0.1
* solving for factor, * => ln(factor) =~ -2.30 / (5 * loadavg)
* ln(factor) =~ (-2.30/5*loadav), or * => factor =~ exp(-1 / ((5 / 2.30) * loadavg))
* factor =~ exp(-1/((5/2.30)*loadav)) =~ exp(-1/(2*loadav)) = * =~ exp(-1 / (2 * loadavg))
* exp(-1/b) =~ (b-1)/b =~ b/(b+1). QED * =~ exp(-1 / b)
* =~ (b - 1) / b
* =~ b / (b + 1)
* =~ (2 * loadavg) / ((2 * loadavg) + 1)
* *
* Proof of (2): * Proof of (2):
* Solve (factor)**(power) =~ .1 given factor == (b/(b+1)): * (b / (b + 1)) ** power =~ .1
* solving for power, * => power * ln(b / (b + 1)) =~ -2.30
* power*ln(b/(b+1)) =~ -2.30, or * => power * (-1 / (b + 1)) =~ -2.30
* power =~ 2.3 * (b + 1) = 4.6*loadav + 2.3 =~ 5*loadav. QED * => power =~ 2.30 * (b + 1)
* => power =~ 4.60 * loadavg + 2.30
* => power =~ 5 * loadavg
* *
* Actual power values for the implemented algorithm are as follows: * Conclusion: decay = (2 * loadavg) / (2 * loadavg + 1)
* loadav: 1 2 3 4
* power: 5.68 10.32 14.94 19.55
*/ */
/* calculations for digital decay to forget 90% of usage in 5*loadav sec */ /* See calculations above */
#define loadfactor(loadav) (2 * (loadav) / ncpu) #define loadfactor(loadavg) (2 * (loadavg) / ncpu)
static fixpt_t static fixpt_t
decay_cpu(fixpt_t loadfac, fixpt_t estcpu) decay_cpu(fixpt_t loadfac, fixpt_t estcpu)
@ -250,22 +266,24 @@ decay_cpu(fixpt_t loadfac, fixpt_t estcpu)
if (__predict_true(loadfac <= FIXPT_MAX / ESTCPU_MAX)) { if (__predict_true(loadfac <= FIXPT_MAX / ESTCPU_MAX)) {
return estcpu * loadfac / (loadfac + FSCALE); return estcpu * loadfac / (loadfac + FSCALE);
} }
#endif /* !defined(_LP64) */ #endif
return (uint64_t)estcpu * loadfac / (loadfac + FSCALE); return (uint64_t)estcpu * loadfac / (loadfac + FSCALE);
} }
/*
* For all load averages >= 1 and max l_estcpu of (255 << ESTCPU_SHIFT),
* sleeping for at least seven times the loadfactor will decay l_estcpu to
* less than (1 << ESTCPU_SHIFT).
*
* note that our ESTCPU_MAX is actually much smaller than (255 << ESTCPU_SHIFT).
*/
static fixpt_t static fixpt_t
decay_cpu_batch(fixpt_t loadfac, fixpt_t estcpu, unsigned int n) decay_cpu_batch(fixpt_t loadfac, fixpt_t estcpu, unsigned int n)
{ {
/*
* For all load averages >= 1 and max l_estcpu of (255 << ESTCPU_SHIFT),
* if we slept for at least seven times the loadfactor, we will decay
* l_estcpu to less than (1 << ESTCPU_SHIFT), and therefore we can
* return zero directly.
*
* Note that our ESTCPU_MAX is actually much smaller than
* (255 << ESTCPU_SHIFT).
*/
if ((n << FSHIFT) >= 7 * loadfac) { if ((n << FSHIFT) >= 7 * loadfac) {
return 0; return 0;
} }
@ -299,13 +317,13 @@ sched_pstats_hook(struct lwp *l, int batch)
return; return;
} }
} }
loadfac = 2 * (averunnable.ldavg[0]); loadfac = 2 * (averunnable.ldavg[0]); /* XXX: should be loadfactor? */
l->l_estcpu = decay_cpu(loadfac, l->l_estcpu); l->l_estcpu = decay_cpu(loadfac, l->l_estcpu);
resetpriority(l); resetpriority(l);
} }
/* /*
* Recalculate the priority of a process after it has slept for a while. * Recalculate the priority of an LWP after it has slept for a while.
*/ */
static void static void
updatepri(struct lwp *l) updatepri(struct lwp *l)
@ -383,10 +401,9 @@ resetpriority(struct lwp *l)
* is running (linearly), and decays away exponentially, at a rate which is * is running (linearly), and decays away exponentially, at a rate which is
* proportionally slower when the system is busy. The basic principle is * proportionally slower when the system is busy. The basic principle is
* that the system will 90% forget that the process used a lot of CPU time * that the system will 90% forget that the process used a lot of CPU time
* in 5 * loadav seconds. This causes the system to favor processes which * in (5 * loadavg) seconds. This causes the system to favor processes which
* haven't run much recently, and to round-robin among other processes. * haven't run much recently, and to round-robin among other processes.
*/ */
void void
sched_schedclock(struct lwp *l) sched_schedclock(struct lwp *l)
{ {