How to completely disable swap?
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27
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I am using Debian sid, hard drive formatted with ext4, running on linux 3.1
I remember on previous linux versions (maybe before 3.0), if I run out of memory, and swap is not enabled, programs will usually crash. This is perfect for my environment: simple web browsing with no critical operations. That is, if I accidentally run across a bad website which uses up too much memory, it just crashes without rendering my terminal unusable.
But in my current setup, the computer hangs with violent I/O throughput in the background. iotop reveals kswapd0 to be the culprit, which means it is due to swapping. After using swapon -s
to determine any swaps that were enabled, I used swapoff -a
to disable all swaps and swapon -s
again to confirm that all swaps were disabled.
Then I tried maximizing my memory usage again. Alas, the behavior I expected didn't happen. Instead, kswapd0 tries over and over to swap out the RAM and fails as there is no swap space. Because it never gives up, my computer is locked in eternal I/O heavy freeze, bad for my disk's health.
Am I doing something wrong in trying to swapoff -a
? Why is the behavior different than what it used to be (probably pre-3.0 times)?
swap freeze
 |Â
show 1 more comment
up vote
27
down vote
favorite
I am using Debian sid, hard drive formatted with ext4, running on linux 3.1
I remember on previous linux versions (maybe before 3.0), if I run out of memory, and swap is not enabled, programs will usually crash. This is perfect for my environment: simple web browsing with no critical operations. That is, if I accidentally run across a bad website which uses up too much memory, it just crashes without rendering my terminal unusable.
But in my current setup, the computer hangs with violent I/O throughput in the background. iotop reveals kswapd0 to be the culprit, which means it is due to swapping. After using swapon -s
to determine any swaps that were enabled, I used swapoff -a
to disable all swaps and swapon -s
again to confirm that all swaps were disabled.
Then I tried maximizing my memory usage again. Alas, the behavior I expected didn't happen. Instead, kswapd0 tries over and over to swap out the RAM and fails as there is no swap space. Because it never gives up, my computer is locked in eternal I/O heavy freeze, bad for my disk's health.
Am I doing something wrong in trying to swapoff -a
? Why is the behavior different than what it used to be (probably pre-3.0 times)?
swap freeze
That doesn't really make sense. Doing theswapoff -a
itself, if there was stuff in the swap, will generate a lot of I/O (and can result in processes getting killed if there is not enough real RAM availabe). Are you sure it's not theswapoff -a
that caused the I/O "storm"?
â Mat
Nov 15 '11 at 11:40
I suppose it is enough to comment thefstab
line about swap. Try if the behavior is the same.
â enzotib
Nov 15 '11 at 11:48
@Matswapoff -a
should disable swap permanently, meaning it should stay disabled after next reboot. I confirmed this. Yet, I/O "storm" still happens during the session after next reboot. For the record, I/O "storm" didn't happen at the moment I didswapoff -a
because swap was 0 at that time.
â syockit
Nov 15 '11 at 11:48
@enzotib I have no swap in my fstab.
â syockit
Nov 15 '11 at 11:49
9
@syockit:swapoff -a
is not permanent.
â Mat
Nov 15 '11 at 11:51
 |Â
show 1 more comment
up vote
27
down vote
favorite
up vote
27
down vote
favorite
I am using Debian sid, hard drive formatted with ext4, running on linux 3.1
I remember on previous linux versions (maybe before 3.0), if I run out of memory, and swap is not enabled, programs will usually crash. This is perfect for my environment: simple web browsing with no critical operations. That is, if I accidentally run across a bad website which uses up too much memory, it just crashes without rendering my terminal unusable.
But in my current setup, the computer hangs with violent I/O throughput in the background. iotop reveals kswapd0 to be the culprit, which means it is due to swapping. After using swapon -s
to determine any swaps that were enabled, I used swapoff -a
to disable all swaps and swapon -s
again to confirm that all swaps were disabled.
Then I tried maximizing my memory usage again. Alas, the behavior I expected didn't happen. Instead, kswapd0 tries over and over to swap out the RAM and fails as there is no swap space. Because it never gives up, my computer is locked in eternal I/O heavy freeze, bad for my disk's health.
Am I doing something wrong in trying to swapoff -a
? Why is the behavior different than what it used to be (probably pre-3.0 times)?
swap freeze
I am using Debian sid, hard drive formatted with ext4, running on linux 3.1
I remember on previous linux versions (maybe before 3.0), if I run out of memory, and swap is not enabled, programs will usually crash. This is perfect for my environment: simple web browsing with no critical operations. That is, if I accidentally run across a bad website which uses up too much memory, it just crashes without rendering my terminal unusable.
But in my current setup, the computer hangs with violent I/O throughput in the background. iotop reveals kswapd0 to be the culprit, which means it is due to swapping. After using swapon -s
to determine any swaps that were enabled, I used swapoff -a
to disable all swaps and swapon -s
again to confirm that all swaps were disabled.
Then I tried maximizing my memory usage again. Alas, the behavior I expected didn't happen. Instead, kswapd0 tries over and over to swap out the RAM and fails as there is no swap space. Because it never gives up, my computer is locked in eternal I/O heavy freeze, bad for my disk's health.
Am I doing something wrong in trying to swapoff -a
? Why is the behavior different than what it used to be (probably pre-3.0 times)?
swap freeze
swap freeze
asked Nov 15 '11 at 11:27
syockit
3841414
3841414
That doesn't really make sense. Doing theswapoff -a
itself, if there was stuff in the swap, will generate a lot of I/O (and can result in processes getting killed if there is not enough real RAM availabe). Are you sure it's not theswapoff -a
that caused the I/O "storm"?
â Mat
Nov 15 '11 at 11:40
I suppose it is enough to comment thefstab
line about swap. Try if the behavior is the same.
â enzotib
Nov 15 '11 at 11:48
@Matswapoff -a
should disable swap permanently, meaning it should stay disabled after next reboot. I confirmed this. Yet, I/O "storm" still happens during the session after next reboot. For the record, I/O "storm" didn't happen at the moment I didswapoff -a
because swap was 0 at that time.
â syockit
Nov 15 '11 at 11:48
@enzotib I have no swap in my fstab.
â syockit
Nov 15 '11 at 11:49
9
@syockit:swapoff -a
is not permanent.
â Mat
Nov 15 '11 at 11:51
 |Â
show 1 more comment
That doesn't really make sense. Doing theswapoff -a
itself, if there was stuff in the swap, will generate a lot of I/O (and can result in processes getting killed if there is not enough real RAM availabe). Are you sure it's not theswapoff -a
that caused the I/O "storm"?
â Mat
Nov 15 '11 at 11:40
I suppose it is enough to comment thefstab
line about swap. Try if the behavior is the same.
â enzotib
Nov 15 '11 at 11:48
@Matswapoff -a
should disable swap permanently, meaning it should stay disabled after next reboot. I confirmed this. Yet, I/O "storm" still happens during the session after next reboot. For the record, I/O "storm" didn't happen at the moment I didswapoff -a
because swap was 0 at that time.
â syockit
Nov 15 '11 at 11:48
@enzotib I have no swap in my fstab.
â syockit
Nov 15 '11 at 11:49
9
@syockit:swapoff -a
is not permanent.
â Mat
Nov 15 '11 at 11:51
That doesn't really make sense. Doing the
swapoff -a
itself, if there was stuff in the swap, will generate a lot of I/O (and can result in processes getting killed if there is not enough real RAM availabe). Are you sure it's not the swapoff -a
that caused the I/O "storm"?â Mat
Nov 15 '11 at 11:40
That doesn't really make sense. Doing the
swapoff -a
itself, if there was stuff in the swap, will generate a lot of I/O (and can result in processes getting killed if there is not enough real RAM availabe). Are you sure it's not the swapoff -a
that caused the I/O "storm"?â Mat
Nov 15 '11 at 11:40
I suppose it is enough to comment the
fstab
line about swap. Try if the behavior is the same.â enzotib
Nov 15 '11 at 11:48
I suppose it is enough to comment the
fstab
line about swap. Try if the behavior is the same.â enzotib
Nov 15 '11 at 11:48
@Mat
swapoff -a
should disable swap permanently, meaning it should stay disabled after next reboot. I confirmed this. Yet, I/O "storm" still happens during the session after next reboot. For the record, I/O "storm" didn't happen at the moment I did swapoff -a
because swap was 0 at that time.â syockit
Nov 15 '11 at 11:48
@Mat
swapoff -a
should disable swap permanently, meaning it should stay disabled after next reboot. I confirmed this. Yet, I/O "storm" still happens during the session after next reboot. For the record, I/O "storm" didn't happen at the moment I did swapoff -a
because swap was 0 at that time.â syockit
Nov 15 '11 at 11:48
@enzotib I have no swap in my fstab.
â syockit
Nov 15 '11 at 11:49
@enzotib I have no swap in my fstab.
â syockit
Nov 15 '11 at 11:49
9
9
@syockit:
swapoff -a
is not permanent.â Mat
Nov 15 '11 at 11:51
@syockit:
swapoff -a
is not permanent.â Mat
Nov 15 '11 at 11:51
 |Â
show 1 more comment
6 Answers
6
active
oldest
votes
up vote
14
down vote
A better solution than turning off swap, which will at best cause random processes to be killed when memory runs low, is to set the per process data segment limit for processes that pull stuff off the net. This way a runaway browser will hit the limit and die, rather than cause the whole system to become unusable. Example, from the shell
(ulimit -d 400000; firefox) &
The number after -d is in kilobytes. You should experiment with this on your system to choose the best value for your browsing habits. The parentheses cause a subshell to be created; the ulimit command only affects that shell and its children, isolating its effects from the parent shell.
Will this work for chromium, say, where we have a bunch ofchromium
processes using small chunks of memory?
â jberryman
Jun 25 '15 at 16:37
@jberryman No, the memory limits are per-process rather than per-user.
â Kyle Jones
Jun 26 '15 at 14:36
Is there a way to send it a specified signal (e.g., SIGHUP) when it reaches the memory limit?
â Geremia
Feb 24 '17 at 16:16
1
@Geremia No. The brk and sbrk system calls stop working, which will make most things curl up and die.
â Kyle Jones
Feb 24 '17 at 16:23
add a comment |Â
up vote
12
down vote
Disabling swap won't do what you want. You will still get violent I/O throughput, but it will be of clean pages rather than dirty ones.
With no swap, the system will compress the cache of clean (unmodified) pages to near zero, because those are the only pages it can evict from physical memory. It can only evict dirty (modified) pages from memory by writing them to swap, with no swap, it has no way to evict dirty pages.
As you run low on physical memory, each process will have to load its code pages from disk as it evicts the previous process code pages. The result will be violent thrashing and excessive work done by the swap subsystem.
This is a special case of a very important principle: For a well-designed system, you can't make it run better by reducing its choices. Linux is a well-designed system. Removing swap just gives it fewer choices, so it's not surprising that it behaves worse.
1
Just short of all memory will pretty much always be allocated. Linux is specifically tuned this way. Do acat /proc/meminfo
on any typical Linux box after a few hours of load.
â David Schwartz
Nov 15 '11 at 18:43
2
@syockit If you disable paging, you can't run any programs. Paging is the mechanism by which files are read in when mapped into memory.
â David Schwartz
Nov 16 '11 at 16:39
1
@psusi : When you have no swap, clean pages are evicted from cache because dirty pages cannot be evicted fast enough (if at all). This is what causes the violent thrashing. Long before any processes are killed, the cache will be squeezed to near zero. This means everything that's not a dirty, anonymous page will be gone. That's what causes the thrashing -- the majority of code pages are not dirty, anonymous pages. So almost every code page faults.
â David Schwartz
Nov 26 '11 at 4:29
2
@psusi : Clean pages will not be reduced to a minimum when you have swap. It will instead swap out dirty, anonymous pages that haven't been recently used. Of course, either way you'll get violent thrashing eventually if the working set exceeds physical memory. The point is, with or without swap, you will get lots of violent thrashing before you actually run out of memory. The difference is, with swap the violent thrashing will be swapping (dirty pages, write and read). Without swap, the violent thrashing will be code faults (clean pages, read only).
â David Schwartz
Nov 28 '11 at 4:57
2
@psusi: You are correct if the concern is a runaway process that rapidly blows up in memory consumption. But that's not what the OP is talking about, which is a process that consumes excessive, but not unbounded or massively excessive, memory. As it grows through the large sweet spot (where the cache is squeezed) it will grow more and more slowly as the system thrashes.
â David Schwartz
Feb 2 '12 at 23:44
 |Â
show 9 more comments
up vote
4
down vote
To make sure that swap is not used, you'd be better off preventing any swap being added at boot. This can be done, depending on the system, by disabling the swap
boot service or just commenting out the swap entry in /etc/fstab
.
As far as your hangup is concerned, the stop()
function in /etc/init.d/swap
might give a clue:
stop()
OpenBSD) swapctl -U -t noblk >/dev/null;;
*) swapoff -a >/dev/null;;
esac
eend 0
Notice the part about deadlock. You can try doing umount -a -t tmpfs
yourself before turning swap off.
Edit:
Probably, you might also achieve your goal by modifying sysctl
settings (see this question).
I don't haveswap
ininit.d
, nor do I have it onfstab
, but I do have/etc/init.d/mountoverflowtmp
that mountstmpfs
for emergency log writes. Does the swap daemon usetmpfs
too?
â syockit
Nov 15 '11 at 11:58
You might have it enabled elsewhere - dogrep -RF swap /etc/
if you wish to find it. But to disable a service, you'd use a command likeservice
(IIRC; I don't use Debian myself).
â rozcietrzewiacz
Nov 15 '11 at 12:02
1
Swap itself does not usetmpfs
, becausetmpfs
is an in-memory (RAM) filesystem. But other services/programs that usetmpfs
might rely on swap in a special manner. I don't really know, but it might have something to do with caching or a special way in whichtmpfs
driver claims access to swap space.
â rozcietrzewiacz
Nov 15 '11 at 12:04
There's something about how Linux handles virtual memory that I don't understand. I've disabled swap in most ways possible: viaswapoff
, and viavm.swappiness=0
. Yetkswapd0
still runs! I wonder if this is a regression from the 2.4 daysâ¦
â syockit
Nov 15 '11 at 15:36
4
@syockit It's expected behavior. The system is still swapping clean pages (pages that contain copies of file data). It requires no swap space to swap clean pages, since they can be read back from sources other than swap.
â David Schwartz
Nov 15 '11 at 16:58
add a comment |Â
up vote
2
down vote
It is better to comment out swap partition entry in /etc/fstab
than running swapoff -a
after each boot.
I have the same issue with kswapd0 on my hardware.
Tuning vm.swappiness
system parameter does not help for me.
sysctl -w vm.swappiness=0
I googled and read a lot of posts, mailing lists, and now I think that this is kernel bug.
When there is no active swap partition and free memory becomes less then some threshold (about 300MB in my case) the system becomes unresponsive due to kswapd0 madness.
Probably it is reproduced with special configuration and conditions.
For somebody it is solved by system re-installation with re-partitioning for others by building custom kernel with kswapd0
disabled.
Ifkswapd0
goes mad and you don't have swap activated you're out of RAM. Your choices are OOM Killer orkswapd0
. Linux goes withkswapd0
because the kernel assumes that it's more important to finish slowly than to abort the process. For casual humans, the threshold where kernel thinks that enough forward progress still does happen is already glacially slow and nearly anybody would rather select OOM Killer.
â Mikko Rantalainen
Aug 31 at 10:35
add a comment |Â
up vote
0
down vote
On my system (debian sid 2016-11-15), I did this:
disable the swap now:
swapoff -a
comment the line with swap partition in /etc/fstab
#### #UUID=c6ddbc95-3bb5-49e1-ab25-b1c505e5360c none swap sw 0 0
disable the mounting of swap in systemd:
systemctl --type swap
systemctl stop dev-sda6.swap
systemctl mask dev-sda6.swap
It would be enought. There is reference of the swap in /etc/initramfs-tools/conf.d/resume
file. I don't know what is the purpose of this. Maybe this file will be a problem on the next reboot (I don't try to reboot yet, my uptime is precious ;)).
add a comment |Â
up vote
0
down vote
the computer hangs with violent I/O throughput in the background.
iotop reveals kswapd0 to be the culprit
I've found one way (so far) to avoid that. If you want to test it and see how it does on your system, see the kernel patch inside this question. Basically, it doesn't evict Active(file)
pages (at least) when under memory pressure, thus the disk thrashing (constant reading) is reduced to almost nothing and OOM-killer is allowed to trigger within 1 second, instead of freezing the OS for what seems like permanently(or at least for many minutes). I am hoping that actual programmers(of which I'm not) would improve the patch and make it into an actual solution, now that they see that what it does is working for these situations.
add a comment |Â
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
14
down vote
A better solution than turning off swap, which will at best cause random processes to be killed when memory runs low, is to set the per process data segment limit for processes that pull stuff off the net. This way a runaway browser will hit the limit and die, rather than cause the whole system to become unusable. Example, from the shell
(ulimit -d 400000; firefox) &
The number after -d is in kilobytes. You should experiment with this on your system to choose the best value for your browsing habits. The parentheses cause a subshell to be created; the ulimit command only affects that shell and its children, isolating its effects from the parent shell.
Will this work for chromium, say, where we have a bunch ofchromium
processes using small chunks of memory?
â jberryman
Jun 25 '15 at 16:37
@jberryman No, the memory limits are per-process rather than per-user.
â Kyle Jones
Jun 26 '15 at 14:36
Is there a way to send it a specified signal (e.g., SIGHUP) when it reaches the memory limit?
â Geremia
Feb 24 '17 at 16:16
1
@Geremia No. The brk and sbrk system calls stop working, which will make most things curl up and die.
â Kyle Jones
Feb 24 '17 at 16:23
add a comment |Â
up vote
14
down vote
A better solution than turning off swap, which will at best cause random processes to be killed when memory runs low, is to set the per process data segment limit for processes that pull stuff off the net. This way a runaway browser will hit the limit and die, rather than cause the whole system to become unusable. Example, from the shell
(ulimit -d 400000; firefox) &
The number after -d is in kilobytes. You should experiment with this on your system to choose the best value for your browsing habits. The parentheses cause a subshell to be created; the ulimit command only affects that shell and its children, isolating its effects from the parent shell.
Will this work for chromium, say, where we have a bunch ofchromium
processes using small chunks of memory?
â jberryman
Jun 25 '15 at 16:37
@jberryman No, the memory limits are per-process rather than per-user.
â Kyle Jones
Jun 26 '15 at 14:36
Is there a way to send it a specified signal (e.g., SIGHUP) when it reaches the memory limit?
â Geremia
Feb 24 '17 at 16:16
1
@Geremia No. The brk and sbrk system calls stop working, which will make most things curl up and die.
â Kyle Jones
Feb 24 '17 at 16:23
add a comment |Â
up vote
14
down vote
up vote
14
down vote
A better solution than turning off swap, which will at best cause random processes to be killed when memory runs low, is to set the per process data segment limit for processes that pull stuff off the net. This way a runaway browser will hit the limit and die, rather than cause the whole system to become unusable. Example, from the shell
(ulimit -d 400000; firefox) &
The number after -d is in kilobytes. You should experiment with this on your system to choose the best value for your browsing habits. The parentheses cause a subshell to be created; the ulimit command only affects that shell and its children, isolating its effects from the parent shell.
A better solution than turning off swap, which will at best cause random processes to be killed when memory runs low, is to set the per process data segment limit for processes that pull stuff off the net. This way a runaway browser will hit the limit and die, rather than cause the whole system to become unusable. Example, from the shell
(ulimit -d 400000; firefox) &
The number after -d is in kilobytes. You should experiment with this on your system to choose the best value for your browsing habits. The parentheses cause a subshell to be created; the ulimit command only affects that shell and its children, isolating its effects from the parent shell.
answered Feb 1 '12 at 19:05
Kyle Jones
11.3k13048
11.3k13048
Will this work for chromium, say, where we have a bunch ofchromium
processes using small chunks of memory?
â jberryman
Jun 25 '15 at 16:37
@jberryman No, the memory limits are per-process rather than per-user.
â Kyle Jones
Jun 26 '15 at 14:36
Is there a way to send it a specified signal (e.g., SIGHUP) when it reaches the memory limit?
â Geremia
Feb 24 '17 at 16:16
1
@Geremia No. The brk and sbrk system calls stop working, which will make most things curl up and die.
â Kyle Jones
Feb 24 '17 at 16:23
add a comment |Â
Will this work for chromium, say, where we have a bunch ofchromium
processes using small chunks of memory?
â jberryman
Jun 25 '15 at 16:37
@jberryman No, the memory limits are per-process rather than per-user.
â Kyle Jones
Jun 26 '15 at 14:36
Is there a way to send it a specified signal (e.g., SIGHUP) when it reaches the memory limit?
â Geremia
Feb 24 '17 at 16:16
1
@Geremia No. The brk and sbrk system calls stop working, which will make most things curl up and die.
â Kyle Jones
Feb 24 '17 at 16:23
Will this work for chromium, say, where we have a bunch of
chromium
processes using small chunks of memory?â jberryman
Jun 25 '15 at 16:37
Will this work for chromium, say, where we have a bunch of
chromium
processes using small chunks of memory?â jberryman
Jun 25 '15 at 16:37
@jberryman No, the memory limits are per-process rather than per-user.
â Kyle Jones
Jun 26 '15 at 14:36
@jberryman No, the memory limits are per-process rather than per-user.
â Kyle Jones
Jun 26 '15 at 14:36
Is there a way to send it a specified signal (e.g., SIGHUP) when it reaches the memory limit?
â Geremia
Feb 24 '17 at 16:16
Is there a way to send it a specified signal (e.g., SIGHUP) when it reaches the memory limit?
â Geremia
Feb 24 '17 at 16:16
1
1
@Geremia No. The brk and sbrk system calls stop working, which will make most things curl up and die.
â Kyle Jones
Feb 24 '17 at 16:23
@Geremia No. The brk and sbrk system calls stop working, which will make most things curl up and die.
â Kyle Jones
Feb 24 '17 at 16:23
add a comment |Â
up vote
12
down vote
Disabling swap won't do what you want. You will still get violent I/O throughput, but it will be of clean pages rather than dirty ones.
With no swap, the system will compress the cache of clean (unmodified) pages to near zero, because those are the only pages it can evict from physical memory. It can only evict dirty (modified) pages from memory by writing them to swap, with no swap, it has no way to evict dirty pages.
As you run low on physical memory, each process will have to load its code pages from disk as it evicts the previous process code pages. The result will be violent thrashing and excessive work done by the swap subsystem.
This is a special case of a very important principle: For a well-designed system, you can't make it run better by reducing its choices. Linux is a well-designed system. Removing swap just gives it fewer choices, so it's not surprising that it behaves worse.
1
Just short of all memory will pretty much always be allocated. Linux is specifically tuned this way. Do acat /proc/meminfo
on any typical Linux box after a few hours of load.
â David Schwartz
Nov 15 '11 at 18:43
2
@syockit If you disable paging, you can't run any programs. Paging is the mechanism by which files are read in when mapped into memory.
â David Schwartz
Nov 16 '11 at 16:39
1
@psusi : When you have no swap, clean pages are evicted from cache because dirty pages cannot be evicted fast enough (if at all). This is what causes the violent thrashing. Long before any processes are killed, the cache will be squeezed to near zero. This means everything that's not a dirty, anonymous page will be gone. That's what causes the thrashing -- the majority of code pages are not dirty, anonymous pages. So almost every code page faults.
â David Schwartz
Nov 26 '11 at 4:29
2
@psusi : Clean pages will not be reduced to a minimum when you have swap. It will instead swap out dirty, anonymous pages that haven't been recently used. Of course, either way you'll get violent thrashing eventually if the working set exceeds physical memory. The point is, with or without swap, you will get lots of violent thrashing before you actually run out of memory. The difference is, with swap the violent thrashing will be swapping (dirty pages, write and read). Without swap, the violent thrashing will be code faults (clean pages, read only).
â David Schwartz
Nov 28 '11 at 4:57
2
@psusi: You are correct if the concern is a runaway process that rapidly blows up in memory consumption. But that's not what the OP is talking about, which is a process that consumes excessive, but not unbounded or massively excessive, memory. As it grows through the large sweet spot (where the cache is squeezed) it will grow more and more slowly as the system thrashes.
â David Schwartz
Feb 2 '12 at 23:44
 |Â
show 9 more comments
up vote
12
down vote
Disabling swap won't do what you want. You will still get violent I/O throughput, but it will be of clean pages rather than dirty ones.
With no swap, the system will compress the cache of clean (unmodified) pages to near zero, because those are the only pages it can evict from physical memory. It can only evict dirty (modified) pages from memory by writing them to swap, with no swap, it has no way to evict dirty pages.
As you run low on physical memory, each process will have to load its code pages from disk as it evicts the previous process code pages. The result will be violent thrashing and excessive work done by the swap subsystem.
This is a special case of a very important principle: For a well-designed system, you can't make it run better by reducing its choices. Linux is a well-designed system. Removing swap just gives it fewer choices, so it's not surprising that it behaves worse.
1
Just short of all memory will pretty much always be allocated. Linux is specifically tuned this way. Do acat /proc/meminfo
on any typical Linux box after a few hours of load.
â David Schwartz
Nov 15 '11 at 18:43
2
@syockit If you disable paging, you can't run any programs. Paging is the mechanism by which files are read in when mapped into memory.
â David Schwartz
Nov 16 '11 at 16:39
1
@psusi : When you have no swap, clean pages are evicted from cache because dirty pages cannot be evicted fast enough (if at all). This is what causes the violent thrashing. Long before any processes are killed, the cache will be squeezed to near zero. This means everything that's not a dirty, anonymous page will be gone. That's what causes the thrashing -- the majority of code pages are not dirty, anonymous pages. So almost every code page faults.
â David Schwartz
Nov 26 '11 at 4:29
2
@psusi : Clean pages will not be reduced to a minimum when you have swap. It will instead swap out dirty, anonymous pages that haven't been recently used. Of course, either way you'll get violent thrashing eventually if the working set exceeds physical memory. The point is, with or without swap, you will get lots of violent thrashing before you actually run out of memory. The difference is, with swap the violent thrashing will be swapping (dirty pages, write and read). Without swap, the violent thrashing will be code faults (clean pages, read only).
â David Schwartz
Nov 28 '11 at 4:57
2
@psusi: You are correct if the concern is a runaway process that rapidly blows up in memory consumption. But that's not what the OP is talking about, which is a process that consumes excessive, but not unbounded or massively excessive, memory. As it grows through the large sweet spot (where the cache is squeezed) it will grow more and more slowly as the system thrashes.
â David Schwartz
Feb 2 '12 at 23:44
 |Â
show 9 more comments
up vote
12
down vote
up vote
12
down vote
Disabling swap won't do what you want. You will still get violent I/O throughput, but it will be of clean pages rather than dirty ones.
With no swap, the system will compress the cache of clean (unmodified) pages to near zero, because those are the only pages it can evict from physical memory. It can only evict dirty (modified) pages from memory by writing them to swap, with no swap, it has no way to evict dirty pages.
As you run low on physical memory, each process will have to load its code pages from disk as it evicts the previous process code pages. The result will be violent thrashing and excessive work done by the swap subsystem.
This is a special case of a very important principle: For a well-designed system, you can't make it run better by reducing its choices. Linux is a well-designed system. Removing swap just gives it fewer choices, so it's not surprising that it behaves worse.
Disabling swap won't do what you want. You will still get violent I/O throughput, but it will be of clean pages rather than dirty ones.
With no swap, the system will compress the cache of clean (unmodified) pages to near zero, because those are the only pages it can evict from physical memory. It can only evict dirty (modified) pages from memory by writing them to swap, with no swap, it has no way to evict dirty pages.
As you run low on physical memory, each process will have to load its code pages from disk as it evicts the previous process code pages. The result will be violent thrashing and excessive work done by the swap subsystem.
This is a special case of a very important principle: For a well-designed system, you can't make it run better by reducing its choices. Linux is a well-designed system. Removing swap just gives it fewer choices, so it's not surprising that it behaves worse.
answered Nov 15 '11 at 16:57
David Schwartz
3,9001325
3,9001325
1
Just short of all memory will pretty much always be allocated. Linux is specifically tuned this way. Do acat /proc/meminfo
on any typical Linux box after a few hours of load.
â David Schwartz
Nov 15 '11 at 18:43
2
@syockit If you disable paging, you can't run any programs. Paging is the mechanism by which files are read in when mapped into memory.
â David Schwartz
Nov 16 '11 at 16:39
1
@psusi : When you have no swap, clean pages are evicted from cache because dirty pages cannot be evicted fast enough (if at all). This is what causes the violent thrashing. Long before any processes are killed, the cache will be squeezed to near zero. This means everything that's not a dirty, anonymous page will be gone. That's what causes the thrashing -- the majority of code pages are not dirty, anonymous pages. So almost every code page faults.
â David Schwartz
Nov 26 '11 at 4:29
2
@psusi : Clean pages will not be reduced to a minimum when you have swap. It will instead swap out dirty, anonymous pages that haven't been recently used. Of course, either way you'll get violent thrashing eventually if the working set exceeds physical memory. The point is, with or without swap, you will get lots of violent thrashing before you actually run out of memory. The difference is, with swap the violent thrashing will be swapping (dirty pages, write and read). Without swap, the violent thrashing will be code faults (clean pages, read only).
â David Schwartz
Nov 28 '11 at 4:57
2
@psusi: You are correct if the concern is a runaway process that rapidly blows up in memory consumption. But that's not what the OP is talking about, which is a process that consumes excessive, but not unbounded or massively excessive, memory. As it grows through the large sweet spot (where the cache is squeezed) it will grow more and more slowly as the system thrashes.
â David Schwartz
Feb 2 '12 at 23:44
 |Â
show 9 more comments
1
Just short of all memory will pretty much always be allocated. Linux is specifically tuned this way. Do acat /proc/meminfo
on any typical Linux box after a few hours of load.
â David Schwartz
Nov 15 '11 at 18:43
2
@syockit If you disable paging, you can't run any programs. Paging is the mechanism by which files are read in when mapped into memory.
â David Schwartz
Nov 16 '11 at 16:39
1
@psusi : When you have no swap, clean pages are evicted from cache because dirty pages cannot be evicted fast enough (if at all). This is what causes the violent thrashing. Long before any processes are killed, the cache will be squeezed to near zero. This means everything that's not a dirty, anonymous page will be gone. That's what causes the thrashing -- the majority of code pages are not dirty, anonymous pages. So almost every code page faults.
â David Schwartz
Nov 26 '11 at 4:29
2
@psusi : Clean pages will not be reduced to a minimum when you have swap. It will instead swap out dirty, anonymous pages that haven't been recently used. Of course, either way you'll get violent thrashing eventually if the working set exceeds physical memory. The point is, with or without swap, you will get lots of violent thrashing before you actually run out of memory. The difference is, with swap the violent thrashing will be swapping (dirty pages, write and read). Without swap, the violent thrashing will be code faults (clean pages, read only).
â David Schwartz
Nov 28 '11 at 4:57
2
@psusi: You are correct if the concern is a runaway process that rapidly blows up in memory consumption. But that's not what the OP is talking about, which is a process that consumes excessive, but not unbounded or massively excessive, memory. As it grows through the large sweet spot (where the cache is squeezed) it will grow more and more slowly as the system thrashes.
â David Schwartz
Feb 2 '12 at 23:44
1
1
Just short of all memory will pretty much always be allocated. Linux is specifically tuned this way. Do a
cat /proc/meminfo
on any typical Linux box after a few hours of load.â David Schwartz
Nov 15 '11 at 18:43
Just short of all memory will pretty much always be allocated. Linux is specifically tuned this way. Do a
cat /proc/meminfo
on any typical Linux box after a few hours of load.â David Schwartz
Nov 15 '11 at 18:43
2
2
@syockit If you disable paging, you can't run any programs. Paging is the mechanism by which files are read in when mapped into memory.
â David Schwartz
Nov 16 '11 at 16:39
@syockit If you disable paging, you can't run any programs. Paging is the mechanism by which files are read in when mapped into memory.
â David Schwartz
Nov 16 '11 at 16:39
1
1
@psusi : When you have no swap, clean pages are evicted from cache because dirty pages cannot be evicted fast enough (if at all). This is what causes the violent thrashing. Long before any processes are killed, the cache will be squeezed to near zero. This means everything that's not a dirty, anonymous page will be gone. That's what causes the thrashing -- the majority of code pages are not dirty, anonymous pages. So almost every code page faults.
â David Schwartz
Nov 26 '11 at 4:29
@psusi : When you have no swap, clean pages are evicted from cache because dirty pages cannot be evicted fast enough (if at all). This is what causes the violent thrashing. Long before any processes are killed, the cache will be squeezed to near zero. This means everything that's not a dirty, anonymous page will be gone. That's what causes the thrashing -- the majority of code pages are not dirty, anonymous pages. So almost every code page faults.
â David Schwartz
Nov 26 '11 at 4:29
2
2
@psusi : Clean pages will not be reduced to a minimum when you have swap. It will instead swap out dirty, anonymous pages that haven't been recently used. Of course, either way you'll get violent thrashing eventually if the working set exceeds physical memory. The point is, with or without swap, you will get lots of violent thrashing before you actually run out of memory. The difference is, with swap the violent thrashing will be swapping (dirty pages, write and read). Without swap, the violent thrashing will be code faults (clean pages, read only).
â David Schwartz
Nov 28 '11 at 4:57
@psusi : Clean pages will not be reduced to a minimum when you have swap. It will instead swap out dirty, anonymous pages that haven't been recently used. Of course, either way you'll get violent thrashing eventually if the working set exceeds physical memory. The point is, with or without swap, you will get lots of violent thrashing before you actually run out of memory. The difference is, with swap the violent thrashing will be swapping (dirty pages, write and read). Without swap, the violent thrashing will be code faults (clean pages, read only).
â David Schwartz
Nov 28 '11 at 4:57
2
2
@psusi: You are correct if the concern is a runaway process that rapidly blows up in memory consumption. But that's not what the OP is talking about, which is a process that consumes excessive, but not unbounded or massively excessive, memory. As it grows through the large sweet spot (where the cache is squeezed) it will grow more and more slowly as the system thrashes.
â David Schwartz
Feb 2 '12 at 23:44
@psusi: You are correct if the concern is a runaway process that rapidly blows up in memory consumption. But that's not what the OP is talking about, which is a process that consumes excessive, but not unbounded or massively excessive, memory. As it grows through the large sweet spot (where the cache is squeezed) it will grow more and more slowly as the system thrashes.
â David Schwartz
Feb 2 '12 at 23:44
 |Â
show 9 more comments
up vote
4
down vote
To make sure that swap is not used, you'd be better off preventing any swap being added at boot. This can be done, depending on the system, by disabling the swap
boot service or just commenting out the swap entry in /etc/fstab
.
As far as your hangup is concerned, the stop()
function in /etc/init.d/swap
might give a clue:
stop()
OpenBSD) swapctl -U -t noblk >/dev/null;;
*) swapoff -a >/dev/null;;
esac
eend 0
Notice the part about deadlock. You can try doing umount -a -t tmpfs
yourself before turning swap off.
Edit:
Probably, you might also achieve your goal by modifying sysctl
settings (see this question).
I don't haveswap
ininit.d
, nor do I have it onfstab
, but I do have/etc/init.d/mountoverflowtmp
that mountstmpfs
for emergency log writes. Does the swap daemon usetmpfs
too?
â syockit
Nov 15 '11 at 11:58
You might have it enabled elsewhere - dogrep -RF swap /etc/
if you wish to find it. But to disable a service, you'd use a command likeservice
(IIRC; I don't use Debian myself).
â rozcietrzewiacz
Nov 15 '11 at 12:02
1
Swap itself does not usetmpfs
, becausetmpfs
is an in-memory (RAM) filesystem. But other services/programs that usetmpfs
might rely on swap in a special manner. I don't really know, but it might have something to do with caching or a special way in whichtmpfs
driver claims access to swap space.
â rozcietrzewiacz
Nov 15 '11 at 12:04
There's something about how Linux handles virtual memory that I don't understand. I've disabled swap in most ways possible: viaswapoff
, and viavm.swappiness=0
. Yetkswapd0
still runs! I wonder if this is a regression from the 2.4 daysâ¦
â syockit
Nov 15 '11 at 15:36
4
@syockit It's expected behavior. The system is still swapping clean pages (pages that contain copies of file data). It requires no swap space to swap clean pages, since they can be read back from sources other than swap.
â David Schwartz
Nov 15 '11 at 16:58
add a comment |Â
up vote
4
down vote
To make sure that swap is not used, you'd be better off preventing any swap being added at boot. This can be done, depending on the system, by disabling the swap
boot service or just commenting out the swap entry in /etc/fstab
.
As far as your hangup is concerned, the stop()
function in /etc/init.d/swap
might give a clue:
stop()
OpenBSD) swapctl -U -t noblk >/dev/null;;
*) swapoff -a >/dev/null;;
esac
eend 0
Notice the part about deadlock. You can try doing umount -a -t tmpfs
yourself before turning swap off.
Edit:
Probably, you might also achieve your goal by modifying sysctl
settings (see this question).
I don't haveswap
ininit.d
, nor do I have it onfstab
, but I do have/etc/init.d/mountoverflowtmp
that mountstmpfs
for emergency log writes. Does the swap daemon usetmpfs
too?
â syockit
Nov 15 '11 at 11:58
You might have it enabled elsewhere - dogrep -RF swap /etc/
if you wish to find it. But to disable a service, you'd use a command likeservice
(IIRC; I don't use Debian myself).
â rozcietrzewiacz
Nov 15 '11 at 12:02
1
Swap itself does not usetmpfs
, becausetmpfs
is an in-memory (RAM) filesystem. But other services/programs that usetmpfs
might rely on swap in a special manner. I don't really know, but it might have something to do with caching or a special way in whichtmpfs
driver claims access to swap space.
â rozcietrzewiacz
Nov 15 '11 at 12:04
There's something about how Linux handles virtual memory that I don't understand. I've disabled swap in most ways possible: viaswapoff
, and viavm.swappiness=0
. Yetkswapd0
still runs! I wonder if this is a regression from the 2.4 daysâ¦
â syockit
Nov 15 '11 at 15:36
4
@syockit It's expected behavior. The system is still swapping clean pages (pages that contain copies of file data). It requires no swap space to swap clean pages, since they can be read back from sources other than swap.
â David Schwartz
Nov 15 '11 at 16:58
add a comment |Â
up vote
4
down vote
up vote
4
down vote
To make sure that swap is not used, you'd be better off preventing any swap being added at boot. This can be done, depending on the system, by disabling the swap
boot service or just commenting out the swap entry in /etc/fstab
.
As far as your hangup is concerned, the stop()
function in /etc/init.d/swap
might give a clue:
stop()
OpenBSD) swapctl -U -t noblk >/dev/null;;
*) swapoff -a >/dev/null;;
esac
eend 0
Notice the part about deadlock. You can try doing umount -a -t tmpfs
yourself before turning swap off.
Edit:
Probably, you might also achieve your goal by modifying sysctl
settings (see this question).
To make sure that swap is not used, you'd be better off preventing any swap being added at boot. This can be done, depending on the system, by disabling the swap
boot service or just commenting out the swap entry in /etc/fstab
.
As far as your hangup is concerned, the stop()
function in /etc/init.d/swap
might give a clue:
stop()
OpenBSD) swapctl -U -t noblk >/dev/null;;
*) swapoff -a >/dev/null;;
esac
eend 0
Notice the part about deadlock. You can try doing umount -a -t tmpfs
yourself before turning swap off.
Edit:
Probably, you might also achieve your goal by modifying sysctl
settings (see this question).
edited Apr 13 '17 at 12:36
Communityâ¦
1
1
answered Nov 15 '11 at 11:49
rozcietrzewiacz
28.2k37191
28.2k37191
I don't haveswap
ininit.d
, nor do I have it onfstab
, but I do have/etc/init.d/mountoverflowtmp
that mountstmpfs
for emergency log writes. Does the swap daemon usetmpfs
too?
â syockit
Nov 15 '11 at 11:58
You might have it enabled elsewhere - dogrep -RF swap /etc/
if you wish to find it. But to disable a service, you'd use a command likeservice
(IIRC; I don't use Debian myself).
â rozcietrzewiacz
Nov 15 '11 at 12:02
1
Swap itself does not usetmpfs
, becausetmpfs
is an in-memory (RAM) filesystem. But other services/programs that usetmpfs
might rely on swap in a special manner. I don't really know, but it might have something to do with caching or a special way in whichtmpfs
driver claims access to swap space.
â rozcietrzewiacz
Nov 15 '11 at 12:04
There's something about how Linux handles virtual memory that I don't understand. I've disabled swap in most ways possible: viaswapoff
, and viavm.swappiness=0
. Yetkswapd0
still runs! I wonder if this is a regression from the 2.4 daysâ¦
â syockit
Nov 15 '11 at 15:36
4
@syockit It's expected behavior. The system is still swapping clean pages (pages that contain copies of file data). It requires no swap space to swap clean pages, since they can be read back from sources other than swap.
â David Schwartz
Nov 15 '11 at 16:58
add a comment |Â
I don't haveswap
ininit.d
, nor do I have it onfstab
, but I do have/etc/init.d/mountoverflowtmp
that mountstmpfs
for emergency log writes. Does the swap daemon usetmpfs
too?
â syockit
Nov 15 '11 at 11:58
You might have it enabled elsewhere - dogrep -RF swap /etc/
if you wish to find it. But to disable a service, you'd use a command likeservice
(IIRC; I don't use Debian myself).
â rozcietrzewiacz
Nov 15 '11 at 12:02
1
Swap itself does not usetmpfs
, becausetmpfs
is an in-memory (RAM) filesystem. But other services/programs that usetmpfs
might rely on swap in a special manner. I don't really know, but it might have something to do with caching or a special way in whichtmpfs
driver claims access to swap space.
â rozcietrzewiacz
Nov 15 '11 at 12:04
There's something about how Linux handles virtual memory that I don't understand. I've disabled swap in most ways possible: viaswapoff
, and viavm.swappiness=0
. Yetkswapd0
still runs! I wonder if this is a regression from the 2.4 daysâ¦
â syockit
Nov 15 '11 at 15:36
4
@syockit It's expected behavior. The system is still swapping clean pages (pages that contain copies of file data). It requires no swap space to swap clean pages, since they can be read back from sources other than swap.
â David Schwartz
Nov 15 '11 at 16:58
I don't have
swap
in init.d
, nor do I have it on fstab
, but I do have /etc/init.d/mountoverflowtmp
that mounts tmpfs
for emergency log writes. Does the swap daemon use tmpfs
too?â syockit
Nov 15 '11 at 11:58
I don't have
swap
in init.d
, nor do I have it on fstab
, but I do have /etc/init.d/mountoverflowtmp
that mounts tmpfs
for emergency log writes. Does the swap daemon use tmpfs
too?â syockit
Nov 15 '11 at 11:58
You might have it enabled elsewhere - do
grep -RF swap /etc/
if you wish to find it. But to disable a service, you'd use a command like service
(IIRC; I don't use Debian myself).â rozcietrzewiacz
Nov 15 '11 at 12:02
You might have it enabled elsewhere - do
grep -RF swap /etc/
if you wish to find it. But to disable a service, you'd use a command like service
(IIRC; I don't use Debian myself).â rozcietrzewiacz
Nov 15 '11 at 12:02
1
1
Swap itself does not use
tmpfs
, because tmpfs
is an in-memory (RAM) filesystem. But other services/programs that use tmpfs
might rely on swap in a special manner. I don't really know, but it might have something to do with caching or a special way in which tmpfs
driver claims access to swap space.â rozcietrzewiacz
Nov 15 '11 at 12:04
Swap itself does not use
tmpfs
, because tmpfs
is an in-memory (RAM) filesystem. But other services/programs that use tmpfs
might rely on swap in a special manner. I don't really know, but it might have something to do with caching or a special way in which tmpfs
driver claims access to swap space.â rozcietrzewiacz
Nov 15 '11 at 12:04
There's something about how Linux handles virtual memory that I don't understand. I've disabled swap in most ways possible: via
swapoff
, and via vm.swappiness=0
. Yet kswapd0
still runs! I wonder if this is a regression from the 2.4 daysâ¦â syockit
Nov 15 '11 at 15:36
There's something about how Linux handles virtual memory that I don't understand. I've disabled swap in most ways possible: via
swapoff
, and via vm.swappiness=0
. Yet kswapd0
still runs! I wonder if this is a regression from the 2.4 daysâ¦â syockit
Nov 15 '11 at 15:36
4
4
@syockit It's expected behavior. The system is still swapping clean pages (pages that contain copies of file data). It requires no swap space to swap clean pages, since they can be read back from sources other than swap.
â David Schwartz
Nov 15 '11 at 16:58
@syockit It's expected behavior. The system is still swapping clean pages (pages that contain copies of file data). It requires no swap space to swap clean pages, since they can be read back from sources other than swap.
â David Schwartz
Nov 15 '11 at 16:58
add a comment |Â
up vote
2
down vote
It is better to comment out swap partition entry in /etc/fstab
than running swapoff -a
after each boot.
I have the same issue with kswapd0 on my hardware.
Tuning vm.swappiness
system parameter does not help for me.
sysctl -w vm.swappiness=0
I googled and read a lot of posts, mailing lists, and now I think that this is kernel bug.
When there is no active swap partition and free memory becomes less then some threshold (about 300MB in my case) the system becomes unresponsive due to kswapd0 madness.
Probably it is reproduced with special configuration and conditions.
For somebody it is solved by system re-installation with re-partitioning for others by building custom kernel with kswapd0
disabled.
Ifkswapd0
goes mad and you don't have swap activated you're out of RAM. Your choices are OOM Killer orkswapd0
. Linux goes withkswapd0
because the kernel assumes that it's more important to finish slowly than to abort the process. For casual humans, the threshold where kernel thinks that enough forward progress still does happen is already glacially slow and nearly anybody would rather select OOM Killer.
â Mikko Rantalainen
Aug 31 at 10:35
add a comment |Â
up vote
2
down vote
It is better to comment out swap partition entry in /etc/fstab
than running swapoff -a
after each boot.
I have the same issue with kswapd0 on my hardware.
Tuning vm.swappiness
system parameter does not help for me.
sysctl -w vm.swappiness=0
I googled and read a lot of posts, mailing lists, and now I think that this is kernel bug.
When there is no active swap partition and free memory becomes less then some threshold (about 300MB in my case) the system becomes unresponsive due to kswapd0 madness.
Probably it is reproduced with special configuration and conditions.
For somebody it is solved by system re-installation with re-partitioning for others by building custom kernel with kswapd0
disabled.
Ifkswapd0
goes mad and you don't have swap activated you're out of RAM. Your choices are OOM Killer orkswapd0
. Linux goes withkswapd0
because the kernel assumes that it's more important to finish slowly than to abort the process. For casual humans, the threshold where kernel thinks that enough forward progress still does happen is already glacially slow and nearly anybody would rather select OOM Killer.
â Mikko Rantalainen
Aug 31 at 10:35
add a comment |Â
up vote
2
down vote
up vote
2
down vote
It is better to comment out swap partition entry in /etc/fstab
than running swapoff -a
after each boot.
I have the same issue with kswapd0 on my hardware.
Tuning vm.swappiness
system parameter does not help for me.
sysctl -w vm.swappiness=0
I googled and read a lot of posts, mailing lists, and now I think that this is kernel bug.
When there is no active swap partition and free memory becomes less then some threshold (about 300MB in my case) the system becomes unresponsive due to kswapd0 madness.
Probably it is reproduced with special configuration and conditions.
For somebody it is solved by system re-installation with re-partitioning for others by building custom kernel with kswapd0
disabled.
It is better to comment out swap partition entry in /etc/fstab
than running swapoff -a
after each boot.
I have the same issue with kswapd0 on my hardware.
Tuning vm.swappiness
system parameter does not help for me.
sysctl -w vm.swappiness=0
I googled and read a lot of posts, mailing lists, and now I think that this is kernel bug.
When there is no active swap partition and free memory becomes less then some threshold (about 300MB in my case) the system becomes unresponsive due to kswapd0 madness.
Probably it is reproduced with special configuration and conditions.
For somebody it is solved by system re-installation with re-partitioning for others by building custom kernel with kswapd0
disabled.
edited May 8 '13 at 11:45
Anthon
58.9k1796160
58.9k1796160
answered Apr 24 '13 at 4:47
gumkins
4121822
4121822
Ifkswapd0
goes mad and you don't have swap activated you're out of RAM. Your choices are OOM Killer orkswapd0
. Linux goes withkswapd0
because the kernel assumes that it's more important to finish slowly than to abort the process. For casual humans, the threshold where kernel thinks that enough forward progress still does happen is already glacially slow and nearly anybody would rather select OOM Killer.
â Mikko Rantalainen
Aug 31 at 10:35
add a comment |Â
Ifkswapd0
goes mad and you don't have swap activated you're out of RAM. Your choices are OOM Killer orkswapd0
. Linux goes withkswapd0
because the kernel assumes that it's more important to finish slowly than to abort the process. For casual humans, the threshold where kernel thinks that enough forward progress still does happen is already glacially slow and nearly anybody would rather select OOM Killer.
â Mikko Rantalainen
Aug 31 at 10:35
If
kswapd0
goes mad and you don't have swap activated you're out of RAM. Your choices are OOM Killer or kswapd0
. Linux goes with kswapd0
because the kernel assumes that it's more important to finish slowly than to abort the process. For casual humans, the threshold where kernel thinks that enough forward progress still does happen is already glacially slow and nearly anybody would rather select OOM Killer.â Mikko Rantalainen
Aug 31 at 10:35
If
kswapd0
goes mad and you don't have swap activated you're out of RAM. Your choices are OOM Killer or kswapd0
. Linux goes with kswapd0
because the kernel assumes that it's more important to finish slowly than to abort the process. For casual humans, the threshold where kernel thinks that enough forward progress still does happen is already glacially slow and nearly anybody would rather select OOM Killer.â Mikko Rantalainen
Aug 31 at 10:35
add a comment |Â
up vote
0
down vote
On my system (debian sid 2016-11-15), I did this:
disable the swap now:
swapoff -a
comment the line with swap partition in /etc/fstab
#### #UUID=c6ddbc95-3bb5-49e1-ab25-b1c505e5360c none swap sw 0 0
disable the mounting of swap in systemd:
systemctl --type swap
systemctl stop dev-sda6.swap
systemctl mask dev-sda6.swap
It would be enought. There is reference of the swap in /etc/initramfs-tools/conf.d/resume
file. I don't know what is the purpose of this. Maybe this file will be a problem on the next reboot (I don't try to reboot yet, my uptime is precious ;)).
add a comment |Â
up vote
0
down vote
On my system (debian sid 2016-11-15), I did this:
disable the swap now:
swapoff -a
comment the line with swap partition in /etc/fstab
#### #UUID=c6ddbc95-3bb5-49e1-ab25-b1c505e5360c none swap sw 0 0
disable the mounting of swap in systemd:
systemctl --type swap
systemctl stop dev-sda6.swap
systemctl mask dev-sda6.swap
It would be enought. There is reference of the swap in /etc/initramfs-tools/conf.d/resume
file. I don't know what is the purpose of this. Maybe this file will be a problem on the next reboot (I don't try to reboot yet, my uptime is precious ;)).
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On my system (debian sid 2016-11-15), I did this:
disable the swap now:
swapoff -a
comment the line with swap partition in /etc/fstab
#### #UUID=c6ddbc95-3bb5-49e1-ab25-b1c505e5360c none swap sw 0 0
disable the mounting of swap in systemd:
systemctl --type swap
systemctl stop dev-sda6.swap
systemctl mask dev-sda6.swap
It would be enought. There is reference of the swap in /etc/initramfs-tools/conf.d/resume
file. I don't know what is the purpose of this. Maybe this file will be a problem on the next reboot (I don't try to reboot yet, my uptime is precious ;)).
On my system (debian sid 2016-11-15), I did this:
disable the swap now:
swapoff -a
comment the line with swap partition in /etc/fstab
#### #UUID=c6ddbc95-3bb5-49e1-ab25-b1c505e5360c none swap sw 0 0
disable the mounting of swap in systemd:
systemctl --type swap
systemctl stop dev-sda6.swap
systemctl mask dev-sda6.swap
It would be enought. There is reference of the swap in /etc/initramfs-tools/conf.d/resume
file. I don't know what is the purpose of this. Maybe this file will be a problem on the next reboot (I don't try to reboot yet, my uptime is precious ;)).
edited Nov 15 '16 at 10:57
Archemar
19.1k93467
19.1k93467
answered Nov 15 '16 at 10:34
homer242
11
11
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0
down vote
the computer hangs with violent I/O throughput in the background.
iotop reveals kswapd0 to be the culprit
I've found one way (so far) to avoid that. If you want to test it and see how it does on your system, see the kernel patch inside this question. Basically, it doesn't evict Active(file)
pages (at least) when under memory pressure, thus the disk thrashing (constant reading) is reduced to almost nothing and OOM-killer is allowed to trigger within 1 second, instead of freezing the OS for what seems like permanently(or at least for many minutes). I am hoping that actual programmers(of which I'm not) would improve the patch and make it into an actual solution, now that they see that what it does is working for these situations.
add a comment |Â
up vote
0
down vote
the computer hangs with violent I/O throughput in the background.
iotop reveals kswapd0 to be the culprit
I've found one way (so far) to avoid that. If you want to test it and see how it does on your system, see the kernel patch inside this question. Basically, it doesn't evict Active(file)
pages (at least) when under memory pressure, thus the disk thrashing (constant reading) is reduced to almost nothing and OOM-killer is allowed to trigger within 1 second, instead of freezing the OS for what seems like permanently(or at least for many minutes). I am hoping that actual programmers(of which I'm not) would improve the patch and make it into an actual solution, now that they see that what it does is working for these situations.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
the computer hangs with violent I/O throughput in the background.
iotop reveals kswapd0 to be the culprit
I've found one way (so far) to avoid that. If you want to test it and see how it does on your system, see the kernel patch inside this question. Basically, it doesn't evict Active(file)
pages (at least) when under memory pressure, thus the disk thrashing (constant reading) is reduced to almost nothing and OOM-killer is allowed to trigger within 1 second, instead of freezing the OS for what seems like permanently(or at least for many minutes). I am hoping that actual programmers(of which I'm not) would improve the patch and make it into an actual solution, now that they see that what it does is working for these situations.
the computer hangs with violent I/O throughput in the background.
iotop reveals kswapd0 to be the culprit
I've found one way (so far) to avoid that. If you want to test it and see how it does on your system, see the kernel patch inside this question. Basically, it doesn't evict Active(file)
pages (at least) when under memory pressure, thus the disk thrashing (constant reading) is reduced to almost nothing and OOM-killer is allowed to trigger within 1 second, instead of freezing the OS for what seems like permanently(or at least for many minutes). I am hoping that actual programmers(of which I'm not) would improve the patch and make it into an actual solution, now that they see that what it does is working for these situations.
answered Aug 31 at 8:56
Marcus Linsner
18414
18414
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That doesn't really make sense. Doing the
swapoff -a
itself, if there was stuff in the swap, will generate a lot of I/O (and can result in processes getting killed if there is not enough real RAM availabe). Are you sure it's not theswapoff -a
that caused the I/O "storm"?â Mat
Nov 15 '11 at 11:40
I suppose it is enough to comment the
fstab
line about swap. Try if the behavior is the same.â enzotib
Nov 15 '11 at 11:48
@Mat
swapoff -a
should disable swap permanently, meaning it should stay disabled after next reboot. I confirmed this. Yet, I/O "storm" still happens during the session after next reboot. For the record, I/O "storm" didn't happen at the moment I didswapoff -a
because swap was 0 at that time.â syockit
Nov 15 '11 at 11:48
@enzotib I have no swap in my fstab.
â syockit
Nov 15 '11 at 11:49
9
@syockit:
swapoff -a
is not permanent.â Mat
Nov 15 '11 at 11:51