Questions[swapping](server)

I have: vm.swappiness = 1 as suggested here to prevent using Swap when there is plenty of available memory. (0 would completely disable swap, as far as I understand)

In frequent cases, I end up with, say, 5GB swap used (total 16GB),
while I have 9GB free available memory (total 32GB)
(only 1GB free if we count cached RAM).

As my application starts to get slow (since most of the Swap is from MariaDB), I run:

swapoff -a
swapon -a

And all the swap memory is moved to the main RAM.
I end up with 5.6GB available RAM, 4GB if we count cached.

There is no apparent reason to have so much Swap.
I've been researching about this, and I cannot find a way to prevent using Swap when it's not needed.
vm.swappiness = 1 is definitely "permanently saved", since it is on the sysctl.conf file, and there have been reboots, etc.

I don't want to disable swap, but the only workaround I can think of is:

1. disable swap
2. have a cron running every 1 minute to check how much free RAM there is
3. enable swap if available RAM is low

But, this is VERY risky to do, and I wish the OS was smart enough to manage this better!

Is there absolutely no way to only use swap when absolutely needed?

So I got the gist of swapping out smaller hard drivers for bigger ones in an MDADM RAID5:

Fail and remove one drive.

mdadm -f /dev/md0 /dev/sdd1
mdadm -r /dev/md0 /dev/sdd1

Shutdown, swap the hard drive, partition and add back to array

mdadm --add /dev/md0 /dev/sdd1

Wait for resync and continue with all the remaining drives.

Finally grow the array

mdadm --grow /dev/md0 --size=max

My question is how do you partition the new bigger drives? I see different instructions around the net and I know most of them are outdated so I wanted to see if anyone has done it recently.

1. Partition the new drives to be the same size as the old drives in the array

2. Partition the new drives to be the maximum size of the drive.

I have a really simple setup, 4 drives RAID5 350GB each. I only have one partition on each drive and build a single array. This is a data archive disk so no need for multiple arrays or partitions.

The other approach I am thinking is backup to one drive, build a 3 drive array, add the backup drive to the array. That is:

• Remove one drive (350GB)
• Add a new drive (1TB)
• Backup all my data into the new drive (my raid data will fit in a new drive)
• Take out the remaining 3 (350GB) old drives
• Put in the rest of the 3 (1TB) new drives
• Build a 3 drive array
• Copy the data to the array from the 1st new drive
• Wipe the backup disk and add to the array to make it 4 drives.

Is this safer? Faster? Thoughts?

When is swap moved back to physical memory in Linux? Is it only on demand, ie. when it's needed? Or is swap slowly transfered to physical memory when the computer is not on high load?

When Linux starts swapping you're basically doomed. Very soon the system won't react to any input any more, but happily swap on until the end of days...

Can you think of a command that holds all processes whatsoever, thus (and while) allowing you to open a clean shell where you can examine the source of the problem and kill the process which ate up all the memory? (I guess this won't be easy, because as the memory is probably completely filled up you'd need to swap out some more memory to gather space for opening a shell, on the other hand all other swapping processes must be stopped.)

If you tied such a command to a hot key then maybe you can use this as an emergency button saving you a lot a time. Any ideas if this is possible at all? Has somebody tried something like this before? If one could realize this it would be a cool feature :)

I have a webserver that has 8GB of ram and is running a pretty intensive php site (1 site) that does file manipulation, graphing, emailing, forums, you name it. The environment is far from static which leads me to believe that very little could be gained from caching anything in ram since almost every request to the server creates new or updated pages. And a lot of caching is done client side so we have a ton of 304 requests when it comes to images, javascript, css.

Additionally I do have language files that are written to flat files on the server where cached ram definitely is good rather than reading from disk. But there are only a handful of files like this.

In about a two weeks I've gone from having 98% free ram to 4% free ram. This has occurred during a time in which we also push several large svn updates onto the server.

My question is whether my server will be better tuned if I periodically clear my cache (I'm aware of Linus Torvalds' feeling about cache) using the following command:

sync; echo 3 > /proc/sys/vm/drop_caches

Or would I be better off editing the following file:

/proc/sys/vm/swappiness  

If I replace the default value of 60 with 30 I should have much less swapping going on and a lot more reuse of stale cache.

It sure feels good to see all that cache freed up using the first command but I'd be lying to you if I told you this was good for the desktop environment. But what about a web server like I've described above? Thoughts?

EDIT: I'm aware that the system will acquire memory as it needs it from the cache memory but thanks for pointing that our for clarity. Am I imagining things when Apache slows down when most of the server memory is stored in cache? Is that a different issue altogether?