Server :: Clear Cache Memory ( RHEL 5.1 ) As It Consumes Almost 100% Physical Memory?
Jan 11, 2010let me know how to clear cache memory ( RHEL 5.1 ) as it consumes almost 100% physical memory.
View 3 Replieslet me know how to clear cache memory ( RHEL 5.1 ) as it consumes almost 100% physical memory.
View 3 RepliesUbuntu 10.10
How do I clear the System cache memory.
For the past few days , I am facing problem in opening websites (firefox & Chrome).
I have to restart the laptop and then access the web sites.
I am monitoring physical memory in a server I administer, and my hardware provider told me they had increased physical memory size to 4Gb... However, using several tools (free -m; top; dmesg | grep Memory; grep MemTotal /proc/meminfo I discovered that I actually have 3Gb, not 4... But, my doubt comes from the fact that dmesg | grem Memory tells me I have 3103396k/4194304k available The first number is effectively 3Gb, but the second one, is 4! so, why I am looking at this two different numbers?
View 1 Replies View RelatedIs that possible that SHM shared memory is counted as cache memory on Linux with kernel 2.6.18?If find it really odd since this memory is not file backed, but I have a piece of code that loads data using shm_open+mmap, and it generates an amount of cache memory in /proc/meminfo that corresponds exactly to the amount of shared memory (I load that data from a file but I am using posix_fadvise(fd,0,0,POSIX_FADV_DONTNEED) to ensure this file is not cached and I made sure that it is working as expected). As far as I know SHM memory was not tagged as cache memory with kernel 2.6.9.If it is the case it is really unfortunate since normally cache memory can be considered to be part of the "available" memory since it can be flushed promptly but this is clearly not the case with SHM memory... Is there an easy way to get the total amount of used SHM memory on a system?
View 4 Replies View RelatedI got the VMrss used by a process as about 2GB, but the physical memory of my computer is only 1G.
View 10 Replies View RelatedMy Ubuntu consumes too much memory. I have checked it with memstat and have found a lot of lines like these:
[Code]...
There are more than 480 lines like this! What are these fontconfigs for? Where do they come from?
Why linux uses swap space, even if there are free physical memory available.
View 4 Replies View RelatedBefore running SOSreport, the free physical memory is 5389Mb out of 8Gb.
After running SOSreport, the free physical memory is 3229Mb.
Why is this so? how we can free up the physical memory.
My computer has 4GB installed but for some reason I could only see the 2 of the 4. At first I thought that this was because the XEN virtual operating system manager was running but even after I turned it off the problem exists. I could only see and use the 2GB of memory.
Code:
free -m
total used free shared buffers cached
Mem: 1914 1874 39 0 9 295
-/+ buffers/cache: 1570 343
Swap: 2062 1934 128
Rhythmbox stalls my computer. I start it and within 10-20 seconds it starts to search something (i.e. in the right bottom there appears a box with a red box sliding left and right). It consumes all my memory (1 giga) and the only thing it does is identify text files in a general directory, NOT in Music. I have completely removed the install, installed it again, the same text files earlier IDed are still there? Is there an alternative?
Computer details: Intel dual core, Nvidia, 64 bit Lucid. The flaw is that it runs on single memory.
I am using ubuntu 10.10 with gnome desktop on a desktop computer (pentium 4 cpu with 1 Gb memory), and when I use OpenOffice Impress, convert opens up and uses most of my cpu power. Even worse, after closing OpenOffice and all instances of nautilus, convert stays in memory and keeps using most of my cpu power (between 30 and 75 %, acccording to system monitor.Is it normal ? How could I set up my computer to limit convert cpu usage, and to unload from memory after being used.
View 3 Replies View RelatedI am writing a script that tells me which process consumes the most memory in the system this is what I have but I keep getting an error:
#! /bin/bash
# Autor: Jose miguel Colella
# Descripcion: Que proceso consume mas memoria
ps -e -o %mem -o args | sort -k 1 | tail -n2 | head -n1 | cut -d -f 3
I keep getting this message: cut: delimiter mist be a single character
I am having a php script which is used for bulk mailing. I run the script every minute through cron job. I have mentioned the path of the php script in a .sh file and execute the .sh file through cron job. Every time i run the script it utilizes high memory which results to server crash. how to restrict memory usage for that process to be a minimum one or how to set priority to be low for that process so that it is executed when there is no high priority jobs, so that the server runs normally without going down when the script runs
View 3 Replies View Relatedwe found that if we use 'top' to show the memory usage of a server (SuSe Linux 10), we can get virtual memory usage as well as 'Resident memory' usage. For virtual mem or a particular process, it is around 1.1GB, which is large but for resident memory, it only consumes 300MB. Are there anyone who knows what the differences are? I would also like to know whether the difference (1.1GB - 300MB) = 800MB are actually available for use by other applications in the system.
View 1 Replies View RelatedI have been setting up a vps I got out with bhost.net, with CentOS installed. I've been learning and have set up everying I need with the exception of ftp/sftp.
Using yum I installed vsftpd and ran into problems, thinking it was something I might of done I did a fresh install of CentOS and I still recieve the same problem on a fresh install so it is nothing I have done to the server.
The problem is when connecting via a sftp client I get an out of memory error. This error is listed in the putty faq ( url ) under A.7.5, there is a brief explaintion of the cure under A.7.6.
there is mention of a login script but I don't know where this is located. I'm a novice at Linux but by no means incompotent when it comes to computing.
I am having a Oracle server installed on Linux server. I want to clear the buffers after a certain time interval. I use the following command for the same.
echo 3 > /proc/sys/vm/drop_caches
Can I have a script which will execute the above script after certain time interval OR a script which will execute the above command when certain memory size is reached.
I'd like to ask you how install new physical memory in my hp ml350 g6 server with linux redhat operating system>
i sugess it is easy like in windows operating system but there is frind told me you must make mount , i am new with linux os .
I am trying to build and bring-up Linux (embedded) for a piece of hardware which have MIPS 74K proccessor 16MB Flash, 128MB DDR and network/usb support. How to configure/set into the kernel the exact addresses of the physical memory map? How does the kernel know where is the system ram, i/o memory, root FS? I have read some book and I found how the applications can go and read some special files like /proc/iomem to find out info about memory but what I need is how to set those addresses at the beginning when I build the kernel and FS in order to boot the kernel on my h/w.
View 3 Replies View RelatedWas wondering if anyone can explain briefly the relationship of "cache" and free memory in the "free" memory command.
View 2 Replies View RelatedWhen we want to setup a linux system, there is a common a suggestion like set the swap space as twice as big than your physical memory, I want to know why do we need this and how is this suggestion come from?
View 4 Replies View RelatedI am doing a test to get the memory used by apache`s apache2 processes. I used a script to get VmSize and VmRss from /proc/pid/status, and loop through that to get the sum of VmSize and VmRss of all the apache2 processes.
I found the VmSize (about 4GB) and VmRss (about 3.4GB) are much larger than the physical memory (1GB) when apache server was saturated. It was said because of the multi-counted libiraries size used by many processes simultaneously. Then , how to get the physical memory used by apache2 processes? Or how to get a more reasonable memory data?
I have a system with 2G of memory and swap memory of 4G.
This is the output from :
PHP Code:
How could they do to the memory cache to be used as much? Because, occasionally, swap is used and note that the system could use the memory cache does not swap ...
Slackware current 64 multilib.
A process is trying one access to memory, for example through an array (ex.: vect[0]=123. What happens?
Here below what I guess but I'm not sure and accept any comment (please, distinguish between "the system" and "the CPU" in case).
Let's suppose swapping to disk disbled.
We have two scenarios: without and with cache.
If no cache is present in the system:
1. The CPU must discover the phys addr of vect[0] virtual addr. To do that, has to read from 3 (or 2 depending on the system?) pages tables, stored in memory as well.
2. The CPU writes to the final address.
These mean 4 memory accesses.
If cache is present:
1. Like above but, if the pages tables are in cache, we have 3 accesses to that.
2. If the req. page is not in cache, it's reads from ram and transferred to it. Afterwards, cache is written.
In the best case we have 4 cache accesses.
I have rhgel5.3 installed on IBM blade server having 16 GB of physical memory. But it showing only 4GB.
#cat /proc/meminfo
#free -m
#top
[code]...
I have a 32 bit Ubuntu installed and my Laptop has 4GB RAM, but only 3GB is considered by Linux. My question is: what is the reason for the upper limit on physical memory ?
Code:
dmesg | grep Memory [0.000000] Memory: 3052428k/3112960k available (4673k kernel code, 56364k reserved, 2121k data, 656k init, 2200904k highmem) I am familiar with the virtual memory concept where linux splits upper 1GB for kernel and lower 3GB for user processes. In total, linux 32bit can address 4GB virtual addresses. Does this meant that 1GB of physical memory is already mapped to 1GB of kernel space and Linux only shows the remaining 3GB physical memory left for the user in the above command.
I did some searching on the internet and found some articles related to this, but it only confused me further since some articles suggest 4GB is the upper limit with mentioning whether it's virtual or physical memory, some bring in the concept of PAE, etc. I'm relative new to Linux's memory management, so it'd be really helpful if someone could answer this.
I read somewhere that i can clean cache memory with this comand:
Code:
david@lap:~$ sudo echo 3 > /proc/sys/vm/drop_caches
bash: /proc/sys/vm/drop_caches: Permission denied
The problem is that permission is denied as you see...
I have a system with 1 GB RAM. I'm running KDE 4. I created a tab to look that the Physical Memory in the System Monitor program, which I assume appears to look at the same stats that "top" looks at. In that Physical Memory tab I have 3 tables: Used Memory, Free Memory, and Application Memory.The Used Memory table shows that the system is using .94 of .98 GiBytes. The Free Memory table shows that the system has .5 GiBytes of RAM free.
However the Application Memory shows that only 339 M-Bytes of RAM is being used.Note that "top" shows the same info.So where is the other .6 GiBytes of RAM that the Used Memory table shows as being used?If I look at the process table which is supposed to encompass all of the processes running, including the ones for the OS, then it appears to add up to the 339 M-Bytes being used in the Application Memory table. Is the rest of the memory being held in reserve by the OS to be used as needed? If so, then why when another application is opened the Free Memory goes down instead of staying constant?I also noticed this memory "black hole" when I was running 11.0 on a system with 4 GB of RAM. The OS appeared to "take up" a large chunk of memory that was NOT being used by any applications and making it "disappear" - meaning that the applications were using about 1.3 GiBytes of RAM and Free Memory was showing only .7 GiBytes instead of the over 2 GB of RAM that should be free.
I allocated a chunk of memory using kmalloc in a Device Driver. Kmalloc provides a pointer to the allocated memory. This is one of my first few drivers.
I assume that the address returned is a Virtual address. I need to find the physical address of the memory location. I am working on an Intel 64 bit Fedora machine. I used the virt_to_phys() routine present in <asm/io_64.h>. I found that this routine returns an unsigned long value (32 bit) instead of an unsigned long long value (64 bit). Moreover, it seems that it simply returns the address - OFFSET instead of extracting the value in the page tables.
So is there any function / system call in Linux which will allow me to see the actual physical address on the Intel 64 arch.
I've installed my debian sid about one month ago (first xfce, next gnome) but noticed that it's kind of really slow. The upgrades take ages, launching (and using) firefox takes so much time,... In comparaison to my ubuntu, archlinux (on the same computer) or previous installation of debian there is clearly a problem somewhere.Today I tried to do a "top" sorted by mem usage : 3.5% xulrunner-stub, 2.1% dropbox, 1.4% aptitude (doing upgrade), 1.4% clementine,... nothing terriblebut still I've 2.7Gb or RAM used (more than 50%)
$ free -m
total used free shared buffers cached
Mem: 3967 26851282 0 79 1938
[code]....
(Ubuntu Linux server, 64-bits)I was troubleshooting a problem with a file (~3.0 GB) which I had just downloaded, but it was failing the integrity test, when I discovered something really unusual.First this is the MD5 of the file after download, which didn't match the expected value:
~% md5sum media.iso
5d74facb904cc1765a468354908a8f34 media.iso
Some time passes, nothing should have changed the file during this time, but when I went to check the file again:
~% md5sum media.iso
a5b97c5016afb39bd67ccfc3fa6ca59e media.iso
This was really unexpected. Since I have a lot of RAM, I suspected this was the effect of caching and something was going awry with it. I decided to retry with the whole file from disk, for my surprise:
~% sudo sysctl -w vm.drop_caches=3 # This linux command invalidates
vm.drop_caches = 3 # everything in the memory cache.
~% md5sum media.iso
2992aa6270f6e1de9154730ed3beedc1 media.iso
I redid it and now it seems to stay consistent, although this still isn't the value I was expecting. Certainly, the contents in memory cache were different from the contents on disk.This is the big problem.To fix the download, I created a torrent on the source machine and opened it in the target machine. Five 1MB chunks out of ~3.0GB failed integrity check. I used the torrent to fix these file chunks and how the file integrity is ok.The problem now is to determine where the data got out-of-sync.
I tested the memory with memtest86+, all but the bit fading test. I was expecting to see some failing memory module, but there wasn't anything. Everything is ok.Filesystem is Ext4, over LVM2, over a 3-disk RAID5 array.Ext4 is considered stable, and if data were inconsistent between disks, mdadm would have warned. But there is nothing in the logs. S.M.A.R.T. error logs are clean, the disks are new (have less than 30 days of "power-on-hours").I'm looking for information about any data-loss bugs in my current kernel (2.6.35), but there doesn't seem to be anything, as far as I looked.what else I could check, or where exactly could be the defect/bug?It is a Ubuntu 10.10 64-bit, Core i7 930, 6 GB non-ECC RAM.
Update: I confirmed that the files are being correctly written to the disk, the pages are being altered after they are read from disk, while in memory. I did a lot more memtests (I left it doing bit fade test overnight),and still nothing. All memory modules seem ok.Some more tests:
~% md5sum media.iso
cc8bcf1ce67ff7704eadc2222650c087 media.iso
~% cp media.iso tmp[code]....(direcat is a version of cat that reads with O_DIRECT, that is, bypassing page cache)There is a clear pattern: it always happens to the 2nd byte in a 16-byte alignment. In that byte, almost always the bit 4 (LSB) flips to one, but there was one instance where bit 2 flipped to zero.