Server :: Why VMrss Is Larger Than The Physical Memory
Aug 30, 2010I got the VMrss used by a process as about 2GB, but the physical memory of my computer is only 1G.
View 10 RepliesI got the VMrss used by a process as about 2GB, but the physical memory of my computer is only 1G.
View 10 Replieslet me know how to clear cache memory ( RHEL 5.1 ) as it consumes almost 100% physical memory.
View 3 Replies View RelatedWhy 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.
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 Relatedhow easy it would be to read the contents of a physical disk that was part of a larger logical volume. The disk contains a "Linux LVM" partition that spans its entire size. My problem is that one of my disks died, and I have to send it back for a warranty replacement. However, the disk is dead, and I can't zero it out. I'm just trying to assess how difficult it would be (or at least how likely it would be) for a tech that's checking out the disk to get at the data.
View 4 Replies View RelatedI have a ubuntu 10.10 server with apaceh2 and php and I want to open a file larger than 2gigs
I've read there is a flag that needs to be compiled into php to do this ?
I want to allocate the continuous memory larger 128k for DMA process. But i can't used kmalloc() function for allocate the memory larger 128KB. allocate the continuous memory larger 128k for DMA process.
View 1 Replies View RelatedI am using Ubuntu and looking for a good editor to edit a file that is > 4GB. I just need to put content at the end and beginning of the file. I suppose I could use something like
cat "text to add" >> huge_file
To append to the file. Is that the route to go? What about prepending? In general, what is the best route if I wanted to edit somewhere in the middle?
I've tried VIM and it fails miserably. I assume emacs and nano would be even worse. What else is there? I assume to accomplish what I am looking for, the editor would have to be specifically designed for this by not keeping the entirety of the file's contents in memory.
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 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 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.
Using Samba I have looked into the file that stores all my web sites, there were a few strange files that get larger and larger all the time. File names are _Za01716 and _Za01820, they are nearly 50mb in size now. I know these are not Log files so what are they and can I delete them?
View 1 Replies View RelatedAs i undertsand - out of 1GB of the virtual Address space for Kernel from 3GB to 4GB of the process address space, Kernel image (code, data, bss, stack, heap) resides staring @0x0 address. Vmalloc area starts either at the end of Physical ram size or at 896M. This 896M cap is mandated to ensure that minimum of 128MB is reserved as vmalloc_reserve for vmalloc,kmap etc.
Is the understanding correct? Now trying to map Physical Zones into this 1GB address space
Initial 16MB is mapped to ZONE_DMA
16MB - 896MB is mapped to ZONE_NORMAL
896MB - 1024MB is mapped to ZONE_HIGHMEM
Does this mean that Kernel image is residing in ZONE_DMA area? Any call to vmalloc() in kernel code will return address beyond 896M? insmod of any LKM will internally invoke vmalloc() to obtain contiguous area - where will this code physically located along with rest of kernel code in ZONE_DMA or in ZONE_HIGHMEM?
Whenever I'm running my application process, I've 1M physical memory usage is increasing for every 2 hours.This I observed using 'free -m' command.But 'top' command did not showing any increase 'RSS' size.It is same as it was started initially.Even though I stopped my process,the increased memory was not released back. If I start my application process then again memory usage start increasing by 1M for every 2 hours. increase of memory usage observer with 'free' and that too when my application is running, but top command is not showing any change in the RSS sizeIf my application is leaking any memory which is allocated by new/malloc, that should be released back whenever my application exit and the size increase will be show through top command for that process, right? This is not happeningThis proves that there is no potential leaks in my process.But why physical memory is increasing when only my process is running?
View 14 Replies View RelatedHow can I get the physical address corresponding to a virtual address in linux by using /proc file system
View 1 Replies View RelatedWhen I try to access at physical address (0xD0000), we known that it is necessary to convert physical address to virtual address using function IOREMAP(0xD0000, 1024) and return me 0xC00D0000.
Now our doubt is when I have a board with I/O in address 0x150, is it necessary to convert this address to other virtual address??? or with inb(0x150) return me state of I/O in this address? How can I known where is this I/O address in my map memory?
i need some information about how Debian 5 manage the physical memory . such as the memory management algorithms. i have googled it a lot but i couldn't find it.
View 1 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.
This may seem like a silly question but I have many servers and sometimes we forget when we login if it physical or virtual running on a VMware system. This makes a diffrence when I try to get a console access etc. So I wanted to know before if its physical or VM.yes I know i can change motd once i get the info or make a list etc. There are many ways not OS related for me to find this info out. But I was wondering if there was a Linux command that I could use when I ssh to a system to check if its physical or logical?I have inventory information etc and vm vsphere to check but that can be time consuming if I just want to check something quick.
uname -a or something like that that would tell me would be cool. I am thinking there is no command as Linux really does not care if its running Vm or physical.
I am trying to figure out the "actual" disk size used by my system. When I run the "df -h" command,I am not taking here into consideration the shared memory of 2Gb as it is a sort of virtual shared memory and is not allocated physically. Is that correct ?
View 8 Replies View RelatedI have a 7.9 TB logical volume I've created from 8 1 TB RAID 0 devices. The volume is formatted with XFS so I can resize when ready. However, I think I want to do something that is not possible. I have 2.5 TB free on my logical volume. I'd like to shrink the volume down to be 6 TB by getting rid of 2 of the 1 TB devices in the physical volume. However pvmove seems to require free extents in order to work. Do I need to add 6 TB of storage, pvmove everything onto it, and then decommission the original 8 1 TB physical devices from the volume?
View 5 Replies View Related