I understood the process of swapping but have a question about the swap space.
As far as I know, if I executes a program, the main memory fetches data from the disk because there is no data in cache and memory. Then what about swap space? Is swap space used only as a back-up storage when page swapping out is needed? or does the program place all of data to swap space when it is executed so that when there is a page fault, it swaps in?
Swapping has largely disappeared. However, M$ has recently reintroduced it to Windoze for certain processes.
In Swapping, the entire process is moved out of memory and stored on disk. The operating system uses swapping to make room in memory for other processes. In the days of 64K address spaces, transferring a process between memory and disk was not that time consuming.
Swapping has been replaced largely by PAGING whereby individual pages of memory are moved to secondary storage rather than the entire process.
Related
When a virtual address space is larger than the physical memory, OS can use swapping to evict page frames (e.g. LRU eviction). CPU generates Page Fault where then page that is in disk is swapped into the main memory. What happens when the virtual address is large enough that neither primary memory or disk have enough storage to hold it? What happens when a page frame is not in the disk either? Is another page fault called?
What happens when the virtual address is large enough that neither primary memory or disk have enough storage to hold it?
A virtual memory system maintains an image of the logical address space in secondary storage. A well-designed operating system is not going to allow a process to map a logical address that does not have a backing already in secondary storage. When your application calls a system service to map pages to the logical address space, the call will fail if there is no secondary storage available for the pages.
What happens when a page frame is not in the disk either?
There are some poorly designed operating systems that will map pages without having secondary storage behind them. You call the system service to map pages, it succeeds even if the pages could not be backed in secondary storage.
In that case, you get a memory exception upon access (and get no hint in your application that the real problem as a memory allocation failure).
Is another page fault called?
No.
In a logical memory system (as supported by most processors) a page has two states:
1. Mapped
2. Unmapped
In a virtual memory system, there are three states:
1. Mapped
2. Unmapped and valid
3. Unmapped and invalid
When a page fault occurs, the processor just knows the page is not mapped to memory. The operating system then has to figure out if the page is in secondary storage somewhere. If it is not, the operating causes the process to see an exception. If it is, the operating system loads and maps the page, the lets the process continue on its merry way.
When a virtual address space is larger than the physical memory, OS can use swapping to evict page frames (e.g. LRU eviction)
Lets assume that a virtual address is 48-bit (so the size of one virtual address space is 256 TiB), and you're running 123 processes where each one has its own virtual address space. This adds up to a total of 31488 TiB of virtual address space. Note: This is "very normal" for a modern 80x86 PC running a modern OS (Windows, Linux, ...).
Out of this 31488 TiB:
almost all of it will be unused and marked as "not present". If software tries to access it you get a page fault, the page fault handler realizes it's a bug, and you probably end up with a SIGSEGV (or "blue screen of death" or ...). Because it isn't being used the OS doesn't need any RAM or any disk space for it.
some of it will be the same things loaded into RAM once and then mapped into many virtual address spaces. This is extremely common for the kernel itself and for shared libraries/DLLs. It also includes cases where the same RAM is used for the virtual file system cache and for memory mapped files, or the same RAM is mapped into 2 or more processes as "shared memory", or when the same RAM is mapped into 2 or more virtual address spaces as "copy on write" (e.g. in the aftermath of fork()).
some will be "allocate on write" - literally the same page full of zeros mapped at many virtual addresses in many virtual address spaces, where if you write to it you get a page fault and the page fault handler allocates a new page of RAM for the page you tried to write to. This allows the OS to pretend that a huge amount of virtual space is allocated and filled with zeros without using any RAM or any disk space (until it actually is modified).
some will be (modified) data that is unique to a specific process.
The end result is that the 31488 TiB of total virtual space might only need a few GiB of RAM (and probably won't use swap space at all).
Over-commit
The OS does a pile of tricks to pretend memory was allocated when it actually wasn't. This creates the potential for a worst case where all the memory the OS pretends is allocated actually does need to be allocated. There are 2 ways to deal with this:
a) Refuse to let processes allocate more if you can't cover the worst case (e.g. return a "not enough memory" error when a process tries to allocate more than the OS can supply). This is bad because the worst case is extremely unlikely and you end up with software failing for no reason ("not enough memory" when there's actually plenty of memory to cover current requirements).
b) Allow processes allocate more (within reason), even if you can't cover the worst case. This works fine most of the time, but if the worst case actually happens something has to break (e.g. the OS terminates a process to free up some RAM).
The best option (in my opinion) is the first option (don't allow over-commit), but to have a large amount of swap space. Essentially; this is like "allow over-commit of RAM, but don't allow over-commit of swap space + RAM"; where the OS will probably be running slowly (due to excessive swap space use) before it has to start telling processes "no more memory"; and where most of the time everything will be in RAM (and ideally swap space is only used to cover the unlikely worst case).
I'm confused by this.
Are pages only memory units that exist in secondary memory or do they also exist in RAM too?
A memory page is the smallest unit of memory used by a virtual memory manager. A page can be backed by physical RAM, or by swap space or a page file on a hard drive. Pages backed by RAM have much faster IO, but as RAM gets full the OS may have to swap out pages to the hard drive.
Pages do not exist [physically] at all. A page is simply a redirection mechanism.
The operating system sets up of linear, logical address space for each process. The logical address space is organized into pages that in turn may map to:
A physical page frame of memory
No where
Somewhere on disk and managed by the operating system.
Paging is a memory management scheme by which a computer stores and retrieves data from secondary storage for use in main memory. Pages are used in RAM too, as a solution of external fragmentation.External fragmentation is a situation when total free space is enough to hold another process but space available is not contiguous. Compaction is one of the solution but for processes which are run-time loaded only. So, Paging is the true solution for external fragmentation where we implement page table which gives illusion that process has been given contiguous memory. Every address from CPU is broken down to page number and offset.
I'm using MongoDB on a 32 bit production system, which sucks but it's out of my control right now. The challenge is to keep the memory usage under ~2.5GB since going over this will cause 32 bit systems to crash.
According to the mongoDB team, the best way to track the memory usage is to use your operating system's process tracking system (i.e. ps or htop on Unix systems; Process Explorer on Windows.) for virtual memory size.
The DB mainly consists of one table which is continually cycling data, i.e. receiving data at regular intervals from sensors, and every day a cron job wipes all data from before the last 3 days. Over a period of time, the memory usage slowly increases. I took some notes over time using db.serverStats(), db.lectura.totalSize() and ps, shown in the chart below. Note that the size of the table in question has reduced in the last month but the memory usage increased nonetheless.
Now, there is some scope for adjustment in how many days of data I store. Today I deleted basically half of the data, and then restarted mongodb, and yet the mem virtual / mem mapped and most importantly memory usage according to ps have hardly changed! Why do these not reduce when I wipe data (and restart)? I read some other questions where people said that mongo isn't really using all the memory that it might appear to be using, and that you can't clear the cache or limit memory use. But then how can I ensure I stay under the 2.5GB limit?
Unless there is a way to stem this dataset-size-irrespective gradual increase in memory usage, it seems to me that the 32-bit version of Mongo is unuseable. Note: I don't mind losing a bit of performance if it solves the problem.
To answer regarding why the mapped and virtual memory usage does not decrease with the deletes, the mapped number is actually what you get when you mmap() the entire set of data files. This does not shrink when you delete records, because although the space is freed up inside the data files, they are not themselves reduced in size - the files are just more empty afterwards.
Virtual will include journal files, and connections, and other non-data related memory usage also, but the same principle applies there. This, and more, is described here:
http://www.mongodb.org/display/DOCS/Checking+Server+Memory+Usage
So, the 2GB storage size limitation on 32-bit will actually apply to the data files whether or not there is data in them. To reclaim deleted space, you will have to run a repair. This is a blocking operation and will require the database to be offline/unavailable while it was run. It will also need up to 2x the original size in terms of free disk space to be able to run the repair, since it essentially represents writing out the files again from scratch.
This limitation, and the problems it causes, is why the 32-bit version should not be run in production, it is just not suitable. I would recommend getting onto a 64-bit version as soon as possible.
By the way, neither of these figures (mapped or virtual) actually represents your resident memory usage, which is what you really want to look at. The best way to do this over time is via MMS, which is the free monitoring service provided by 10gen - it will graph virtual, mapped and resident memory for you over time as well as plenty of other stats.
If you want an immediate view, run mongostat and check out the corresponding memory columns (res, mapped, virtual).
In general, when using 64-bit builds with essentially unlimited storage, the data will usually greatly exceed the available memory. Therefore, mongod will use all of the available memory it can in terms of resident memory (which is why you should always have swap configured to the OOM Killer does not come into play).
Once that is used, the OS does not stop allocating memory, it will just have the oldest items paged out to make room for the new data (LRU). In other words, the recycling of memory will be done for you, and the resident memory level will remain fairly constant.
Your options for stretching 32-bit are limited, but you can try some things. The thing that you run out of is address space, and the increases in the sizes of additional database files mean that you would like to avoid crossing over the boundary from "n" files to "n+1". It may be worth structuring your data into more or fewer databases so that you can get the maximum amount of actual data into memory and as little as possible "dead space".
For example, if your database named "mydatabase" consists of the files mydatabase.ns (the namespace file) at 16 MB, mydatabase.0 at 64 MB, mydatabase.1 at 128 MB and mydatabase.2 at 256 MB, then the next file created for this database will be mydatabase.3 at 512 MB. If instead of adding to mydatabase you instead created an additional database "mynewdatabase" it would start life with mynewdatabase.ns at 16 MB and mynewdatabase.0 at 64 MB ... quite a bit smaller than the 512 MB that adding to the original database would be. In fact, you could create 4 new databases for less space than would be consumed by adding a new file to the original database, and because the files are smaller they would be easier to fit into contiguous blocks of memory.
It is a well-known message that 32-bit should not be used for production.
Use 64-bit systems.
Point.
Can any one please make me clear what is the difference between virtual memory and swap space?
And why do we say that for a 32-bit machine the maximum virtual memory accessible is 4 GB only?
There's an excellent explantation of virtual memory over on superuser.
Simply put, virtual memory is a combination of RAM and disk space that running processes can use.
Swap space is the portion of virtual memory that is on the hard disk, used when RAM is full.
As for why 32bit CPU is limited to 4gb virtual memory, it's addressed well here:
By definition, a 32-bit processor uses
32 bits to refer to the location of
each byte of memory. 2^32 = 4.2
billion, which means a memory address
that's 32 bits long can only refer to
4.2 billion unique locations (i.e. 4 GB).
There is some confusion regarding the term Virtual Memory, and it actually refers to the following two very different concepts
Using disk pages to extend the conceptual amount of physical memory a computer has - The correct term for this is actually Paging
An abstraction used by various OS/CPUs to create the illusion of each process running in a separate contiguous address space.
Swap space, OTOH, is the name of the portion of disk used to store additional RAM pages when not in use.
An important realization to make is that the former is transparently possible due to the hardware and OS support of the latter.
In order to make better sense of all this, you should consider how the "Virtual Memory" (as in definition 2) is supported by the CPU and OS.
Suppose you have a 32 bit pointer (64 bit points are similar, but use slightly different mechanisms). Once "Virtual Memory" has been enabled, the processor considers this pointer to be made as three parts.
The highest 10 bits are a Page Directory Entry
The following 10 bits are a Page Table Entry
The last 12 bits make up the Page Offset
Now, when the CPU tries to access the contents of a pointer, it first consults the Page Directory table - a table consisting of 1024 entries (in the X86 architecture the location of which is pointed to by the CR3 register). The 10 bits Page Directory Entry is an index in this table, which points to the physical location of the Page Table. This, in turn, is another table of 1024 entries each of which is a pointer in physical memory, and several important control bits. (We'll get back to these later). Once a page has been found, the last 12 bits are used to find an address within that page.
There are many more details (TLBs, Large Pages, PAE, Selectors, Page Protection) but the short explanation above captures the gist of things.
Using this translation mechanism, an OS can use a different set of physical pages for each process, thus giving each process the illusion of having all the memory for itself (as each process gets its own Page Directory)
On top of this Virtual Memory the OS may also add the concept of Paging. One of the control bits discussed earlier allows to specify whether an entry is "Present". If it isn't present, an attempt to access that entry would result in a Page Fault exception. The OS can capture this exception and act accordingly. OSs supporting swapping/paging can thus decide to load a page from the Swap Space, fix the translation tables, and then issue the memory access again.
This is where the two terms combine, an OS supporting Virtual Memory and Paging can give processes the illusion of having more memory than actually present by paging (swapping) pages in and out of the swap area.
As to your last question (Why is it said 32 bit CPU is limited to 4GB Virtual Memory). This refers to the "Virtual Memory" of definition 2, and is an immediate result of the pointer size. If the CPU can only use 32 bit pointers, you have only 32 bit to express different addresses, this gives you 2^32 = 4GB of addressable memory.
Hope this makes things a bit clearer.
IMHO it is terribly misleading to use the concept of swap space as equivalent to virtual memory. VM is a concept much more general than swap space. Among other things, VM allows processes to reference virtual addresses during execution, which are translated into physical addresses with the support of hardware and page tables. Thus processes do not concern about how much physical memory the system has, or where the instruction or data is actually resident in the physical memory hierarchy. VM allows this mapping. The referenced item (instruction or data) may be resident in L1, or L2, or RAM, or finally on disk, in which case it is loaded into main memory.
Swap space it is just a place on secondary memory where pages are stored when they are inactive. If there is no sufficient RAM, the OS may decide to swap-out pages of a process, to make room for other process pages. The processor never ever executes instruction or read/write data directly from swap space.
Notice that it would be possible to have swap space in a system with no VM. That is, processes that directly access physical addresses, still could have portions of it on
disk.
Though the thread is quite old and has already been answered. Still would like to share this link as this is the simplest explanation I have found so far. Below link has got diagrams for better visualization.
Key Difference: Virtual memory is an abstraction of the main memory. It extends the available memory of the computer by storing the inactive parts of the content RAM on a disk. Whenever the content is required, it fetches it back to the RAM. Swap memory or swap space is a part of the hard disk drive that is used for virtual memory. Thus, both are also used interchangeably.
Virtual memory is quiet different from the physical memory. Programmers get direct access to the virtual memory rather than physical memory. Virtual memory is an abstraction of the main memory. It is used to hide the information of the real physical memory of the system. It extends the available memory of the computer by storing the inactive parts of the RAM's content on a disk. When the content is required, it fetches it back to the RAM. Virtual memory creates an illusion of a whole address space with addresses beginning with zero. It is mainly preferred for its optimization feature by which it reduces the space requirements. It is composed of the available RAM and disk space.
Swap memory is generally called as swap space. Swap space refers to the portion of the virtual memory which is reserved as a temporary storage location. Swap space is utilized when available RAM is not able to meet the requirement of the system’s memory. For example, in Linux memory system, the kernel locates each page in the physical memory or in the swap space. The kernel also maintains a table in which the information regarding the swapped out pages and pages in physical memory is kept.
The pages that have not been accessed since a long time are sent to the swap space area. The process is referred to as swapping out. In case the same page is required, it is swapped in physical memory by swapping out a different page. Thus, one can conclude that swap memory and virtual memory are interconnected as swap memory is used for the technique of virtual memory.
difference-between-virtual-memory-and-swap-memory
"Virtual memory" is a generic term. In Windows, it is called as Paging or pagination. In Linux, it is called as Swap.
Or it would be faster to re-read that data from mapped memory once again, since the OS might implement its own cache?
The nature of data is not known in advance, it is assumed that file reads are random.
i wanted to mention a few things i've read on the subject. The answer is no, you don't want to second guess the operating system's memory manager.
The first comes from the idea that you want your program (e.g. MongoDB, SQL Server) to try to limit your memory based on a percentage of free RAM:
Don't try to allocate memory until there is only x% free
Occasionally, a customer will ask for a way to design their program so it continues consuming RAM until there is only x% free. The idea is that their program should use RAM aggressively, while still leaving enough RAM available (x%) for other use. Unless you are designing a system where you are the only program running on the computer, this is a bad idea.
(read the article for the explanation of why it's bad, including pictures)
Next comes from some notes from the author of Varnish, and reverse proxy:
Varnish Cache - Notes from the architect
So what happens with squids elaborate memory management is that it gets into fights with the kernels elaborate memory management, and like any civil war, that never gets anything done.
What happens is this: Squid creates a HTTP object in "RAM" and it gets used some times rapidly after creation. Then after some time it get no more hits and the kernel notices this. Then somebody tries to get memory from the kernel for something and the kernel decides to push those unused pages of memory out to swap space and use the (cache-RAM) more sensibly for some data which is actually used by a program. This however, is done without squid knowing about it. Squid still thinks that these http objects are in RAM, and they will be, the very second it tries to access them, but until then, the RAM is used for something productive.
Imagine you do cache something from a memory-mapped file. At some point in the future that memory holding that "cache" will be swapped out to disk.
the OS has written to the hard-drive something which already exists on the hard drive
Next comes a time when you want to perform a lookup from your "cache" memory, rather than the "real" memory. You attempt to access the "cache", and since it has been swapped out of RAM the hardware raises a PAGE FAULT, and cache is swapped back into RAM.
your cache memory is just as slow as the "real" memory, since both are no longer in RAM
Finally, you want to free your cache (perhaps your program is shutting down). If the "cache" has been swapped out, the OS must first swap it back in so that it can be freed. If instead you just unmapped your memory-mapped file, everything is gone (nothing needs to be swapped in).
in this case your cache makes things slower
Again from Raymon Chen: If your application is closing - close already:
When DLL_PROCESS_DETACH tells you that the process is exiting, your best bet is just to return without doing anything
I regularly use a program that doesn't follow this rule. The program
allocates a lot of memory during the course of its life, and when I
exit the program, it just sits there for several minutes, sometimes
spinning at 100% CPU, sometimes churning the hard drive (sometimes
both). When I break in with the debugger to see what's going on, I
discover that the program isn't doing anything productive. It's just
methodically freeing every last byte of memory it had allocated during
its lifetime.
If my computer wasn't under a lot of memory pressure, then most of the
memory the program had allocated during its lifetime hasn't yet been
paged out, so freeing every last drop of memory is a CPU-bound
operation. On the other hand, if I had kicked off a build or done
something else memory-intensive, then most of the memory the program
had allocated during its lifetime has been paged out, which means that
the program pages all that memory back in from the hard drive, just so
it could call free on it. Sounds kind of spiteful, actually. "Come
here so I can tell you to go away."
All this anal-rententive memory management is pointless. The process
is exiting. All that memory will be freed when the address space is
destroyed. Stop wasting time and just exit already.
The reality is that programs no longer run in "RAM", they run in memory - virtual memory.
You can make use of a cache, but you have to work with the operating system's virtual memory manager:
you want to keep your cache within as few pages as possible
you want to ensure they stay in RAM, by the virtue of them being accessed a lot (i.e. actually being a useful cache)
Accessing:
a thousand 1-byte locations around a 400GB file
is much more expensive than accessing
a single 1000-byte location in a 400GB file
In other words: you don't really need to cache data, you need a more localized data structure.
If you keep your important data confined to a single 4k page, you will play much nicer with the VMM; Windows is your cache.
When you add 64-byte quad-word aligned cache-lines, there's even more incentive to adjust your data structure layout. But then you don't want it too compact, or you'll start suffering performance penalties of cache flushes from False Sharing.
The answer is highly OS-specific. Generally speaking, there will be no sense in caching this data. Both the "cached" data as well as the memory-mapped can be paged away at any time.
If there will be any difference it will be specific to an OS - unless you need that granularity, there is no sense in caching the data.