When a page miss occurs in the main memory, the particular page is then retrieved from the disk.
My question to the community, how is the memory in the disk indexed with the 32-bit physical address we have after the translation?
Related
My question is (as stated) is there an access to the disk/virtual memory every time there is a page fault?
If not, in what cases is there no disk access?
(this is a general question about paging and not implementation dependent)
Thank you
There are some cases where a page fault will not trigger a disk access.
A page fault is an exception triggered when a process requests a virtual address that is not mapped by the MMU (memory-management unit) to a physical page. If the page is not resident anywhere in physical memory (RAM), the virtual address is valid, and page data that exists in the backing store (i.e. disk) is required, the page fault handler will trigger an access to the disk. This situation is usually classified as a hard page fault or major page fault.
During a minor page fault or soft page fault, the requested physical page is not mapped in the virtual address space of the process by the MMU, but no disk access is necessary, for a few possible reasons:
The page could be resident in physical memory, even if it isn’t mapped by the process. The page may be part of shared memory and mapped by another process, or the page could not currently be mapped by any process, but in physical memory and not yet written to disk or erased.
Some operating systems have the concept of a “working set” of pages that are mapped at a given time for a process, and will set an upper limit on the number of pages that can be part of the working set. Pages may be removed from the working set of a process when it is determined that the working set of the process has grown too large, but remain unmodified in the free page list long enough to be requested and mapped again.
The page fault could also be a demand-zero fault — a demand-zero fault can be satisfied by allocating a free page in physical memory and setting its contents to all 0’s, so no disk access is necessary.
Cases where an address that is not a part of the virtual address space is requested can also trigger a page fault, usually called an invalid page fault. If a process attempts to reference a NULL pointer, for example, the page fault handler will usually trigger a segmentation fault. (Invalid page faults don't necessarily always occur due to a page protection violation -- this is just any example of one case).
(I referenced the Wikipedia page for Page fault and would definitely recommend it as a concise introduction to page faults!)
I observe substantial speedups in data transfer when I use pinned memory for CUDA data transfers. On linux, the underlying system call for achieving this is mlock. From the man page of mlock, it states that locking the page prevents it from being swapped out:
mlock() locks pages in the address range starting at addr and continuing for len bytes. All pages that contain a part of the specified address range are guaranteed to be resident in RAM when the call returns successfully;
In my tests, I had a fews gigs of free memory on my system so there was never any risk that the memory pages could've been swapped out yet I still observed the speedup. Can anyone explain what's really going on here?, any insight or info is much appreciated.
CUDA Driver checks, if the memory range is locked or not and then it will use a different codepath. Locked memory is stored in the physical memory (RAM), so device can fetch it w/o help from CPU (DMA, aka Async copy; device only need list of physical pages). Not-locked memory can generate a page fault on access, and it is stored not only in memory (e.g. it can be in swap), so driver need to access every page of non-locked memory, copy it into pinned buffer and pass it to DMA (Syncronious, page-by-page copy).
As described here http://forums.nvidia.com/index.php?showtopic=164661
host memory used by the asynchronous mem copy call needs to be page locked through cudaMallocHost or cudaHostAlloc.
I can also recommend to check cudaMemcpyAsync and cudaHostAlloc manuals at developer.download.nvidia.com. HostAlloc says that cuda driver can detect pinned memory:
The driver tracks the virtual memory ranges allocated with this(cudaHostAlloc) function and automatically accelerates calls to functions such as cudaMemcpy().
CUDA use DMA to transfer pinned memory to GPU. Pageable host memory cannot be used with DMA because they may reside on the disk.
If the memory is not pinned (i.e. page-locked), it's first copied to a page-locked "staging" buffer and then copied to GPU through DMA.
So using the pinned memory you save the time to copy from pageable host memory to page-locked host memory.
If the memory pages had not been accessed yet, they were probably never swapped in to begin with. In particular, newly allocated pages will be virtual copies of the universal "zero page" and don't have a physical instantiation until they're written to. New maps of files on disk will likewise remain purely on disk until they're read or written.
A verbose note on copying non-locked pages to locked pages.
It could be extremely expensive if non-locked pages are swapped out by OS on a busy system with limited CPU RAM. Then page fault will be triggered to load pages into CPU RAM through expensive disk IO operations.
Pinning pages can also cause virtual memory thrashing on a system where CPU RAM is precious. If thrashing happens, the throughput of CPU can be degraded a lot.
I have an application that loads 170 files (let’s say they are text files) from disk in individual objects and kept in memory all the time. The memory is allocated once when I load those files from disk. So, there is no memory fragmentation involved. I also use FastMM to make sure my applications never leaks memory.
The application compares all these files with each other to find similarities. Over-simplified we can say that we compare text strings but the algorithm is way more complex as I have to allow some differences between strings. Each file is about 300KB. Loaded in memory (the object that holds it) it takes about 0.4MB of RAM. So, the running app takes about 60MB or RAM (working set). It processes the data for about 15 minutes. The thing is that it generates over 40 million page faults.
Why? I have about 2GB of free RAM. From what I know Page Faults are slow. How much they are slowing down my program?
How can I optimize the program to reduce these page faults? I guess it has something to do with data locality. Does anybody know some example algorithms for this (Delphi)?
Update:
But looking at the number of page faults (no other application in Task Manager comes close to mine, not even by far) I guess that I could increase the speed of my application IF I manage to optimize memory layout (reduce the page faults).
Delphi 7, Win 7 32 bit, RAM 4GB (3GB visible, 2GB free).
Caveat - I'm only addressing the page faulting issue.
I cannot be sure but have you considered using Memory Mapped files? In this way windows will use the files themselves as the paging file (rather than the main paging file pagrefile.sys). If the files are read only then the number of page faults should theoretically decrease as the pages won't need to written out to disk via the paging file as windows will just load the data from the file itself as needed.
Now to reduce files from paging in and out you need to try and go through the data in one direction so that as new data is read, older pages can be discarded for ever. Here is where you trade off going over the files again and caching data - the cache has to be stored somewhere.
Note that Memory Mapped files is how windows loads .dlls and .exes amongst other things. I've used them to scan though gigabyte files without hitting memory limits (we had MBs in those days and not GBs of ram).
However from the data you describe I'd suggest the ability to not go back ovver files will reduce the amount of repaging going on.
On my machine most pagefaults are reported for developer studio which is reported to have 4M page faults after 30+ minutes total CPU time. You get 10 times more, in half the time. And memory is scarce on my system. So 40M faults seems like a lot.
It could just maybe be you have a memory leak.
the working set is only the physical memory in use for your application. If you leak memory, and don't touch it, it will get paged out. You will see the virtual memory useage (or page file use) increase. These pages might be swapped back in when the heap memory walks the heap, to get swapped out again by windows.
Because you have a lot of RAM, the swapped out pages will stay in physical memory, as nobody else needs them. (a page recovered from RAM counts as a soft fault, from disk as a hard one)
Do you use an exponential resize system ?
If you grow the block of memory in too small increments while loading, it might constantly request large blocks from the system, copy the data over, and then release the old block (assuming that fastmm (de)allocates very large blocks directly from the OS).
Maybe somehow this causes a loop where the OS releases memory from your app's process, and then adds it again, causing page faults on first write.
Also avoid Tstringlist.load* methods for very large files, IIRC these consume twice the space needed.
i just have a general question about cache memory. How would a program perform badly on a cache based system ? , since cache memory stores adresses from main memory that is requested, aswell as adresses that ranges around the same adress as the one copied from the main memory.
Read this: What every programmer should know about memory (PDF).
In my Windows XP Task Manager, some processes display a higher value in the Mem Usage column than the VMSize. My Firefox instance, for example shows 111544 K as mem usage and 100576 K as VMSize.
According to the help file of Task Manager Mem Usage is the working set of the process and VMSize is the committed memory in the Virtual address space.
My question is, if the number of committed pages for a process is A and the number of pages in physical memory for the same process is B, shouldn't it always be B ≤ A? Isn't the number of pages in physical memory per process a subset of the committed pages?
Or is this something to do with sharing of memory among processes? Please explain. (Perhaps my definition of 'Working Set' is off the mark).
Thanks.
Virtual Memory
Assume that your program (eg Oracle) allocated 100 MB of memory upon startup - your VM size goes up by 100 MB though no additional physical / disk pages are touched. ie VM is nothing but memory book keeping.
The total available physical memory + paging file memory is the maximum memory that ALL the processes in the system can allocate. The system does this so that it can ensure that at any point time if the processes actually start consuming all that memory it allocated the OS can supply the actual physical pages required.
Private Memory
If the program copies 10 MB of data into that 100 MB, OS senses that no pages have been allocated to the process corresponding to those addresses and assigns 10 MB worth of physical pages into your process's private memory. (This process is called page fault)
Working Set
Definition : Working set is the set of memory pages that have been recently touched by a program.
At this point these 10 pages are added to the working set of the process. If the process then goes and copies this data into another 10 MB cache previously allocated, everything else remains the same but the Working Set goes up again by 10 Mb if those old pages where not in the working set. But if those pages where already in the working set, then everything is good and the programs working set remains the same.
Working Set behaviour
Imagine your process never touches the first 10 pages ever again, in which case these pages are trimmed off from your process's working set and possibly sent to the page file so that the OS can bring in other pages that are more frequently used. However if there are no urgent low memory requirements, then this act of paging need not be done and OS can act as if its rich in memory. In this case the working set simply lets these pages remain.
When is Working Set > Virtual Memory
Now imagine the same program de-allocates all the 100 Mb of memory. The programs VM size is immediately reduced by 100 MB (remember VM = book keeping of all memory allocation requests)
The working set need not be affected by this, since that doesn't change the fact that those 10 Mb worth of pages where recently touched. Therefore those pages still remain in the working set of the process though the OS can reclaim them whenever it requires.
This would effectively make the VM < working set. However this will rectify if you start another process that consumes more memory and the working set pages are reclaimed by the OS.
XP's Task Manager is simply wrong. EDIT: If you don't believe me (and someone doesn't, because they voted this down), read Firefox 3 Memory Usage. I quote:
If you’re looking at Memory Usage
under Windows XP, your numbers aren’t
going to be so great. The reason:
Microsoft changed the meaning of
“private bytes” between XP and Vista
(for the better).
Sounds like MS got confused. You only change something like that if it's broken.
Try Process Explorer instead. What Task Manager labels "VM Size", Process Explorer (more correctly) labels "Private Bytes". And in Process Explorer, Working Set (and Private Bytes) are always less than or equal to Virtual Size, as you would expect.
File mapping
Very common way how Mem Usage can be higher than VM Size is by using file mapping objects (hence it can be related to shared memory, as file mapping is used to share memory). With file mapping you can have a memory which is committed (either in page file or in physical memory, you do not know), but has no virtual address assigned to it. The committed memory appears in Mem Usage, while used virtual addresses usage is tracked by VM Size.
See also:
What does “VM Size” mean in the Windows Task Manager? on Stackoverflow
Breaking the 32 bit Barrier in my developer blog
Usenet discussion Still confused why working set larger than virtual memory
Memory usage is the amount of electronic memory currently allocated to the process.
VM Size is the amount of virtual memory currently allocated to the process.
so ...
A page that exists only electronically will increase only Memory Usage.
A page that exists only on disk will increase only VM Size.
A page that exists both in memory and on disk will increase both.
Some examples to illustrate:
Currently on my machine, iexplore has 16,000K Memory Usage and 194,916 VM Size. This means that most of the memory used by Internet Explorer is idle and has been swapped out to disk, and only a fraction is being kept in main memory.
Contrast with mcshield.exe with has 98,984K memory usage and 98,168K VM Size. My conclusion here is that McAfee AntiVirus is active, with at lot of memory in use. Since it's been running for quite some time (all day, since booting), I expect that most of the 98,168K VM Size is copies of the electronic memory - though there's nothing in Task Manager to confirm this.
You might find some explaination in The Memory Shell Game
Working Set (A) – This is a set of virtual memory pages (that are committed) for a process and are located in physical RAM. These pages fully belong to the process. A working set is like a "currently/recently working on these pages" list.
Virtual Memory – This is a memory that an operating system can address. Regardless of the amount of physical RAM or hard drive space, this number is limited by your processor architecture.
Committed Memory – When an application touches a virtual memory page (reads/write/programmatically commits) the page becomes a committed page. It is now backed by a physical memory page. This will usually be a physical RAM page, but could eventually be a page in the page file on the hard disk, or it could be a page in a memory mapped file on the hard disk. The memory manager handles the translations from the virtual memory page to the physical page. A virtual page could be in located in physical RAM, while the page next to it could be on the hard drive in the page file.
BUT: PF (Page File) Usage - This is the total number of committed pages on the system. It does not tell you how many are actually written to the page file. It only tells you how much of the page file would be used if all committed pages had to be written out to the page file at the same time.
Hence B > A...
If we agree that B represents "mem usage" or also PF usage, the problem comes from the fact it actually represents potential page usages: in Xp, this potential file space can be used as a place to assign those virtual memory pages that programs have asked for, but never brought into use...
Memory fragmentation is probably the reason:
If the process allocates 1 octet, it counts for 1 octet in the VMSize, but this 1 octet requires a physical page (4K on windows operating system).
If after allocating/freeing memory, the process has a second octet that is separated by more than 4K from the first one, this second octet will always be stored on a separate physical page than the 1 one.
So the VM Size count is 2 octets but the Memory Usage is 2 pages== 8K
So the fact that MemUsage is greater than VMSize shows that process does a lot of allocation and deallocation and fragments the memory.
This could be because the process is started a long time ago.
Or else there is place for optimization ;-)