I am currently taking a course in Operating Systems and I came across address virtualization. I will give a brief about what I know and follow that with my question.
Basically, the CPU(modern microprocessors) generates virtual addresses and then an MMU(memory management unit) takes care of translating those virtual address to their corresponding physical addresses in the RAM. The example that was given by the professor is there is a need for virtualization because say for example: You compile a C program. You run it. And then you compile another C program. You try to run it but the resident running program in memory prevents loading a newer program even when space is available.
From my understanding, I think having no virtualization, if the compiler generates two physical addresses that are the same, the second won't run because it thinks there isn't enough space for it. When we virtualize this, as in the CPU generates only virtual addresses, the MMU will deal with this "collision" and find a spot for the other program in RAM.(Our professor gave the example of the MMU being a mapping table, that takes a virtual address and maps it to a physical address). I thought of that idea to be very similar to say resolving collisions in a hash table.
Could I please get some input on my understanding and any further clarification is appreciated.
Could I please get some input on my understanding and any further clarification is appreciated.
Your understanding is roughly correct.
Clarifications:
The data structures are nothing like a hash table.
If anything, the data structures are closer to a BTree, but even there are important differences with that as well. It is really closest to a (Java) N-dimensional array which has been sparsely allocated.
It is mapping pages rather than complete virtual / physical addresses. (A complete address is a page address + an offset within the page.).
There is no issue with collision. At any point in time, the virtual -> physical mappings for all users / processes give a one-to-one mapping from (process id + virtual page) to a either a physical RAM page or a disk page (or both).
The reasons we use virtual memory are:
process isolation; i.e. one process can't see or interfere with another processes memory
simplifying application writing; i.e. each process thinks it has a contiguous set off memory addresses, and the same set each time. (To a first approximation ...)
simplifying compilation, linking, loading; i.e. the compilers, etc there is no need to "relocate" code at compile time or run time to take into account other.
to allow the system to accommodate more processes than it has physical RAM for ... though this comes with potential risks and performance penalties.
I think you have a fundamental misconception about what goes on in an operating system in regard to memory.
(1) You are describing logical memory, not virtual memory. Virtual memory refers to the use of disk storage to simulate memory. Unmapped pages of logical memory get mapped to disk space.
Sadly, the terms logical memory and virtual memory get conflated but they are distinct concepts the the distinction is becoming increasingly important.
(2) Programs run in a PROCESS. A process only runs one program at a time (in unix each process generally only runs one program (two if you count the cloned caller) in its life.
In modern systems each process process gets a logical address space (sequential addresses) that can be mapped to physical locations or no location at all. Generally, part of that logical address space is mapped to a kernel area that is shared by all processes. The logical address space is create with the process. No address space—no process.
In a 32-bit system, addresses 0-7FFFFFFF might be user address that are (generally) mapped to unique physical locations while 80000000-FFFFFFFFmight be mapped to a system address space that is the same for all processes.
(3) Logical memory management primarily serves as a means of security; not as a means for program loading (although it does help in that regard).
(4) This example makes no sense to me:
You compile a C program. You run it. And then you compile another C program. You try to run it but the resident running program in memory prevents loading a newer program even when space is available.
You are ignoring the concept of a PROCESS. A process can only have one program running at a time. In systems that do permit serial running of programs with the same process (e.g., VMS) the executing program prevents loading another program (or the loading of another program causes the termination of the running program). It is not a memory issue.
(5) This is not correct at all:
From my understanding, I think having no virtualization, if the compiler generates two physical addresses that are the same, the second won't run because it thinks there isn't enough space for it. When we virtualize this, as in the CPU generates only virtual addresses, the MMU will deal with this "collision" and find a spot for the other program in RAM.
The MMU does not deal with collisions. The operating system sets up tables that define the logical address space when the process start. Logical memory has nothing to do with hash tables.
When a program accesses logical memory the rough sequence is:
Break down the address into a page and an offset within the page.
Does the page have an corresponding entry in the page table? If not FAULT.
Is the entry in the page table valid? If not FAULT.
Does the page table entry allow the type of access (read/write/execute) requested in the current operating mode (kernel/user/...)? If not FAULT.
Does the entry map to a physical page? If not PAGE FAULT (go load the page from disk--virtual memory––and try again).
Access the physical memory referenced by the page table.
Related
with reference to following questions
1 How are the different segments like heap, stack, text related to the physical memory?
2 memory allocation in data/bss/heap and stack
3 how does a program runs in memory and the way memory is handled by Operating system
4 How convert address in elf to physical address
especially in question 1 what is asked to clear still needs experts comments.
at the same time different answers/comments in opposite directions making this more confusing. what we know is that elf is loaded in virtual address space and when needed the actual physical page, process is provided with that page by MMU. I have severe doubts about this. if mappings are only made in virtual address then how execution is started? one possibility is that segments from elf binary are loaded into virtual address space and an empty pagetable is also created which is populated on page faults on each memory access. And in this population there is no role of ELF binary stored on disk.I can delete my elf stored on disk with out effecting execution?
if above understanding is right confusion lies in lack of knowledge on our side about virtual memory. As VM is considered deception but it seems it is more than that. lets say int x=12 when this line is loaded in virtual space it means there is record keeping for value of x (12) in VM. when some instruction mov x,register is executed for the first time, a page is created in RAM only to give actual space to make this instruction run?
After noticing the reputation scores of mine and others with similar question even if this is basic question, this concept is creating a lot of confusions for me at least.
if above understanding is right confusion lies in lack of knowledge on our side about virtual memory.
The understanding you state is not right, but your confusion certainly does seem to be largely about virtual memory. The point of a virtual memory system is to allow each process to access the whole address space (more or less) as if it were the only process running, and without regard to how much physical RAM is available. "Virtual" does not mean "fake" or any such thing -- with one or two caveats, if a program has virtual memory assigned to it, then it has bona fide storage. The "virtual" part has to do with the real location at any given time of the storage associated with any particular virtual address.
You also seem to be confusing yourself by associating ELF too closely with virtual memory. Certainly the OS will normally load an ELF executable into virtual memory (backed by some combination of physical RAM and disk that may vary over time), but it would do the same with a program in any other executable format it supports, and there are others.
I have severe doubts about this. if mappings are only made in virtual address then how execution is started?
The operating system kernel is the exception -- it runs in real memory. Managing virtual memory for all running processes is one of its jobs, and the page table for each process belongs to it, not to the process. The kernel handles setting up and starting processes, including their page tables, and at all times it knows the actual location of all storage assigned to each one, and to what virtual addresses that storage is mapped.
And in this population there is no role of ELF binary stored on disk. I can delete my elf stored on disk with out [a]ffecting execution?
The ELF binary stored on disk contains executable code and data for the program that will be loaded into the process's (virtual) memory when the program is launched. This is the job of the ELF dynamic linker (ld.linux.so.2, for example). Certainly the ELF file is needed to launch the program. After the process has started, then sure, the on-disk ELF binary can be unlinked (via the rm command, for example) without affecting the execution of the program, since it has been loaded into memory (on Unix-like systems; Windows is different, but does not support ELF). That binary will then be unavailable to launch any new instance of the program.
Note, however, that unlinking is not necessarily the same thing as deleting. As long as any process has the unlinked file open, it will remain allocated on the file system. This may well be the case for a running ELF program, as one good loading strategy is to map parts of the binary file into memory (the .text and .rodata sections, for example). This will hold the file open as long any process with such mappings is alive, even if the binary is unlinked from the file system.
Virtual Memory is a quite complex topic for me. I am trying to understand it. Here is my understanding for a 32-bit system. Example RAM is just 2GB. I have tried reading many links, and I am not confident at the moment. I would like you people to help me in clearing up my concepts. Please acknowledge my points, and also please answer for what you feel is wrong. I have also a confused section in my points. So, here starts the summary.
Every process thinks it is only running. It can access the 4GB of memory - virtual address space.
When a process access a virtual address it is translated to physical address via MMU.
This MMU is a part of a CPU - a hardware.
When the MMU cannot translate the address to a physical one, it raises a page fault.
On page fault, the kernel is notified. The kernel check the VM area struct. If it can find it - may be on disk. It will do some page-in /page-out. And get this memory on the RAM.
Now MMU will again try and will succeed this time.
In case the kernel cannot find the address, it will raise a signal. For example, invalid access will raise a SIGSEGV.
Confused points.
Does Page table is maintained in Kernel? This VM area struct has a page table ?
How MMU cannot find the address in physical RAM. Let's say it translates to some wrong address in RAM. Still the code will execute, but it will be a bad address. How MMU ensures that it is reading a right data? Does it consult Kernel VM area everytime?
Is the Mapping table - virtual to physical is inside a MMU. I have read it that is maintained by an individual process. If it is inside a process, why I can't see it.
Or if it is MMU, how MMU generates the address - is it that Segment + 12-bit shift -> Page frame number, and then the addition of offset (bits -1 to 10) -> gives a physical address.
Does it mean that for a 32-bit architecture, with this calculation in my mind. I can determine the physical address from a virtual address.
cat /proc/pid_value/maps. This shows me the current mapping of the vmarea. Basically, it reads the Vmarea struct and prints it. That means that this is important. I am not able to fit this piece in the complete picture. When the program is executed does the vmarea struct is generated. Is VMAREA comes only into the picture when the MMU cannnot translate the address i.e. Page fault? When I print the vmarea it displays the address range , permission and mapped to file descriptor, and offset. I am sure this file descriptor is the one in the hard-disk and the offset is for that file.
The high-mem concept is that kernel cannot directly access the Memory region greater than 1 GB(approx). Thus, it needs a page table to indirectly map it. Thus, it will temporarily load some page table to map the address. Does HIGH MEM will come into the picture everytime. Because Userspace can directly translate the address via MMU. On what scenario, does kernel really want to access the High MEM. I believe the kernel drivers will mostly be using kmalloc. This is a direct memory + offset address. In this case no mapping is really required. So, the question is on what scenario a kernel needs to access the High Mem.
Does the processor specifically comes with the MMU support. Those who doesn't have MMU support cannot run LInux?
Does Page table is maintained in Kernel? This VM area struct has a page table ?
Yes. Not exactly: each process has a mm_struct, which contains a list of vm_area_struct's (which represent abstract, processor-independent memory regions, aka mappings), and a field called pgd, which is a pointer to the processor-specific page table (which contains the current state of each page: valid, readable, writable, dirty, ...).
The page table doesn't need to be complete, the OS can generate each part of it from the VMAs.
How MMU cannot find the address in physical RAM. Let's say it translates to some wrong address in RAM. Still the code will execute, but it will be a bad address. How MMU ensures that it is reading a right data? Does it consult Kernel VM area everytime?
The translation fails, e.g. because the page was marked as invalid, or a write access was attempted against a readonly page.
Is the Mapping table - virtual to physical is inside a MMU. I have read it that is maintained by an individual process. If it is inside a process, why I can't see it.
Or if it is MMU, how MMU generates the address - is it that Segment + 12-bit shift -> Page frame number, and then the addition of offset (bits -1 to 10) -> gives a physical address.
Does it mean that for a 32-bit architecture, with this calculation in my mind. I can determine the physical address from a virtual address.
There are two kinds of MMUs in common use. One of them only has a TLB (Translation Lookaside Buffer), which is a cache of the page table. When the TLB doesn't have a translation for an attempted access, a TLB miss is generated, the OS does a page table walk, and puts the translation in the TLB.
The other kind of MMU does the page table walk in hardware.
In any case, the OS maintains a page table per process, this maps Virtual Page Numbers to Physical Frame Numbers. This mapping can change at any moment, when a page is paged-in, the physical frame it is mapped to depends on the availability of free memory.
cat /proc/pid_value/maps. This shows me the current mapping of the vmarea. Basically, it reads the Vmarea struct and prints it. That means that this is important. I am not able to fit this piece in the complete picture. When the program is executed does the vmarea struct is generated. Is VMAREA comes only into the picture when the MMU cannnot translate the address i.e. Page fault? When I print the vmarea it displays the address range , permission and mapped to file descriptor, and offset. I am sure this file descriptor is the one in the hard-disk and the offset is for that file.
To a first approximation, yes. Beyond that, there are many reasons why the kernel may decide to fiddle with a process' memory, e.g: if there is memory pressure it may decide to page out some rarely used pages from some random process. User space can also manipulate the mappings via mmap(), execve() and other system calls.
The high-mem concept is that kernel cannot directly access the Memory region greater than 1 GB(approx). Thus, it needs a page table to indirectly map it. Thus, it will temporarily load some page table to map the address. Does HIGH MEM will come into the picture everytime. Because Userspace can directly translate the address via MMU. On what scenario, does kernel really want to access the High MEM. I believe the kernel drivers will mostly be using kmalloc. This is a direct memory + offset address. In this case no mapping is really required. So, the question is on what scenario a kernel needs to access the High Mem.
Totally unrelated to the other questions. In summary, high memory is a hack to be able to access lots of memory in a limited address space computer.
Basically, the kernel has a limited address space reserved to it (on x86, a typical user/kernel split is 3Gb/1Gb [processes can run in user space or kernel space. A process runs in kernel space when a syscall is invoked. To avoid having to switch the page table on every context-switch, on x86 typically the address space is split between user-space and kernel-space]). So the kernel can directly access up to ~1Gb of memory. To access more physical memory, there is some indirection involved, which is what high memory is all about.
Does the processor specifically comes with the MMU support. Those who doesn't have MMU support cannot run Linux?
Laptop/desktop processors come with an MMU. x86 supports paging since the 386.
Linux, specially the variant called µCLinux, supports processors without MMUs (!MMU). Many embedded systems (ADSL routers, ...) use processors without an MMU. There are some important restrictions, among them:
Some syscalls don't work at all: e.g fork().
Some syscalls work with restrictions and non-POSIX conforming behavior: e.g mmap()
The executable file format is different: e.g bFLT or ELF-FDPIC instead of ELF.
The stack cannot grow, and its size has to be set at link-time.
When a program is loaded first the kernel will setup a kernel VM-Area for that process is it? This Kernel VM Area actually holds where the program sections are there in the memory/HDD. Then the entire story of updating CR3 register, and page walkthrough or TLB comes into the picture right? So, whenever there is a pagefault - Kernel will update the page table by looking at Kernel virtual memory area is it? But they say Kernel VM area keeps updating. How this is possible, since cat /proc/pid_value/map will keep updating.The map won't be constant from start to end. SO, the real information is available in the Kernel VM area struct is it? This is the acutal information where the section of program lies, it could be HDD or physical memory -- RAM? So, this is filled during process loading is it, the first job? Kernel does the page in page out on page fault, and will update the Kernel VM area is it? So, it should also know the entire program location on the HDD for page-in / page out right? Please correct me here. This is in continuation to my first question of the previous comment.
When the kernel loads a program, it will setup several VMAs (mappings), according to the segments in the executable file (which on ELF files you can see with readelf --segments), which will be text/code segment, data segment, etc... During the lifetime of the program, additional mappings may be created by the dynamic/runtime linkers, by the memory allocator (malloc(), which may also extend the data segment via brk()), or directly by the program via mmap(),shm_open(), etc..
The VMAs contain the necessary information to generate the page table, e.g. they tell whether that memory is backed by a file or by swap (anonymous memory). So, yes, the kernel will update the page table by looking at the VMAs. The kernel will page in memory in response to page faults, and will page out memory in response to memory pressure.
Using x86 no PAE as an example:
On x86 with no PAE, a linear address can be split into 3 parts: the top 10 bits point to an entry in the page directory, the middle 10 bits point to an entry in the page table pointed to by the aforementioned page directory entry. The page table entry may contain a valid physical frame number: the top 22 bits of a physical address. The bottom 12 bits of the virtual address is an offset into the page that goes untranslated into the physical address.
Each time the kernel schedules a different process, the CR3 register is written to with a pointer to the page directory for the current process. Then, each time a memory access is made, the MMU tries to look for a translation cached in the TLB, if it doesn't find one, it looks for one doing a page table walk starting from CR3. If it still doesn't find one, a GPF fault is raised, the CPU switches to Ring 0 (kernel mode), and the kernel tries to find one in the VMAs.
Also, I believe this reading from CR, page directory->page-table->Page frame number-memory address this all done by MMU. Am I correct?
On x86, yes, the MMU does the page table walk. On other systems (e.g: MIPS), the MMU is little more than the TLB, and on TLB miss exceptions the kernel does the page table walk by software.
Though this is not going to be the best answer, iw ould like to share my thoughts on confused points.
1. Does Page table is maintained...
Yes. kernel maintains the page tables. In fact it maintains nested page tables. And top of the page tables is stored in top_pmd. pmd i suppose it is page mapping directory. You can traverse through all the page tables using this structure.
2. How MMU cannot find the address in physical RAM.....
I am not sure i understood the question. But in case because of some problem, the instruction is faulted or out of its instruction area is being accessed, you generally get undefined instruction exception resulting in undefined exception abort. If you look at the crash dumps, you can see it in the kernel log.
3. Is the Mapping table - virtual to physical is inside a MMU...
Yes. MMU is SW+HW. HW is like TLB and all. The mapping tables are stored here. For instructions, that is for code section i always converted the physical-virtual address and always they matched. And almost all the times it matches for Data sections as well.
4. cat /proc/pid_value/maps. This shows me the current mapping of the vmarea....
This is more used for analyzing the virtual addresses of user space stacks. As you know virtually all the user space programs can have 4 GB of virtual address. So unlike kernel if i say 0xc0100234. You cannot directly go and point to the istruction. So you need this mapping and the virtual address to point the instruction based on the data you have.
5. The high-mem concept is that kernel cannot directly access the Memory...
High-mem corresponds to user space memory(some one correct me if i am wrong). When kernel wants to read some data from a address at user space you will be accessing the HIGHMEM.
6. Does the processor specifically comes with the MMU support. Those who doesn't have MMU support cannot run LInux?
MMU as i mentioned is HW + SW. So mostly it would be coming with the chipset. and the SW would be generally architecture dependent. You can disable MMU from kernel config and build. I have never tried it though. Mostly these days allthe chipsets have it. But small boards i think they disable MMU. I am not entirely sure though.
As all these are conceptual questions, i may be lacking some knowledge and be wrong at places. If so others please correct me.
We have address translation table to translate virtual address (VA) of a process to its corresponding physical address in RAM, but if the table does not have any entry for a VA , it results in page fault and kernal goes to backing store (often a hard drive) and fetch the corresponding data and update the RAM and address translation table. So my question is how does the OS come to know what is the address corresponding to a VA in backing store ? Does it have a separate translation table for that?
A process starts by allocating virtual memory. That eventually will cause a page fault when the program starts actually addressing the virtual memory address. The OS knows that the memory access is valid. Since it was allocated explicitly.
So no harm done, the OS simply maps the VM address to a physical address.
If the page fault is for an address that was not previously requested to be a valid VM address then the processor will discover that there is no page table entry for the address. And will instead raise an GP fault, an AccessViolation or segfault in your program. Kaboom, program over.
There is no direct correlation, at least not in the way that you suppose.
The operating system divides virtual and phsyical RAM as well as swap space (backing store) and mapped files into pages, most commonly 4096 bytes.
When your code accesses a certain address, this is always a virtual address within a page that is either valid-in-core, valid-not-accessed, valid-out-of-core, or invalid. The OS may have other properties (such as "has been written to") in its books, but they're irrelevant for us here.
If the page is in-core, then it has a physical address, otherwise it does not. When swapped out and in again, the same identical page could in theory very well land in a different physical region of memory. Similarly, the page after some other page in memory (virtual or physical) could be before that page in the swap file or in a memory-mapped file. There's no guarantee for that.
Thus, there is no such thing as translating a virtual address to a physical address in backing store. There is only a translation from a virtual address to a page which may temporarily have a physical address. In the easiest case, "translating" means dividing by 4096, but of course other schemes are possible.
Further, every time your code accesses a memory location, the virtual address must be translated to a physical one. There exists dedicated logic inside a CPU to do this translation fully automatically (for a very small subset of "hot" pages, often as few as 64), or in a hardware-assisted way, which usually involves a lookup in a more or less complicated hierarchical structure.
This is also a fault, but it's one that you don't see. The only faults that you get to see are the ones when the OS doesn't have a valid page (or can't supply it for some reason), and thus can't assign a physical address to the to-be-translated virtual one.
When your program asks for memory, the OS remembers that certain pages are valid, but they do not exist yet because you have never accessed them.
The first time you access a page, a fault happens and obviously its address is nowhere in the translation tables (how could it be, it doesn't exist!). Thus the OS looks into its books and (assuming the page is valid) it either loads the page from disk or assigns the address of a zero page otherwise.
Not rarely, the OS will cheat and all zero pages are the same write-protected zero page until you actually write to it (at which point, a fault occurs and you are secretly redirected to a different physical memory area, one which you can write to, too.
Otherwise, that is if you haven't reserved memory, the OS sends a signal (or an equivalent, Windows calls it "exception") which will terminate your process unless handled.
For a multitude of reasons, the OS may later decide to remove one or several pages from your working set. This normally does not immediately remove them, but maked them candidates for being swapped (for non-mapped data) or discarded (for mapped data) in case more memory is needed. When you access an address in one of these pages again, it is either re-added to your working set (likely pushing another one out) or reloaded from disk.
In either case, all the OS needs to know is how to translate your virtual address to a page identifier of some sort (e.g. "page frame number"), and whether the page is resident (and at what address).
I think your question answer is issue about interrupt table.
Page fault is a kind of software interrupt, and operating system must have some solution to that interrupt.And the solution code is already in the os kernel, and that piece of code address is right at the interrupt table.So the page fault happen, os will go to that piece of code to get the unmapped page into the physical memory.
This is OS-specific, but many implementations share logic with memory-mapped file features (so that anonymous pages actually are memory-mapped views of the pagefile, flagged so that the content can be discards at unmapping instead of flushed).
For Windows, much of this is documented here, on the CreateFileMapping page
I'd like to share certain memory areas around multiple computers, that is, for a C/C++ project. When something on computer B accesses some memory area which is currently on computer A, that has to be locked on A and sent to B. I'm fine when its only linux compitable.
Thanks in advance :D
You cannot do this for a simple C/C++ project.
Common computer hardware does not have the physical properties that support this directly: Memory on one system cannot be read by another system.
In order to make it appear to C/C++ programs on different machines that they are sharing memory, you have to write software that provides this function. Typically, you would need to do something like this:
Allocate some pages in the virtual memory address space (of each process).
Mark those pages read-only.
Set a handler to receive the exception that occurs when the process attempts to write to the read-only memory. (This handler might be in the operating system, as some sort of kernel extension, or it might be a signal handler in your process.)
When the exception is received, determine what the process was attempting to write to memory. Write that to the page (perhaps by writing it through a separate mapping in virtual memory to the same physical memory, with this extra mapping marked writeable).
Send a message by network communications to the other machine telling it that memory has changed.
Resume execution in the process after the instruction that wrote to memory.
Additionally, you need to determine what to do about memory coherence: If two processes write to the same address in memory at nearly the same time, what happens? If process A writes to location X and then reads location Y while, at nearly the same time, process B writes to location Y and reads X, what do they see? Is it okay if the two processes see data that cannot possibly be the result of a single time sequence of writes to memory?
On top of all that, this is hugely expensive in time: Stores to memory that require exception handling and network operations take many thousands, likely hundreds of thousands, times as long as normal stores to memory. Your processes will execute excruciatingly slowly whenever they write to this shared memory.
There are software solutions, as noted in the comments. These use the paging hardware in the processors on a node to detect access, and use your local network fabric to disseminate the changes to the memory. One hardware alternative is reflective memory - you can read more about it here:
https://en.wikipedia.org/wiki/Reflective_memory
http://www.ecrin.com/embedded/downloads/reflectiveMemory.pdf
Old page was broken
http://www.dolphinics.com/solutions/embedded-system-reflective-memory.html
Reflective memory provides low latency (about one microsecond per hop) in either a ring or tree configuration.
I have a doubt when each process has it's own separate page table then why is there s system wide page table required ? Also if Page table is such that it maps virtual address to a physical address then I think two process may map to same physical address because all process have same virtual address space . Any good link on system wide page table will also solve my problem?
Each process has its own independent virtual address space - two processes can have virtpage 1 map to different physpages. Processes can participate in shared memory, in which case they each have some virtpage mapping to the same physpage.
The virtual address space of a process can be used to map virtpages to physpages, to memory mapped files, devices, etc. Virtpages don't have to be wired to RAM. A process could memory-map an entire 1GB file - in which case, its physical memory usage might only be a couple megs, but its virtual address space usage would be 1GB or more. Many processes could do this, in which case the sum of virtual address space usage across all processes might be, say, 40 GB, while the total physical memory usage might be only, say, 100 megs; this is very easy to do on 32-bit systems.
Since lots of processes load the same libraries, the OS typically puts the libs in one set of read-only executable pages, and then loads mappings in the virtpage space for each process to point to that one set of pages, to save on physical memory.
Processes may have virtpage mappings that don't point to anything, for instance if part of the process's memory got written to the page file - the process will try to access that page, the CPU will trigger a page fault, the OS will see the page fault and handle it by suspending the process, reading the pages back into ram from the page file and then resuming the process.
There are typically 3 types of page faults. The first type is when the CPU does not have the virtual-physical mapping in the TLB - the processor invokes the pagefault software interrupt in the OS, the OS puts the mapping into the processor for that process, then the proc re-runs the offending instructions. These happen thousands of times a second.
The second type is when the OS has no mapping because, say, the memory for the process has been swapped to disk, as explained above. These happen infrequently on a lightly loaded machine, but happen more often as memory pressure is increased, up to 100s to 1000s of times per second, maybe even more.
The third type is when the OS has no mapping because the mapping does not exist - the process is trying to access memory that does not belong to it. This generates a segfault, and typically, the process is killed. These aren't supposed to happen often, and solely depend on how well written the software is on the machine, and does not have anything to do with scheduling or machine load.
Even if you already knew that, I figured I throw that in for the community.