How smart is mmap? - macos

mmap can be used to share read-only memory between processes, reducing the memory foot print:
process P1 mmaps a file, uses the mapped memory -> data gets loaded into RAM
process P2 mmaps a file, uses the mapped memory -> OS re-uses the same memory
But how about this:
process P1 mmaps a file, loads it into memory, then exits.
another process P2 mmaps the same file, accesses the memory that is still hot from P1's access.
Is the data loaded again from disk? Is the OS smart enough to re-use the virtual memory even if "mmap count" dropped to zero temporarily?
Does the behaviour differ between different OS? (I'm mostly interested in Linux/OS X)
EDIT: In case the OS is not smart enough -- would it help if there is one "background process", keeping the file mmaped, so it never leaves the address space of at least one process?
I am of course interested in performance when I mmap and munmap the same file successively and rapidly, possibly (but not necessarily) within the same process.
EDIT2: I see answers describing completely irrelevant points at great length. To reiterate the point -- can I rely on Linux/OS X to not re-load data that already resides in memory, from previous page hits within mmaped memory segments, even though the particular region is no longer mmaped by any process?


The presence or absence of the contents of a file in memory is much less coupled to mmap system calls than you think. When you mmap a file, it doesn't necessarily load it into memory. When you munmap it (or if the process exits), it doesn't necessarily discard the pages.
There are many different things that could trigger the contents of a file to be loaded into memory: mapping it, reading it normally, executing it, attempting to access memory that is mapped to the file. Similarily, there are different things that could cause the file's contents to be removed from memory, mostly related to the OS deciding it wants the memory for something more important.
In the two scenarios from your question, consider inserting a step between steps 1 and 2:
1.5. another process allocates and uses a large amount of memory -> the mmaped file is evicted from memory to make room.
In this case the file's contents will probably have to get reloaded into memory if they are mapped again and used again in step 2.
versus:
1.5. nothing happens -> the contents of the mmaped file hang around in memory.
In this case the file's contents don't need to be reloaded in step 2.
In terms of what happens to the contents of your file, your two scenarios aren't much different. It's something like this step 1.5 that would make a much more important difference.
As for a background process that is constantly accessing the file in order to ensure it's kept in memory (for example, by scanning the file and then sleeping for a short amount of time in a loop), this would of course force the file to remain in memory. but you're probably better off just letting the OS make its own decision about when to evict the file and when not to evict it.


The second process likely finds the data from the first process in the buffer cache. So in most cases the data will not be loaded again from disk. But since the buffer cache is a cache, there are no guarantees that the pages don't get evicted inbetween.
You could start a third process and use mmap(2) and mlock(2) to fix the pages in ram. But this will probably cause more trouble than it is worth.
Linux substituted the UNIX buffer cache for a page cache. But the principle is still the same. The Mac OS X equivalent is called Unified Buffer Cache (UBC).

Related

Does GetWriteWatch work with Memory-Mapped FIles?

outI'm working with memory mapped files (MMF) with very large datasets (depending on the input file), where each file has ~50GB and there are around 40 files open at the same time. Of course this depends, I can also have smaller files, but I can also have larger files - so the system should scale itself.
The MMF is acting as a backing buffer, so as long as I have enough free memory there shoud occur no paging. The problem is that the windows memory manager and my application are two autonomous processes. In good conditions everything is working fine, but the memory manager obviously is too slow in conditions where I'm entering low memory conditions, the memory is full and then the system starts to page (which is good), but I'm still allocating memory, because I don't get any information about the paging.
In the end I'm entering a state where the system stalls, the memory manager pages and I'm allocating.
So I came to the point where I need to advice the memory manager, check current memory conditions and invoke the paging myself. For that reason I wanted to use the GetWriteWatch to inspect the memory region I can flush.
Interestingly the GetWriteWatch does not work in my situation, it returns a -1 without filling the structures. So my question is does GetWriteWatch work with MMFs?

Does GetWriteWatch work with Memory-Mapped Files?
I don't think so.
GetWriteWatch accepts memory allocated via VirtualAlloc function using MEM_WRITE_WATCH.
File mapping are mapped using MapViewOfFile* functions that do not have this flag.

Does mmap directly access the page cache, or a copy of the page cache?

To ask the question another way, can you confirm that when you mmap() a file that you do in fact access the exact physical pages that are already in the page cache?
I ask because I’m doing testing on a 192 core machine with 1TB of RAM, on a 400GB data file that is pre-cached into the page cache prior to the test (by just dropping the cache, then doing md5sum on the file).
Initially, I had all 192 threads each mmap the file separately, on the assumption that they would all get (basically) the same memory region back (or perhaps the same memory region but somehow mapped multiple times). Accordingly, I assumed two threads using two different mappings to the same file would both have direct access to the same pages. (Let’s ignore NUMA for this example, though obviously it’s significant at higher thread counts.)
However, in practice I found performance would get terrible at higher thread counts when each thread separately mmapped the file. When we removed that and instead just did a single mmap that was passed into the thread (such that all threads just directly access the same memory region), then performance improved dramatically.
That’s all great, but I’m trying to figure out why. If in fact mmapping a file just grants direct access to the existing page cache, then I would think that it shouldn’t matter how many times you map it — it should all go to the exact same place.
But given that there was such a performance cost, it seemed to me that in fact each mmap was being independently and redundantly populated (perhaps by copying from the page cache, or perhaps by reading again from disk).
Can you comment on why I was seeing such different performance between shared access to the same memory, versus mmapping the same file?
Thanks, I appreciate your help!

I think I found my answer, and it deals with the page directory. The answer is yes, two mmapped regions of the same file will access the same underlying page cache data. However, each mapping needs to independently map each of the virtual pages to the physical pages -- meaning 2x as many entries in the page directory to access the same RAM.
Basically, each mmap() creates a new range in virtual memory. Every page of that range corresponds to a page of physical memory, and that mapping is stored in a hierarchical page directory -- with one entry per 4KB page. So every mmap() of a large region generates a huge number of entries in the page directory.
My guess is it doesn't actually define them all up front, which is why mmap() is instant to call even for a giant file. But over time it probably has to establish those entries as there are faults on the mmapped range, meaning over the course of time it gets filled out. This extra work to populate the page directory is probably why threads using different mmaps are slower than threads sharing the same mmap. And I bet the kernel needs to erase all those entries when unmapping the range -- which is why unmmap() is so slow.
(There's also the translation lookaside buffer, but that's per-CPU, and so small I don't think that matters much here.)
Anyway, it sounds like re-mapping the same region just adds extra overhead, for what seems to me like no gain.

What is the cache's role when writing to memory?

I have a function that does very little reading, but a lot of writing to RAM. When I run it multiple times on the same core (the main thread), it runs about 5x as fast than if I launch the function on a new thread every run (which doesn't guarantee the same core is used between runs), as I launch and join between runs.
This suggests the cache is being used heavily for the write process, but I don't understand how. I thought the cache was only useful for reads.

Modern processors usually have write-buffers. The reason is that writes are, to a first approximation, pure sinks. The processor doesn't usually have to wait for the store to reach the coherent memory hierarchy before it executes the next instruction.
(Aside: Obviously stores are not pure sinks. A later read from the written-to memory location should return the written value, so the processor must snoop the write-buffer, and either stall the read or forward the written value to it)
Obviously such buffer(s) are of finite size, so when the buffers are full the next store in the program can't be executed and stalls until a slot in the buffer is made available by an older store becoming architecturally visible.
Ordinarily, the way a write leaves the buffer is when the value is written to cache (since a lot of writes are actually read back again quickly, think of the program stack as an example). If the write only sets part of the cacheline, the rest of the cacheline must remain unmodified, so consequently it must be loaded from the memory hierarchy.
There are ways to avoid loading the old cacheline, like non-temporal stores, write-combining memory or cacheline-zeroing instructions.
Non-temporal stores and write-combining memory combine adjacent writes to fill a whole cacheline, sending the new cacheline to the memory hierarchy to replace the old one.
POWER has an instruction that zeroes a full cacheline (dcbz), which also removes the need to load the old value from memory.
x86 with AVX512 has cacheline-sized registers, which suggests that an aligned zmm-register store could avoid loading the old cacheline (though I do not know whether it does).
Note that many of these techniques are not consistent with the usual memory-ordering of the respective processor architectures. Using them may require additional fences/barriers in multi-threaded operation.

Transferring 1-2 megabytes of data through regular files in Windows - is it slower than through RAM?

I'm passing 1-2 MB of data from one process to another, using a plain old file. Is it significantly slower than going through RAM entirely?
Before answering yes, please keep in mind that in modern Linux at least, when writing a file it is actually written to RAM, and then a daemon syncs the data to disk from time to time. So in that way, if process A writes a 1-2 MB into a file, then process B reads them within 1-2 seconds, process B would simply read the cached memory. It gets even better than that, because in Linux, there is a grace period of a few seconds before a new file is written to the hard disk, so if the file is deleted, it's not written at all to the hard disk. This makes passing data through files as fast as passing them through RAM.
Now that is Linux, is it so in Windows?
Edit: Just to lay out some assumptions:
The OS is reasonably new - Windows XP or newer for desktops, Windows Server 2003 or newer for servers.
The file is significantly smaller than available RAM - let's say less than 1% of available RAM.
The file is read and deleted a few seconds after it has been written.

When you read or write to a file Windows will often keep some or all of the file resident in memory (in the Standby List). So that if it is needed again, it is just a soft-page fault to map it into the processes' memory space.
The algorithm for what pages of a file will be kept around (and for how long) isn't publicly documented. So the short answer is that if you are lucky some or all of it may still be in memory. You can use the SysInternals tool VMmap to see what of your file is still in memory during testing.
If you want to increase your chances of the data remaining resident, then you should use Memory Mapped Files to pass the data between the two processes.
Good reading on Windows memory management:
Mysteries of Windows Memory Management Revealed

You can use FILE_ATTRIBUTE_TEMPORARY to hint that this data is never needed on disk:
A file that is being used for temporary storage. File systems avoid writing data back to mass storage if sufficient cache memory is available, because typically, an application deletes a temporary file after the handle is closed. In that scenario, the system can entirely avoid writing the data. Otherwise, the data is written after the handle is closed.
(i.e. you need use that flag with CreateFile, and DeleteFile immediately after closing that handle).
But even if the file remains cached, you still have to copy it twice: from your process A to the cache (the WriteFile call), and from cache to the proces B (ReadFile call).
Using memory mapped files (MMF, as josh poley already suggested) has the primary advantage of avoiding one copy: the same physical memory pages are mapped into both processes.
A MMF can be backed by virtual memory, which means basically that it always stays in memory unless swapping becomes necessary.
The major downside is that you can't easily grow the memory mapping to changing demands, you are stuck with the initial size.
Whether that matters for an 1-2 MB data transfer depends mostly on how you acquire and what you do with the data, in many scenarios the additional copy doesn't really matter.

What are the differences between virtual memory and physical memory?

I am often confused with the concept of virtualization in operating systems. Considering RAM as the physical memory, why do we need the virtual memory for executing a process?
Where does this virtual memory stand when the process (program) from the external hard drive is brought to the main memory (physical memory) for the execution.
Who takes care of the virtual memory and what is the size of the virtual memory?
Suppose if the size of the RAM is 4GB (i.e. 2^32-1 address spaces) what is the size of the virtual memory?

Softwares run on the OS on a very simple premise - they require memory. The device OS provides it in the form of RAM. The amount of memory required may vary - some softwares need huge memory, some require paltry memory. Most (if not all) users run multiple applications on the OS simultaneously, and given that memory is expensive (and device size is finite), the amount of memory available is always limited. So given that all softwares require a certain amount of RAM, and all of them can be made to run at the same time, OS has to take care of two things:
That the software always runs until user aborts it, i.e. it should not auto-abort because OS has run out of memory.
The above activity, while maintaining a respectable performance for the softwares running.
Now the main question boils down to how the memory is being managed. What exactly governs where in the memory will the data belonging to a given software reside?
Possible solution 1: Let individual softwares specify explicitly the memory address they will use in the device. Suppose Photoshop declares that it will always use memory addresses ranging from 0 to 1023 (imagine the memory as a linear array of bytes, so first byte is at location 0, 1024th byte is at location 1023) - i.e. occupying 1 GB memory. Similarly, VLC declares that it will occupy memory range 1244 to 1876, etc.
Advantages:
Every application is pre-assigned a memory slot, so when it is installed and executed, it just stores its data in that memory area, and everything works fine.
Disadvantages:
This does not scale. Theoretically, an app may require a huge amount of memory when it is doing something really heavy-duty. So to ensure that it never runs out of memory, the memory area allocated to it must always be more than or equal to that amount of memory. What if a software, whose maximal theoretical memory usage is 2 GB (hence requiring 2 GB memory allocation from RAM), is installed in a machine with only 1 GB memory? Should the software just abort on startup, saying that the available RAM is less than 2 GB? Or should it continue, and the moment the memory required exceeds 2 GB, just abort and bail out with the message that not enough memory is available?
It is not possible to prevent memory mangling. There are millions of softwares out there, even if each of them was allotted just 1 kB memory, the total memory required would exceed 16 GB, which is more than most devices offer. How can, then, different softwares be allotted memory slots that do not encroach upon each other's areas? Firstly, there is no centralized software market which can regulate that when a new software is being released, it must assign itself this much memory from this yet unoccupied area, and secondly, even if there were, it is not possible to do it because the no. of softwares is practically infinite (thus requiring infinite memory to accommodate all of them), and the total RAM available on any device is not sufficient to accommodate even a fraction of what is required, thus making inevitable the encroaching of the memory bounds of one software upon that of another. So what happens when Photoshop is assigned memory locations 1 to 1023 and VLC is assigned 1000 to 1676? What if Photoshop stores some data at location 1008, then VLC overwrites that with its own data, and later Photoshop accesses it thinking that it is the same data is had stored there previously? As you can imagine, bad things will happen.
So clearly, as you can see, this idea is rather naive.
Possible solution 2: Let's try another scheme - where OS will do majority of the memory management. Softwares, whenever they require any memory, will just request the OS, and the OS will accommodate accordingly. Say OS ensures that whenever a new process is requesting for memory, it will allocate the memory from the lowest byte address possible (as said earlier, RAM can be imagined as a linear array of bytes, so for a 4 GB RAM, the addresses range for a byte from 0 to 2^32-1) if the process is starting, else if it is a running process requesting the memory, it will allocate from the last memory location where that process still resides. Since the softwares will be emitting addresses without considering what the actual memory address is going to be where that data is stored, OS will have to maintain a mapping, per software, of the address emitted by the software to the actual physical address (Note: that is one of the two reasons we call this concept Virtual Memory. Softwares are not caring about the real memory address where their data are getting stored, they just spit out addresses on the fly, and the OS finds the right place to fit it and find it later if required).
Say the device has just been turned on, OS has just launched, right now there is no other process running (ignoring the OS, which is also a process!), and you decide to launch VLC. So VLC is allocated a part of the RAM from the lowest byte addresses. Good. Now while the video is running, you need to start your browser to view some webpage. Then you need to launch Notepad to scribble some text. And then Eclipse to do some coding.. Pretty soon your memory of 4 GB is all used up, and the RAM looks like this:
Problem 1: Now you cannot start any other process, for all RAM is used up. Thus programs have to be written keeping the maximum memory available in mind (practically even less will be available, as other softwares will be running parallelly as well!). In other words, you cannot run a high-memory consuming app in your ramshackle 1 GB PC.
Okay, so now you decide that you no longer need to keep Eclipse and Chrome open, you close them to free up some memory. The space occupied in RAM by those processes is reclaimed by OS, and it looks like this now:
Suppose that closing these two frees up 700 MB space - (400 + 300) MB. Now you need to launch Opera, which will take up 450 MB space. Well, you do have more than 450 MB space available in total, but...it is not contiguous, it is divided into individual chunks, none of which is big enough to fit 450 MB. So you hit upon a brilliant idea, let's move all the processes below to as much above as possible, which will leave the 700 MB empty space in one chunk at the bottom. This is called compaction. Great, except that...all the processes which are there are running. Moving them will mean moving the address of all their contents (remember, OS maintains a mapping of the memory spat out by the software to the actual memory address. Imagine software had spat out an address of 45 with data 123, and OS had stored it in location 2012 and created an entry in the map, mapping 45 to 2012. If the software is now moved in memory, what used to be at location 2012 will no longer be at 2012, but in a new location, and OS has to update the map accordingly to map 45 to the new address, so that the software can get the expected data (123) when it queries for memory location 45. As far as the software is concerned, all it knows is that address 45 contains the data 123!)! Imagine a process that is referencing a local variable i. By the time it is accessed again, its address has changed, and it won't be able to find it any more. The same will hold for all functions, objects, variables, basically everything has an address, and moving a process will mean changing the address of all of them. Which leads us to:
Problem 2: You cannot move a process. The values of all variables, functions and objects within that process have hardcoded values as
spat out by the compiler during compilation, the process depends on
them being at the same location during its lifetime, and changing them is expensive. As a result,
processes leave behind big "holes" when they exit. This is called
External Fragmentation.
Fine. Suppose somehow, by some miraculous manner, you do manage to move the processes up. Now there is 700 MB of free space at the bottom:
Opera smoothly fits in at the bottom. Now your RAM looks like this:
Good. Everything is looking fine. However, there is not much space left, and now you need to launch Chrome again, a known memory-hog! It needs lots of memory to start, and you have hardly any left...Except.. you now notice that some of the processes, which were initially occupying large space, now is not needing much space. May be you have stopped your video in VLC, hence it is still occupying some space, but not as much as it required while running a high resolution video. Similarly for Notepad and Photos. Your RAM now looks like this:
Holes, once again! Back to square one! Except, previously, the holes occurred due to processes terminating, now it is due to processes requiring less space than before! And you again have the same problem, the holes combined yield more space than required, but they are scattered around, not much of use in isolation. So you have to move those processes again, an expensive operation, and a very frequent one at that, since processes will frequently reduce in size over their lifetime.
Problem 3: Processes, over their lifetime, may reduce in size, leaving behind unused space, which if needed to be used, will require
the expensive operation of moving many processes. This is called
Internal Fragmentation.
Fine, so now, your OS does the required thing, moves processes around and start Chrome and after some time, your RAM looks like this:
Cool. Now suppose you again resume watching Avatar in VLC. Its memory requirement will shoot up! But...there is no space left for it to grow, as Notepad is snuggled at its bottom. So, again, all processes has to move below until VLC has found sufficient space!
Problem 4: If processes needs to grow, it will be a very expensive operation
Fine. Now suppose, Photos is being used to load some photos from an external hard disk. Accessing hard-disk takes you from the realm of caches and RAM to that of disk, which is slower by orders of magnitudes. Painfully, irrevocably, transcendentally slower. It is an I/O operation, which means it is not CPU bound (it is rather the exact opposite), which means it does not need to occupy RAM right now. However, it still occupies RAM stubbornly. If you want to launch Firefox in the meantime, you can't, because there is not much memory available, whereas if Photos was taken out of memory for the duration of its I/O bound activity, it would have freed lot of memory, followed by (expensive) compaction, followed by Firefox fitting in.
Problem 5: I/O bound jobs keep on occupying RAM, leading to under-utilization of RAM, which could have been used by CPU bound jobs in the meantime.
So, as we can see, we have so many problems even with the approach of virtual memory.
There are two approaches to tackle these problems - paging and segmentation. Let us discuss paging. In this approach, the virtual address space of a process is mapped to the physical memory in chunks - called pages. A typical page size is 4 kB. The mapping is maintained by something called a page table, given a virtual address, all now we have to do is find out which page the address belong to, then from the page table, find the corresponding location for that page in actual physical memory (known as frame), and given that the offset of the virtual address within the page is same for the page as well as the frame, find out the actual address by adding that offset to the address returned by the page table. For example:
On the left is the virtual address space of a process. Say the virtual address space requires 40 units of memory. If the physical address space (on the right) had 40 units of memory as well, it would have been possible to map all location from the left to a location on the right, and we would have been so happy. But as ill luck would have it, not only does the physical memory have less (24 here) memory units available, it has to be shared between multiple processes as well! Fine, let's see how we make do with it.
When the process starts, say a memory access request for location 35 is made. Here the page size is 8 (each page contains 8 locations, the entire virtual address space of 40 locations thus contains 5 pages). So this location belongs to page no. 4 (35/8). Within this page, this location has an offset of 3 (35%8). So this location can be specified by the tuple (pageIndex, offset) = (4,3). This is just the starting, so no part of the process is stored in the actual physical memory yet. So the page table, which maintains a mapping of the pages on the left to the actual pages on the right (where they are called frames) is currently empty. So OS relinquishes the CPU, lets a device driver access the disk and fetch the page no. 4 for this process (basically a memory chunk from the program on the disk whose addresses range from 32 to 39). When it arrives, OS allocates the page somewhere in the RAM, say first frame itself, and the page table for this process takes note that page 4 maps to frame 0 in the RAM. Now the data is finally there in the physical memory. OS again queries the page table for the tuple (4,3), and this time, page table says that page 4 is already mapped to frame 0 in the RAM. So OS simply goes to the 0th frame in RAM, accesses the data at offset 3 in that frame (Take a moment to understand this. The entire page, which was fetched from disk, is moved to frame. So whatever the offset of an individual memory location in a page was, it will be the same in the frame as well, since within the page/frame, the memory unit still resides at the same place relatively!), and returns the data! Because the data was not found in memory at first query itself, but rather had to be fetched from disk to be loaded into memory, it constitutes a miss.
Fine. Now suppose, a memory access for location 28 is made. It boils down to (3,4). Page table right now has only one entry, mapping page 4 to frame 0. So this is again a miss, the process relinquishes the CPU, device driver fetches the page from disk, process regains control of CPU again, and its page table is updated. Say now the page 3 is mapped to frame 1 in the RAM. So (3,4) becomes (1,4), and the data at that location in RAM is returned. Good. In this way, suppose the next memory access is for location 8, which translates to (1,0). Page 1 is not in memory yet, the same procedure is repeated, and the page is allocated at frame 2 in RAM. Now the RAM-process mapping looks like the picture above. At this point in time, the RAM, which had only 24 units of memory available, is filled up. Suppose the next memory access request for this process is from address 30. It maps to (3,6), and page table says that page 3 is in RAM, and it maps to frame 1. Yay! So the data is fetched from RAM location (1,6), and returned. This constitutes a hit, as data required can be obtained directly from RAM, thus being very fast. Similarly, the next few access requests, say for locations 11, 32, 26, 27 all are hits, i.e. data requested by the process is found directly in the RAM without needing to look elsewhere.
Now suppose a memory access request for location 3 comes. It translates to (0,3), and page table for this process, which currently has 3 entries, for pages 1, 3 and 4 says that this page is not in memory. Like previous cases, it is fetched from disk, however, unlike previous cases, RAM is filled up! So what to do now? Here lies the beauty of virtual memory, a frame from the RAM is evicted! (Various factors govern which frame is to be evicted. It may be LRU based, where the frame which was least recently accessed for a process is to be evicted. It may be first-come-first-evicted basis, where the frame which allocated longest time ago, is evicted, etc.) So some frame is evicted. Say frame 1 (just randomly choosing it). However, that frame is mapped to some page! (Currently, it is mapped by the page table to page 3 of our one and only one process). So that process has to be told this tragic news, that one frame, which unfortunate belongs to you, is to be evicted from RAM to make room for another pages. The process has to ensure that it updates its page table with this information, that is, removing the entry for that page-frame duo, so that the next time a request is made for that page, it right tells the process that this page is no longer in memory, and has to be fetched from disk. Good. So frame 1 is evicted, page 0 is brought in and placed there in the RAM, and the entry for page 3 is removed, and replaced by page 0 mapping to the same frame 1. So now our mapping looks like this (note the colour change in the second frame on the right side):
Saw what just happened? The process had to grow, it needed more space than the available RAM, but unlike our earlier scenario where every process in the RAM had to move to accommodate a growing process, here it happened by just one page replacement! This was made possible by the fact that the memory for a process no longer needs to be contiguous, it can reside at different places in chunks, OS maintains the information as to where they are, and when required, they are appropriately queried. Note: you might be thinking, huh, what if most of the times it is a miss, and the data has to be constantly loaded from disk into memory? Yes, theoretically, it is possible, but most compilers are designed in such a manner that follows locality of reference, i.e. if data from some memory location is used, the next data needed will be located somewhere very close, perhaps from the same page, the page which was just loaded into memory. As a result, the next miss will happen after quite some time, most of the upcoming memory requirements will be met by the page just brought in, or the pages already in memory which were recently used. The exact same principle allows us to evict the least recently used page as well, with the logic that what has not been used in a while, is not likely to be used in a while as well. However, it is not always so, and in exceptional cases, yes, performance may suffer. More about it later.
Solution to Problem 4: Processes can now grow easily, if space problem is faced, all it requires is to do a simple page replacement, without moving any other process.
Solution to Problem 1: A process can access unlimited memory. When more memory than available is needed, the disk is used as backup, the new data required is loaded into memory from the disk, and the least recently used data frame (or page) is moved to disk. This can go on infinitely, and since disk space is cheap and virtually unlimited, it gives an illusion of unlimited memory. Another reason for the name Virtual Memory, it gives you illusion of memory which is not really available!
Cool. Earlier we were facing a problem where even though a process reduces in size, the empty space is difficult to be reclaimed by other processes (because it would require costly compaction). Now it is easy, when a process becomes smaller in size, many of its pages are no longer used, so when other processes need more memory, a simple LRU based eviction automatically evicts those less-used pages from RAM, and replaces them with the new pages from the other processes (and of course updating the page tables of all those processes as well as the original process which now requires less space), all these without any costly compaction operation!
Solution to Problem 3: Whenever processes reduce in size, its frames in RAM will be less used, so a simple LRU based eviction can evict those pages out and replace them with pages required by new processes, thus avoiding Internal Fragmentation without need for compaction.
As for problem 2, take a moment to understand this, the scenario itself is completely removed! There is no need to move a process to accommodate a new process, because now the entire process never needs to fit at once, only certain pages of it need to fit ad hoc, that happens by evicting frames from RAM. Everything happens in units of pages, thus there is no concept of hole now, and hence no question of anything moving! May be 10 pages had to be moved because of this new requirement, there are thousands of pages which are left untouched. Whereas, earlier, all processes (every bit of them) had to be moved!
Solution to Problem 2: To accommodate a new process, data from only less recently used parts of other processes have to be evicted as required, and this happens in fixed size units called pages. Thus there is no possibility of hole or External Fragmentation with this system.
Now when the process needs to do some I/O operation, it can relinquish CPU easily! OS simply evicts all its pages from the RAM (perhaps store it in some cache) while new processes occupy the RAM in the meantime. When the I/O operation is done, OS simply restores those pages to the RAM (of course by replacing the pages from some other processes, may be from the ones which replaced the original process, or may be from some which themselves need to do I/O now, and hence can relinquish the memory!)
Solution to Problem 5: When a process is doing I/O operations, it can easily give up RAM usage, which can be utilized by other processes. This leads to proper utilization of RAM.
And of course, now no process is accessing the RAM directly. Each process is accessing a virtual memory location, which is mapped to a physical RAM address and maintained by the page-table of that process. The mapping is OS-backed, OS lets the process know which frame is empty so that a new page for a process can be fitted there. Since this memory allocation is overseen by the OS itself, it can easily ensure that no process encroaches upon the contents of another process by allocating only empty frames from RAM, or upon encroaching upon the contents of another process in the RAM, communicate to the process to update it page-table.
Solution to Original Problem: There is no possibility of a process accessing the contents of another process, since the entire allocation is managed by the OS itself, and every process runs in its own sandboxed virtual address space.
So paging (among other techniques), in conjunction with virtual memory, is what powers today's softwares running on OS-es! This frees the software developer from worrying about how much memory is available on the user's device, where to store the data, how to prevent other processes from corrupting their software's data, etc. However, it is of course, not full-proof. There are flaws:
Paging is, ultimately, giving user the illusion of infinite memory by using disk as secondary backup. Retrieving data from secondary storage to fit into memory (called page swap, and the event of not finding the desired page in RAM is called page fault) is expensive as it is an IO operation. This slows down the process. Several such page swaps happen in succession, and the process becomes painfully slow. Ever seen your software running fine and dandy, and suddenly it becomes so slow that it nearly hangs, or leaves you with no option that to restart it? Possibly too many page swaps were happening, making it slow (called thrashing).
So coming back to OP,
Why do we need the virtual memory for executing a process? - As the answer explains at length, to give softwares the illusion of the device/OS having infinite memory, so that any software, big or small, can be run, without worrying about memory allocation, or other processes corrupting its data, even when running in parallel. It is a concept, implemented in practice through various techniques, one of which, as described here, is Paging. It may also be Segmentation.
Where does this virtual memory stand when the process (program) from the external hard drive is brought to the main memory (physical memory) for the execution? - Virtual memory doesn't stand anywhere per se, it is an abstraction, always present, when the software/process/program is booted, a new page table is created for it, and it contains the mapping from the addresses spat out by that process to the actual physical address in RAM. Since the addresses spat out by the process are not real addresses, in one sense, they are, actually, what you can say, the virtual memory.
Who takes care of the virtual memory and what is the size of the virtual memory? - It is taken care of by, in tandem, the OS and the software. Imagine a function in your code (which eventually compiled and made into the executable that spawned the process) which contains a local variable - an int i. When the code executes, i gets a memory address within the stack of the function. That function is itself stored as an object somewhere else. These addresses are compiler generated (the compiler which compiled your code into the executable) - virtual addresses. When executed, i has to reside somewhere in actual physical address for duration of that function at least (unless it is a static variable!), so OS maps the compiler generated virtual address of i into an actual physical address, so that whenever, within that function, some code requires the value of i, that process can query the OS for that virtual address, and OS in turn can query the physical address for the value stored, and return it.
Suppose if the size of the RAM is 4GB (i.e. 2^32-1 address spaces) what is the size of the virtual memory? - The size of the RAM is not related to the size of virtual memory, it depends upon the OS. For example, on 32 bit Windows, it is 16 TB, on 64 bit Windows, it is 256 TB. Of course, it is also limited by the disk size, since that is where the memory is backed up.

Virtual memory is, among other things, an abstraction to give the programmer the illusion of having infinite memory available on their system.
Virtual memory mappings are made to correspond to actual physical addresses. The operating system creates and deals with these mappings - utilizing the page table, among other data structures to maintain the mappings. Virtual memory mappings are always found in the page table or some similar data structure (in case of other implementations of virtual memory, we maybe shouldn't call it the "page table"). The page table is in physical memory as well - often in kernel-reserved spaces that user programs cannot write over.
Virtual memory is typically larger than physical memory - there wouldn't be much reason for virtual memory mappings if virtual memory and physical memory were the same size.
Only the needed part of a program is resident in memory, typically - this is a topic called "paging". Virtual memory and paging are tightly related, but not the same topic. There are other implementations of virtual memory, such as segmentation.
I could be assuming wrong here, but I'd bet the things you are finding hard to wrap your head around have to do with specific implementations of virtual memory, most likely paging. There is no one way to do paging - there are many implementations and the one your textbook describes is likely not the same as the one that appears in real OSes like Linux/Windows - there are probably subtle differences.
I could blab a thousand paragraphs about paging... but I think that is better left to a different question targeting specifically that topic.

I am shamelessly copying the excerpts from man page of top
VIRT -- Virtual Image (kb)
The total amount of virtual memory used by the task. It includes all code, data and shared libraries plus pages that have been
swapped out and pages that have been mapped but not used.
SWAP -- Swapped size (kb)
Memory that is not resident but is present in a task. This is memory that has been swapped out but could include additional non-
resident memory. This column is calculated by subtracting physical memory from virtual memory

See here: Physical Vs Virtual Memory
Virtual memory is stored on the hard drive and is used when the RAM is filled. Physical memory is limited to the size of the RAM chips installed in the computer. Virtual memory is limited by the size of the hard drive, so virtual memory has the capability for more storage.

Physical memory: Physical memory refers to the RAM or the primary memory in the computer. Physical memory is a volatile memory. Therefore, it requires a continuous flow of power to retain data.
Virtual memory is a logical memory. In other words, it is a memory management technique performed by the operating system. Virtual memory allows the programmer to use more memory for the programs than the available physical memory. If the physical memory is 4GB and the virtual memory is 16GB, the programmer can use the 16GB virtual memory to execute the program. Using virtual memory, he can execute complex programs that require more memory than the physical memory.
The main difference between physical and virtual memory is that the physical memory refers to the actual RAM of the system attached to the motherboard, but the virtual memory is a memory management technique that allows the users to execute programs larger than the actual physical memory.

So, any program typically needs to store/read data from memory (physical address). CPU generates address space that is referred as Logical Address Apace, which is a set of addresses that can be used to access the program in memory, since logical address space corresponds to actual addresses there.
Addresses in logical address space are referred to as Logical or Virtual Addresses, whereas the set of addresses that correspond to them in physical memory are referred to as Physical Address Space, each of them called Physical Address, which is actual address in memory.
Note: MMU or Memory Management Unit is responsible for the address binding that takes place.
Source: "Operating System Concepts" by Abraham Silberschatz (Chapter8 Memory Management Strategies)

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