the name of state when data reading from hard drive - cpu-architecture

In computer organization and architecture, what is the name of the state when data must be read from hard drive? I have tried to search it on StackOverflow and textbooks but could not find the answer.

In common usage people often say a process is in the disk wait or more generally the i/o wait state. In Linux the systat command uses the iowait label.

Related

What if my mmap virtual memory exceeds my computer’s RAM?

Background and Use Case
I have around 30 GB of data that never changes, specifically, every dictionary of every language.
Client requests to see the definition of a word, I simply respond with it.
On every request I have to conduct an algorithmic search of my choice so I don’t have to loop through the over two hundred million words I have stored in my .txt file.
If I open the txt file and read it so I can search for the word, it would take forever due to the size of the file (even if that file is broken down into smaller files, it is not feasible nor it is what I want to do).
I came across the concept of mmap, mentioned to me as a possible solution to my problem by a very kind gentleman on discord.
Problem
As I was learning about mmap I came across the fact that mmap does not store the data on the RAM but rather on a virtual RAM… well regardless of which it is, my server or docker instances may have no more than 64 GB of RAM and that chunk of data taking 30 of them is quite painful and makes me feel like there needs to be an alternative that is better. Even on a worst case scenario, if my server or docker container does not have enough RAM for the data stored on mmap, then it is not feasible, unless I am wrong as to how this works, which is why I am asking this question.
Questions
Is there better solution for my use case than mmap?
Will having to access such a large amount of data through mmap so I don’t have to open and read the file every time allocate RAM memory of the amount of the file that I am accessing?
Lastly, if I was wrong about a specific statement I made on what I have written so far, please do correct me as I am learning lots about mmap still.
Requirements For My Specific Use Case
I may get a request from one client that has tens of words that I have to look up, so I need to be able to retrieve lots of data from the txt file effectively.
The response to the client has to be as quick as possible, the quicker the better, I am talking ideally a less than three seconds, or if impossible, then as quick as it can be.

system hardware related

hi every body there
first of all thanks for your support and help .Now i want two know how to calculate the load on a computer system. i heard about load sensor and other utility which mostly provide the option of finding the temperature of hard disk or system , like top, system monitor but i don't want to use them.
all i want to know simply load on CPU like CPU is 40% free, load on memory means total usages of memory or amount of free memory ,free disk space etc.
i don't need soft ware tool for finding these thing .
all i want to know is there any program in c or in other language or script which can find out these thing one by one or simultaneously?
are there any commands to find this?
or can anybody explain how system monitor work?
waiting for your help
Depends on your operating system. On Unix/Linux, there is the directory /proc which contains a lot of files/directories with system information such as /proc/loadvg or /proc/cpuinfo
These aren't real files on your disk, but virtual files containing system information. But you can just open them and read them with standard file functions, so it works with any programming language, from C to Python.

How to quickly get directory (and contents) size in cygwin perl

I have a perl script which monitors several windows network share drive usages. It currently monitors the free space of several network drives using the cygwin df command. I'd like to add the individual drive usages as well. When I use the
du -s | grep total
command, it takes for ever. I need to look at the shared drive usages because there are several network drives that are shared from a single drive on the server. Thus, filling one network drive fills them all (yes I know, not the best solution, not my choice).
So, I'd like to know if there is a quicker way to get the folder usage that doesn't take for ever.
du -s works by recursively querying the size of every directory and file. If your filesystem implementation doesn't store this total value somewhere, this is the only way to determine disk usage. Therefore, you should investigate which filesystem and drivers you are using, and see if there is a way to directly query for this data. Otherwise, you're probably SOL and will have to suck up the time it takes to run du.
1) The problem possibly lies in the fact that they are network drives - local du is acceptably fast in most cases. Are you doing du on the exact server where the disk is housed? If not, try to approach the problem from a different angle - run an agent on every server hosting the drives which calculates the local du summaries and then report the totals to a central process (either IPC or heck, by writing a report into a file on that same share filesystem).
2) If one of the drives is taking a significantly larger share of space (on average) than the rest of them, you can optimize by doing du on all but the "biggest" one and then calculate the biggest one by subtracting the sum of others from df results
3) Also, to be perfectly honest, it sounds like a suboptimal solution from design standpoint - while you indicated that it's not your choice, I'd strongly recommend that you post a question on how you can improve the design within the parameters you were given (to ServerFault website, not SO)

Why can't DMBSes rely on the OS buffer pool?

Stonebraker's paper (Operating System Support for Database Management) explains that, "the overhead to fetch a block from the buffer pool manager usually includes that of a system call and a core-to-core move." Forget about the buffer-replacement strategy, etc. The only point I question is the quoted.
My understanding is that when a DBMS wants to read a block x it issues a common read instruction. There should be no difference from that of any other application requesting a read.
I'm not looking for generic answers (I got them, and read papers). I seek a detailed answer of the described problem.
See Does a file read from a Java application invoke a system call?
Reading from your other question, and working forward:
When the DBMS must bring a page from disk it will involve at least one system call. At his point most DBMSs place the page into their own buffer. (They also end up in the OS' buffer, but that's unimportant).
So, we have one system call. However, we can avoid any further system calls. This is possible because the DBMS is caching pages in its own memory space. The first thing the DBMS will do when it decides it needs a page is check and see if it has it in its cache. If it does, it retrieves it from there without ever invoking a system call.
The DBMS is free to expire pages in its cache in whatever way is most beneficial for its IO needs. The OS's cache is expired in a more general way since the OS has other things to worry about. One example of this is that a DBMS will typically use a great deal of memory to cache pages as it knows that disk IO is one of the most expensive things it can do. The OS won't do this as it has to balance the cost of disk IO against having memory for other applications to use.
The operating system disk i/o must be generalised to work for a variety of situations. The DBMS can sometimes gain significant performance using less general code that is optimised to its own needs.
The DBMS does its own caching, so doesn't want to work through the O/S caching. It "owns" the patch of disk, so it doesn't need to worry about sharing with other processes.
Update
The link to the paper is a help.
Firstly, the paper is almost thirty years old and is referring to long-obsolete hardware. Notwithstanding that, it makes quite interesting reading.
Firstly, understand that disk i/o is a layered process. It was in 1981 and is even more so now. At the lowest point, a device driver will issue physical read/write instructions to the hardware. Above that may be the o/s kernel code then the o/s user space code then the application. Between a C program's fread() and the disk heads moving, there are at least three or four levels and might be considerably more. The DBMS may seek to improve performance might seek to bypass some layers and talk directly with the kernel, or even lower.
I recall some years ago installing Oracle on a Sun box. It had an option to dedicate a disk as a "raw" partition, where Oracle would format the disk in its own manner and then talk straight to the device driver. The O/S had no access to the disk at all.
It's mainly a performance issue. A dbms has highly specific and unusual I/O demands.
The OS may have any number of processes doing I/O and filling its buffers with the assorted cached data that this produces.
And of course there is the issue of size and what gets cached (a dbms may be able to peform better cache for its needs than the more generic device buffer caching).
And then there is the issue that a generic “block” may in fact amount to a considerably larger I/O burden (this depends on partitioning and such like) than what a dbms ideally would like to bear; its own cache may be tuned to work better with the layout of the data on the disk and thereby able to minimise I/O.
A further thing is the issue of indexes and similar means to speed up queries, which of course works rather better if the cache actually knows what these mean in the first place.
The real issue is that the file buffer cache is not in the filesystem used by the DBMS; it's in the kernel and shared by all of the filesystems resident in the system. Any memory read out of the kernel must be copied into user space: this is the core-to-core move you read about.
Beyond this, some other reasons you can't rely on the system buffer pool:
Often, DBMS's have a really good idea about its upcoming access patterns, and it can't communicate these patterns to the kernel. This can lead to lower performance.
The buffer cache is traditional stored in a fixed-size kernel memory range, so it cannot grow or shrink. That also means the cache is much smaller than main memory, so by using the buffer cache a DBMS would be unable to take advantage of system resources.
I know this is old, but it came up as unanswered.
Essentially:
The OS uses a separate address spaces for every process.
Retrieving information from any other address space requires a system call or page fault. **(see below)
The DBMS is a process with its own address space.
The OS buffer pool Stonebraker describes is in the kernel address space.
So ... to get data from the kernel address space to the DBMS's address space, a system call or page fault is unavoidable.
You're correct that accessing data from the OS buffer pool manager is no more expensive than a normal read() call. (In fact, it's done with a normal read call.) However, Stonebraker is not talking about that. He's specifically discussing the caching needs of DBMSes, after the data has been read from the disk and is present in RAM.
In essence, he's saying that the OS's buffer pool cache is too slow for the DBMS to use because it's stored in a different address space. He's suggesting using a local cache in the same process (and therefore same address space), which can give you a significant speedup for applications like DBMSes which hit the cache heavily, because it will eliminate that syscall overhead.
Here's the exact paragraph where he discusses using a local cache in the same process:
However, many DBMSs including INGRES
[20] and System R [4] choose to put a
DBMS managed buffer pool in user space
to reduce overhead. Hence, each of
these systems has gone to the
trouble of constructing its own
buffer pool manager to enhance
performance.
He also mentions multi-core issues in the excerpt you quote above. Similar effects apply here, because if you can have just one cache per core, you may be able to avoid the slowdowns from CPU cache flushes when multiple CPUs are reading and writing the same data.
** BTW, I believe Stonebraker's 1981 paper is actually pre-mmap. He mentions it as future work. "The trend toward providing the file system as a part of shared virtual memory (e.g., Pilot [16]) may provide a solution to this problem."

Do any common OS file systems use hashes to avoid storing the same content data more than once?

Many file storage systems use hashes to avoid duplication of the same file content data (among other reasons), e.g., Git and Dropbox both use SHA256. The file names and dates can be different, but as long as the content gets the same hash generated, it never gets stored more than once.
It seems this would be a sensible thing to do in a OS file system in order to save space. Are there any file systems for Windows or *nix that do this, or is there a good reason why none of them do?
This would, for the most part, eliminate the need for duplicate file finder utilities, because at that point the only space you would be saving would be for the file entry in the file system, which for most users is not enough to matter.
Edit: Arguably this could go on serverfault, but I feel developers are more likely to understand the issues and trade-offs involved.
ZFS supports deduplication since last month: http://blogs.oracle.com/bonwick/en_US/entry/zfs_dedup
Though I wouldn't call this a "common" filesystem (afaik, it is currently only supported by *BSD), it is definitely one worth looking at.
It would save space, but the time cost is prohibitive. The products you mention are already io bound, so the computational cost of hashing is not a bottleneck. If you hashed at the filesystem level, all io operations which are already slow will get worse.
NTFS has single instance storage.
NetApp has supported deduplication (that's what its called in the storage industry) in the WAFL filesystem (yeah, not your common filesystem) for a few years now. This is one of the most important features found in the enterprise filesystems today (and NetApp stands out because they support this on their primary storage also as compared to other similar products which support it only on their backup or secondary storage; they are too slow for primary storage).
The amount of data which is duplicate in a large enterprise with thousands of users is staggering. A lot of those users store the same documents, source code, etc. across their home directories. Reports of 50-70% data deduplicated have been seen often, saving lots of space and tons of money for large enterprises.
All of this means that if you create any common filesystem on a LUN exported by a NetApp filer, then you get deduplication for free, no matter what the filesystem created in that LUN. Cheers. Find out how it works here and here.
btrfs supports online de-duplication of data at the block level. I'd recommend duperemove as an external tool is needed.
It would require a fair amount of work to make this work in a file system. First of all, a user might be creating a copy of a file, planning to edit one copy, while the other remains intact -- so when you eliminate the duplication, the hard link you created that way would have to give COW semantics.
Second, the permissions on a file are often based on the directory into which that file's name is placed. You'd have to ensure that when you create your hidden hard link, that the permissions were correctly applied based on the link, not just the location of the actual content.
Third, users are likely to be upset if they make (say) three copies of a file on physically separate media to ensure against data loss from hardware failure, then find out that there was really only one copy of the file, so when that hardware failed, all three copies disappeared.
This strikes me as a bit like a second-system effect -- a solution to a problem long after the problem ceased to exist (or at least matter). With hard drives current running less than $100US/terabyte, I find it hard to believe that this would save most people a whole dollar worth of hard drive space. At that point, it's hard to imagine most people caring much.
There are file systems that do deduplication, which is sort of like this, but still noticeably different. In particular, deduplication is typically done on a basis of relatively small blocks of a file, not on complete files. Under such a system, a "file" basically just becomes a collection of pointers to de-duplicated blocks. Along with the data, each block will typically have some metadata for the block itself, that's separate from the metadata for the file(s) that refer to that block (e.g., it'll typically include at least a reference count). Any block that has a reference count greater than 1 will be treated as copy on write. That is, any attempt at writing to that block will typically create a copy, write to the copy, then store the copy of the block to the pool (so if the result comes out the same as some other block, deduplication will coalesce it with the existing block with the same content).
Many of the same considerations still apply though--most people don't have enough duplication to start with for deduplication to help a lot.
At the same time, especially on servers, deduplication at a block level can serve a real purpose. One really common case is dealing with multiple VM images, each running one of only a few choices of operating systems. If we look at the VM image as a whole, each is usually unique, so file-level deduplication would do no good. But they still frequently have a large chunk of data devoted to storing the operating system for that VM, and it's pretty common to have many VMs running only a few operating systems. With block-level deduplication, we can eliminate most of that redundancy. For a cloud server system like AWS or Azure, this can produce really serious savings.