Block I/O scheduler with sorting and power failure - no-op

While reading about I/O schedulers (CFQ, noop, deadline) I found that some of them (for example deadline) sorts blocks before writing.
Could anybody give a hint how file system survives power failures when blocks are sorted before writing? Let's say journal has to go before writing data on file system level, then i/o scheduler reordered blocks and in the middle of writing this batch to a block device we have a power failure?

An I/O barrier is issued just after the journal block write requests, to ensure ordering between the journal and final write. The barrier operation is respected by all layers in the storage system, for example it ensures cache flush at the Disk or switch etc.

Related

Round Robin scheduling and IO

I'm unsure how Round Robin scheduling works with I/O Operations. I've learned that CPU bound processes are favoured by Round Robin scheduling, but what happens if a process finishes its time slice early?
Say we neglect the dispatching process itself and a process finishes its time slice early, will the scheduler schedule another process if its CPU bound, or will the current process start its IO operation, and since that isn't CPU bound, will immediately switch to another (CPU bound) process after? And if CPU bound processes are favoured, will the scheduler schedule ALL CPU bound process until they are finished and only afterwards schedule the I/O processes?
Please help me understand.
There are two distinct schedulers: the CPU (process/thread ...) scheduler, and the I/O scheduler(s).
CPU schedulers typically employ some hybrid algorithms, because they certainly do regularly encounter both pre-emption and processes which voluntarily give up part of their time-slice. They must service higher-priority work quickly, while not "starving" anyone. (A study of the current Linux scheduler is most interesting. There have been several.)
CPU schedulers identify processes as being either "primarily 'I/O-bound'" or "primarily 'CPU-bound'" at this particular time, knowing that their characteristics can and do change. If your process repeatedly consumes full time slices, it is seen as CPU-bound.
I/O schedulers seek to order and re-order the I/O request queues for maximum efficiency. For instance, to keep the read/write head of a physical disk-drive moving efficiently in a single direction. (The two components of disk-drive delay are "seek time" and "rotational latency," with "seek time" being by-far the worst of the two. Per contra, solid-state drives have very different timing.) I/O-schedulers also have to be aware of the channels (disk interface cards, cabling, etc.) that provide access to each device: they can't simply watch what any one drive is doing. As with the CPU-scheduler, requests must be efficiently handled but never "starved." Linux's I/O-schedulers are also readily available for your study.
"Pure round-robin," as a scheduling discipline, simply means that all requests have equal priority and will be serviced sequentially in the order that they were originally submitted. Very pretty birds though they are, you rarely encounter Pure Robins in real life.

Scala concurrency performance issues

I have a data mining app.
There is 1 Mining Actor which receives and processes a Json containing 1000 objects. I put this into a list and foreach, I log the data by sending it to 1 Logger Actor which logs data into many files.
Processing the list sequentially, my app uses 700MB and takes ~15 seconds of 20% cpu power to process (4 core cpu). When I parallelize the list, my app uses 2GB and ~ the same amount of time and cpu to process.
My questions are:
Since I parallelized the list and thus the computation, shouldn't the compute-time decrease?
I think having only one Logger Actor is a bottleneck in this case. The computation may be faster but the bottleneck hides the speed increase. So if I add more Loggers to the pool, the app time should decrease?
Why does the memory usage jump to 2GB? Does the JVM have to store the entire collection in memory to parallelize it? And after the computation is done, the JVM garbage collector should deal with it?
Without more details, any answer is a guess. However, even a guess might point you to the right direction.
Parallelized execution should decrease the running time but your problem might lie elsewhere. For some reason, your CPU is idling a lot even in the single-threaded mode. You do not specify whether you read the input from disk or the network or where you write your output to. You explicitly say that you write logs to a lot of files. Disk and network reading/writing might in your case take much longer than data processing. Most probably your process is idle due to this I/O waiting. You should not expect any speedups from parallelizing a job that spends 80% of its time waiting on I/O. I therefore also suspect that loggers are not the bottleneck here.
The memory usage might jump if your threads allocate a lot of memory each. In that case, the more threads you have the more memory will be required. I don't know what kind of collection you are parallelizing on, but most are stored in memory, completely. Yes, the garbage collector will free any resources that do not require you to explicitly free them, such as files.
How many threads for reading and writing to the hard disk?
The memory increases because I send messages faster than the Logger can write, so the Mailbox balloons in size until the Logger has processed the messages and the GC kicks in.
I solved this by writing state to a protocol buffer file. Before doing any writes, I compare with the protobuf file because reads are significantly cheaper than writes. My resource usage is now 10% for 2 seconds, and less than 400MB RAM.

Confirm basic understanding of MongoDB's acknowledged write concern

Using MongoDB (via PyMongo) in the default "acknowledged" write concern mode, is it the case that if I have a line that writes to the DB (e.g. a mapReduce that outputs a new collection) followed by a line that reads from the DB, the read will always see the changes from the write?
Further, is the above true for all stricter write concerns than "acknowledged," i.e. "journaled" and "replica acknowledged," but not true in the case of "unacknowledged"?
If the write has been acknowledged, it should have been written to memory, thus any subsequent query should get the current data. This won't work if you have a replica set and allow reads from secondaries.
Journaled writes are written to the journal file on disk, which protects your data in case of power / hardware failures, etc. This shouldn't have an impact on consistency, which is covered as soon as the data is in memory.
Any replica configuration in the write concern will ensure that writes need to be acknowledged by the majority / all nodes in the replica set. This will only make a difference if you read from replicas or to protect your data against unreachable / dead servers.
For example in case of WiredTiger database engine, there'll be a cache of pages inside memory that are periodically written and read from disk, depending on memory pressure. And, in case of MMAPV1 storage engine, there would be a memory mapped address space that would correspond to pages on the disk. Now, the secondary structure that's called a journal. And a journal is a log of every single thing that the database processes - notice that the journal is also in memory.
When does the journal gets written to the disk?
When the app request something to the mongodb server via a TCP connection - and the server is gonna process the request. And it's going to write it into the memory pages. But they may not write to the disk for quite a while, depending on the memory pressure. It's also going to update request into the journal. By default, in the MongoDB driver, when we make a database request, we wait for the response. Say an acknowledged insert/update. But we don't wait for the journal to be written to the disk. The value that represents - whether we're going to wait for this write to be acknowledged by the server is called w.
w = 1
j = false
And by default, it's set to 1. 1 means, wait for this server to respond to the write. By default, j equals false, and j which stands for journal, represents whether or not we wait for this journal to be written to be written to the disk before we continue. So, what are the implications of these defaults? Well, the implications are that when we do an update/insert - we're really doing the operation in memory and not necessarily to the disk. This means, of course, it's very fast. And periodically (every few seconds) the journal gets written to the disk. It won't be long, but during this window of vulnerability when the data has been written into the server's memory into the pages, but the journal has not yet been persisted to the disk, if the server crashed, we could lose the data. We also have to realize that, as a programmer just because the write came back as good and it was written successfully to the memory. It may not ever persist to disk if the server subsequently crashes. And whether or not this is the problem depends on the application. For some applications, where there are lots of writes and logging small amount of data, we might find that it's very hard to even keep up with the data stream, if we wait for the journal to get written to the disk, because the disk is going to be 100 times, 1,000 times slower than memory for every single write. But for other applications, we may find that it's completely necessary for us to wait for this to be journaled and to know that it's been persisted to the disk before we continue. So, it's really upto us.
The w and j value together are called write concern. They can be set in the driver, at the collection level, database level or a client level.
1 : wait for the write to be acknowledged
0 : do not wait for the write to be acknowledged
TRUE : sync to journal
FALSE : do not sync to journal
There are also other values for w as well that also have some significance. With w=1 & j=true we could make sure that those writes have been persisted to disk. Now, if the writes have been written to the journal, then what happens is if the server crashes, then even though the pages may not be written back to disk yet, on recovery, the server can look at the journal on the disk - the mongod process and recreate all the writes that were not yet persisted to the pages. Because, they've been written to the journal. So, that's why this gives us a greater level of safety.

MSI: Why do we need to write the line back when other CPU is going to override it?

In the book "Computer Architecture", by Hennessy/Patterson, 5th ed, on page 360 they describe MSI protocol, and write something like:
If the line is in state "Exclusive" (Modified), then on receiving "Write Miss" from the bus the current CPU 1) writes back the line into the bus, and then 2) goes into "Invalid" state.
Why do we need to write-back the line, if it will be overwritten anyway by the successive write by the other CPU?
Is it connected with the fact that every CPU should see the same writes? (but I don't see why is it a problem not see this particular write by some other CPU)
Here is the protocol from their book (question in green, in purple it is clear: we need to write-back in order to supply the line to requesting CPU):
Writing back the modified data to memory is not strictly necessary in a MSI protocol. The state diagrams also seem to assume a system with low cost memory access (data is supplied by memory even when found in the shared state in another cache) and a shared bus connecting to the memory interface.
However, the modified data cannot simply be dropped as in shared state since the requesting processor might only be modifying part of the cache block (e.g., only one byte). Whatever portions of the block that are not modified by the requesting processor must still be available either in memory or at the requesting processor (the other processor has already invalidated its copy). With a shared bus and low cost memory access the cost difference of adding a write-back to memory over just communicating the data to the other processor is small.
In addition, even on a word-addressed system with word-sized cache blocks, keeping the old data available allows the write miss request to be sent speculatively (as with out-of-order execution or prefetch-for-write) without correctness issues.
(Per-byte tracking of modified [or valid as a superset of modified] state would allow some data communication to be avoided at the cost of extra state bits and a more complex communication system.)

The implementation of read: Is copying memory to kernel space firstly better than user space directly?

When a process is blocked after the calling of "read", the kernel reads data from ios and coping it to the buffer, but where is the buffer, in the kernel or in the user space(which is the parameter of the "read" function).It saves coping from kernel space to user space for the later choice, furthermore, there is much more buffer in the user space. But it should change the cr3 every time when coping, which flushes all the TLB data. That is what I know for the two choices, Is there anything else?
One common pattern that I see in the kernel is that a buffer is generally allocated in kernel space - kzalloc(PAGE_SIZE, GFP_KERNEL) (not necessarily PAGE_SIZE), then the read happens to that buffer. Then it is copied to the userspace using simple_read_from_buffer(..) (fs/libfs.c) - which internally uses copy_to_user(). Though this is generally done for simple I/O operations or other (eg. debugfs) reads.
One of the way is to ask device driver to map its kernel buffer to process's user address space via mmap and in turns uses something like remap_pfn_range.
When driver finishes the I/O operation, the copying to its kernel buffer doesn't need cr3 change or shoot down the TLB.
While the user process is waiting for the I/O, it is highly possible that it will be scheduled out for another new process to run, and then the cr3 has to change (+ TLB flush) for a entire I/O operation to be completed.