Caffeine AsyncLoadingCache and thundering herd - caffeine

Does .get() on a Caffeine AsyncLoadingCache prevent concurrent loads, by delaying subsequent threads which are calling .get() until the first one completes? Or that it can be configured to return a stale value while a self-populating load request is occurring?
This is so that a thundering herd can be prevented by using the cache.
I am seeing behavior which indicates that the thundering herd is not handled even though I am using a cache.
I create the cache like so:
val queryResponseCache: AsyncLoadingCache<Request, Response> = Caffeine.newBuilder()
.maximumSize(1000)
.expireAfterWrite(5, TimeUnit.SECONDS)
.recordStats()
.buildAsync(queryLoader)
And use it in conjunction with a L2 cache in redis like so (kotlin elvis operator):
queryResponseCache.getIfPresent(key) ?: fetchFromRedis(key) ?: queryResponseCache.get(key)
I understand that getIfPresent is concurrent, but the subsequent calls which end up calling fetchFromRedis() / get() seem to have problems. I guess moving the fetchFromRedis into the asyncLoad() function might be better for load tolerance.

A cache stampede is supported when you load through the cache. In your example using getIfPresent and loading the value, then I assume you put it into the cache explicitly inside of fetchFromRedis. Either way, you are ensuring a racy get-load-put due to bypassing the cache except when absent in Redis.
If you move the logic into asyncLoad, as you surmised, it would let the cache handle the stampede. The redis lookup, db query, and storing back into redis can all be performed as a chain of asynchronous tasks where the final future is returned to asyncLoad. Then the cache will compute the future once and return it to all subsequent calls until the entry is evicted.

Related

GCP dataflow - processing JSON takes too long

I am trying to process json files in a bucket and write the results into a bucket:
DataflowPipelineOptions options = PipelineOptionsFactory.create()
.as(DataflowPipelineOptions.class);
options.setRunner(BlockingDataflowPipelineRunner.class);
options.setProject("the-project");
options.setStagingLocation("gs://some-bucket/temp/");
Pipeline p = Pipeline.create(options);
p.apply(TextIO.Read.from("gs://some-bucket/2016/04/28/*/*.json"))
.apply(ParDo.named("SanitizeJson").of(new DoFn<String, String>() {
#Override
public void processElement(ProcessContext c) {
try {
JsonFactory factory = JacksonFactory.getDefaultInstance();
String json = c.element();
SomeClass e = factory.fromString(json, SomeClass.class);
// manipulate the object a bit...
c.output(factory.toString(e));
} catch (Exception err) {
LOG.error("Failed to process element: " + c.element(), err);
}
}
}))
.apply(TextIO.Write.to("gs://some-bucket/output/"));
p.run();
I have around 50,000 files under the path gs://some-bucket/2016/04/28/ (in sub-directories).
My question is: does it make sense that this takes more than an hour to complete? Doing something similar on a Spark cluster in amazon takes about 15-20 minutes. I suspect that I might be doing something inefficiently.
EDIT:
In my Spark job I aggregate all the results in a DataFrame and only then write the output, all at once. I noticed that my pipeline here writes each file separately, I assume that is why it's taking much longer. Is there a way to change this behavior?
Your jobs are hitting a couple of performance issues in Dataflow, caused by the fact that it is more optimized for executing work in larger increments, while your job is processing lots of very small files. As a result, some aspects of the job's execution end up dominated by per-file overhead. Here's some details and suggestions.
The job is limited rather by writing output than by reading input (though reading input is also a significant part). You can significantly cut that overhead by specifying withNumShards on your TextIO.Write, depending on how many files you want in the output. E.g. 100 could be a reasonable value. By default you're getting an unspecified number of files which in this case, given current behavior of the Dataflow optimizer, matches number of input files: usually it is a good idea because it allows us to not materialize the intermediate data, but in this case it's not a good idea because the input files are so small and per-file overhead is more important.
I recommend to set maxNumWorkers to a value like e.g. 12 - currently the second job is autoscaling to an excessively large number of workers. This is caused by Dataflow's autoscaling currently being geared toward jobs that process data in larger increments - it currently doesn't take into account per-file overhead and behaves not so well in your case.
The second job is also hitting a bug because of which it fails to finalize the written output. We're investigating, however setting maxNumWorkers should also make it complete successfully.
To put it shortly:
set maxNumWorkers=12
set TextIO.Write.to("...").withNumShards(100)
and it should run much better.

RxJava: retryWhen with retry limit

I am new to ReactiveX and reactive programming in general. I need to implement a retry mechanism for Couchbase CAS operations, but the example on the Couchbase website shows a retryWhen which seems to retry indefinitely. I need to have a retry limit and retry count somewhere in there.
The simple retry() would work, since it accepts a retryLimit, but I don't want it to retry on every exception, only on CASMismatchException.
Any ideas? I'm using the RxJava library.
In addition to what Simon Basle said, here is a quick version with linear backoff:
.retryWhen(notification ->
notification
.zipWith(Observable.range(1, 5), Tuple::create)
.flatMap(att ->
att.value2() == 3 ? Observable.error(att.value1()) : Observable.timer(att.value2(), TimeUnit.SECONDS)
)
)
Note that "att" here is a tuple which consists of both the throwable and the number of retries, so you can very specifically implement a return logic based on those two params.
If you want to learn even more, you can peek at the resilient doc I'm currently writing: https://gist.github.com/daschl/db9fcc9d2b932115b679#retry-with-delay
retryWhen is clearly a little bit more complicated than simple retry, but here's the gist of it:
you pass a notificationHandler function to retryWhen which takes an Observable<Throwable> and outputs an Observable<?>
the emission of this returned Observable determine when retry should occur or stop
so, for each occurring Exception in the original stream, if the handler's one emits 1 item, there'll be 1 retry. If it emits 2 items, there'll be 2...
as soon as the handler's stream emits an error, retry is aborted.
Using this, you can both:
work only on CasMismatchExceptions: just have your function return an Observable.error(t) in other cases
retry only for a specific number of times: for each exception, flatMap from an Observable.range representing the max number of retries, have it return an Observable.timer using the retry # if you need increasing delays.
Your use case is pretty close to the one in RxJava doc here
reviving this thread since in the Couchbase Java SDK 2.1.2 there's a new simpler way to do that: use the RetryBuilder:
Observable<Something> retryingObservable =
sourceObservable.retryWhen(
RetryBuilder
//will limit to the relevant exception
.anyOf(CASMismatchException.class)
//will retry only 5 times
.max(5)
//delay doubling each time, from 100ms to 2s
.delay(Delay.linear(TimeUnit.MILLISECONDS, 2000, 100, 2.0))
.build()
);

Returning values from InputFormat via the Hadoop Configuration object

Consider a running Hadoop job, in which a custom InputFormat needs to communicate ("return", similarly to a callback) a few simple values to the driver class (i.e., to the class that has launched the job), from within its overriden getSplits() method, using the new mapreduce API (as opposed to mapred).
These values should ideally be returned in-memory (as opposed to saving them to HDFS or to the DistributedCache).
If these values were only numbers, one could be tempted to use Hadoop counters. However, in numerous tests counters do not seem to be available at the getSplits() phase and anyway they are restricted to numbers.
An alternative could be to use the Configuration object of the job, which, as the source code reveals, should be the same object in memory for both the getSplits() and the driver class.
In such a scenario, if the InputFormat wants to "return" a (say) positive long value to the driver class, the code would look something like:
// In the custom InputFormat.
public List<InputSplit> getSplits(JobContext job) throws IOException
{
...
long value = ... // A value >= 0
job.getConfiguration().setLong("value", value);
...
}
// In the Hadoop driver class.
Job job = ... // Get the job to be launched
...
job.submit(); // Start running the job
...
while (!job.isComplete())
{
...
if (job.getConfiguration().getLong("value", -1))
{
...
}
else
{
continue; // Wait for the value to be set by getSplits()
}
...
}
The above works in tests, but is it a "safe" way of communicating values?
Or is there a better approach for such in-memory "callbacks"?
UPDATE
The "in-memory callback" technique may not work in all Hadoop distributions, so, as mentioned above, a safer way is, instead of saving the values to be passed back in the Configuration object, create a custom object, serialize it (e.g., as JSON), saved it (in HDFS or in the distributed cache) and have it read in the driver class. I have also tested this approach and it works as expected.
Using the configuration is a perfectly suitable solution (admittedly for a problem I'm not sure I understand), but once the job has actually been submitted to the Job tracker, you will not be able to amend this value (client side or task side) and expect to see the change on the opposite side of the comms (setting configuration values in a map task for example will not be persisted to the other mappers, nor to the reducers, nor will be visible to the job tracker).
So to communicate information back from within getSplits back to your client polling loop (to see when the job has actually finished defining the input splits) is fine in your example.
What's your greater aim or use case for using this?

Correct way to use InternetReadFile() asynchronously

I've got code that's performing HTTP requests using WinInet API's asynchronously. In general, my code works, but I'm confused about the 'right' way to do things. In the documentation for InternetReadFile(), it states:
To ensure all data is retrieved, an application must continue to call
the InternetReadFile function until the function returns TRUE and the
lpdwNumberOfBytesRead parameter equals zero.
but in asynchronous mode, it may (or may not) return false, and an error of ERROR_IO_PENDING, indicating it'll do the work asynchronously, and call my callback when finished. If I read the documentation literally, it seems that the asynchronous calls could also just do a partial read of the requested buffer, and require the caller to keep calling InternetReadFile until a read of 0 bytes is encountered.
A typical implementation using InternetReadFile() synchronously would look something like this:
while(InternetReadFile(Request, Buffer, BufferSize, &BytesRead) && BytesRead != 0)
{
// do something with Buffer
}
but with the possibility that any one call to InternetReadFile() could signal that it's going to do the work asynchronously (and perhaps read part, but not all of your request), it becomes much more complicated. If I turn to MSDN sample code for guidance, the implementation is simple, simply calling InternetReadFile() once, and expecting a single return having read the entire requested buffer either instantly or asynchronously. Is this the correct way to use this function, or is MSDN Sample Code ignoring the possibility that InternetReadFile() will only read part of the requested buffer?
After a more careful reading of the asynchronous example, I see now that it is reading repeatedly until a successful read of 0 bytes is encountered. So to answer my own question, you must call InternetReadFile() over and over again, and be prepared for either a synchronous or asynchronous response.
Reading InternetReadFile() repeatedly until it returns TRUE and BytesRead is 0 is a correct way to use InternetReadFile(), but not enough if you work asynchronously.
As MSDN says
When running asynchronously, if a call to InternetReadFile does not result in a completed transaction, it will return FALSE and a subsequent call to GetLastError will return ERROR_IO_PENDING. When the transaction is completed the InternetStatusCallback specified in a previous call to InternetSetStatusCallback will be called with INTERNET_STATUS_REQUEST_COMPLETE.
So InternetReadFile() may return FALSE and set the last error to ERROR_IO_PENDING value if you work in asynchronous mode.
When InternetSetStatusCallback will be called again with INTERNET_STATUS_REQUEST_COMPLETE, the lpvStatusInformation parameter will contain the address of an INTERNET_ASYNC_RESULT structure (see InternetStatusCallback callback function). The INTERNET_ASYNC_RESULT.dwResult member will contain the result of asynchronous operation (TRUE or FALSE since you called InternetReadFile) and the INTERNET_ASYNC_RESULT.dwError will contain a error code only if dwResult is FALSE.
If dwResult is TRUE then your Buffer contains data read from Internet, and the BytesRead contains the number of bytes read asynchronously.
So one of the most important things when you work asynchronously, the Buffer and the BytesRead must be persistent between InternetStatusCallback calls, i.e. must not be allocated on the stack. Otherwise it has undefined behaviour, causes memory corruption, etc.

What is an idempotent operation?

What is an idempotent operation?
In computing, an idempotent operation is one that has no additional effect if it is called more than once with the same input parameters. For example, removing an item from a set can be considered an idempotent operation on the set.
In mathematics, an idempotent operation is one where f(f(x)) = f(x). For example, the abs() function is idempotent because abs(abs(x)) = abs(x) for all x.
These slightly different definitions can be reconciled by considering that x in the mathematical definition represents the state of an object, and f is an operation that may mutate that object. For example, consider the Python set and its discard method. The discard method removes an element from a set, and does nothing if the element does not exist. So:
my_set.discard(x)
has exactly the same effect as doing the same operation twice:
my_set.discard(x)
my_set.discard(x)
Idempotent operations are often used in the design of network protocols, where a request to perform an operation is guaranteed to happen at least once, but might also happen more than once. If the operation is idempotent, then there is no harm in performing the operation two or more times.
See the Wikipedia article on idempotence for more information.
The above answer previously had some incorrect and misleading examples. Comments below written before April 2014 refer to an older revision.
An idempotent operation can be repeated an arbitrary number of times and the result will be the same as if it had been done only once. In arithmetic, adding zero to a number is idempotent.
Idempotence is talked about a lot in the context of "RESTful" web services. REST seeks to maximally leverage HTTP to give programs access to web content, and is usually set in contrast to SOAP-based web services, which just tunnel remote procedure call style services inside HTTP requests and responses.
REST organizes a web application into "resources" (like a Twitter user, or a Flickr image) and then uses the HTTP verbs of POST, PUT, GET, and DELETE to create, update, read, and delete those resources.
Idempotence plays an important role in REST. If you GET a representation of a REST resource (eg, GET a jpeg image from Flickr), and the operation fails, you can just repeat the GET again and again until the operation succeeds. To the web service, it doesn't matter how many times the image is gotten. Likewise, if you use a RESTful web service to update your Twitter account information, you can PUT the new information as many times as it takes in order to get confirmation from the web service. PUT-ing it a thousand times is the same as PUT-ing it once. Similarly DELETE-ing a REST resource a thousand times is the same as deleting it once. Idempotence thus makes it a lot easier to construct a web service that's resilient to communication errors.
Further reading: RESTful Web Services, by Richardson and Ruby (idempotence is discussed on page 103-104), and Roy Fielding's PhD dissertation on REST. Fielding was one of the authors of HTTP 1.1, RFC-2616, which talks about idempotence in section 9.1.2.
No matter how many times you call the operation, the result will be the same.
Idempotence means that applying an operation once or applying it multiple times has the same effect.
Examples:
Multiplication by zero. No matter how many times you do it, the result is still zero.
Setting a boolean flag. No matter how many times you do it, the flag stays set.
Deleting a row from a database with a given ID. If you try it again, the row is still gone.
For pure functions (functions with no side effects) then idempotency implies that f(x) = f(f(x)) = f(f(f(x))) = f(f(f(f(x)))) = ...... for all values of x
For functions with side effects, idempotency furthermore implies that no additional side effects will be caused after the first application. You can consider the state of the world to be an additional "hidden" parameter to the function if you like.
Note that in a world where you have concurrent actions going on, you may find that operations you thought were idempotent cease to be so (for example, another thread could unset the value of the boolean flag in the example above). Basically whenever you have concurrency and mutable state, you need to think much more carefully about idempotency.
Idempotency is often a useful property in building robust systems. For example, if there is a risk that you may receive a duplicate message from a third party, it is helpful to have the message handler act as an idempotent operation so that the message effect only happens once.
A good example of understanding an idempotent operation might be locking a car with remote key.
log(Car.state) // unlocked
Remote.lock();
log(Car.state) // locked
Remote.lock();
Remote.lock();
Remote.lock();
log(Car.state) // locked
lock is an idempotent operation. Even if there are some side effect each time you run lock, like blinking, the car is still in the same locked state, no matter how many times you run lock operation.
An idempotent operation produces the result in the same state even if you call it more than once, provided you pass in the same parameters.
An idempotent operation is an operation, action, or request that can be applied multiple times without changing the result, i.e. the state of the system, beyond the initial application.
EXAMPLES (WEB APP CONTEXT):
IDEMPOTENT:
Making multiple identical requests has the same effect as making a single request. A message in an email messaging system is opened and marked as "opened" in the database. One can open the message many times but this repeated action will only ever result in that message being in the "opened" state. This is an idempotent operation. The first time one PUTs an update to a resource using information that does not match the resource (the state of the system), the state of the system will change as the resource is updated. If one PUTs the same update to a resource repeatedly then the information in the update will match the information already in the system upon every PUT, and no change to the state of the system will occur. Repeated PUTs with the same information are idempotent: the first PUT may change the state of the system, subsequent PUTs should not.
NON-IDEMPOTENT:
If an operation always causes a change in state, like POSTing the same message to a user over and over, resulting in a new message sent and stored in the database every time, we say that the operation is NON-IDEMPOTENT.
NULLIPOTENT:
If an operation has no side effects, like purely displaying information on a web page without any change in a database (in other words you are only reading the database), we say the operation is NULLIPOTENT. All GETs should be nullipotent.
When talking about the state of the system we are obviously ignoring hopefully harmless and inevitable effects like logging and diagnostics.
Just wanted to throw out a real use case that demonstrates idempotence. In JavaScript, say you are defining a bunch of model classes (as in MVC model). The way this is often implemented is functionally equivalent to something like this (basic example):
function model(name) {
function Model() {
this.name = name;
}
return Model;
}
You could then define new classes like this:
var User = model('user');
var Article = model('article');
But if you were to try to get the User class via model('user'), from somewhere else in the code, it would fail:
var User = model('user');
// ... then somewhere else in the code (in a different scope)
var User = model('user');
Those two User constructors would be different. That is,
model('user') !== model('user');
To make it idempotent, you would just add some sort of caching mechanism, like this:
var collection = {};
function model(name) {
if (collection[name])
return collection[name];
function Model() {
this.name = name;
}
collection[name] = Model;
return Model;
}
By adding caching, every time you did model('user') it will be the same object, and so it's idempotent. So:
model('user') === model('user');
Quite a detailed and technical answers. Just adding a simple definition.
Idempotent = Re-runnable
For example,
Create operation in itself is not guaranteed to run without error if executed more than once.
But if there is an operation CreateOrUpdate then it states re-runnability (Idempotency).
Idempotent Operations: Operations that have no side-effects if executed multiple times.
Example: An operation that retrieves values from a data resource and say, prints it
Non-Idempotent Operations: Operations that would cause some harm if executed multiple times. (As they change some values or states)
Example: An operation that withdraws from a bank account
It is any operation that every nth result will result in an output matching the value of the 1st result. For instance the absolute value of -1 is 1. The absolute value of the absolute value of -1 is 1. The absolute value of the absolute value of absolute value of -1 is 1. And so on. See also: When would be a really silly time to use recursion?
An idempotent operation over a set leaves its members unchanged when applied one or more times.
It can be a unary operation like absolute(x) where x belongs to a set of positive integers. Here absolute(absolute(x)) = x.
It can be a binary operation like union of a set with itself would always return the same set.
cheers
In short, Idempotent operations means that the operation will not result in different results no matter how many times you operate the idempotent operations.
For example, according to the definition of the spec of HTTP, GET, HEAD, PUT, and DELETE are idempotent operations; however POST and PATCH are not. That's why sometimes POST is replaced by PUT.
An operation is said to be idempotent if executing it multiple times is equivalent to executing it once.
For eg: setting volume to 20.
No matter how many times the volume of TV is set to 20, end result will be that volume is 20. Even if a process executes the operation 50/100 times or more, at the end of the process the volume will be 20.
Counter example: increasing the volume by 1. If a process executes this operation 50 times, at the end volume will be initial Volume + 50 and if a process executes the operation 100 times, at the end volume will be initial Volume + 100. As you can clearly see that the end result varies based upon how many times the operation was executed. Hence, we can conclude that this operation is NOT idempotent.
I have highlighted the end result in bold.
If you think in terms of programming, let's say that I have an operation in which a function f takes foo as the input and the output of f is set to foo back. If at the end of the process (that executes this operation 50/100 times or more), my foo variable holds the value that it did when the operation was executed only ONCE, then the operation is idempotent, otherwise NOT.
foo = <some random value here, let's say -2>
{ foo = f( foo ) }   curly brackets outline the operation
if f returns the square of the input then the operation is NOT idempotent. Because foo at the end will be (-2) raised to the power (number of times operation is executed)
if f returns the absolute of the input then the operation is idempotent because no matter how many multiple times the operation is executed foo will be abs(-2).
Here, end result is defined as the final value of variable foo.
In mathematical sense, idempotence has a slightly different meaning of:
f(f(....f(x))) = f(x)
here output of f(x) is passed as input to f again which doesn't need to be the case always with programming.
my 5c:
In integration and networking the idempotency is very important.
Several examples from real-life:
Imagine, we deliver data to the target system. Data delivered by a sequence of messages.
1. What would happen if the sequence is mixed in channel? (As network packages always do :) ). If the target system is idempotent, the result will not be different. If the target system depends of the right order in the sequence, we have to implement resequencer on the target site, which would restore the right order.
2. What would happen if there are the message duplicates? If the channel of target system does not acknowledge timely, the source system (or channel itself) usually sends another copy of the message. As a result we can have duplicate message on the target system side.
If the target system is idempotent, it takes care of it and result will not be different.
If the target system is not idempotent, we have to implement deduplicator on the target system side of the channel.
For a workflow manager (as Apache Airflow) if an idempotency operation fails in your pipeline the system can retry the task automatically without affecting the system. Even if the logs change, that is good because you can see the incident.
The most important in this case is that your system can retry the task that failed and doesn't mess up the pipeline (e.g. appending the same data in a table each retry)
Let's say the client makes a request to "IstanceA" service which process the request, passes it to DB, and shuts down before sending the response. since the client does not see that it was processed and it will retry the same request. Load balancer will forward the request to another service instance, "InstanceB", which will make the same change on the same DB item.
We should use idempotent tokens. When a client sends a request to a service, it should have some kind of request-id that can be saved in DB to show that we have already executed the request. if the client retries the request, "InstanceB" will check the requestId. Since that particular request already has been executed, it will not make any change to the DB item. Those kinds of requests are called idempotent requests. So we send the same request multiple times, but we won't make any change