I am trying to submit a job to a cluster [running Sun Grid Engine (SGE)]. The job kept being terminated with the following report:
Job 780603 (temp_new) Aborted
Exit Status = 137
Signal = KILL
User = heaswara
Queue = std.q#comp-0-8.local
Host = comp-0-8.local
Start Time = 08/24/2013 13:49:05
End Time = 08/24/2013 16:26:38
CPU = 02:46:38
Max vmem = 12.055G
failed assumedly after job because:
job 780603.1 died through signal KILL (9)
The resource requirements I had set were:
#$ -l mem_free=10G
#$ -l h_vmem=12G
mem_free is the amount of memory my job requires and h_vmem is the is the upper bound on the amount of memory the job is allowed to use. I wonder my job is being terminated because it requires more than that threshold (12G).
Is there a way to estimate how much memory will be required for my operation? I am trying to figure out what should be the upper bound.
Thanks in advance.
It depends on the nature of the job itself. If you know anything about the program that is being run (i.e., you wrote it), you should be able to make an estimate on how much memory it is going to want. If not, your only recourse is to run it without the limit and see how much it actually uses.
I have a bunch of FPGA build and simulation jobs that I run. After each job, I track how much memory was actually used. I can use this historical information to make an estimate on how much it might use in the future (I pad by 10% in case there are some weird changes in the source). I still have to redo the calculations whenever the vendor delivers a new version of the tools, though, as quite often the memory footprint changes dramatically.
Related
I'm using Splunk to monitor my applications.
I also store resource statistics in my Splunk too.
Goal: I want to find the optimum CPU limit for each container.
How to I write a query that finds an optimum CPU limit? Or the other question is Should I?
Concern1: When I start customizing my query and let's say that I have used MAX(CPU) command. It doesn't mean that my container will be running at level most of the time. So, I might set an unnecessary high limit for my containers.
Let me explain, when I find a CPU limit value via MAX(CPU) command as 10, this top value might be happened because of a bulk operation. So, my container's expected resource may be around 1.2 all the time, except this single 1 operation that one. So, using MAX value won't work.
Concern2: Let's say that I have used the value of AVG(CPU) value and used it. And that is 2, So how many of my operations will be waited for how many minutes after this change? Or how many of them are going to be timed out? It may create a lot of side-effects. How will I decide the real average value? What parameters should be used?
Is it possible to include such conditions in the query? Or do I need an AI to decide it? :)
Here are my givin parameters:
path=statistics.cpus_system_time_secs
path=statistics.cpus_user_time_secs
path=statistics.cpus_nr_periods
path=statistics.cpus_nr_throttled
path=statistics.cpus_throttled_time_secs
path=statistics.cpus_limit
I bet you can ask better questions than me. Let's discuss.
"Optimum" is going to depend greatly on your own environment (resources available, application priority, etc)
You probably want to look at a combination of the following factors:
avg(CPU)
max(CPU) (and time spent there)
min(CPU) (and time spent there)
I suspect your "optimum" limit is going to be a % below your max...but only if you're spending 'a lot' of time maxxed-out
And, of course, being "maxed" may not matter, if other containers are running acceptably
Keep in mind, once you set that limit, your max will drop (as, likely, will your avg)
I am running a pipeline on multiple images. The pipeline consist of reading the images from file system, doing so processing on each of them, then saving the images to file system. However the dask worker fails due to MemoryError.
Is there a way to assure the dask workers don't load too many images in memory? i.e. Wait until there is enough space on a worker before starting the processing pipeline on a new image.
I have one scheduler and 40 workers with 4 cores, 15GB ram and running Centos7. I am trying to process 125 images in a batch; each image is fairly large but small enough to fit on a worker; around 3GB require for the whole process.
I tried to process a smaller amount of images and it works great.
EDITED
from dask.distributed import Client, LocalCluster
# LocalCluster is used to show the config of the workers on the actual cluster
client = Client(LocalCluster(n_workers=2, resources={'process': 1}))
paths = ['list', 'of', 'paths']
# Read the file data from each path
data = client.map(read, path, resources={'process': 1)
# Apply foo to the data n times
for _ in range(n):
data = client.map(foo, x, resources={'process': 1)
# Save the processed data
data.map(save, x, resources={'process': 1)
# Retrieve results
client.gather(data)
I expected the images to be process as space was available on the workers but it seems like the images are all loaded simultaneously on the different workers.
EDIT:
My issues is that all task get assigned to workers and they don't have enough memory. I found how to limit the number of task a worker handle at a single moment [https://distributed.readthedocs.io/en/latest/resources.html#resources-are-applied-separately-to-each-worker-process](see here).
However, with that limit, when I execute my task they all finish the read step, then the process step and finally the save step. This is an issues since the image are spilled to disk.
Would there be a way to make every task finish before starting a new one?
e.g. on Worker-1: read(img1)->process(img1)->save(img1)->read(img2)->...
Dask does not generally know how much memory a task will need, it can only know the size of the outputs, and that, only once they are finished. This is because Dask simply executes a pthon function and then waits for it to complete; but all osrts of things can happen within a python function. You should generally expect as many tasks to begin as you have available worker cores - as you are finding.
If you want a smaller total memory load, then your solution should be simple: have a small enough number of workers, so that if all of them are using the maximum memory that you can expect, you still have some spare in the system to cope.
To EDIT: you may want to try running optimize on the graph before submission (although this should happen anyway, I think), as it sounds like your linear chains of tasks should be "fused". http://docs.dask.org/en/latest/optimize.html
Let's say I have a function which does some work by spawning multiple processes. I want to compare CPU time vs real time taken by this function.
def test do
prev_real = System.monotonic_time(:millisecond)
# Code to complete some task
# Spawn different processes & give each process some task
# Receive result
# Finish task
current_real = System.monotonic_time(:millisecond)
diff_real = current_real - prev_real
IO.puts "Real time " <> to_string(diff_real)
IO.puts "CPU time ?????"
end
How to calculate CPU time required by the given function? I am interested in calculating CPU time/Real time ratio.
If you are just trying to profile your code rather than implement your own profiling framework I would recommend using already existing tools like:
fprof which will give you information about time spent in functions (real and own)
percept which will provide you information about which processes in your system ware working at any given time and on what
xprof which is design to help you find which calls to your function will cause it to take more time (trigger inefficient branch of code).
They take advantage of both erlang:trace to figure out which function is being executed and for how long and erlang:system_profile with runnable_procs to determine which processes are currently running. You might start a function, hit a receive or be preemptive rescheduled and wait without doing any actual work. Combining those two might be complicated, and I would recommend using already existing tools before trying glue together your own.
You could also look into tools like erlgrind and eflame if you are looking for more visual representations of your calls.
How do I improve performance when writing to neo4j. I currently have neo4j set up on a server and I am currently running it in embedded more. I believe my configurations are storing all the content of my graph database in memory based upon configurations I've found online
neostore.nodestore.db.mapped_memory=0
neostore.relationship.db.mapped_memory=0
neostore.propertystore.db.mapped_memory=0
neostore.propertystore.db.strings.mapped_memory=0
neostore.propertystore.db.arrays.mapped_memory=0
neostore.propertystore.db.index.keys.mapped_memory=0
neostore.propertystore.db.index.mapped_memory=0
node_auto_indexing=true
node_keys_indexable=type,id
cache_type=strong
use_memory_mapped_buffers=false
node_cache_size=12G
relationship_cache_size=12G
node_cache_array_fraction=10
relationship_cache_array_fraction=10
Please let me know if this is incorrect. The problem that I am encountering is that when I try to persist information to the graph database. It appears that those times are not very quick in comparison to our MYSQL times of the samething(ex. to add 250 items would take about 3sec and in MYSQL it takes 1sec) . I read online that when you have multiple indexes that that can slow down performance on persisting data so I am working on that right now to see if that is my culprit. But, I just wanted to make sure that my configurations seem to be inline when it comes to running your graph database in memory.
Second question to this topic. Okay, if my configurations are good and my database is indeed in memory, then is there a way to optimize persisting data just in case this isn't the silver bullet. If we ran one thread against our test that executes this functionality, oppose to 10 threads, its seems like the times for execution bubbles up
ex.( thread 1 finishes 1s, thread 2 finishes 2s, thread 3 finishes 3s,etc). Is there some special multithreaded configuration that I am missing to improve the performance when mulitple threads are hitting it at one time.
Neo4J version
1.9.1-enterprise
My Jvm configs are
-Xms25G -Xmx25G -XX:+UseNUMA -XX:+UseSerialGC
My Machine Specs:
File system type ext3
You cache arguments are invalid.
node_cache_size=12G
relationship_cache_size=12G
node_cache_array_fraction=10
relationship_cache_array_fraction=10
These can only be used with the GCR cache. Setting the cache isn't going to put everything in memory for you at start up, you will have to write code to do this for you. Something like this:
GlobalGraphOperations ggo = GlobalGraphOperations.at(graphDatabaseFactory);
for (Node n : ggo.getAllNodes()) {
for (String propertyKey : n.getPropertyKeys()) {
n.getProperty(propertyKey);
}
for (Relationship relationship : n.getRelationships()) {
}
}
Beware with the strong cache, if you have a lot of nodes/relationships, eventually your cache will become large and performing GC against it will cause long pauses in your system.
My recommendation would be to use the memory mapped files, as this is an OS handled and will be outside of heap space. It doesn't provide near the speed of caching, but it will provide a speed up if you have to read from the neo store.
Update: The while() condition below gets optimized out by the compiler, so both threads just skip the condition and enter the C.S. even with -O0 flag. Does anyone know why the compiler is doing this? By the way, declaring the global variables volatile causes the program to hang for some odd reason...
I read the CUDA programming guide but I'm still a bit unclear on how CUDA handles memory consistency with respect to global memory. (This is different from the memory hierarchy) Basically, I am running tests trying to break sequential consistency. The algorithm I am using is Peterson's algorithm for mutual exclusion between two threads inside the kernel function:
flag[threadIdx.x] = 1; // both these are global
turn = 1-threadIdx.x;
while(flag[1-threadIdx.x] == 1 && turn == (1- threadIdx.x));
shared_gloabl_variable_x ++;
flag[threadIdx.x] = 0;
This is fairly straightforward. Each thread asks for the critical section by setting its flag to one and by being nice by giving the turn to the other thread. At the evaluation of the while(), if the other thread did not set its flag, the requesting thread can then enter the critical section safely. Now a subtle problem with this approach is that if the compiler re-orders the writes so that the write to turn executes before the write to flag. If this happens both threads will end up in the C.S. at the same time. This fairly easy to prove with normal Pthreads, since most processors don't implement sequential consistency. But what about GPUs?
Both of these threads will be in the same warp. And they will execute their statements in lock-step mode. But when they reach the turn variable they are writing to the same variable so the intra-warp execution becomes serialized (doesn't matter what the order is). Now at this point, does the thread that wins proceed onto the while condition, or does it wait for the other thread to finish its write, so that both can then evaluate the while() at the same time? The paths again will diverge at the while(), because only one of them will win while the other waits.
After running the code, I am getting it to consistently break SC. The value I read is ALWAYS 1, which means that both threads somehow are entering the C.S. every single time. How is this possible (GPUs execute instructions in order)? (Note: I have compiled it with -O0, so no compiler optimization, and hence no use of volatile).
Edit: since you have only two threads and 1-threadIdx.x works, then you must be using thread IDs 0 and 1. Threads 0 and 1 will always be part of the same warp on all current NVIDIA GPUs. Warps execute instructions SIMD fashion, with a thread execution mask for divergent conditions. Your while loop is a divergent condition.
When turn and flags are not volatile, the compiler probably reorders the instructions and you see the behavior of both threads entering the C.S.
When turn and flags are volatile, you see a hang. The reason is that one of the threads will succeed at writing turn, so turn will be either 0 or 1. Let's assume turn==0: If the hardware chooses to execute thread 0's part of the divergent branch, then all is OK. But if it chooses to execute thread 1's part of the divergent branch, then it will spin on the while loop and thread 0 will never get its turn, hence the hang.
You can probably avoid the hang by ensuring that your two threads are in different warps, but I think that the warps must be concurrently resident on the SM so that instructions can issue from both and progress can be made. (Might work with concurrent warps on different SMs, since this is global memory; but that might require __threadfence() and not just __threadfence_block().)
In general this is a great example of why code like this is unsafe on GPUs and should not be used. I realize though that this is just an investigative experiment. In general CUDA GPUs do not—as you mention most processors do not—implement sequential consistency.
Original Answer
the variables turn and flag need to be volatile, otherwise the load of flag will not be repeated and the condition turn == 1-threadIdx.X will not be re-evaluated but instead will be taken as true.
There should be a __threadfence_block() between the store to flag and store to turn to get the right ordering.
There should be a __threadfence_block() before the shared variable increment (which should also be declared volatile). You may also want a __syncthreads() or at least __threadfence_block() after the increment to ensure it is visible to other threads.
I have a hunch that even after making these fixes you may still run into trouble, though. Let us know how it goes.
BTW, you have a syntax error in this line, so it's clear this isn't exactly your real code:
while(flag[1-threadIdx.x] == 1 and turn==[1- threadIdx.x]);
In the absence of extra memory barriers such as __threadfence(), sequential consistency of global memory is enforced only within a given thread.