I'd like to know if my pytorch code is fully utilizing the GPU SMs. According to this question gpu-util in nvidia-smi only shows how time at least one SM was used.
I also saw that typing nvidia-smi dmon gives the following table:
# gpu pwr gtemp mtemp sm mem enc dec mclk pclk
# Idx W C C % % % % MHz MHz
0 132 71 - 58 18 0 0 6800 1830
Where one would think that sm% would be SM utilization, but I couldn't find any documentation on what sm% means. The number given is exactly the same as gpu-util in nvidia-smi.
Is there any way to check the SM utilization?
On a side note, is there any way to check memory bandwidth utilization?
Related
We are doing some performance measurements including some memory footprint measurements. We've been doing this with GNU time.
But, I cannot tell if they are measuring in kilobytes (1000 bytes) or kibibytes (1024 bytes).
The man page for my system says of the %M format key (which we are using to measure peak memory usage): "Maximum resident set size of the process during its lifetime, in Kbytes."
I assume K here means the SI "Kilo" prefix, and thus kilobytes.
But having looked at a few other memory measurements of various things through various tools, I trust that assumption like I'd trust a starved lion to watch my dogs during a week-long vacation.
I need to know, because for our tests 1000 vs 1024 Kbytes adds up to a difference of nearly 8 gigabytes, and I'd like to think I can cut down the potential error in our measurements by a few billion.
Using the below testing setup, I have determined that GNU time on my system measures in Kibibytes.
The below program (allocator.c) allocates data and touches each of it 1 KiB at a time to ensure that it all gets paged in. Note: This test only works if you can page in the entirety of the allocated data, otherwise time's measurement will only be the largest resident collection of memory.
allocator.c:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define min(a,b) ( ( (a)>(b) )? (b) : (a) )
volatile char access;
volatile char* data;
const int step = 128;
int main(int argc, char** argv ){
unsigned long k = strtoul( argv[1], NULL, 10 );
if( k >= 0 ){
printf( "Allocating %lu (%s) bytes\n", k, argv[1] );
data = (char*) malloc( k );
for( int i = 0; i < k; i += step ){
data[min(i,k-1)] = (char) i;
}
free( data );
} else {
printf("Bad size: %s => %lu\n", argv[1], k );
}
return 0;
}
compile with: gcc -O3 allocator.c -o allocator
Runner Bash Script:
kibibyte=1024
kilobyte=1000
mebibyte=$(expr 1024 \* ${kibibyte})
megabyte=$(expr 1000 \* ${kilobyte})
gibibyte=$(expr 1024 \* ${mebibyte})
gigabyte=$(expr 1000 \* ${megabyte})
for mult in $(seq 1 3);
do
bytes=$(expr ${gibibyte} \* ${mult} )
echo ${mult} GiB \(${bytes} bytes\)
echo "... in kibibytes: $(expr ${bytes} / ${kibibyte})"
echo "... in kilobytes: $(expr ${bytes} / ${kilobyte})"
/usr/bin/time -v ./allocator ${bytes}
echo "===================================================="
done
For me this produces the following output:
1 GiB (1073741824 bytes)
... in kibibytes: 1048576
... in kilobytes: 1073741
Allocating 1073741824 (1073741824) bytes
Command being timed: "./a.out 1073741824"
User time (seconds): 0.12
System time (seconds): 0.52
Percent of CPU this job got: 75%
Elapsed (wall clock) time (h:mm:ss or m:ss): 0:00.86
Average shared text size (kbytes): 0
Average unshared data size (kbytes): 0
Average stack size (kbytes): 0
Average total size (kbytes): 0
Maximum resident set size (kbytes): 1049068
Average resident set size (kbytes): 0
Major (requiring I/O) page faults: 0
Minor (reclaiming a frame) page faults: 262309
Voluntary context switches: 7
Involuntary context switches: 2
Swaps: 0
File system inputs: 16
File system outputs: 8
Socket messages sent: 0
Socket messages received: 0
Signals delivered: 0
Page size (bytes): 4096
Exit status: 0
====================================================
2 GiB (2147483648 bytes)
... in kibibytes: 2097152
... in kilobytes: 2147483
Allocating 2147483648 (2147483648) bytes
Command being timed: "./a.out 2147483648"
User time (seconds): 0.21
System time (seconds): 1.09
Percent of CPU this job got: 99%
Elapsed (wall clock) time (h:mm:ss or m:ss): 0:01.31
Average shared text size (kbytes): 0
Average unshared data size (kbytes): 0
Average stack size (kbytes): 0
Average total size (kbytes): 0
Maximum resident set size (kbytes): 2097644
Average resident set size (kbytes): 0
Major (requiring I/O) page faults: 0
Minor (reclaiming a frame) page faults: 524453
Voluntary context switches: 4
Involuntary context switches: 3
Swaps: 0
File system inputs: 0
File system outputs: 8
Socket messages sent: 0
Socket messages received: 0
Signals delivered: 0
Page size (bytes): 4096
Exit status: 0
====================================================
3 GiB (3221225472 bytes)
... in kibibytes: 3145728
... in kilobytes: 3221225
Allocating 3221225472 (3221225472) bytes
Command being timed: "./a.out 3221225472"
User time (seconds): 0.38
System time (seconds): 1.60
Percent of CPU this job got: 99%
Elapsed (wall clock) time (h:mm:ss or m:ss): 0:01.98
Average shared text size (kbytes): 0
Average unshared data size (kbytes): 0
Average stack size (kbytes): 0
Average total size (kbytes): 0
Maximum resident set size (kbytes): 3146220
Average resident set size (kbytes): 0
Major (requiring I/O) page faults: 0
Minor (reclaiming a frame) page faults: 786597
Voluntary context switches: 4
Involuntary context switches: 3
Swaps: 0
File system inputs: 0
File system outputs: 8
Socket messages sent: 0
Socket messages received: 0
Signals delivered: 0
Page size (bytes): 4096
Exit status: 0
====================================================
In the "Maximum resident set size" entry, I see values that are closest to the kibibytes value I expect from that raw byte count. There is some difference because its possible that some memory is being paged out (in cases where it is lower, which none of them are here) and because there is more memory being consumed than what the program allocates (namely, the stack and the actual binary image itself).
Versions on my system:
> gcc --version
gcc (GCC) 6.1.0
Copyright (C) 2016 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
> /usr/bin/time --version
GNU time 1.7
> lsb_release -a
LSB Version: :base-4.0-amd64:base-4.0-noarch:core-4.0-amd64:core-4.0-noarch:graphics-4.0-amd64:graphics-4.0-noarch:printing-4.0-amd64:printing-4.0-noarch
Distributor ID: CentOS
Description: CentOS release 6.10 (Final)
Release: 6.10
Codename: Final
I am using two graphic cards and the GeForce gtx980 with 4GB, where I compute my neuronal network is always jumping from 0 to 99% and from 99% to 0% (repeating) at the last line of the pasted shell output.
After around 90seconds it did the first calculation. I put my images one after another into the neuronal network (for-loop). And the following calculations only need 20 seconds (3 epochs) and the GPU jumps between 96 and 100%.
Why is it jumping at the beginning?
I use the flag:
config.gpu_options.allow_growth = True
with tf.Session(config=config) as sess:
Can I be sure that is really using not less megabytes than nvidia-smi -lms 50 is showing me?
2017-08-10 16:33:24.836084: W tensorflow/core/platform/cpu_feature_guard.cc:45] The TensorFlow library wasn't compiled to use SSE4.1 instructions, but these are available on your machine and could speed up CPU computations.
2017-08-10 16:33:24.836100: W tensorflow/core/platform/cpu_feature_guard.cc:45] The TensorFlow library wasn't compiled to use SSE4.2 instructions, but these are available on your machine and could speed up CPU computations.
2017-08-10 16:33:25.052501: I tensorflow/stream_executor/cuda/cuda_gpu_executor.cc:893] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero
2017-08-10 16:33:25.052861: I tensorflow/core/common_runtime/gpu/gpu_device.cc:940] Found device 0 with properties:
name: GeForce GTX 980
major: 5 minor: 2 memoryClockRate (GHz) 1.2155
pciBusID 0000:03:00.0
Total memory: 3.94GiB
Free memory: 3.87GiB
2017-08-10 16:33:25.187760: W tensorflow/stream_executor/cuda/cuda_driver.cc:523] A non-primary context 0x8532640 exists before initializing the StreamExecutor. We haven't verified StreamExecutor works with that.
2017-08-10 16:33:25.188006: I tensorflow/stream_executor/cuda/cuda_gpu_executor.cc:893] successful NUMA node read from SysFS had negative value (-1), but there must be at least one NUMA node, so returning NUMA node zero
2017-08-10 16:33:25.188291: I tensorflow/core/common_runtime/gpu/gpu_device.cc:940] Found device 1 with properties:
name: GeForce GT 730
major: 3 minor: 5 memoryClockRate (GHz) 0.9015
pciBusID 0000:02:00.0
Total memory: 1.95GiB
Free memory: 1.45GiB
2017-08-10 16:33:25.188312: I tensorflow/core/common_runtime/gpu/gpu_device.cc:832] Peer access not supported between device ordinals 0 and 1
2017-08-10 16:33:25.188319: I tensorflow/core/common_runtime/gpu/gpu_device.cc:832] Peer access not supported between device ordinals 1 and 0
2017-08-10 16:33:25.188329: I tensorflow/core/common_runtime/gpu/gpu_device.cc:961] DMA: 0 1
2017-08-10 16:33:25.188335: I tensorflow/core/common_runtime/gpu/gpu_device.cc:971] 0: Y N
2017-08-10 16:33:25.188339: I tensorflow/core/common_runtime/gpu/gpu_device.cc:971] 1: N Y
2017-08-10 16:33:25.188348: I tensorflow/core/common_runtime/gpu/gpu_device.cc:1030] Creating TensorFlow device (/gpu:0) -> (device: 0, name: GeForce GTX 980, pci bus id: 0000:03:00.0)
Epoche: 0001 cost= 0.620101001 time= 115.366318226
Epoche: 0004 cost= 0.335480299 time= 19.4528050423
We have a java application running on Mule. We have the XMX value configured for 6144M, but are routinely seeing the overall memory usage climb and climb. It was getting close to 20 GB the other day before we proactively restarted it.
Thu Jun 30 03:05:57 CDT 2016
top - 03:05:58 up 149 days, 6:19, 0 users, load average: 0.04, 0.04, 0.00
Tasks: 164 total, 1 running, 163 sleeping, 0 stopped, 0 zombie
Cpu(s): 4.2%us, 1.7%sy, 0.0%ni, 93.9%id, 0.2%wa, 0.0%hi, 0.0%si, 0.0%st
Mem: 24600552k total, 21654876k used, 2945676k free, 440828k buffers
Swap: 2097144k total, 84256k used, 2012888k free, 1047316k cached
PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND
3840 myuser 20 0 23.9g 18g 53m S 0.0 79.9 375:30.02 java
The jps command shows:
10671 Jps
3840 MuleContainerBootstrap
The jstat command shows:
S0C S1C S0U S1U EC EU OC OU PC PU YGC YGCT FGC FGCT GCT
37376.0 36864.0 16160.0 0.0 2022912.0 1941418.4 4194304.0 445432.2 78336.0 66776.7 232 7.044 17 17.403 24.447
The startup arguments are (sensitive bits have been changed):
3840 MuleContainerBootstrap -Dmule.home=/mule -Dmule.base=/mule -Djava.net.preferIPv4Stack=TRUE -XX:MaxPermSize=256m -Djava.endorsed.dirs=/mule/lib/endorsed -XX:+HeapDumpOnOutOfMemoryError -Dmyapp.lib.path=/datalake/app/ext_lib/ -DTARGET_ENV=prod -Djava.library.path=/opt/mapr/lib -DksPass=mypass -DsecretKey=aeskey -DencryptMode=AES -Dkeystore=/mule/myStore -DkeystoreInstance=JCEKS -Djava.security.auth.login.config=/opt/mapr/conf/mapr.login.conf -Dmule.mmc.bind.port=1521 -Xms6144m -Xmx6144m -Djava.library.path=%LD_LIBRARY_PATH%:/mule/lib/boot -Dwrapper.key=a_guid -Dwrapper.port=32000 -Dwrapper.jvm.port.min=31000 -Dwrapper.jvm.port.max=31999 -Dwrapper.disable_console_input=TRUE -Dwrapper.pid=10744 -Dwrapper.version=3.5.19-st -Dwrapper.native_library=wrapper -Dwrapper.arch=x86 -Dwrapper.service=TRUE -Dwrapper.cpu.timeout=10 -Dwrapper.jvmid=1 -Dwrapper.lang.domain=wrapper -Dwrapper.lang.folder=../lang
Adding up the "capacity" items from jps shows that only my 6144m is being used for java heap. Where the heck is the rest of the memory being used? Stack memory? Native heap? I'm not even sure how to proceed.
If left to continue growing, it will consume all memory on the system and we will eventually see the system freeze up throwing swap space errors.
I have another process that is starting to grow. Currently at about 11g resident memory.
pmap 10746 > pmap_10746.txt
cat pmap_10746.txt | grep anon | cut -c18-25 | sort -h | uniq -c | sort -rn | less
Top 10 entries by count:
119 12K
112 1016K
56 4K
38 131072K
20 65532K
15 131068K
14 65536K
10 132K
8 65404K
7 128K
Top 10 entries by allocation size:
1 6291456K
1 205816K
1 155648K
38 131072K
15 131068K
1 108772K
1 71680K
14 65536K
20 65532K
1 65512K
And top 10 by total size:
Count Size Aggregate
1 6291456K 6291456K
38 131072K 4980736K
15 131068K 1966020K
20 65532K 1310640K
14 65536K 917504K
8 65404K 523232K
1 205816K 205816K
1 155648K 155648K
112 1016K 113792K
This seems to be telling me that because the Xmx and Xms are set to the same value, there is a single allocation of 6291456K for the java heap. Other allocations are NOT java heap memory. What are they? They are getting allocated in rather large chunks.
Expanding a bit more details on Peter's answer.
You can take a binary heap dump from within VisualVM (right click on the process in the left-hand side list, and then on heap dump - it'll appear right below shortly after). If you can't attach VisualVM to your JVM, you can also generate the dump with this:
jmap -dump:format=b,file=heap.hprof $PID
Then copy the file and open it with Visual VM (File, Load, select type heap dump, find the file.)
As Peter notes, a likely cause for the leak may be non collected DirectByteBuffers (e.g.: some instance of another class is not properly de-referencing buffers, so they are never GC'd).
To identify where are these references coming from, you can use Visual VM to examine the heap and find all instances of DirectByteByffer in the "Classes" tab. Find the DBB class, right click, go to instances view.
This will give you a list of instances. You can click on one and see who's keeping a reference each one:
Note the bottom pane, we have "referent" of type Cleaner and 2 "mybuffer". These would be properties in other classes that are referencing the instance of DirectByteBuffer we drilled into (it should be ok if you ignore the Cleaner and focus on the others).
From this point on you need to proceed based on your application.
Another equivalent way to get the list of DBB instances is from the OQL tab. This query:
select x from java.nio.DirectByteBuffer x
Gives us the same list as before. The benefit of using OQL is that you can execute more more complex queries. For example, this gets all the instances that are keeping a reference to a DirectByteBuffer:
select referrers(x) from java.nio.DirectByteBuffer x
What you can do is take a heap dump and look for object which are storing data off heap such as ByteBuffers. Those objects will appear small but are a proxy for larger off heap memory areas. See if you can determine why lots of those might be retained.
I'm using Ipython parallel in an optimisation algorithm that loops a large number of times. Parallelism is invoked in the loop using the map method of a LoadBalancedView (twice), a DirectView's dictionary interface and an invocation of a %px magic. I'm running the algorithm in an Ipython notebook.
I find that the memory consumed by both the kernel running the algorithm and one of the controllers increases steadily over time, limiting the number of loops I can execute (since available memory is limited).
Using heapy, I profiled memory use after a run of about 38 thousand loops:
Partition of a set of 98385344 objects. Total size = 18016840352 bytes.
Index Count % Size % Cumulative % Kind (class / dict of class)
0 5059553 5 9269101096 51 9269101096 51 IPython.parallel.client.client.Metadata
1 19795077 20 2915510312 16 12184611408 68 list
2 24030949 24 1641114880 9 13825726288 77 str
3 5062764 5 1424092704 8 15249818992 85 dict (no owner)
4 20238219 21 971434512 5 16221253504 90 datetime.datetime
5 401177 0 426782056 2 16648035560 92 scipy.optimize.optimize.OptimizeResult
6 3 0 402654816 2 17050690376 95 collections.defaultdict
7 4359721 4 323814160 2 17374504536 96 tuple
8 8166865 8 196004760 1 17570509296 98 numpy.float64
9 5488027 6 131712648 1 17702221944 98 int
<1582 more rows. Type e.g. '_.more' to view.>
You can see that about half the memory is used by IPython.parallel.client.client.Metadata instances. A good indicator that results from the map invocations are being cached is the 401177 OptimizeResult instances, the same number as the number of optimize invocations via lbview.map - I am not caching them in my code.
Is there a way I can control this memory usage on both the kernel and the Ipython parallel controller (who'se memory consumption is comparable to the kernel)?
Ipython parallel clients and controllers store past results and other metadata from past transactions.
The IPython.parallel.Client class provides a method for clearing this data:
Client.purge_everything()
documented here. There is also purge_results() and purge_local_results() methods that give you some control over what gets purged.
I am currently trying to connect my DIY DC77 clock to ntpd (using Ubuntu). I followed the instructions here: http://wiki.ubuntuusers.de/Systemzeit.
With ntpq I can see the DCF77 clock
~$ ntpq -c peers
remote refid st t when poll reach delay offset jitter
==============================================================================
+dispatch.mxjs.d 192.53.103.104 2 u 6 64 377 13.380 12.608 4.663
+main.macht.org 192.53.103.108 2 u 12 64 377 33.167 5.008 4.769
+alvo.fungus.at 91.195.238.4 3 u 15 64 377 16.949 7.454 28.075
-ns1.blazing.de 213.172.96.14 2 u - 64 377 10.072 14.170 2.335
*GENERIC(0) .DCFa. 0 l 31 64 377 0.000 5.362 4.621
LOCAL(0) .LOCL. 12 l 927 64 0 0.000 0.000 0.000
So far this looks OK. However I have two questions.
What exactly is the sign of the offset? Is .DCFa. ahead of the system clock or behind the system clock?
.DCFa. points to refclock-0 which is a DIY DCF77 clock emulating a Meinberg clock. It is connected to my Ubuntu Linux box with an FTDI usb-serial adapter running at 9600 7e2. I verified with a DSO that it emits the time with jitter significantly below 1ms. So I assume the jitter is introduced by either the FTDI adapter or the kernel. How would I find out and how can I reduce it?
Part One:
Positive offsets indicate time in the client is behind time on the server.
Negative offsets indicate that time in the client is ahead of time on the server.
I always remember this as "what needs to happen to my clock?"
+0.123 = Add 0.123 to me
-0.123 = Subtract 0.123 from me
Part Two:
Yes the USB serial converters add jitter. Get a real serial port:) You can also use setserial and tell it that the serial port needs to be low_latency. Just apt-get setserial.
Bonus Points:
Lose the unreferenced local clock entry. NO LOCL!!!!