I have been using CatBoost on CPU and got good results, but wanted to speed it up by using GPU. However, all metrics from GPU are worse than those from CPU. I did search around and found a suggestion that one could try to increase border_count to 255. I tried that but it did not help.
Some have claimed that GPU output would yield variations. While it might be true, I have found these variations are always worse than the CPU metrics.
Any pointer that I can try? It is a binary classification project. It has 1.2M rows for training and 120K rows for validattion. It has 218 features and 42 of them are categorical. The rest are floats or integers.
Related
I am just creating a normal image classifier for rock-paper-scissors.I am using my local gpu itself and it isnt a high end gpu. When i began training the model it kept giving the error:
ResourceExhaustedError: OOM when allocating tensor with shape.
I googled this error and they suggested I decrease my batch size which i did. It still did not solve anything however later I changed my image size to 50*50 initially it was 200*200 and then it started training with an accuracy of 99%.
Later i wanted to see if i could do it with 150*150 sized images as i found a tutorial on the official tensorflow channel on youtube I followed their exact code and it still did not work. I reduced the batch size, still no solution. Later I changed the no. of units in the dense layer initially it was 512 and then i decreased it to 200 and it worked fine but now the accuracy is pretty trash. I was just wondering is there anyway I could tune my model according to my gpu without affecting my accuracy? So I was just wondering how does the no. of units in the dense layer matter? It would really help me alot.
i=Input(shape=X_train[0].shape)
x=Conv2D(64,(3,3),padding='same',activation='relu')(i)
x=BatchNormalization()(x)
x=Conv2D(64,(3,3),padding='same',activation='relu')(x)
x=BatchNormalization()(x)
x=MaxPool2D((2,2))(x)
x=Conv2D(128,(3,3),padding='same',activation='relu')(x)
x=BatchNormalization()(x)
x=Conv2D(128,(3,3),padding='same',activation='relu')(x)
x=BatchNormalization()(x)
x=MaxPool2D((2,2))(x)
x=Flatten()(x)
x=Dropout(0.2)(x)
x=Dense(512,activation='relu')(x)
x=Dropout(0.2)(x)
x=Dense(3,activation='softmax')(x)
model=Model(i,x)
okay now when I run this with image size of 150*150 it throws that error,
if I change the size of the image to 50*50 and reduce batch size to 8 it works and gives me an accuracy of 99. but if I use 150*150 and reduce the no. of units in the dense layer to 200(random) it works fine but accuracy is very very bad.
I am using a low end nvidia geforce mx 230 gpu.
And my vram is 4 gigs
For 200x200 images the output of the last MaxPool has a shape of (50,50,128) which is then flattened and serves as the input of the Dense layer giving you in total of 50*50*128*512=163840000 parameters. This is a lot.
To reduce the amount of parameters you can do one of the following:
- reduce the amount of filters in the last Conv2D layer
- do a MaxPool of more than 2x2
- reduce the size of the Dense layer
- reduce the size of the input images.
You have already tried the two latter options. You will only find out by trial and error which method ultimately gives you the best accuracy. You were already at 99%, which is good.
If you want a platform with more VRAM available, you can use Google Colab https://colab.research.google.com/
I'm testing and comparing GPU speed up with different numbers of work-items (no work-groups). The kernel I'm using is a very simple but long operation. When I test with multiple work-items, I use a barrier function and split the work in smaller chunks to get the same result as with just one work-item. I measure the kernel execution time using cl_event and the results are the following:
1 work-item: 35735 ms
2 work-items: 11822 ms (3 times faster than with 1 work-item)
10 work-items: 2380 ms (5 times faster than with 2 work-items)
100 work-items: 239 ms (10 times faster than with 10 work-items)
200 work-items: 122 ms (2 times faster than with 100 work-items)
CPU takes about 580 ms on average to do the same operation.
The only result I don't understand and can't explain is the one with 2 work items. I would expect the speed up to be about 2 times faster compared to the result with just one work item, so why is it 3?
I'm trying to make sense of these numbers by looking at how these work-items were distributed on processing elements. I'm assuming if I have just one kernel, only one compute unit (or multiprocessor) will be activated and the work items distributed on all processing elements (or CUDA cores) of that compute unit. What I'm also not sure about is whether a processing element can process multiple work-items at the same time, or is it just one work-item per processing element?
CL_DEVICE_MAX_WORK_ITEM_SIZES are 1024 / 1024 / 64 and CL_DEVICE_MAX_WORK_GROUP_SIZE 1024. Since I'm using just one dimension, does that mean I can have 1024 work-items running at the same time per processing element or per compute unit? When I tried with 1000 work-items, the result was a smaller number so I figured not all of them got executed, but why would that be?
My GPU info: Nvidia GeForce GT 525M, 96 CUDA cores (2 compute units, 48 CUDA cores per unit)
The only result I don't understand and can't explain is the one with 2
work items. I would expect the speed up to be about 2 times faster
compared to the result with just one work item, so why is it 3?
The exact reasons will probably be hard to pin down, but here are a few suggestions:
GPUs aren't optimised at all for small numbers of work items. Benchmarking that end of the scale isn't especially useful.
35 seconds is a very long time for a GPU. Your GPU probably has other things to do, so your work-item is probably being interrupted many times, with its context saved and resumed every time.
It will depend very much on your algorithm. For example, if your kernel uses local memory, or a work-size dependent amount of private memory, it might "spill" to global memory, which will slow things down.
Depending on your kernel's memory access patterns, you might be running into the effects of read/write coalescing. More work items means fewer memory accesses.
What I'm also not sure about is whether a processing element can process multiple work-items at the same time, or is it just one work-item per processing element?
Most GPU hardware supports a form of SMT to hide memory access latency. So a compute core will have up to some fixed number of work items in-flight at a time, and if one of them is blocked waiting for a memory access or barrier, the core will continue executing commands on another work item. Note that the maximum number of simultaneous threads can be further limited if your kernel uses a lot of local memory or private registers, because those are a finite resource shared by all cores in a compute unit.
Work-groups will normally run on only one compute unit at a time, because local memory and barriers don't work across units. So you don't want to make your groups too large.
One final note: compute hardware tends to be grouped in powers of 2, so it's usually a good idea to make your work group sizes a multiple of e.g. 16 or 64. 1000 is neither, which usually means some cores will be doing nothing.
When I tried with 1000 work-items, the result was a smaller number so I figured not all of them got executed, but why would that be?
Please be more precise in this question, it's not clear what you're asking.
I am running a Python program that calls H2O for deep learning (training and testing). The program runs in a loop of 20 iterations and in each loop calls H2ODeepLearningEstimator() 4 times and associated predict() and model_performance(). I am doing h2o.remove_all() and cleaning up all data-related Python objects after each iteration.
Data size: training set 80,000 with 122 features (all float) with 20% for validation (10-fold CV). test set 20,000. Doing binary classification.
Machine 1: Windows 7, 4 core, Xeon, each core 3.5GHz, Memory 32 GB
Takes about 24 hours to complete
Machine 2: CentOS 7, 20 core, Xeon, each core 2.0GHz, Memory 128 GB
Takes about 17 hours to complete
I am using h2o.init(nthreads=-1, max_mem_size = 96)
So, the speed-up is not that much.
My questions:
1) Is the speed-up typical?
2) What can I do to achieve substantial speed-up?
2.1) Will adding more cores help?
2.2) Are there any H2O configuration or tips that I am missing?
Thanks very much.
- Mohammad,
Graduate student
If the training time is the main effort, and you have enough memory, then the speed up will be proportional to cores times core-speed. So, you might have expected a 40/14 = 2.85 speed-up (i.e. your 24hrs coming down to the 8-10 hour range).
There is a typo in your h2o.init(): 96 should be "96g". However, I think that was a typo when writing the question, as h2o.init() would return an error message. (And H2O would fail to start if you'd tried "96", with the quotes but without the "g".)
You didn't show your h2o.deeplearning() command, but I am guessing you are using early stopping. And that can be unpredictable. So, what might have happened is that your first 24hr run did, say, 1000 epochs, but your second 17hr run did 2000 epochs. (1000 vs. 2000 would be quite an extreme difference, though.)
It might be that you are spending too much time scoring. If you've not touched the defaults, this is unlikely. But you could experiment with train_samples_per_iteration (e.g. set it to 10 times the number of your training rows).
What can I do to achieve substantial speed-up?
Stop using cross-validation. That might be a bit controversial, but personally I think 80,000 training rows is going to be enough to do an 80%/10%/10% split into train/valid/test. That will be 5-10 times quicker.
If it is for a paper, and you want to show more confidence in the results, once you have your final model, and you've checked that test score is close to valid score, then rebuild it a couple of times using a different seed for the 80/10/10 split, and confirm you end up with the same metrics. (*)
*: By the way, take a look at the score for each of the 10 cv models you've already made; if they are fairly close to each other, then this approach should work well. If they are all over the place, you might have to re-consider the train/valid/test splits - or just think about what it is in your data that might be causing that sensitivity.
I'm using the new TensorFlow profiler to profile memory usage in my neural net, which I'm running on a Titan X GPU with 12GB RAM. Here's some example output when I profile my main training loop:
==================Model Analysis Report======================
node name | requested bytes | ...
Conv2DBackpropInput 10227.69MB (100.00%, 35.34%), ...
Conv2D 9679.95MB (64.66%, 33.45%), ...
Conv2DBackpropFilter 8073.89MB (31.21%, 27.90%), ...
Obviously this adds up to more than 12GB, so some of these matrices must be in main memory while others are on the GPU. I'd love to see a detailed breakdown of what variables are where at a given step. Is it possible to get more detailed information on where various parameters are stored (main or GPU memory), either with the profiler or otherwise?
"Requested bytes" shows a sum over all memory allocations, but that memory can be allocated and de-allocated. So just because "requested bytes" exceeds GPU RAM doesn't necessarily mean that memory is being transferred to CPU.
In particular, for a feedforward neural network, TF will normally keep around the forward activations, to make backprop efficient, but doesn't need to keep the intermediate backprop activations, i.e. dL/dh at each layer, so it can just throw away these intermediates after it's done with these. So I think in this case what you care about is the memory used by Conv2D, which is less than 12 GB.
You can also use the timeline to verify that total memory usage never exceeds 12 GB.
simple problem from Wilkinson and Allen's Parallel Programming: Techniques and Applications Using Networked Workstations and Parallel Computers. Working through the exercises at the end of the first chapter and want to make sure that I'm on the right track. The full question is:
1-11 A multiprocessor consists of 10 processors, each capable of a peak execution rate of 200 MFLOPs (millions of floating point operations per second). What is the performance of the system as measured in MFLOPs when 10% of the code is sequential and 90% is parallelizable?
I assume the question wants me to find the number of operations per second of a serial processor which would take the same amount of time to run the program as the multiprocessor.
I think I'm right in thinking that 10% of the program is run at 200 MFLOPs, and 90% is run at 2,000 MFLOPs, and that I can average these speeds to find the performance of the multiprocessor in MFLOPs:
1/10 * 200 + 9/10 * 2000 = 1820 MFLOPs
So when running a program which is 10% serial and 90% parallelizable the performance of the multiprocessor is 1820 MFLOPs.
Is my approach correct?
ps: I understand that this isn't exactly how this would work in reality because it's far more complex, but I would like to know if I'm grasping the concepts.
Your calculation would be fine if 90% of the time, all 10 processors were fully utilized, and 10% of the time, just 1 processor was in use. However, I don't think that is a reasonable interpretation of the problem. I think it is more reasonable to assume that if a single processor were used, 10% of its computations would be on the sequential part, and 90% of its computations would be on the parallelizable part.
One possibility is that the sequential part and parallelizable parts can be run in parallel. Then one processor could run the sequential part, and the other 9 processors could do the parallelizable part. All processors would be fully used, and the result would be 2000 MFLOPS.
Another possibility is that the sequential part needs to be run first, and then the parallelizable part. If a single processor needed 1 hour to do the first part, and 9 hours to do the second, then it would take 10 processors 1 + 0.9 = 1.9 hours total, for an average of about (1*200 + 0.9*2000)/1.9 ~ 1053 MFLOPS.