I'd like to be able to estimate whether a proposed model is small enough to be trained on a GPU with a given amount of memory

If I have a simple CNN architecture like this:

  • Input: 50x50x3
  • C1: 32 3x3 kernels, with padding (I guess in reality theyre actually 3x3x3 given the input depth?)
  • P1: 2x2 with stride 2
  • C2: 64 3x3 kernels, with padding
  • P2: 2x2 with stride 2
  • FC: 500 neurons
  • Output: softmax 10 classes
  • Mini batch size of 64

Assuming 32bit floating point values, how do you calculate the memory cost of each layer of the network during training? and then the total memory required to train such a model?


3 Answers 3


I will assume by C1, C2, etc, you mean convolutional layers, and by P1 ,P2 you mean pooling layers, and FC means fully connected layers.

We can calculate the memory required for a forward pass like this:

One image

If you're working with float32 values, then following the link provided above by @Alexandru Burlacu you have:

Input: 50x50x3 = 7,500 = 7.5K

C1: 50x50x32 = 80,000 = 80K

P1: 25x25x32 = 20,000 = 20K

C2: 25x25x64 = 40,000 = 40K

P2: 12x12x64 = 9,216 = 9.2K <- This is a problem (and my approximation is a very hand-wavy guess here). Instead of working with 50, 25, '12.5', it would make more sense to work with multiples of 32. I've heard working with multiples of 32 is also more efficient from a memory standpoint. The reason this is a bad idea is 2x2 pooling doesn't divide the space properly, as far as I can tell. Feel free to correct me if I'm wrong.

FC: 1x500 = 500 = 0.5K

Output: 1 x 10 = 10 = 0.01K (next to nothing)

Total memory: 7.5K + 80K + 20K + 40K + 0.5K = 157.2K * 4 bytes = 628.8 KB

That's for one image.


If you're working with a minibatch size of 64, then you're reading 64 of these into memory at once and performing the operations all together, scaling everything up like this:

Input: 64x50x50x3 = 480,000 = 480K = 0.48M

C1: 64x50x50x32 = 5,120,000 = 5.12M

P1: 64x25x25x32 = 1,280,000 = 1.28M

C2: 64x25x25x64 = 2,560,000 = 2.56M

P2: 64x12x12x64 = 589,824 = 590K = 0.59M

FC: 64x500 = 32,000 = 32K = 0.032M

Output: 1x10x64 = 640 = 0.64K = 0.00064M (we don't care, this is tiny)

Total memory: 10M x 4 bytes ~ 40MB (I'm saying approximate because the website also says an approximate value)

EDIT: I misread the website, sorry.

According to the website, a backward pass requires about triple this, because of the need to store:

  • the activations and associated gradients for each neuron - these are of equal size;

  • the gradients of the weights (parameters) which are the same size as the parameters;

  • the value of the momentum, if you're using it;

  • some kind of miscellaneous memory (I don't understand this part)

  • $\begingroup$ Do we need to also consider the memory required for weights/parameters? For example, for your layer C1 above you have 50x50x32 but that would just be the values for the layer activation maps, what about the learned values (32x3x3) within the kernel itself? $\endgroup$
    – Simon
    Commented May 27, 2017 at 13:10
  • $\begingroup$ Yes, you do, according to cs.stanford.edu/~quocle/tutorial2.pdf. The backprop algorithm using SGD needs the previous values of the weights. But with max pooling (are you doing max pooling?), you don't need to keep track of the gradient with respect to each connection. You need to keep track of which connection the max value came from, and only compute the gradient for that branch. $\endgroup$ Commented May 27, 2017 at 16:26
  • $\begingroup$ @StatsSorceress, thanks for this answer but one question appeared in my mind, what happen if convolution layer size is 11x11 instead of 3x3? and what happen if a ReLU followed by a convolutional layer ? $\endgroup$ Commented Feb 12, 2018 at 10:24
  • $\begingroup$ Is that the total memory used for the network? Or, that is the considering only the outputs of each layer and not the results of the internal steps of the computations? $\endgroup$ Commented Aug 10, 2021 at 14:28
  • $\begingroup$ @EduardoReis I don't think this takes any internal steps into account, no. That was partially discussed in my comment above. I don't know anything about how much memory that would require, sorry. $\endgroup$ Commented Aug 10, 2021 at 15:38

Maybe this link will give you an explanation on how to compute the memory usage of an arbitrary neural network. Bellow in the link is explained the memory usage of the VGGNet model. Click here and scroll down a bit))


While training a convNet, total memory required include following:

  • Memory for parameters
  • Memory for the output of intermediate layers
  • Memory for the gradient of each parameter
  • Extra memory needed if you are using optimizer like Momentum, RMSprop, Adams etc
  • Miscellaneous memory for implementation

A good rough approximation is number of parameters x 3 x 4(if you are using 32-bit float) bytes

Well, now this is how you calculate the number of parameters:

  • Conv layer: (kernel width x kernel height) x number of channels x depth + depth (add depth only if bias is there)
  • FC layer: numb of input*numb of output + output (output is added to include the number of bias)
  • Max pool layer: no parameter

Now just sum the number of all the parameters and use the formula I mentioned.


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