python - Calculate the Output size in Convolution layer

Question

What will be the output size, if the input to convolution layer of neural network is an image of size 128X128X3 and 40 filters of size 5X5 are applied to it?

Answer 1

you can use this formula [(W−K+2P)/S]+1.

• W is the input volume - in your case 128
• K is the Kernel size - in your case 5
• P is the padding - in your case 0 i believe
• S is the stride - which you have not provided.

So, we input into the formula:

Output_Shape = (128-5+0)/1+1

Output_Shape = (124,124,40)

NOTE: Stride defaults to 1 if not provided and the 40 in (124, 124, 40) is the number of filters provided by the user.

Answer 2

You can find it in two ways: simple method: inpput_size - (filter_size - 1) But the second method is the standard to find the output size.

Second method: (((W - K + 2P)/S) + 1)
        Here W = Input size
        K = Filter size
        S = Stride
        P = Padding 

Answer 3

Best resource:

Dimension calculation

Parameters in CNN

Answer 4

Formula : n[i]=(n[i-1]−f[i]+2p[i])/s[i]+1

where,

n[i-1]=128

f[i]=5

p[i]=0

s[i]=1

so,

n[i]=(128-5+0)/1+1 =124

so the size of the output layer is: 124x124x40 Where '40' is the number of filters

Answer 5

(124*124*3)*40 = 1845120 width = 124 height = 124 depth = 3 no. of filters = 40 stride = 1 padding = 0

Answer 6

Let me start simple; since you have square matrices for both input and filter let me get one dimension. Then you can apply the same for other dimension(s). Imagine your are building fences between trees, if there are N trees, you have to build N-1 fences. Now apply that analogy to convolution layers.

Your output size will be: input size - filter size + 1

Because your filter can only have n-1 steps as fences I mentioned.

Let's calculate your output with that idea. 128 - 5 + 1 = 124 Same for other dimension too. So now you have a 124 x 124 image.

That is for one filter.

If you apply this 40 times you will have another dimension: 124 x 124 x 40

Here is a great guide if you want to know more about advanced convolution arithmetic: https://arxiv.org/pdf/1603.07285.pdf