Deep Learning · TEE · section

VGG16: small filters in a regular sequence

CNN architectures and transfer learning

Follow the five blocks and their output sizes. The 16 counts learned layers, not pooling or flattening. Tap the diagram to enlarge.

Definition. VGG16 is a convolutional architecture with 13 convolutional layers and three dense layers. Its standard convolution blocks contain 2, 2, 3, 3 and 3 convolutions. Pooling, activations and flattening are not counted in the name's 16 weight layers.

Each convolution uses a 3 × 3 kernel, stride 1 and padding 1, preserving height and width. Each block ends with 2 × 2 max pooling of stride 2. The final classifier is 4,096 → 4,096 → 1,000 for the original 1,000-class task.

Block	Convolutions	Channels after its convolutions	Size after its pooling
1	Two	64	112 × 112 × 64
2	Two	128	56 × 56 × 128
3	Three	256	28 × 28 × 256
4	Three	512	14 × 14 × 512
5	Three	512	7 × 7 × 512

This table starts with 224 × 224 × 3. The flattened length is 7 × 7 × 512 = 25,088. Flattening has zero learned parameters. Stacking small filters provides repeated nonlinear transformations while progressively combining spatial information. A limitation is the large dense classifier and its memory cost.

Question (practice, 3 marks). VGG16's first block contains two same-padded 3 × 3 convolutions with 64 output filters each, followed by 2 × 2 max pooling with stride 2. Input is 224 × 224 × 3. Find both convolution parameter counts including bias and the pooled output shape.

Given

Conv1: 3 input channels, 64 output channels. Conv2: 64 input channels, 64 output channels. Both have kernel 3, stride 1 and padding 1. Pooling has kernel 2, stride 2 and no padding.

To find

Trainable parameters in each convolution and the output after pooling.

Formula — which one and why

$P_{\mathrm{conv}}=(K^2C_{\mathrm{in}}+1)C_{\mathrm{out}}$ LaTeX: P_{\mathrm{conv}}=(K^2C_{\mathrm{in}}+1)C_{\mathrm{out}}

Use each layer's own input depth; the second layer receives the first layer's 64 maps. Use the spatial-output formula for pooling. Max pooling introduces no trainable parameters.

Solution — step by step

Step 1: first convolution. Each filter has 3 × 3 × 3 = 27 weights and one bias.

$P_1=(27+1)(64)=1792$ LaTeX: P_1=(27+1)(64)=1792

Step 2: second convolution. Each filter now spans 64 channels, giving 3 × 3 × 64 = 576 weights.

$P_2=(576+1)(64)=36928$ LaTeX: P_2=(576+1)(64)=36928

Step 3: add parameters.

$P_{\mathrm{block}}=1792+36928=38720$ LaTeX: P_{\mathrm{block}}=1792+36928=38720

Step 4: calculate pooled side length. The same-padded convolutions retained side length 224.

$\frac{224-2}{2}+1=111+1=112$ LaTeX: \frac{224-2}{2}+1=111+1=112

Final answer: 1,792 and 36,928 parameters; total = 38,720; pooled shape = 112 × 112 × 64.

Exam convention: use the architecture explicitly supplied in a calculation. For the standard VGG16 architecture, remember 2–2–3–3–3 convolutions and 2 × 2 pooling; 25,088 is the flattened feature length.