# R26 Q6a Subject: Deep Learning Exam: MT2 Reader: https://deep-learning-study-room.vercel.app/#mt2/read/12/mt2-c12-s7 Canonical: https://deep-learning-study-room.vercel.app/read/dl/mt2/section/mt2-c12-s7/ ## Question **Question (10 marks).** For the binary CNN below, calculate output dimensions and trainable parameters after every layer. Input 256 × 256 × 3. → Conv K = 5, filters = 32, stride = 1, padding = 0. → Pool K = 2, stride = 2, padding = 0. → Conv K = 5, filters = 64, stride = 1, padding = 0. → Pool K = 2, stride = 2, padding = 0. → Conv K = 5, filters = 128, stride = 1, padding = 1. → Pool K = 2, stride = 2, padding = 0. → Flatten. → Dense 84 units. → Dense 32 units. → Dense 1 units. The printed output size is unspecified; assume one sigmoid output. Unspecified convolution padding is taken as valid. ## Answer — step by step ### Given Input 256 × 256 × 3. Conv K = 5, filters = 32, stride = 1, padding = 0. Pool K = 2, stride = 2, padding = 0. Conv K = 5, filters = 64, stride = 1, padding = 0. Pool K = 2, stride = 2, padding = 0. Conv K = 5, filters = 128, stride = 1, padding = 1. Pool K = 2, stride = 2, padding = 0. Dense 84 units. Dense 32 units. Dense 1 units. assume one sigmoid output. ### To find Find every layer shape, its parameter count and the network total. ### Formula — which one and why **Complete CNN calculation.** Carry each output shape into the next layer. Convolution changes spatial size and channel count; pooling changes spatial size; flattening converts all activations into dense-layer inputs. H,W,C = feature-map height, width and channels; K = kernel size; S = stride; P\_pad = padding. One bias is assumed for each learned filter or dense neuron unless specified otherwise. $$ N_{\mathrm{out}}=\left\lfloor\frac{N+2P_{\mathrm{pad}}-K}{S}\right\rfloor+1 $$ $$ P_{\mathrm{conv}}=(K^2C_{\mathrm{in}}+1)C_{\mathrm{out}} $$ $$ n_{\mathrm{flat}}=HWC $$ $$ P_{\mathrm{dense}}=(n_{\mathrm{in}}+1)n_{\mathrm{out}} $$ $$ P_{\mathrm{pool}}=P_{\mathrm{flatten}}=0 $$ **Yaad rakhne ka cue:** Ek layer ka output, next layer ka input; shape aur parameters alag count. ### Solution — step by step Use output size $\left\lfloor\frac{N+2P-K}{S}\right\rfloor+1$ , Conv weights K² × Cin × Cout and biases Cout, Dense weights Nin × Nout and biases Nout. Assume a bias per learned filter/unit, full input-channel connectivity and no trainable pooling parameters. ReLU and softmax activations themselves add no trainable parameters. ### Layer-by-layer calculation Carry the complete output shape of one layer into the next. Spatial size determines the number of activations; it does not multiply the number of shared convolution parameters. ### Layer 1 — convolution Layer 1: what changes and what is learned Input shape**256 × 256 × 3**height × width × channels → Convolution**32 filters · 5 × 5 · S = 1 · P = 0** → Output shape**252 × 252 × 32**height × width × channels Weights**2,400** + Biases**32** = Trainable parameters**2,432** Spatial positions share the filter weights. Output depth equals the number of filters. Input: **256 × 256 × 3**. Use 32 filters of spatial size 5 × 5, stride 1, and padding 0 on each side. **1. Calculate height and width.** $$ N_{\mathrm{out}}=\left\lfloor\frac{256+2(0)-5}{1}\right\rfloor+1=252 $$ **2. Determine output channels.** Each filter produces one feature map. Therefore the output is **252 × 252 × 32**. **3. Count trainable values.** Each filter receives all 3 input channels. Its weights are reused at every spatial position. $$ N_{\mathrm{weights}}=5\times5\times3\times32=2{,}400 $$ $$ N_{\mathrm{biases}}=32 $$ $$ P_1=2{,}400+32=2{,}432 $$ ### Layer 2 — pooling Layer 2: what changes and what is learned Input shape**252 × 252 × 32**height × width × channels → Pooling**2 × 2 window · S = 2** → Output shape**126 × 126 × 32**height × width × channels Weights**0** + Biases**0** = Trainable parameters**0** Spatial dimensions change; the channel count stays the same. Pooling learns no weights. Input: **252 × 252 × 32**. Use a 2 × 2 window, stride 2, and zero padding. $$ N_{\mathrm{out}}=\left\lfloor\frac{252-2}{2}\right\rfloor+1=126 $$ Pooling acts independently on each channel. The output is **126 × 126 × 32**. **Trainable parameters: 0.** Max/average pooling has no learned filter weights or biases. ### Layer 3 — convolution Layer 3: what changes and what is learned Input shape**126 × 126 × 32**height × width × channels → Convolution**64 filters · 5 × 5 · S = 1 · P = 0** → Output shape**122 × 122 × 64**height × width × channels Weights**51,200** + Biases**64** = Trainable parameters**51,264** Spatial positions share the filter weights. Output depth equals the number of filters. Input: **126 × 126 × 32**. Use 64 filters of spatial size 5 × 5, stride 1, and padding 0 on each side. **1. Calculate height and width.** $$ N_{\mathrm{out}}=\left\lfloor\frac{126+2(0)-5}{1}\right\rfloor+1=122 $$ **2. Determine output channels.** Each filter produces one feature map. Therefore the output is **122 × 122 × 64**. **3. Count trainable values.** Each filter receives all 32 input channels. Its weights are reused at every spatial position. $$ N_{\mathrm{weights}}=5\times5\times32\times64=51{,}200 $$ $$ N_{\mathrm{biases}}=64 $$ $$ P_3=51{,}200+64=51{,}264 $$ ### Layer 4 — pooling Layer 4: what changes and what is learned Input shape**122 × 122 × 64**height × width × channels → Pooling**2 × 2 window · S = 2** → Output shape**61 × 61 × 64**height × width × channels Weights**0** + Biases**0** = Trainable parameters**0** Spatial dimensions change; the channel count stays the same. Pooling learns no weights. Input: **122 × 122 × 64**. Use a 2 × 2 window, stride 2, and zero padding. $$ N_{\mathrm{out}}=\left\lfloor\frac{122-2}{2}\right\rfloor+1=61 $$ Pooling acts independently on each channel. The output is **61 × 61 × 64**. **Trainable parameters: 0.** Max/average pooling has no learned filter weights or biases. ### Layer 5 — convolution Layer 5: what changes and what is learned Input shape**61 × 61 × 64**height × width × channels → Convolution**128 filters · 5 × 5 · S = 1 · P = 1** → Output shape**59 × 59 × 128**height × width × channels Weights**204,800** + Biases**128** = Trainable parameters**204,928** Spatial positions share the filter weights. Output depth equals the number of filters. Input: **61 × 61 × 64**. Use 128 filters of spatial size 5 × 5, stride 1, and padding 1 on each side. **1. Calculate height and width.** $$ N_{\mathrm{out}}=\left\lfloor\frac{61+2(1)-5}{1}\right\rfloor+1=59 $$ **2. Determine output channels.** Each filter produces one feature map. Therefore the output is **59 × 59 × 128**. **3. Count trainable values.** Each filter receives all 64 input channels. Its weights are reused at every spatial position. $$ N_{\mathrm{weights}}=5\times5\times64\times128=204{,}800 $$ $$ N_{\mathrm{biases}}=128 $$ $$ P_5=204{,}800+128=204{,}928 $$ ### Layer 6 — pooling Layer 6: what changes and what is learned Input shape**59 × 59 × 128**height × width × channels → Pooling**2 × 2 window · S = 2** → Output shape**29 × 29 × 128**height × width × channels Weights**0** + Biases**0** = Trainable parameters**0** Spatial dimensions change; the channel count stays the same. Pooling learns no weights. Input: **59 × 59 × 128**. Use a 2 × 2 window, stride 2, and zero padding. $$ N_{\mathrm{out}}=\left\lfloor\frac{59-2}{2}\right\rfloor+1=29 $$ Pooling acts independently on each channel. The output is **29 × 29 × 128**. **Trainable parameters: 0.** Max/average pooling has no learned filter weights or biases. The division gives 28.5; take its floor before adding 1. A partial window outside the unpadded input is not included. ### Layer 7 — flatten Layer 7: what changes and what is learned Input shape**29 × 29 × 128**height × width × channels → Flatten**Rearrange activations** → Output shape**107,648**output features Weights**0** + Biases**0** = Trainable parameters**0** The same activation values are rearranged into one vector. No parameter is learned. Input: **29 × 29 × 128**. Rearrange all activation values into one vector. $$ n_{\mathrm{flat}}=29\times29\times128=107{,}648 $$ Output: a vector with **107,648 entries**. Flattening changes the arrangement, so it adds **0 trainable parameters**. ### Layer 8 — fully connected layer Layer 8: what changes and what is learned Input shape**107,648**input features → Dense layer**84 neurons** → Output shape**84**output features Weights**9,042,432** + Biases**84** = Trainable parameters**9,042,516** Every output neuron receives every input feature, with one bias per output neuron. Input: **107,648 features**. Output: **84 neurons**. Every output neuron connects to every input feature and has one bias. $$ N_{\mathrm{weights}}=107{,}648\times84=9{,}042{,}432 $$ $$ N_{\mathrm{biases}}=84 $$ $$ P_8=9{,}042{,}432+84=9{,}042{,}516 $$ ### Layer 9 — fully connected layer Layer 9: what changes and what is learned Input shape**84**input features → Dense layer**32 neurons** → Output shape**32**output features Weights**2,688** + Biases**32** = Trainable parameters**2,720** Every output neuron receives every input feature, with one bias per output neuron. Input: **84 features**. Output: **32 neurons**. Every output neuron connects to every input feature and has one bias. $$ N_{\mathrm{weights}}=84\times32=2{,}688 $$ $$ N_{\mathrm{biases}}=32 $$ $$ P_9=2{,}688+32=2{,}720 $$ ### Layer 10 — fully connected layer Layer 10: what changes and what is learned Input shape**32**input features → Dense layer**1 neurons** → Output shape**1**output features Weights**32** + Biases**1** = Trainable parameters**33** Every output neuron receives every input feature, with one bias per output neuron. Input: **32 features**. Output: **1 neurons**. Every output neuron connects to every input feature and has one bias. $$ N_{\mathrm{weights}}=32\times1=32 $$ $$ N_{\mathrm{biases}}=1 $$ $$ P_10=32+1=33 $$ ### Summary of the complete network | Layer | Spatial or flatten calculation | Output | Weights | Biases | Total | | --- | --- | --- | --- | --- | --- | | 1 conv | $\left\lfloor\frac{256+2 \times 0-5}{1}\right\rfloor+1$ | 252 x 252 x 32 | 2,400 | 32 | 2,432 | | 2 pool | $\left\lfloor\frac{252-2}{2}\right\rfloor+1$ | 126 x 126 x 32 | 0 | 0 | 0 | | 3 conv | $\left\lfloor\frac{126+2 \times 0-5}{1}\right\rfloor+1$ | 122 x 122 x 64 | 51,200 | 64 | 51,264 | | 4 pool | $\left\lfloor\frac{122-2}{2}\right\rfloor+1$ | 61 x 61 x 64 | 0 | 0 | 0 | | 5 conv | $\left\lfloor\frac{61+2 \times 1-5}{1}\right\rfloor+1$ | 59 x 59 x 128 | 204,800 | 128 | 204,928 | | 6 pool | $\left\lfloor\frac{59-2}{2}\right\rfloor+1$ | 29 x 29 x 128 | 0 | 0 | 0 | | 7 flat | 29 × 29 × 128 | 107648 | 0 | 0 | 0 | | 8 dense | 107648 → 84 | 84 | 9,042,432 | 84 | 9,042,516 | | 9 dense | 84 → 32 | 32 | 2,688 | 32 | 2,720 | | 10 dense | 32 → 1 | 1 | 32 | 1 | 33 | **Parameter substitutions, in layer order:** Layer 1: (5 x 5 x 3 + 1) x 32 = 2,432. Layer 3: (5 x 5 x 32 + 1) x 64 = 51,264. Layer 5: (5 x 5 x 64 + 1) x 128 = 204,928. Layer 8: (107648 + 1) x 84 = 9,042,516. Layer 9: (84 + 1) x 32 = 2,720. Layer 10: (32 + 1) x 1 = 33. **Total:** 2,432 + 51,264 + 204,928 + 9,042,516 + 2,720 + 33 = **9,303,893 trainable parameters**. Pooling changes activation size; it does not itself add learned weights. Always floor noninteger spatial dimensions. **Correction:** Conv 3 has (5 × 5 × 64+1) × 128 = **204,928** parameters, not 73,856 as printed in the key. Correct total is **9,303,893**. If binary classification is instead implemented with two softmax outputs, the last dense layer has(32+1) × 2 = 66 parameters and the total becomes 9,303,926; explicitly state the output convention.