# Global average pooling

Subject: Deep Learning

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/#tee/formulas/47

Canonical: https://deep-learning-study-room.vercel.app/read/dl/tee/formulas/47/

Global average pooling takes one mean per complete spatial channel. The worked example compares the resulting dense classifier with a classifier after flattening.



$$
g_c=\frac{1}{HW}\sum_{i=1}^{H}\sum_{j=1}^{W}a_{i,j,c}
$$



## Global average pooling: one average for each channel

![Average each channel separately: these are the exact values used in the worked example.](https://deep-learning-study-room.vercel.app/assets/figures/u4_gap.svg)

Diagram: Average each channel separately: these are the exact values used in the worked example.. Labels: u4 gap | Global average pooling keeps channels separate | Channel A | 1 | 3 | 5 | 7 | → | 4 | 16 ÷ 4 | Channel B | 2 | 4 | 6 | 8 | → | 5 | 20 ÷ 4 | 2 × 2 × 2 input → two channel means → [4, 5]

Average each channel separately: these are the exact values used in the worked example. Tap the diagram to enlarge.

**Definition.** Global average pooling, or GAP, replaces every complete spatial feature map by its mean. An H × W × C tensor becomes C numbers. The channels remain separate; we do not average all channels into one number.



$$
g_c=\frac{1}{HW}\sum_{i=1}^{H}\sum_{j=1}^{W}a_{i,j,c}
$$



The letter c selects a channel; i and j select its row and column; a is the value at that location. Add all H × W entries in that channel and divide by H × W. GAP has no learned weights or biases. A classifier after GAP can still have trainable parameters. In GoogLeNet, 7 × 7 × 1,024 becomes 1 × 1 × 1,024 before the final 1,000-class linear classifier.

**Question (practice, 3 marks). A 2 × 2 × 2 tensor has channel A =  $\begin{bmatrix}1 & 3\\5 & 7\end{bmatrix}$  and channel B =  $\begin{bmatrix}2 & 4\\6 & 8\end{bmatrix}$ . Apply GAP, then find the parameters of a three-output dense classifier receiving the result, including one bias per output.**

### Given

H = 2, W = 2, C = 2. Each of the two channels contains four values. The classifier has three outputs.

### To find

The GAP result and dense-head parameter count.

### Formula — which one and why

Use one mean per channel. The dense layer receives two numbers, so each of its three outputs needs two weights and one bias.



$$
P_{\mathrm{dense}}=(n_{\mathrm{in}}+1)n_{\mathrm{out}}
$$



### Solution — step by step

**Step 1: add channel A.** 1 + 3 + 5 + 7 = 16.

**Step 2: average channel A.** 16 ÷ 4 = 4.

**Step 3: add channel B.** 2 + 4 + 6 + 8 = 20.

**Step 4: average channel B.** 20 ÷ 4 = 5. Thus GAP outputs **[4,5]**.

**Step 5: count the classifier's weights.** 2 inputs × 3 outputs = 6 weights.

**Step 6: add three biases.** 6 + 3 = **9 parameters**. GAP itself contributes zero.

**Final answer:** GAP = [4,5]; GAP shape = 1 × 1 × 2; classifier parameters = 9. If we flattened all eight inputs instead, that head would require (8 + 1) × 3 = 27 parameters.

**Yaad rakho:** ek channel ka poora average; channels ko aapas mein mat milao.