# The faculty's 480-channel bottleneck numerical

Subject: Deep Learning

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/#tee/read/8/tee-c8-s14

Canonical: https://deep-learning-study-room.vercel.app/read/dl/tee/section/tee-c8-s14/

![Both paths start from 14 × 14 × 480 and end at 14 × 14 × 48. The bottleneck cost includes both layers.](https://deep-learning-study-room.vercel.app/assets/figures/u4_bottleneck_480.svg)

Diagram: Both paths start from 14 × 14 × 480 and end at 14 × 14 × 48. The bottleneck cost includes both layers.. Labels: u4 bottleneck 480 | Same output shape, different computation | Input | 14 × 14 × 480 | → | Direct 5 × 5 | 48 filters | → | Output | 14 × 14 × 48 | 112,896,000 MACs | 1 × 1 reduction | 480 → 16 channels | → | Reduced 5 × 5 | 16 → 48 channels | → | Output | 14 × 14 × 48 | 1,505,280 + 3,763,200 = 5,268,480 MACs | The lower path also starts from 14 × 14 × 480. Count both layers.

Both paths start from 14 × 14 × 480 and end at 14 × 14 × 48. The bottleneck cost includes both layers. Tap the diagram to enlarge.

**Question (practice, 3 marks). Compare a direct same-padded 5 × 5 convolution with 48 filters on a 14 × 14 × 480 input against a 1 × 1 reduction to 16 channels followed by the same 5 × 5, 48-filter layer. Use stride 1. Count MACs, ignore bias additions, and find the percentage reduction.**

### Given

Both paths produce 14 × 14 × 48. Direct spatial filters span 480 channels; reduced spatial filters span 16. The reduction has 16 output filters.

### To find

MACs for the direct path, both bottleneck stages, their total and percentage saved.

### Formula — which one and why



$$
M=H_oW_oC_oK^2C_i
$$



M is the MAC count. H\_oW\_oC\_o counts output values; K²C\_i is the products used for each output. One MAC means one multiply-accumulate pair. Count both layers of the bottleneck.



$$
\mathrm{saving}\%=100\frac{M_{\mathrm{direct}}-M_{\mathrm{bottleneck}}}{M_{\mathrm{direct}}}
$$



### Solution — step by step

**Step 1: direct output count.** 14 × 14 × 48 = 9,408.

**Step 2: products per direct output.** 5 × 5 × 480 = 12,000.

**Step 3: direct MACs.**



$$
9408(12000)=112896000
$$



**Step 4: reduction-layer MACs.** Each of 14 × 14 × 16 outputs combines 480 values.



$$
14(14)(16)(480)=1505280
$$



**Step 5: reduced spatial-layer MACs.** Each of 9,408 outputs now combines 5 × 5 × 16 = 400 values.



$$
9408(400)=3763200
$$



**Step 6: add both bottleneck costs.**



$$
1505280+3763200=5268480
$$



**Step 7: calculate the saving.**



$$
112896000-5268480=107627520
$$





$$
100\frac{107627520}{112896000}\approx95.333333\%
$$



**Final answer:** direct = **112.896 million MACs**; bottleneck = **5.26848 million MACs**; saving ≈ **95.33%**. The lecture's 112.9M and 5.3M are rounded versions. This arithmetic proves reduced computation; equal prediction accuracy is not guaranteed by the count alone.