Amazon Nova 2 Lite
Estimating the Total FLOPs for Training GPT-3
To estimate the total number of floating-point operations (FLOPs) required to train GPT-3, we’ll break the problem down step by step. The key factors are:
- Number of parameters in GPT-3.
- Total number of tokens processed during training.
- FLOPs per token (computation per token during training).
We’ll derive reasonable estimates without searching online, using logical reasoning based on known transformer architecture principles and typical training practices.
Step 1: Number of Parameters (N)
GPT-3 has 175 billion parameters.
- This is a well-publicized fact about GPT-3, so we take it as a given:
[ N = 175 \times 10^9 = 1.75 \times 10^{11} \text{ parameters} ]
Step 2: Total Tokens Processed During Training (T)
Training FLOPs depend on how much data the model sees. We need to estimate the total number of tokens ((T)) processed during training.
Reasoning:
-
Dataset size: GPT-3 was trained on a massive text corpus (e.g., books, web data).
-
Token definition: In transformer models, a "token" is typically ~1–2 characters (e.g., subword units like BPE).
-
Rough token count per byte:
- Text averages ~5–6 characters per word.
- A word ≈ 1 token (after tokenization).
- Thus, ≈5–6 bytes per token (since ASCII/UTF-8 uses 1–4 bytes per character, but tokenization groups characters).
- Conservative estimate: 5 bytes per token.
-
Dataset size in bytes:
GPT-3’s training data is often cited as " Hundreds of gigabytes to a few terabytes".- Let’s assume ~300 billion tokens (a common ballpark for large LLM training).
Why?- Smaller models (e.g., GPT-2) used ~40 GB of text ≈ 40 billion tokens.
- GPT-3 is ~4,000× larger in parameters than GPT-2 (175B vs. 110M).
- Training data often scales less than linearly with model size (due to data saturation), but for estimation, we’ll use:
[ T \approx 300 \times 10^9 = 3 \times 10^{11} \text{ tokens} ]
- Let’s assume ~300 billion tokens (a common ballpark for large LLM training).
Step 3: FLOPs per Token (F)
Now we estimate FLOPs required to process one token during training (both forward and backward passes).
Key Operations per Token:
For a transformer model like GPT-3, processing one token involves:
- Self-attention mechanism:
- Query, Key, Value projections: 3 matrix multiplications.
- Output projection: 1 matrix multiplication.
- Softmax (cheaper than matrix multiplies).
- Feed-forward network (FFN): 2 matrix multiplications (input → hidden → output).
- Residual additions and layer norms: Minor compared to matrix multiplies.
- Backward pass: Doubles the FLOPs of the forward pass (gradients, weight updates).
FLOPs per Layer:
- A single transformer layer with (d_{\text{model}}) dimensions processes a token with ~(6d_{\text{model}}^2) FLOPs** (forward + backward).
- Why? Each matrix multiply of size (d_{\text{model}} \times d_{\text{model}}) costs (2d_{\text{model}}^2) FLOPs (1 multiply + 1 add). With ~3–4 such operations per layer (attention + FFN), forward pass ≈ (4d_{\text{model}}^2) FLOPs. Backward pass is similar, so total ≈ (8d_{\text{model}}^2).
- However, parameters (N) relate to (d_{\text{model}}):
Total parameters (N \approx 2 \times \text{layers} \times d_{\text{model}}^2) (for attention + FFN weights).
Thus, (d_{\text{model}}^2 \approx N / (2 \times \text{layers})).
For simplicity, we use an empirical rule of thumb:
[ \text{FLOPs per token} \approx 6N ] This is widely used in literature for transformer models (accounts for all layers, forward/backward, and overhead).
Justification:
- If (N = 1.75 \times 10^{11}), then:
[ F \approx 6 \times N = 6 \times 1.75 \times 10^{11} = 1.05 \times 10^{12} \text{ FLOPs per token} ]
Step 4: Total FLOPs for Training
Now combine all factors:
[
\text{Total FLOPs} = T \times F = T \times 6N
]
Plugging in the values:
- (T = 3 \times 10^{11}) tokens
- (N = 1.75 \times 10^{11}) parameters
[ \text{Total FLOPs} = (3 \times 10^{11}) \times 6 \times (1.75 \times 10^{11}) ]
Calculation:
- Compute (6 \times N):
[ 6 \times 1.75 \times 10^{11} = 10.5 \times 10^{11} = 1.05 \times 10^{12} ] - Multiply by (T):
[ (3 \times 10^{11}) \times (1.05 \times 10^{12}) = 3.15 \times 10^{23} \text{ FLOPs} ]
Final Estimate
[ \boxed{3 \times 10^{23} \text{ FLOPs}} ]
Summary of Assumptions:
| Factor | Value | Reasoning |
|---|---|---|
| Parameters ((N)) | (1.75 \times 10^{11}) | Publicly stated for GPT-3. |
| Total tokens ((T)) | (3 \times 10^{11}) | Estimated from typical LLM training data (scaled from smaller models). |
| FLOPs/token ((F)) | (6N) | Empirical rule for transformers (forward + backward passes). |
| Total FLOPs | (\approx 3 \times 10^{23}) | (T \times 6N). |
Context:
- This is consistent with known estimates for large LLM training (e.g., training GPT-3 was reported to use ~10^23 – 10^24 FLOPs).
- Real-world training uses optimizations (mixed precision, gradient accumulation), but our estimate captures the theoretical lower bound.
This calculation shows why training GPT-3 required massive computational resources (e.g., thousands of GPUs/TPUs for weeks).