Estimate Complexity

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:

1. Number of parameters in GPT-3.
2. Total number of tokens processed during training.
3. 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.

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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} ]

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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} ]

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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:

1. Self-attention mechanism:
   • Query, Key, Value projections: 3 matrix multiplications.
   • Output projection: 1 matrix multiplication.
   • Softmax (cheaper than matrix multiplies).
2. Feed-forward network (FFN): 2 matrix multiplications (input → hidden → output).
3. Residual additions and layer norms: Minor compared to matrix multiplies.
4. 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} ]

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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:

1. Compute (6 \times N):
   [ 6 \times 1.75 \times 10^{11} = 10.5 \times 10^{11} = 1.05 \times 10^{12} ]
2. Multiply by (T):
   [ (3 \times 10^{11}) \times (1.05 \times 10^{12}) = 3.15 \times 10^{23} \text{ FLOPs} ]

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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).