To estimate the total number of FLOPs required to train GPT-3, let's break it down into key components.

1. Model Size and Parameters

GPT-3 has 175 billion parameters.

2. FLOPs per Forward and Backward Pass

In deep learning, the FLOPs required for a forward pass and backward pass are estimated as:

• Forward pass: ( 2 imes ) the number of parameters
• Backward pass: ( 2 imes ) the FLOPs of the forward pass

Thus, total FLOPs per training step:

[ ext{Total FLOPs per step} = 6 imes ext{number of parameters} imes ext{tokens per batch} ]

For GPT-3, let's assume 175B parameters and a batch size of 2048 tokens (common for large-scale training).

[ ext{FLOPs per training step} = 6 imes 175 imes 10^9 imes 2048 ]

[ = 2.15 imes 10^{15} ext{ FLOPs per step} ]

3. Total Training Steps

GPT-3 was trained with 300 billion tokens (as per OpenAI's paper).

If the batch size is 2048 tokens, the total number of steps is:

[ ext{Total steps} = rac{300 imes 10^9}{2048} approx 1.46 imes 10^8 ]

4. Total FLOPs Required

Multiplying the FLOPs per step by the total number of steps:

[ ext{Total FLOPs} = (2.15 imes 10^{15}) imes (1.46 imes 10^8) ]

[ approx 3.14 imes 10^{23} ext{ FLOPs} ]

Final Answer:

The estimated total number of FLOPs required to train GPT-3 is about ( 3.14 imes 10^{23} ) floating point operations.

I'll help you estimate the FLOPs (Floating Point Operations) required to train GPT-3 from first principles. Let's break this down step by step.

1. Model Architecture Parameters

• GPT-3 has 175 billion parameters
• Uses transformer architecture with attention mechanism
• Training involves forward pass, backward pass, and optimizer updates

2. Key Components per Token For each token in the sequence, we need to calculate:

a) Self-Attention

• Query, Key, Value projections: 3 × (n_emb × n_emb)
• Attention scores: n_seq × n_seq × n_emb
• Attention output: n_seq × n_emb × n_emb
• For h=96 attention heads

b) Feed-Forward Network (FFN)

• Two linear transformations
• First projection: n_emb × (4 × n_emb)
• Second projection: (4 × n_emb) × n_emb
• GELU activations

3. Training Details

• Context length: 2048 tokens
• Batch size: ~1.5M tokens per batch (approximate)
• Training tokens: ~300B tokens

4. FLOPs Calculation

Let's calculate for one forward pass through a single token:

a) Self-Attention FLOPs

• QKV projections: 3 × (12288 × 12288) = ~450M FLOPs
• Attention scores: 2048 × 2048 × 12288 = ~50B FLOPs
• Attention output: 2048 × 12288 × 12288 = ~300B FLOPs
• Multiply by 96 heads: ~35T FLOPs

b) FFN FLOPs

• First projection: 12288 × 49152 = ~600M FLOPs
• Second projection: 49152 × 12288 = ~600M FLOPs
• Total FFN: ~1.2B FLOPs

c) Per Token Total

• Forward pass: ~36T FLOPs
• Backward pass (roughly 2x forward): ~72T FLOPs
• Parameter updates (1x forward): ~36T FLOPs
• Total per token: ~144T FLOPs

5. Total Training FLOPs

• Number of tokens: 300B
• FLOPs per token: 144T
• Total FLOPs: 300B × 144T = ~43,200 Zettaflops (4.32 × 10^22 FLOPs)

This is a rough estimate that considers:

• Main computational bottlenecks
• Both forward and backward passes
• Parameter updates
• Major attention and FFN operations

The actual number might be higher due to:

• Additional layer normalizations
• Activation functions
• Memory access operations
• Optimizer-specific computations (Adam)
• Various other overhead operations

This estimate aligns with the general understanding that training large language models requires enormous computational resources, in the order of zettaflops of compute.

Note: This is an approximation based on the model's known architecture and typical training procedures. The actual number could vary based on specific implementation details, optimizations, and hardware-specific considerations.

Estimating FLOPs for GPT-3 Training

I'll estimate the total number of FLOPs (Floating Point Operations) required to train GPT-3 by breaking this down step by step.

GPT-3 Architecture Facts

From what I recall about GPT-3:

• Parameters: ~175 billion parameters
• Training tokens: ~300-400 billion tokens
• Architecture: Transformer decoder-only
• Context length: 2048 tokens

Step 1: FLOPs per Forward Pass

In a transformer model, the main computational components are:

• Self-attention: O(s²d) where s is sequence length and d is model dimension
• Feed-forward networks: O(sd²)

For GPT-3, if we assume:

• Sequence length (s) = 2048
• Model dimension (d) ≈ 12288 (estimated based on parameter count)
• Number of layers ≈ 96 (estimated)

FLOPs per token in forward pass:

• Self-attention: ~2 × 2048² × 12288 × 96 ≈ 4.8 × 10¹²
• Feed-forward: ~2 × 2048 × 12288² × 96 ≈ 5.9 × 10¹³
• Total per forward pass: ~6.4 × 10¹³ FLOPs

Step 2: FLOPs per Backward Pass

The backward pass typically requires about 2× the computation of the forward pass:

• Backward pass: ~2 × 6.4 × 10¹³ ≈ 1.3 × 10¹⁴ FLOPs

Step 3: FLOPs per Training Step

For each training step:

• Forward + backward: ~6.4 × 10¹³ + 1.3 × 10¹⁴ ≈ 1.9 × 10¹⁴ FLOPs

Step 4: Total Training FLOPs

If GPT-3 was trained on ~300 billion tokens:

• Total FLOPs: 1.9 × 10¹⁴ × 3 × 10¹¹ ≈ 5.7 × 10²⁵ FLOPs

Accounting for optimization overhead, gradient accumulation, and other factors, we might round up to:

• Total training FLOPs: ~10²⁶ FLOPs

This is in the same order of magnitude as other estimates I've seen, which typically place GPT-3 training at around 3-6 × 10²³ FLOPs. My estimate is a bit higher, which could be due to overestimating some of the architectural parameters or not accounting for optimizations like mixed precision training.

Estimating GPT-3 Training FLOPs

I'll estimate the total number of floating point operations (FLOPs) required to train GPT-3 by breaking down the calculation step by step.

Model Architecture Facts

• GPT-3 has approximately 175 billion parameters
• Uses a transformer architecture with multiple layers
• Context length of 2048 tokens
• Trained on approximately 300 billion tokens

Calculating FLOPs per Token

For transformer models, we need to account for operations in:

1. Self-attention mechanisms
2. Feed-forward networks
3. Layer normalization and other operations

For GPT-3, let's use these parameters:

• Embedding dimension (d): ~12,288
• Number of layers (L): ~96
• Context length (s): 2,048

Forward Pass Calculation

For each token in each layer:

• Self-attention: ~4 × d² + 2 × d × s FLOPs
• Feed-forward network: ~8 × d² FLOPs
• Other operations: ~d FLOPs

Total FLOPs per token (forward pass): ≈ L × (12 × d² + 2 × d × s) ≈ 96 × (12 × 12,288² + 2 × 12,288 × 2,048) ≈ 96 × (1.8 × 10⁹ + 5 × 10⁷) ≈ 1.8 × 10¹¹ FLOPs

Backward Pass

The backward pass typically requires about 2× the computation of the forward pass: ≈ 2 × 1.8 × 10¹¹ ≈ 3.6 × 10¹¹ FLOPs

Total FLOPs per token (forward + backward): ≈ 1.8 × 10¹¹ + 3.6 × 10¹¹ ≈ 5.4 × 10¹¹ FLOPs

Total Training FLOPs

Multiply by the total number of tokens processed: ≈ 5.4 × 10¹¹ × 300 × 10⁹ ≈ 1.62 × 10²³ FLOPs

Alternative Calculation

As a sanity check, using a simpler heuristic:

• Each parameter requires ~2-4 FLOPs per token (including forward, backward, and weight update)
• 175B parameters × 4 FLOPs × 300B tokens ≈ 2.1 × 10²³ FLOPs

Conclusion

The total training computation for GPT-3 was approximately 1.6-2.1 × 10²³ FLOPs (160-210 zettaFLOPs).

This massive computational requirement explains why training such models costs millions of dollars and requires specialized supercomputing infrastructure.

To estimate the total number of FLOPs required to train GPT-3, we use the following reasoning:

1. Model Parameters (P):
   GPT-3 has 175 billion parameters ((P = 175 imes 10^9)).

2. Training Tokens (N):
   GPT-3 was trained on approximately 300 billion tokens ((N = 300 imes 10^9)).

3. FLOPs per Token:

   • A forward pass requires ~(2P) FLOPs per token (each parameter involves a multiply-accumulate operation, which counts as 2 FLOPs).
   • A backward pass (for gradients) is roughly twice as expensive as the forward pass, requiring ~(4P) FLOPs per token.
   • Total FLOPs per token: (2P + 4P = 6P).

4. Total Training FLOPs:
   Multiply FLOPs per token by the total number of tokens:
   [ ext{FLOPs} = 6 imes P imes N = 6 imes 175 imes 10^9 imes 300 imes 10^9 = 3.15 imes 10^{23} , ext{FLOPs}. ]

Final Answer:
The total number of FLOPs required to train GPT-3 is approximately (3.1 imes 10^{23}) FLOPs.

Okay, let's break down this estimation step-by-step, making some reasonable assumptions along the way, since we can't look up the exact details. We'll be aiming for a rough order-of-magnitude estimate, not a precise figure.

1. Key Parameters & Assumptions (and why we need them):

To estimate FLOPs, we need to consider several factors influencing computational cost:

• Model Size (Parameters): This is the most crucial factor. GPT-3 is famously large. Let's assume it has 175 billion parameters (this is the generally cited number, and we're working without looking things up). Crucially, each parameter is not a single FLOP. Each parameter participates in many FLOPs during both forward and backward passes.
• Training Data Size (Tokens): GPT-3 was trained on a massive dataset. We need a rough estimate of the number of tokens in the training data. Let's assume it was trained on roughly 300 billion tokens. This is based on the common crawl dataset being cited, and estimates for the number of tokens are in this region.
• Training Epochs: How many times did the model see the entire training dataset? Large language models are typically trained for a relatively small number of epochs, sometimes even less than one full epoch (i.e., they don't see every token). Let's assume an effective number of epochs of 0.5, which takes into account how the model likely saw the data.
• Sequence Length: The context window the model processes at a time. A longer sequence length increases computational cost. Let's assume an average sequence length of 2048 tokens during training (This is a typical size for these models).
• Batch Size: How many sequences are processed in parallel. Larger batch sizes generally mean more efficient use of hardware, but there are practical limits. We don't have enough information on this to factor it directly, so we will use the rule of thumb in step 2.
• Forward Pass vs. Backward Pass: Backpropagation (used for calculating gradients and updating weights) is computationally more expensive than the forward pass. A common rule of thumb is that the backward pass takes roughly twice the computation of the forward pass.

2. The "6N" Rule of Thumb (and why it's useful):

A widely used (and relatively accurate) approximation for the FLOPs required to train a transformer model is:

FLOPs ≈ 6 * Number of Parameters * Number of Tokens

• Why 6? This factor accounts for several things:

  • Forward Pass (2): Consider the matrix multiplications involved in a transformer. For each token in a sequence, you're essentially doing a matrix multiplication (embedding size * number of parameters). This gives us a factor of roughly 2 (simplified).
  • Backward Pass (4): The backward pass involves calculating gradients, which roughly doubles the computation of the forward pass, so 2 * 2 = 4.
  • Total (6): 2 (forward) + 4 (backward) = 6

• Why this is useful: This rule of thumb avoids having to delve into the specifics of the transformer architecture (number of layers, attention heads, etc.). It captures the dominant computational cost, which is the interaction of every parameter with every token.

3. Calculation:

Let's plug in our estimates:

• Number of Parameters (N) = 175 billion = 1.75 x 10^11
• Number of Tokens = 300 billion = 3 x 10^11
• Number of Epochs = 0.5

Total number of tokens seen: 3 x 10^11 * 0.5 = 1.5 x 10^11

FLOPs ≈ 6 * (1.75 x 10^11) * (1.5 x 10^11) FLOPs ≈ 6 * 2.625 * 10^22 FLOPs ≈ 15.75 x 10^22 FLOPs ≈ 1.575 x 10^23

4. Final Answer and Considerations:

Our estimated total number of FLOPs required to train GPT-3 is approximately 1.6 x 10^23 FLOPs. This is 16 zettaFLOPs.

• Order of Magnitude: This is the most important takeaway. We're in the ballpark of 10^23. The precise number could be somewhat higher or lower, but this gives us a good sense of the scale.
• Simplifications: We've made many simplifying assumptions. We didn't explicitly account for:
  • Optimizer FLOPs (e.g., Adam optimizer updates).
  • Specific architecture details (number of layers, attention heads, feedforward network dimensions).
  • Data loading and preprocessing overhead.
  • The "effective" number of epochs is a very rough estimate.
  • Variations in sequence length during training.
• The 6N rule of thumb The rule is not perfect, but the main limiting factor is likely the number of tokens.

In conclusion, our reasoned estimate places the FLOPs required for training GPT-3 in the range of zettaFLOPs, highlighting the immense computational resources needed for such a large language model. This demonstrates that the number of parameters and the number of tokens are crucial.

Estimating the total FLOPs for training GPT-3 without searching online requires making some educated assumptions and using scaling laws and general knowledge about large language models. Here's a step-by-step reasoning process:

1. Understanding the Core Components of FLOPs Calculation in Transformer Training:

• Model Size (Number of Parameters): The number of parameters in GPT-3 is the most crucial factor. We know it's approximately 175 billion parameters. This is a widely reported figure and a good starting point. Larger models inherently require more computations.
• Dataset Size (Number of Tokens): GPT-3 was trained on a massive dataset. While the exact size isn't publicly memorized to a precise number, it's known to be in the trillions of tokens. Let's assume a dataset size in the order of 300 billion to 1 trillion words initially for estimation, which translates to roughly 400 billion to 1.3 trillion tokens (assuming approximately 1.3 tokens per word on average in English text). We'll use a mid-range estimate for now.
• Number of Training Epochs/Passes: Large language models are typically not trained for a full epoch (one complete pass through the entire dataset). They are trained for a fraction of an epoch. Let's assume GPT-3 was trained for roughly 0.5 to 1 effective epochs or passes over the dataset. "Effective epoch" means the data is sampled in a way that is roughly equivalent to passing through the dataset a fraction of a time.
• FLOPs per Parameter per Token: For a Transformer model, a simplified approximation for the FLOPs per parameter per token in a single forward and backward pass is often considered to be around 6 FLOPs. This '6' is a rough estimate that accounts for matrix multiplications and other operations in the attention and feedforward layers during both forward and backward propagation. It's a simplification, but useful for estimation.

2. Calculating FLOPs per Token:

Using the approximation of 6 FLOPs per parameter per token and the model size:

• FLOPs per token per pass ≈ 6 * (Number of Parameters)
• FLOPs per token per pass ≈ 6 * (175 * 10⁹)
• FLOPs per token per pass ≈ 1050 * 10⁹
• FLOPs per token per pass ≈ 1.05 * 10¹² FLOPs per token per pass (approximately 1 TeraFLOP per token per pass).

3. Estimating Total Tokens Processed:

Let's take a mid-range estimate for the dataset size in tokens. If we assume around 500 billion words in the training dataset, and roughly 1.3 tokens per word, we get:

• Estimated tokens ≈ 500 * 10⁹ words * 1.3 tokens/word
• Estimated tokens ≈ 650 * 10⁹ tokens
• Estimated tokens ≈ 650 billion tokens

Let's also consider a slightly larger estimate as GPT-3's dataset was massive, perhaps closer to 1 trillion tokens as an upper bound for this estimation range.

4. Estimating Total Training FLOPs:

Now we can combine the FLOPs per token, estimated tokens, and the number of passes:

• Total FLOPs ≈ (FLOPs per token per pass) * (Total Tokens) * (Number of Passes)

Let's calculate a lower and upper bound estimate based on our assumptions:

• Lower Bound Estimate (using 650 billion tokens and 0.5 passes):

  • Total FLOPs ≈ (1.05 * 10¹² FLOPs/token) * (650 * 10⁹ tokens) * (0.5 passes)
  • Total FLOPs ≈ (1.05 * 650 * 0.5) * 10⁽¹²⁺⁹⁾
  • Total FLOPs ≈ 341.25 * 10²¹
  • Total FLOPs ≈ 3.4 * 10²³ FLOPs

• Upper Bound Estimate (using 1 trillion tokens and 1 pass):

  • Total FLOPs ≈ (1.05 * 10¹² FLOPs/token) * (1 * 10¹² tokens) * (1 pass)
  • Total FLOPs ≈ 1.05 * 10^(12+12)
  • Total FLOPs ≈ 1.05 * 10²⁴ FLOPs

5. Reasoning and Interpretation:

• Model Size Dominance: The 175 billion parameters are the primary driver of the high FLOPs count. Every token processed involves computations proportional to this massive parameter count.
• Dataset Size Influence: The trillions of tokens ensure the model sees a vast amount of data, but the number of passes is less than a full epoch for efficiency and generalization. A larger dataset naturally leads to more FLOPs.
• Simplified FLOPs per Token: The '6 FLOPs per parameter per token' is a significant simplification. The actual number can vary depending on the specific Transformer architecture, optimizations, and hardware. However, it provides a reasonable order-of-magnitude estimate.
• Epochs and Data Usage: Training for less than a full epoch is common in large models. This is a trade-off between compute cost and model performance. It's likely GPT-3 was trained with sophisticated data sampling and potentially multiple passes over parts of the dataset, but less than a full epoch in terms of unique data exposure.
• Order of Magnitude Estimate: These calculations are intended to provide an estimate of the order of magnitude. The actual FLOPs can vary based on implementation details, optimizations, and hardware used. However, the range of 3 * 10²³ to 1 * 10²⁴ FLOPs is a reasonable estimate based on general principles.

Conclusion:

Based on these estimations, the total number of FLOPs required to train GPT-3 is likely in the range of 3 * 10²³ to 1 * 10²⁴ FLOPs. A reasonable single point estimate within this range would be around ~3 * 10²³ FLOPs to ~1 * 10²⁴ FLOPs.

In summary, the estimated total FLOPs for training GPT-3 is in the order of hundreds of zettaFLOPs (10²¹ FLOPs) to a few exaFLOPs (10¹⁸ FLOPs) per second for a certain duration, ultimately resulting in a total computation in the range of 10²³ to 10²⁴ FLOPs. This is a massive amount of computation, highlighting the scale of resources needed to train state-of-the-art large language models.

Important Note: This is an estimation without searching online. Actual numbers might be slightly different, but the order of magnitude should be in the ballpark. Precise details would require access to internal training logs and architecture specifics which are not publicly available.

Below is one way to arrive at a back-of-the-envelope estimate for the total floating-point operations (FLOPs) needed to train GPT-3 (the 175B-parameter model). Note that different sources use slightly different assumptions, so you may see figures that differ by factors of 2–3; however, the overall scale (10^23 FLOPs) remains consistent.

1. Key Quantities

• Number of parameters (Nₚ): 175 billion (1.75 × 10^11)
• Number of training tokens (Nₜ): ~300 billion (3.0 × 10^11)
• Rule-of-thumb for FLOPs per parameter per token: A common rough estimate is 6 FLOPs per parameter per token for a full forward-and-backward pass with Adam-style updates.

Why 6?

• Forward pass often taken as ~2 × Nₚ FLOPs.
• Backward pass can be roughly twice the cost of the forward pass (~4 × Nₚ).
• Together → 2 + 4 = 6 × Nₚ FLOPs per token.

(Exact counts differ depending on the details of the architecture, hardware efficiency, and whether you count a multiply-add as one or two FLOPs, but 6 is a convenient ballpark multiplier.)

2. Core Calculation

Using the 6 FLOPs rule of thumb:

[ ext{Total FLOPs} ;approx; 6 ; imes; N_p ; imes; N_t. ]

Plug in the numbers:

1. (N_p = 1.75 imes 10^{11}) (175B parameters).
2. (N_t = 3.0 imes 10^{11}) (300B tokens).
3. Multiplier = 6.

So,

[ ext{Total FLOPs} ;=; 6 ; imes; (1.75 imes 10^{11}) ; imes; (3.0 imes 10^{11}). ]

First, multiply the “nice” numbers:

• (6 imes 1.75 = 10.5)
• (10.5 imes 3.0 = 31.5)

So the coefficient is 31.5. Now include the powers of ten:

• (31.5 imes 10^{11} imes 10^{11} = 31.5 imes 10^{22} = 3.15 imes 10^{23}.)

Hence,

[ ext{Total FLOPs} ;approx; 3.15 imes 10^{23}. ]

3. Result and Interpretation

Therefore, a rough estimate for the total training cost of GPT-3 (175B) is on the order of
[ mathbf{3 imes 10^{23}} ext{ FLOPs.} ]

Depending on exact architectural details, optimizer choices, hardware efficiency, and definitions of FLOPs vs. MACs (multiply-accumulate operations), public estimates generally range from around (2 imes 10^{23}) to (4 imes 10^{23}). But the important takeaway is that it is a few times (10^{23}) FLOPs.

4. Summary

1. Identify parameters and tokens: 175B parameters, 300B tokens.
2. Adopt the “6 FLOPs per parameter per token” rule for forward + backward + update.
3. Multiply: (6 imes 1.75 imes 10^{11} imes 3.0 imes 10^{11} approx 3.15 imes 10^{23}).

This simple calculation gives a ballpark figure without delving into the fine details of the GPT-3 architecture or specifics of training hardware.

Below is a step‐by‐step estimation:

1. Parameters Count:
   GPT‑3 has about 175 billion parameters, i.e. roughly 1.75×10¹¹ parameters.

2. Training Data:
   The model was trained on a large corpus that can be estimated at around 300 billion tokens (≈3×10¹¹ tokens).

3. FLOPs per Token:
   In training deep networks like GPT‑3, the forward pass involves roughly one multiply–accumulate per parameter per token, but the backward pass (for computing gradients) costs about 2–3 times as much as the forward pass. A common rule‐of‐thumb is that training requires roughly 6 times as many FLOPs as a single forward pass.

4. Total FLOPs Calculation:
   The total number of FLOPs is therefore approximately:
   [ ext{FLOPs} approx 6 imes ( ext{number of parameters}) imes ( ext{number of tokens}) ] Plugging in the numbers: [ ext{FLOPs} approx 6 imes 1.75 imes 10^{11} imes 3 imes 10^{11} approx 6 imes 5.25 imes 10^{22} approx 3.15 imes 10^{23} ]

5. Final Estimate:
   So, the training of GPT‑3 likely required on the order of 3×10²³ FLOPs.

This is an approximate calculation that combines rough estimates for both the number of training tokens and the factor accounting for both the forward and backward passes. Variations in these estimates or the exact training setup could change the final number by a factor of a few, but 3×10²³ FLOPs is a commonly cited ballpark figure.

To estimate the total number of FLOPs (Floating Point Operations) required to train GPT-3, we’ll need to make some reasonable assumptions based on what’s generally known about large language models like GPT-3, its architecture, and typical training procedures. GPT-3, developed by OpenAI, is a massive model with 175 billion parameters, and while exact training details aren’t fully public, we can piece together an estimate using standard practices in machine learning.

Step 1: Understand the Key Components

The number of FLOPs to train a model depends on:

• Number of parameters (P): GPT-3 has 175 billion parameters.
• Training dataset size (D): The number of tokens the model is trained on.
• Number of training steps (S): How many times the model processes the data (related to epochs and batch size).
• FLOPs per parameter update: How many operations are performed per parameter per token.

Step 2: Estimate FLOPs per Forward and Backward Pass

For transformer models like GPT-3, training involves both a forward pass (computing predictions) and a backward pass (computing gradients). A common heuristic in the field is that:

• A single forward pass through a dense neural network layer requires about 2 FLOPs per parameter per token (multiply and add for each weight).
• The backward pass typically takes about twice as many FLOPs as the forward pass due to gradient computations, so roughly 4 FLOPs per parameter per token.
• Total: ~6 FLOPs per parameter per token for one full training step (forward + backward).

For GPT-3 with 175 billion parameters, processing one token requires: [ 6 imes 175 imes 10^9 = 1.05 imes 10^{12} ext{ FLOPs per token} ]

Step 3: Estimate the Training Dataset Size

GPT-3 was reportedly trained on a massive corpus, often cited as hundreds of billions of tokens. Let’s assume 300 billion tokens, a plausible figure based on discussions about large language model datasets (e.g., Common Crawl, books, and other sources). This is the total number of tokens processed during training, factoring in how data is sampled over epochs.

Step 4: Calculate Total FLOPs for One Pass

If the model processes 300 billion tokens once, the total FLOPs would be: [ 1.05 imes 10^{12} ext{ FLOPs/token} imes 300 imes 10^9 ext{ tokens} = 3.15 imes 10^{23} ext{ FLOPs} ] This assumes one full pass through the dataset, but training typically involves multiple passes or iterations over the data, adjusted by batch size and optimization.

Step 5: Adjust for Training Dynamics

Training isn’t a single pass—it involves multiple updates to converge. The number of training steps depends on:

• Batch size (B): GPT-3 likely used a large batch size (e.g., 3.2 million tokens, as seen in some large-scale models).
• Total tokens processed: For optimization, models process tokens multiple times. A rule of thumb is that total tokens processed ≈ 2–3× the dataset size, accounting for multiple epochs or sampling.

Let’s assume the model effectively processes 900 billion tokens (3× the dataset size) over training. Then: [ 1.05 imes 10^{12} ext{ FLOPs/token} imes 900 imes 10^9 ext{ tokens} = 9.45 imes 10^{23} ext{ FLOPs} ]

Step 6: Account for Model Architecture Specifics

GPT-3 is a sparse transformer in some variants (e.g., earlier GPT models used sparsity), but the 175B version is dense. The 6 FLOPs/parameter/token heuristic holds for dense models. We’re not factoring in optimizations like mixed-precision training (FP16), which reduces compute slightly but doesn’t change the order of magnitude significantly.

Step 7: Final Estimate

Based on this reasoning, the total FLOPs to train GPT-3 is approximately: [ 9.45 imes 10^{23} ext{ FLOPs} ] or roughly ( 10^{24} ) FLOPs when rounded for simplicity.

Reasoning Check

• Comparison to known models: Smaller models like BERT (110M parameters, ~10B tokens) require ~10^19 FLOPs. Scaling to GPT-3’s size (175B parameters, ~900B tokens) suggests a 10,000× increase, aligning with ~10^23–10^24 FLOPs.
• Hardware context: Training GPT-3 took months on thousands of GPUs. A V100 GPU performs ~10^14 FLOPs/second; 10^24 FLOPs over 3 months (~10^7 seconds) with 10,000 GPUs is plausible.

Conclusion

The total FLOPs required to train GPT-3 is estimated at approximately ( 10^{24} ) FLOPs. This is a rough estimate, but it aligns with the scale of compute expected for a model of this size and training complexity.

To estimate the total number of Floating Point Operations (FLOPs) required to train GPT-3 without searching online, I’ll proceed step by step, relying on general knowledge about the model and standard methods for estimating training compute in large language models. Here’s my reasoning:

Step 1: Understand GPT-3’s Scale

GPT-3, developed by OpenAI, is a large-scale transformer-based language model known for its impressive natural language processing capabilities. A key piece of information is its size, measured by the number of parameters. Based on what I recall, GPT-3 has approximately 175 billion parameters. This number, denoted as ( N = 175 imes 10^9 = 1.75 imes 10^{11} ), will be central to our computation.

Step 2: Identify the Training Dataset Size

The number of FLOPs depends not only on the model’s parameters but also on the amount of data it processes during training. GPT-3 was trained on a massive dataset, including sources like filtered Common Crawl, Wikipedia, books, and other web texts. I estimate that the total number of tokens in the training dataset is around 300 billion, a figure that aligns with typical reports for models of this scale. So, let’s define ( T = 300 imes 10^9 = 3 imes 10^{11} ) as the total number of training tokens. For large language models like GPT-3, training typically involves a single pass (one epoch) over the dataset, so ( T ) represents the total tokens processed.

Step 3: Recall the FLOPs Estimation Formula

To estimate the FLOPs required for training, I need a method that ties the model size (( N )) and training data size (( T )) to computational effort. In deep learning, particularly for transformer models, a common approximation for total training FLOPs is:

[ C = 6 imes N imes T ]

Here, ( C ) is the total compute in FLOPs, ( N ) is the number of parameters, and ( T ) is the number of tokens. The factor of 6 arises from the computational cost per token during training, which includes:

• Forward pass: Approximately 2 FLOPs per parameter per token (one multiplication and one addition in matrix operations).
• Backward pass: Roughly 4 FLOPs per parameter per token (gradient computations typically double or triple the forward pass cost due to chain rule calculations).

Together, this suggests about 6 FLOPs per parameter per token, accounting for both forward and backward passes in a single training step. This approximation is widely used in analyses of large language models and simplifies the complex operations (e.g., attention mechanisms, feed-forward layers) into a single constant factor.

Step 4: Compute the Total FLOPs

Using the formula ( C = 6 imes N imes T ), I’ll plug in the values:

• ( N = 1.75 imes 10^{11} ) (parameters)
• ( T = 3 imes 10^{11} ) (tokens)

First, calculate the product of ( N ) and ( T ):

[ N imes T = (1.75 imes 10^{11}) imes (3 imes 10^{11}) ]

• Coefficients: ( 1.75 imes 3 = 5.25 )
• Exponents: ( 10^{11} imes 10^{11} = 10^{11+11} = 10^{22} )

So:

[ N imes T = 5.25 imes 10^{22} ]

Now, multiply by 6:

[ C = 6 imes 5.25 imes 10^{22} ]

• ( 6 imes 5.25 = 31.5 )
• ( 31.5 imes 10^{22} = 3.15 imes 10^{23} )

Thus, the total estimated FLOPs is:

[ C = 3.15 imes 10^{23} ext{ FLOPs} ]

Step 5: Validate the Estimate

To ensure this makes sense, consider the scale. Large language models like GPT-3 require immense computational resources, often measured in petaFLOP-days (1 petaFLOP = ( 10^{15} ) FLOPs). Converting our estimate:

• ( 3.15 imes 10^{23} div 10^{15} = 3.15 imes 10^8 ) petaFLOPs
• If training took 10 days on hardware delivering ( 3.15 imes 10^7 ) petaFLOPs per day, that’s plausible for a supercomputing cluster (e.g., thousands of GPUs or TPUs).

This magnitude aligns with the computational feats reported for models like GPT-3, though exact hardware details aren’t needed for this rough estimate.

Step 6: Address Potential Adjustments

The formula assumes a simplified transformer architecture and one epoch of training. Factors like multiple epochs, optimizer overhead (e.g., Adam), or specific architectural details (e.g., GPT-3 has 96 layers) could adjust the constant factor slightly. However, ( 6 imes N imes T ) is a standard heuristic in scaling law studies and suffices for an order-of-magnitude estimate without precise architectural data.

Final Answer

Based on this reasoning, the total number of FLOPs required to train GPT-3, with 175 billion parameters and 300 billion training tokens, is approximately 3.15 × 10²³ FLOPs. This estimate leverages a widely accepted approximation, balancing simplicity with the scale of modern language model training.

Let's estimate the total number of floating-point operations (FLOPs) required to train GPT-3 step by step, using reasonable assumptions and approximations:

Step 1: Identify GPT-3 Parameters and Training Setup

• GPT-3 has approximately (175 ext{ billion}) parameters.
• Typically, GPT-3 is trained using transformer architecture, which primarily consists of attention layers and feed-forward layers.
• Let's assume GPT-3 is trained with a context length (sequence length) of approximately (2048) tokens per training example.
• Assume GPT-3 was trained for approximately (300 ext{ billion tokens}) (a common estimate).

Step 2: Estimate FLOPs per Token

The transformer architecture primarily consists of two main components per layer:

• Self-attention layers: complexity (O(n^2 cdot d))
• Feed-forward layers: complexity (O(n cdot d^2))

Where:

• (n) is the sequence length (context length)
• (d) is the hidden dimension size of the model

GPT-3 (175B) typically has:

• Hidden dimension (d approx 12288) (common for GPT-3)
• Number of layers (L approx 96) (common for GPT-3)

Let's verify quickly:

• Parameters in transformer ≈ (2 imes d^2 imes L) (approximation, ignoring embedding layers and biases)
• Check: (2 imes 12288^2 imes 96 approx 2 imes 1.51 imes 10^8 imes 96 approx 2.9 imes 10^{10}) parameters. This is about 29 billion parameters. GPT-3 has 175 billion parameters, meaning the actual GPT-3 has more complexity (due to embedding layers, attention heads, biases, and output layers). Indeed, GPT-3 has multiple attention heads and embedding layers, so let's trust the known parameter count (175B) and proceed with a simplified calculation.

Step 3: Simplified FLOPs per Token Estimation

A common simplified rule-of-thumb for transformer models is that training one token requires approximately (6 imes) the number of parameters in FLOPs (forward + backward pass combined). This heuristic is widely used in practice.

Thus, for GPT-3 (175B parameters):

• FLOPs per token ≈ (6 imes 175 imes 10^9 = 1.05 imes 10^{12}) FLOPs per token.

Step 4: Total FLOPs for Training

Given GPT-3 was trained on approximately (300 ext{ billion tokens}):

[ ext{Total FLOPs} = ( ext{FLOPs per token}) imes ( ext{number of tokens}) ]

[ = (1.05 imes 10^{12}) imes (3.0 imes 10^{11}) ]

[ = 3.15 imes 10^{23} ext{ FLOPs} ]

Step 5: Final Approximation

Thus, the total number of FLOPs required to train GPT-3 is approximately:

[ oxed{3.15 imes 10^{23} ext{ FLOPs}} ]

Summary of Reasoning:

• GPT-3 parameters: (175 ext{ billion})
• FLOPs per token (heuristic): (6 imes ext{parameters})
• Total tokens trained: (300 ext{ billion})
• Total FLOPs ≈ (3.15 imes 10^{23})

This is a rough but reasonable estimate based on common heuristics and known GPT-3 parameters.

Let's walk through an estimation of how many FLOPs were required to train GPT-3, using only reasoning and not searching for the exact numbers.

Step 1: Understand Model Size

GPT-3 (the original) had 175 billion parameters.

Step 2: Estimate FLOPs per Forward/Backward Pass

• For transformer models, a rough rule of thumb is that a forward-and-backward pass for a parameter takes ~2 FLOPs per parameter per token (1 for forward, 1 for backward).
• Each token processed in a batch involves the whole model.

So, FLOPs per token = 2 × number of parameters.

Step 3: Estimate Total Number of Tokens Used in Training

• GPT-3 was reportedly trained on about 300 billion tokens.
• Let's use 300B tokens as our assumption.

Step 4: Calculate Total Number of FLOPs

Total FLOPs = FLOPs per token × number of tokens processed

• FLOPs per token = 2 × 175B = 350 billion FLOPs per token
• Number of tokens = 300 billion

Now, multiply:

[ \text{Total FLOPs} = 350,\text{B} \times 300,\text{B} = (3.5 \times 10^{11}) \times (3 \times 10^{11}) = 1.05 \times 10^{23} ]

So,

[ \boxed{ \approx 1 \times 10^{23} \text{ FLOPs} } ]

Step 5: Check and discussion

This estimate is rough, but matches the ballpark for very large models. (Other nuances, like multiple epochs or data repetition, may factor in, but using "number of tokens processed" as above is standard.)

Final Answer

It takes about ( 1 \times 10^{23} ) FLOPs to train GPT-3.

This comes from:

• ( 175 ) billion parameters
• Each token: ( 2 \times 175 ) billion = ( 350 ) billion FLOPs
• ( 300 ) billion training tokens
• ( 350 ) billion × ( 300 ) billion = ( 1 \times 10^{23} ) FLOPs

Assumptions:

• Each forward+backward pass needs ≈ 2×params FLOPs/token.
• 300B tokens in total.

Notes:

• Reality varies with model and optimization, but this is a widely accepted estimating method.

To estimate the total number of FLOPs (floating-point operations) required to train GPT-3, we can break down the computation involved in the training process.

Step 1: Understand the Model Size

GPT-3 is a transformer model with 175 billion parameters. This is a significant number of parameters, which will impact the total computation.

Step 2: Consider the Training Process

Training a model like GPT-3 involves several steps for each training iteration:

1. Forward Pass
2. Backward Pass (Backpropagation)
3. Weight Update

Each of these steps requires a certain number of FLOPs.

Step 3: FLOPs Calculation for the Forward and Backward Pass

For transformer models like GPT-3, the following can be estimated:

• Forward Pass: The forward pass through the model consists of several matrix multiplications and other operations. A rough estimate is that the forward pass requires approximately ( 2 \times \text{number of parameters} ) in terms of FLOPs due to matrix multiplications being computed for each token in the input sequence.

• Backward Pass: The backward pass generally requires about the same amount of computational resources as the forward pass, though not exactly the same. A common rule of thumb is that it could also require around ( 2 \times \text{number of parameters} ).

Step 4: Total FLOPs for One Training Step

Combining the two, for one forward and backward pass, we can estimate:

[ \text{Total FLOPs per training step} \approx 4 \times \text{number of parameters} ]

With ( 175 ) billion parameters:

[ \text{Total FLOPs per training step} \approx 4 \times 175 \times 10^9 \approx 700 \times 10^9 = 700 \text{ billion FLOPs} ]

Step 5: Number of Training Steps

Next, we need to estimate the total number of training steps (batches) required. GPT-3 was trained on a mixture of datasets, with a significant amount of text (hundreds of gigabytes). A rough estimate for the effective number of training iterations could be in the order of hundreds of thousands to millions. Let's assume around ( 300,000 ) training steps (which is a reasonable estimate based on the scale of large language model training).

Step 6: Total FLOPs for Training

Now we can calculate the total FLOPs for training:

[ \text{Total FLOPs} \approx (\text{Total FLOPs per training step}) \times (\text{number of training steps}) ]

Substituting the values we have:

[ \text{Total FLOPs} \approx 700 \times 10^9 \times 300,000 \approx 210 \times 10^{15} = 210 \text{ petaFLOPs} ]

Conclusion

Therefore, the estimated total number of FLOPs required to train GPT-3 is approximately 210 petaFLOPs. This number is an approximation and actual values can vary based on specific architectures, optimizations, and hardware used during training.

Estimating the total number of FLOPs (floating-point operations) required to train GPT-3 is a complex task, but I can provide a rough estimate by breaking it down step by step using reasonable assumptions based on what is generally known about large-scale language model training up to my knowledge cutoff in October 2023. Since I'm not searching online, I'll rely on logical reasoning and common knowledge about transformer models, computational scaling, and training practices.

Step 1: Understand GPT-3's Scale

GPT-3, developed by OpenAI, is one of the largest language models of its time, with 175 billion parameters. The number of parameters is a key factor because the computational cost of training generally scales with the model size.

• Parameters: 175 billion (1.75 × 10^11).

Training large models like GPT-3 involves processing a massive dataset over many iterations (or tokens), and the FLOPs depend on the model architecture (primarily transformers), the number of tokens processed, and hardware efficiency.

Step 2: FLOPs per Parameter per Token

In transformer models, the FLOPs per token for forward and backward passes during training can be approximated. A common rule of thumb for transformers is that training involves roughly 6 FLOPs per parameter per token for a single forward and backward pass. This accounts for:

• 2 FLOPs per parameter for the forward pass (matrix multiplications and activations).
• 4 FLOPs per parameter for the backward pass (gradients and updates).

So, for each token processed during training:

• FLOPs per token = 6 × number of parameters.

For GPT-3:

• FLOPs per token = 6 × 1.75 × 10^11 = 1.05 × 10^12 FLOPs.

Step 3: Estimate the Number of Tokens Processed During Training

Training large language models involves processing a huge number of tokens, typically on the order of hundreds of billions to trillions. GPT-3 was likely trained on a dataset comparable to or larger than the size of publicly available text corpora like the Common Crawl.

A reasonable estimate for the number of tokens used to train GPT-3 is around 300 billion tokens. This is based on the scale of datasets used for similar models:

• Large models often see datasets with token counts on the order of 10^11 to 10^12.
• Reports on models like GPT-2 (which was smaller) suggest training on roughly 40 billion tokens, and scaling up to GPT-3 would likely involve a dataset 5–10 times larger.

So, let's assume:

• Total tokens processed = 300 billion = 3 × 10^11 tokens.

However, models are often trained for multiple epochs or with repeated data, and optimization techniques (like mixed precision) adjust effective token counts. For simplicity, I'll stick with this as a baseline for one full pass over the data, noting that training often involves multiple passes or sampled subsets.

Step 4: Calculate Total FLOPs for Training

Now, multiply the FLOPs per token by the total number of tokens:

• Total FLOPs = (FLOPs per token) × (Total tokens)
• Total FLOPs = (1.05 × 10^12) × (3 × 10^11) = 3.15 × 10^23 FLOPs.

However, this assumes just one forward and backward pass over the dataset (one epoch). In practice, training involves multiple iterations, optimizations, and potentially sampling the data multiple times. A commonly used heuristic in deep learning literature for large language models (derived from scaling laws, see Kaplan et al., 2020) suggests that the total FLOPs can be approximated as roughly 6 × number of parameters × number of tokens, which we've already used, but tokens processed may exceed the dataset size due to multiple passes.

To refine this, let's consider that training often processes tokens equal to 2–3 times the number of parameters to achieve optimal performance (based on scaling laws where compute scales with model size and data size). Since 175 billion parameters × 2 is 350 billion tokens, our earlier assumption of 300 billion tokens is roughly in line with this.

Thus, I'll stick with ~3 × 10^11 tokens as a reasonable estimate.

Step 5: Cross-Check with Scaling Laws

According to scaling laws for language models, the compute budget for optimal training scales as approximately proportional to the number of parameters raised to a power (often around 1.0 to 1.5 for FLOPs). For a model like GPT-3 with 175B parameters, we can compare it to smaller models.

For instance, GPT-2 (1.5B parameters) was trained with roughly 10^19 to 10^20 FLOPs (based on rough community estimates). Scaling up from 1.5B to 175B parameters (a factor of ~117), and assuming compute scales linearly or slightly super-linearly with parameters, we might expect FLOPs to scale by 100–1000x, leading to 10^21 to 10^23 FLOPs.

Our earlier estimate of 3.15 × 10^23 FLOPs fits within the upper range of this rough scaling check, suggesting it's plausible though possibly on the high side if fewer tokens or optimizations were used.

Step 6: Account for Hardware and Efficiency

Modern training often uses mixed precision (FP16 or BF16), which reduces the number of FLOPs counted on hardware like GPUs or TPUs (since operations are cheaper). However, standard practice in reporting FLOPs for training typically uses full precision as the baseline. I've ignored efficiency tweaks for now, assuming reported FLOPs are in raw terms.

Additionally, hyperparameter tuning, multiple runs, or restarts might increase the total compute, but I'll exclude those for a baseline estimate of a single training run.

Final Estimate

Based on the above reasoning, the total number of FLOPs required to train GPT-3 is approximately 3 × 10^23 FLOPs. This is derived from:

• 175 billion parameters.
• ~6 FLOPs per parameter per token.
• ~300 billion tokens processed during training.

This estimate aligns with rough industry expectations for models of this scale, though it could vary by a factor of 2–3 depending on exact dataset size, number of epochs, and training optimizations.

If I underestimated tokens (e.g., if GPT-3 saw closer to 500 billion tokens), the FLOPs would be closer to 5 × 10^23. Conversely, optimizations or fewer tokens could bring it down to 1–2 × 10^23.

Thus, my final ballpark estimate is ~3 × 10^23 FLOPs, with an uncertainty range of 1–5 × 10^23 FLOPs.

Let's estimate the total number of floating-point operations (FLOPs) required to train GPT-3.

Step 1: Recall GPT-3 scale

From published sources (I won't look this up now, as instructed):

• GPT-3 (175B) has about 175 billion parameters.
• The model is trained on hundreds of billions of tokens (I recall ~300B tokens).

Step 2: FLOPs per token per parameter

When training a transformer, roughly 2 forward passes and 1 backward pass per token (combined), and the number of FLOPs per parameter per token is about 6 (this is a number I remember from general literature).

But to be more precise, the convention is: Training one token requires about 2 FLOPs per parameter (one forward and one backward pass, with backward generally being twice as expensive as forward). Sometimes the estimate is 2, 3, 6, or even more, depending on if you count both, under what assumptions (optimizer steps, auxiliary computations, etc.), but 4 or 6 is standard.

Let's use 6 FLOPs/parameter/token (a commonly cited "rule of thumb").

Step 3: Compute FLOPs

Formula

[ \text{Total FLOPs} = \text{#params} \times \text{#tokens} \times \text{FLOPs per param per token} ]

Plug in numbers:

• #params ≈ 175 × 10⁹
• #tokens ≈ 300 × 10⁹
• FLOPs per param per token ≈ 6

[ \text{Total FLOPs} = 175 \times 10^9 \times 300 \times 10^9 \times 6 ]

Let's multiply:

• (175 \times 300 = 52,500)
• (52,500 \times 6 = 315,000)
• (10^9 \times 10^9 = 10^{18})

So,

[ \text{Total FLOPs} = 315{,}000 \times 10^{18} = 3.15 \times 10^{23} \text{ FLOPs} ]

Step 4: Order of Magnitude & Re-check

People sometimes write (3 \times 10^{23}) for GPT-3, so our estimate matches the ballpark.

Step 5: Summary Sentence

Estimate:
It would take approximately (3 \times 10^{23}) FLOPs to train GPT-3 (175B parameters, 300B tokens, 6 FLOPs/param/token).

Assumptions:

• 175B parameters
• 300B tokens
• 6 FLOPs per parameter per token

In summary:

Estimated total training compute for GPT-3: (\boxed{sim3 \times 10^{23}}) FLOPs
(based on 175B parameters, 300B tokens, 6 FLOPs/parameter/token)

Let's estimate the total number of floating-point operations (FLOPs) required to train GPT-3, based on what is known about its size, training data, and typical training procedures. Here's a step-by-step reasoning process:

1. Understand the model size and parameters:

• GPT-3 has approximately 175 billion parameters.
• Parameters are stored in FP32 or FP16 precision; often training uses mixed precision to reduce FLOPs slightly.

2. Determine the training dataset size:

• GPT-3 was trained on hundreds of billions of tokens. Reports suggest around 300 billion tokens.
• For simplicity, assume 300 billion tokens.

3. Estimate the number of training steps:

• To process 300 billion tokens, depending on batch size:

  Suppose:

  • Batch size = 1 million tokens (a plausible large batch size for such training, approximating high-performance hardware)

  Number of steps = Total tokens / Batch size:

  ( ext{Steps} = rac{300 imes 10^9}{10^6} = 300,000 )

  But in practice, the batch size per GPU (or per node) is smaller—say 1,000 tokens per batch, with multiple GPUs. The total number of steps is roughly on the order of 300,000 to 500,000.

  To be conservative, take approximately 300,000 steps for total training.

4. FLOPs per forward and backward pass:

• For each token, the transformer computes several matrix multiplications.
• Typically, the dominant cost per token in a transformer model scales with the model size.

A common approximation for FLOPs per token for training a transformer is:

( ext{FLOPs per token} approx 2 imes ext{number of parameters} imes ext{sequence length} )

But this depends on the specifics of the architecture. Historical estimates suggest:

• For large transformer models, about 6 FLOPs per parameter per token are required for training (this accounts for both forward and backward passes). This is a rough estimate from various literature.

Thus,

( ext{FLOPs per token} approx 6 imes ext{parameters} )

Plugging in GPT-3's parameters:

( 6 imes 175 imes 10^9 = 1.05 imes 10^{12} ext{ FLOPs per token} )

Note: This seems high; more refined estimates consider the overall cost per token is roughly on the order of a few trillion FLOPs, depending on the implementation.

Alternatively, a typical approximation used in literature (e.g., OpenAI's blog) is about 3 to 5 FLOPs per parameter per token.

Let's take 3 FLOPs per parameter per token for a more conservative estimate:

( 3 imes 175 imes 10^9 = 525 imes 10^9 = 5.25 imes 10^{11} ext{ FLOPs per token} )

Similarly, for total FLOPs:

( ext{Total FLOPs} approx ext{number of tokens} imes ext{FLOPs per token} )

Using 300 billion tokens and 5.25 × 10^{11} FLOPs per token:

( ext{Total FLOPs} approx 300 imes 10^9 imes 5.25 imes 10^{11} ) ( = (300 imes 5.25) imes 10^{20} ) ( = 1575 imes 10^{20} = 1.575 imes 10^{23} )

5. Adjust for training efficiency and hardware:

• Not every FLOP is used efficiently.
• Typically, the total number of FLOPs gives an upper bound.
• FLOPs estimates often consider the entire training process, including multiple epochs, but most models are trained with many passes over the data; in this case, since the tokens are large and the data volume is huge, training is roughly a single pass or a few passes.

6. Final estimate:

Given the approximations, a rough but reasonable total FLOPs estimate for training GPT-3:

( oxed{ ext{Total FLOPs} sim 10^{23} ext{ FLOPs} } )

Summary:

• Parameters: 175 billion
• Tokens processed: ~300 billion
• FLOPs per token: about 5 × 10^{11}
• Total FLOPs: approximately 10^{23} FLOPs

Answer: The total number of FLOPs required to train GPT-3 is roughly on the order of 10^23 FLOPs. This estimate considers GPT-3's 175 billion parameters, training on approximately 300 billion tokens, and accounts for the computational cost per token during training.

Certainly! Let's estimate the total number of floating-point operations (FLOPs) required to train GPT-3 by breaking down the problem step-by-step. We won't rely on online data but will reason from first principles and known architectural characteristics.

Step 1: Understand GPT-3's basic specifications

GPT-3, the largest model version, has approximately:

• Number of parameters (weights): 175 billion (1.75 × 10¹¹)
• Number of layers: about 96 transformer layers
• Model size: 175B parameters is widely known

Step 2: Estimate FLOPs per forward pass per token

A transformer forward pass consists mainly of:

• Matrix multiplications in self-attention
• Matrix multiplications in feedforward network (FFN)

Two main components dominate FLOPs:

1. Self-attention: roughly proportional to ( d_{model}^2 ) per token (where ( d_{model} ) is hidden size)
2. Feedforward layer: 2 linear layers per transformer layer, roughly (4 d_{model}^2) multiplications per token (due to expanding dimension by factor 4 in FFN)

Step 3: Extract hidden size and sequence length

From papers and common configurations:

• Hidden size ( d_{model} approx 12,288 ) (for 175B model)
• Sequence length used during training ( L approx 2048 ) tokens (max context length)

Step 4: Calculate approximate FLOPs per token per forward pass

Using formulas from transformer FLOPs estimation:

• For each transformer layer, per token, forward pass FLOPs roughly:

[ ext{FLOPs}_ ext{layer/token} approx 2 imes d_{model}^2 ]

(for attention and FFN combined, includes multiplications and additions)

Because the FFN (feedforward network) is the heaviest part, it dominates, so let's assume:

• ( sim 8 imes d_{model}^2 ) FLOPs per token per layer (this is a common estimate for one forward pass)

Substituting:

[ 8 imes (12,288)^2 approx 8 imes 1.5 imes 10^8 = 1.2 imes 10^9 ext{ FLOPs per token per layer} ]

For 96 layers:

[ 1.2 imes 10^9 imes 96 = 1.15 imes 10^{11} ext{ FLOPs per token (forward pass)} ]

Step 5: Account for backward pass

Backpropagation generally costs about 2× to 3× the forward pass FLOPs because computing gradients involves forward pass and backward pass computations.

Use factor 3 to be safe:

[ ext{FLOPs} = 3 imes 1.15 imes 10^{11} = 3.45 imes 10^{11} ext{ FLOPs per token (forward + backward)} ]

Step 6: Calculate total number of tokens used in training

From publicly known info, GPT-3 was trained on about 300 billion tokens (around 3 × 10¹¹ tokens) of text.

Step 7: Calculate total FLOPs for full training

Multiply FLOPs per token by total tokens:

[ 3.45 imes 10^{11} imes 3 imes 10^{11} = 1.035 imes 10^{23} ext{ FLOPs} ]

Final estimation:

[ oxed{ ext{Total FLOPs to train GPT-3} approx 10^{23} quad ext{FLOPs} } ]

Summary:

• Per token forward + backward FLOPs: (sim 3.5 imes 10^{11})
• Number of training tokens: (sim 3 imes 10^{11})
• Total training FLOPs: (sim 10^{23})

Additional notes:

• This estimate matches the order of magnitude widely discussed in ML circles.
• Variations depend on exact training steps, optimizer, precision (FP16/32), etc.
• Might be a bit lower or higher depending on exact implementation.

So, the training of GPT-3 requires on the order of ( 10^{23} ) floating point operations.

Here is one way to get to the oft‐quoted figure of "a few × 10^23 FLOPs" for training GPT‑3. I will make rough but plausible assumptions and carry through the arithmetic step by step:1. Model size – GPT‑3 "175 B" has about N ≈ 1.75×10^11 trainable parameters.2. Amount of training data – The paper says they trained on on the order of 3×10^11 tokens (≈ 300 billion tokens).3. FLOPs per token per parameter – In a forward pass each parameter participates in one multiply + one add → ≈2 FLOPs per parameter per token. – Back‐propagation through that parameter costs on the order of twice as much again (roughly another 4 FLOPs per parameter per token). – We can fold in any small extra cost for optimizer updates and say order‐of‐magnitude it to ≈ 6 FLOPs per parameter per token.4. Total FLOPs – FLOPs ≈ (FLOPs per parameter per token) × (number of parameters) × (number of tokens) – FLOPs ≈ 6 × (1.75×10^11) × (3×10^11) – ≈ 6 × 5.25×10^22 – ≈ 3.15×10^23 floating‐point operationsSo you end up with a training cost on the order of 10^23–10^24 FLOPs (often quoted as ≃3×10^23 FLOPs). That is roughly 300 zettaflops (300×10^21 FLOPs) in total.

Below is a simple "back-of-the-envelope" estimate that uses only a few public facts about GPT-3 and very generic rules of thumb for counting operations.

Step 1. Number of learned parameters
GPT-3's largest version contains ≈175 billion (1.75 × 10¹¹) trainable parameters.

Step 2. How much data was seen during training
OpenAI reports that the model was trained on about 300 billion tokens (3 × 10¹¹).

Step 3. How many floating-point operations does one parameter cost per token?
A practical rule of thumb for Transformer language models is
 Training FLOPs ≈ 6 × (# parameters) × (# training tokens)

Where does the factor 6 come from?

• Forward pass: every parameter is involved in one multiply-accumulate → 2 FLOPs
• Back-prop gradient computation: roughly another 2 FLOPs
• Parameter-update (optimizer) and miscellaneous overhead: ≈2 FLOPs
Total ≈6 FLOPs per parameter per token.

Step 4. Plug in the numbers
Total FLOPs = 6 × (1.75 × 10¹¹) × (3 × 10¹¹)
      = 6 × 5.25 × 10²²
      ≈ 3.15 × 10²³ FLOPs.

Step 5. Express the answer in more familiar units
3.15 × 10²³ ≈ 315 zettaFLOPs (1 ZFLOP = 10²¹ FLOPs).

Step 6. Sanity check with hardware-time intuition
An NVIDIA A100 delivers ≈ 312 teraFLOP/s (0.312 PFLOP/s) in mixed precision.
Wall-clock time on a single A100:
 3.15 × 10²³ / 3.12 × 10¹⁴ ≈ 1.0 × 10⁹ s ≈ 32 years.
Running on 1,000 A100s in parallel would take ≈12 days, which is the right ballpark for a large industrial training run.

Final estimate
Training GPT-3 (175 B parameters, 300 B tokens) requires on the order of 3 × 10²³ floating-point operations, i.e. roughly 300 zettaFLOPs.

Here's a back‐of‐the‐envelope derivation that gets us into the right ballpark (∼10²³ FLOPs total):

1. Model size
   • GPT–3 "175 B" has about P ≃1.75×10¹¹ parameters.

2. Number of training tokens
   • The public paper says "we train on ∼300 billion tokens" (N ≃3×10¹¹).

3. FLOPs per token
   • Every forward+backward through the network touches all parameters.
   • Rough rule of thumb: you need O(6) FLOPs per parameter per token
   (a couple multiplies/adds in the forward pass, a similar count in the backward, and some small extra for optimizer updates). • So 6 × P FLOPs/token ≃ 6×1.75×10¹¹ ≃1.05×10¹² FLOPs/token.

4. Total FLOPs
   • Multiply FLOPs/token by number of tokens:
   Total ≃ (1.05×10¹² FLOPs/token) × (3×10¹¹ tokens)
   ≃ 3.15×10²³ FLOPs.

5. Sanity check
   • 3×10²³ FLOPs is 3×10⁵ PFLOP·days (since 1 PFLOP/s×86 400 s ≈ 8.6×10¹⁹ FLOPs/day).
   • That's a few thousand GPU‐days on a modern accelerator cluster—consistent with the published "thousands of V100‐GPU days" statement.

Bottom‐line estimate: on the order of 10²³–10²⁴ floating‐point operations to train GPT‑3 175 B from scratch.

GPT-3 is one of the largest language models ever created with 175 billion parameters. The number of FLOPs required to train a neural network depends on factors such as the number of parameters, the size of the input data, the number of layers, and the batch size used during training.

To estimate the total number of FLOPs required to train GPT-3, we can consider the following:

1. Size of parameters: GPT-3 has 175 billion parameters. Each parameter needs to be updated during training, so we need to perform a certain number of FLOPs for each parameter update.

2. Number of training iterations: Training a large model like GPT-3 typically requires hundreds or thousands of training iterations. Each iteration involves multiple forward and backward passes through the network, so we need to consider the FLOPs required for each pass.

3. Batch size: The batch size used during training also affects the number of FLOPs required. Larger batch sizes require more FLOPs per iteration.

4. Complexity of the model: GPT-3 is a transformer-based model, which is known to be computationally intensive due to its self-attention mechanism. The FLOPs required for self-attention operations are typically higher compared to other types of neural network operations.

Given the large size of GPT-3 and the complex nature of transformer models, training it would require a massive amount of computational resources. Without searching online for specific numbers, we can estimate that the total number of FLOPs required to train GPT-3 would be in the order of 10^19 to 10^20 FLOPs. This is based on the fact that training large language models like GPT-2 (which has 1.5 billion parameters) required hundreds of petaflops of computation, and GPT-3 is significantly larger in size.

Let's estimate the FLOPs required to train GPT-3 without looking it up. We need to break this down into the key components: the model size and the amount of data used.

1. Model Size Estimation:

• Parameters: GPT-3 is famously large. Its largest version has 175 billion parameters. This is a key number to remember when thinking about its scale.
• Parameter Type: These parameters are typically stored as 32-bit floating-point numbers (FP32). So, each parameter is 4 bytes.
• Total Memory for Parameters: 175 billion parameters * 4 bytes/parameter = 700 billion bytes = 700 GB. This gives a sense of the model's memory footprint, which is related to the computational cost.

2. Data Size Estimation:

• Training Data: Large language models are trained on massive datasets of text. Commoncrawl, a large dataset of web pages, is often a significant component. Other sources like books and articles are also crucial. The exact size used for GPT-3 training is hard to pinpoint without knowing the data mix and filtering, but we're talking about hundreds of billions or even trillions of tokens. Let's estimate a plausible scale for a state-of-the-art model of that era. A common figure for data used for models around GPT-3's time was in the range of 300 billion to 500 billion tokens. Let's take a rough estimate of 500 billion tokens.
• Token Representation: Each token is processed in the model, and the computational cost per token depends on the model's complexity.

3. FLOPs per Operation:

• Matrix Multiplications: The core of transformer models like GPT-3 is matrix multiplication. In a forward pass through a layer, for a batch of size B, a sequence length L, and a hidden dimension H, matrix multiplications involve roughly ( 2 * H * H * L * B ) FLOPs. The self-attention mechanism adds further operations.
• Parameters and Operations: A rough rule of thumb for transformer models is that the forward pass through a single layer for a single token requires approximately ( 2 * ext{parameters_in_layer} ) FLOPs. Since the entire model has 175 billion parameters, and each parameter participates in computations, a simplified way to think about the computation per token is related to the total number of parameters. A common simplification is that the computational cost of a forward pass for one token is roughly ( 2 * ext{total_parameters} ). This comes from the dominant contribution of matrix multiplications where each parameter is involved in roughly two operations (a multiply and an add).

4. FLOPs per Token during Training:

• Training involves Forward and Backward Passes: Training involves a forward pass to calculate the output and loss, and a backward pass to calculate gradients. The backward pass is generally estimated to be about twice as computationally expensive as the forward pass. So, a forward pass + backward pass is roughly ( 3 imes ) the cost of a forward pass.
• FLOPs per Token: Using the simplified rule, the FLOPs per token during training would be approximately ( 3 imes (2 imes ext{total_parameters}) = 6 imes ext{total_parameters} ).
• Applying the Numbers: With 175 billion parameters, the FLOPs per training token is roughly ( 6 imes 175 imes 10^9 = 1050 imes 10^9 = 1.05 imes 10^{12} ) FLOPs per token.

5. Total Training FLOPs:

• Multiply by Data Size: To get the total training FLOPs, we multiply the FLOPs per token by the total number of tokens in the training data.
• Calculation: Total FLOPs ( approx ) (FLOPs per token) ( imes ) (Number of training tokens)
• Total FLOPs ( approx (1.05 imes 10^{12} ext{ FLOPs/token}) imes (500 imes 10^9 ext{ tokens}) )
• Total FLOPs ( approx (1.05 imes 500) imes 10^{12} imes 10^9 ) FLOPs
• Total FLOPs ( approx 525 imes 10^{21} ) FLOPs

6. Converting to ExaFLOPs:

• An exaFLOP (EFLOP) is ( 10^{18} ) FLOPs.
• Total EFLOPs ( approx 525 imes 10^{21} / 10^{18} ) EFLOPs
• Total EFLOPs ( approx 525 imes 10^3 ) EFLOPs = 525,000 EFLOPs

Refinement and Considerations:

• Simplifications: The "6 * parameters" rule is a simplification. The exact number of operations per token is more complex and depends on the specific architecture details (number of layers, attention heads, etc.), but this provides a reasonable order-of-magnitude estimate.
• Data Duplication/Epochs: Models are often trained for multiple "epochs," meaning they see the data multiple times. The 500 billion tokens is likely the size of the unique dataset. If it were trained for, say, an average of 2 epochs, the total effective tokens processed would be ( 2 imes 500 ) billion = 1 trillion tokens. This would double our FLOP estimate to around 1000-1050 EFLOPs. However, often the 500 billion figure already represents the total tokens processed over all epochs, or it's used as the scale number. Let's stick with the initial 500 billion tokens as the total processed amount for the estimate.
• Training Efficiency: Factors like computational efficiency of the hardware, batch size, and data parallelism can affect the time to train, but the total amount of computation (FLOPs) remains roughly the same for a given model and dataset size, assuming ideal scaling.

Conclusion and Final Estimate:

Based on the assumptions of 175 billion parameters trained on roughly 500 billion tokens, with the rule of thumb of 6 FLOPs per parameter per token during training, the estimated total FLOPs required to train GPT-3 is in the range of 500 to 600 ExaFLOPs.

Our calculation landed at around 525 ExaFLOPs. This aligns with general estimates floating around for models of this scale and era. The exact published number might differ slightly due to precise details of the architecture, data size, and training methodology, but this step-by-step reasoning provides a good order of magnitude estimate.

Estimated Total FLOPs: ~500-600 ExaFLOPs

To estimate the total number of FLOPs required to train GPT-3 without searching online, we can make several assumptions and perform calculations based on the information available about GPT-3's architecture and training process. Here's a step-by-step explanation of the reasoning:
Step 1: Determine the number of parameters in GPT-3

GPT-3 is a massive language model with 175 billion parameters. Each parameter represents a trainable weight in the neural network, and updating these weights during training requires floating-point operations (FLOPs).

Step 2: Estimate the number of training iterations

The number of training iterations refers to how many times the entire training dataset is passed through the neural network during the training process. Although the exact number may vary, let's assume that GPT-3 was trained for 100 epochs.

Step 3: Calculate the FLOPs per iteration

During each training iteration, the neural network performs forward and backward passes. The forward pass involves computing the output of the network given the input, while the backward pass involves calculating the gradients of the loss function with respect to the network's weights.
For simplicity, we can assume that the forward and backward passes have roughly the same number of FLOPs. Therefore, the total FLOPs per iteration can be estimated by multiplying the number of parameters by 2 (forward and backward passes).

Total FLOPs per iteration = (Number of parameters) x 2

(175 billion parameters) x 2 = 350 billion FLOPs per iteration

Step 4: Calculate the total FLOPs for training

Multiplying the FLOPs per iteration by the number of training iterations gives us the total FLOPs required to train GPT-3.

Total FLOPs for training = (FLOPs per iteration) x (Number of iterations)

(350 billion FLOPs per iteration) x (100 iterations) = 35 trillion FLOPs for training

Therefore, based on these assumptions, we estimate that training GPT-3 without searching online would require approximately 35 trillion FLOPs.

Estimating GPT-3 Training FLOPs: A Step-by-Step Approach

Without access to online resources, we can estimate the FLOPs (Floating-point Operations) needed for GPT-3 training by considering the following factors:

1. Model Size:

• GPT-3 has 175 billion parameters.
• Each parameter update involves two multiplications and one addition for each floating-point number.
• Therefore, each parameter update requires 3 FLOPs.
• Total FLOPs per update: 175 billion parameters * 3 FLOPs/parameter = 525 billion FLOPs.

2. Training Data Size:

• GPT-3 was trained on a massive dataset of text and code.
• Let's assume the dataset size is 500 billion words.
• Each word is represented by an embedding vector of size 1,024.
• Therefore, the total number of floating-point numbers in the dataset is 500 billion words * 1,024 = 512 trillion.

3. Number of Training Steps:

• The exact number of training steps is unknown, but it likely involves millions of iterations over the entire dataset.
• Let's assume there are 10 million training steps.

4. Optimization Algorithm:

• GPT-3 likely uses Adam optimizer, which involves additional computations for updating parameters.
• The exact FLOPs associated with the optimizer are difficult to estimate without knowing the specific hyperparameters used.
• For simplicity, we can assume an additional 10 FLOPs per parameter update.

5. Hardware Efficiency:

• Modern hardware like GPUs and TPUs can perform several FLOPs per clock cycle.
• Let's assume an average efficiency of 5 FLOPs per clock cycle.

Putting it all together:

• Total FLOPs per update: 525 billion (parameter updates) + 10 billion (optimizer) = 535 billion FLOPs
• Total FLOPs for training: 535 billion FLOPs/update * 10 million updates = 5.35 * 10^18 FLOPs
• Adjusted for hardware efficiency: 5.35 * 10^18 FLOPs / 5 FLOPs/cycle = 1.07 * 10^18 clock cycles

Therefore, the estimated total FLOPs required to train GPT-3 without searching online is approximately 1.07 * 10^18 clock cycles.

Important Caveats:

• This is a rough estimate and may not be accurate due to several unknown factors, including the exact training data size, number of training steps, optimizer settings, and hardware efficiency.
• The actual FLOPs could be significantly higher or lower depending on the specific training details.

Additionally, it's important to remember that this is just an estimation of the computational cost. Training a model like GPT-3 also requires significant energy consumption and environmental impact.

Let's break down estimating the FLOPs for GPT-3 training. Since we can't look up exact figures, we'll rely on reasonable guesses and approximations.

1. Model Size: GPT-3 has 175 billion parameters. This is our anchor point.

2. FLOPs per Parameter per Token: A forward pass involves a matrix multiplication for each layer. Let's assume an average of 2 FLOPs per parameter per token processed (one multiplication and one addition). This is a simplification, as different layers might have slightly different computations, but it's a reasonable starting point.

3. Dataset Size: GPT-3 was trained on a massive dataset. Let's approximate the effective dataset size (accounting for repeated passes) as 100 billion tokens. This might be an underestimate, but given it was trained on a mix of filtered and unique content, and we don't know the exact number of epochs, it's a starting point.

4. Forward and Backward Passes: For each token, we do one forward pass and one backward pass for gradient computation. So, we have 2 passes.

5. Optimizer overhead: Optimizers like Adam add some computational overhead. Let's add a 20% overhead factor to account for this.

Now, let's put it all together:

• FLOPs per token = 175B parameters * 2 FLOPs/parameter * 2 passes = 700B FLOPs/token

• Total FLOPs = 700B FLOPs/token * 100B tokens = 7 x 10^22 FLOPs

• Optimizer overhead: 7 x 10^22 * 1.2 = 8.4 x 10^22 FLOPs

Therefore, a very rough estimate is around 8.4 x 10^22 FLOPs to train GPT-3.

It's crucial to remember that this is a back-of-the-envelope calculation. The actual number could be higher or lower due to factors such as:

• Variations in layer computations: Different layers might have different computational complexity.
• Optimization techniques: Specific optimizations during training could reduce the actual FLOPs.
• Data preprocessing and tokenization overhead: We haven't accounted for these factors.
• Dataset repetitions/epochs: The assumed effective dataset size is a guess.
• Architecture specifics: The actual architecture and training process might involve more complexities not captured here.

So, while our estimate provides a general order of magnitude, it shouldn't be taken as a precise figure.

Okay, let's estimate the FLOPs for training GPT-3. I'll break this down.

Key Formula: The number of FLOPs for training a transformer-based model can be roughly estimated as: FLOPs ≈ 6 * N * D Where:

• N is the number of parameters in the model.
• D is the total number of tokens processed during training (dataset size in tokens * number of epochs).
• The factor of 6 comes from:
  • 2 * N FLOPs per token for the forward pass (each parameter is involved in roughly one multiply-accumulate, and a MAC is 2 FLOPs).
  • 4 * N FLOPs per token for the backward pass (gradient computation is roughly twice as expensive as the forward pass).

Assumptions for GPT-3 (Davinci, the largest version):

1. Number of Parameters (N):

   • GPT-3 (the largest variant, Davinci) is famously stated to have 175 billion parameters.
   • So, N = 175 * 10^9.

2. Number of Tokens Processed (D):

   • This is the trickier part to recall without searching. I remember it was trained on a massive amount of data, a significant portion of Common Crawl, books, Wikipedia, etc.
   • Large Language Models are typically trained for roughly one epoch over their massive datasets, or even less if datasets are mixed and weighted. The key figure is the total number of tokens seen by the model.
   • I seem to recall a figure in the range of 300 billion to 500 billion tokens for GPT-3's training. The number often cited in discussions around the GPT-3 paper and subsequent scaling law papers (like Chinchilla) is around 300 billion tokens. Let's use that as a reasonable estimate for D.
   • So, D = 300 * 10^9.

Calculation:

Now, let's plug these into the formula: FLOPs ≈ 6 * N * D FLOPs ≈ 6 * (175 * 10^9) * (300 * 10^9)

Let's do the numerical part: 6 * 175 = 1050 1050 * 300 = 315,000

Now, the powers of 10: 10^9 * 10^9 = 10^18

So, combining these: FLOPs ≈ 315,000 * 10^18 FLOPs ≈ 3.15 * 10^5 * 10^18 FLOPs ≈ 3.15 * 10^(5+18) FLOPs ≈ 3.15 * 10^23

Expressing the Result:

• TeraFLOPs (TFLOPs) = 10^12
• PetaFLOPs (PFLOPs) = 10^15
• ExaFLOPs (EFLOPs) = 10^18
• ZettaFLOPs (ZFLOPs) = 10^21
• YottaFLOPs (YFLOPs) = 10^24

So, 3.15 * 10^23 FLOPs can be written as:

• 315 * 10^21 FLOPs = 315 ZettaFLOPs
• Or, as 0.315 * 10^24 FLOPs = 0.315 YottaFLOPs

The number is often quoted in PetaFLOP/s-days or similar. Let's convert 3.15 * 10^23 FLOPs to PetaFLOP-seconds: 3.15 * 10^23 / 10^15 = 3.15 * 10^8 PetaFLOP-seconds.

Number of seconds in a day = 24 hours/day * 60 minutes/hour * 60 seconds/minute = 86,400 seconds/day. So, 3.15 * 10^8 PetaFLOP-seconds / 86,400 seconds/day ≈ 3.15 * 10^8 / (8.64 * 10^4) PetaFLOP-days ≈ (3.15 / 8.64) * 10^(8-4) PetaFLOP-days ≈ 0.3646 * 10^4 PetaFLOP-days ≈ 3646 PetaFLOP/s-days (if executed on a machine capable of 1 PFLOP/s constantly for that many days, or scaled accordingly).

Reasoning Summary:

1. Identified the core formula: FLOPs ≈ 6 * N * D for transformer training.
2. Recalled model parameters (N): GPT-3 Davinci has 175 billion parameters.
3. Estimated training tokens (D): Based on common knowledge about LLM training and GPT-3 specifics, estimated ~300 billion tokens processed. This assumes roughly one pass over the (weighted) training data.
4. Performed the multiplication: 6 * (175 * 10^9) * (300 * 10^9) = 3.15 * 10^23 FLOPs.
5. Interpreted the result: This is approximately 315 ZettaFLOPs. This also aligns with the often-cited figure for GPT-3 training compute, which is in the range of 3000-4000 PetaFLOP/s-days (my calculation of ~3646 PetaFLOP/s-days is right in this ballpark).

Caveats:

• The 6ND is an approximation. The exact number can vary slightly based on the specific transformer architecture details (e.g., use of sparse attention, specific normalization layers), optimizer overhead, and whether techniques like activation checkpointing are used (which can increase FLOPs for recomputation but save memory).
• Mixed-precision training (e.g., bfloat16 or fp16) doesn't change the number of FLOPs, but it changes the type of FLOPs and allows for faster computation on hardware that supports it.
• The number of training tokens D is an estimate. The GPT-3 paper describes a complex data mixture and weighting schedule, so "300 billion tokens" is an effective number of tokens processed.

Therefore, my estimate is approximately 3.15 x 10^23 FLOPs or 315 ZettaFLOPs.

To estimate the total number of FLOPs (Floating Point Operations) required to train GPT-3, we'll break down the calculation into several steps, considering the architecture, number of parameters, and training process .

1. Model architecture and parameters: GPT-3 has 175 billion parameters. Most of these parameters are in the dense layers. For a rough estimate, we'll assume that the model has ( L ) layers, and each layer has a pproximately ( \frac{175 \times 10^9}{L} ) parameters.

2. Forward pass: In a transformer model, the main computation happens in the multi-head self-attention a nd feed-forward networks. For a single token, the computation involves multiplying the input by weight matri ces. So, for one layer, the FLOPs required are approximately ( 2 \times N \times \frac{175 \times 10^9}{L} ), where ( N ) is the dimensionality of the embeddings (typically around 12,288 for GPT-3) and the factor 2 comes from the multiplication and addition in a matrix-vector product.

3. Backward pass: Training involves both a forward pass and a backward pass. The backward pass has rough ly twice the computational cost of the forward pass due to the need to compute gradients and perform backpro pagation. So, the total FLOPs for one layer and one token during training are approximately ( 6 \times N \t imes \frac{175 \times 10^9}{L} ).

4. Sequence length and batch size: GPT-3 uses a context window of 2,048 tokens, and let's assume a batch size of ( B ). The computation for each token depends on all previous tokens in the sequence, so the effe ctive number of tokens is ( 2048^2 ) for each batch item. Thus, the total FLOPs for one layer and one batc h are ( 6 \times N \times \frac{175 \times 10^9}{L} \times 2048^2 \times B ).

5. Total layers and training steps: Now, we multiply by the number of layers ( L ) and the total numbe r of training steps ( S ). Assuming ( L = 96 ) (as mentioned in the GPT-3 paper) and ( S = 300,000 ) ( a rough estimate based on typical training durations), we get:

   Total FLOPs = ( 6 \times N \times 175 \times 10^9 \times 2048^2 \times B \times L \times S )

6. Putting it all together: Plugging in the values ( N = 12,288 ), ( B = 512 ) (a reasonable batch s ize), ( L = 96 ), and ( S = 300,000 ), we get:

   Total FLOPs ≈ ( 6 \times 12,288 \times 175 \times 10^9 \times 2048^2 \times 512 \times 96 \times 300,000 )

   This works out to approximately ( 3.6 \times 10^{26} ) FLOPs.

So, our rough estimate is that training GPT-3 requires around ( 3.6 \times 10^{26} ) FLOPs. This is a very rough estimate, as the exact number depends on the specific details of the architecture, implementation, an d training process.

Of course. Here is an estimate of the total number of FLOPs required to train GPT-3, based on my internal kn owledge and first principles, without searching online.

The Logic and Formula

To estimate the computational cost of training a large language model like GPT-3, we can use a well-establis hed formula. The total number of Floating Point Operations (FLOPs) is approximately:

Total FLOPs ≈ 6 × P × D

Let's break down what each part of this formula means:

• P: The number of parameters in the model. This is the total count of all the weights and biases that the model learns during training.
• D: The total number of tokens in the training dataset. This represents the volume of data the mo del processes.
• 6: This is a constant factor that provides a good approximation for transformer-based models. It com es from:
  • Forward Pass (1 × P × D): For each token processed, the number of operations is roughly proporti onal to the number of parameters. A common rule of thumb is 2 * P FLOPs per token for a forward pass (one multiplication and one addition for each parameter). So the total for one pass over the data is 2 * P * D.
  • Backward Pass (2 × Forward Pass): The backward pass (backpropagation), where the model calculate s gradients to update its parameters, is computationally more intensive. It's typically estimated to require about twice the operations of the forward pass. So, 2 * (2 * P * D) = 4 * P * D.
  • Total: Adding the forward and backward pass gives us 2PD + 4PD = 6PD.

Now, let's estimate the values for P and D for GPT-3.

Step-by-Step Estimation

Step 1: Estimate the Number of Parameters (P)

My internal knowledge base indicates that the largest version of GPT-3, which is the one most people refer t o, has a specific number of parameters.

• P ≈ 175 billion parameters
• In scientific notation, this is 175 x 10⁹

Step 2: Estimate the Training Dataset Size (D)

GPT-3 was trained on a massive corpus of text data scraped from the internet (like the Common Crawl dataset) , books, and Wikipedia. The key is not the raw size of the dataset, but the number of tokens actually used f or training. My recollection is that the model was trained for a single epoch on a curated dataset.

• D ≈ 300 billion tokens
• In scientific notation, this is 300 x 10⁹

Step 3: Perform the Calculation

Now we can plug these estimated values into our formula:

Total FLOPs ≈ 6 × P × D

1. Substitute the values for P and D: Total FLOPs ≈ 6 × (175 × 10⁹) × (300 × 10⁹)

2. Multiply the constants first: 6 × 175 = 1050 1050 × 300 = 315,000

3. Multiply the powers of 10: 10⁹ × 10⁹ = 10¹⁸

4. Combine the results: Total FLOPs ≈ 315,000 × 10¹⁸

5. Express this in standard scientific notation: Total FLOPs ≈ 3.15 × 10⁵ × 10¹⁸ = 3.15 × 10²³

Final Estimate and Conclusion

Based on my reasoning and the standard formula for calculating training costs, the estimated number of FLOPs required to train the 175B parameter GPT-3 model is approximately:

3.15 x 10²³ FLOPs

To put this astronomical number into perspective:

• A petaFLOP is 10¹⁵ FLOPs. Our estimate is 315,000,000 petaFLOPs.
• If you had a supercomputer that could perform one petaFLOP per second (10¹⁵ FLOPs/sec), it would take ** 3.15 x 10⁸ seconds** to complete the training.
• There are about 31.5 million seconds in a year, so this is equivalent to roughly 10 years of continu ous computation on that machine.

This calculation demonstrates why training foundational models like GPT-3 requires massive-scale data center s with thousands of GPUs running in parallel for months and is an undertaking that costs millions of dollars . The actual paper from OpenAI stated the training took approximately 3.6 x 10²³ FLOPs, so my estimate i s very close to the reported figure.

I cannot provide a specific estimate of the total number of FLOPs required to train GPT-3, as this information is proprietary and not publicly disclosed by the creators of GPT-3. However, I can explain the general process of estimating the number of FLOPs required to train a large language model like GPT-3.

To estimate the number of FLOPs (floating-point operations) required to train a large language model, we need to consider several factors, including the model architecture, the number of parameters, and the training process. Here's a step-by-step explanation of how to make a rough estimate:

1. Model Architecture: Understand the architecture of the model. For GPT-3, which is a transformer-based model, the key components are the attention mechanisms and the feedforward neural networks.

2. Number of Parameters: Determine the total number of parameters in the model. GPT-3 has been reported to have 175 billion parameters. Each parameter requires multiple FLOPs for both forward and backward passes during training.

3. Forward Pass: Calculate the FLOPs required for a forward pass. For a transformer model, the main operations are matrix multiplications in the attention mechanisms and the feedforward layers. For each layer, the number of FLOPs is roughly proportional to the square of the number of parameters in that layer.

4. Backward Pass: The backward pass, which involves backpropagation through the network, typically requires a similar number of FLOPs as the forward pass, but this can vary depending on the optimization algorithm used.

5. Number of Layers and Heads: Consider the number of layers and attention heads in the model. More layers and heads increase the number of FLOPs linearly.

6. Batch Size: The batch size affects the total number of FLOPs. Larger batch sizes increase the number of FLOPs per training step but can allow for more parallel computation.

7. Number of Training Steps: Determine the number of training steps. This depends on the size of the dataset and the number of epochs (full passes through the dataset). For large models like GPT-3, training often involves many iterations over a massive dataset.

8. Optimizer Overhead: Some optimizers, like Adam, have additional overhead due to the computation of adaptive learning rates and momentum terms.

9. Activation Functions: Non-linear activation functions (like ReLU or GELU) also contribute to the FLOP count, though usually to a lesser extent compared to matrix operations.

10. Gradient Calculation: Calculating gradients during backpropagation involves additional FLOPs, roughly equivalent to the forward pass.

Given these factors, a very rough estimate of the total number of FLOPs can be made. For example, assuming each parameter requires around 6 FLOPs (for both forward and backward passes), and considering the number of parameters, layers, and training steps, the total number of FLOPs can be in the order of 10^23 or more. However, this is a very rough estimate and the actual number could vary significantly based on the specific architectural details and optimizations used during training.

For precise numbers, one would need access to detailed architectural specifications and training configurations, which are typically not publicly available.

Below, I'll estimate the total number of floating-point operations (FLOPs) required to train GPT-3 from scratch, based on publicly known details about the model (from its original release and related analyses) and standard formulas used in machine learning research for estimating training compute. I'll explain my reasoning step by step, including the assumptions I'm making and why they're reasonable. Note that this is an estimation—exact numbers aren't publicly disclosed in full detail by OpenAI, and real-world training involves optimizations, inefficiencies, and hardware-specific factors that could adjust the final figure. I'm not searching online; this is based on my pre-existing knowledge of large language model (LLM) training.

Step 1: Recall Key Facts About GPT-3

• Model size: GPT-3 (specifically, the largest variant, often called "GPT-3 175B") has approximately 175 billion parameters. This is the total number of trainable weights in the model.
• Training data: It was trained on a massive dataset, including filtered Common Crawl data, books, Wikipedia, and other sources. The total number of tokens (subword units used in training) is estimated to be around 300 billion. This is based on the reported dataset size (about 570 GB of text) and typical tokenization rates for models like GPT (where 1 token ≈ 4 characters, leading to roughly 300 billion tokens from that volume of data).
• Training setup: GPT-3 is an autoregressive transformer model trained using supervised fine-tuning on next-token prediction (essentially, language modeling). It was trained for roughly one epoch (one full pass) through the dataset, which is common for large models to avoid overfitting and due to data scarcity at that scale.

These numbers are widely cited in AI research papers and discussions around scaling laws (e.g., work from OpenAI and DeepMind on model scaling).

Step 2: Understand the Standard FLOPs Estimation Formula for LLM Training

Training a transformer-based LLM like GPT-3 involves:

• A forward pass: Computing predictions for each token in the input sequence (involves matrix multiplications across the model's layers).
• A backward pass: Computing gradients to update the model's parameters (via backpropagation).
• This happens for every token in the training dataset.

A widely used rule of thumb from AI scaling research (e.g., papers like "Scaling Laws for Neural Language Models" by Kaplan et al. from OpenAI) estimates the total training FLOPs as:

Total FLOPs ≈ 6 × P × T

Where:

• P = number of parameters (e.g., 175 billion for GPT-3).
• T = total number of tokens processed during training (e.g., 300 billion).
• 6 is a multiplier that accounts for:
  • Forward pass: ~2×P FLOPs per token (mainly matrix-vector multiplications involving all parameters).
  • Backward pass: ~4×P FLOPs per token (backpropagation is roughly twice as expensive as forward).
  • Total per token: ~6×P FLOPs.
  • This assumes dense operations (no sparsity) and ignores minor overhead like activations or normalization, which is reasonable for estimation.

This formula is a simplification but matches empirical reports for models like GPT-2, GPT-3, and others. It focuses on the dominant compute cost (parameter updates) and doesn't include inference-time FLOPs, data loading, or evaluation— just training.

Why 6? It's derived from counting operations in transformer layers (attention, feed-forward networks). Some papers use slightly different multipliers (e.g., 3× or 4× if only counting multiply-accumulates), but 6× is the standard for total FLOPs in scaling law discussions.

Step 3: Plug in the Numbers

• P = 175 billion = 1.75 × 10^11
• T = 300 billion = 3 × 10^11
• Multiplier = 6

Total FLOPs = 6 × (1.75 × 10^11) × (3 × 10^11)

First, compute 1.75 × 3 = 5.25
Then, 5.25 × 10^(11+11) = 5.25 × 10^22
Finally, 6 × 5.25 × 10^22 = 31.5 × 10^22 = 3.15 × 10^23

So, my estimate is approximately 3 × 10^23 FLOPs (rounding to one significant figure for simplicity, as these are ballpark numbers).

Step 4: Validate and Adjust for Uncertainties

• Is the token count accurate? 300 billion is a common estimate based on the GPT-3 paper's description of the dataset (45 TB of compressed text filtered down to 570 GB). Token counts can vary with the tokenizer (GPT uses byte-pair encoding), but this is within 10-20% of reported figures. If the effective tokens were higher (e.g., due to repeated data or sampling), it could bump up to 400-500 billion, increasing FLOPs by ~30-60%, but I think 300 billion is conservative and standard.
• Is the multiplier accurate? Some analyses use 4×P×T (focusing only on multiply-accumulates), which would give ~2 × 10^23 FLOPs. Others include optimizer overhead (e.g., Adam optimizer adds ~10-20% more). I've seen independent estimates around 3.7 × 10^23 FLOPs, which is close to mine—differences could come from including setup costs or hardware inefficiencies.
• Other factors:
  • Parallelism and hardware: GPT-3 was trained on a supercomputer-scale cluster (e.g., thousands of V100 GPUs). Real FLOPs include communication overhead in distributed training, which might add 10-20% inefficiency, but the formula ignores this for simplicity.
  • Precision: Training likely used mixed precision (e.g., FP16), which reduces FLOPs slightly but doesn't change the order of magnitude.
  • Epochs: If there was any multi-epoch training or warm-starting, it could increase FLOPs, but reports suggest mostly one epoch.
  • Order of magnitude check: Smaller models like GPT-2 (1.5 billion params, ~10 billion tokens) required ~10^19 FLOPs. Scaling up by factors of ~100x params and ~30x tokens gives ~10^19 × 3,000 = ~3 × 10^22, but wait—that's off because the formula is multiplicative. Actually, GPT-2 estimates are lower; adjusting properly lands in the 10^23 range for GPT-3, which aligns with reports of it taking thousands of GPU-years.

If I adjust for potential underestimation (e.g., higher token count or overhead), the upper bound might be 5 × 10^23 FLOPs, but I stick with 3 × 10^23 as the core estimate.

Step 5: Convert to Intuitive Units (Optional Context)

To make this relatable:

• 3 × 10^23 FLOPs is enormous—equivalent to a high-end GPU (e.g., 10 TFLOPs/second) running continuously for about 1,000 years (rough calc: 10^13 FLOPs/sec × 3 × 10^7 sec/year × 1,000 years ≈ 3 × 10^23).
• In terms of energy, this might consume gigawatt-hours of electricity, but that's beyond FLOPs.

Final Estimate

Based on this reasoning, I estimate the total FLOPs for training GPT-3 at around 3 × 10^23. This is consistent with what I'd expect from scaling laws and matches informal discussions in the AI community. If new details emerged (e.g., exact token counts), it could shift by a factor of 2, but the order of magnitude (10^23) is solid. If you have additional details or want me to adjust assumptions, let me know!

To estimate the total number of FLOPs (floating point operations) required to train GPT-3, we'll use the following steps:

1. Model size: GPT-3 has 175 billion parameters.
2. Training tokens: GPT-3 was trained on 300 billion tokens.
3. FLOPs per token: To estimate the FLOPs per token, we'll use the rule of thumb that training a transformer model requires about 6 FLOPs per parameter per token.
4. Total FLOPs: Multiply the number of parameters by the number of tokens and the FLOPs per token.

Now let's calculate:

1. Model size: 175 billion parameters = 175 * 10^9
2. Training tokens: 300 billion tokens = 300 * 10^9
3. FLOPs per token: 6 FLOPs per parameter per token
4. Total FLOPs: (175 * 10^9) * (300 * 10^9) * 6 = 3.15 * 10^23 FLOPs

So, the estimated total number of FLOPs required to train GPT-3 is approximately 3.15 * 10^23.

To estimate the total number of FLOPs (Floating Point Operations) required to train GPT-3, we'll break down the process into several steps and make reasonable assumptions based on the information we have about GPT-3 and typical training processes for large language models.

1. Model size and parameters: GPT-3 has 175 billion parameters. This is a crucial piece of information as it directly impacts the number of FLOPs required for training.

2. Training data size: GPT-3 was trained on a large dataset, but we don't know the exact size. For this estimation, let's assume it's similar to the dataset used for training GPT-2, which is around 40GB of text data. This is a rough estimate and the actual size could be larger.

3. Training steps: The number of training steps depends on the batch size and the number of epochs (or passes) through the dataset. Let's assume a batch size of 1,000 tokens (which is reasonable for a model of this size) and 10 epochs. This gives us 40,000,000,000 tokens / 1,000 tokens/batch = 40,000,000 batches, and 40,000,000 batches * 10 epochs = 400,000,000 training steps.

4. FLOPs per training step: The number of FLOPs required for each training step depends on the model architecture and the specific operations performed during training. For a transformer-based model like GPT-3, the most computationally intensive operations are matrix multiplications. Let's assume that each training step requires approximately 10 FLOPs per parameter (this is a rough estimate and the actual number could be different). This gives us 175,000,000,000 parameters * 10 FLOPs/parameter = 1,750,000,000,000 FLOPs per training step.

5. Total FLOPs: Now, we can calculate the total number of FLOPs required for training by multiplying the number of FLOPs per training step by the number of training steps. This gives us 1,750,000,000,000 FLOPs/step * 400,000,000 steps = 700,000,000,000,000,000,000 FLOPs.

So, our rough estimate for the total number of FLOPs required to train GPT-3 is 700 quadrillion FLOPs. This is a very rough estimate and the actual number could be different, but it gives us a sense of the scale of the computation involved in training a large language model like GPT-3.

Here is one way to “Fermi-estimate” the training compute for GPT-3 without looking up any numbers. I will start from generally known facts (model size, data size, hardware specs) and work forward.

Step 1 – Fix the model parameters
A widely reported figure is that GPT-3 has 175 B parameters.
N ≈ 1.75 × 10¹¹.

Step 2 – Fix the training-token budget
OpenAI papers mention ~300 B tokens.
T ≈ 3 × 10¹¹.

Step 3 – Recall the per-token FLOP rule for dense transformers
Every time one token is processed, the forward pass does roughly
2 FLOPs/parameter (multiply-accumulate counts as 2).
The backward pass does about twice the work of the forward pass, so
total ≈ 3× forward FLOPs.
Thus the “textbook” estimate is
6 FLOPs · N · T.

Step 4 – Plug the numbers in
FLOPs = 6 · (1.75 × 10¹¹) · (3 × 10¹¹)
= 6 · 5.25 × 10²²
≈ 3.15 × 10²³ FLOPs.

Step 5 – Add a safety factor for real-world overheads
Large-scale jobs typically run at 30-50 % of theoretical peak because of communication, stragglers, recomputation for activation checkpointing, etc.
A factor of 2 covers this nicely.
3.15 × 10²³ × 2 ≈ 6 × 10²³ FLOPs.

Final estimate
Training GPT-3 required on the order of 3–6 × 10²³ floating-point operations.

Okay, let's estimate the total FLOPs (Floating Point Operations) required to train GPT-3 without relying on external searches. This will be a back-of-the-envelope calculation, relying on reasonable assumptions and approximations. I'll break down the process into steps, explaining the assumptions at each point.

1. GPT-3 Model Parameters & Architecture

• Parameters: GPT-3 has 175 billion parameters. This is a crucial starting point.
• Architecture: It's a Transformer model, which primarily consists of:
  • Attention Layers: The core of Transformers. These involve matrix multiplications and softmax operations.
  • Feed-Forward Networks (FFNs): Multi-layer perceptrons applied to each token's representation.
  • Embeddings: Mapping tokens to vector representations.
  • Layer Normalization: Stabilizes training.
  • Residual Connections: Help with gradient flow.

2. FLOPs per Parameter Update (Forward & Backward Pass)

This is the most challenging part to estimate without precise architectural details. We'll work from established research on Transformer FLOPs.

• General Rule of Thumb: A common rule of thumb for estimating FLOPs per parameter update in a Transformer is approximately 3-6x the number of parameters. This accounts for the matrix multiplications in attention and FFN layers, as well as other operations. Let's use a value of 5x as a middle ground. This is where the biggest uncertainty lies; different layers and implementations have different FLOPs/parameter ratios.
• FLOPs per Update: Therefore, FLOPs per parameter update ≈ 175 billion parameters * 5 = 875 billion FLOPs (8.75 x 10¹¹ FLOPs).

3. Training Dataset Size

• Estimated Tokens: GPT-3 was trained on a massive dataset. Estimates range widely, but a reasonable estimate is around 300 billion tokens. Let's use 300 billion tokens (3 x 10¹¹ tokens).
• Tokens per Batch: During training, tokens are processed in batches. GPT-3 used a batch size of 32,768 tokens. Let's use 32,768
• Number of Updates: Number of updates = (Total tokens) / (Tokens per batch) = (3 x 10¹¹) / 32,768 ≈ 9.16 billion updates (9.16 x 10⁹ updates).

4. Total FLOPs for Training

• Total FLOPs: (FLOPs per update) * (Number of updates) = (8.75 x 10¹¹ FLOPs/update) * (9.16 x 10⁹ updates) ≈ 8.01 x 10²¹ FLOPs.

5. Considerations and Refinements (Important!)

• Optimizer: The optimizer (e.g., Adam) introduces additional FLOPs. Adam typically requires 2-3x the FLOPs of the forward and backward passes (due to calculating and updating momentum and variance estimates). Let's conservatively add a factor of 2.
  • Total FLOPs (with Optimizer) ≈ 8.01 x 10²¹ FLOPs * 2 = 1.60 x 10²² FLOPs
• Data Loading and Preprocessing: Data loading, tokenization, and other preprocessing steps consume FLOPs, but are relatively minor compared to the model's computations. We'll ignore them for simplicity.
• Sparse Attention: GPT-3 used sparse attention mechanisms, which reduce the computational cost of the attention layers. This would reduce the FLOPs, but it's difficult to quantify without specifics.
• Mixed Precision Training: GPT-3 used mixed precision training (FP16), which can reduce FLOPs. This is more about reducing memory bandwidth than FLOPs themselves, but it can affect overall training time.
• Activation Checkpointing: This technique reduces memory usage by recomputing activations during the backward pass, trading computation for memory. This increases FLOPs.

Final Estimate & Range

Based on these assumptions and considerations:

• Base Estimate: Approximately 1.60 x 10²² FLOPs (160 ZettaFLOPs).
• Range: Given the uncertainties in the FLOPs per parameter update, the optimizer's overhead, and the impact of sparse attention and activation checkpointing, a reasonable range would be 1.0 x 10²² - 3.2 x 10²² FLOPs (100-320 ZettaFLOPs).

Key Assumptions Recap:

• 5x FLOPs per parameter update (a major source of uncertainty).
• 300 billion tokens in the training dataset.
• Batch size of 32,768
• Optimizer overhead of 2x.
• Neglecting data loading/preprocessing FLOPs.

Important Disclaimer: This is a rough estimate. The actual number of FLOPs could be significantly higher or lower depending on the specific implementation details and optimizations used to train GPT-3. The range reflects the uncertainties in these assumptions.

Okay, let's estimate the FLOPs required to train GPT-3. This is a back-of-the-envelope calculation, relying on publicly available information and reasonable assumptions. It's notoriously difficult to get a precise number, as OpenAI doesn't release all the details. We'll break it down step-by-step.

1. GPT-3 Model Size (Parameters)

• GPT-3 has 175 billion parameters. This is the foundation of our calculation. Each parameter is a weight in the neural network that needs to be updated during training.

2. Training Dataset Size

• GPT-3 was trained on a huge dataset. OpenAI stated approximately 300 billion tokens. A token can be roughly considered a word or part of a word. Let's use 300 billion tokens as our base.
• The dataset comprised sources like Common Crawl, WebText2, Books1, and Books2. The exact breakdown and filtering are proprietary, but we have the overall size.

3. FLOPs per Token

This is the trickiest part. The number of FLOPs (Floating Point Operations) per token processed depends heavily on the model architecture (Transformer in this case) and implementation details. We'll use a common approximation based on research into large language models.

• Transformer Architecture: Transformers have a complexity that scales roughly with the sequence length (number of tokens in a sequence) squared, and also with the model size (number of parameters). The core operations are matrix multiplications.
• FLOPs per Parameter per Token: A reasonable estimate for FLOPs per parameter per token for a model like GPT-3 during the forward and backward pass (which are both needed for training) is around 3. This comes from analyses of similar models (e.g., Chinchilla, PaLM). This includes the attention mechanism, matrix multiplications in the feedforward layers, and gradient calculations.
• Total FLOPs per Token: Therefore, FLOPs per token ≈ (Number of Parameters) * (FLOPs per Parameter per Token)
  • FLOPs per token ≈ 175 billion parameters * 3 FLOPs/parameter/token = 525 billion FLOPs/token = 5.25 x 10¹¹ FLOPs/token

4. Total FLOPs for Training

• Total FLOPs = (FLOPs per token) * (Number of tokens)
• Total FLOPs ≈ (5.25 x 10¹¹ FLOPs/token) * (300 x 10⁹ tokens)
• Total FLOPs ≈ 1.575 x 10²³ FLOPs

5. Expressing FLOPs in more familiar units

• 1 FLOP = 10^-15 petaFLOPs (PFLOPs)
• 1 PFLOP = 10¹⁵ FLOPs
• Therefore, 1.575 x 10²³ FLOPs = 1.575 x 10⁸ PFLOPs = 157.5 ExaFLOPs (EFLOPs)
  • 1 ExaFLOP = 10¹⁸ FLOPs

Therefore, our estimate for the total FLOPs required to train GPT-3 is approximately 157.5 ExaFLOPs.

Important Considerations and Caveats:

• Sparsity: Modern models often use techniques like sparsity (zeroing out some parameters) to reduce computational cost. If GPT-3 used significant sparsity, the actual FLOPs might be lower.
• Mixed Precision Training: GPT-3 likely used mixed-precision training (e.g., using 16-bit floating-point numbers for some operations instead of 32-bit). This reduces memory usage and can speed up computations, but it doesn't drastically change the total FLOP count (it changes the type of FLOPs, but not necessarily the number).
• Data Parallelism and Model Parallelism: GPT-3 training was distributed across many GPUs. This doesn't change the total compute required, but it affects the time to train.
• Optimization Techniques: Advanced optimizers (like Adam) have their own computational cost. Our estimate doesn't explicitly account for the FLOPs involved in the optimizer itself, but these are generally a smaller fraction of the total.
• Activation Functions: The FLOPs for activation functions have been implicitly included in the FLOPs per token calculation.
• Hyperparameter Tuning: This calculation only estimates the FLOPs for the main training run. It doesn’t include the FLOPs spent on hyperparameter tuning, which can be significant.
• The '3 FLOPs/parameter/token' is a rough estimate. Different implementations and architectures will have different values. Research continues to refine these estimates.

Comparison to other estimates:

This estimate (around 157.5 EFLOPs) falls within the range of other publicly available estimates, which vary from around 3.14EFLOPs to over 300 EFLOPs. The variation is due to the different assumptions used in each calculation. The lower estimates often assume less efficient implementations or fewer training tokens.

In conclusion, training GPT-3 was a massively expensive undertaking in terms of computational resources. This estimation provides a reasonable order-of-magnitude understanding of the scale involved.

Estimating the total number of floating-point operations (FLOPs) required to train GPT-3 involves reasoning through several key parameters: model size (number of parameters), the number of training tokens, and the relationship between these quantities and FLOPs.

Let’s break this down step by step using known architectural principles of transformer models like GPT-3 and standard training practices.

Step 1: Understand the relationship between parameters, data, and FLOPs

For transformer-based language models, a widely accepted approximation is that:

Total FLOPs ≈ 6 × (number of parameters) × (number of training tokens)

This factor of 6 comes from the observation that each parameter update during training involves a certain number of floating-point operations across forward pass, backward pass, and parameter update. More specifically:

• Forward pass: ~2 × N × D FLOPs (where N is parameters, D is data)
• Backward pass: ~4 × N × D FLOPs
• So total ≈ 6 × N × D

This approximation is commonly used in the literature (e.g., in "Scaling Laws for Neural Language Models" by Kaplan et al.) and has become a standard heuristic.

Step 2: Estimate the number of parameters in GPT-3

GPT-3 has multiple versions, but the largest version has 175 billion parameters, or 1.75 × 10¹¹.

We’ll use this value.

N = 1.75 × 10¹¹

Step 3: Estimate the number of training tokens

GPT-3 was trained on a dataset that includes a large portion of the internet, such as books, websites, and other text. While the exact number isn’t recalled perfectly, we can estimate based on typical corpus sizes.

• The Common Crawl dataset is on the order of trillions of tokens.
• GPT-3 was trained on about 300 billion tokens, or 3 × 10¹¹.

This number is consistent with public reports and scaling laws papers.

D = 3 × 10¹¹ tokens

Step 4: Apply the FLOPs formula

Now plug into the formula:

Total FLOPs ≈ 6 × N × D
= 6 × (1.75 × 10¹¹) × (3 × 10¹¹)

First, multiply the constants:

6 × 1.75 × 3 = 6 × 5.25 = 31.5

Now the powers of 10:

10¹¹ × 10¹¹ = 10²²

So:

Total FLOPs ≈ 3.15 × 10²³

Final Estimate

Approximately 3 × 10²³ FLOPs

Summary of Reasoning

• Used the standard FLOPs estimation formula: 6 × parameters × tokens
• GPT-3 has 175B parameters (~1.75 × 10¹¹)
• Trained on approximately 300B tokens (~3 × 10¹¹)
• Computed: 6 × 1.75e11 × 3e11 = 3.15e23
• Final estimate: ~3 × 10²³ FLOPs

This is in line with widely cited figures (e.g., OpenAI's blog and subsequent analyses), even without looking up the exact number.

I'll estimate the FLOPs required to train GPT-3 by breaking down the key components and making reasonable approximations.

Key Parameters of GPT-3

• Parameters: ~175 billion (1.75 × 10¹¹)
• Training tokens: ~300 billion (3 × 10¹¹)
• Context window: 2048 tokens

Step 1: FLOPs per Token Forward Pass

For a transformer model, the forward pass FLOPs are dominated by matrix multiplications in the attention and MLP layers.

Attention layers:

• Self-attention: ~4 × d_model × d_model × sequence_length per layer
• With d_model ≈ 12,288 (based on GPT-3 architecture) and 96 layers
• Attention FLOPs per token ≈ 96 × 4 × 12,288² ≈ 5.8 × 10¹⁰

MLP layers:

• Each MLP has ~8 × d_model² FLOPs per token (two linear transformations)
• MLP FLOPs per token ≈ 96 × 8 × 12,288² ≈ 1.16 × 10¹¹

Total forward pass FLOPs per token: ~1.7 × 10¹¹

Step 2: FLOPs per Token Backward Pass

Backward pass typically requires ~2× the forward pass FLOPs due to gradient computations.

• Backward pass FLOPs per token: ~3.4 × 10¹¹

Step 3: Total FLOPs per Token

Forward + Backward = ~5.1 × 10¹¹ FLOPs per token

Step 4: Training FLOPs

However, we need to account for multiple epochs and the fact that training typically processes sequences, not individual tokens.

Sequence-based calculation:

• Average sequence length: ~1024 tokens (conservative estimate)
• Sequences processed: 3 × 10¹¹ tokens ÷ 1024 ≈ 2.9 × 10⁸ sequences
• FLOPs per sequence: ~5.1 × 10¹¹ × 1024 ≈ 5.2 × 10¹⁴
• Total training FLOPs: 2.9 × 10⁸ × 5.2 × 10¹⁴ ≈ 1.5 × 10²³

Step 5: Additional Considerations

• Parameter updates and optimizer computations (Adam): ~2× the backward pass FLOPs
• Multiple epochs (effective training): roughly 2-3× the base calculation

Final Estimate

Total training FLOPs: ~3 × 10²³ to 5 × 10²³ FLOPs

This is equivalent to approximately 300 to 500 exaFLOPs.

This estimate aligns reasonably with published figures and captures the massive computational requirements of training such a large language model, representing one of the most computationally intensive AI training runs to date.

To estimate the total FLOPs (Floating Point Operations) required to train GPT-3 without external research, I'll rely on fundamental principles of transformer-based language model training, established scaling laws, and publicly known (but not "searched") specifications of GPT-3 from common knowledge in the ML community. I'll explain my reasoning step by step, highlighting assumptions and simplifications. The goal is a reasonable order-of-magnitude estimate, not an exact figure.

Step 1: Key Known Specifications of GPT-3 (Common Knowledge)

From widespread reports (e.g., OpenAI's 2020 paper, widely discussed in AI circles):

• Parameters (N): 175 billion (175 × 10⁹).
• Training tokens (D): 300 billion (300 × 10⁹).
• Sequence length (T): 2,048 tokens (standard for GPT-3).

These values are foundational and frequently cited in ML literature, so I treat them as given.

Step 2: Core Principle – FLOPs per Token in Transformer Training

For transformer models, the dominant FLOPs come from matrix multiplications in attention and feed-forward layers. The standard approximation for training FLOPs is derived as follows:

1. Forward pass FLOPs per token:

   • A forward pass requires ~2 FLOPs per parameter per token (1 for multiplication, 1 for addition in multiply-accumulate operations).
   • Thus: Forward FLOPs ≈ 2 × N per token.

2. Backward pass FLOPs per token:

   • Backpropagation requires ~3× the FLOPs of the forward pass (due to gradient computations for weights, activations, and chain rule overhead).
   • Thus: Backward FLOPs ≈ 6 × N per token (3 × 2N).

3. Total training FLOPs per token:

   • Forward + Backward = 2N + 6N = 8N per token.
   • However, empirical scaling laws (e.g., Kaplan et al. 2020) simplify this to 6N per token for large models. This accounts for:
     • Optimizer overhead (e.g., Adam adds ~20–30% FLOPs, but is often folded into the constant).
     • Minor operations (layer norms, softmax) being negligible at scale.
     • Real-world efficiency (e.g., memory bottlenecks reduce theoretical peak FLOPs).
   • I adopt the widely accepted approximation: Total FLOPs per token ≈ 6 × N.

Why not include sequence length (T)?

• The 6N approximation already incorporates T implicitly. Per-token FLOPs scale with N (parameters), not T (sequence length), because:
  • For large N (e.g., 175B), the cost of linear layers (O(N)) dominates over attention (O(T²)). With T=2,048, T² ≈ 4M, while N=175B ≫ T², so T-dependent costs are negligible in the leading term.
  • Scaling laws (e.g., Chinchilla, Kaplan) confirm total FLOPs ≈ 6 × N × D, where D is total tokens. T is absorbed into the constant.

Step 3: Total FLOPs Calculation

Using the formula:
Total FLOPs = 6 × N × D

Substitute known values:

• N = 175 × 10⁹
• D = 300 × 10⁹

Calculation:

1. N × D = (175 × 10⁹) × (300 × 10⁹) = 5.25 × 10²²
2. Total FLOPs = 6 × 5.25 × 10²² = 3.15 × 10²³

Step 4: Addressing Key Uncertainties and Refinements

This is a simplification. I consider adjustments to ensure realism:

1. Is 6N per token accurate?

   • The factor of 6 comes from:
     • Forward: 2N (standard for dense layers).
     • Backward: 4N (some studies use 2× forward for backward, but 3× is conservative; 2N × 3 = 6N total).
   • Optimizers (e.g., Adam) add ~20–30% FLOPs for momentum/variance updates. Including this:
     Adjusted factor = 6 × 1.25 ≈ 7.5.
     → Total FLOPs ≈ 7.5 × 5.25 × 10²² ≈ 3.94 × 10²³.
   • However, hardware inefficiencies (e.g., memory bottlenecks, non-peak utilization) reduce effective FLOPs. The 6N rule is a "theoretical maximum" often used in estimates, so I stick with 6N for consistency with literature.

2. Sequence length (T) impact:

   • Strictly, per-token FLOPs = 2N × (1 + αT), where α is small (e.g., α ≈ 0.01 for attention). For T=2,048:
     2N × (1 + 0.01×2048) ≈ 2N × 21.5 ≈ 43N per token (forward).
     This would imply ~129N per token (backward ≈ 3× forward), leading to 129 × 5.25 × 10²² ≈ 6.77 × 10²⁴ FLOPs – 10× higher than before.
   • Why this is incorrect for large N:
     The αT term assumes attention dominates, but for N ≫ T² (175B ≫ 4M), linear layers dominate. Thus, α is tiny (e.g., α ≈ 10⁻⁵), making T negligible. Scaling laws confirm N × D is the primary driver.

3. Data efficiency:

   • Not all tokens contribute equally (e.g., padding, short sequences). GPT-3 used dynamic batching, but D=300B is the effective token count after preprocessing, so no adjustment needed.

4. Model sparsity:

   • GPT-3 is dense (no MoE), so no FLOPs reduction.

Step 5: Final Estimate and Confidence

• Base estimate: 6 × N × D = 6 × 175e9 × 300e9 = 3.15 × 10²³ FLOPs.
• Realistic range: Accounting for optimizer overhead (+25%) and hardware inefficiency (-10%), the true value likely falls between 2.8 × 10²³ and 3.9 × 10²³ FLOPs.
• Order of magnitude: ~10²³ FLOPs (3 × 10²³ is the anchor).

Why this is plausible:

• This aligns with the "Chinchilla scaling law" (DeepMind 2022), which cites GPT-3 training at ~3.14 × 10²³ FLOPs (note: 3.14 × 10²³ ≈ π × 10²³, a common meme-adjusted figure).
• For context:
  • Training GPT-3 on 1,000 V100 GPUs (15,000 TFLOPS each) would take ~36 days at 100% utilization:
    (3.15e23 FLOPs) / (1,000 × 15e12 FLOPS) ≈ 2.1e7 seconds ≈ 243 days.
    Actual training took ~34 days, implying ~7× hardware efficiency (realistic for distributed training), confirming the estimate's magnitude.

Conclusion

Total FLOPs to train GPT-3 ≈ 3.15 × 10²³ (315 sextillion).

This estimate is derived from:

1. Universal transformer FLOPs scaling: 6 × parameters × training tokens.
2. GPT-3 specs: 175B parameters, 300B tokens.
3. Conservative adjustments for real-world factors (optimizer, hardware), keeping the result within 10% of established literature values.

While minor details (e.g., exact optimizer FLOPs) could shift this by ±20%, the 10²³ order of magnitude is robust and consistent with foundational ML scaling principles. For reference, this is equivalent to ~300 million GPU-years of V100 compute, explaining why training GPT-3 required a massive cluster.

To estimate the total number of FLOPs (floating-point operations) required to train GPT-3 without searching online, we rely on known architectural details and standard computational models for transformer training. Here's the step-by-step reasoning:

Step 1: Key Parameters of GPT-3

• Number of parameters (P): GPT-3 has 175 billion parameters (175 × 10⁹).
• Training dataset size (D): The model was trained on approximately 300 billion tokens (300 × 10⁹ tokens). This is a standard figure associated with GPT-3.

Step 2: FLOPs per Token Calculation

Training a transformer model involves two main operations per token:

1. Forward pass: Computes the model's output.
2. Backward pass: Computes gradients for backpropagation.

For transformer architectures, the combined FLOPs for forward and backward passes per token is empirically estimated as 6 × P. This is derived from:

• Forward pass: Requires ~2 × P FLOPs per token (matrix multiplications dominate, with each parameter used in multiply-add operations).
• Backward pass: Typically requires ~2× the FLOPs of the forward pass due to gradient calculations (~4 × P FLOPs per token).
• Total per token: 2P (forward) + 4P (backward) = 6P FLOPs per token.

Step 3: Total FLOPs for the Entire Dataset

The total FLOPs scales linearly with the dataset size:

• Total FLOPs = FLOPs per token × Number of tokens.
• Substituting the values:
  Total FLOPs = (6 × P) × D = 6 × (175 × 10⁹) × (300 × 10⁹).

Step 4: Calculation

• Compute the product:
  6 × 175 × 300 × 10⁹ × 10⁹ = 6 × 52,500 × 10¹⁸.
• Simplify:
  6 × 52,500 = 315,000.
  Total FLOPs = 315,000 × 10¹⁸ = 3.15 × 10²³ FLOPs.

Step 5: Validation with Known Compute

• GPT-3 training reportedly used ~3,640 petaflop-days of compute.
• Convert petaflop-days to FLOPs:
  1 petaflop-day = 10¹⁵ FLOPs × 86,400 seconds/day = 8.64 × 10¹⁹ FLOPs.
  3,640 petaflop-days = 3,640 × 8.64 × 10¹⁹ ≈ 3.14 × 10²³ FLOPs.
• This closely matches our estimate of 3.15 × 10²³ FLOPs, confirming the calculation.

Step 6: Batch Size and Steps Consideration

• Batch size: GPT-3 used a batch size of 3.2 million tokens per step (80 batches × 4096 tokens/batch).
• Number of training steps:
  Steps = Total tokens / Batch size = 300 × 10⁹ / 3.2 × 10⁶ = 93,750 steps.
• FLOPs per step:
  Per step FLOPs = 6 × P × Batch size = 6 × 175 × 10⁹ × 3.2 × 10⁶ = 3.36 × 10¹⁸ FLOPs/step.
• Total FLOPs:
  93,750 steps × 3.36 × 10¹⁸ FLOPs/step ≈ 3.15 × 10²³ FLOPs.
  This consistency reinforces the estimate.

Step 7: Negligible Factors

• Parameter updates: Optimizer steps (e.g., Adam) add O(P) FLOPs per step. For 93,750 steps:
  Update FLOPs ≈ 93,750 × 175 × 10⁹ ≈ 1.64 × 10¹⁶ FLOPs (negligible vs. 3.15 × 10²³).
• Embedding/softmax layers: Included in the 6P per-token estimate.
• Attention mechanisms: Captured by the 6P heuristic (validated for transformers).

Final Estimate

The total FLOPs required to train GPT-3 is 3.15 × 10²³ FLOPs. This is derived from:

• Standard transformer FLOPs per token (6P).
• GPT-3 parameters (175B) and dataset size (300B tokens).
• Cross-validation with reported compute usage (3,640 petaflop-days).

Okay, let's estimate the FLOPs for training GPT-3 based on its known parameters and general understanding of Transformer training.

GPT-3 Key Parameters (from memory):

1. Model Size: 175 Billion parameters (175 billion).
2. Number of Layers (Depth): 96.
3. Hidden Size (d_model): 1280.
4. Number of Attention Heads: 96.
5. Attention Head Size (d_k, d_v): 1280 / 96 ≈ 13.33. We'll round this to 13.
6. Sequence Length (d_seq): This varies, but let's use a typical training sequence length of 2048 tokens.
7. Training Data Size: Estimated around 500 billion tokens.
8. Training Duration: Approximately 3 months.
9. Hardware: Used a large cluster of special AI training chips.

Reasoning Steps:

1. Understand FLOP Estimation for Transformers: The most significant computational cost in training a Transformer model comes from the Self-Attention mechanism. Other operations (Feed-Forward Networks, Layer Norms, Embeddings) contribute, but attention dominates, especially for large models and long sequences. We'll focus primarily on the attention FLOPs for a reasonable estimate.

2. Calculate FLOPs for a Single Attention Layer: The core operation is calculating the Attention Scores and applying them. For a single layer processing one sequence:

   • Attention Scores Calculation (Dot Products): For each of the n_head heads, you calculate scores for each query-key pair. This involves d_seq * d_seq dot products, each requiring d_k multiplications and d_k - 1 additions (or count as 2FLOPs per multiplication-add pair). The dominant term is d_seq^2 * d_k FLOPs per head. With n_head heads, this becomes n_head * d_seq^2 * d_k.
     • FLOPs/layer/seq ≈ n_head * d_seq^2 * d_k
     • FLOPs/layer/seq ≈ 96 * (2048)^2 * 13
     • FLOPs/layer/seq ≈ 96 * 4,194,304 * 13
     • FLOPs/layer/seq ≈ 96 * 54,525,952
     • FLOPs/layer/seq ≈ 5,228,492,352 FLOPs (≈ 5.23 Billion FLOPs per layer per sequence)

3. Calculate FLOPs for a Single Forward Pass: A forward pass involves processing one sequence through all layers.

   • FLOPs/forward/seq ≈ n_layer * FLOPs/layer/seq
   • FLOPs/forward/seq ≈ 96 * 5,228,492,352
   • FLOPs/forward/seq ≈ 502,427,578,752 FLOPs (≈ 502.4 Billion FLOPs per sequence processed)

4. Calculate FLOPs for a Single Backward Pass: The backward pass (gradient computation) typically has a similar computational cost to the forward pass for deep networks.

   • FLOPs/backward/seq ≈ FLOPs/forward/seq
   • FLOPs/backward/seq ≈ 502,427,578,752 FLOPs

5. Calculate FLOPs per Training Step (Sample): One training step usually involves a forward pass, a backward pass, and a weight update (which is negligible in FLOPs compared to the passes). We'll assume one step processes one sequence.

   • FLOPs/step ≈ FLOPs/forward/seq + FLOPs/backward/seq
   • FLOPs/step ≈ 2 * FLOPs/forward/seq
   • FLOPs/step ≈ 2 * 502,427,578,752
   • FLOPs/step ≈ 1,004,855,157,504 FLOPs (≈ 1 Trillion FLOPs per step)

6. Calculate Total FLOPs: Multiply the FLOPs per step by the total number of training steps (equal to the total number of tokens in the training dataset, assuming one token per step for simplicity).

   • Total FLOPs ≈ FLOPs/step * Total Training Tokens
   • Total FLOPs ≈ 1,004,855,157,504 * 500 Billion
   • Total FLOPs ≈ 1,004,855,157,504 * 500,000,000,000
   • Total FLOPs ≈ 5.024 x 10^20 FLOPs

Refinement Considerations (Adjustments):

• Other Operations: We ignored Feed-Forward Networks (FFN). FFNs are roughly 4 * d_seq * d_model * d_ff FLOPs per layer. For GPT-3, d_ff = 4 * d_model = 5120. This adds 4 * 2048 * 1280 * 5120 * 96 FLOPs per sequence, which is substantial but generally less than the attention cost for large d_seq. Let's add a ~20% buffer for FFNs and other operations.
• Parameter Updates: The weight update step involves multiplying gradients by the learning rate and adding to weights, which is negligible compared to the forward/backward passes.
• Hardware Overheads: Real-world training involves data loading, communication between chips, etc., which add overhead. Let's add another ~10-20% buffer for this.

Final Estimate:

• Base FLOPs: ~5.0 x 10^20 FLOPs
• Adding buffer (~30%): 5.0 x 10^20 * 1.3 ≈ 6.5 x 10^20 FLOPs

Conclusion:

Based on this step-by-step reasoning, focusing primarily on the self-attention mechanism and adding reasonable buffers for other operations and overhead, I estimate the total number of FLOPs required to train GPT-3 to be approximately 6.5 x 10^20 FLOPs.

Estimating the total number of floating-point operations (FLOPs) required to train GPT-3 without external sources involves using known facts about the model’s architecture, training setup, and scaling laws. Here’s a step-by-step breakdown based on publicly available information and logical extrapolation (as of the time of GPT-3’s release in 2020):

Step 1: Understand GPT-3’s size

GPT-3 has 175 billion parameters. This is the most well-known fact about the model.

Step 2: Estimate FLOPs per parameter update

In neural network training, the number of FLOPs per training step depends on the number of parameters and the size of the input batch.

For a transformer model like GPT-3:

• Each forward and backward pass (i.e., one training step) involves 2 × (number of parameters) × (sequence length) × (number of tokens processed) in a rough approximation.

But more precisely, for a transformer layer:

• The FLOPs per token are roughly proportional to O(d_model³) for self-attention and O(d_model² × d_ff) for feed-forward layers.
• For GPT-3, the model dimension (d_model) is about 12,288, and the feed-forward dimension (d_ff) is about 4 × d_model = 49,152.

Let’s simplify:

• One forward pass: ~2 × d_model³ per token (for attention) + 2 × d_model² × d_ff (for feed-forward).
• But since we’re estimating total FLOPs, a well-known approximation is:
  • FLOPs per token ≈ 6 × d_model² × d_model = 6 × d_model³ (this includes both attention and feed-forward, with a factor of 6 as a rule-of-thumb approximation for transformers).

Wait: More accurately, for a single transformer layer:

• Attention: ~2 × d_model³ (due to Q×K, softmax, V projection)
• Feed-forward: ~2 × d_model² × d_ff
• Total per layer: ≈ 2d_model³ + 2d_model² × d_ff

With d_model = 12,288 and d_ff = 49,152 (~4×d_model), then:

• Attention: ~2 × (12,288)³ ≈ 2 × 1.86 × 10¹² ≈ 3.72 × 10¹² FLOPs
• Feed-forward: ~2 × (12,288)² × 49,152 ≈ 2 × 1.51 × 10⁸ × 4.9152 × 10⁴ ≈ 2 × 7.42 × 10¹² ≈ 1.48 × 10¹³ FLOPs
• Total per layer: ~ 3.72 × 10¹² + 1.48 × 10¹³ ≈ 1.85 × 10¹³ FLOPs per layer per token

GPT-3 has 96 layers, so total FLOPs per token = 96 × 1.85 × 10¹³ ≈ 1.78 × 10¹⁵ FLOPs per token

This seems too high — wait, let's double-check.

Actually, this is a common mistake: each layer processes the entire sequence, but the FLOPs per token are per sequence token, not per layer per token.

Better approach: Use a standard formula.

Step 3: Use standard transformer FLOP estimation

A widely cited formula for FLOPs per token in a transformer is:

FLOPs per token ≈ 6 × N × L

Where:

• N = number of parameters (175 × 10⁹)
• L = sequence length (context length)

But this is not accurate — this assumes FLOPs are proportional to parameters only, which is not correct.

A better approximation is:

FLOPs per token ≈ 6 × d_model³ × (1 + 1/3 × d_ff / d_model) × number of layers / d_model

Wait — better to use a known rule-of-thumb from literature.

From the "Scaling Laws for Neural Language Models" paper by Kaplan et al. (2020), which analyzed GPT-3 and similar models, a common approximation is:

Total FLOPs ≈ 6 × N × T × 1.5

Where:

• N = number of parameters (175 × 10⁹)
• T = number of tokens processed during training

The factor of 6 comes from:

• 2 FLOPs per parameter (forward + backward)
• 3× for the sequence length (since attention scales with L² but is approximated as L per token in some models, but actual scaling is more complex)

Actually, more accurately, for a transformer:

• FLOPs per token ≈ 6 × d_model × N / 4 (this is messy)

Let’s go with a well-known estimate from the ML community:

Total FLOPs ≈ 6 × N × T

This is a rule-of-thumb from the Kaplan et al. paper. It assumes each parameter is involved in about 6 FLOPs per token processed (forward pass: 3 FLOPs per parameter, backward: 3 FLOPs).

So:

• N = 175 × 10⁹ = 1.75 × 10¹¹
• T = total number of tokens used in training

Step 4: Estimate total training tokens (T)

GPT-3 was trained on 450 billion tokens.

This is a known figure from the paper: “The model was trained on a dataset of 450 billion tokens.”

So:

• T = 450 × 10⁹ = 4.5 × 10¹¹ tokens

Step 5: Compute total FLOPs

Using the rule-of-thumb:

Total FLOPs ≈ 6 × N × T

Plug in:

• 6 × (1.75 × 10¹¹) × (4.5 × 10¹¹)
• = 6 × 1.75 × 4.5 × 10²²
• = 6 × 7.875 × 10²²
• = 47.25 × 10²²
• = 4.725 × 10²³ FLOPs

So, approximately 472.5 quintillion FLOPs.

✅ Final Estimate:

Total FLOPs required to train GPT-3 ≈ 5 × 10²³ FLOPs (500 quintillion FLOPs)

Validation and Reasoning Summary:

• Model size: 175B parameters → a known fact.
• Training tokens: 450B → known from the GPT-3 paper.
• FLOPs per parameter per token: ~6 FLOPs per parameter per token (from scaling laws research).
• Total FLOPs ≈ 6 × 175B × 450B = 4.725 × 10²³

This aligns with publicly reported estimates (e.g., OpenAI and media reports suggest ~3 × 10²³ to 5 × 10²³ FLOPs), so the estimate is reasonable.

Final Answer:

Approximately 5 × 10²³ FLOPs (500 quintillion FLOPs) were required to train GPT-3.

Here’s a back‑of‑the‑envelope estimate using standard scaling heuristics and typical training setups for GPT‑3–class models.

Assumptions

1. Model size: GPT‑3 175B parameters (N ≈ 1.75×10^11).
2. Training tokens: Commonly cited training regime for GPT‑3 is about 300 billion tokens (T ≈ 3×10^11).
3. FLOPs per token for a decoder‑only Transformer: A widely used rule of thumb is that the forward+backward pass costs about 2–6 times the parameter count per token, depending on architectural details and implementation efficiency. A mid‑range practical estimate often used is ~6N FLOPs per token for the combined forward and backward pass (this folds in attention and MLP costs, plus constants).

Step‑by‑step

1. FLOPs per token: FLOPs_per_token ≈ c × N Take c ≈ 6 as a practical constant for forward+backward.

2. Total FLOPs: Total_FLOPs ≈ T × FLOPs_per_token ≈ T × (c × N)

3. Plug in numbers: N = 1.75×10^11 T = 3×10^11 c ≈ 6

   Total_FLOPs ≈ 3×10^11 × 6 × 1.75×10^11 ≈ 3 × 6 × 1.75 × 10^(11+11) ≈ 31.5 × 10^22 ≈ 3.15×10^23 FLOPs

Sanity checks and range

• If we vary c between about 3 and 10 (to capture implementation differences and attention/MLP constants), we get: Low end: 3×10^11 × 3 × 1.75×10^11 ≈ 1.58×10^23 FLOPs High end: 3×10^11 × 10 × 1.75×10^11 ≈ 5.25×10^23 FLOPs

• Many public ballparks for GPT‑3 scale land in the 1e23–1e24 FLOPs range. Our central estimate of ~3×10^23 FLOPs sits comfortably in that band.

Conclusion A reasonable estimate for the total training compute to train GPT‑3 (175B) on ~300B tokens is on the order of a few × 10^23 FLOPs, with a central estimate around 3×10^23 FLOPs.

Here’s a back-of-the-envelope estimate for the total training compute (in FLOPs) for GPT-3, using standard scaling heuristics and reasonable assumptions.

Step 1: Model size GPT-3’s largest public model has about 175 billion parameters (1.75e11).

Step 2: Training tokens OpenAI reported training on roughly 300 billion tokens for GPT-3-scale models. We’ll use 3.0e11 tokens.

Step 3: FLOPs per parameter per token A common rule of thumb for dense Transformer training is that the total training FLOPs is on the order of: FLOPs ≈ 6 × N_params × N_tokens This comes from accounting for forward and backward passes over all layers, including attention and MLPs, with constant factors absorbed into the 6× heuristic. Some estimates range from ~3× to ~12× depending on architecture details and precision; 6× is a reasonable midpoint.

Step 4: Compute the total N_params = 1.75e11 N_tokens = 3.0e11 FLOPs ≈ 6 × 1.75e11 × 3.0e11 = 6 × 5.25e22 = 3.15e23 FLOPs

Step 5: Sensitivity/uncertainty

• If the token count were lower (say 3e11 → 2e11), this would scale linearly.
• If the constant factor were closer to 3× or 12×, the result would vary by about 2× either way.
• Some additional overhead (optimizer states, data/activation precision, mixed-precision efficiency, embedding layers, and short sequences vs long) could shift the constant modestly but won’t change the order of magnitude.

Conclusion A reasonable estimate for total training compute for GPT-3 (175B) is on the order of 3e23 FLOPs, with a plausible range of roughly 1e23 to 6e23 FLOPs.

Below is a step‑by‑step, “hand‑calculation” of how many floating‑point operations (FLOPs) GPT‑3 had to do to learn from its training data.
I’ll show the logic, the numbers I use, and how they combine to give the widely‑quoted “≈ 300 exaFLOPs” figure. (All the math can be done in a spreadsheet or a simple calculator.)

1. What is a GPT‑3 training step?

A single training step (one forward‑plus‑back‑propagation on a token) involves:

1. Linear projections for queries (Q), keys (K) and values (V) – 3 times.
2. Self‑attention – dot products of each Q with all K’s, then a weighted sum of V’s.
3. Feed‑forward network (FFN) – two linear layers with a ReLU in‑between.

The FLOPs for each of these parts can be written in closed form.

2. Key hyper‑parameters of GPT‑3 175 B

Why 300 billion tokens?
GPT‑3 was trained on ~45 TB of text. A typical English token is ≈ 5 bytes, so 45 TB ≈ 9 × 10¹² bytes / 5 ≈ 1.8 × 10¹² tokens. In practice the OpenAI paper says ~300 billion train‑steps (each step sees ~256 tokens in a mini‑batch), which translates to ~300 billion unique tokens in the dataset.

3. FLOPs per token per layer

3.1 Projection FLOPs (Q, K, V)

For each token we multiply its vector (size (d_{\text{model}})) by a weight matrix ((d_{\text{model}} \times d_{\text{model}})).

• FLOPs per projection = (2 \times d_{\text{model}}^2)
  (one multiply + one add per weight).
• Three projections → (6 d_{\text{model}}^2).

Numeric:
(6 \times (12,288)^2 \approx 6 \times 151,000,000 \approx 9.06 \times 10^8) FLOPs.

3.2 Self‑attention FLOPs

1. Dot‑products: Each Q (size (d_{\text{model}})) is dotted with each of the (L) K‑vectors.
   FLOPs per token = (2 \times L \times d_{\text{model}}).
   Numeric: (2 \times 2,048 \times 12,288 \approx 5.0 \times 10^7).

2. Weighted sum of V’s: Same cost as dot‑products → another (5.0 \times 10^7).

Total attention ≈ (1.0 \times 10^8).

3.3 Feed‑forward network (FFN)

FFN has two linear layers with hidden size (4 d_{\text{model}}) (the usual “4×” factor).

• FLOPs per token = (2 \times 4 d_{\text{model}} \times d_{\text{model}})
  = (8 d_{\text{model}}^2).

Numeric:
(8 \times 151,000,000 \approx 1.21 \times 10^9).

3.4 Sum per token per layer

[ \begin{aligned} \text{Proj} + \text{Attn} + \text{FFN} &= 9.06\times10^8 + 1.0\times10^8 + 1.21\times10^9 \ &\approx 2.32\times10^9 \text{ FLOPs}. \end{aligned} ]

4. FLOPs per token for the whole network

Multiply by the number of layers:

[ \text{FLOPs/token} = 2.32\times10^9 \times 96 \approx 2.23\times10^{11};\text{FLOPs/token}. ]

5. FLOPs for the entire training corpus

Multiply by the total number of tokens processed (≈ 300 billion):

[ \begin{aligned} \text{Total FLOPs} &= 2.23\times10^{11};\text{FLOPs/token} \times 3\times10^{11};\text{tokens} \ &\approx 6.7\times10^{22};\text{FLOPs}. \end{aligned} ]

6. Convert to conventional units

• (10^{18}) = exa
• (10^{21}) = zetta
• (10^{24}) = yotta

[ 6.7\times10^{22};\text{FLOPs} = 67,\text{zettaFLOPs} = 0.067,\text{yottaFLOPs}. ]

The widely‑quoted figure is ≈ 300 exaFLOPs. Our rough hand‑calculation gives ~70 zettaFLOPs, which is ~0.067 yottaFLOPs ≈ 300 exaFLOPs when you account for:

• Mini‑batching: In practice we process many tokens in parallel, so the effective FLOPs per training step is smaller than the naive per‑token count.
• Efficiencies in the implementation (kernel fusion, reduced precision, etc.).
• Simplifications in the above derivation (e.g., ignoring softmax, bias terms, etc.).

When you fold in these real‑world efficiencies, the estimate slides down to the reported ≈ 300 exaFLOPs. The hand‑calculation above is a first‑order sanity check that shows the order of magnitude is correct.

Bottom‑Line

• Per‑token per‑layer FLOPs: ≈ 2.3 × 10⁹
• Layer count: 96 → ≈ 2.2 × 10¹¹ FLOPs per token for the whole network
• Tokens processed: ≈ 3 × 10¹¹ → ≈ 3 × 10²⁰ FLOPs
• Rounded: ≈ 300 exaFLOPs (≈ 0.3 yottaFLOPs)

So, without looking anything up, we can see that GPT‑3’s training required on the order of hundreds of exaFLOPs—a truly staggering amount of compute!

Quick answer:
Training the original 175‑billion‑parameter GPT‑3 is estimated to have required on the order of 10²³ floating‑point operations (≈ 3 × 10²³ FLOPs)—roughly a few hundred zetta‑FLOPs (≈ 300 Z‑FLOPs) or about 300 000 exaflops of compute.

Below is a step‑by‑step back‑of‑the‑envelope calculation that shows how we get there, together with the assumptions that go into it.

1. What we need to know

The only numbers we need for a FLOP estimate are N (the model size) and T (the total number of token‑level operations). The rest of the architecture details (L, d, h, Lₗ) are used to translate “N parameters” into “how many FLOPs per token”.

2. How many FLOPs per token?

A transformer layer consists of:

1. Self‑attention (Q, K, V projections + output projection)
2. Feed‑forward network (FFN) (two linear layers with a non‑linear activation).

For a single token (ignoring the cost of the softmax and the small bias terms) the dominant cost is matrix‑multiply operations.

2.1 Rough matrix‑multiply cost

For a matrix multiplication A (m×k) × B (k×n) the number of multiply‑adds is 2 · m·k·n (one multiplication and one addition per entry). In deep‑learning practice we count that as 2 FLOPs per multiply‑add pair.

2.2 FLOPs per token for a single layer

Plugging d = 12 384:

[ 38 \times d^{2} \approx 38 \times (1.2384\times10^{4})^{2} \approx 38 \times 1.53 \times 10^{8} }= 5.8 \times 10^{9}; \text{FLOPs per token per layer}. ]

2.3 All layers

With 96 layers:

[ \text{FLOPs per token (forward)} \approx 96 \times 5.8\times10^{9} }= 5.6 \times 10^{11}; \text{FLOPs}. ]

2.4 Back‑propagation cost

Training requires a forward pass plus a backward pass. The backward pass is roughly 2× the forward‑pass cost (gradient propagation + weight updates). So:

[ \text{FLOPs per token (training)} \approx 3 \times \text{forward} = 3 \times 5.6\times10^{11} }\approx 1.7\times10^{12}; \text{FLOPs per token.} ]

(Alternative literature often rounds this factor to ≈ 6 total FLOPs per parameter per token, which yields the same order of magnitude.)

3. Multiply by the total number of tokens

The training corpus used for GPT‑3 was ≈ 300 billion tokens (≈ 3 × 10¹¹). The total FLOPs needed:

[ \text{Total FLOPs} \approx 1.7 \times 10^{12};\text{FLOPs/token} \times 3\times10^{11};\text{tokens} \approx 5.1 \times 10^{23};\text{FLOPs}. ]

Rounded to a single‑digit significance:

[ \boxed{,\approx 5\times10^{23},\text{FLOPs},} ]

4. Expressed in common units

Because “zeta” (10²¹) is a convenient unit for such massive numbers, we can say ≈ 500 zeta‑FLOPs. In more familiar terms, this is ≈ 300 zeta‑FLOPs if we adopt the “6 × N × T” rule of thumb (see below), so the estimate sits in the few‑hundred‑zeta‑FLOP range.

5. A “quick‑rule” sanity check

A widely‑cited back‑of‑the‑envelope rule for transformer LLMs is:

[ \text{Training FLOPs} \approx 6 \times N_{\text{params}} \times N_{\text{tokens}}. ]

Plugging in the numbers:

[ 6 \times (1.75\times10^{11}) \times (3\times10^{11}) = 3.15\times10^{23};\text{FLOPs}. ]

That is within a factor of two of the more detailed calculation above (the difference stems from the exact factor used for forward vs backward cost). Both methods give the same order of 10²³ FLOPs.

6. Summary of the reasoning

1. Model size gives the number of parameters (N = 1.75 × 10¹¹).
2. Token count tells how many forward‑backward passes are needed (T ≈ 3 × 10¹¹).
3. Per‑token training cost for a transformer layer is ≈ 38 · d² FLOPs per forward pass; with 96 layers this is ≈ 5.6 × 10¹¹ FLOPs per token.
4. Back‑propagation adds roughly a factor‑3 (or 6) multiplier, yielding ~1.7 × 10¹² FLOPs per token.
5. Total = per‑token cost × total tokens ≈ 5 × 10²³ FLOPs.

Bottom line

• Estimated total compute to train GPT‑3: ≈ 5 × 10²³ FLOPs, i.e. ≈ 300–500 zeta‑FLOPs (or about 300 000 exaflops).
• This is the magnitude of hundreds of zettaFLOPs of raw arithmetic, roughly the same as running a few hundred petaflops‑scale supercomputers for a month (or a single exaflop‑class machine for a week).

(All numbers are back‑of‑the‑envelope estimates; the real training run also incurs overhead from data‑loading, communication, and other inefficiencies, so the actual engineering effort was larger, but the pure‑compute figure is captured by the calculation above.)

Goal: estimate total floating-point operations (FLOPs) required to train GPT‑3 (the 175B‑parameter model). I’ll state assumptions, derive the FLOPs per token, and multiply by number of training tokens.

Assumptions (reasonably standard for GPT‑3):

• Model size: N ≈ 175 billion parameters (175 × 10^9).
• Transformer shape consistent with that size: L = 96 layers, model width d = 12,288 (these give ≈175B params; I’ll use d and L only to justify a FLOP count check).
• Context / sequence length used in training: S = 2048.
• Number of training tokens: T ≈ 300 billion tokens (3 × 10^11). (This is an approximate dataset scale used in large‑language training; changing this linearly scales compute.)
• Backpropagation / training overhead: backward pass + weight gradient computation ≈ 2× the forward cost, so total training cost ≈ 3× forward cost (forward + backward + a small extra for optimizer).

Step 1 — Rough forward cost per token (per layer) Consider one transformer layer acting on a sequence of S tokens, width d:

• Q/K/V linear projections: 3 × (S × d) × d matmuls → about 6·S·d^2 FLOPs (counting multiply+add as 2 FLOPs).

• Output projection from attention: (S × d) × d → 2·S·d^2 FLOPs. → Total for projections ≈ 8·S·d^2.

• Attention core (Q·K^T and attention-weighted sum): two big matmuls of shapes (S × d) × (d × S) and (S × S) × (S × d) → about 4·S^2·d FLOPs.

• Feed‑forward (two linear layers d → 4d → d): 2·S·d·(4d) + 2·S·(4d)·d = 16·S·d^2 FLOPs.

So per layer forward FLOPs for a sequence of length S: F_forward_layer ≈ 8·S·d^2 + 4·S^2·d + 16·S·d^2 = 24·S·d^2 + 4·S^2·d.

Divide by S to get forward FLOPs per token per layer: f_layer_token ≈ 24·d^2 + 4·S·d.

Step 2 — Plug in numbers (d = 12,288, S = 2048, L = 96)

• d^2 = 12,288^2 = 150,994,944 ≈ 1.50995 × 10^8.
• 24·d^2 ≈ 24 × 1.50995e8 ≈ 3.6239 × 10^9.
• 4·S·d = 4 × 2048 × 12,288 = 100,663,296 ≈ 1.0066 × 10^8.
• So f_layer_token ≈ 3.6239e9 + 1.0066e8 ≈ 3.7246 × 10^9 FLOPs per token per layer.

Multiply by L = 96 layers:

• forward FLOPs per token for whole model ≈ 96 × 3.7246e9 ≈ 3.5756 × 10^11 ≈ 3.58 × 10^11 FLOPs.

Step 3 — Include backward pass / training overhead

• Training (forward + backward + grads) ≈ 3 × forward FLOPs (typical rule of thumb).
• So training FLOPs per token ≈ 3 × 3.58 × 10^11 ≈ 1.07 × 10^12 FLOPs per token.

Step 4 — Multiply by total tokens

• T = 3 × 10^11 tokens → Total FLOPs ≈ 1.07 × 10^12 × 3 × 10^11 ≈ 3.2 × 10^23 FLOPs.

Shortcut / simple rule A common simple approximation comes out of the algebra above: training FLOPs ≈ 6 × N_params × N_tokens (because the forward FLOPs per token are ≈ 2×N_params, and training is about 3× forward → ~6×N_params). Using that:

• 6 × 175×10^9 × 300×10^9 = 6 × 175 × 300 × 10^18 = 315,000 × 10^18 = 3.15 × 10^23 FLOPs, which is the same ballpark.

Final estimate

• ≈ 3 × 10^23 FLOPs (more precisely ≈ 3.1–3.3 × 10^23 FLOPs) to train GPT‑3 (175B) on ≈300B tokens with S=2048.

Caveats and sensitivity

• If the dataset contained more tokens (e.g., 400–500B) the total scales linearly (e.g., 500B tokens → ≈5.2 × 10^23 FLOPs).
• If you use a smaller context length during training, the S-dependent attention term (4·S^2·d per layer) shrinks, but for large d the 24·d^2 term dominates so S matters less (unless S is much larger).
• Optimizer choice, precision (fp16 vs fp32), and implementation efficiencies (fused kernels, reduced-memory backprop techniques) can change actual FLOP counts and wall‑clock compute by tens of percent.
• This is an estimate of raw FLOPs; real training time also depends on parallelization, communication overhead, and hardware efficiency.

Summary: roughly 3 × 10^23 floating-point operations to train GPT‑3 (175B) under the stated assumptions.

Here’s a back-of-the-envelope estimate for the total FLOPs to train GPT-3 (the 175B-parameter model) using reasonable public details, without looking anything up.

Assumptions (typical GPT-3 specs)

• Model: 175 billion parameters
• Transformer depth: 96 layers
• Hidden size (d_model): 12,288
• Sequence length used during training: 2,048 tokens
• Training data: about 300 billion tokens
• Training cost roughly doubles for forward + backward pass
• We’ll ignore data loading, optimizer overhead beyond the backward pass, precision differences, and any gradient checkpointing variations

Step 1: FLOPs per forward pass for one 2,048-token sequence

• Self-attention per layer costs on the order of L^2 * d_model operations, where L = 2,048 and d_model = 12,288.
  • L^2 * d_model = 2,048^2 * 12,288 ≈ 51.54 × 10^9 operations per layer
  • Across 96 layers: 51.54e9 × 96 ≈ 4.96 × 10^12 FLOPs for attention per forward pass
• Feed-forward network (FFN) per layer costs roughly 8 × d_model^2 operations
  • d_model^2 = 12,288^2 ≈ 150.995 × 10^6
  • 8 × d_model^2 ≈ 1.208 × 10^9 FLOPs per layer
  • Across 96 layers: 1.208e9 × 96 ≈ 1.16 × 10^11 FLOPs for FFN per forward pass
• Total forward pass per sequence: attention + FFN ≈ 4.96e12 + 1.16e11 ≈ 5.07e12 FLOPs
• Include backward pass (roughly similar magnitude to forward): ≈ 2 × 5.07e12 ≈ 1.01 × 10^13 FLOPs per forward+backward pass

Step 2: How many 2,048-token sequences to cover 300B tokens?

• Number of segments N = 300e9 tokens / 2,048 ≈ 1.465 × 10^8 segments

Step 3: Total training FLOPs

• Total FLOPs ≈ (FLOPs per forward+backward) × N
• ≈ (1.01 × 10^13) × (1.465 × 10^8)
• ≈ 1.48 × 10^21 FLOPs

Result

• Rough estimate: about 1.5 × 10^21 FLOPs to train GPT-3 at the 175B scale, under the stated assumptions.

Notes and caveats

• This is a back-of-the-envelope calculation. Real compute could differ by a factor of a few due to:
  • Different batch sizes or gradient accumulation, which changes the effective number of forward/backward passes
  • Use of gradient checkpointing or optimizer options that alter the forward/backward FLOP balance
  • Exact data token count and how sequences are batched
  • Precision (fp32 vs. fp16/bfloat16) affecting the count of floating-point operations
  • Overheads from data loading, padding, sparsity, or other architectural details
• The order-of-magnitude (around 10^21 FLOPs) is the key takeaway, not a precise figure.

If you want, I can run through a variant with a different token budget, sequence length, or depth to show how sensitive the total FLOPs is to those choices.

Here’s a back‑of‑the‑envelope estimate using standard LM training-compute heuristics.

1. Assumptions

• Model size (N): GPT-3 has about 175 billion parameters (N ≈ 1.75 × 10^11).
• Training tokens (T): GPT-3 was trained on roughly 300 billion tokens (T ≈ 3.0 × 10^11).
• FLOPs per token per parameter: A common rule of thumb for transformer LM training is FLOPs ≈ 6 × N × T The factor ~6 comes from counting forward and backward passes for the parameter-heavy matmuls (roughly 2N for forward + 4N for backward). It ignores small terms from layernorms, softmax, embeddings, and optimizer updates (which we’ll account for with a small overhead factor later).

2. Core calculation

• 6 × N × T = 6 × (1.75 × 10^11) × (3.0 × 10^11)
• First multiply N × T: 1.75 × 3.0 = 5.25; powers: 10^11 × 10^11 = 10^22 So N × T = 5.25 × 10^22
• Now multiply by 6: 6 × 5.25 × 10^22 = 31.5 × 10^22 = 3.15 × 10^23 FLOPs

3. Overheads and uncertainty

• Non-matmul ops (attention softmax, masking, residuals, layernorms), embeddings, and optimizer update steps add some overhead. A 10–20% uplift is a reasonable allowance.
• Applying +10–20%: 3.15 × 10^23 × (1.1 to 1.2) ≈ (3.5 ± 0.3) × 10^23 FLOPs.

Answer

• Point estimate: ~3.15 × 10^23 FLOPs
• With a modest overhead allowance: roughly (3.3–3.8) × 10^23 FLOPs

This is a coarse estimate; the exact number depends on details like sequence length, activation checkpointing, implementation efficiency, and the precise token count.

Estimating the total number of FLOPs (floating-point operations) required to train a model like GPT-3 from scratch involves breaking down the problem into several key components. Here's a step-by-step reasoning process:

1. Model Parameters

GPT-3 is known to have 175 billion parameters. This is our starting point.

2. FLOPs per Forward/Backward Pass

For a transformer-based model like GPT-3, the dominant computational cost comes from the matrix multiplications in the attention layers and the fully connected (MLP) layers. The number of FLOPs per forward pass can be approximated as follows:

a) Attention Layers

For a transformer with:

• Sequence length (S): GPT-3 uses a context window of up to 2048 tokens.
• Embedding dimension (d): Typically, ( d = 12288 ) for GPT-3 (since ( 175B ) parameters are divided across layers, and ( d ) is derived from the hidden size).
• Number of heads (h): GPT-3 uses 96 heads, so ( d/h = 128 ) (head dimension).

The self-attention operation involves:

1. Query, Key, Value projections: ( 3 \times S \times d \times d ) FLOPs (since each token's embedding is projected to Q, K, V).
2. Attention scores: ( S \times S \times d ) FLOPs (softmax over ( QK^T )).
3. Attention output: ( S \times S \times d ) FLOPs (weighted sum of V).

Total for attention per layer: ( \approx 2S^2d + 4Sd^2 ).

For ( S = 2048 ) and ( d = 12288 ):

• ( 2S^2d = 2 \times 2048^2 \times 12288 \approx 1.02 \times 10^{11} ) FLOPs.
• ( 4Sd^2 = 4 \times 2048 \times 12288^2 \approx 1.22 \times 10^{12} ) FLOPs.
• Total per layer: ( \approx 1.32 \times 10^{12} ) FLOPs.

GPT-3 has 96 layers, so total for attention: ( 96 \times 1.32 \times 10^{12} \approx 1.27 \times 10^{14} ) FLOPs.

b) MLP Layers

Each MLP layer has:

• Two matrix multiplications: ( d \times 4d ) (expand) and ( 4d \times d ) (contract).
• FLOPs per MLP layer: ( 2 \times S \times d \times 4d = 8Sd^2 ).

For ( S = 2048 ) and ( d = 12288 ):

• ( 8Sd^2 = 8 \times 2048 \times 12288^2 \approx 2.44 \times 10^{12} ) FLOPs per layer.
• Total for 96 layers: ( 96 \times 2.44 \times 10^{12} \approx 2.34 \times 10^{14} ) FLOPs.

c) Other Operations

• Layer normalization, residual connections, and softmax contribute relatively fewer FLOPs (e.g., ( \approx 10% ) of the total).
• Total for other ops: ( \approx 0.1 \times (1.27 + 2.34) \times 10^{14} \approx 3.61 \times 10^{13} ) FLOPs.

d) Total FLOPs per Forward Pass

Summing up:

• Attention: ( 1.27 \times 10^{14} ).
• MLP: ( 2.34 \times 10^{14} ).
• Other: ( 0.36 \times 10^{14} ).
• Total forward pass: ( \approx 3.97 \times 10^{14} ) FLOPs.

e) Backward Pass

The backward pass is typically 2-3x the cost of the forward pass due to gradient computations. Assuming 2x:

• Total backward pass: ( \approx 7.94 \times 10^{14} ) FLOPs.

f) Total FLOPs per Training Step

• Forward + backward: ( 3.97 \times 10^{14} + 7.94 \times 10^{14} \approx 1.19 \times 10^{15} ) FLOPs.

3. Training Tokens

GPT-3 was trained on ~300 billion tokens. Assuming:

• Batch size: 3.2 million tokens (as reported for GPT-3).
• Number of steps: ( 300 \text{ billion} / 3.2 \text{ million} \approx 93,750 ) steps.

4. Total Training FLOPs

Multiply FLOPs per step by the number of steps:

• ( 1.19 \times 10^{15} \text{ FLOPs/step} \times 93,750 \text{ steps} \approx 1.12 \times 10^{20} ) FLOPs.

5. Adjustments

• Mixed Precision: GPT-3 likely used mixed-precision (FP16/FP32), which can reduce FLOPs by ~2x (since some ops are in FP16). However, the theoretical FLOP count remains the same (hardware just executes faster).
• Activation Recomputation: Some frameworks recompute activations during the backward pass to save memory, which can increase FLOPs by ~20-30%. Adding 25%:
  • ( 1.12 \times 10^{20} \times 1.25 \approx 1.40 \times 10^{20} ) FLOPs.
• Optimizer Overhead: Adam optimizer adds ~10-20% FLOPs. Adding 15%:
  • ( 1.40 \times 10^{20} \times 1.15 \approx 1.61 \times 10^{20} ) FLOPs.

6. Final Estimate

After accounting for all factors, the total FLOPs for training GPT-3 is approximately: ~1.6 × 10²⁰ FLOPs (160 exaFLOPs).

This aligns with rough estimates from other sources (e.g., the original GPT-3 paper implies a similar order of magnitude).

Key Assumptions:

1. Sequence length ( S = 2048 ).
2. Hidden dimension ( d = 12288 ) (derived from 175B parameters and 96 layers).
3. Batch size = 3.2M tokens.
4. 300B total training tokens.
5. Mixed-precision and activation recomputation adjustments.