Mistral Large
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 .
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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.
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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.
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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} ).
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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 ).
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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 )
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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.
