Equation 6 · The Deployment Envelope: Small Models Where the Power Is Not
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the halving the bits per weight halves. Read the equation part by part below; each part has a contextual explanation and a link to its mathematical background.
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Autoregressive decoding sits at the far left of that graph. Generating one token requires reading essentially every weight and the accumulated key-value cache once, and performing roughly two arithmetic operations per parameter. At one byte per weight the intensity is about two operations per byte; at four bits per weight, about four. Pope and colleagues formalised this partitioning problem for large transformers and showed how latency, throughput and cost trade off under different sharding strategies, with generation and prefill behaving as different regimes [ 3 ] . The bound that matters on a device follows immediately: if W bytes of weights and cache must cross the memory bus for each…
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Autoregressive decoding sits at the far left of that graph. Generating one token requires reading essentially every weight and the accumulated key-value cache once, and performing roughly two arithmetic operations per parameter. At one byte per weight the intensity is about two operations per byte; at four bits per weight, about four. Pope and colleagues formalised this partitioning problem for large transformers and showed how latency, throughput and cost trade off under different sharding strategies, with generation and prefill behaving as different regimes [ 3 ] . The bound that matters on a device follows immediately: if W bytes of weights and cache must cross the memory bus for each token and the sustained bandwidth is B , then the time per token cannot be less than
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