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Lifecycle-Optimal Tokenization: Vocabulary Size as a Deployment-Regime-Dependent Infrastructure Parameter

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arXiv cs.LG

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arXiv:2608.11361v1 Announce Type: new Abstract: Tokenizer vocabulary size is a foundational design choice in large language model (LLM) infrastructure, yet it is typically fixed at training time based on convention rather than deployment analysis. We show that the cost-optimal vocabulary is not a constant but a function of the serving regime. We formalize total deployment cost as $C_{lifecycle}(V) = C_{train}(V) + \lambda \cdot C_{infer}(V, B)$, where $\lambda$ is inference volume and $B$ is the serving batch size. Through controlled experiments on two GPU families spanning the memory-bound to compute-bound regimes (A10G, ridge $\approx$ 117 FLOP/byte; A100, ridge $\approx$ 183 FLOP/byte), we demonstrate: (1) the inference-optimal vocabulary shifts 16x with serving batch, from 32k at $B=1$ to 524k at $B=64+$, driven by amortization of the $V \times d$ unembedding matrix read; (2) at 1.3-2.3B model scale, quality (bits per byte, BPB) is optimized at $V=65$k, confirming scale-dependent vocabulary preference; (3) the lifecycle-optimal vocabulary diverges from training-optimal by up to 16x for production deployments. Quality is approximately invariant across the optimal range ($<$2% BPB spread), making vocabulary a pure systems optimization with no quality penalty in the measured range. Our results provide actionable capacity planning guidance: on-device deployments ($B=1$) should use $V \approx 32$k; datacenter serving ($B \geq 64$, $\lambda \geq 10$) should use $V \approx 131$-262k.

Key takeaways

  • 01arXiv:2608.11361v1 Announce Type: new Abstract: Tokenizer vocabulary size is a foundational design choice in large language model (LLM) infrastructure, yet it is typically fixed at training time based on convention rather than deployment analysis.
  • 02We show that the cost-optimal vocabulary is not a constant but a function of the serving regime.
  • 03We formalize total deployment cost as $C_{lifecycle}(V) = C_{train}(V) + \lambda \cdot C_{infer}(V, B)$, where $\lambda$ is inference volume and $B$ is the serving batch size.
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