Scaling laws in large language models and toy models

Nicolas Zucchet – nzucchet@stanford.edu

MLSS Tübingen Tutorial – September 8th 2026

Why scaling laws?

The early successes of large language models relied on two main observations:

  1. As we scale models and data, the next-token prediction improves in a predictable way.
  2. As loss improves, the ability of the model to solve tasks we ultimately care about increases.

Scaling laws are the science behind 1., are what justified the huge investments in compute and are key to the pretraining of current models.

They are interesting from an engineering perspective (useful tools to design models), and from a scientific standpoint (suggest some universal principles underlying learning to be discovered!).

Goal of the tutorial

Build intuition on why scaling laws are useful and why they hold.

Agenda

Refresher on language modeling

Part I: scaling laws in practice

  • What scaling laws are and what do they tell us
  • How scaling laws guide model and learning recipe development

Part II: understanding where scaling laws come from with toy models

  • Scaling laws arise in a toy associative memory model
  • What did we learn from the toy model?

Part III: train our own model using scaling laws

Refresher on language modeling

Language model pretraining

In modern large language models, there are many training stages; we focus on pretraining.

Context in, next-token distribution out.

the cat sat on the context sequence model embedding prediction head matcouchfloorroof next-token distribution cross-entropy loss mat observed token

At inference: sample, append, repeat.

Part I: scaling laws in practice

A short history

before 2019 loss sometimes falls as a power law in data and model size 2020 the same laws hold at LLM scale, over seven decades of compute 2022 how to spend a budget: scale N and D together the Chinchilla rule since 2022 extended in many directions: mixture of experts, data limits, inference, recipes, stages now how every frontier run is planned

We review some of the results from (Kaplan et al., 2020) and (Hoffmann et al., 2022) in this part and provide some pointers to extensions.

Loss, model size, data and compute

The variables we care about are:

  • Loss L. Next-token prediction cross-entropy, which is the objective of learning (in the pretraining phase).
  • Model size N. Number of parameters the neural network has (w/o embedding matrices).
  • Number of tokens D. Total number of tokens trained on. Each token usually serves as target of the next token objective once, but can appear in the context much more often.
  • Compute C. Total number of elementary operations (e.g., addition or multiplication in a certain numerical precision like FP16) performed to train the entire model. Proxy for how much it cost to train a model.

N, D and C are actually coupled: we will now show that C = 6ND.

How much does one matrix multiplication cost?

A linear layer does y = Wx, where W\in\mathbb{R}^{m\times n} has N=mn parameters.

weights W · 4 × 3 213 021 112 201 × input x · 3 421 = output y · 4 13 5 8 9 213421 one row, one dot product 02 × 41 × 23 × 1 +++= 13 3 multiplications + 3 additions = 6 FLOPs n multiplications + n additions = 2n FLOPs repeat the same dot product for every row this 4 × 3 example: 4 rows × 6 FLOPs per row = 24 FLOPs in general: 2N FLOPs

From one multiplication to training compute

Learning matrix weights for each new token in the context requires the same number of flops.

For Transformers, matrix multiplications dominate the FLOPs cost (but not memory!), so C = 6ND approximately holds; see (Kaplan et al., 2020) for details.

Side note. This holds for dense Transformers; only the number of activated parameters matters in Mixture of Experts, they increase the number of parameters while keeping FLOPs constant.

Scaling laws

How does the loss L evolve as a function of D, N, and C?

parameters N 1.8k 26k 466k 8.2M 145M 1.3B
106 108 1010 2.5 3 4 6 8 10 tokens processed D loss
Training curves from (Kaplan et al., 2020).

A larger model reaches any given loss after fewer tokens.

Performance improves as we scale models up, but this does not look as smooth as what we were promised…

Scaling laws

How does the loss L evolve as a function of D, N, and C?

parameters N 1.8k 26k 466k 8.2M 145M 1.3B
109 1012 1015 1018 1021 2.5 3 4 6 8 10 compute C loss compute-efficient frontier C-0.050
Training curves from (Kaplan et al., 2020).

The same runs against compute. The envelope (best loss for a given budget) decreases smoothly with compute.

A compute-optimal model is undertrained: optimal training stops with loss still ~8\% above convergence level.

Scaling laws to train the best large language model

In traditional deep learning

  1. tune hyperparameters of the model (learning rate, optimizer, architecture, data…)
  2. see how it affects validation loss
  3. pick the final model using the best hyperparameters

Problem: we can only train our large language model once; how should we pick the hyperparameters?

We can leverage scaling laws to do that!

Compute-optimal model and dataset sizes

We now know what the compute-optimal loss looks like, but not how to achieve it (i.e. how to scale N and D).

Different strategies exist:

  • Pareto front
  • IsoFLOP curves
  • Parametric fit

All should end up with the same result, but they manipulate the data in different ways; it is great to build intuition and will be useful for Part III, so we review them next following (Hoffmann et al., 2022).

Important note. Learning rates are tailored to each model size, and are close to optimal.

Pareto front

For each model size, compare loss against compute (exactly as we did before, but this time with other data).

model size N 57M 175M 552M 1.7B 4.5B 16.2B
1019 1020 1021 1022 2.2 2.5 3 4 5 compute C (flops) loss C-0.052 100M 1B 10B model size N* C0.518 1019 1020 1021 1022 10B 100B compute C (flops) dataset size D* C0.482

All 43 model sizes of (Hoffmann et al., 2022) present in the reconstruction of its Figure 4 by (Besiroglu et al., 2024); the legend names six of them.

IsoFLOP curves

At fixed compute budgets, sweep the model size.

C (×1018) 6 10 30 60 100 300 600 1k 3k
100M 1B 10B 2.2 2.4 2.6 3 3.4 model size N loss 300M 1B 3B model size N* C0.513 1019 1020 1021 10B 30B 100B compute C (flops) dataset size D* C0.487

One \log L parabola in \log N fitted per compute budget, over all nine budgets of (Hoffmann et al., 2022).

Parametric fit

One law for all the runs:

L=E + \frac{A}{N^{a}}+\frac{B}{D^{b}}

E=1.81, A=476, a=0.35, B=1799, b=0.36

Note. Only this construction models the irreducible loss E, the entropy of the data.

Optimizing the loss under the constraint C=6ND gives N^*\propto C^{0.51} and D^*\propto C^{0.49}.

Sanity check: the three methods gave very similar exponents.

loss L 3.0 2.8 2.6 2.4 2.2 2.1
108 109 1010 109 1010 1011 model size N training tokens D C = 1020 1021 1022 compute-optimal

Data from (Hoffmann et al., 2022), fitting from (Besiroglu et al., 2024) (all data points except ones that are too overparametrized).

Chinchilla rule of thumb

Look at the Chinchilla law more closely:

L = E + A \, N^{-a} + B \, D^{-b}.
The two exponents, a=0.35 and b=0.36, are almost the same.

As a result, the ratio between the compute-optimal model and dataset size remains roughly constant.

Rule of thumb: 20 tokens per parameter. Way more than what (Kaplan et al., 2020) advocated for before (Besiroglu et al., 2024), and the current trend is towards a larger number for other reasons, e.g. inference cost (Gadre et al., 2025) (Sardana et al., 2024).

Exercise. Derive the optimal model size and dataset size from the parametric fit and justify the claim that we have made here.

One layer deeper

So far, we have only discussed two parameters N and D that we can play with.

What about more specific architecture details? e.g., number of layers, width of the model, the number of attention heads…

And optimization details? e.g., batch size, sequence length…

Too expensive to retune everything every time, so there are some choices that everyone is making (e.g., what expansion ratio should an MLP have, ratio width / depth of the network, what learning rates should be etc…)

As a result, scaling laws are recipe-specific (on top of being data specific)!

That said, there are some principles that one could use: compare recipe-specific scaling laws.

What about other hyperparameters?

How to develop a recipe (model / optimizer / data…) when we only have limited compute? What matters is how good it will be at large scale, but we can only afford small runs.

A single model size cannot tell us: compare scaling laws.

method A method B
one size: B looks better 100M 1B 10B 100B model size N loss (lower is better) … but the trends meet 100M 1B 10B 100B model size N … or the gap widens 100M 1B 10B 100B model size N

PSA: Only fully trust methods that use baselines with properly tuned learning rates!!

Conclusion and takeaways (Part I)

In next-token prediction language modeling, performance (and other observables) increases smoothly as a function of compute.

Scaling laws provide us with a framework to design the best large model possible when we cannot afford to train it multiple times.

This paradigm applies elsewhere, as long as

  1. model is powerful enough (Transformers usually are!)
  2. data is rich enough and the long tail captures behavior we care about

Such conditions hold in other domains, like vision (Zhai et al., 2022) or time series (Edwards et al., 2024).

To learn more

We have only scratched the surface of scaling laws…

As large language models are becoming more and more complex and used in different practical scenarios, many questions arise:

To see scaling laws in action: (Hall et al., 2025) has fully public LLM development.

Part II: understanding where scaling laws come from with toy models

Where we stand

In Part I, we have discussed what scaling laws are and how they are useful in practice.

There are many open questions though:

  • why power laws (and not, e.g., exponential laws)?
  • where do the exponents come from? how do they depend on the data?

Goals of this part:

  • introduce a toy model of natural language and LLMs
  • get some theoretical intuition on how LLMs behave and how pretraining yields scaling laws

Simplifying language modeling

Assumption 1. Language modeling as context to next-token memories

Next-token prediction requires learning a mapping from contexts to next-token probability distributions

the cat sat on the context
 mat
couch
floor
roof
model0.31
0.24
0.19
0.02
data0.5
0.18
0.32
0

The model explicitly represents next-token probability distributions, while the data only provides samples from an implicit target distribution.

Assumption. No structure shortcuts the learning; nothing generalizes from one context to another, so the model has no choice but to memorize the mapping, context by context.

Simplifying language modeling

Assumption 1. Language modeling as context to next-token memories

Assumption 2. Sequence models as random embedding generation and prediction

contexts the cat sat on the … Paris is the capital of … water freezes at … encode embedding space e₁ e₂ e₃ unit circle predict next token mat.52 couch.23 floor.15 roof.07

We assume that the embeddings are iid distributed uniformly on the unit sphere.

Simplifying language modeling

Assumption 1. Language modeling as context to next-token memories

Assumption 2. Sequence models as random embedding generation and prediction

Assumption 3. Contexts follow a power law distribution

Power laws appear in many places (Newman, 2005). Zipf’s law for word frequencies is the famous example for language: frequency is approximately inverse to rank (Zipf, 1949).

We assume that contexts follow a power law, that is

p(i) \propto i^{-\alpha}.

with i the index of the context.

10⁻¹ 10⁻² 10⁻³ 10⁻⁴ 10⁻⁵ 10⁻⁶ 10⁻⁷ 1 10 100 1k 10k 100k frequency p(i) context rank i few frequent contexts many rare contexts i−1.2

Simplifying language modeling

Assumption 1. Language modeling as context to next-token memories

Assumption 2. Sequence models as random embedding generation and prediction

Assumption 3. Contexts follow a power law distribution

Assumption 4. There is only one likely next token in the data

Let’s keep our life simple, at least for now :)

The toy model

Assumption 1. Language modeling as context to next-token memories

Assumption 2. Sequence models as random embedding generation and prediction

Assumption 3. Contexts follow a power law distribution

Assumption 4. There is only one likely next token in the data

More formally, we have

  • Infinitely many different contexts, with context i appearing with probability p(i) \propto i^{-\alpha}.
  • Each context i has a random embedding e_i drawn from the unit sphere of dimension h.
  • The model predicts the next-token distribution through
    p(\cdot | i) = \mathrm{softmax}(We_i).
  • Only one out of the d tokens exists in the data (we have some multi-class prediction problem).

Some theoretical intuition

Model. p(\cdot | i) = \mathrm{softmax}(We_i), with p(i) \propto i^{-\alpha}. Embedding dimension h, d possible next tokens.

Can we guess (without running experiments!)?

  • How does the optimal loss scale as a function of the model size?
  • How does the optimal loss scale as a function of the dataset size?

In the following:

  1. Derivation of the capacity of the model (number of contexts that can be perfectly predicted) and use it to derive scaling as a function of model size N.
  2. Heuristic derivation for the scaling as a function of dataset size D.

Some theoretical intuition

Model. p(\cdot | i) = \mathrm{softmax}(We_i), with p(i) \propto i^{-\alpha}. Embedding dimension h, d possible next tokens.

We can get some intuition of the storage capacity with the following Hebbian model

W = \sum_i z_i e_i^\top \quad \text{with } z_{ij} = 1 \text{ if } j = y_i \text{ and } 0 \text{ otherwise}.

Corresponds to taking one gradient descent step, starting from W=0, on the objective

\frac{1}{2}\lVert W e_i - z_i \rVert^2.

Contexts with embeddings close to e_i will be pushed towards outputting z_i (and thus to predict the correct next token).

Some theoretical intuition

Model. p(\cdot | i) = \mathrm{softmax}(We_i), with p(i) \propto i^{-\alpha}. Embedding dimension h, d possible next tokens.

We can get some intuition of the storage capacity with the following Hebbian model

W = \sum_i z_i e_i^\top \quad \text{with } z_{ij} = 1 \text{ if } j = y_i \text{ and } 0 \text{ otherwise}.

Predicting the next-token for context i:

We_i = \underset{\textbf{signal}}{\underline{z_i e_i^\top e_i \vphantom{\sum_{j\neq i}}}} + \underset{\textbf{noise}}{\underline{\sum_{j\neq i} z_j e_j^\top e_i}}

The prediction for context i is perturbed by the other embeddings close to e_i.

Some theoretical intuition

Hebbian model. W = \sum_i z_i e_i^\top.

contexts the cat sat on the … wipe your shoes on the … the coin rolled on the … she mopped the kitchen … next token mat mat floor floor encode embeddings embedding output

Querying. (W e_i)_y = \sum_{j \,:\, y_j = y} e_j^\top e_i: compare the query e_i with all the other embeddings and see which color dominates.

Some theoretical intuition

Interference. The prediction for context i is perturbed by the other embeddings close to e_i.

prediction: incorrect query prediction: correct increasing h makes embeddings more orthogonal prediction: correct increasing d spreads noise over more tokens

As a result, capacity increases with d and h.

Exercise. Demonstrate that it is proportional to hd up to logarithmic factors.

Capacity, in theory and in practice

one memory per parameter trained Hebbian

In theory, capacity is proportional to the number of parameters, up to logarithmic factors. We fix d=256 and vary h.

At large h the trained model is exactly parallel to N (slope 1.01) and stores 0.16 memories per parameter.

Hebbian gets 0.02, an order of magnitude less.

To learn more. (Cabannes et al., 2024) for the full analysis, (Nichani et al., 2025) for other architectures, (Zucchet et al., 2026) for several possible next tokens.

Capacity, in theory and in practice

In practice, capacity is also proportional to the number of parameters. For Mixture of Experts, the total number of parameters matters, so they store more knowledge per active parameter.

50K 100K 200K 500K 1M 2M facts 2 bits/param

Stored synthetic facts, 1000 exposures each (Allen-Zhu & Li, 2024)

0.17M 0.5M 2.5M 7M params bits in the data

Random bitstrings memorized (Morris et al., 2025)

From capacity to scaling laws

Now that we have some idea of how the capacity scales as a function of the model size, how do we get scaling laws?

For the model size: assume infinite data, and that the first contexts below capacity get 0 loss and some value l afterwards.

L(i) l 0 capacity(N) context index i

From capacity to scaling laws

Now that we have some idea of how the capacity scales as a function of the model size, how do we get scaling laws?

For the model size: assume infinite data, and that the first contexts below capacity get 0 loss and some value l afterwards.

As a result:

L(N) = \sum_i p(i) \, L(i)

Average the per-context loss over how often each context appears.

\phantom{L(N)} \propto \sum_{i \leq \mathrm{capacity}(N)} i^{-\alpha} \times 0 \; + \sum_{i > \mathrm{capacity}(N)} i^{-\alpha} \times l

Contexts below capacity are free; each one above costs l.

\phantom{L(N)} \propto \mathrm{capacity}(N)^{1 - \alpha} \propto N^{1-\alpha}.

The tail sum is set by where it starts, and capacity grows like N.

From capacity to scaling laws

For the model size: the loss is a step function of the context index, and the step moves right as N grows.

L(i) l 0 capacity(N) context index i L(N) L(N) 1 0.1 0.01 1k 10k 100k 1m number of parameters N N1−α

Note that in this case there is no residual entropy and the loss will converge to 0 under infinite compute and data.

From capacity to scaling laws

For the dataset size: assume instead that the model gets right every context it has seen at least once.

Context i has a high chance to appear in the first D tokens as soon as

p(i) \, D \geq 1 \qquad \Longleftrightarrow \qquad i \leq D^{1/\alpha}.

Only the first D^{1/\alpha} contexts are ever seen.

Using the same argument as before,

L(D) = \sum_{i > D^{1/\alpha}} p(i) \, l \; \propto \; \left(D^{1/\alpha}\right)^{1-\alpha} = D^{\frac{1}{\alpha} - 1}.

Same tail sum, cut at D^{1/\alpha} instead of the capacity.

From capacity to scaling laws

We now have intuition for how the loss scales as a function of model size N and data D.

L 1 0.1 0.01 1k 10k 100k 1M N1−α D1/α−1 model size N dataset size D

Power laws in the data become power laws in the loss, in both model size N and tokens D.

Whatever the value of \alpha, the bottleneck is the data D.

As \alpha increases and the tail becomes rarer, the data becomes even more of a bottleneck.

Note. Faster loss decay is not necessarily a good thing here, as the long tail takes longer to learn.

Are the toy model predictions good enough?

If we further assume Chinchilla scaling, we get:

L(N, D) = \frac{A}{N^{\alpha - 1}} + \frac{B}{D^{\frac{1}{\alpha}-1}}.

Optimizing under the constraint C=6ND gives

\begin{align*} N^*(C) &\propto C^{\frac{1}{1+\alpha}}\\ D^*(C) &\propto C^{\frac{\alpha}{1+\alpha}}\\ L(C) &\propto C^{\frac{1 - \alpha }{1+\alpha}} \end{align*}

Does this back of the envelope calculation predict what is happening in simulation?

Are the toy model predictions good enough?

At fixed compute budgets, we sweep the model size and fit a parabola in \log N.

C (×109) 10 40 160 640 2.6k 10k
10k 100k 1M 0.8 1 1.2 1.5 model size N loss 10 100 1k 10k 100k 1M compute C (×109) model size N* C0.449

Six IsoFLOP profiles. Learning rates are tuned for each run.

The loss decreases as C^{-0.091} (predicted C^{-0.091}), the model size grows as C^{0.449} (vs. C^{0.455}).

Are the toy model predictions good enough?

We assumed the Chinchilla scaling and recovered similar compute exponents as in theory, but how good is the full law?

L= E + A N^{-a}+B D^{-b}.

N and data D treated as independent.

Fit on 75 low-compute runs, then test on 4 held-out runs at 816\times the fitting budget.

fit region
extrapolation
Chinchilla
3.2%
4.8%
Skaling
0.4%
0.6%

Relative rms error on L-E; E is known here (0).

Additive scaling laws do not extrapolate well.

Idea: couple model size and data.

L = E + \left(\frac{A}{N^a}+\frac{B}{D^b}\right)^k

To learn more. (Videau et al., 2026) on why this functional form makes sense and how it better extrapolates.

Do the exponents follow the theory?

How good are the scaling exponents compared to the theory as we change the data distribution, in particular \alpha?

To estimate those coefficients, we train models of a given size until the loss plateaus (model coefficient), and train the largest reasonable model varying the dataset size (data coefficient).

Our super simple theory is not too bad!

model limited data limited theory

What if data was finite?

Same experiment as before, but only keeping first 2,000 contexts (instead of infinitely many).

The loss starts decreasing as a power law and finishes as an exponential one.

This is the regime deep learning used to be in: with a bounded amount to learn, held-out loss saturates.

But score those same models on the full distribution and the gain largely vanishes: the held-out loss flattens, and stays worse than training on everything.

first 2k contexts tested on all infinite context pool

Digression: emergent behavior under the hood

At the macro level, performance improves smoothly, but at the task level it can suddenly jump with scale: emergent abilities (Wei et al., 2022) (Olsson et al., 2022).

Our toy model can capture some of this behavior!

To learn more. This toy model does not capture emergence through learning, see (Zucchet et al., 2025) for an example of such a toy model.

ranks 1-1k 2-3k 5-6k 10-11k 20-30k

Conclusion and takeaways (Part II)

We modeled language modeling with an associative memory exposed to a power law data distribution.

With simple theory, we were able to accurately predict scaling law exponents.

Takeaway

Scaling power laws naturally arise when the data itself follows power laws.

Warning. We cannot say anything on whether this mechanism is the bottleneck that yields specific scaling behavior in actual LLMs, we can just say how changing the data affects scaling (which would be much harder in practice!).

To learn more

This is only one possible theory of scaling laws and we do not yet understand which mechanisms yield the scaling laws.

Some cool work in the field:

Part III: let’s train our own model using scaling laws

Training a frontier toy model

Problem. We have access to new important data and want to train the best possible (toy) model on it.

Constraints. You don’t know what the model is and can control the number of parameters N, the number of tokens D, as well as the learning rate, and what are the tuning experiments you do. The total amount of compute for both hyperparameter optimization and the final run is fixed.

Your job. Come up with the best model!

The code

Notebook available here, underlying model and data are a black box on purpose.

lab = Lab("me", budget=1e13, rounds=3)  # contains task (max 3 rounds), DON'T TOUCH
sweep = Sweep(
     c=[4e9, 12e9],
     n=[2**(13 + i) for i in range(5)],  # fix C+N, C+D, or N+D
     lr=[0.001, 0.01, 0.1],  # either array, or a function of c, n, d  
)  # Cartesian product, test all combinations
sweep + Sweep(n=[2**14], d=[2**14], lr=[0.01])  # Sweeps can be added!
sweep.estimate(lab)  # estimates FLOPs used by the sweep

round_results = lab.run_round("first round", sweep)  # runs the sweep
# Fits are method specific, see notebook
result = lab.hero(n=n, d=d, lr=lr, predicted=loss)  # final run!!

Additional details

  • Total compute budget is 10^{13} FLOPs
  • At most 3 rounds, running everything should take 5–10 minutes
  • No AI help for strategy
  • Only restart experiments in case of bugs
  • Only run Part 0 and Part 2 of the code (Part 1 is just demo)
  • Choose Colab runtime to be GPU (will be faster!)
  • Report your results at the end using the link that will be given!

How did the room do?

0 %100 %of the budget spent on the hero run
loading the sheet ...

From the notebook. lab.hero(...) prints a form link with your three numbers already in it – click it, submit.

Or fill the form yourself.

References (1/4)

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