How do you know a fine-tuned AI model is actually better?

You measure it. Hold out a portion of the client's real examples so the model never sees them during training, score the plain base model on those held-out questions first, fine-tune, then score again on the identical questions with identical scoring. The difference is the lift. Because the only thing that changed is the fine-tune, the improvement is attributable to it.

Does fine-tuning actually improve accuracy, or is the base model already good enough?

It depends on the task and, critically, on how much data you have. On a public benchmark of 510 lawyer-annotated commercial contracts, a base Qwen3-4B extracted short contract fields at 62.2% accuracy. Fine-tuned on 396 examples it reached 83.5-85.4%, a lift of 21.3 to 23.2 points across three runs. The same model trained on only 48 examples produced a lift inside the noise floor, so the amount of data was the deciding factor rather than the model's ceiling.

How much of our own data do we need to fine-tune a model?

Fewer examples than most people expect, but more than a handful. In our contract-extraction measurements, 48 training examples produced no measurable improvement while roughly 400 produced a 22-point gain. The honest answer for any specific case is that it depends on how varied and how consistent the examples are, which is what a paid readiness assessment measures on the client's own corpus.

Why can't you just use a public AI benchmark score?

Public benchmarks measure general ability on standardised tests, not your work. A model that scores well in the abstract can still be mediocre at reading your contracts or following your policies. Public benchmarks are also increasingly targeted during training, so a high score can reflect test preparation as much as capability. The only benchmark that answers your question is built from your own tasks.

Research record 43 of 67

Predicting The Cost From The Task

By Joshua Bolick, Awareness Software Group ~13 min read August 2026

Ongoing research. This is an experimental result from active work, not a settled conclusion. The numbers are what we measured and the method is described so you can judge it, but the programme is still running and later experiments may revise what it means. More about this programme.

In plain English

What we asked. In a previous experiment we denied a model each earlier word in turn and measured how much longer it then took to learn. Those costs fell into neat groups, and the groups lined up with the length of the repeating pattern in the exercise rather than with how far back the word sat. Words the same distance apart in the pattern cost the same to take away, even when one was much further back than the other. That was a striking observation and it was also, on one exercise, indistinguishable from a coincidence about the particular numbers involved.

What we found. So we changed the exercise. We ran the whole thing again with repeat lengths of three, four and five, and for each one we scored how cleanly the costs group under every possible grouping, not just the one we expected to win. That last part matters: a grouping that looks tidy is worth nothing if a wrong grouping looks equally tidy. The correct grouping won in every exercise, and every wrong grouping came out at about the value that means no grouping at all. Where the split was even, two words in the same group cost exactly the same to the decimal place.

Why it matters. This is the first thing we have found on this project that can be worked out from the exercise alone, without looking at the model. Everything else we publish measures what a model does; this says what the task itself dictates about the price of interfering with it. Two consequences are worth knowing. Taking away a word that holds no part of the current answer can cost more than taking away the word that *is* the answer, because a repeating exercise gives the model a spare copy of the answer to fall back on. And a few removals actually speed the model up, because they save it the work of learning to ignore a distraction. What we still cannot say is *which* group will turn out to be the expensive one -- that changes between exercises, and we do not yet know why.

The rest of this page is the technical record: the design, every number, and the limits. It is written for a reviewer, and you do not need it to have understood the result above.

New here? How to read a research record

Start at the verdict. Every record states, before the experiment was run, what result would have made us abandon the idea. That is the "kill test". Then it says whether the test fired. Nothing gets reinterpreted after the fact.
Numbers in square brackets are uncertainty. 23.4 [18.1, 28.7] means our best estimate is 23.4 and the true value is probably somewhere in that range. If a range includes zero, we cannot claim an effect.
Negative results are kept. Roughly half of what is published here says an idea did not work, including several of our own. Those pages are not failures, they are the output. Work that only publishes what worked is not measuring anything.
Read the Limits section. Every record ends with what it does not show. It is the most honest part of any experiment and usually the shortest.
Pro tip: the figures near the top are designed to carry the result on their own. If you read nothing else, read the caption under each one, which says what it shows and what to take from it.

EXPLORATORY. Not a preregistered study. Local CPU, 108 training runs, no GPU, no cost.

Program v2 Bucket N, item N12. Decisive computation: analysis/cost_classes_move.py. Output: analysis/cost_classes_move.json. Reproduce with python analysis/cost_classes_move.py, or re-derive every endpoint with --reuse.

The question

N4 denied a model each earlier position in turn and found the costs cluster into clean pairs, lags 1 and 5 agreeing to 0.8 steps, lags 2 and 6 to 5.0, a between-class range of 53.3 against a within-class spread of 8.1, ratio 6.56. On delayed-copy at lag 4 those pairs are the residue classes modulo 4.

**How strongly the cost of an intervention falls into groups.** In an earlier result we denied a model each earlier word in turn, and the costs fell into neat groups matching the repeat length of the exercise. That could have been a coincidence about the numbers involved. So we changed the repeat length -- three, four and five -- and for each scored how cleanly the costs group under every candidate grouping, not just the right one. The orange starred bar in each set is the grouping that matches the exercise. The correct grouping wins in every exercise, and every wrong one sits at about 1, which is what no grouping at all looks like. So the cost of taking something away from the model is set by that thing's role in the structure of the exercise, and it moves when the exercise moves. This is the first thing we have found that lets you predict what an intervention will cost from the task alone, without looking at the model. The effect is weakest on the shortest repeat, where the groups are less even.

That reading was descriptive, and one task cannot tell it apart from a coincidence about the numbers 1 to 8. If the grouping is really about which prefix element a position carries, then changing the task's lag must move the classes with it.

Kill test, fixed in advance: the classes do not move with the lag.

The endpoint is a ranking, so no threshold can manufacture it. For each task lag the clustering ratio is computed under every candidate modulus, not only the true one. A ratio that is large under the true modulus but equally large under a wrong one says nothing; the true modulus has to win.

Result: the classes move, at every lag tested

task lag	modulus 2	modulus 3	modulus 4	modulus 5	modulus 6
3	`0.12`	`2.04` `[1.55, 2.53]`	`0.64`	`1.05`	`0.43`
4	`0.18`	`0.99`	`6.99` `[5.36, 8.62]`	`0.61`	`0.16`
5	`0.15`	`0.59`	`1.02`	`11.20` `[4.60, 17.79]`	`0.96`

The task's own lag wins every row, and no runner-up interval overlaps the winner's: at lag 3 [1.55, 2.53] against [0.99, 1.12]; at lag 4 [5.36, 8.62] against [0.86, 1.12]; at lag 5 [4.60, 17.79] against [0.97, 1.07]. Every wrong modulus sits at or below 1.05, which is what "no clustering at all" looks like on this statistic.

The kill test does not fire. N4's anchor reproduced: its 6.56 sits inside the lag-4 interval [5.36, 8.62], on different seeds and with the sign convention imported rather than re-derived.

The maps

Mean delay in steps, by the position denied, grouped by residue class.

task lag	class	positions and delays
3	0	lag 3 `−18.8`, lag 6 `−23.8`
3	1	lag 1 `+11.2`, lag 4 `+13.8`, lag 7 `+3.8`
3	2 (the answer)	lag 2 `+35.0`, lag 5 `+6.2`, lag 8 `−1.2`
4	0	lag 4 `+3.8`, lag 8 `−11.2`
4	1	lag 1 `+6.2`, lag 5 `+6.2`
4	2	lag 2 `+47.5`, lag 6 `+52.5`
4	3 (the answer)	lag 3 `+30.0`, lag 7 `+18.8`
5	1	lag 1 `+2.5`, lag 6 `+2.5`
5	2	lag 2 `+30.0`, lag 7 `+31.2`
5	3	lag 3 `+82.5`, lag 8 `+62.5`

At lag 4 the class-1 pair is identical to the decimal place; at lag 5 so is its class-1 pair, and its class-2 pair agrees to 1.2 steps. Positions the same distance apart in the task's own periodic structure cost the same thing to remove, however far back they sit.

What is weaker, and it is the lag-3 row

Lag 3's ratio is 2.04 against lag 4's 6.99 and lag 5's 11.20, and its classes are visibly noisier, class 2 spans +35.0 to −1.2. The verdict there rests on a clear but much smaller margin. Two readings are available and this record cannot separate them: shorter lags make an easier task, so blocking any one position matters less; or three classes over eight positions gives an uneven 3/3/2 split where four classes over eight gives a clean 2/2/2/2. The trend is that the effect strengthens with lag, on three points.

What this establishes

The cost of an intervention is predictable from the task's structure alone, without reference to the model, the run, or which position happens to be involved. That is the first result in this programme with that shape: everything else here measures what a model does, and this measures what the task dictates about what doing something to the model will cost.

Two consequences carried over from N4 and now known to be structural rather than incidental: denying a class that carries no part of the current answer can cost more than denying the class that is the answer (lag 4: +49.2 against +23.3), because a periodic task supplies more than one copy of its answer; and denying some classes accelerates training, lag 3's class 0 by −18.8 and −23.8 steps, lag 4's lag 8 by −11.2.

It does not explain why one class is the expensive one. The measurement says cost is a function of residue; it does not say which residue will be costly, and the ordering is not the same across lags. That is the next question and it is a genuinely open one.

Verdict

The kill test does not fire. The true modulus wins at all three task lags, with no runner-up interval overlapping and every wrong modulus at or below 1.05.
N4's anchor reproduced on different seeds.
The effect strengthens with lag, 2.04, 6.99, 11.20, on three points.
Within-class agreement is exact where the split is even: +6.2/+6.2 at lag 4, +2.5/+2.5 at lag 5.
Which residue is expensive is not predicted, only that cost is a function of residue.

Limits

Four seeds per cell, three task lags, one width, one learning rate, one block window and strength. The window (100 steps) and strength (10.0) are N4's and were not swept.
Three points is a trend, not a law. "Strengthens with lag" is a description of 2.04, 6.99, 11.20, and lags 6 and above are untested.
The class split is uneven at some lags. Over positions 1 to 8, lag 3 gives 3/3/2 and lag 5 gives 2/2/2/1/1, and classes with one member are excluded from the ratio by construction. That is a defensible rule but it means the lags are not compared on identical footing.
One task family. delayed-copy is periodic by construction, which is exactly what makes residue classes meaningful. Nothing here says the result generalises to a task without that structure, and M1 showed this task is solved outright by a lookup.
The intervention is on the training objective, not on the architecture or the data, so "cost" means "delay to the accuracy midpoint under this particular block".
Why a given residue is costly is unexplained, and the ordering differs between lags, so the obvious story: that the expensive class is the one needed soonest, is not supported by these three maps.

Terms on this page

Every piece of vocabulary this record uses, in plain language. Generated from the text above, so it cannot drift out of step with it.

accuracy: The fraction of answers a model gets right on questions it was not trained on.
architecture: The blueprint of a model: what components it has and how they connect. Two models can be the same size and completely different architectures.
kill test: A condition written down before running the experiment that says what result would make us abandon the idea. Fixing it in advance is what stops a disappointing result being reinterpreted as an encouraging one.
learning rate: How big a step training takes each time it updates the model. Too small and nothing happens; too big and it never settles.
seed: The number that fixes all the randomness in a training run. Same seed, same run. Running several seeds is how you tell a real effect from a lucky one.
width: How many internal numbers a model uses at each layer. The usual way we vary model size in these experiments.