Gradient accumulation

Gradient accumulation combines gradient contributions from several processed mini-batches before one optimizer update. Parameters stay fixed while those contributions are evaluated. With consistent normalization, an additive per-example loss function can therefore use an effective batch size larger than the group processed at once.^[ref]

For an original example, let per-example scalar gradients be $1,3,8$ at the same parameter value. Process the first two together and the last separately. Their means are 2 and 8. The combined per-example mean isThe unweighted mean of group means is 5 and gives the smaller group too much weight. One implementation accumulates each group's summed loss divided by the total example count, then performs one update.

Accumulation is also different from consecutive updates. Let $f_i(\theta )=(\theta -a_i{)}^2/2$, with $a=(-1,-3,-8)$, starting at $\theta =0$ and using step size $0.1$. One accumulated mean-gradient step gives $\theta =-0.4$. Updating first on the two-example mean gives $\theta =-0.2$; the last example then has gradient $7.8$, and a second step gives $\theta =-0.98$.

The distinction is where the gradients are evaluated. Agreement with one larger group requires fixed parameters and compatible model state, additive losses and consistent weighting. A method whose calculations depend on the whole group can violate those assumptions. State whether batch size denotes a processed group or the complete set contributing to an update.

The effective batch size counts the example contributions combined in one optimizer update, including gradient accumulation and contributions from parallel workers. With $P$ workers, $a$ accumulated groups per worker and $b$ examples in each full group,The PyTorch documentation uses this count to specify when accumulated gradients are unscaled and the optimizer is stepped.^[ref]

For an original example, two workers each process three groups of four examples before updating: $b_{\mathrm{eff}}=2\cdot 3\cdot 4=24$. For an additive per-example loss function, the desired gradient is $24^{-1}{\sum }_{i=1}^{24}\nabla f_i(\theta )$, evaluated at unchanged parameters $\theta$. If each worker accumulates the three group-mean gradients divided by three, then averaging the two workers' results gives that normalization. Summing the workers instead would double the gradient.

Incomplete groups require their actual counts. Group sizes $4,4,2$ on each of two workers give $b_{\mathrm{eff}}=20$, and the combined mini-batch gradient weights each group mean by its example count. Dividing all six group means by six would overweight the two-example groups.

The count describes contributions, not necessarily distinct examples. If both workers process the same twelve example identifiers, there are 24 contributions but only twelve distinct examples. The usual inverse-count reduction of variance requires assumptions about independent random variables; duplication can violate them. Nor does this bookkeeping by itself establish a rule for changing the learning rate.