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Batch size counts the examples combined in a mini-batch. In the simple stochastic gradient descent loop where each group produces one parameter update, it controls how many examples contribute to that update. It differs from a training epoch, which counts a pass over the available training data.^[ref]

For an original counting example, partition ten examples with nominal batch size $b=4$ and retain the remainder. The actual sizes are $4,4,2$, giving $\lceil 10/4\rceil =3$ updates per training epoch. Discarding the remainder instead gives two updates and uses only eight examples. More generally the retained-remainder update count is $\lceil n/b\rceil$ for $n$ examples, provided there is one update per group.

Batch size also changes gradient sampling noise. If independently sampled component gradients at a fixed parameter have scalar variance ${\sigma }^2$, their sample mean hasFor an original exact example, suppose one component gradient is equally likely to be 0 or 2. Its mean is 1 and variance is 1. A two-example mini-batch gradient takes values $0,1,2$ with probabilities $1/4,1/2,1/4$, giving variance $1/2$. This calculation holds parameters fixed and assumes independent sampling; sampling without replacement has a finite population correction.

With gradient accumulation, several groups may contribute before an update. Two full four-example groups then involve eight examples per update if their gradients are normalized consistently. State whether a reported size counts the processed group or all examples contributing to an update. Lower sampling variance alone does not establish faster training for a fixed computation budget.

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.

An optimizer update changes the adjustable parameters of an optimization problem according to an algorithm's current information. Evaluating a loss function, computing its gradient and changing the parameters are distinct operations. A general update may also change an optimizer state, such as stored information from previous steps:Here $g_k$ is the supplied gradient and ${\eta }_k$ a learning rate. The PyTorch interface makes this separation explicit with gradient computation followed by optimizer.step().^[ref]

For plain gradient descent with no extra state, ${\theta }_{k+1}={\theta }_k-\eta g_k$. In an original scalar example, choose $f(\theta )={\theta }^2/2$, ${\theta }_0=2$ and $\eta =0.1$. The gradient is 2, so the update gives ${\theta }_1=1.8$ and lowers the loss from 2 to 1.62. Recompute the gradient at the new parameter value to obtain ${\theta }_2=1.62$. Reusing the old gradient 2 instead would give 1.6. This explicitly locates where the algorithm obtains its information.^[ref]

An update is not automatically an improvement: using $\eta =3$ for the first step above gives ${\theta }_1=-4$ and loss 8. With gradient accumulation, several gradient calculations instead contribute before a single optimizer update; their weighting sets the effective batch size. Track the update count separately from the number of processed groups, particularly when applying a learning rate schedule.