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Data parallel replicas

Compare kernel, parallelism and disaggregated-serving choices using memory, communication and latency constraints.

Data-parallel replicas

The simplest way to serve more traffic is to run several complete copies of the model, replicas, each handling its own requests. Aggregate capacity grows with the number of replicas (three replicas of 70 tokens/s give 210 tokens/s). But a single request still runs on one replica and passes through every layer, so its own speed does not improve.

replicas3target aggregate tokens/s500
replica 1
32 layers
replica 2
32 layers
replica 3
32 layers
aggregate capacity = 3 × 70 = 210 tokens/s replicas needed for 500 tokens/s: ceil(500/70) = 8 one 100-token request: 1.43 s whatever the replica count, and it still passes through all 32 layers

Replicas serve separate requests with their own copy of the model. They raise how much work the service can take at once; they do not make one request's forward path shorter.

Add replicas and see capacity rise while the single-request time stays the same.

Predict first. Does a second identical replica halve one request's layer count?

Chapter context

Scaling serving changes how work is placed and moved. An operation can be mathematically unchanged while its memory traffic changes dramatically. Device count alone does not describe the performance of the resulting system.

Chapter reference: math, worked example & code

This section belongs to the whole chapter, so it is the same on every lesson in it and may cover a different lesson's topic. The code for the experiment you just ran is inside the experiment itself.

Math and notation

A weight-traffic lower-bound estimate is

tstep≳Mread/BW.t_{step}\gtrsim M_{read}/BW.

This assumes the stated bytes really traverse the bottleneck bandwidth. KV transfer adds roughly bytes_to_move/network_bandwidth before protocol and synchronization overhead. Batching and caching can change the bytes per produced token.

Worked example

A fictional 14-GB dense weight scan across a sustained 700-GB/s memory path takes at least 14/700=0.02 seconds, giving a 50-step/s bandwidth ceiling under the one-scan-per-step assumption. Real execution includes other memory traffic and work. A 512-MiB KV transfer over 32 GiB/s has a payload-only lower bound of 1/64 second.

Reference code

The following example uses the Python standard library and does not call a remote service.

from math import isclose
assert isclose(14/700,.02)
assert isclose(1/(14/700),50)
transfer=(512*1024**2)/(32*1024**3)
assert isclose(transfer,1/64)

Python examples are provided for study and local execution; this page does not execute Python.

Common mistakes

Peak device bandwidth is not guaranteed sustained application bandwidth. Exact attention can be computed with different memory layouts. Additional GPUs can add communication bottlenecks.