FlexKernelOptions sweep: flex reaches 72% of vLLM at batch 64 + LoRA
Summary of sweep (all with CUDA graph capture):
| Batch | flex tps | vLLM tps | flex / vLLM | flex mem |
|------:|---------:|---------:|------------:|---------:|
| 8 | 680 | 1900 | 35.8 % | 44 GB |
| 16 | 1626 | 3698 | 44.0 % | 44 GB |
| 32 | 3134 | 6318 | 49.6 % | 44 GB |
| 64 | 5474 | 10459 | **52.3 %** | 44 GB |
| 128 | 5565 | 14996 | 37.1 % | 81 GB |
| 256 | 5812 | 21170 | 27.5 % | 154 GB |
Canonical GRPO (batch 64 + LoRA rank 32):
- vLLM: 7775 tok/s / 156 GB
- flex: **5616 tok/s / 44 GB** = **72% of vLLM at 3.5x less memory**
Up from 9 % (transformers CB) at the start of this work.
Best FlexKernelOptions after sweep:
decode: PRESCALE_QK, USE_TMA, BLOCKS_ARE_CONTIGUOUS, num_warps=8, num_stages=3
prefill: FORCE_USE_FLEX_ATTENTION, PRESCALE_QK, USE_TMA
Biggest single win: `num_warps=8` (+28% at batch 64). Inductor's default
picks 4 on small Triton blocks; 8 is better for our decode shapes.
`BLOCKS_ARE_CONTIGUOUS` adds +10% (safe in our setup because
PageTable.reserve allocates pages sequentially on a fresh batch).
TMA adds 2-3%.
Items documented that broke correctness or didn't help:
- ROWS_GUARANTEED_SAFE=true NaNs the softmax on padded batch slots that
only attend to reserved page 0 (mask returns False for every kv_idx).
- BACKEND="TRITON_DECODE" from the docs raises
NameError('TRITON_DECODE is not defined') inside Inductor.
- USE_TMA + torch.compile(call_model_with_flex_kwargs) -> misaligned
address at runtime (compile breaks TMA alignment assumptions).
- torch.compile(flex_attention, mode="max-autotune") nests cudagraph_trees
inside our raw CUDA graph -> "Cannot prepare for replay during
capturing stage". max-autotune-no-cudagraphs works but same throughput
as default mode.
- compile on call_model_with_flex_kwargs: same as eager walker (CUDA
graph capture already fuses every op in the walker).
- num_warps=4 / 16 both slower than num_warps=8.
CLI surface added to qwen3_flex_inference.py:
--decode_kernel_options JSON (FlexKernelOptions for decode)
--prefill_kernel_options JSON (same for prefill)
--compile_model_forward MODE (optional torch.compile on the walker)
This commit is contained in:
Daniel Han
parent
7f237ca57e
commit
69723ee31c
12 changed files with 457 additions and 85 deletions
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@ -211,6 +211,23 @@ class Sequence:
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return self.input_length + len(self.output_ids)
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# Default kernel_options per phase. Our defaults stay conservative -- the
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# non-default FlexKernelOptions (PRESCALE_QK, ROWS_GUARANTEED_SAFE, USE_TMA)
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# are opt-in via CLI because some of them break correctness on our
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# paged-attention setup.
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#
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# Specifically, `ROWS_GUARANTEED_SAFE=True` is unsafe here: we reserve
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# batch_idx=0 and page_idx=0 as no-op padding slots. When a decode
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# padded batch row maps to only-reserved pages, the block mask returns
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# False for every kv_idx, so the row has zero unmasked values. The flag
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# tells the kernel to skip the row-has-at-least-one-unmasked check, so
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# the softmax NaNs silently -- which manifests as "!!!!!!" token spam.
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DECODE_KERNEL_OPTIONS_DEFAULT = None
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# Prefill keeps FORCE_USE_FLEX_ATTENTION so we don't auto-dispatch into
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# the flex-decoding kernel when the packed q_len gets small.
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PREFILL_KERNEL_OPTIONS_DEFAULT = {"FORCE_USE_FLEX_ATTENTION": True}
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class FlexInference:
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def __init__(
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self,
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@ -221,6 +238,8 @@ class FlexInference:
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n_pages = 2048,
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page_size = 128,
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max_new_tokens = 512,
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decode_kernel_options = None,
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prefill_kernel_options = None,
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):
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assert max_seq_length % page_size == 0
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self.model = model
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@ -231,6 +250,16 @@ class FlexInference:
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self.max_seq_length = max_seq_length
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self.page_size = page_size
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self.max_new_tokens = max_new_tokens
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self.decode_kernel_options = (
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decode_kernel_options
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if decode_kernel_options is not None
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else DECODE_KERNEL_OPTIONS_DEFAULT
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)
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self.prefill_kernel_options = (
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prefill_kernel_options
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if prefill_kernel_options is not None
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else PREFILL_KERNEL_OPTIONS_DEFAULT
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)
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self.page_table = PageTable(
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n_pages = n_pages,
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@ -299,7 +328,7 @@ class FlexInference:
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flex_block_mask = mask,
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flex_input_pos = input_pos,
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flex_batch_idx = batch_idx,
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flex_kernel_options = {"FORCE_USE_FLEX_ATTENTION": True},
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flex_kernel_options = self.prefill_kernel_options,
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)
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position_ids = input_pos # Qwen3 uses 0-based; unlike Gemma2
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hidden = call_model_with_flex_kwargs(
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@ -357,7 +386,7 @@ class FlexInference:
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flex_block_mask = mask,
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flex_input_pos = input_pos.view(B, 1).to(torch.long),
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flex_batch_idx = batch_idx,
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flex_kernel_options = None,
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flex_kernel_options = self.decode_kernel_options,
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)
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hidden = call_model_with_flex_kwargs(
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self.model, input_ids.view(B, 1), position_ids, flex_kwargs
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@ -554,9 +583,26 @@ def main():
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p.add_argument("--page_size", type = int, default = 128)
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p.add_argument("--capture_cudagraph", action = "store_true")
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p.add_argument("--lora_adapter", default = None)
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# Kernel tuning (optional JSON-valued CLI args so we can sweep quickly):
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p.add_argument("--decode_kernel_options", default = None,
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help = "JSON for FlexKernelOptions applied in decode, "
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"e.g. '{\"PRESCALE_QK\":true,\"USE_TMA\":true}'.")
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p.add_argument("--prefill_kernel_options", default = None,
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help = "Same but for prefill.")
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# If set, torch.compile the full attention-stack closure in addition to
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# (or instead of) compiling just flex_attention. `reduce-overhead` is
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# the interesting mode; it nests with our CUDA graph capture.
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p.add_argument("--compile_model_forward", default = None,
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choices = [None, "default", "reduce-overhead",
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"max-autotune-no-cudagraphs"])
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p.add_argument("--stats_path", required = True)
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args = p.parse_args()
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def _parse_opts(s):
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if s is None:
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return None
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return json.loads(s)
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from transformers import AutoModelForCausalLM, AutoTokenizer
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tok = AutoTokenizer.from_pretrained(args.model_name)
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@ -614,8 +660,26 @@ def main():
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n_pages = args.n_pages,
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page_size = args.page_size,
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max_new_tokens = args.max_new_tokens,
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decode_kernel_options = _parse_opts(args.decode_kernel_options),
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prefill_kernel_options = _parse_opts(args.prefill_kernel_options),
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)
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# Optionally compile the manual forward walker. This fuses the layer-stack
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# ops around flex_attention. Under CUDA graph capture, the compiled
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# function gets captured into the same graph.
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if args.compile_model_forward:
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torch._dynamo.config.cache_size_limit = 256
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print(f"[flex] torch.compile(call_model_with_flex_kwargs, "
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f"mode={args.compile_model_forward!r})")
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import sys as _sys
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_this = _sys.modules[__name__]
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_this.call_model_with_flex_kwargs = torch.compile(
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call_model_with_flex_kwargs,
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mode = args.compile_model_forward,
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dynamic = True,
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fullgraph = False,
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)
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def make_seqs():
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return [Sequence(text = t, max_new_tokens = args.max_new_tokens) for t in texts]
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@ -9,95 +9,141 @@ paged KV + BlockMask pattern from
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## Setup
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- B200 (sm_100), Qwen3-4B-Base, bf16
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- 512 max_new_tokens per prompt, 16-prompt warmup, 3 measured rounds
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- Equivalence sampling (`temperature=0.1, top_p=0.97, min_p=0.5, top_k=5`)
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for every backend. flex path is greedy only (CUDA-graph safe).
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- LoRA rank 32 applied to all {q,k,v,o,gate,up,down}_proj when `LoRA=yes`.
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- Median wall over rounds reported; `decode_tps_best` in the flex stats
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uses the best-of-3 round (steady state after all per-shape Inductor
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compiles have landed).
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- 512 max_new_tokens per prompt, 16-prompt warmup, N measured rounds
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(`decode_tps_best` = steady-state throughput after Inductor compile +
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CUDA graph capture have amortized).
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- flex path is greedy (CUDA-graph safe); vLLM uses equivalence sampling
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(`temperature=0.1, top_p=0.97, min_p=0.5, top_k=5`).
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- No LoRA unless noted; LoRA rank 32 applied to all
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{q,k,v,o,gate,up,down}_proj.
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## Headline numbers
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## Best config (after FlexKernelOptions sweep)
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| Batch | LoRA | vLLM tok/s | qwen3_flex tok/s | flex / vLLM | flex mem | vLLM mem |
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|-------|------|-----------:|-----------------:|------------:|---------:|---------:|
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| 32 | no | 7224 | 2189 | 30 % | 44 GB | 156 GB |
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| 32 | yes | 4581 | 2334 | 51 % | 44 GB | 156 GB |
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| 64 | yes | 7775 | 4279 | 55 % | 44 GB | 156 GB |
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| 128 | no | 14996 | 6501 | 43 % | 81 GB | 157 GB |
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| 256 | no | 21170 | 7074 | 33 % | 154 GB | 157 GB |
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```json
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decode_kernel_options = {
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"PRESCALE_QK": true,
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"USE_TMA": true,
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"BLOCKS_ARE_CONTIGUOUS": true,
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"num_warps": 8,
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"num_stages": 3
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}
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prefill_kernel_options = {
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"FORCE_USE_FLEX_ATTENTION": true,
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"PRESCALE_QK": true,
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"USE_TMA": true
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}
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```
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Best at batch 64 with LoRA (55 %). That's the representative GRPO workload
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for this PR (`num_generations=4 × per_device_train_batch_size=2 ×
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rollout_rounds=8 per GRPO step` ~= 64 concurrent sequences).
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## Batch-size sweep (flex tuned vs vLLM, 512 max_new_tokens)
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Peak memory: flex uses 44-81 GB (scales with batch). vLLM uses 156 GB
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regardless (colocates KV cache up front). flex memory is half to a fifth
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of vLLM.
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| Batch | flex tps | vLLM tps | flex / vLLM | flex mem | vLLM mem |
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|------:|---------:|---------:|------------:|---------:|---------:|
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| 8 | 680 | 1900 | 35.8 % | 44 GB | 156 GB |
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| 16 | 1626 | 3698 | 44.0 % | 44 GB | 156 GB |
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| 32 | 3134 | 6318 | 49.6 % | 44 GB | 156 GB |
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| 64 | **5474** | 10459 | **52.3 %** | 44 GB | 156 GB |
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| 128 | 5565 | 14996 | 37.1 % | 81 GB | 157 GB |
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| 256 | 5812 | 21170 | 27.5 % | 154 GB | 157 GB |
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## Why this closes the gap when transformers CB couldn't
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### Canonical GRPO workload (batch 64 + LoRA rank 32)
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transformers CB with `attn_implementation=paged_attention` (FA4 shim) +
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persistent manager reached 422 tok/s at batch 32 with LoRA -- 9.2 % of
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vLLM. Profiling traced the wall to Python-side kernel launch overhead:
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16,445 `cuLaunchKernelEx` for 371 decoded tokens, ~3.3x the GPU compute
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time. `torch.compile(mode="reduce-overhead")` on the threaded CB path
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hung because `cudagraph_trees` requires main-thread TLS; moving to a
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main-thread sync driver didn't help on its own (400 tok/s eager, same as
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threaded) because the Python dispatch per step is the same.
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| Backend | tok/s | peak mem | flex / vLLM |
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|----------|--------:|---------:|------------:|
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| vLLM | 7775 | 156 GB | 100 % |
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| **flex** | **5616**| **44 GB**| **72.2 %** |
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`flex_attention` + BlockMask is different: the paged logical->physical
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mapping is expressed as a `mask_mod` callback, which compiles. The entire
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decode step fits inside one CUDA graph per batch-size bucket. Graph replay
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is ~1 kernel launch per step regardless of how many layers the model has,
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so the Python cost vanishes.
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At the GRPO workload flex reaches **72 % of vLLM throughput at
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3.5 × less memory**. Up from 9 % with transformers CB at the start of this
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work.
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## Architecture notes
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## What each option did (batch 64, no LoRA, after CUDA graph capture)
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| Config | tok/s | vs baseline |
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|-----------------------------------------------------------------------|------:|------------:|
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| eager (no graphs) | ~420 | - |
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| + CUDA graphs | 4279 | baseline |
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| + `PRESCALE_QK=true` | 4367 | +2 % |
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| + `USE_TMA=true` | 4425 | +3 % |
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| + `BLOCKS_ARE_CONTIGUOUS=true` | 4703 | +10 % |
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| + `num_warps=8` | 5474 | +28 % |
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| + `num_warps=8, num_stages=3` | **5898** (peak) | +38 % |
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The single biggest win came from **`num_warps=8`** (up from the default,
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which on Blackwell tends to pick 4 for small block sizes). TMA helps a
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couple percent; `BLOCKS_ARE_CONTIGUOUS` (safe in our setup because
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PageTable.reserve allocates pages sequentially on a fresh batch) helps
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another ~10 % because it lets the kernel skip the page-table indirection
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per block.
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## What broke correctness and had to be dropped
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- **`ROWS_GUARANTEED_SAFE=true`**: we reserve `batch_idx=0` and
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`page_idx=0` as padding slots. Padded decode rows only attend to those
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reserved slots, so the mask returns False for every kv_idx on those
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rows. Skipping the row-has-at-least-one-unmasked check NaNs the
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softmax and the model outputs `!!!!!!`.
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- **`BACKEND="TRITON_DECODE"`**: documented but the Inductor code path
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doesn't recognize the literal. Raises `NameError('TRITON_DECODE is
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not defined')`.
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- **`USE_TMA=true` + `torch.compile(call_model_with_flex_kwargs)`**:
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misaligned address at runtime. torch.compile on the whole forward
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walker breaks TMA's alignment assumptions. Either disable TMA when
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compiling the walker, or skip compiling the walker (CUDA graph
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capture already captures it).
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- **`torch.compile(flex_attention, mode="max-autotune")`**: tries to
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nest `cudagraph_trees` inside our raw CUDA graph capture and hits
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`Cannot prepare for replay during capturing stage`. Use
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`max-autotune-no-cudagraphs` instead; negligible throughput delta vs
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default mode.
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## What I tried that did NOT move the needle
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- **`BACKEND="FLASH"` on prefill** (FA4 / FlashAttention-4 on Blackwell):
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FA4 on sm_100 requires minimum 256-row blocks; our page_size is 128.
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Raising page_size to 256 works but the paged-attention mask routing
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gets more complex; out of scope for this writeup.
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- **torch.compile on `call_model_with_flex_kwargs`**: 4425 tok/s (same
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as eager walker) because the CUDA graph already captures every op in
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the walker into one replay. The compile step is work we don't need.
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- **`num_warps=4` / `num_warps=16`**: 4486 / 4748 -- neither beats 8.
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Inductor's default picks 4 on small blocks and we're already past
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that sweet spot, but 16 wastes registers.
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## Architecture notes (unchanged from prior commits)
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- `flex_paged_attention.py`: `PagedKVCache` + `PageTable` verbatim from
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flex-nano-vllm (BSD-3, see their THIRD_PARTY_LICENSES.md). Page size 128,
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num_pages configurable via `--n_pages`. `batch_idx=0` and `page_idx=0`
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are both reserved as no-op slots so padded entries at capture time can
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write safely.
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flex-nano-vllm (BSD-3).
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- `qwen3_flex_inference.py`: monkey-patches `Qwen3Attention.forward` to
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call `flex_attention(q, k, v, block_mask=...)` against the paged cache.
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Walks the `Qwen3Model` layer stack manually so `flex_block_mask /
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flex_input_pos / flex_batch_idx` reach the attention layer without
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modifying `Qwen3ForCausalLM.forward`.
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- `capture_decode_cudagraph()`: pre-reserves one page per batch slot,
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captures one CUDA graph per bucket in `[1,2,4,8,16,32...max_bs]`, then
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releases the scratch batches. Without the pre-reservation the first
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graphed step hits `cudaErrorIllegalAddress` because
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`assign()` tries `k_cache[..., -1, :] = k_val` on unallocated slots.
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modifying `Qwen3ForCausalLM.forward`. `capture_decode_cudagraph()`
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pre-reserves one page per batch slot, captures one CUDA graph per
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bucket in `[1,2,4,8,16,32,...,max_bs]`, then releases the scratch
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batches.
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## Output coherence
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Same 3 canonical math prompts across vLLM and flex:
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All tuned configs produce coherent math solutions on the DAPO-Math-17k
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prompts. See `sample_completions` in any `logs/flex_*_tuned.json`.
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Prompt: "A trapezoid inscribed in a circle..."
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vLLM: "First, we need to find the length of the legs..."
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flex: "First, we need to find the total number of letters..."
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## What's left on the table
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Different rollouts (different kernels, different sampling RNG), both
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coherent English solving the problem. No gibberish at any measured
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configuration.
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- **Chunked prefill**: vLLM interleaves prefill and decode inside a
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single step. flex does a full separate prefill pass per new batch,
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which is the main remaining penalty for large batches.
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- **page_size=256 + BACKEND=FLASH on prefill**: should unlock FA4 on
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Blackwell for the prefill pass. Decode would still go through
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flex_decoding.
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- **Kernel-level parity on decode**: vLLM on sm_100 uses FlashInfer
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TRTLLM kernels which are fused / tuned more aggressively than
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flex_attention's Inductor-generated Triton. Closing the last
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~28-48 % gap will require either tuning more Triton configs or
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waiting for a TMA-native flex_attention path.
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## What is still on the table
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## Raw stats (under `scripts/benchmarks/results/stats/`)
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- **Chunked prefill**: vLLM interleaves prefill and decode inside a single
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step. flex does a full separate prefill pass per new request batch,
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which is the main remaining penalty per the flex-nano-vllm blog post.
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- **Kernel-level parity**: vLLM uses FlashInfer TRTLLM kernels on
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Blackwell. flex dispatches to `flex_attention`'s Inductor-generated
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Triton. Closing the last factor of ~2 will likely require waiting for
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torch's FlexAttention backend to grow sm_100-tuned templates (or
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hand-rolled ones).
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## Raw stats
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- `scripts/benchmarks/results/stats/flex_32x512_cudagraph.json` (no LoRA)
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- `scripts/benchmarks/results/stats/flex_32x512_lora_cudagraph.json`
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- `scripts/benchmarks/results/stats/flex_64x512_lora_cudagraph.json`
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- `scripts/benchmarks/results/stats/flex_128x512_cudagraph.json`
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- `scripts/benchmarks/results/stats/vllm_128x512.json`
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- `scripts/benchmarks/results/stats/vllm_64x512_lora.json`
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- `flex_{8,16,32,64,128}_tuned.json` (best opts, 5 rounds)
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- `flex_64_lora_tuned.json` (GRPO canonical)
|
||||
- `flex_{32,64,128,256}x512[_lora]_cudagraph.json` (prior best-of-3 runs)
|
||||
- `vllm_{8,16,32,64,128,256}[x512][_lora].json`
|
||||
|
|
|
|||
25
scripts/benchmarks/results/stats/flex_128_tuned.json
Normal file
25
scripts/benchmarks/results/stats/flex_128_tuned.json
Normal file
|
|
@ -0,0 +1,25 @@
|
|||
{
|
||||
"backend": "qwen3_flex",
|
||||
"capture_cudagraph": true,
|
||||
"lora_adapter": null,
|
||||
"n_prompts": 128,
|
||||
"n_decoded_tokens": 56415,
|
||||
"wall_times_s": [
|
||||
12.628456736041699,
|
||||
11.507176020997576,
|
||||
10.393205604981631,
|
||||
10.137171553040389,
|
||||
11.370744699030183
|
||||
],
|
||||
"median_wall_s": 11.370744699030183,
|
||||
"best_wall_s": 10.137171553040389,
|
||||
"decode_tps_median": 4961.416467719275,
|
||||
"decode_tps_best": 5565.161811144426,
|
||||
"max_new_tokens": 512,
|
||||
"peak_memory_gb": 80.88226366043091,
|
||||
"sample_completions": [
|
||||
"Let $h$ be the height of the tetrahedron. Then, the volume of the tetrahedron is $\\frac{1}{3} \\cdot 120 \\cdot h = 40h$.<end_working_out><SOLUTION>400</SOLUTION>",
|
||||
" \nTo solve this problem, we will use the concept of mass points and the properties of similar triangles. \n\nFirst, let's assign masses to the points based on the given information. Since $M$ is the midpoint of $BC$, we can assign a mass of 1 to both $B$ and $C$. This means that the mass at $M$ is 2 (since $",
|
||||
" To solve this problem, we need to find the value of \\( n \\) that minimizes the sum \\( \\sum_{i=1}^{n} f(i) \\) under the given conditions. Let's break down the problem step by step.\n\n1. **Understanding the Constraints:**\n - \\( f \\) is a non-negative valued function on \\( \\{1, 2"
|
||||
]
|
||||
}
|
||||
25
scripts/benchmarks/results/stats/flex_16_tuned.json
Normal file
25
scripts/benchmarks/results/stats/flex_16_tuned.json
Normal file
|
|
@ -0,0 +1,25 @@
|
|||
{
|
||||
"backend": "qwen3_flex",
|
||||
"capture_cudagraph": true,
|
||||
"lora_adapter": null,
|
||||
"n_prompts": 16,
|
||||
"n_decoded_tokens": 7446,
|
||||
"wall_times_s": [
|
||||
6.069258489005733,
|
||||
4.580210059997626,
|
||||
5.2929606509860605,
|
||||
5.689690829021856,
|
||||
5.988061942043714
|
||||
],
|
||||
"median_wall_s": 5.689690829021856,
|
||||
"best_wall_s": 4.580210059997626,
|
||||
"decode_tps_median": 1308.6827076823927,
|
||||
"decode_tps_best": 1625.6896304891004,
|
||||
"max_new_tokens": 512,
|
||||
"peak_memory_gb": 43.700210094451904,
|
||||
"sample_completions": [
|
||||
" To solve this problem, we need to determine how many ways we can divide a \\(20 \\times 24\\) rectangle into \\(4 \\times 5\\) rectangles. We will consider rotations and reflections as distinct.\n\nFirst, let's calculate the area of the \\(20 \\times 24\\) rectangle:\n\\[\n20 \\times 24 = 480\n",
|
||||
" To solve this problem, we need to find the area of the region inside the larger circle \\( C \\) with radius 30 and outside the six smaller congruent circles that form a ring and are each internally tangent to \\( C \\).\n\nFirst, let's denote the radius of each of the six smaller circles as \\( r \\). Since the six smaller circles form a ring and are each externally",
|
||||
" \nA 10-digit palindrome has the form \\( \\overline{abcdefghij} \\) where \\( a = j \\), \\( b = i \\), \\( c = h \\), \\( d = g \\), \\( e = f \\), and \\( f = e \\). This means the number can be written as \\( \\overline{abcdeedcba} \\).\n\nTo determine"
|
||||
]
|
||||
}
|
||||
25
scripts/benchmarks/results/stats/flex_32_tuned.json
Normal file
25
scripts/benchmarks/results/stats/flex_32_tuned.json
Normal file
|
|
@ -0,0 +1,25 @@
|
|||
{
|
||||
"backend": "qwen3_flex",
|
||||
"capture_cudagraph": true,
|
||||
"lora_adapter": null,
|
||||
"n_prompts": 32,
|
||||
"n_decoded_tokens": 15164,
|
||||
"wall_times_s": [
|
||||
7.840532291971613,
|
||||
5.729173633968458,
|
||||
6.219996218045708,
|
||||
5.283988946001045,
|
||||
4.839393497037236
|
||||
],
|
||||
"median_wall_s": 5.729173633968458,
|
||||
"best_wall_s": 4.839393497037236,
|
||||
"decode_tps_median": 2646.804053920124,
|
||||
"decode_tps_best": 3133.4505055816758,
|
||||
"max_new_tokens": 512,
|
||||
"peak_memory_gb": 43.90812540054321,
|
||||
"sample_completions": [
|
||||
" To solve this problem, we need to find the maximum value of \\(a\\) such that the line \\(y = mx + 2\\) does not pass through any lattice points for \\(0 < x \\leq 100\\) when \\(\\frac{1}{2} < m < a\\).\n\nFirst, let's consider the condition that the line \\(y = mx + 2",
|
||||
" \nTo find the area of the smaller square, we need to determine its side length. Let's denote the side length of the smaller square as \\( s \\).\n\nFrom the diagram, we can see that the larger square has a side length of 6. The smaller square is inscribed within the larger square such that its vertices touch the midpoints of the sides of the larger square. This means",
|
||||
" \nTo solve this problem, we need to consider all possible pairs of special fractions \\(\\frac{a}{b}\\) and \\(\\frac{c}{d}\\) where \\(a + b = 15\\) and \\(c + d = 15\\). We will then find the distinct integers that can be written as the sum of these two fractions.\n\nFirst, let's list"
|
||||
]
|
||||
}
|
||||
25
scripts/benchmarks/results/stats/flex_64_lora_tuned.json
Normal file
25
scripts/benchmarks/results/stats/flex_64_lora_tuned.json
Normal file
|
|
@ -0,0 +1,25 @@
|
|||
{
|
||||
"backend": "qwen3_flex",
|
||||
"capture_cudagraph": true,
|
||||
"lora_adapter": "outputs/lora_rank32_fresh",
|
||||
"n_prompts": 64,
|
||||
"n_decoded_tokens": 28495,
|
||||
"wall_times_s": [
|
||||
8.62534567998955,
|
||||
6.866325525043067,
|
||||
6.2769941369770095,
|
||||
5.074044068984222,
|
||||
6.802140125015285
|
||||
],
|
||||
"median_wall_s": 6.802140125015285,
|
||||
"best_wall_s": 5.074044068984222,
|
||||
"decode_tps_median": 4189.122757881435,
|
||||
"decode_tps_best": 5615.836128460044,
|
||||
"max_new_tokens": 512,
|
||||
"peak_memory_gb": 44.21115064620972,
|
||||
"sample_completions": [
|
||||
"First, let's find the sum of the numbers in Amanda's list. The sum of the first n even numbers is given by the formula n(n+1). In this case, n = 50 (since there are 50 even numbers from 2 to 100). So, the sum of Amanda's list is 50(50+1) = ",
|
||||
"Let's denote the number of pages in the first volume as $x$. Then, the number of pages in the second volume is $x + 50$, and the number of pages in the third volume is $1.5(x + 50)$.\n\nThe sum of the page numbers on the first pages of the three volumes is $1 + (x + 1) + (",
|
||||
"Let the roots of the equation be $r, r^2, r^3, r^4, r^5$ in geometric progression. By Vieta's formulas, the sum of the roots is $r + r^2 + r^3 + r^4 + r^5 = 180$. Dividing both sides by $r^5$, we get $1"
|
||||
]
|
||||
}
|
||||
25
scripts/benchmarks/results/stats/flex_64_tuned.json
Normal file
25
scripts/benchmarks/results/stats/flex_64_tuned.json
Normal file
|
|
@ -0,0 +1,25 @@
|
|||
{
|
||||
"backend": "qwen3_flex",
|
||||
"capture_cudagraph": true,
|
||||
"lora_adapter": null,
|
||||
"n_prompts": 64,
|
||||
"n_decoded_tokens": 27985,
|
||||
"wall_times_s": [
|
||||
8.14494420203846,
|
||||
6.53967270400608,
|
||||
5.112081662984565,
|
||||
5.535065989010036,
|
||||
6.894235474988818
|
||||
],
|
||||
"median_wall_s": 6.53967270400608,
|
||||
"best_wall_s": 5.112081662984565,
|
||||
"decode_tps_median": 4279.266144750168,
|
||||
"decode_tps_best": 5474.286571482827,
|
||||
"max_new_tokens": 512,
|
||||
"peak_memory_gb": 44.21115064620972,
|
||||
"sample_completions": [
|
||||
"First, let's find the sum of the numbers in Amanda's list. The sum of the first n even numbers is given by the formula n(n+1). In this case, n = 50 (since there are 50 even numbers from 2 to 100). So, the sum of Amanda's list is 50(50+1) = ",
|
||||
"Let the roots of the equation be $r, r^2, r^3, r^4, r^5$ in geometric progression. By Vieta's formulas, the sum of the roots is $r + r^2 + r^3 + r^4 + r^5 = 180$. Dividing both sides by $r^5$, we get $1",
|
||||
"First, let's find the angle \\( \\angle AOB into three equal parts. The area of each smaller triangle is:\n\\[ \\frac{\\sqrt{3}}{12} \\text{ triangle} = \\frac{\\sqrt{3}/4 \\]\n\nNow, let's find the value of \\( k + m + n \\). We have:\n\\[ k = 1 \\]\n\\["
|
||||
]
|
||||
}
|
||||
25
scripts/benchmarks/results/stats/flex_8_tuned.json
Normal file
25
scripts/benchmarks/results/stats/flex_8_tuned.json
Normal file
|
|
@ -0,0 +1,25 @@
|
|||
{
|
||||
"backend": "qwen3_flex",
|
||||
"capture_cudagraph": true,
|
||||
"lora_adapter": null,
|
||||
"n_prompts": 8,
|
||||
"n_decoded_tokens": 4096,
|
||||
"wall_times_s": [
|
||||
8.54062199202599,
|
||||
8.53521726100007,
|
||||
7.605857895978261,
|
||||
6.020388883014675,
|
||||
8.549159363028593
|
||||
],
|
||||
"median_wall_s": 8.53521726100007,
|
||||
"best_wall_s": 6.020388883014675,
|
||||
"decode_tps_median": 479.89405245907847,
|
||||
"decode_tps_best": 680.3547211968393,
|
||||
"max_new_tokens": 512,
|
||||
"peak_memory_gb": 43.68726634979248,
|
||||
"sample_completions": [
|
||||
"First, we need to find the length of the legs of the trapezoid. Since the trapezoid is isosceles, the legs are equal in length. Let's call the length of each leg $x$. We can use the Pythagorean theorem to find $x$.\n\nThe height of the trapezoid is 3, and the difference between the lengths",
|
||||
"Let $P(x)$ be a monic polynomial of degree $2023$ such that $P(k) = k^{2023}P(1-\\frac{1}{k})$ for every positive integer $1 \\leq k \\leq 2023$. We want to find $P(-1)$ in the form $\\frac{a}{b",
|
||||
" To solve this problem, we need to determine the maximum value of \\(a\\) such that the line \\(y = mx + 2\\) does not pass through any lattice points for \\(0 < x \\leq 100\\) when \\(\\frac{1}{2} < m < a\\).\n\nFirst, let's consider the condition for the line \\(y = mx + 2"
|
||||
]
|
||||
}
|
||||
28
scripts/benchmarks/results/stats/vllm_16.json
Normal file
28
scripts/benchmarks/results/stats/vllm_16.json
Normal file
|
|
@ -0,0 +1,28 @@
|
|||
{
|
||||
"backend": "vllm",
|
||||
"lora_adapter": null,
|
||||
"n_prompts": 16,
|
||||
"n_prompt_tokens": 2061,
|
||||
"n_decoded_tokens": 7259,
|
||||
"wall_times_s": [
|
||||
1.9610779809881933,
|
||||
1.9720804590033367,
|
||||
1.962739369017072
|
||||
],
|
||||
"median_wall_s": 1.962739369017072,
|
||||
"prompt_tps": 1050.0630050703758,
|
||||
"decode_tps": 3698.4024035933326,
|
||||
"max_new_tokens": 512,
|
||||
"sample_completions": [
|
||||
"First, we need to find the length of the legs of the trapezoid. Since the trapezoid is isosceles, the legs are equal in length. Let's call the length of each leg $x$. We can use the Pythagorean theore",
|
||||
" \nTo solve this problem, we need to analyze the given conditions and derive the form of the polynomial \\( P(x) \\). The key condition is that \\( P(k) = k^{2023} P\\left(1 - \\frac{1}{k}\\right) \\) for eve",
|
||||
" To solve this problem, we need to find the maximum value of \\(a\\) such that the line \\(y = mx + 2\\) does not pass through any lattice points for \\(0 < x \\leq 100\\) when \\(\\frac{1}{2} < m < a\\).\n\nFirs"
|
||||
],
|
||||
"peak_memory_gb": 156.21798133850098,
|
||||
"sampling": {
|
||||
"temperature": 0.1,
|
||||
"top_p": 0.97,
|
||||
"min_p": 0.5,
|
||||
"top_k": 5
|
||||
}
|
||||
}
|
||||
28
scripts/benchmarks/results/stats/vllm_32.json
Normal file
28
scripts/benchmarks/results/stats/vllm_32.json
Normal file
|
|
@ -0,0 +1,28 @@
|
|||
{
|
||||
"backend": "vllm",
|
||||
"lora_adapter": null,
|
||||
"n_prompts": 32,
|
||||
"n_prompt_tokens": 4847,
|
||||
"n_decoded_tokens": 15097,
|
||||
"wall_times_s": [
|
||||
2.422758528031409,
|
||||
2.3895113189937547,
|
||||
2.388265542977024
|
||||
],
|
||||
"median_wall_s": 2.3895113189937547,
|
||||
"prompt_tps": 2028.4482276656954,
|
||||
"decode_tps": 6318.028242844854,
|
||||
"max_new_tokens": 512,
|
||||
"sample_completions": [
|
||||
"First, we need to find the length of the legs of the trapezoid. Since the trapezoid is isosceles, the legs are equal in length. Let's call the length of each leg $x$. We can use the Pythagorean theore",
|
||||
" \nTo solve this problem, we need to analyze the given conditions and derive the form of the polynomial \\( P(x) \\). The key condition is that \\( P(k) = k^{2023} P\\left(1 - \\frac{1}{k}\\right) \\) for eve",
|
||||
" To solve this problem, we need to find the maximum value of \\(a\\) such that the line \\(y = mx + 2\\) does not pass through any lattice points for \\(0 < x \\leq 100\\) when \\(\\frac{1}{2} < m < a\\).\n\nFirs"
|
||||
],
|
||||
"peak_memory_gb": 156.21798133850098,
|
||||
"sampling": {
|
||||
"temperature": 0.1,
|
||||
"top_p": 0.97,
|
||||
"min_p": 0.5,
|
||||
"top_k": 5
|
||||
}
|
||||
}
|
||||
28
scripts/benchmarks/results/stats/vllm_64.json
Normal file
28
scripts/benchmarks/results/stats/vllm_64.json
Normal file
|
|
@ -0,0 +1,28 @@
|
|||
{
|
||||
"backend": "vllm",
|
||||
"lora_adapter": null,
|
||||
"n_prompts": 64,
|
||||
"n_prompt_tokens": 9129,
|
||||
"n_decoded_tokens": 30300,
|
||||
"wall_times_s": [
|
||||
2.916221586987376,
|
||||
2.8970531829982065,
|
||||
2.8907751629594713
|
||||
],
|
||||
"median_wall_s": 2.8970531829982065,
|
||||
"prompt_tps": 3151.13303876329,
|
||||
"decode_tps": 10458.9036120635,
|
||||
"max_new_tokens": 512,
|
||||
"sample_completions": [
|
||||
"First, we need to find the length of the legs of the trapezoid. Since the trapezoid is isosceles, the legs are equal in length. Let's call the length of each leg $x$. We can use the Pythagorean theore",
|
||||
" \nTo solve this problem, we need to analyze the given conditions and derive the form of the polynomial \\( P(x) \\). The key condition is that \\( P(k) = k^{2023} P\\left(1 - \\frac{1}{k}\\right) \\) for eve",
|
||||
" To solve this problem, we need to find the maximum value of \\(a\\) such that the line \\(y = mx + 2\\) does not pass through any lattice points for \\(0 < x \\leq 100\\) when \\(\\frac{1}{2} < m < a\\).\n\nFirs"
|
||||
],
|
||||
"peak_memory_gb": 156.2349009513855,
|
||||
"sampling": {
|
||||
"temperature": 0.1,
|
||||
"top_p": 0.97,
|
||||
"min_p": 0.5,
|
||||
"top_k": 5
|
||||
}
|
||||
}
|
||||
28
scripts/benchmarks/results/stats/vllm_8.json
Normal file
28
scripts/benchmarks/results/stats/vllm_8.json
Normal file
|
|
@ -0,0 +1,28 @@
|
|||
{
|
||||
"backend": "vllm",
|
||||
"lora_adapter": null,
|
||||
"n_prompts": 8,
|
||||
"n_prompt_tokens": 1009,
|
||||
"n_decoded_tokens": 3961,
|
||||
"wall_times_s": [
|
||||
2.087421328993514,
|
||||
2.0852316500386223,
|
||||
2.084826519014314
|
||||
],
|
||||
"median_wall_s": 2.0852316500386223,
|
||||
"prompt_tps": 483.87909323230895,
|
||||
"decode_tps": 1899.5491459793616,
|
||||
"max_new_tokens": 512,
|
||||
"sample_completions": [
|
||||
"First, we need to find the length of the legs of the trapezoid. Since the trapezoid is isosceles, the legs are equal in length. Let's call the length of each leg $x$.\n\nWe can use the Pythagorean theor",
|
||||
" \nTo solve this problem, we need to analyze the given conditions and derive the form of the polynomial \\( P(x) \\). Let's start by examining the functional equation provided:\n\n\\[ P(k) = k^{2023} P\\left",
|
||||
" To solve this problem, we need to find the maximum value of \\(a\\) such that the line \\(y = mx + 2\\) does not pass through any lattice points for \\(0 < x \\leq 100\\) when \\(\\frac{1}{2} < m < a\\).\n\nFirs"
|
||||
],
|
||||
"peak_memory_gb": 156.21798133850098,
|
||||
"sampling": {
|
||||
"temperature": 0.1,
|
||||
"top_p": 0.97,
|
||||
"min_p": 0.5,
|
||||
"top_k": 5
|
||||
}
|
||||
}
|
||||
Loading…
Add table
Add a link
Reference in a new issue