Computed

Qwen2.5 0.5B Instruct on an Apple M3 Ultra (512GB)

Yes — the unquantized weights fit with room for 32k tokens of context.

Change the numbers →Qwen2.5 0.5B Instruct on every card Apple M3 Ultra (512GB) across every model

Verdict

Fits at FP16

384.0 GiB usable of 512 GB

Weights (Q4_K_M)

379 MiB

942 MiB at FP16 · 494M params · 6 of 6 quant rows are measured files

KV cache

12.0 KiB per token at FP16

24 layers × 2 KV heads × 64 dimensions

Speed ceiling

752 tok/s

819 GB/s ÷ bytes read per token

Every quant, against this card

Weight bytes are the size of the real published file wherever one exists — 6 of these6 rows are measured from bartowski/Qwen2.5-0.5B-Instruct-GGUF, the rest computed from the parameter count. Max context is what the KV cache can grow to in whatever memory the weights leave behind, capped at the 32k tokens this model was trained to address.

Bars are the weight bytes at each precision; the dashed line is the usable memory of an Apple M3 Ultra (512GB). Drawn from the computed byte counts, not sketched.

Precision	Bits/weight	Weights	Source	Fits	Max context	Ceiling
FP16 / BF16	16.00	942 MiB	measured file	yes	32,768 (model cap)	752 tok/s
Q8_0	8.60	506 MiB	measured file	yes	32,768 (model cap)	1296 tok/s
Q6_K	8.19	482 MiB	measured file	yes	32,768 (model cap)	1351 tok/s
Q5_K_M	6.80	401 MiB	measured file	yes	32,768 (model cap)	1573 tok/s
Q4_K_M	6.44	379 MiB	measured file	yes	32,768 (model cap)	1643 tok/s
Q3_K_M	5.76	339 MiB	measured file	yes	32,768 (model cap)	1796 tok/s

What the context actually costs

The KV cache is 2 · layers · kv_heads · head_dim bytes per token per element — 2 · 24 · 2 · 64 · 2 B = 12.0 KiB. It depends on the key/value head count, not the attention-head count and not the parameter count. This model shares 2 KV heads across 14 query heads, which divides the cache by 7 against multi-head attention.

Context	KV cache (FP16)	KV cache (8-bit)	Plus Q4_K_M weights
4,096	48 MiB	24 MiB	427 MiB
8,192	96 MiB	48 MiB	475 MiB
32,768	384 MiB	192 MiB	763 MiB

The speed ceiling, and where it comes from

Generating one token reads every active weight from memory once. At FP16 / BF16 that is 942 MiB, plus a pass over the KV cache. An Apple M3 Ultra (512GB) moves 819 GB/s, so the arithmetic ceiling is 752 tok/s. Treat it as a bound, not an estimate: attention overhead, kernel launches and imperfect memory access keep real runtimes at roughly 60–80% of it, and nothing pushes past it.

Where these numbers come from

The model

Parameters: 494,032,768
Layers: 24
Attention / KV heads: 14 / 2
Head dimension: 64
Trained context: 32,768
Checkpoint as published: 942 MiB

Read from Qwen/Qwen2.5-0.5B-Instruct. The parameter count is the Hub's own total over the tensor shapes, not a figure taken from the model's name.

The accelerator

Memory: 512 GB LPDDR5 unified
Bandwidth: 819 GB/s
Assumed usable: 75% → 384.0 GiB

Capacity and bandwidth from the vendor's specification. The usable fraction is an assumption, not a spec: unified memory is shared with the OS and the display, and the GPU working-set cap is raisable on Apple silicon with `sudo sysctl iogpu.wired_limit_mb`.

Questions this pairing answers

How much VRAM does Qwen2.5 0.5B Instruct need?

942 MiB for the weights at FP16 — 494M parameters at two bytes each — and 379 MiB at Q4_K_M. The KV cache is on top of that and is not a fixed number: this model spends 12.0 KiB per token of context, so 8,192 tokens costs a further 96 MiB. An Apple M3 Ultra (512GB) makes 384.0 GiB of its 512 GB available on the assumption below.

Can an Apple M3 Ultra (512GB) run Qwen2.5 0.5B Instruct?

Yes — the unquantized weights fit with room for 32k tokens of context. That is the weights and the KV cache together, against 384.0 GiB of usable memory.

How fast will Qwen2.5 0.5B Instruct run on an Apple M3 Ultra (512GB)?

No faster than 752 tokens/second at FP16 / BF16, and in practice below it. Decoding is memory-bound: every token reads all 942 MiB of weights plus the cache, and this card moves 819 GB/s. That division is the ceiling — no kernel, runtime or driver beats it, and a real runtime typically reaches 60–80% of it.

Why does the context length change how much memory Qwen2.5 0.5B Instruct needs?

Because the KV cache holds one key and one value vector per token, per layer, for the whole conversation, and it is allocated separately from the weights. This model has 24 layers and 2 key/value heads of 64 dimensions, shared across 14 query heads — grouped-query attention, which divides the cache by 7. That works out at 12.0 KiB per token. Parameter count tells you nothing about this number.

The same model on a different card

A different model on the same card

All 62 models on Apple M3 Ultra (512GB) →

Method and limits. Weight bytes are the byte size of the real published file wherever one exists, and the model's exact parameter count times the published llama.cpp bits-per-weight where it does not. That distinction is on every row above and it matters at both ends: a sub-1B model's Q4_K_M file runs a third larger than the nominal figure because k-quants keep its embedding tables at higher precision, and an already-4-bit release cannot be quantized upward at all. The KV cache is 2 · layers · kv_heads · head_dim · bytes per token, summed over layers with each sliding-window layer capped at its window. The speed figure is a roofline bound, not a benchmark: bandwidth divided by bytes read per token, which no runtime exceeds and every runtime falls short of. The usable fraction of card memory is an assumption: 75% here, which is what this lane assumes for unified memory, where the OS and window server share the same pool — the other class assumes 92%, so it is not one number applied to every device. That is the only assumed input on this page; every other figure is computed from the model config and the card's published specification. Nothing on this page is written by a language model. Architecture from the model's published config, fetched 2026-08-07.