Compare GPT-4.1 Mini and GPT OSS 120B, both from OpenAI, in 4 community votes, gpt oss 120b wins 100% of head-to-head duels, context windows of 1.0M vs 131K, tested across 38 shared challenges. Updated February 2026.
In 4 community votes, GPT OSS 120B wins 100% of head-to-head duels. Based on blind community voting from the RIVAL open dataset of 4+ human preference judgments for this pair.
38 challenges
Tests an AI's ability to make educated estimates based on technical knowledge
Quick answer:
Training the original 175‑billion‑parameter GPT‑3 is estimated to have required on the order of 10²³ floating‑point operations (≈ 3 × 10²³ FLOPs)—roughly a few hundred zetta‑FLOPs (≈ 300 Z‑FLOPs) or about 300 000 exaflops of compute.
Below is a step‑by‑step back‑of‑the‑envelope calculation that shows how we get there, together with the assumptions that go into it.
| Quantity | Typical reported value for GPT‑3 | Why it matters |
|---|---|---|
| Number of parameters, N | ≈ 175 billion (1.75 × 10¹¹) | Determines the size of each matrix‑multiply in the model. |
| Training token count, T | ≈ 300 billion tokens (3 × 10¹¹) | Total number of token‑level forward‑passes the model sees. |
| Sequence length, L | ≈ 2048 tokens per example (the context window). | Determines how many per‑token matrix‑products are needed per forward pass. |
| Number of layers, Lₗ | 96 transformer blocks. | |
| Hidden dimension, d | 12 384 (the width of each linear projection). | |
| Number of attention heads, h | 96 (so each head has size d/h = 128). | |
| Training passes | 1 epoch (the published training used roughly 1 × the dataset; we treat the 300 B tokens as the total “token‑steps” already). |
The only numbers we need for a FLOP estimate are N (the model size) and T (the total number of token‑level operations). The rest of the architecture details (L, d, h, Lₗ) are used to translate “N parameters” into “how many FLOPs per token”.
A transformer layer consists of:
For a single token (ignoring the cost of the softmax and the small bias terms) the dominant cost is matrix‑multiply operations.
For a matrix multiplication A (m×k) × B (k×n) the number of multiply‑adds is 2 · m·k·n (one multiplication and one addition per entry). In deep‑learning practice we count that as 2 FLOPs per multiply‑add pair.
| Component | Approx. dimensions | FLOPs (per token) |
|---|---|---|
| Q, K, V projections (3× per token) | d × d → 3·(2·d·d) = 6·d² | |
| Attention scores (Q·Kᵀ) | L·d → 2·L·d² | |
| Weighted sum (A·V) | L·d → 2·L·d² | |
| Output projection | d × d → 2·d² | |
| FFN first linear (d → 4d) | 2·d·4d = 8·d² | |
| FFN second linear (4d → d) | 8·d² | |
| Total per layer | ≈ 38 · d² FLOPs per token (rounded up). |
Plugging d = 12 384:
[ 38 \times d^{2} \approx 38 \times (1.2384\times10^{4})^{2} \approx 38 \times 1.53 \times 10^{8} }= 5.8 \times 10^{9}; \text{FLOPs per token per layer}. ]
With 96 layers:
[ \text{FLOPs per token (forward)} \approx 96 \times 5.8\times10^{9} }= 5.6 \times 10^{11}; \text{FLOPs}. ]
Training requires a forward pass plus a backward pass. The backward pass is roughly 2× the forward‑pass cost (gradient propagation + weight updates). So:
[ \text{FLOPs per token (training)} \approx 3 \times \text{forward} = 3 \times 5.6\times10^{11} }\approx 1.7\times10^{12}; \text{FLOPs per token.} ]
(Alternative literature often rounds this factor to ≈ 6 total FLOPs per parameter per token, which yields the same order of magnitude.)
The training corpus used for GPT‑3 was ≈ 300 billion tokens (≈ 3 × 10¹¹). The total FLOPs needed:
[ \text{Total FLOPs} \approx 1.7 \times 10^{12};\text{FLOPs/token} \times 3\times10^{11};\text{tokens} \approx 5.1 \times 10^{23};\text{FLOPs}. ]
Rounded to a single‑digit significance:
[ \boxed{,\approx 5\times10^{23},\text{FLOPs},} ]
| Unit | Value |
|---|---|
| FLOPs | ≈ 5 × 10²³ |
| Peta‑FLOPs (10¹⁵) | ≈ 5 × 10⁸ PFLOPs |
| Exa‑FLOPs (10¹⁸) | ≈ 5 × 10⁵ EFLOPs |
| Zetta‑FLOPs (10²¹) | ≈ 5 × 10² ZFLOPs (≈ 500 ZFLOPs) |
Because “zeta” (10²¹) is a convenient unit for such massive numbers, we can say ≈ 500 zeta‑FLOPs. In more familiar terms, this is ≈ 300 zeta‑FLOPs if we adopt the “6 × N × T” rule of thumb (see below), so the estimate sits in the few‑hundred‑zeta‑FLOP range.
A widely‑cited back‑of‑the‑envelope rule for transformer LLMs is:
[ \text{Training FLOPs} \approx 6 \times N_{\text{params}} \times N_{\text{tokens}}. ]
Plugging in the numbers:
[ 6 \times (1.75\times10^{11}) \times (3\times10^{11}) = 3.15\times10^{23};\text{FLOPs}. ]
That is within a factor of two of the more detailed calculation above (the difference stems from the exact factor used for forward vs backward cost). Both methods give the same order of 10²³ FLOPs.
(All numbers are back‑of‑the‑envelope estimates; the real training run also incurs overhead from data‑loading, communication, and other inefficiencies, so the actual engineering effort was larger, but the pure‑compute figure is captured by the calculation above.)
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