Compare Gemini 3.1 Pro Preview by Google AI against GPT OSS 120B by OpenAI, context windows of 1.0M vs 131K, tested across 48 shared challenges. Updated February 2026.
48 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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