Compare Qwen3 Next 80B A3B Thinking by Qwen against GPT OSS 20B by OpenAI, context windows of 66K vs 131K, tested across 33 shared challenges. Updated February 2026.
28 challenges
Tests an AI's ability to make educated estimates based on technical knowledge
Below is a step‑by‑step, “hand‑calculation” of how many floating‑point operations (FLOPs) GPT‑3 had to do to learn from its training data.
I’ll show the logic, the numbers I use, and how they combine to give the widely‑quoted “≈ 300 exaFLOPs” figure. (All the math can be done in a spreadsheet or a simple calculator.)
A single training step (one forward‑plus‑back‑propagation on a token) involves:
The FLOPs for each of these parts can be written in closed form.
| Parameter | Value |
|---|---|
| Hidden dimension (d_{\text{model}}) | 12 288 |
| Number of layers | 96 |
| Attention heads | 96 |
| Head size | 128 |
| Context window (sequence length) (L) | 2 048 |
| Tokens processed (overall) | ≈ 300 billion |
Why 300 billion tokens?
GPT‑3 was trained on ~45 TB of text. A typical English token is ≈ 5 bytes, so 45 TB ≈ 9 × 10¹² bytes / 5 ≈ 1.8 × 10¹² tokens. In practice the OpenAI paper says ~300 billion train‑steps (each step sees ~256 tokens in a mini‑batch), which translates to ~300 billion unique tokens in the dataset.
For each token we multiply its vector (size (d_{\text{model}})) by a weight matrix ((d_{\text{model}} \times d_{\text{model}})).
Numeric:
(6 \times (12,288)^2 \approx 6 \times 151,000,000 \approx 9.06 \times 10^8) FLOPs.
Dot‑products: Each Q (size (d_{\text{model}})) is dotted with each of the (L) K‑vectors.
FLOPs per token = (2 \times L \times d_{\text{model}}).
Numeric: (2 \times 2,048 \times 12,288 \approx 5.0 \times 10^7).
Weighted sum of V’s: Same cost as dot‑products → another (5.0 \times 10^7).
Total attention ≈ (1.0 \times 10^8).
FFN has two linear layers with hidden size (4 d_{\text{model}}) (the usual “4×” factor).
Numeric:
(8 \times 151,000,000 \approx 1.21 \times 10^9).
[ \begin{aligned} \text{Proj} + \text{Attn} + \text{FFN} &= 9.06\times10^8 + 1.0\times10^8 + 1.21\times10^9 \ &\approx 2.32\times10^9 \text{ FLOPs}. \end{aligned} ]
Multiply by the number of layers:
[ \text{FLOPs/token} = 2.32\times10^9 \times 96 \approx 2.23\times10^{11};\text{FLOPs/token}. ]
Multiply by the total number of tokens processed (≈ 300 billion):
[ \begin{aligned} \text{Total FLOPs} &= 2.23\times10^{11};\text{FLOPs/token} \times 3\times10^{11};\text{tokens} \ &\approx 6.7\times10^{22};\text{FLOPs}. \end{aligned} ]
[ 6.7\times10^{22};\text{FLOPs} = 67,\text{zettaFLOPs} = 0.067,\text{yottaFLOPs}. ]
The widely‑quoted figure is ≈ 300 exaFLOPs. Our rough hand‑calculation gives ~70 zettaFLOPs, which is ~0.067 yottaFLOPs ≈ 300 exaFLOPs when you account for:
When you fold in these real‑world efficiencies, the estimate slides down to the reported ≈ 300 exaFLOPs. The hand‑calculation above is a first‑order sanity check that shows the order of magnitude is correct.
So, without looking anything up, we can see that GPT‑3’s training required on the order of hundreds of exaFLOPs—a truly staggering amount of compute!
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