What is the difference between Qwen: Qwen3 Max and GPT OSS 120B?

Qwen: Qwen3 Max is developed by Qwen while GPT OSS 120B is developed by OpenAI. Qwen: Qwen3 Max has a 256K token context window vs GPT OSS 120B's 131K. You can compare their actual outputs across 46 challenges on RIVAL to see how they differ in practice.

Which is better, Qwen: Qwen3 Max or GPT OSS 120B?

It depends on your use case. Qwen: Qwen3 Max and GPT OSS 120B each have strengths in different areas. RIVAL lets you compare their real outputs side-by-side across 46 challenges so you can judge which fits your needs best.

How much does Qwen: Qwen3 Max cost compared to GPT OSS 120B?

Qwen: Qwen3 Max costs $1.2/M input tokens and GPT OSS 120B costs $0.18/M input tokens. GPT OSS 120B is $1.02/M cheaper per input. Check their side-by-side outputs on RIVAL to see if the price difference is justified by quality.

How can I compare Qwen: Qwen3 Max and GPT OSS 120B on RIVAL?

This page shows a side-by-side comparison of Qwen: Qwen3 Max and GPT OSS 120B across shared challenges. You can vote on which model produced the better output, and Pro users can create custom challenges to test both models with their own prompts.

Rival

Qwen: Qwen3 MaxvsGPT OSS 120B

46 challenges

Why Qwen: Qwen3 Max?

2x more context256K

Why GPT OSS 120B?

7.4x cheaper overall$0.18/M in · $0.80/M out

Favorites

Qwen: Qwen3 MaxGPT OSS 120B

Movie

The Shawshank Redemption

19948.7

The Godfather

19728.7

Album

Kind of Blue

Miles Davis

—

Book

Dune

Frank Herbert

—

City

Kyoto

Japan

Tokyo

Japan

Game

The Legend of Zelda: Ocarina of Time

Adventure, Action4.4

Minecraft

Action, Arcade4.4

12 / 46

Estimate Complexityreasoning

Tests an AI's ability to make educated estimates based on technical knowledge

Qwen: Qwen3 Max

Qwen · Sep 2025text

No content available for this output.

GPT OSS 120B

Openai · Aug 2025text

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.

1. What we need to know

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”.

2. How many FLOPs per token?

A transformer layer consists of:

Self‑attention (Q, K, V projections + output projection)
Feed‑forward network (FFN) (two linear layers with a non‑linear activation).

For a single token (ignoring the cost of the softmax and the small bias terms) the dominant cost is matrix‑multiply operations.

2.1 Rough matrix‑multiply cost

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.

2.2 FLOPs per token for a single layer

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}. ]

2.3 All layers

With 96 layers:

[ \text{FLOPs per token (forward)} \approx 96 \times 5.8\times10^{9} }= 5.6 \times 10^{11}; \text{FLOPs}. ]

2.4 Back‑propagation cost

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.)

3. Multiply by the total number of tokens

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},} ]

4. Expressed in common units

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.

5. A “quick‑rule” sanity check

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.

6. Summary of the reasoning

Model size gives the number of parameters (N = 1.75 × 10¹¹).
Token count tells how many forward‑backward passes are needed (T ≈ 3 × 10¹¹).
Per‑token training cost for a transformer layer is ≈ 38 · d² FLOPs per forward pass; with 96 layers this is ≈ 5.6 × 10¹¹ FLOPs per token.
Back‑propagation adds roughly a factor‑3 (or 6) multiplier, yielding ~1.7 × 10¹² FLOPs per token.
Total = per‑token cost × total tokens ≈ 5 × 10²³ FLOPs.

Bottom line

Estimated total compute to train GPT‑3: ≈ 5 × 10²³ FLOPs, i.e. ≈ 300–500 zeta‑FLOPs (or about 300 000 exaflops).
This is the magnitude of hundreds of zettaFLOPs of raw arithmetic, roughly the same as running a few hundred petaflops‑scale supercomputers for a month (or a single exaflop‑class machine for a week).

(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.)

Vote

2 / 46

AI Board Game Logicreasoning

Tests an AI's ability to understand game rules and strategy

Qwen: Qwen3 Max

Qwen · Sep 2025svg

No content available for this output.

Openai · Aug 2025text

No content available for this output.

Vote

12 / 46

Minimalist Landing Pageweb design

Tests an AI's ability to generate a complete, working landing page

Qwen: Qwen3 Max

Qwen · Sep 2025website

GPT OSS 120B

Openai · Aug 2025website

Vote

12 of 46

FAQ

Continue exploring

Qwen: Qwen3 Max vs Pony AlphaNew provider

Qwen: Qwen3 Max vs MiniMax M2.5New provider

Qwen: Qwen3 Max vs GPT OSS 20BCross-provider

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.

1. What we need to know

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).

2. How many FLOPs per token?

A transformer layer consists of:

Self‑attention (Q, K, V projections + output projection)
Feed‑forward network (FFN) (two linear layers with a non‑linear activation).

For a single token (ignoring the cost of the softmax and the small bias terms) the dominant cost is matrix‑multiply operations.

2.1 Rough matrix‑multiply cost

2.2 FLOPs per token for a single layer

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}. ]

2.3 All layers

With 96 layers:

[ \text{FLOPs per token (forward)} \approx 96 \times 5.8\times10^{9} }= 5.6 \times 10^{11}; \text{FLOPs}. ]

2.4 Back‑propagation cost

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.)

3. Multiply by the total number of tokens

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},} ]

4. Expressed in common units

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)

5. A “quick‑rule” sanity check

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}. ]

6. Summary of the reasoning

Model size gives the number of parameters (N = 1.75 × 10¹¹).
Token count tells how many forward‑backward passes are needed (T ≈ 3 × 10¹¹).
Per‑token training cost for a transformer layer is ≈ 38 · d² FLOPs per forward pass; with 96 layers this is ≈ 5.6 × 10¹¹ FLOPs per token.
Back‑propagation adds roughly a factor‑3 (or 6) multiplier, yielding ~1.7 × 10¹² FLOPs per token.
Total = per‑token cost × total tokens ≈ 5 × 10²³ FLOPs.

Bottom line

Estimated total compute to train GPT‑3: ≈ 5 × 10²³ FLOPs, i.e. ≈ 300–500 zeta‑FLOPs (or about 300 000 exaflops).
This is the magnitude of hundreds of zettaFLOPs of raw arithmetic, roughly the same as running a few hundred petaflops‑scale supercomputers for a month (or a single exaflop‑class machine for a week).