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  1. Home
  2. Models
  3. Kimi K2.5
  4. Estimate Complexity

See how Kimi K2.5 compares on this challenge:

  • Kimi K2.5 vs Claude Sonnet 4.6
  • Kimi K2.5 vs Gemini 3.1 Pro Preview
  • Kimi K2.5 vs Z.ai: GLM 5

4 models have responded to the "Estimate Complexity" challenge. Compare their approaches side-by-side on RIVAL. This response is part of RIVAL's open dataset of 5,600+ AI model responses.

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Back to Kimi K2.5

Estimate Complexity

Kimi K2.5's response to the 'Estimate Complexity' challenge.

Kimi K2.5
January 27, 2026
Text

?
Prompt

Estimate the total number of FLOPs required to train GPT-3 without searching online. Explain your reasoning step by step.

Kimi K2.5
Response

kimi-k2.5 logo
Kimi K2.5Estimate Complexity
kimi-k2.5 logo
Kimi K2.5

Here is a step-by-step estimate for the total training FLOPs of GPT-3.

Step 1: Identify the key variables

To estimate training compute for a dense transformer, we need two numbers:

  • $N$ (Parameters): GPT-3 has 175 billion (1.75 × 10¹¹) parameters.
  • $D$ (Tokens): GPT-3 was trained on roughly 300 billion (3 × 10¹¹) tokens. (This is the widely cited figure from the original paper for the final training corpus).

Step 2: Use the standard training FLOPs formula

For decoder-only transformers trained with backpropagation, the total floating-point operations (FLOPs) can be approximated as:

$$\text{FLOPs} \approx 6 \times N \times D$$

Why 6?

  • Forward pass: Processing one token requires roughly 2 FLOPs per parameter (1 multiply + 1 accumulate for the matrix operations).
  • Backward pass: Computing gradients requires roughly 4 FLOPs per parameter (approximately 2× the forward cost to calculate gradients with respect to both inputs and weights).
  • Total: $2 + 4 = 6$ FLOPs per parameter per token.

(Note: This formula dominates the cost; the quadratic attention cost is negligible compared to the linear parameter cost for a model of this size).

Step 3: Calculate

$$\text{FLOPs} \approx 6 \times (1.75 \times 10^{11}) \times (3 \times 10^{11})$$

$$= 6 \times 1.75 \times 3 \times 10^{22}$$

$$= 6 \times 5.25 \times 10^{22}$$

$$= 31.5 \times 10^{22}$$

$$= \mathbf{3.15 \times 10^{23} \text{ FLOPs}}$$

Step 4: Account for uncertainty and overhead

The raw calculation gives ~3.15 × 10²³ FLOPs. However, we should adjust for real-world factors:

  • Dataset size variance: If GPT-3 was trained on closer to 400–500B tokens (including potential epochs or different data mixes), the number could be 4–5 × 10²³ FLOPs.
  • Implementation overhead: Data loading, checkpointing, communication in distributed training, and the Adam optimizer updates add 10–20% overhead.

Final Estimate

Approximately $3 \times 10^{23}$ to $5 \times 10^{23}$ FLOPs (roughly 300 to 500 sextillion FLOPs).

For context, this is equivalent to about 3,000 to 5,000 petaflop-seconds (or PFLOP-seconds), which aligns with the estimated thousands of petaflop-days often cited for GPT-3's training run on V100 GPUs.

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About Kimi K2.5

Capabilities

ConversationReasoningCode GenerationAnalysis

Categories

TextCodeMultimodal

Specifications

Provider
Moonshotai
Released
2026-01-27
Size
LARGE
Context
262,144 tokens

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SAME PROMPT

Claude Sonnet 4.6's version

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