memory benchmark
CI Memories measures privacy failures in memory-enabled agents using contextual-integrity scenarios. This metric is the violation rate; lower is better.
Updated Aug 17, 2026
Higher score ranks better on this benchmark.
Rank | Model | Score | Percentile | Participants | Evidence | Evaluated |
|---|
| Rank01 | ModelME | Score26.4% | Percentile100.0% | Participants1 | EvidenceC | Evaluated |
The leading models and scores on this benchmark.
The first five results on this benchmark, with official price and output speed added where the model identity can be matched.
Ranking basisThis ci memories violation rate AI model leaderboard uses descending score in the benchmark's original unit. The leaderboard ranking keeps matched price and speed data separate from benchmark evidence.
Selection summary
Muse Glimmer-30B currently leads CI Memories Violation Rate with 26.4%. It is the top model on this specific benchmark, while the best LLM for the broader task should also be checked against other benchmarks, price and runtime.
Use this leaderboard with the supporting benchmark results and coverage details above. A leaderboard position summarizes the selected ranking signal; it does not replace workload-specific testing.
What CI Memories Violation Rate measures and how its scores work.
CI Memories measures privacy failures in memory-enabled agents using contextual-integrity scenarios. This metric is the violation rate; lower is better.
Scores are shown in ratio. This benchmark is not independently verified and has an evidence level of B.
Benchmark scores retain their original unit. Overall score eligibility is shown separately.
Common questions about CI Memories Violation Rate.
Muse Glimmer-30B is currently ranked first with 26.4%.
CI Memories measures privacy failures in memory-enabled agents using contextual-integrity scenarios. This metric is the violation rate; lower is better.
Yes. Higher values rank better for this benchmark.
1 model results are currently shown.
No. This benchmark is shown for reference but does not contribute to the overall score.