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LLM ticket deduplication costs skyrocket with quadratic scaling, but categorization offers savings

Using large language models to identify duplicate support tickets can become prohibitively expensive due to the quadratic growth in comparisons required. A naive approach of comparing every ticket with every other ticket results in a cost that scales with the square of the number of tickets. A more efficient method involves first categorizing tickets with an LLM and then comparing only tickets within the same category, which can reduce costs by up to 20 times. However, uneven distribution of tickets across categories can lessen these savings, suggesting further subdivision of large categories may be necessary. AI

IMPACT Highlights the significant cost implications of using LLMs for pairwise comparisons, driving the need for efficient categorization strategies in AI-powered support systems.

RANK_REASON Article discusses the cost-effectiveness and scaling challenges of using LLMs for a specific application (ticket deduplication), rather than a new model release or core research.

Read on dev.to — LLM tag →

AI-generated summary · Google Gemini · from 1 sources. How we write summaries →

LLM ticket deduplication costs skyrocket with quadratic scaling, but categorization offers savings

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33 / 100
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Tool
Article discusses the cost-effectiveness and scaling challenges of using LLMs for a specific application (ticket deduplication), rather than a new model release or core research.
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Only one publisher covered this so far. Single-source stories can still rank when the publisher is high-authority, but they lack cross-source corroboration.
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product, infra
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High
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Breaking (< 6h)
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COVERAGE [1]

  1. dev.to — LLM tag TIER_1 English(EN) · Andrey Altrouter ·

    Asking an LLM "are these two tickets duplicates?" costs $111 for 1,000 tickets and $11,130 for 10,000

    <p>You have a pile of support tickets and want to find the ones that describe the same problem. Models are good at this: show one two tickets, ask "same issue, yes or no?", and the answer is usually right. So you write a loop that compares each ticket with every other one. It cos…