Two systems: one has a mean of and a p99.9 of ms, the other a mean of and a p99.9 of . Which do you want on a market-data path?
- AThe first, since it is faster on average
- BThe second, because its tail is far tighter
- CEither – the difference is only one in a thousand messages
- DThe first, then add a retry when it is slow
Show the answer and worked solution
Answer: B – The second, because its tail is far tighter
Take the second. A market-data path is judged on jitter rather than on the average, because slow paths are correlated with bursts and bursts are exactly when the messages matter: a two-millisecond stall lands during the news event you were positioned for. A predictable five microseconds is worth far more than an unpredictable two. "Only one message in a thousand" is precisely the wrong way to think about it, since that message is disproportionately the one that counted, and a retry does not help because by the time you know the path was slow the opportunity is gone and you have added load to a system already struggling.
Worked solution
- Formula
- Substitute
- Solve
- Which is when the messages matter most.
- Answer
Sanity check. A retry does not help, because by the time you know the path was slow the opportunity is gone.
- A. Its tail is a hundred times worse, and the tail lands during the bursts that follow news.
- B. Correct. Jitter is the target: a predictable 5 µs beats an unpredictable 2 µs.
- C. That one message is disproportionately the one that mattered.
- D. By the time you know it was slow, the opportunity has gone and the retry adds load.
Takeaway: A trading system is judged on its tail, not its mean.
Answer it in practice – your answer is marked and recorded.
Learn the method
Latency, the memory hierarchy and why the tail is the number
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