3 ms·
No, that's not Little's Law. Little's Law describes the depth of a queue, not it's throughput. In the original article, this assertion is wrong except under s
by honkhonkpants 10y ago
No, that's not Little's Law. Little's Law describes the depth of a queue, not it's throughput.
In the original article, this assertion is wrong except under special conditions that aren't described in the article. In computer systems you can have a latency of X and a throughput of Y, Z, or any number with no particular relation to X. Consider for example if I have a system with a throughput of 1000/s and a latency of 1ms, which satisfies the given relation but for no particular reason. Now suppose I add at the beginning of every request a 1-hour delay. Now my latency is 3600001ms, but my throughput is still 1000/s. The only thing that has changed is the number of requests in flight.
- rawnlq 10y agoIf you assume that queue is bounded (which is true for real systems), then Little's Law does hold: Number of requests in queue = latency * throughput E.g., left hand side is a constant if you assume the system is fully utilized or constantly under max load.