4 ms·

This relies on the assumption that work time per ticket follows a distribution described by the second M. If you get tickets that require much more work, i.e. t

by sampo 3y ago

This relies on the assumption that work time per ticket follows a distribution described by the second M. If you get tickets that require much more work, i.e. tickets that are outliers and not from the distribution described by the M, then your estimates will end up wrong, too.

pjot 3y ago
Eh, not necessarily. Queuing is a function of throughput, which by definition takes processing time into account. Queue length = arrival rate * processing time Transitive properties allow you to solve for one variable, given you have the other two. This is known as Little’s Law[0]. With this, you can now deduce/estimate how long a ticket will be in the backlog, how fast you need to complete a task, how long until everything is finished. [0]: https://en.m.wikipedia.org/wiki/Little%27s_law https://en.m.wikipedia.org/wiki/Little%27s_law
- sampo 3y ago
  > This is known as Little’s Law Little's law lets you calculate the third variable easily, given that you know the other two. You can calculate from existing data. As such, it doesn't help you calculate forecasts. Or you can, if you make assumptions on the distributions. But some outliers can then turn the reality very different from what your forecast was. > how fast you need to complete a task You can calculate how fast you need to complete a task. But if an outlier task comes your way, it doesn't help you to actually complete that particular task in time.
kqr 3y ago
If you're able to multiplex a little bit (which most humans naturally do) the shape of the service time distribution does not matter as much as it may seem. (In the limit, perfect timesharing across tasks makes even the fattest of tails look M/M/k.) The more severe constraint is actually the first M. Fortunately, that holds true in practise very often.

4 ms·

This relies on the assumption that work time per ticket follows a distribution described by the second M. If you get tickets that require much more work, i.e. t

by sampo 3y ago

pjot 3y ago
Eh, not necessarily. Queuing is a function of throughput, which by definition takes processing time into account. Queue length = arrival rate * processing time Transitive properties allow you to solve for one variable, given you have the other two. This is known as Little’s Law[0]. With this, you can now deduce/estimate how long a ticket will be in the backlog, how fast you need to complete a task, how long until everything is finished. [0]: https://en.m.wikipedia.org/wiki/Little%27s_law https://en.m.wikipedia.org/wiki/Little%27s_law
- sampo 3y ago
  > This is known as Little’s Law Little's law lets you calculate the third variable easily, given that you know the other two. You can calculate from existing data. As such, it doesn't help you calculate forecasts. Or you can, if you make assumptions on the distributions. But some outliers can then turn the reality very different from what your forecast was. > how fast you need to complete a task You can calculate how fast you need to complete a task. But if an outlier task comes your way, it doesn't help you to actually complete that particular task in time.
kqr 3y ago
If you're able to multiplex a little bit (which most humans naturally do) the shape of the service time distribution does not matter as much as it may seem. (In the limit, perfect timesharing across tasks makes even the fattest of tails look M/M/k.) The more severe constraint is actually the first M. Fortunately, that holds true in practise very often.