Little’s Law#
SLOP3068, week 2
Amos Redfern
The claim#
You can work out how long people are waiting
without timing a single one of them
L = λW#
What the letters are#
| Symbol | What it is | How you get it |
|---|---|---|
| L | number in the system | count heads |
| λ | arrival rate | count arrivals, divide by time |
| W | time in the system | the thing you wanted |
Two of these are easy from thirty metres away.
The third one is not, so we solve for it.
Do it now#
Look at the queue downstairs.
- twelve people in it
- four joined in the last minute
W = 12 / 4 = three minutes
You have not followed anybody.
Why it is strange#
It does not care about
- the order you serve people in
- how many servers there are
- the shape of the arrival distribution
- whether the thing queueing is a person, a packet or a kidney
No assumptions about any of it.
Any system. Any discipline. Any distribution.
The price is that every letter is an average.
The price#
Stable. Not filling up, not draining.
Point it at a growing backlog and it returns a number that is true and useless.
That number is a property of the snapshot, not of the queue.
Failure one, instability#
A passport office with a two-year backlog
- L is enormous
- λ looks normal
- W comes out as two years
Correct. Also not a wait; it is a debt.
Failure two, the boundary#
Draw the box before you count.
- who balked and never joined
- who reneged and walked out
- who was served without joining
An arrival in one count and not the other breaks the law quietly.
Failure three, the mismatch#
Arrivals counted during the rush.
Residence time measured in the lull.
Both numbers correct. The ratio is fiction.
Wednesday#
Two people, one queue, thirty minutes, separate sheets.
You will disagree with your partner on about one arrival in six.
The disagreements are the session.
Before you go#
Count a queue. Any queue.
Get a W without a stopwatch.
Bring the number and how you got it.