Three averages
Little's law is about a system: anything that things enter and later leave. Our box holds a router's queue and its link, the server that sends one packet at a time. Packets arrive, maybe wait, get sent, then leave.
Packets arrive at λ (the Greek letter lambda) = 1,000 per second: one every millisecond on average. They come in bursts, not evenly spaced. That is fine.
W is the average time one packet spends inside the box: waiting in line plus being sent. Our packets stay between 0.8 and 2.8 ms each; their average is W = 2 ms.
Now count the packets inside the box at each moment. The count jumps between 0 and 4 as packets come and go. Averaged over time, it is L = 2 packets.
The law
Little's law ties the three together: L = λ × W. Here 1,000 packets per second × 0.002 seconds = 2 packets. Units check: per second times seconds leaves just packets.
Each packet adds one packet-millisecond to the chart for every millisecond it is inside. In 8 ms, 8 packets stay 2 ms each on average: 16 packet-ms. Spread over 8 ms, that is 2 inside on average.
The law does not care how arrivals are spaced, how long each service takes, or which packet goes first. It only needs a stable system: over a long time, as many leave as arrive (Little, 1961).
Using it
Draw the box around the link alone. Each packet spends S = 0.8 ms there, so on average the link holds λ × S = 1,000 × 0.0008 = 0.8 packets. It is busy 80% of the time: its utilisation.
Now the box is only the line. Packets wait Wq = 2 − 0.8 = 1.2 ms on average, so Lq = 1,000 × 0.0012 = 1.2 packets are queued. Queue plus link: 1.2 + 0.8 = 2.
Same law, different box, any units. A café gets 30 customers an hour, and each stays 20 minutes on average. λ = 0.5 per minute, so L = 0.5 × 20 = 10 customers inside at a typical moment.
Same law, different box: know two, get the third. A web server handles 200 requests per second and has 50 open at once on average. So each request takes W = L ÷ λ = 50 ÷ 200 = 0.25 s.
The arithmetic
;; step 1/11 · the-system ;; at the system a "system" is any box with a way in and a way out: a router queue, a whole network, a web server, a café, a warehouse here: one FIFO queue + one link that sends 0.8 ms per packet 1,000 bytes × 8 / 10 Mb/s = 0.8 ms the drawing runs 1,000× slower than the network: 1 ms takes 1 s
The arrival pattern (8 packets repeating every 8 ms) is illustrative; every average shown is computed from it exactly. The drawing runs 1,000 times slower than the network, so one millisecond takes one second on screen.