One queue
A router's outgoing link is a server with a queue. Packets arrive at random moments (a Poisson process) and take a random time to send (exponential). That is an M/M/1 queue. Here 8,000 packets/s arrive; the link sends 10,000/s.
Utilisation ρ = λ ÷ μ = 0.80. On average L = ρ/(1−ρ) = 4 packets sit at the hop, and each spends T = 1/(μ−λ) = 0.50 ms there: 5 times its 0.10 ms sending time.
A chain of queues
On the way to a server, packets cross many routers, and every hop has its own queue. Three hops in a row form a tandem network: what leaves one queue is what arrives at the next.
Jackson (1957) proved that with Poisson arrivals and exponential service, each hop behaves like its own M/M/1 queue, fed by its total arrival rate. Burke's theorem is the key: packets leave an M/M/1 queue as a Poisson stream again.
So the end-to-end delay is a simple sum. Each hop holds a packet 0.50 ms on average, so three hops take 0.50 + 0.50 + 0.50 = 1.50 ms of waiting and sending. Travel time on the cables comes on top.
Real traffic
Averages hide unlucky packets. At one hop the time is exponential, so 1 packet in 100 takes over ln(100) × 0.50 = 2.30 ms. Over three hops the 99th percentile (p99) is 4.20 ms, almost three times the mean.
Real traffic is not Poisson: it comes in bursts at every time scale (Leland et al., 1994). Kingman's formula estimates the cost. With gaps between packets twice as uneven, the wait at ρ 0.80 is 2.5 times longer: 1.00 ms, not 0.40.
Using the results
Traffic grows from 8,000 to 9,500 packets/s, only 19% more. ρ climbs to 0.95 and each hop now takes 1/(10,000 − 9,500) s = 2.00 ms. End to end: 6.00 ms, four times as long.
Now give each link 20% more capacity: 120 Mb/s. ρ falls to 0.79 and each hop takes 1/(12,000 − 9,500) s = 0.40 ms. End to end: 1.20 ms instead of 6.00. A little spare capacity removes most of the delay.
Engineers run the maths backwards. Goal: p99 under 5 ms over three hops at 9,500 packets/s. The model says each link needs at least 111.8 Mb/s. Bursts need more, so they pick 120 Mb/s, then measure the real p99.
The arithmetic
;; step 1/10 · one-queue · send ;; senders → hop 1 M/M/1 = Markov (Poisson) arrivals / Markov (exponential) service / 1 server arrivals λ = 8,000 packets/s, gaps exponential, mean 125 µs link 100 Mb/s, mean packet 1,250 bytes = 10,000 bits service μ = 100,000,000 / 10,000 = 10,000 packets/s, mean 0.10 ms each first in, first out; the buffer is assumed big enough never to drop
Each tray draws at most 5 envelopes; the meter gives the real average number waiting. The delays here are waiting plus sending at each hop; propagation along the cables and processing inside the routers add a fixed amount on top.