Ethernet throughput
Let
- k be the number of nodes trying to send,
- p is the probability of one station sending, and
- r is the round trip time
Then the probability that one of the nodes will succeed is A = kp(1-p)k-1
This has a maximum at p=1/k, (when?) and its limit for large k is 1/e = 0.37
- If the packets are all of minimum length this is the efficiency
- Expected number of tries is 1/A = e = 2.7 at this maximum, including the successful transmission
- The waste, also called the ‘contention interval’, is therefore 1.7r
- For packets of length L the efficiency is L/(1 + 1.7r)=1/(1 + 1.7r/L) ~ 1 - 1.7r/L when 1.7r/L is small
The biggest packet is 1.5 Kbytes = 20 r, and this yields an efficiency of 91.5%.