Ethernet Performance
probability A that some station acquires the channel:
maximized when p = 1/k, with A → 1/e as k →∞
The probability that the contention interval has exactly j slots:
mean number of slots per contention:
slot duration = 2τ, the mean contention interval, w = 2τ/A.
w is at most 2τe
∼∼
5.4τ.
mean frame takes P sec
The longer the cable, the longer the contention interval