This analysis assumes "Early Release" of the token. Let:

µ

be mean frame service rate (1/*D _{T}*)

*lambda*

be mean aggregate frame arrival rate at all stations

*U*

be channel efficiency; at
equilibrium, *U* = *rho* = *lambda*/µ

*q*

be mean frame queue length at a station when the token arrives

*N*

be number of stations on the ring

*D _{P}*

be propagation delay around cable

*R*

be data rate

*TRT*

be Token Rotation Time, the period between appearances of the token at a point in the ring

Token Rotation Time, *TRT* = cable propagation delay (*D _{P}*)
+ (at least) 1-bit delay (1/

*TRT = D _{P} + N/R + qN*/µ

assuming that the station service time, q/µ, is less than the Token Holding
Time (*THT*) the Lightly Loaded case. The queue of frames which
accumulates at a station in a period *TRT* is 1 *N*'th of the total
number of frames which arrive at all stations:

*q = TRT × lambda / N*

Thus, substituting for *q* in the expression for *TRT*:

*TRT = D _{P} + N/R + TRT* ×

which can be re-arranged to give an expression for Token Rotation Time, *TRT*:

*TRT* =

When a frame arrives at a station, it can expect to be about half way
through a period of Token Rotation Time, and thus delay is on average, about ½*TRT*.
As efficiency `-->`
1, *TRT* `-->`
*infinity*.

In case of a system at equilibrium, *q/µ<THT* because each station
clears its entire backlog of accumulated frames before the Token Holding Timer
expires. Then:

*U* = = *rho* **
Lightly Loaded Case**

In case of a temporarily overloaded (i.e. non-equilibrium) system, where,
q/µ>*THT*, channel efficiency, *U* < *rho*, because each
station can not clear its entire backlog of accumulated frames. Then:

*U* = **Heavily
Loaded Case**

e.g. In a system at equilibrium, let *N*=100, *R*=10 Mbps, frame
size = 1000 bits, *D _{P}*=5 µsec, and mean aggregate frame arrival
rate at all stations,