Analysis of IEEE 802.5 Token Ring Performance

© Copyright University of New Haven 1998

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


be mean frame service rate (1/DT)


be mean aggregate frame arrival rate at all stations


be channel efficiency; at equilibrium, U = rho = lambda


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


be number of stations on the ring


be propagation delay around cable


be data rate


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

Token Rotation Time, TRT = cable propagation delay (DP) + (at least) 1-bit delay (1/R) at each of N stations + Station Service Time (q/µ) at each of N stations:

TRT = DP + N/R + qN

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 = DP + N/R + TRT × lambda × N/(µ×N) = DP + N/R + rho × TRT

which can be re-arranged to give an expression for Token Rotation Time, 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 ½TRT. 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, DP=5 µsec, and mean aggregate frame arrival rate at all stations, lambda = 5000 frames/sec. Then, frame service time, 1/µ = DT = 1000/R = 100 µsecs and efficiency, rho = lambda / µ = 5000 × 100×10-6 = ½. Delay = ½TRT = ½ (5 + 100/10)/(1-½) = 15 µsec. A station's queue of backlogged frames, q = 30×10-6×5000/100 = 0.0015, giving station service time, q/µ = .0015×100 = 0.15 µsecs. Assuming that THT exceeds 0.15 µsecs, then the Lightly Loaded equation gives: efficiency, U = 0.15/(0.15+5/100+1/10) = ½. Assuming THT=1 millisecond, then the Heavily Loaded equation gives U = 1000/1000.15, very close to 1.

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