Ethernet Throughput

Throughput of a system refers to the rate of processing of a task thereby generating results. Ethernet is a set of technologies primarily used in LANs, whose primary data units are frames. The throughput of Ethernet is measured by the rate of successful delivery of frames over a communication channel.

There are several methods for representing Ethernet throughput. The least ambiguous among them is calculation of channel efficiency. Channel efficiency is the percentage of the net bit rate (in bits per second) of a channel that is actually communicated. Suppose that an Ethernet connection has a speed of 100 Mbps. But, it is seen that effectively 80 Megabits of data are transmitted per second. Here, the channel efficiency is 80% or 0.8.

Ethernet Channel Efficiency Nominal Speed: 100 Mbps Effective Throughput: 80 Mbps Channel Efficiency = 80/100 = 80% Due to collisions, retransmissions, and protocol overhead 0% Efficiency 100% Efficiency

Formulation of Channel Efficiency for Ethernet Throughput

Let us assume an Ethernet network has k stations and each station transmits with a probability p during a contention slot. Let A be the probability that some station acquires the channel. A is calculated as:

A = kp (1?p)^(k-1)

The value of A is maximized at p = 1/k. If there can be innumerable stations connected to the Ethernet network, i.e. k ? ?, the maximum value of A will be 1/e ? 0.368.

Let Q be the probability that the contention period has exactly j slots. Q is calculated as:

Q = A (1?A)^(j?1)

Let M be the mean number of slots per contention. The mean number of contention slots is:

M = 1/A

Given that ? is the propagation time, each slot has duration 2?. Hence the mean contention interval, w, will be 2?/A.

Let P be the time in seconds for a frame to propagate.

Channel Efficiency Formula

The channel efficiency, when a number of stations want to send frames, can be calculated as:

Channel Efficiency = P / (P + w)
                   = P / (P + 2?/A)

Let F be the length of frame, L be the cable length, c be the speed of signal propagation and e be the contention slots per frame. The channel efficiency in terms of these parameters is:

Channel Efficiency = 1 / (1 + 2?L/(cF) × e)

Key Factors Affecting Throughput

Factor Effect on Throughput
Number of stations (k) Higher k increases collision probability, reducing efficiency
Frame length (F) Longer frames improve efficiency by reducing overhead ratio
Cable length (L) Longer cables increase propagation delay, reducing efficiency
Transmission probability (p) Optimal value is 1/k for maximum channel acquisition

Conclusion

Ethernet throughput depends on channel efficiency, which is affected by the number of stations, frame length, and network propagation delays. The theoretical maximum efficiency approaches 36.8% under ideal conditions with optimal transmission probabilities.

Updated on: 2026-03-16T23:36:12+05:30

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