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The theoretical maximum utilization (percentage of capacity used) of a segment goes down as the value of a (the length of the medium in relation to the frame size) goes up How does this work The media must be clear before another frame can be transmitted If a is large, then the stations must wait for the frame to be sent the length of the medium before it is clear It is important to remember that when data is on the wire, utilization is 100% When no data is on the wire, utilization is 0% and cuts into the overall utilization of the medium If, for example a = 1, then the maximum utilization of the medium can be at best only 50% As soon as the start of the frame hits the end of the wire, the end of the frame is being transmitted All devices must wait for the end of the frame to leave the wire Since the length of the wire in bits is the length of the frame, the same amount of time used to transmit the frame is now spent waiting for the frame to propagate all the way down the wire Thus the wire is only used to transmit data half of the time, so utilization is at best only 50% This is the best-case scenario; there is no overhead, no collisions, no errors, no lost data, and another frame is there to be sent at the exact point the first frame leaves the wire The following table illustrates several cases of a 10 Mbps segment that uses 5000-bit frames As the length of the cable increases, a goes up and the theoretical maximum utilization starts to go down dramatically You can see why it is important to keep a as low as possible
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Since the maximum length for many media is about 100 meters, it is impossible for some LAN protocols (including Ethernet) to even approach the length required to get a large value for a Keep in mind that this calculation is theoretical and should be used to understand how networks work In the real world, network performance can be influenced by a variety of factors
Relative Throughput for Token-Based Protocols
When considering network technologies, it's useful to define the relative throughput value Relative throughput is the ratio of the achieved data rate over the theoretical limit of the throughput For example, if you measure only 20 Mbps of data transfer on your 100 Mbps segment, your relative throughput is 02 The token-passing and collision-based protocols each have their own calculations to determine the relative throughput of a given type of LAN These equations are different because the two LAN protocols have different methods of carrying data Let's start with token-based arrangements You'll remember that in token-based protocols, a frame is transmitted from the token-holding station to the medium The frame travels from station to station until it comes full circle and is received by the sending station This station passes the token along to the station's neighbor, so that it can place a frame on the network if it needs to If this station doesn't have any frames to send, it passes the token farther along Token Rings that carry minimal amounts of network traffic have low utilization values because the station wishing to send a frame must wait for the token The time spent waiting for the token is time when data is not being sent This reduces the utilization time Token passing works better on a busy segment than do collision-based protocols, however, because the token is passed in sequence; every station has an equal chance to place a frame on the network Since there is little time spent waiting for the token (between two stations that have data to send), the realized throughput goes up
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