7.7 A Complex Model 2 Configuration Example

Let's continue the process of finding the total round trip delay time by doing the calculations for the middle segments. In the worst-case path for the network in Figure 7.3 there are three middle segments composed of a maximum length 10BASE5 segment, and two 500 meter long 10BASE-FL fiber optic segments. By looking in the table under mid-segments we find that the 10BASE5 segment has a Max delay value of 89.8.

Note that the repeaters are connected to the 10BASE5 with AUI cables and outboard MAUs. That means we need to add the delay for two AUI cables. Let's assume that we don't know how long the AUI cables are, therefore we'll use the value for two maximum-length AUI cables in the segment, one at each connection to a repeater. That gives us an AUI cable delay of 9.76 to add to the total path delay.

We can calculate the segment delay value for the 10BASE-FL mid-segments by multiplying the 500 meter length of each segment times the RT Delay/meter, which is 0.1, which gives us a result of 50. We then add 50 to the mid-segment base value for a 10BASE-FL segment, which is 33.5, for a total segment delay of 83.5.

Even though it's not shown in Figure 7.3, fiber optic links often use outboard fiber optic MAUs and AUI cables. Just to make things a little harder, let's assume that we used two AUI cables of 25 meters each to make a connection from the repeaters to outboard fiber optic MAUs on the 10BASE-FL segments. That gives us a total of 50 meters of AUI cable on each 10BASE-FL segment. Since we have two such middle segments we can represent the total AUI cable length for both segments by adding 9.76 extra bit times to the total path delay.

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