5-4-3 Rule am i right? (Media)

mobri09

Correct me if i am wrong:

When working with coaxil ethernet cable:
10base2
10base5 5 segments can be connected using no more than 4 repeaters and only three of those segments can be populated.


When working with Utp ethernet Networks:

A total of 5 segments can be used on the network with a total of 4 repeaters, but all 5 segments can be populated.

10base networks only 2 segments can be populated?

My question is are these right and should i know any more 5-4-3 rules for any other media? Let me know if so.

Second question is does 1000 base cx use stp or coaxil...Exam cram saids Stp and Techexams saids Coaxil?

Technotes were extremely helpful! Making sure i am on the right track though.

aznluvsmc

The 5-4-3 rule really only applies to networks using Coaxial cable in a bus topology with repeaters. This rule is basically obselete now with use of switches.

You need to know that 5-4-3 rule states you can only have 5 segments separated by 4 repeaters with no more than 3 populated segments. If you know that, then the questions the exam will ask you will be a cake walk.

Webmaster

Second question is does 1000 base cx use stp or coaxil...Exam cram saids Stp and Techexams saids Coaxil?

it is a coaxial type of cable. (not the same as coaxial cables used for 10base2 or 5) To be exact, it is a shielded 150 Ohm pair of coaxial cable, also known as Twinax (twin from pair, ax from coax )

illogos

The 5-4-3 rule can be summarized as follows:

First, it applies to the longest path between two systems on a network. It has to do with the timing on ethernet networks. It has to do with collision domains. If the path between any two systems in the domain is too long then the following can happen: Computer A starts to transmit a frame. Computer B starts to transmit a frame after Computer A started but before the frame reached Computer B(thus B thought the line was free). Now, a collision will occur. When the path is too long, A can finish transmiting before the frame from B starts to reach it. Since collision detection is only active during transmission, A will not hear the collision.

The 5-4-3 rule is a generalization of the complex calculations to determine exactly the size a certain network can be. It works as 5 segments, 4 repeaters, and 3 populated segments between the farthest two systems in the collision domain. So, you *can* have more in the network as long as the farthest two are within the 5-4-3 rule.

Since this applies to collision domains, if you segment a network with a bridge, a system must be within 5-4-3 of the bridge. Populated segments are defined as segments with computers directly attached to it. Remember that in 10BaseT a hub counts as both a segment and a repeater. However, if you just have a interconnecting hub(only hubs are plugging into a specific hub) it does not count as populated.

On my test, yesterday, I do not remember getting tested directly on the 5-4-3 rule. However, on pratice tests I took there were questions but they were almost all in concern to physical bus topologies. I hope this helps clear things up a little.

If you have access to the Meyer's all in one second edition, look to page 141 for the explanation of 5-4-3. There are some helpful diagrams to go along with it.

For 1000baseCX there is a picture on page 152 of the Meyer's book. It is indeed 150 ohm Twinaxial cable. It has Twinaxial printed on the outer insulation. On the inside it looks like standard coax: it has the shielding, and the inner insulation, but instead of a single main wire there are two.

mobri09

Everyone has anwsered my questions like always! Thank you for you time!