Some notes :
Keep in mind that dual speed hub (not switch) is actually two hubs in one box with a bridge in between. The bridge is useful for limiting traffic from the 10 -> 100 and from 100 -> 10. If the destination address is running at the same speed as the source, the other side of the hub never gets involved and sees no traffic. Nifty idea with a catch. Because it's actually two hubs in one, the 5-4-3 rule only allows one more hub in the system. If you take two of these hubs, and connect them together using just the normal hub ports, you have the potential of having 4 hubs in series.
====>===[10 -> 100]====>====[100 -> 10]====>=== hub 1 link hub 2Going from 10baseT to 10baseT via a 100base-TX link, you have the equivalent of 4 hubs in series. This will NOT work reliably, but can be helped along if the delays in the 100base-TX link are reduced. Therefore, the short cable lengths are needed.
The 5-4-3 rule states that you can have no more than 5 total segments, 4 repeaters and 3 populated segments (implying that 2 of the 5 total segments are inter-repeater link segments). This is a generalized rule of thumb, not a hard rule.
The reason behind this limit is based is based on the fact that the maximum round-trip signal propagation delay can be no greater than 51.2 microseconds (512 bit times, which is 64 bytes, the minimum length of an Ethernet frame). The idea here is that if one of the two most distant device begins sending a frame and just before that frame reaches the other of the two most distant devices, the other device starts sending a frame, the first device sees the resulting collision before it finishes sending its frame.
Whether a collision domain is constructed of copper or fiber, or a mixture of the two, the same 51.2 micorosecond timing constraint must be observed if you are using layer 1 devices such as repeaters and/or media converters
And for what its worth, transceivers don't count as repeaters.
And the rules for twisted pair is as follows:
Maximum nodes per segment : 512 Maximum populated segments : 1024 Maximum number of repeaters : 4 Maximum # of segments : 1024 Maximum segment length : 100 meters Maximum total LAN length : N/A Minimum cabling distances : .06 meters
Now the FAQ section :
Q: How do you define segment?
A: A segment is an electrically (or optically) continuous conductor. Coax, twisted pair, or fiber. A segment connect a computer to a hub and one hub to another. It may be a single length of cable or several sections of cable connected together.
Q: Could a segment be, say 150 PC's with 32 HUBS or something?
A: The 5-4-3 rule is not the size of the network. There is no small limit to the size of a network. The legal limit is 1024 devices and more practically about 750.
Q: Does segment apply to the physical length of the wire on that segment, which you can only extend using repeaters?
A: 5-4-3 limits the maximum path length between the two most distant nodes in the collision domain.
And remember, a switch or router starts a new collision domain, so everything starts counting again.
I highly recommend this site for many reasons. It has 10BaseT cable rules, 100BaseTX cable rules, the 5-4-3 rule diagram.
Now onto 100BaseTX :
100baseTX is not 10baseT. The 5-4-3 rule does not apply. The "rule" for 100baseTX is not more than three segments and two repeaters, with not more than 5 meters of cable between repeaters. If you pull a copy of the spec and dig into it you'll find ways to tweak this a bit, but there's generally not much profit in trying.
If you need more than 5 meters between hubs, best thing to do is either put a two-port switch in the path between the two hubs or use a switch instead of a hub at one end or another.
The "bridging port" would in more current terminology be described as a "switching port" and it starts the cabling rules over--you can go 100-5-100 from the port.