5-4-3 Rule?
Whats the deal with this 5-4-3 rule? is it the same as this statement from one of the technotes:
That is from the 10Base2 section of the Network Topologies Technote
Maximum length of collision domain is 2500 meters (5 segments, 4 repeaters, max. 3 segments populated)
That is from the 10Base2 section of the Network Topologies Technote
CQ Services, Inc.
ISP, Repair, Web Hosting, Web Design, Online Advertising, & Computer Sales
ISP, Repair, Web Hosting, Web Design, Online Advertising, & Computer Sales
Comments
-
Webmaster Admin Posts: 10,292 AdminYes. www.techexams.net/technotes/networkplus/networkcomponents.shtml
Includes a network diagram as well... -
jeff CQ Member Posts: 20 ■□□□□□□□□□how does that get you 2500 meters if the thinnet max is 185?CQ Services, Inc.
ISP, Repair, Web Hosting, Web Design, Online Advertising, & Computer Sales -
Drakonblayde Member Posts: 542If I'm not mistaken, the Ethernet maximum is 2500 meters, not necessarily 10base2. 10base5 would give you a maximum of 2500 meters under the 5-4-3 rule. 10baseF breaks that rule if I remember correctly, being able to do 2000 meters per segment, but also being limited to no more than 2 repeaters. 100BaseFX is limited to 400 meters per segment, and is also limited to 2 repeaters I think, but even if it's not, it still falls within that maximum cabling distance.= Marcus Drakonblayde
================
CCNP-O-Meter:
=[0%]==[25%]==[50%]==[75%]==[100%]
==[X]===[X]====[ ]=====[ ]====[ ]==
=CCNA==BSCI==BCMSN==BCRAN==CIT= -
Webmaster Admin Posts: 10,292 AdminIndeed the maximum for Ethernet, strictly taken, meaning 10Mbps Ethernet. In 100BaseT Fast Ethernet the maximums collision domain ranges from 100 to 412 m depending on the type of media (copper, fiber, multimode fiber) and the type repeaters being used (Maximum one Class 1 repeater, or two class 2 repeaters.)
The 5-4-3 diagram in the TechNotes is logical, as it is a general Ethernet design constraint.
In 10Mbps Ethernet (also 10Base2), no two stations can be separated by more than 2500 meters of cable (or 4 repeaters).
Only important for the CCDA exam, but here's a quote from some .edu site, which describes the why:The most significant design rule for Ethernet is that the round-trip propagation delay in one collision domain must not exceed 512 bit times. This is a requirement for collision detection to work correctly. This rule means that the maximum round-trip delay for 10-Mbps Ethernet is 51.2 microseconds. The maximum round-trip delay for 100-Mbps Ethernet is only 5.12 microseconds because the bit time on 100-Mbps Ethernet is 0.01 microseconds as opposed to 0.1 microseconds on 10-Mbps Ethernet.
To make 100-Mbps Ethernet work, there are much more severe distance limitations than those required for 10-Mbps Ethernet. The general rule is that a 100-Mbps Ethernet has a maximum diameter of 205 meters when UTP cabling is used, whereas 10-Mbps Ethernet has a maximum diameter of 2500 meters. -
chmars53 Member Posts: 10 ■□□□□□□□□□Great quote, webmaster!
Jeff, remember too that any collision domain stops at devices like Bridges and Routers. And that STACKED hubs, routers, & bridges, act as a SINGLE device.
The entire idea of the 5-4-3 rule is that it takes TIME for a signal to get from one endpoint to another, and that to avoid UNKNOWN ERRORS you must not have a device farther away than the TIME it takes to transmit the packet.
Ever talked on a phone where there is a big delay (you see it all the time on CNN and other satellite link-ups) ? People talk over each other because they have such a delay in transmission that they don't know the other person has decided to add another comment instead of just waiting for a response to the first comment. In the Ethernet world, 5-4-3 stops that.Repairing Mainframes since 1978 - & still learning