Tiongster wrote: > > Hi to all pros, > > I would like to ask a very simple which is also very confusing. > If I > connected three hubs together have I break a 5-4-3 rule?
No, 3 hubs connected together does not break the 5-4-3 "rule." > But I > don't > think I have break the 5-4-3 rule because I believe the wiring > of the > hub is a bus circuit which is a segment The 5-4-3 "rule" already takes into account that repeaters (hubs) introduce some delay. You don't have to consider their bus architecture, or if they even have a bus architecture. The "rule" includes soem default amount of delay for the repeaters (hubs). > and with three hubs > connected > together it consist of three segments. Four segments, actually, but that's OK. PC---HUB---HUB---HUB---PC Each --- represents a segment. > I hope someone will > correct me if > I'm wrong as I have been debating with my peers for past few > days. My > peers insisted four hubs connected is the correct one which is > the > maximum network of the 5-4-3 rule. Aren't four hubs consist of > four > segments which have broke the 5-4-3 rule? Four hubs is still OK too. PC---HUB---HUB---HUB---HUB---PC The 5-4-3 topology is just one possibility. Other topologies are possible as long as round-trip delay doesn't exceed the time to send the minimum-sized frame (512 bits). Priscilla >Thanks in advance. > > Chiam Chin Tiong CCNA > ITE Dover student > ICQ 153179194 > > [GroupStudy removed an attachment of type > application/octet-stream which had a name of image001.jpg] > > [GroupStudy removed an attachment of type > application/octet-stream which had a name of image001.jpg] > > Message Posted at: http://www.groupstudy.com/form/read.php?f=7&i=65235&t=65140 -------------------------------------------------- FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
