Newell Ryan D SrA 18 CS/SCBT wrote: > > > 500 Meters?? It's 2500 meters. In one example of such a > network, there can > > be 5 segments, 4 repeaters (hubs), but only 3 segments can > have end > > systems. That's the infamous 5-4-3 "rule." It makes a lot of > > assumptions. Really, the > > size of the network depends on round-trip propagation delay > for the > > particular equipment, cables, and cable lengths. > > Maybe I was wrong for thinking that. If my net was all 10 Base > T, then with > max 5 segments...500 meters. That's were I got that number > from. Measuring > the size of the collision domain is well under slot time. So I > could > technically extend the size of the network.
The segment from the hub to the end station might be 100 meters, as that's how structured cabling is usually done. Between hubs probably isn't 100 meters, for what it's worth. In fact, it might be fiber-optic cabling. > > One of the things I ran into was the formula to use to > calculate the round > trip delay. With the formula in your book I came up with 210 > bit times round > trip for 500 meter 4 hub network. But with the definitive > guide's method I > got 362 bit times. When I was going back and forth between > books I think I > got lost somewhere. For a 100 meter cable they suggest 11.3 bit > times. While > you suggest 5 one-way or 10 round trip...very close. But they > start with a > base value. > Example First segment would be 26.55 bit times instead of 11.3. > The base > value is 15.25. 15.25+11.3=26.55 bit times for the first > segment. Technically, IEEE does say to add some DTE delay time, i.e. time at the stations themselves, both the sender and receiver. This is all documented in IEEE 802.3 documents, which are available for free from IEEE. It's not worth reading though (for this purpose I mean.) > > I think I understand the theory behind slot time. It takes a > station 51.2 > micro seconds to transmit the smallest frame. So station a > needs to be > notified by any other station if a collision was to happen > while it was > still transmitting. That's it. > So when the first bit of station a's > preamble hits > station z (at the other side of the network) rx pins while > station z was > transmitting, it's first bit hits the repeater. The repeater is > going to use > collision enforcement to make all stations including station a > is aware of > the collision. This must happen before station a finishes > transmitting the > smallest Ethernet frame. I think that is it. > > So should bit time be the time it takes to transmit the > preamble and 512 > bits? The preamble doesn't count. It's used to recover timing. A station or repeater might not catch all of the preamble. It just has to see the pattern and the start of frame delimiter. A repeater regenerates the preamble, by the way. > > One more thing... > > A proper preamble should look like 10101010 or AA. I'm sure I > read somewhere > that a collision would appear with all 5's or C's. We used to see 55s on old coax networks. Never saw Cs though. > How would > that be > possible if as soon as the repeater detects a collision it > sends out a jam > signal out all its ports? Then you would see alternating ones and zeros on the end of a frame. I have seen this, but not recently. My current NIC won't give me bad frames so even a sniffer doesn't give them to me. > Also a frame with a bad CRC is > suspect of a > collision. The frame got damaged when the collision occured. > How? If you know where I could get more reading on > this that > would be great! IEEE 802.3. > > Thanks for answering my questions! > > > > "We are what we repeatedly do. Excellence, then, is not an act, > but a > habit."--Aristotle > > Message Posted at: http://www.groupstudy.com/form/read.php?f=7&i=63669&t=63659 -------------------------------------------------- FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
