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slot time and min frame size
#1
could someone make me clear with this plz..
Slot time is the time for a signal to reach the farthest end and get a jam signal from there back .. for traditional 10mbps ethernet using coax cable max cable lebgth was 500m..so using max 4 repeaters total network length was 2500m..so two times it is 5000m..and with 2*10^8 as the propagation delay..
it would take 25microsec..but considering 3microsec delay/repeater it would be 4 repeaters*3microsec *2(for both direction) ie another 24microsec..total of approx 50 microsec..during which our station can transmit 500 bits..rounded off to nearest power of two..that would be 512 bits..
Now if that was the correct or approx correct calculation for 10mbps..
Consider this for std ethernet 100mbps..
100mbps uses utp cables..whose max length is about 100m..so using 4 repeaters thats 500 m..giving a total of 1000m both way..which takes 5microsec..using same signal speed..and during this time our station can send 500 bits again..so approx 512 bits.. so that makes the slot time in both the same .. but why don't we consider repeater delay here ?? or am i getting it wrong..if so plz correct me .. thanks in advance ..
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slot time and min frame size - by ggauravr - 02-07-2010, 03:01 PM

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