03-18-2006 12:06 PM - edited 03-03-2019 02:21 AM
can anyone tell me why we using 51.2microseconds for time slots?
or 512bit as the minimum frame size?
i've tried to calculate that
x=(2500*2(distance))/(2*10(8))
it get 25micsecond
03-18-2006 04:27 PM
It comes from the following:
Minimum size of ethernet frame = 64 bytes = 512 bits
Time taken to transmit a minimum size ethernet frame on a 10Mbits link = 512/10,000,000 = 51.2 microseconds.
Hope that helps - pls rate the post if it does.
Paresh
03-18-2006 10:46 PM
yes , but why 512bits?
why IEEE802.3 choosed this number to be the minimum frame size? what was the calculation ?
i couldnt find any normal answer on google.
the one-trip propogation delay is 12.5uSeconds (for 2500M cable)..
two round-trips 25uSeconds....? why not 250bit?
03-19-2006 03:11 PM
As you worked out, the round-trip time would be around 25 microseconds. From what I understand, the IEEE decided to build in a safety margin when deciding on the 512 bits minimum frame size.
Hope that helps - pls rate the post if it does.
Paresh
03-19-2006 03:17 PM
if you are talking about the 64-byte ethernet frame, it's the smallest allowable frame size; 18-byte ethernet header + 46-bytes of data.
03-19-2006 07:24 PM
yeah that's the minimum frame, but why .
i made some research on that
here's the results.
we need to discovre the total path delay.
25us for 2500 cable
4 repeaters class I add 0.7us for each repeater
DTE delay
and safety margin
so
total delay=link delay+repater delay+DTE delay+saftey margin
that should come out 51.2us :))
still need to get the numbers thu
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