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?

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

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