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time slot proofs

Unanswered Question
Mar 18th, 2006
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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



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pkhatri Sat, 03/18/2006 - 16:27
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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

haplo1213 Sat, 03/18/2006 - 22:46
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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?


pkhatri Sun, 03/19/2006 - 15:11
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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

haplo1213 Sun, 03/19/2006 - 19:24
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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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