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Community Member

## time slot proofs

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

5 REPLIES
Purple

## Re: time slot proofs

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

Community Member

## Re: time slot proofs

yes , but why 512bits?

why IEEE802.3 choosed this number to be the minimum frame size? what was the calculation ?

the one-trip propogation delay is 12.5uSeconds (for 2500M cable)..

two round-trips 25uSeconds....? why not 250bit?

Purple

## Re: time slot proofs

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

Community Member

## Re: time slot proofs

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.

Community Member

## Re: time slot proofs

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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