# hhhheeeellllpppppppppp

Discussion in 'Cisco' started by g-ca I, Mar 31, 2006.

1. ### g-ca IGuest

Can somebody PLEASE write the formula for calculating "minimum frame
size in Ethernrt"??? it's of crucial importance...

g-ca I, Mar 31, 2006

2. ### MervGuest

Merv, Mar 31, 2006

3. ### g-ca IGuest

Thank you so much.
However, the page discusses about maximum frame size, not minimum one
:-(
Can you help me about the minumum????

g-ca I, Apr 3, 2006
4. ### MervGuest

> write the formula for calculating "minimum frame size in Ethernrt"???

see

www.erg.abdn.ac.uk/users/gorry/course/lan-pages/enet-calc.html

Please look at the above article VERY CAREFULLY.

Notice the column heading thatsays "Minimum Size Frame"

In memory teh minimum size Ethernet frame will be 64 bytes ( 60 bytes
without CRC)

Minimum on the wire is as the article indicates 84 bytes.

Merv, Apr 3, 2006
5. ### g-ca IGuest

Yes, but the question I should answer asks for explanation why minumum
frame size is needed and it also indicates that there was a formula
from which the standard size of 64 bits was derived. And finaly it asks
for the formula. With that formula I sholud also prove in a way that it
is not posssible for smaller size... Somebody told me that he found the
formula in Internet. So, now Im sure that the question is not tricky
and that the formula exists. But I must found myself. And I can not :-(
:-( :-(

g-ca I, Apr 3, 2006
6. ### MervGuest

The reason a minimum frame size was/is to allow enough time for two
stations at maximum distance from each other on an Ethernet bus-segment
(read coax cable) so that the trasnmitting station can transmit a
minimum size frame and the "receiving" station can detect a collision
and have enough time to send a jam signal that the transmitting
stations will receive (i.e know that a collision occurred).

This is known as the Ethernet slot time = 2 x 232 bits + 48 bits = 512
bits / 8 = 64 bytes

Merv, Apr 3, 2006