I'm going to chip in both with some simplifying rules and some
additional information. I suggest you also read RFCs 2679 and 2544.
I will have the first white paper of a performance series available
free on Certzone on May 1.
>now if he recited that from memory, we should all just throw in the
>towel.... hehe :) I certainly suck at math anyway...reciting some crazy
>formula to calculate latency would send me over the edge!
>
>>>> "Priscilla Oppenheimer" 04/11/02 02:44PM >>>
>Great answer!
>
>Priscilla
>
>At 02:01 PM 4/11/02, Kent Hundley wrote:
>>There are several factors:
>>
>>1) Clock rate of the line
>>2) Buffering delay by any intermediary devices such as ATM/FR switches
Both actual forwarding by the intermediary device plus internal
buffering delays.
> >3) Speed of light
>>
>>If we take a simple case and say that there are no layer 2 devices in the
>>path and only digital cross-connects. I have read (somewhere) that the
>>speed of electron transference in copper is a little faster than the speed
>>of light in fiber over short distance, so use the speed of light in fiber
>>(roughly .7 X 186,000 miles per second) as the baseline. (note that the
>>reference given by another poster says the speed of electromagnetic signals
>>in copper is .66 of the speed of light, which would mean it is slightly
>>slower than speed of light in fiber, either way its pretty close to a wash)
>>Given these assumptions you get:
It actually varies among copper cable types. Thick Ethernet is about
.66c, while thinwire is about .5 c.
A practical approximation for WANs is six microseconds per kilometer
of airline distance. In cities, multiply that distance by 3.
> >
>>speed of a single bit = speed of line insertion for 1 bit + speed of light
>>delay + speed of line removal for 1 bit
>>
>>speed of line insertion for 1 bit = speed of line removal for 1 bit =
>>1/clock rate
>>
>>speed of light delay = number of miles / (.7 * 186000 miles per second)
>>
>>
>>
>>As an example, for a clock rate of 128Kbps and a distance of 1000 miles:
>>
>>speed of line insertion and removal for 1 bit = 2 * (1/128000) = .000015625
>>sec = .015625 ms
>>
>>speed of light delay = 1000 / (.7 * 186000) = .00768 sec = 7.68 ms
>>
>>7.68 ms + .015625 ms = 7.7 ms (roughly)
>>
>>Again, this assumes no delay in buffering in the path of any kind. It also
>>assumes that there is no congestion at either end of the link. Bottom
line,
> >keep in mind these are rough numbers, but I think you get the idea.
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