From: Jonny English (redkidneybeans@gmail.com)
Date: Sat Dec 20 2008 - 10:02:55 ARST
My name is not Anthony,
but here's an answer anyway...
you have bc = 2560 and this means Tc=10ms
Therefore we have 2560 bits per interval. This means the fragment size has
to be 2560/8 which is where there 320 bytes comes from.
On Sat, Dec 20, 2008 at 11:19 PM, David Mammadov <dato193@rambler.ru> wrote:
> Hello, Anthony.
>
> CIR = 256000
> bc = 2560
>
> Task: ensure that a single packet cannot take more than one interval to be
> transmitted.
> Solution: frame-relay fragment 320
>
> If I wll just set mtu to 320 or less on the interface, will it be valid
> solution?
>
>
> Best Regards,
> David
>
>
> Blogs and organic groups at http://www.ccie.net
>
> _______________________________________________________________________
> Subscription information may be found at:
> http://www.groupstudy.com/list/CCIELab.html
Blogs and organic groups at http://www.ccie.net
This archive was generated by hypermail 2.1.4 : Thu Jan 01 2009 - 12:53:09 ARST