From: Jonny English (redkidneybeans@gmail.com)
Date: Sat Dec 20 2008 - 10:07:20 ARST
Also note your fragment is in bytes which is why we divide by 8
On Sun, Dec 21, 2008 at 1:02 AM, Jonny English <redkidneybeans@gmail.com>wrote:
> 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
>>
>>
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