From: Chris McGuire (cmcguire@firstdigital.com)
Date: Thu Apr 10 2008 - 12:50:17 ART
Possibly, however the Traffic Shaping should be able to buffer any excess
and this is only for one interval if excess bandwidth can be used. If this
is a lab question, then the answer would most likely be the 1536-512K
answer, just from my experience. The general rule of thumb is to not over
think a question. You are definitely on the right track...
Thanks,
Chris
On 4/10/08 9:47 AM, "John" <jgarrison1@austin.rr.com> wrote:
> Why? Don't you run the risk of oversubscribing the circut during some
> intervals?
>
> I should have asked for the logic behind the answer given with responses my
> bad
> ----- Original Message -----
> From: "Chris McGuire" <cmcguire@firstdigital.com>
> To: "John" <jgarrison1@austin.rr.com>; <ccielab@groupstudy.com>
> Sent: Thursday, April 10, 2008 9:22 AM
> Subject: Re: Frame traffic shaping
>
>
>> I would say you would choose 1536 - 512K for the BE on DLCI 2.
>>
>> Thanks
>> Chris
>>
>>
>> On 4/10/08 9:20 AM, "John" <jgarrison1@austin.rr.com> wrote:
>>
>>> Lets say you have an interface with two pvc's, and a port speed of 1536k.
>>> You
>>> want DLCI 1 to have a cir of 128k and no BE and a TC of 50 ms. DLCI 2
>>> has a
>>> cir of 512k with the BE equal to the remaining bandwidth of the
>>> interface(TC
>>> is the same as DLCI 1). Does the BE for DLCI 2 equal 1536k-640k(the sum
>>> of
>>> the cir's for DLCI 1 and 2)/20 or does it equal 1536k-512k (the speed of
>>> DLCI
>>> 2)/20
>>>
>>>
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>>
>>
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