From: prashant shukla (shukla_cisco@yahoo.co.in)
Date: Wed Jan 17 2007 - 10:54:41 ART
Hi Edison,
we guys still have some issues, let me explain the scenarios.
Case1# Port speed/AR= 512 Kbps, CIR= 128 Kbps. Excess Burst= "portspeed"
here finding the Bc with a default Tc=125 value is no issue.
Be= (512-128)xTc / 1000......this formula i wil use when the questions says "allow the VC to burst to port speed".
Case2# Port speed/AR= 512 Kbps, CIR= 128 Kbps. EIR= 300Kbps.
If the questions says " Allow the Vc to burst upto 300Kbps for 32ms".
Then in this case Tc=32ms. ( U mentioned 512Kbps is to confuse )
Be= (300-128)x32/1000 -------------->> (Correct me Incase)
Case3# Port speed/AR= 512 Kbps, CIR= 128 Kbps. EIR= 64Kbps.
NOW this is Where i get thinking, as i need to burst to an excess of 64K for say 100ms, so Tc=100, then in the Be formula what do i use.
I think it shud be like this: AR-CIR=64Kbps.
Be= 64x100/1000
My justification is , for an excess burst of 64K the total wil be 128+64=192Kbps
and when i put the same for my Be equation.
192-128=64.
Pls Correct/Clarify my issues.
I hope i aint bothering you much.
Edison Ortiz <edisonmortiz@gmail.com> wrote: > If the packet is larger than Bc, will this packet be fragmented to fit
> into Bc?
Yes.
> If not, how the Tc can affect the voice latency?
You need to avoid fragmentation on voice traffic. That's the reason you need
to lower Tc - send packets faster.
----- Original Message -----
From: "dampened"
To: ""Edison Ortiz""
Cc: "Cisco certification" ; ;
""prashant shukla""
Sent: Tuesday, January 16, 2007 5:40 AM
Subject: Re: IEWB--- Be calculations.
>
> Some confusion here. I am not sure what is the relation between
> fragmentation and Bc value.
>
> If the packet is larger than Bc, will this packet be fragmented to fit
> into Bc?
> If not, how the Tc can affect the voice latency?
>
> Please comment.
>
>
>
>
>
>
>
> "Edison Ortiz"
> Sent by: nobody@groupstudy.com
> 01/15/2007 12:14 AM
> Please respond to
> "Edison Ortiz"
>
> To
> "prashant shukla" ,
> cc
>
> Subject
> Re: IEWB--- Be calculations.
>
>
>
>
>
> #1
> 128k or 384k now becomes your AR and the Tc is given for the circuit on
> that
> case.
>
> Be = (128-64)x32/1000
> Be = (384-64)x64/1000
>
> Forget about the 512k AR, it's there to confuse you.
>
> #2
>
> Other signs to look for, if you have voice on the circuit. Voice latency
> needs the minimum
> Tc value which is 10ms. Then you need to fragmented, example:
>
> Bc = (CIR*Tc)/1000
> Bc = (512000*10)/1000 which is Bc 5120 (640 bytes) - passes fragmentation
> rule
>
> Whereas if you have
>
> Bc = (512000*32)/1000 which is Bc 16384 (2048 bytes) - fails fragmentation
> rule and this
> packet must be fragmented.
>
>
>
>
>
> ----- Original Message -----
> From: "prashant shukla"
> To:
> Sent: Sunday, January 14, 2007 2:31 AM
> Subject: IEWB--- Be calculations.
>
>
>> Gurus,
>>
>> This is what i understand and need clarification on.
>> AR= 512Kbps ; CIR = 64Kbps.
>>
>> Bc= CIR x Tc / 1000
>> Be= (AR-CIR) x Tc / 1000
>>
>> #1. So im able to the above calculation if the question mentions "
>> allow
>> DLCi 103 to burst upto port speed " then the Be formula works with
>> (512-64)kbps x TC/1000, my doubt comes when the question says, " Allow
>> the
>> DLCI to burst upto 128 /384 Kbps for 32ms or 64ms etc etc." here how to
>> look for the Be, as the burst is not upto port speed, so im struggling to
>> find a "Universal rule".
>>
>> #2.Secondly; in questions where its not mentioned any Tc value I would
>> like to
>> go for 125ms as default, what other "Signs" i should look for where the
>> Tc value can be influenced, e.g. In one IEWB it mentions " The packet
>> above 960 bytes should be fragmented". Is this a catch!! as the the Tc
>> chosen made .
>>
>> hope i was clear enough... :-)
>>
>> Shukla.
>>
>>
>>
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