From: Swaroop Potdar (swarooppotdar@hotmail.com)
Date: Sat Oct 30 2004 - 11:26:47 GMT-3
ok but by the method u mentioned i would have 16 time intervals for the
transmission.
ie; as per the soln
bc * tc = CIR
64512 /62.5 = 1032.192
or even
64512 * 16 = 1032192
but the data to be transmitted per second is 1024000 bits
>From: "ccie2be" <ccie2be@nyc.rr.com>
>To: "Swaroop Potdar" <swarooppotdar@hotmail.com>, <ccielab@groupstudy.com>
>Subject: Re: IEWB Lab 16 - 8.1-4 QOS Doubt.
>Date: Sat, 30 Oct 2004 10:15:58 -0400
>
>-Tc is interval determined by Bc / CIR and defaults to 1/8 or .125 seconds
>for cir's at or below 650kbps and 1/16 (or 63ms) for cir's greater than
>650kbps. The smallest it can be is 10ms or 1/100th of a second. (See IE
>lab
>7, QOS tasks)
>
>
>
>
>
>----- Original Message -----
>From: "Swaroop Potdar" <swarooppotdar@hotmail.com>
>To: <ccielab@groupstudy.com>
>Sent: Saturday, October 30, 2004 8:50 AM
>Subject: IEWB Lab 16 - 8.1-4 QOS Doubt.
>
>
> > Hi Team,
> >
> > I had a query regarding the question 8. (section QOS ) in IEWB lab 16.
> >
> > I didnt understand the values given in the solution guide.
> >
> > R3 is the hub and R4 and R5 are the spokes R3 has a E1 capacity and the
> > requirement says that we should use the half of that for each circuit
> > between the spokes, with fram relay traffic shaping and we should burst
>upto
> > 75% of the access rate if there is accumulated credit using default TC
>fo
> > rthe CIR.
> >
> > so with that on the spokes..the configuration looks llike below on each
>of
> > the spokes:
> >
> > map-class frame-relay CIR
> > frame-relay cir 1024000
> > frame-relay bc 64512
> > frame-relay be 32256
> >
> >
> > Now the question is the default TC is 125 ms which gives a BC of 128000
> >
> > but the solution looks different: your all inputs would be valuable.
> >
> >
> > Thanks in advance,
> >
> > Swaroop.
> >
> > _________________________________________________________________
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>
Regards,
Swaroop.
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