From: zhuqingliu (zql@xxxxxxxxxx)
Date: Fri Feb 16 2001 - 00:10:32 GMT-3
1Mbps=1024Kbps=1024*1024bps
If limit the broadcast to 3Mbps, the packet size is 1024B,
1024*1024*3/(8*1024)=384
set port broadcast 2/9 384
/Perry
----- Original Message -----
From: Hardin Les - SMTP <hardinl@bah.com>
To: Derek Buelna <dameon@aracnet.com>; <ccielab@groupstudy.com>
Sent: Friday, February 16, 2001 12:00 AM
Subject: Re: Bootcamp lab 18: Broadcast Supression
>
> FYI.
>
> Actually, 3Mbps = 3,072,000 bits per sec.
>
> Les
>
> At 02:37 AM 2/15/2001 -0800, Derek Buelna wrote:
> >Hi,
> >
> >I was thinking that if I was asked to configure this I'd probably do it
> >one way but they would have wanted it the other way.
> >
> >In lab 18 I believe it says to limit broadcasts on a port, on the
> >catalyst, to 3Mb/s.
> >
> >Ok, so I read that the percentage of bandwidth method is more accurate
> >and also hardware based but the answer uses the per packet method.
> >
> >I'm taking it that 3,000,000 bits/s / 8 = 375,000 bytes/s. I'm
> >assuming that an average packet size of 1024 would make for an average
> >throughput of 366.21 of these kinds of packets/s.
> >
> >Am I understanding this correctly?
> >
> >Wouldn't it be easier to just say 30%, since it's a 10,000,000 bits/s
> >link?
> >
> >set port broadcast 2/9 367
> >set port broadcast 2/9 30%
> >
> >Thanks,
> >
> >-Derek
> >
> >
> >
> >
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