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Re: Re: Urgent Help Needed...

  • From: "john smith" <johnsmith0302@hotmail.com>
  • Date: Tue, 18 Mar 2003 21:19:16 +0530
  • Cc: <mpls-ops@mplsrc.com>
  • Resent-Date: Tue, 18 Mar 2003 11:44:25 -0500
  • To: "Rik Wade" <rik@tcpip.fsnet.co.uk>
  • X-OriginalArrivalTime: 18 Mar 2003 15:46:40.0602 (UTC) FILETIME=[915C4BA0:01C2ED65]
  • X-Originating-IP: [203.124.140.38]


> On Tue, 18 Mar 2003, john smith wrote:
>
> > Throughput = average number of bits/time (if u dont mind using L2
overheads)
>
> I'd be wary of getting in to using an "average number of bits" in this
> calculation. By definition, an "average" is already calculated over a
> period of time. We are then dividing it by a further component of time.

hmmm u answered that "average" one urself below......

>
> I'd keep it simple by just saying:
>
> Throughput (bps) = total number of bits transferred / time in seconds
>
> As John says though, including large quiet periods in the selected
> time period may adversely affect the result you're after.

so average over smaller periods and then add sum_average/time

apologies....i missed the "sum" bit .......



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