So 64 byte packets account for 31.3% of traffic

96 byte 25.4%

128 byte 6.8%

160 byte 15.9%

And so on 

If you add those percentages up you get about 99.7%.  So the other .003
percent are mixed in with the rest.

 

 

Regards,

 

Tyson Scott - CCIE #13513 R&S, Security, and SP

Managing Partner / Sr. Instructor - IPexpert, Inc.

Mailto: [email protected]

Telephone: +1.810.326.1444, ext. 208

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eFax: +1.810.454.0130

 

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From: [email protected]
[mailto:[email protected]] On Behalf Of Kingsley
Charles
Sent: Monday, July 12, 2010 7:44 AM
To: [email protected]
Subject: [OSL | CCIE_Security] sh ip cache verbose flow

 

Hi all

I am trying to interpret the following:

router#sh ip cache verbose flow
IP packet size distribution (95662 total packets):
   1-32   64   96  128  160  192  224  256  288  320  352  384  416  448
480
   .000 .313 .254 .068 .159 .079 .063 .061 .000 .000 .000 .000 .000 .000
.000

    512  544  576 1024 1536 2048 2560 3072 3584 4096 4608
   .000 .000 .000 .000 .000 .000 .000 .000 .000 .000 .000


The packet size distribution is given which in total should be 95662
packets. I am not getting a doc that explains how to interpret the above
given O/P.

For 64 it is .313 and 96 is .254. Are .313 and .254 the percentage of 64 and
96 sized packet?

Can someone, please let me know logic and if possible share me reference
link for it.



With regards
Kings

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