From: Joseph Brunner (joe@affirmedsystems.com)
Date: Tue Sep 18 2007 - 13:32:46 ART
For policing its
Bits |bytes for bc (which sets tc)|bytes for be
For shaping its
Bits |BITS for bc (which sets tc|BITS for be
So if the task stated
2Mbps cir with the MINIMUM tc
For policing its
Police 2000000 2500
For shaping its
Shave average 2000000 20000
Check this policing config of the same task
!
!
class-map match-all http
match protocol http
!
!
policy-map httppolicy
class http
police 2000000 2500
class class-default
fair-queue
interface GigabitEthernet0/0
service-policy output httppolicy
!
!
Verification;
rack1r6#sh policy-map int g0/0
GigabitEthernet0/0
Service-policy output: httppolicy
Class-map: http (match-all)
0 packets, 0 bytes
5 minute offered rate 0 bps, drop rate 0 bps
Match: protocol http
police:
cir 2000000 bps, bc 2500 bytes
conformed 0 packets, 0 bytes; actions:
transmit
exceeded 0 packets, 0 bytes; actions:
drop
conformed 0 bps, exceed 0 bps
Class-map: class-default (match-any)
22 packets, 2254 bytes
5 minute offered rate 0 bps, drop rate 0 bps
Match: any
Queueing
Flow Based Fair Queueing
Maximum Number of Hashed Queues 256
(total queued/total drops/no-buffer drops) 0/0/0
Joe
-----Original Message-----
From: nobody@groupstudy.com [mailto:nobody@groupstudy.com] On Behalf Of
Mohmmad, Imran
Sent: Tuesday, September 18, 2007 11:23 AM
To: ccielab@groupstudy.com
Subject: Traffic policing
Hi Experts,
Plz help me I am struggling from yesterday to identify the formula for
calculating the Bc and Be, while policing the traffic, I have also
checked the Cisco and from earlier response of my question I got the
following formula to calculate the Bc and I am not able to digest this
equation, can anyone Plz explain me how to interpret the same
Normal Burst = CIR [BPS] *(1 byte)/ (8 bits) *1.5 seconds
-Imran
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