# 6.1 ZBFW - why burst 8000?

Hi, anyone can explain why the burst value is 8000 (in bytes)?

Thanks

• I would like to know this as well. We are presently doing this task, shouldn't the value be 128000 (125/1000) = 16000

• I agree, policy rate value is entered as bps not kbps.

• I am having the same problem - should it not be 128000 x 0.125 /8 = 2000 btyes ?

IS the 16000 you refer to in bits or bytes ?

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Hi, anyone can explain why the burst value is 8000 (in bytes)?

Thanks

This is an old thread, but I am hoping someone can give the answer this question.

I am getting better at the calculations for  CIR  Be Bc ,  however I am stumped on this one.  I have been trying every formula for Be, but I am not understanding how BURST 8000 is created.

class CMAP_OTHER_PROTOCOLS

inspect

police rate 128000 burst 8000

Scenario says:

Limit the aggregate rate of DNS and ICMP packets inbound to 128Kbps

My Attempt for burst:

128,000 * 125/1000/8 = 2000

edit:   this is wb-II Lab 4 (dynamips and non-dynamips)

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Worbook Feedback Ticket# YAF-495286

• Have you labbed this up and tested 2000 vs 8000?

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I have labbed it and played with the calculations.  The lowest number allowed for BURST was 1000 as I recall.

I will experiment more right now.

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I just redid the config with 1000, 2000, and 8000, and the output changes as expected.

The Solution Guide's only verification is  show policy-map type inspect zone-pair

with police rate 128000 burst 2000  <-- 128000 * 125/1000 /8

Zone-pair: ZP_OUTSIDE_TO_INSIDE
Police
rate 128000 bps,2000 limit
conformed 0 packets, 0 bytes; actions: transmit
exceeded 0 packets, 0 bytes; actions: drop
conformed 0 bps, exceed 0 bps

with police rate 128000 burst 8000 <--- SolutionGuide

Zone-pair: ZP_OUTSIDE_TO_INSIDE
Police
rate 128000 bps,8000 limit
conformed 0 packets, 0 bytes; actions: transmit
exceeded 0 packets, 0 bytes; actions: drop
conformed 0 bps, exceed 0 bps

EDIT:   Version 12.4(15)T7

• The latest version which is web based for labs 1 to 10 has a note above "class CMAP_OTHER_PROTOCOLS"stating that the "burst value is randomly selected".  If the task doesn't specify a value to enter but the configuration requires a value then it doesn't really matter.  Technically I would select a value of 16000 or below.

This new web version will be released next week along with an updated PDF version for labs 1 to 10 in Vol 2.

--
Brian Dennis, CCIEx5 #2210 (R&S/ISP-Dial/Security/SP/Voice)

INE, Inc.

From: JoeM <[email protected]>
Date: Friday, August 2, 2013 7:22 PM
To: Brian Dennis <[email protected]>
Subject: Re: [iewb-rs-vol2-v5-lab4] 6.1 ZBFW - why burst 8000?

I have just redone the config with 1000 and 2000 (and 8000), and the output changes as expected.

In the Solution Guide, the only verification for this is  show policy-map type inspect zone-pair

with police rate 128000 burst 2000  <-- 128000 * 125/1000 /8

Zone-pair: ZP_OUTSIDE_TO_INSIDE
Police
rate 128000 bps,2000 limit
conformed 0 packets, 0 bytes; actions: transmit
exceeded 0 packets, 0 bytes; actions: drop
conformed 0 bps, exceed 0 bps

with police rate 128000 burst 8000 <--- SolutionGuide

Zone-pair: ZP_OUTSIDE_TO_INSIDE
Police
rate 128000 bps,8000 limit
conformed 0 packets, 0 bytes; actions: transmit
exceeded 0 packets, 0 bytes; actions: drop
conformed 0 bps, exceed 0 bps

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The latest version which is web based for labs 1 to 10 has a note above "class CMAP_OTHER_PROTOCOLS"stating that the "burst value is randomly selected".  If the task doesn't specify a value to enter but the configuration requires a value then it doesn't really matter.  Technically I would select a value of 16000 or below.

This new web version will be released next week along with an updated PDF version for labs 1 to 10 in Vol 2.

Brian, thanks for taking the time to respond to this task.   I admit I have been a little confused about the equations and when to use them.  It is sometimes nice to see the numbers in the scenario, so that they can be plugged in -- if my equation is correct.   Thanks again.

• The key point is that the burst value does not matter if the lab does not specify a value should be entered, but the command requires them.  They are going to be looking for 128k and not what burst value entered when grading.  Additionally it is good if people start to think about the scenario and technology and come to this conclusion on their own that the value is irrelevant.