11-13-2007 03:58 AM
Hi,
I have an interface with 2 rate limiting statement:
!
interface FastEthernetX/X/X
rate-limit output access-group 102 2000000 375000 750000 conform-action transmit exceed-action drop
rate-limit output access-group 101 10000000 1875000 3750000 conform-action transmit exceed-action drop
!
and 2 access lists
access-list 101 permit ip any any
access-list 102 permit ip any host X.X.X.Y
access-list 102 permit ip any host X.X.X.Z
My question is the following :
What is the total bandwidth available 10 or 12 Mb?
I suspect it is 10 as access list 102 is part of access list 101 but and traffic from 101 is limited to 10 M but I would like a confirmation.
Thanks
Solved! Go to Solution.
11-29-2007 08:00 AM
Ok, after trying to research the burst function of rate-limit on the web, I finally asked someone smart and found that my previous answer was grossly inaccurate and for that I apologize.
So here's the short answer - you will have the sum of the two rate-limit statements for total bandwidth. You could conceivably use the entire 2Mbps bandwidth + more (see below) sending from the two hosts specified by ACL102 PLUS 10Mbps bandwidth + more (see below) being sent from other sources permitted by ACL101.
And then here's the long answer. For the first rate-limit statement, you can send 375000 bytes in a 1.5 second timselot - You can determine the timeslot by taking ((Bc)*8)/avg-rate - in this case, (375000*8)/2000000. Where it gets complicated is calculating the effects of idle time. If the previous 1.5 second timeslot was completely idle, you can use all of those tokens in the current timeslot which would still give you an average of 2Mbps over 3 seconds. But you also have excess burst configured at twice the normal burst. Excess burst uses the previous two timeslots, so if both previous timeslots (the previous 3 seconds) were idle, you could send an additional 750000 bytes in the current timeslot. So, over the current 1.5 seconds (the first 3 seconds being idle), you could send up to 1,500,000 bytes(12Mb). Over the entire 4.5 seconds, you average 2.67Mbps. This means that those two hosts can send a maximum average output of 2.67Mbps.
Using the same calculations, the second rate-limit statement would afford you a max average output of 12.72Mbps.
This means that your maximum rate-limited interface output would be 15.39Mbps.
Does that help?
11-17-2007 05:33 PM
Could you provide a little more detail as to what your are trying to accomplish? If you are trying to limit certain traffic would you not be better served using a policy map?
11-27-2007 01:26 PM
My understanding is that you would have a total maximum bandwidth of 24Mb/s counting the available excess burst. You could conceivably have 4Mb/s worth of bursted traffic from the defined hosts in ACL 102 in addition to 20Mb/s worth of bursted traffic from other sources from ACL 101.
Someone please correct me if I'm wrong.
11-29-2007 08:00 AM
Ok, after trying to research the burst function of rate-limit on the web, I finally asked someone smart and found that my previous answer was grossly inaccurate and for that I apologize.
So here's the short answer - you will have the sum of the two rate-limit statements for total bandwidth. You could conceivably use the entire 2Mbps bandwidth + more (see below) sending from the two hosts specified by ACL102 PLUS 10Mbps bandwidth + more (see below) being sent from other sources permitted by ACL101.
And then here's the long answer. For the first rate-limit statement, you can send 375000 bytes in a 1.5 second timselot - You can determine the timeslot by taking ((Bc)*8)/avg-rate - in this case, (375000*8)/2000000. Where it gets complicated is calculating the effects of idle time. If the previous 1.5 second timeslot was completely idle, you can use all of those tokens in the current timeslot which would still give you an average of 2Mbps over 3 seconds. But you also have excess burst configured at twice the normal burst. Excess burst uses the previous two timeslots, so if both previous timeslots (the previous 3 seconds) were idle, you could send an additional 750000 bytes in the current timeslot. So, over the current 1.5 seconds (the first 3 seconds being idle), you could send up to 1,500,000 bytes(12Mb). Over the entire 4.5 seconds, you average 2.67Mbps. This means that those two hosts can send a maximum average output of 2.67Mbps.
Using the same calculations, the second rate-limit statement would afford you a max average output of 12.72Mbps.
This means that your maximum rate-limited interface output would be 15.39Mbps.
Does that help?
11-29-2007 09:20 AM
Thanks for the response. it helps.
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