Re: Re: setting the FR-de bit on exceeding bandwidth posted 11/23/2004
Actually, I'm NOT sure that's true, "if you do not specify Be, Be is 0".
What I've done is use the police command to only specify bps and then
checked what values IOS automatically computed for bc and be with the show
policy-map <name> int X command.
What I saw were non-zero values for both bc and be. And bc and be were the
same value. For example, the other day, I used this command, police 64000.
Then, in the output of show policy-map TEST int s0, it showed bc and be both
equal to 2000 - if I remember accurately which implies a Tc that's not equal
to 1/8 of a second. Go figure.
What I don't know is this. It could be that the value for be shown in the
output is actually bc + be in which case, be is actually 0.
Why, whoever wrote the code for this would require us, the users, to know
that sometimes be is actually bc + be is orders of magnitude above my pay
----- Original Message -----
Sent: Tuesday, November 23, 2004 4:46 AM
Subject: Re: Re: setting the FR-de bit on exceeding bandwidth
> So, hopefully, I helped clear up a little confusion instead of adding to
> That is indeed easy to get confusing. I have been fighing with these terms
for a time.
> Sally, do you undertood the values involved? I am asking because it is
easy to forget some points. Your question asks for very interesting thoughts
and after trying to simulate your exercise I went back to the books. Wendell
is a great book.
> The IOS calculates how many bytes you it can send to met the shape rate.
This is Bc. If you do not specify Bc, IOS will do it to reach the shape
rate. Now, if you do not specify Be, Be is 0, which means you are just
sending Bc. Well, if you are justing sending Bc, you are commited to the
contracted (considering your CIR=shape rate). You see? There is no reason to
set packets with DE because no packets will be discarded.
> On the other hand, if you specify Be, what ever goes above Bc is not
commited, and the IOS will set DE if you configure it to do so.
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