Re: [LARTC] HTB, MPU, and suitable values

[Ed Wildgoose <lists@wildgooses.com>]

Jason Boxman wrote:

>On Monday 17 May 2004 17:23, Ed Wildgoose wrote:
><snip>
>  
>
>>Read the follows to that post as well.  Basically it's only an
>>approximation.  The "MPU" is basically pointing out that your ADSL
>>stream is encapsulated in an ATM stream.  ATM uses fixed size 64 byte
>>packets.  You need at least 2 of these, hence the 108 figure for MPU.
>>Now you also need to estimate overhead which is going to be the size of
>>the header on those ATM packets.
>>    
>>
>
>Now I'm confused.  Is it 53 bytes or 64 bytes?
>
>http://www.faqs.org/docs/Linux-HOWTO/ADSL-Bandwidth-Management-HOWTO.html
>  
>

You are right.  Something happened and I somehow failed to divide 106 by 
2 and get 53...  I have been doing a load of code using 2^n all day, and 
32/64, etc were really on my mind just then.  Sorry....

>>However, that still leaves the "wasted space" on the end of small
>>packets (eg those that take up 2.5 ATM cells, how much does the 0.5 take
>>up).
>>
>>I suggested a crude way to tweak that patch (easy to see how it works if
>>you look at the relevant lines in the orig file). However, I dont even
>>have a working QOS system so I haven't even compiled it!  Look up the
>>specs for ATM though and you should be able to tweak that suggested line
>>change and get something.
>>    
>>
>
>So the patch is supposed to increase the cost of dequeuing packets, then, 
>provided you know what numbers to use?
>  
>

Well, I haven't taken the time to trace that code, but with a 10 sec 
look at it, it appears to be simply accumulating the size of incoming 
packets based on the actual size of the data.  So I simply suggested 
dividing by 53, rounding up, then adding on the "overhead" on a per 
packet basis, rather than a per data block basis

Actually having looked at your ADSL HOWTO link, of course the best 
calculation would be to simply divide the amount of data by 48 (the data 
size of ATM packets).  Then round up (since 0.5 packets means needing 1 
whole packet).  Then multiply this number by 53 (size of atm packet 
including its header).  This would give the exact amount of bandwidth.

I would code this as:

size = (  (int)((datasize-1)/48) + 1) * 53

You could hardcode something similar into your tc and see if it helps 
(just remove PMU and overhead code added by the existing patch).

If you are scared of looking at code.  Don't be.  It really isn't as 
scary as it might look!

Good luck.  Interested to hear if it works...

Ed W
_______________________________________________
LARTC mailing list / LARTC@mailman.ds9a.nl
http://mailman.ds9a.nl/mailman/listinfo/lartc HOWTO: http://lartc.org/