Re: Simple DSCP question

From: Saul Arjona (saul.arjona.bueno@gmail.com)
Date: Tue Oct 02 2007 - 04:20:26 ART


Here is the link where you can find this information on the univercd

http://www.cisco.com/univercd/cc/td/doc/product/software/ios122/122cgcr/fqos_c/fqcprt7/qcfdfsrv.htm

However if you find the logic on how to convert any value to binary you can
get the DSCP without any reference:

think the 6 digit value as to different three digit parts , and also the AF
as two parts xy ,for example AF11 (think as it not as AF eleven but as AF
one one)

For the first part (first three digits)

replace the first three digits normally from decimal to Binary so from
left to right the first digit would be a decimal 4 the second digit would
be 2 and the third digit would be 1

so for AF1x it would be 001, for AF2x 010, for AF3x 011, for AF4x 100

Now for the second part (the last three digits)

From left to right the third digit is going to be always 0, and the first
and second digit replace them normally from decimal to binary (the first
digit is going to be equivalent to decimal 2 and the second digit is going
to be equivalent to decimal 1)

So for AFx1 010 , AFx2 100 AFx3 110

then just couple the two parts and voila! you have the DSCP

Now, for the CS follow the same logic but the last three digits will be
always 0

So for the first part (first three digits)

replace the first three digits normally from decimal to Binary so from
left to right the first digit would be a decimal 4 the second digit would
be 2 and the third digit would be 1

So for CS1 001, CS2 010, CS3 011, CS4 100, CS5 101, CS6 110, CS7 111

then just add the last three 0 and you have the DSCP.

Hope this makes sense.

Regards
-Saul



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