OK, here is my response, to Marko's mini-task....

  Already specified groups:
>
> 226.0.0.0/7
> 229.0.0.0/8
>  I was thinking that for these (two examples) it would be on R4:
> 229.0.x.x/ wild card 0.0.255.255  and 226.0.0.0 0.0.255.255
> Here I am interpreting as "second half of octets of already specified
> addresses".
>
> ========
> alternate logic
> 224-239 (first MCAST Octet) = 16 first octet options and half of that is 8
>
> first half: 224,225,226,227,228,229,230,231
>
> second half: 232,233,234,235,236,237,238,239
> group-list (acl): 232.0.0.0 7.255.255.255
>
> this is the interpretation of the solution.


> :-)
>
> I guess if I wrote it out like that the proctor would be happy to clear
> things up.  It shows I know what I am talking about just confused about
> syntax.
>
> thanks for the help Marko (with a K).  ;-)
>
_______________________________________________
For more information regarding industry leading CCIE Lab training, please visit 
www.ipexpert.com

Reply via email to