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Greetings all,

Q9.1 in IEWB VolII lab20 is asking on the minimum entries for the below IP address:

51.3.0.1

51.5.0.1

51.7.0.1

51.3.0.9

51.5.0.9

51.7.0.9

and the answer was

access-list 1 deny 51.3.0.1 0.0.0.8

access-list 1 deny 51.5.0.1 0.2.0.8

how we reahced such entries?

kindly assist

Hi  khalidjorf

Check out http://blog.internetworkexpert.com/2007/12/26/q-how-do-i-compute-complex-wildcard-masks-for-access-lists/ for an explaination of wildcard masks in access lists...

Cheers

Dan

ok, lets group a little bit:

51.3.0.1
51.3.0.9


51.5.0.1
51.5.0.9

51.7.0.1
51.7.0.9


and start some binary:

51.3.0.1 -> 1 -> last octet: 0000 0001
51.3.0.9 -> 9 -> last octet: 0000 1001
---------------------------------------
                             0000 1000 = decimal 8
(0 = care, 1 = dont care bits)

reference is: 51.3.0.1 and mask 0.0.0.8


next one:
51.5.0.1
51.5.0.9

51.7.0.1
51.7.0.9
we still have 1/9 in the last octet, same mask as above.

2nd octet:

51.5.0.0 -> 5 -> 0000 0101
51.7.0.0 -> 7 -> 0000 0111
------------------------------
                 0000 0010 = decimal 2

lets put it together:
51.5.0.1 and mask 0.2.0.8


everything clear?
just keep in mind to create the mask:
0 means that the bits are fixed (dont change) and 1 is dynamic (can change).


greetings

-steven

Appriciate your assistance Steve.

regards

khalidjorf

Hey,

do check these links, they surly will help.

http://blog.internetworkexpert.com/2008/09/15/binary-math-part-i/

http://blog.internetworkexpert.com/2008/11/03/binary-math-part-ii/

Scott has really explained it will... I read it once and the idea is in my head now...

however this will take some time and a lot of practice......