Hi Khalid,
You don't care about the 1st, 2nd and 4th octet, because that will be 0
as it doesn't change.
You will have the 3rd octet being
1
3
5
7
9
If you write that in binary you will get
1 00000001
3 00000011
5 00000101
7 00000111
9 00001001
So in the wildcard mask we know that a fixed bit will be marked as a 0
and a wildcard bit will be marked as a 1.
1 00000001
3 00000011
5 00000101
7 00000111
9 00001001
00001110 <-- Bit 5,6,7 (from left to right) is wildcard bits. And
converting 00001110 to decimal will give you 14.
The wildcard mask being 0.0.14.0.
Hint: Use Windows notepad and calculator and write all IP address' in
binary. That will give you a better visual overview.
/Ralph Olsen
Hi all,
we have 2 routers R1.......> R2 and R1 has 10 Lo interfaces:
Lo1: 192.1.1.1/24 <http://192.1.1.1/24>
Lo2: 192.1.2.1/24 <http://192.1.2.1/24>
Lo3L 192.1.3.1/24 <http://192.1.3.1/24> upto 192.1.10.1/24
<http://192.1.10.1/24>
we need to configure R2 to ONLY receive the odd addresses from R1
the answer was:
R2:
access-list 1 permit 192.1.1.0 <http://192.1.1.0> 0.0.14.0
<http://0.0.14.0>
router rip
distance 255 150.50.17.1 <http://150.50.17.1> 255.255.255.255
<http://255.255.255.255> 1
Note: 150.50.17.1 <http://150.50.17.1> is R1
this Q is from the workbook. Can you give me a full explanation
about 0.0.14.0 <http://0.0.14.0>
thnx
