Supernetting question

that IP is valid as it would fall into 10.22.128.0 /22 which encompasses 10.22.128.1 to 10.22.131.254

Rack1SW3(config)#int lo2
Rack1SW3(config-if)#ip address 10.22.128.1 255.255.252.0
Rack1SW3(config-if)#int lo3
Rack1SW3(config-if)#
*Mar 1 01:04:32.848: %LINK-3-UPDOWN: Interface Loopback3, changed state to up
*Mar 1 01:04:33.850: %LINEPROTO-5-UPDOWN: Line protocol on Interface Loopback3,
changed state to up
Rack1SW3(config-if)#ip address 10.22.131.254 255.255.252.0
% 10.22.128.0 overlaps with Loopback2
Rack1SW3(config-if)#

Since we're on the topic of subnetting, is it possible to know the subnet of an IP addr by just looking at a IP?

ex: 10.10.10.14
Tell me what subnet this belongs to.

With no SM mask given is it possible? All of the subnetting I've done gives you the SM to work with.
Thanks

You obviously know the defaults for Class A, B, and C, so you could guess one of those, but you wouldn't be able to tell without the subnet mask.

nope all you can do is assume it is the classful mask then it wont be a subnet but a network

^
Thats what I thought. Even knowing the default SM for class A doesnt really help since obviously the other octects have been messed with.

Quoting from Todd Lammle's ICDN1 text..

Based on the output of the next routing table, which interface will a packet with a destination address of 10.10.10.14 be forwared to?

C 10.10.10.16/28
C 10.10.10.8/29
C 10.10.10.4/30
C 10.10.10.0/30

First, you can see that the network is subnetted and each interface has a different subnetm ask. And I have to tell you -- you cant just answer this question if you cant subnet! 10.10.10.14 would be a host in the 10.10.10.8/29 subnet....if you dont get it...Just go back and reread Chapter 3..." pg 342,243

Huh??? So does he expect you to find the subnet 10.10.10.14 belongs to based on using the subnet masks given on the routing output (basically trying 28,28, and 30 to see which fits?). I dont really see his logic there by him just giving you an IP and telling you to re-read the entire chapter if you dont get it.

Yea, he does. I did it just by glancing at it. .16 starts too high, and the /30s don't give you enough bits to reach .14, so by process of elimination, it's /29. Also, you know /29 gives you three bits, which gives you a value of eight, so right away you know that range is .8-.15.

WillTech105 wrote: »

^
Thats what I thought. Even knowing the default SM for class A doesnt really help since obviously the other octects have been messed with.

Quoting from Todd Lammle's ICDN1 text..

Based on the output of the next routing table, which interface will a packet with a destination address of 10.10.10.14 be forwared to?

C 10.10.10.16/28
C 10.10.10.8/29
C 10.10.10.4/30
C 10.10.10.0/30

First, you can see that the network is subnetted and each interface has a different subnetm ask. And I have to tell you -- you cant just answer this question if you cant subnet! 10.10.10.14 would be a host in the 10.10.10.8/29 subnet....if you dont get it...Just go back and reread Chapter 3..." pg 342,243

Huh??? So does he expect you to find the subnet 10.10.10.14 belongs to based on using the subnet masks given on the routing output (basically trying 28,28, and 30 to see which fits?). I dont really see his logic there by him just giving you an IP and telling you to re-read the entire chapter if you dont get it.

routers look for the most specific match in the routing table to pick the outgoing interface

I wrote the entire thing out so I was able to figure out it was 10.10.10.8 after getting the SM of 29.

00001010.00001010.00001010.00001110
11111111.11111111.11111111.11111000

10 . 10 . 10 . 8 Subnet

I'm sure theres the quicker method to get this rather than write it all out (I just wrote out the last octet since I know all 1s will give me the same either way). I'll take a look at

http://www.techexams.net/forums/ccna-ccent/38772-subnetting-made-easy.html

See if that'll help me speed up my subnetting instead of writing it all out.

Thanks again guys!

what if you can't use process of elimination and it matches 4 entries in the routing table

WillTech105 wrote: »

I'm sure theres the quicker method to get this rather than write it all out

Yea there is. Since you know you have three bits to work with, you know the networks are going to increment by 8 each time.

Therefore, you get
10.10.10.0-7
10.10.10.8-15
10.10.10.16-23
etc.

kryolla wrote: »

what if you can't use process of elimination and it matches 4 entries in the routing table

Was that in response to me? I had my post in first!

I agree with you; I was just telling him how to work out that one

no it wasnt directed at you. I wasnt sure if he understood longest prefix routing but he might not be at that point in his study

"longest prefix routing" sorry doesnt ring a bell (still studying for ICDN1).

Thanks again for your the help!!