Help with VLSM and Route summariztion with and without contigious networks.

Tommy2727

Greeting all, I'm back and I'm very upset once again. For some reason I can not grasp the concept and fundamentals of route summarization & VLSM. I'm pretty comfortable with the basic fundamentals but implementing and finding a discontigious summary route and when to use VLSM to find specific networks / hosts has me stumped.. Can some one please dummy down the formula, process or post some resources why and how? these q & a's came from Boson but I can not see how and why . Please..... any assistance is greatly appreciated.

Cheers!

1. What is the most efficient summarization that can be used to advertise for the following networks?

172.1.4.0/25

172.1.4.128/25

172.1.5.0/24

172.1.6.0/24

172.1.7.0/24

Answer 172.1.4.0/22

Can someone please dummy down this?

2. A summarized route for a router is 192.1.144.0/20 which of the following packet destinations addresses will be forwarded?

answer?
192.1.151.254
192.1.59.2


Can someone tell me why please? my formula is flailed

BrizoH

Bit of a rushed reply below, apologies if anything isn't too clear.

1. What is the most efficient summarization that can be used to advertise for the following networks?

172.1.4.0/25

172.1.4.128/25

172.1.5.0/24

172.1.6.0/24

172.1.7.0/24

Answer 172.1.4.0/22

Can someone please dummy down this?

A /22 subnet mask is 255.255.252.0 - giving you an increment of 4 in the 3rd octet, which means valid networks would include:

172.1.0.0 - 172.1.3.255
172.1.4.0 - 172.1.7.255*
172.1.8.0 - 172.1.11.255
etc

The second network includes all of your more specific subnets in need of summarisation so it's a perfect summary route.

2. A summarized route for a router is 192.1.144.0/20 which of the following packet destinations addresses will be forwarded?

answer?
192.1.151.254
192.1.59.2

Can someone tell me why please? my formula is flailed

Again, once you have your increment it's easier.

A /20 subnet is 255.255.240.0 which gives you an increment of 16 in the 3rd octet, so your valid subnets would be:

192.1.0.0 - 192.1.15.255
192.1.16.0 - 192.1.31.255
192.1.32.0 - 192.1.47.255
..
..
192.1.128.0 - 192.1.143.255
192.1.144.0 - 192.1.159.255 *
192.1.160.0 - 192.1.175.255

Of the two possible answers only 192.1.151.254 belongs to the given subnet.

I'm a fan of Jeremy Ciora's subnetting method - it's all about getting your increment.

https://learningnetwork.cisco.com/servlet/JiveServlet/previewBody/6014-102-1-19236/Subnetting%20Examples.pdf

Once I had this nailed (with lot's of practise on http://www.subnettingquestions.com) it really helped in the exam.