10.0.0.0

10.0.1.0

10.0.2.0

10.0.3.0

what is the most appropriate summarization for these routes?

a) 10.0.0.0/21

b) 10.0.0.0/22

c) 10.0.0.0/23

d) 10.0.0.0/24

well i think i got it. the answer is (b).

The faster way with no Binary Math is: This will blow your mind how easy it is.

1) Find the Block size to cover all of the networks in the 3rd octet= Networks 0,1,2,3 answer is 4 block size.

2) Now count on your fingers backwards: 128,64,32,16,8,4, Should have 6 fingers sticking up.

3) How many common bits in the first 2 octets, 8+8= 16, Now add your like bits from your finger counting 16+6=22

4) Answer is 10.0.0.0/22

5) Again, no binary math and it all works from your 128,64,32,16,8,4,2,1 chart

For example: Summarize these networks?

172.16.16.0

172.16.17.0

................

................

172.16.25.0

1) What block size covers the all of the 3rd octets? Answer is 16

2) Count backwards on your fingers: 128,64,32,16

3) 4 fingers are up, added to the first 16bits. 8+8+4=20

4) Block size started at 16 so the answer is 172.16.16.0/20

@ Hondabuff

Could you use your method on

10.1.2.0

10.1.7.0

10.1.9.0

Do you have to count block from subnet boundary?

opps..

block that covers meaning > than not = to.

thanks.

Could you use your method on

10.1.2.0

10.1.7.0

10.1.9.0

10.1.{0-8 not enough because of the 9, 0-16 will cover all of 0-2-9-16} .0 All routes from 1-15 will be included because of the overage.

Technically it would be considered a poorly designed network.

0-8, 0-16, 0-32, looks like all we need is 16 places to cover them all. We will be wasting places because the 10.1.9.0 is in the next block of 8.

So a summarized network of 10.1.0.0/20. A block size of 16 is a 240 mask.

The 10.1 make up 16 bits allready and 240 mask uses 4 bits. 16+4=/20