Tman87 wrote: » Hey, So the book is right in saying the block size in third octet is 4. Because they all can be grouped together and summarized as 172.1.4.0/22 which would mean the third octet moves up in increments of 4. So first block would be 0 - 3 then the next block, which is the one you have listed is 4 to 7. Hope that helps.
however there is two network which is the 172.1.4.128/25 and 172.1.4.0/25 what about those one?
Nans wrote: » Let me put it like this for you to better understand. I cannot ask you to follow this, but will give you some edge until you master these. So as you see 172.1 remains constant and the rest keep changing. Since summarization help you achieve less advertisements you are going to do an address that fits everything under one hood. So in this case the third octet should catch your eye and as the previous post said yes the Block size is 4, since it keep all the addresses under one hood (4-7). Block size and interesting octet are good. Now lets get across the subnet mask. So you need a a block size of 4 which require 2 bits and that is in the third octet. So the last two bits will be borrowed and the rest all are 1's in the third octet (11111100=252). So the subnet mask is 255.255.252.0. Now the network address is 172.1.4.0 and the subnet mask is 255.255.252.0 (/22). Dude, that's the part of the address range. If you write 172.1.4.0/24 (172.1.4.0/25 + 172.1.4.128/25) both the groups will be a part of this one, instead the author wanted to give you more understanding and divided a single network into two by borowing a bit from hosts and and breaking it into two.Practice Subnetting which is the only way you can be perfect All the best.
