Hi all ,
By starting with a given IP address and prefix (subnet mask ) 172.16.0.0 255.255.252.0 assigned by the net administrator ,you can begin creating your network documentation
This is calculating and assigning address without VLSM .
He needs 4 subnetworks with 481,69,23,2 hosts so 9 host bits at a minimum to cover the subnet with 481 hosts and therefore you run out of addressing as you will only end up with 2 subnetworks each with 510 as calculated as follows .
The way I would interpret this is as follows :
172.16.0.0/22 gives 2 subnets 172.16.0.0 /23 and 172.16.2.0/23
172.16.0.0 /22
11111111.11111111.1111110.00000000
the first 22 bits Network section
the 23rd bit the bit borrowed for subnetting the subnet
the last 9 bits host bits which gives 2^9 –2 hosts
ie 510 hosts per each of two subnets
Therefore , 172.16.0.0/22 gives the following 2 subnets
Subnet 0 172.16.0.0/23
Subnet 1 172.16.2.0/23
You are borrowing 1 bit from the host bits as opposed to 7 bits which the following book interpretation does .
The Book interprets it as follows
In the book he starts with 172.16.0.0 /23
255.255.254.0
11111111.11111111.11111110.00000000
the first 16 bits are the Network section
the next 7 bits are the bits borrowed for subnetting ie 2^7 subnets =128 subnets
The last 9 bits are host bits which gives 2^9 –2 hosts ie 510 hosts per each of 128 subnets
172.16.0.0/22
gives
subnet 0 172.16.0.0/23
subnet 1 172.16.2.0/23
subnet 2 172.16.4.0/23
.
.
.
.
subnet128 172.16.254.0/23
He seems to “chop up “ the main address 172.16.0.0/16 and create 172.16.0.0/23 subnetworks each with 510 usable hosts (2^9-2) .
He borrows 7 bits from the third octet 2^7 =128
This will give enough subnets with enough hosts per subnet even though it is wasteful of addresses .VLSM would be more efficient.
Can you advise me on the interpretation of the above ?
Thanks
Stephen
