IP allocation CIDR

Question

In an engineering college, the network administrator has IP $208.10.128.0/24$. He needs to divide this network into $4$ segments $(ECE = 60, CSE = 120, EEE = 30, MEC = 30)$. ECE block got $208.10.128.0/26$ as its address. Then what will be MEC block address?

Comments

its a  rule for CIDR blocks...first thing block size should be in powers of two because we have to divide them and second rule says first IP address should be divisible by block size...if we dont have block size in 2 powers how we will divide them by reserving bits?

​​​​​​​and u can get different answers here depending u have chosen 00 01 11 or 10 for subnet ID

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yea habib i think question is wrong

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@Debashish bro , plz consider the link :

http://superuser.com/questions/628227/network-addressing-without-wasting-ip-addresses

There is an example where wastage of address space is allowed since when we have no of addresses in a subnet say 60 , we have to choose /26 in CIDR notation since 64 is the next higher number which is a power of 2 due to subnet id which is chosen say 193.28.76.0/26 so range of addresses that will come under this will be from 193.28.76.0/26 to 193.28.76.63/26 so although the last 4 addresses are wasted but this will belong to this subnet only due to the subnet mask chosen.But we can not assign these addresses to other subnets.So other subnet address has to begin with 193.28.76.64/26 if next subnet also contains 60 addresses ,say and not 193.28.76.60/26

Answer 1

CSE must be alloted 208.10.128.1000 0000/25, since it has 120 hosts. We are now left with a block 208.10.128.0100 0000/26 because a block 208.10.128.0000 0000/26 is already alloted to ECE. Block 208.10.128.0100 0000/26 can further be divided into 2 blocks 208.10.128.0100 0000/27 and 208.10.128.0110 0000/27 to accomodate 30 hosts each, so anyone of them can be alloted to MEC.

Comments

@amitlko please confirm these three points,

• ECE block last address is a.b.c.00111111, and for CSE we can not continue from next valid ip 64 onwards because then we cannot accommodate 120 users in 7 bits (Or, 25 mask)
• And we can not start from zero using 7 bits (or 25 mask), because then, IP's will be overlapped between ECE and CSE.
• So we must start from a.b.c.10000000/25

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@Debashish Deka

Yes, we can't start with 0 for CSE. The reason is that only a.b.c.0100 0000/26 is left (a.b.c.0000 0000/26 is already alloted to ECE) and it can accomodate maximum of 62 hosts. We have to use a.b.c.1000 0000/25 for CSE in order to accomodate 120 hosts.

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I guess this is ok!

Answer 2

There are 2^(232 − 24) = 256 addresses in this block. The first address is 208.10.128.0/24; the last address is 14.24.74.255/24

[A] ECE :The number of addresses in the second subblock is not  a power of 2 either. We allocate 64 addresses.The subnet mask is 26. The first address in this block is   208.10.128.0/26; the last address is 208.10.128.59/26

[B] CSE :The number of addresses in the first subblock is not a power of 2. We allocate 128 addresses.The subnet  mask is 25.The       first address is 208.10.128.60/25; the last address is 208.10.128.179/25

[C] EEE : First address 208.10.128.180/27 the last address 208.10.128.209/27

[D] MEC :First address 208.10.128.210/27 the last address 208.10.128.239/27