An organization is granted the block 130.56.0.0/16. The administrator wants to create 1024 subnets. 1. Find the subnet mask. 2. Find the number of addresses in each subnet. 3. Find the first and the last address in the first subnet. 4. Find the first and the last address in the last subnet (subnet 1024).
Answers
Answer:
Explanation:
An organization is granted the block 130.56.0.0/16. The administrator wants to create 1024 subnets.
1) subnet address -> 182.44.82.0. Subnet mask will be 255.255.255.192
2) the number of addresses in each subnet - 62 valid addresses can exist in each subnet.
3) The first and the last address in the first subnet - 130.56.0.1 & 130.56.0.62 respectively.
4) The first and the last address in the last subnet - 130.56.255.193 & 130.56.255.254.
Answer:
1) 255.255.255.192 will be the subnet mask ,
2) 62 valid addresses can exist in each subnet,
3) First address in subnet 1 will be 130.56.0.1 and last address is sibnet 1 will be 130.56.0.62 ,
4) First address is 1024 subnet will be 130.56.255.193 and last address in 1024 subnet will be 130.56.255.254.
Explanation :
1)2^n = 1024
Therefore, n = 10.
The address is here is a Class B ,
so, the default mask is /16. 10 bits are necessary for subnets and
hence the correct subnet mask is /(16+10) = /26.
In the format of dotted decimal format, it shall be 192 and therefore, the subnet mask will be 255.255.255.192 .
2) The rest of the bits must be used for the addressed i.e. there will be 32 - 26 = 6 bits available for the address component.
So, a total of 2^6=64 bits shall be available. Also, 2 bits per subnet cannot be allocated and subnet mask will be able to maintain 62 valid addresses
3)First address in subnet 1 = 130.56.0.1
Last address is subnet 1 = 130.56.0.62
The first address can be estimated by ANDing the address 130.56.0.0 with the subnet mask /26 like below: 10000010 00111000 00000000 00000000 (130.56.0.0)
This address can not be allocated, so we will consider the next address: 10000010 00111000 00000000 00000001 (130.56.0.1).
In Similar manner, the last address that can be allocated before the broadcast address will be 130.56.0.62.
4)First address in the 1024 subnet = 130.56.255.193
Last address in 1024 subnet = 130.56.255.254
Total number of addresses