Question

An organization is granted the block 130.56.0.0/16. The administrator wants to create 1024 subnets.

Answer 1

a) 255.255.255.192 will be the subnet mask

 b) 62 valid addresses can exist in each subnet

 c) First address in subnet 1 will be 130.56.0.1 and last address is sibnet 1 will be 130.56.0.62

 d) First address is 1024 subnet will be 130.56.255.193 and last address in 1024 subnet will be 130.56.255.254

Answer 2

Answer:

Explanation:

A) 2^n = 1024 Therefore, n = 10. The address given here is a Class B address, therefore the default mask is /16. 10 bits are required for subnets and hence the required subnet mask is /(16+10) = /26. In dotted decimal format, it shall be 192 and therefore, the subnet mask will be 255.255.255.192 . B) The remaining bits must be used for the address in each subnet i.e. there will be 32 - 26 = 6 bits available for the address component. Therefore, there will be a total of 2^6=64 bits available. Also, 2 bits per subnet cannot be allocated and hence, each subnet mask will be able to maintain 62 valid addresses . C)First address in subnet 1 = 130.56.0.1 Last address is subnet 1 = 130.56.0.62 The first address for the block can be calculated 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 the similar manner, the last address that can be allocated before the broadcast address will be 130.56.0.62. D)First address in the 1024 subnet = 130.56.255.193 Last address in 1024 subnet = 130.56.255.254 Total number of addresses =...