1  You are given an IP address for a host /20  What is/are the  Subnet address?  Broadcast address?  The number of useable hosts available for this subnet?  The number of useable subnets available for this network?  The assignable address range for this subnet?

2  Given an IP address for a host /20  The address is of class B.  /20  subnet mask: = = 16 Multiples of 16: 16, 32, 48, 64, 80, ….  Subnet address? The third octet is 35 which is between 32 and 48, so we choose the minimum value: 32 therefore, subnet address is

3  Broadcast address? subnet address is and the next subnet address is We know that the broadcast address is that of the next subnet address minus 1 

4  The number of useable hosts available for this subnet? The address is of class B and subnet mask is , so we borrowed 4 bits to make subnetting and 12 bits are left for hosts  x = 12 the number of useable hosts = 2 x -2 = = 4094

5  The number of useable subnets available for this network?  subnet mask is , so we borrowed 4 bits to make subnetting  x = 4 the number of useable subnets = 2 x -2 = = 14

6  The assignable address range for this subnet? The subnet address is and its broadcast address is The assignable address range is all addresses between the subnet address and broadcast address: …

7  Your organisation has been assigned a class B IP address of  You require about 2000 subnetworks  Work out the  Subnet mask required for this subnet  The network and broadcast addresses for the first 5 useable subnets  The number of hosts for each subnet  The assignable address range of the first 5 useable subnets

8  Your organisation has been assigned a class B IP address of  You require about 2000 subnetworks Number of subnets = 2 x -2, so we have to find the value of x so as to get a value closer or equal to If we choose x = 11, we get = 2048 – 2 = 2046.

9  Subnet mask required for this subnet The address is of class B and the default mask is But, we need to borrow 11 (eleven) bits for subnetting   subnet mask is

10  The network and broadcast addresses for the first 5 useable subnets: 256 – 224 = 32 Multiples of 32: 32, 64, 96, 128, 160, 192, ….. Subnet address Broadcast address  1 st subnet:  2 nd subnet:  3 rd subnet:  4 th subnet:  5 th subnet:

11  The number of hosts for each subnet We borrowed 11 (eleven) bits for subnetting, so we are left with 5 bits for host.  x = 5  number of hosts in each subnets = 2 x -2 = = 30

12  The assignable address range of the first 5 useable subnets: The assignable address range is all addresses between the subnet address and broadcast address.  1 st subnet:  2 nd subnet: …  3 rd subnet: …  4 th subnet: …  5 th subnet: …