1

You are given an IP address for a host 172.168.35.10/20

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 172.168.35.10/20 The address is of class B. /20 subnet mask:

11111111.11111111.11110000.00000000 = 255.255.240.0

256-240 = 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 172.168.32.0

3

Broadcast address? subnet address is 172.168.32.0 and the next

subnet address is 172.168.48.0 . We know that the broadcast address is that of the next subnet address minus 1

172.168.48.0- 1------------------------ 172.168.47.255

4

The number of useable hosts available for this subnet?

The address is of class B and subnet mask is 11111111.11111111.11110000.00000000, so we borrowed 4 bits to make subnetting and 12 bits are left for hosts

x = 12 the number of useable hosts = 2x-2 = 212-2 =

4094

5

The number of useable subnets available for this network?

subnet mask is 11111111.11111111.11110000.00000000, so we borrowed 4 bits to make subnetting x = 4

the number of useable subnets = 2x-2 = 24-2 = 14

6

The assignable address range for this subnet?The subnet address is 172.168.32.0 and its

broadcast address is 172.168.47.255The assignable address range is all addresses

between the subnet address and broadcast address:

172.168.32.1172.168.32.2… …174.168.47.254

7

Your organisation has been assigned a class B IP address of 130.10.0.0 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 130.10.0.0 You require about 2000 subnetworksNumber of subnets = 2x-2, so we have to find

the value of x so as to get a value closer or equal to 2000.

If we choose x = 11, we get 211 - 2 = 2048 – 2 = 2046.

9

Subnet mask required for this subnetThe address is of class B and the default mask

is 255.255.0.0. But, we need to borrow 11

(eleven) bits for subnetting 11111111.11111111.11111111.11100000

subnet mask is 255.255.255.224

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

1st subnet: 130.10.0.32 130.10.0.63 2nd subnet: 130.10.0.64 130.10.0.95 3rd subnet: 130.10.0.96 130.10.0.127 4th subnet: 130.10.0.128 130.10.0.159 5th subnet: 130.10.0.160 130.10.0.191

11

The number of hosts for each subnetWe borrowed 11 (eleven) bits for subnetting,

so we are left with 5 bits for host.x = 5 number of hosts in each subnets = 2x-2 =

25-2 = 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.

1st subnet: 130.10.0.33 ..... 130.10.0.62 2nd subnet: 130.10.0.65 ….. 130.10.0.94 3rd subnet: 130.10.0.97 ….. 130.10.0.126 4th subnet: 130.10.0.129 ….. 130.10.0.158 5th subnet: 130.10.0.161 ..… 130.10.0.190