Provide detailed answer and explanation to each questions Co

Provide detailed answer and explanation to each questions.

Consider a network with the IP address 192.168.1.1/16

Part I

1. What is the network prefix for this network, in the form a.b.c.d/x? [2 points]

2. Assuming that this network is to be broken down into two subnets of equal size, write down the network prefix of each subnetwork. [3 points]

3. Assume that the 1st subnet is divided into 5 equal subnetworks, and write down the network prefix of each subnet. [10 points]

4. Of the 5 subnets, select the 1st one and write down the first usable address, the last usable address, and the broadcast address. [6 points]

Part II

For the other subnet of the original address range above, assume that 4 subnetworks need to be created with the following details A - 300 hosts, B - 500 hosts, C - 88 hosts and D - 60 hosts.

1. Using the concept of Variable Length Subnetting, derive and write down the network prefixed of A, B, C and D. [16 points]

2. Assuming that network A is further divided into two subnetworks (A1 and A2) of size 30 and 200, write down the network prefixed of these subnets. [8 points]

Solution

Consider a network with the IP address 192.168.1.1/16
Part I
What is the network prefix for this network, in the form a.b.c.d/x?
192.168.1.1/16
Assuming that this network is to be broken down into two subnets of equal size, write down the network prefix of each subnetwork.
here we have to divide into 2 subnet so that 1 but will be increase in mask.
192.168.1.1/17
192.168.3.1/17

Assume that the 1st subnet is divided into 5 equal subnetworks, and write down the network prefix of each subnet.
network is divided into 5 subnet so that atleast 3 bit will be used in subnet mask
192.168.1.1/20
192.168.2.1/20
192.168.3.1/20
192.168.4.1/20
192.168.5.1/20
  
Of the 5 subnets, select the 1st one and write down the first usable address, the last usable address, and the broadcast address.
1. 192.168.1.1/20
broadcast address : 192.168.15.255

192.168.2.1/20
broadcast address : 192.168.15.255

192.168.3.1/20
broadcast address : 192.168.15.255

192.168.4.1/20
broadcast address : 192.168.15.255

192.168.5.1/20
broadcast address : 192.168.15.255

Part II
For the other subnet of the original address range above, assume that 4 subnetworks need to be created with the following details A - 300 hosts, B - 500 hosts, C - 88 hosts and D - 60 hosts.
Using the concept of Variable Length Subnetting, derive and write down the network prefixed of A, B, C and D.
Assuming that network A is further divided into two subnetworks (A1 and A2) of size 30 and 200, write down the network prefixed of these subnets.
sol:
A - 300 hosts, B - 500 hosts, C - 88 hosts and D - 60 hosts.
A need 300 host so that it need 9 bit address from host whose range is max 2^9 - 2 = 510 host
so that subnet mask will be
A : 192.168.1.1/18
now A will be divided into 2 subnet of A1:30 , A2:200
A1 : 192.168.2.1/19
A1 : 192.168.6.1/19
B : 192.168.2.1/18
C : 192.168.3.1/18
D : 192.168.4.1/18