Consider a network with the IP address 192.168.1.1/16 Part I 1. What is the netw

Question

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]

Explanation / Answer

Answer:

We have given IP address : 192.168.1.1/16

1. The network prefix of this network is :

Network mask of this IP is : 255.255.0.0

Now convert into binary : 11111111.11111111.00000000.00000000 count number of 1s comes 16 . Therefore prefix network is /16

2. The prefix network of the IP address after subnetting is :

Since we choose two bits , network mask = 255.255.255.192 = 11111111.11111111.11111111.11000000 , count number of 1s = 26

Therefore prefix network = /26

3. First subnet is divided into 5 subnets, we are choosing 3 bits = 255.255.255.224 = 11111111.11111111.11111111.11100000 = 27

Therefore prefix network = /27

4. First usable address = 192.168.0.0/27

Last usable address = 192.168.255.255/27

Broadcast address = 192.168.1.255

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