Question

FILL IN THE TABLE BY SUBNETTING 192.168.1.0 SO THAT ALL SUBNETS CREATED HAVE THE NECESSARY IP'S

TO ACCOMODATE 15 DEVICES ON THE NETWORK WHICH INCLUDES THE ROUTER GATEWAY INTERFACE. CHOOSE

THE MOST EFFICIENT SUBNET MASK AND FILL IN THE SUBNET ID'S (NETWORK IP), HOST RANGE IP'S, BROADCAST

IP AND SUBNET MASK

THE FOLLOWING HEADS ARE NEEDED

SUBNET IDENTIFIER VALID HOST RANGE BROADCAST ADDRESS

Answer 1

Hi, I would love to help you out with this questions answer and hope you get it and like an answer, Please give a thums up if you get something to learn. So not wasting much time let us start with an answer:

Firstly we need to understand how the subnet is been divided for the better understanding of the table that is created below , so there are two different formulas for the number of subnets and number of hosts per subnet. One thing you must know that the network bit in the ip address is represented by 1 and the host bit is represented by 0. So here we need 15 hosts per subnet that must be according to the formula 2ⁿ-2 where n = number of host bits so n=5 then the hosts in the subnet will be 30 this is becuse if we put n=4 then the total hosts per subnet will become 14 which is less than required. So now there are 5 host bits and 3 network bit as the last block of ip address has total 8 bits .

Now the number of subnets will be 2ⁿ where n = network bits so acc to 3 network bit number of subnets will be 8 .

Now the first ip in the range is network or subnet id and the last ip in the range is the broadcast address and the subnet mask is accordingly. Now let us fill the table using the above concept

Network Address IP	Host Range	Broadcast Address
192.168.1.0	192.168.1.1 - 192.168.1.30	192.168.1.31
192.168.1.32	192.168.1.33 - 192.168.1.62	192.168.1.63
192.168.1.64	192.168.1.65 - 192.168.1.94	192.168.1.95
192.168.1.96	192.168.1.97 - 192.168.1.126	192.168.1.127
192.168.1.128	192.168.1.129 - 192.168.1.158	192.168.1.159
192.168.1.160	192.168.1.161 - 192.168.1.190	192.168.1.191
192.168.1.192	192.168.1.193 - 192.168.1.222	192.168.1.223
192.168.1.224	192.168.1.225 - 192.168.1.254	192.168.1.255

So this is the desire table for the question.

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