Computer Networks | Network Layer | Subnetting

Question

X Institute has been assigned the Class B network address 165.87.0.0. X needs to divide the network into eight subnets. What subnet mask should be applied to the network to provide the most hosts subnet?

(a) 255.255.192.0

(c) 255.255.240.0

(b) 255.255.224.0

(d) 255.255.248.0

Answer 1

 

IP of X is 165.87.0.0.  Which is a class B IP-address, hence 16 bits are for network-ID(NID) and 16 bits are for host-ID(HID).

Now to divide network into 8 subnets we will have to take 3 bits from hosts so that $2^{3}$ =8 subets possible (from 000 to 111).

Number of bits taken from hosts to do subnetting= 3 bits.

Now IP address of any subnet has 19 bits for Subnet-ID part (16 NID bits + 3 extra bits taken from HID to do subnetting).

So subnet mask will have all 1’s in subnet ID part and all zeroes in host ID part

11111111.11111111.11100000.00000000 =255.255.224.0

 

hence, subnet mask will be 255.255.224.0

 

 

 

Answer 2

To divide a network into subnets while maximizing the number of hosts per subnet, we need to find the subnet mask that allows for the largest number of host addresses within each subnet. This involves borrowing some bits from the host portion of the IP address to create subnets. Let's calculate the required subnet mask step by step:

1. Determine the number of bits required to represent 8 subnets: 2^3 = 8, so we need 3 bits.

2. Determine the number of bits used for the default Class B network address: 16 bits (first two octets).

3. Subtract the number of bits used for subnets from the total number of bits for the network address to find the remaining bits for hosts: 16 - 3 = 13 bits.

4. Convert the remaining 13 bits to decimal to find the host addresses per subnet: 2^13 - 2 = 8190 (subtracting 2 for the network and broadcast addresses).

5. Now we have the subnet mask in CIDR notation: /29 (since we used 3 bits for subnets).

6. Convert the subnet mask to dotted decimal notation: 255.255.255.248.

So, the subnet mask that should be applied to the network to provide the most hosts per subnet is 255.255.255.248 (/29 in CIDR notation). This allows for 8 subnets, each with 6 usable host addresses.

 

Answer 3

in class b 16NID and 16HID

2^3 subnet we find SID= 3 bit

No of 1’s in subnet mask =NID+SID=16+3=19

No of 0’s in subnet mask= HID =32-19=13

so, 11111111.11111111.11100000.00000000

i.e 255.255.224.0

option (b) is correct

Answer 4

Ans:  255.255.224.0

We Have Class B addressing so first 16 bit are for net-id, and next 16 bit are for host id and we are gonna use host bit for subnetting .

Note one thing that first address in subnet is also reserve for net-id and last address for broadcast , 

If we use 3 bit for subnetting we will have (2^3) subnets so 3 bits will give us 8 subnet address
t

so 11111111. 11111111. 11100000.00000000  is subnet mask.

that will be 255.255.224.0 is correct answer.