Solution

For class B address, the size of the network field is 16 bits and 16 bits for HID. But in the network field, the first 2 bits are fixed as a prefix 10; hence the total number of networks possible is $2^{14}$ , and the number of addresses possible is $2^{16}$ in each network.

Solution

Given, Subnet Mask = 255.255.255.224 => 11111111.11111111.11111111.11100000

In Subnet mask number of 1’s represent NID+SID bits, and Number of 0’s represent HID bits

So, NID = 27 bits and HID = 32 - 27 = 5 bits

To calculate no. of subnets we need no. of subnet bits. The given IP address is of class B and the subnet mask is 255.255.255.224. So no. of subnet bits is 27-16 = 11.

Hence no. of subnets are 211 = 2048.

Given IP address is, 162.112.25.35

=> 162.112.25.00100011

NID HID

As we know that, a first network address for NID, all 0’s in HID part and Last address for DBA, all 1’s in HID part.

Therefore, Possible range is between

162.112.25.00100000(first address for NID 162.112.25.32) to

162.112.25.00111111(Last address for broadcast address 162.112.25.63)

Solution

In given IP address NID = 22 bit, HID = 32 - 22 = 10 bit

There number of hosts possible = 210 - 2 = 1022 host

Solution

Following are the valid private IP address ranges.

For Class A - 10.0.0.0 - 10.255.255.255 (16,777,216 IP addresses)

For Class B - 172.16.0.0 - 172.31.255.255 (1,048,576 IP addresses)

For Class C - 192.168.0.0 - 192.168.255.255 (65,536 IP addresses)

Solution

No. of bit in HID part = 32-27 = 5 bits so that the subnet mask will be 255.255.255.224

Do Bitwise AND with given IP and subnet mask, we will get NID 215.15.25.128

In the fourth octet, first three-bit will be fixed for the subnet bits, and the remaining 5 bits are for HID, so the maximum possible value in binary is 11111.

As we know, the addresses with all 1’s in the host part are the broadcast address and can't be assigned to a host.

So the maximum possible value in the last octet, for a host IP is 10011110, which is 158.

Solution

Take each IP and do bitwise AND with the given Subnet Mask. If we get the same network ID for the given IP's, then it will belong to the same subnet.

Solution

Do the Bitwise AND operation of List -I IP address with Netmask given in the table.

Compare the result with a network ID from the given table.

The network ID is matched with the result, and the packet will be forwarded to the corresponding interface.

Example: 129.196.167.151 AND 255.255.254.0 = 128.96.166.0, Therefore packet will forward through Interface 2.

Suppose 129.196.171.92 AND 255.255.254.0 = 128.96.170.0. So it will go to interface 0. And If it does not match with any NID, it will go to the default interface.

Solution

When we do bitwise AND between IP Address and Subnet Mask, it gives Network ID. If both IP addresses provide the same Network ID. It means both IP addresses belong to the same network else it's on a different network.

When we do bitwise AND between IP address 215.24.101.113 and subnet mask 255.255.255.224 result is 215.24.101.96 and

when we do bitwise AND between IP address 215.24.101.91 and subnet mask 255.255.255.224 result is 215.24.101.64.

Therefore, 215.24.101.96 and 215.24.101.64 are different network IDs, so option D is the correct answer.

Solution

In Class C network first three octets are reserved for NID, so we have total 8 bits

for host and subnets. If we want to distribute addresses in subnets so first we should consider the subnet with maximum host, here which is a subnet with 110 hosts.

For 110 hosts we require 7 bits for host id, one bit remains for the subnet, which we fix to 1 for this subnet, now subnet mask for this subnet is 255.255.255.10000000 = 255.255.255.128 and subnet address is 200.200.200.128

For the first subnet, we have fixed the last octet 8th bit to 1, and for the second and third subnet, 8th bit will be 0.

Now for 60 hosts, we require 6 bit for host id, two-bit remain for the subnet, in which 8th bit is already fixed to 0; we can configure 7th bit only for these subnets;

For the second and third subnet, we fix the 7th bit to 0 and 1 respectively.

Subnet address for the second and third subnet is 200.200.200.0 and 200.200.200.64 respectively.

Subnet mask for the second and third subnet is 255.255.255.11000000 = 255.255.255.192

For class C subnet mask is 255.255.255.0

As its class C address, we have to take two-bit from the last octet.

Therefore, Possible subnet bits would be 00,01,10,11

MSB in last 8 bits helps us to get two subnets

10000000 →→ subnet1

00000000 →→ subnet2

subnet2 is divided into 2 more subnets using 7th bit

00000000 →→ subnet2(0)

01000000 →→ subnet2(1)

Solution

The size of network ID is 24 bit in class C networks. So bits after the 24th bit can be used to create 12 departments. Total 4 bits are needed to create 12 different departments.

As we know, in the subnet mask, NID+SID (24+4 = 28 bits) part is all 1‘s.

Therefore, the subnet mask will be 255.255.255.240.

Solution

As the given address is a class B IP address. So, NID = 16 bit and HID = 16 bits.

As we know that, a first network address for NID, all 0’s in HID part and Last address for DBA, all 1’s in HID part.

Therefore, NID = 181.66.0.0

And DBA = 181.66.255.255

Solution

Data | sender address | destination address

I - 2 (DBA can never be used as source IP)

II - 3 (Limited broadcast address 255.255.255.255, Limited Broadcasting data reaches from source to all the hosts in the same network).

III - 1 (Unicast packet within the network)

Solution

To create 1024 subnets, 10 bits required. ($2^{10}$ = 1024)

The given IP address, NID bit is 16 bits.

The subnet mask will be /16+10 = /26 (So, the default subnet mask is 255.255.255.192).

The remaining bits are 32 – 26 = 6 bits for the HID.

Therefore, possible number of host / subnet = $2^6$

Following will be the possible ranges :

1st subnet = 125.74.0.0/26 to 125.74.0.63/26

2nd subnet = 125.74.0.64/26 to 125.74.0.127/26

3rd subnet = 125.74.0.128/26 to 125.74.0.191/26

---- so on ----

1023rd sunbet = 125.74.255.128/26 to 125.74.255.191/26

1024th subnet = 125.74.255.192/26 to 125.74.255.255/26

Solution

IPv4 multicast addresses are defined by the most-significant bit pattern of 1110, which is a Class D address. The multicast addresses are in the range from 224.0.0.0 to 239.255.255.255.

221.125.105.55 is class C IP address; It can not be used for multicasting.

Solution

Given IP address 157.26.07.255 is DBA. and it's classless.

As we know, in DBA, the HID part is all 1's

Therefore, 168.17.00000111.11111111

NID HID

As we know, For the Subnet mask, NID part is all 1's and HID part is all 0's

Since the number of bits in NID or HID is not known, we can take any length for NID.

So the last 11 bits can be HID (bcz DBA contains all 1's in HID).

Similarly, the last 10 bits can be HID, or 9 bits or 8 bits and so on.

Depending on the number of bits, the Subnet mask will vary.

The range of Netmask can be 255.255.248.0 to 255.255.255. 254