Question

a. Explain with an example a scenario where both the class full and classless IP addressing scheme can be used to configure the IP address for the network.

b. Suppose that a network has the IP network 192.168.10.0/27. Answer the following questions based on the IP arrangement.

1. Find the IP address of the second subnet.

2. Compute the number of valid hosts.

3. Find the usable IP address

4. Find the broadcast address.

Answer 1

a. Explain with an example a scenario where both the class full and classless IP addressing scheme can be used to configure the IP address for the network.

Let us consider a situation where we need to provide communication for 1000 hosts.

Using a classful addressing we can only select the range of ipaddresses according to predefined classes like Class A,class B, class C, class D. The respective range of hosts and networks for a given class is as follows

class	networks	hosts
A	126	16777214
B	16384	65534
C	2097512	254

we cannot select class C because the no of max hosts for a given network is 254 only. Hence wee need to select class B addresses only which has 65534 hosts capacity . But for our network we only require 1000 hosts so remaining 55534 no of ipaddress will be unfilled and the addresses will not be used efficiently. even if we consider of taking 4 class C networks that will be very hard to maintain in a single network. .

So if we use classless network for this we can take a class B network and subnet into networks with 1024 hosts capacity. By this only 1024 ipaddress will be filled and remaining addresses can be used by other networks.

b. Suppose that a network has the IP network 192.168.10.0/27. Answer the following questions based on the IP arrangement.

192.168.10.0/27 can be written in binary as 11000000 . 101010000 . 00001010.00000000 and subnet is 11111111.11111111.11111111.11100000. so in order to get network address we need to apply and between subnet and an ip address. for this case it will be same as 192.168.10.0( 1000000 . 101010000 . 00001010.00000000 ) and the total no hosts will be

2^(32-subnetbits) =2^(32-27) = 2^5 =32

from the above we can conculde main network is divided into subnets each of

out of 32 first one will be network address and last will broadcast address.

1. Find the IP address of the second subnet.

ip address of second subnet will be next of 1000000 . 101010000 . 00001010.00000000 ie 1000000 . 101010000 . 00001010.00100000 in decimal it is (192.168.10.32).

2. Compute the number of valid hosts.

valid number of hosts =32-2 =30

3. Find the usable IP address

usable ip addresses =192.168.10.1 to 192.168.10.30    (32 - 1 n/w address -1 broadcast address)

4. Find the broadcast address.

broadcast address = last address of network= 192.168.10.31

Thank You....!!!!
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