Question

Network Subnetting 3

3. On a Class B network, how many hosts per subnet will you get if you have a subnet mask of 255.255.240.0  On how many subnets was divided this class B address?  Show your calculations

4. Your company has leased a Class C network whose network ID is 200. 200. 200.0. You want to create 8 subnets within this network.  Show how many bits, you would need to change, and the range of addresses in one of the subnets.

Answer 1

3) Class B:

Subnet mask is 255.255.255.240

Convert the Subnet mask into binary values

We get 11111111.11111111.11111111.11110000

As this is a Class B network, the first 16 bits represents Network ID and last 16 bits represents Host ID.

We need to concentrate on Host ID to calculate number of subnets and number of hosts in each subnet.

First we need to find out no. of subnet bits and host bits in Host ID.

Now calculate the number of 1's and 0's in Host ID

Number of 1's=12(also called as subnet bits)

Number of 0's=4(also called as host bits)

We can calculate no. of Subnets by using the formula 2^{(no.of 1's)}

i.e 2¹²=4096 subnets

We can calculate no.of Hosts in each subnet using the formula 2^{(no.of 0's)}-2

i.e 2⁴-2    [we have to subtract 2 because 1 address is fixed for subnet and 1 address is fixed for broadcast]

=> 16-2

=> 14 Hosts in each subnet

4) Class C network:

Given Network ID as 200.200.200.0 and subnet mask would be 255.255.255.0(for class C)

Represent the subnet mask in Binary format.

We get 11111111.11111111.11111111.00000000

As this is a Class C network,the first 24 bits represent Network ID and last 8 bits represent Host ID.

We need to concentrate on Host ID to calculate number of subnets.

Given that we need to create 8 Subnets.

We need to find the no. of subnet bits in Host ID

We know the formula to find no.of subnets i.e 2^{(no.of subnet bits in Host ID)}

Let 'x' be no.of subnet bits in Host ID

=> 2^x = 8

=> 2^x  = 2³ [ if a^m=aⁿ , then m=n]

=> x=3

So we need to change first 3 bits of Host ID to create 8 Subnets.

Range of addresses:

After changing the 3 bits in Host ID,we get the new subnet mask as follows

11111111.11111111.11111111.11100000

i.e new subnet mask is 255.255.255.224

no.of subnets=8

no.of hosts=2⁵-2 [we have to subtract 2 because 1 address is fixed for subnet and 1 address is fixed for broadcast]

=> 32-2

=>30 hosts in each subnet

To calculate range of subnet, we need to include subnet address and also broadcast address

so we get 32 different addresses for each subnet

so range is from 200.200.200.0 to 200.200.200.31 for first subnet and 200.200.200.32 to 200.200.200.63 for second subnet and so on till 8th subnet.