CS 250 Computer Networks Fundamentals Fall 2016 Homework 3 A

CS 250 Computer Networks Fundamentals Fall 2016 Homework 3 A network administrator is trying to redesign the network by dividing it into smaller subnets. The following are the requirements for these subnets: Subnet 1: 50 PCs Subnet 2: 60 PCs Subnet 3: 30 PCs Subnet 4: 30 PCs Subnet 5: 30 PCs Subnet 6: 10 PCs Subnet 7: 10 PCs Using 192.168.43.0/24 to create these subnets, complete the following The network address of Subnet 1 is: The first usable address in Subnet lis: The last usable address in in Subnet l is: The broadcast address of in Subnet 1 is: The network address ofSubnet 2 is: The first usable address in Subnet 2 is: The last usable address in in Subnet 2 is: The broadcast address of in Subnet 2 is: The network address ofSubnet 3 is: The first usable address in Subnet 3 is: The last usable address in in Subnet 3 is: The broadcast address of in Subnet 3 is:

Dear Student here is the complete answer of the above asked question...

The given IP address = 192.168.43.0/24

The given IP address belong to Class C address.

The default subnet mask for class C is 255.255.255.0.

Binary notation of 255.255.255.0 is 11111111.11111111.11111111.00000000.

It means that 24 bit for network and rest bit for host(bit=1 i.e Network, if bit=0 i.e Host).

Now we want to create 7 subnet from the given address.And subnet 1 and 2 has more than 50 PC connected to it.And sunetwork 3,4 and 5 have 30 PCs connected to it respectively.And subnet 6, 7 has 10 Pcs connected to it.

Below is the solution...

Answer....

For Subnet-1 and Subnet -2

In the subnet 1 and subnet 2 administrator want to connect 50 PC, 60 PC respectively. So Host bit must be more. So create a network in which we can connet 50 and 60 PC.We can borrow 2 bits from the Host bit to create subnetwork.

IF i borrow 2 bit then the mask will be 11111111.11111111.11111111.11000000

Subnet = 2^N=2²=4 subnet
Block size (256 - subnet mask) = 256 - 192 = 64.
Valid subnets ( Count blocks from 0) = 0,64,128,192
Total hosts (2^H) =2⁶= 64
Valid hosts per subnet ( Total host - 2 ) = 64 - 2 = 62

Subnet-1

The network address for Subnet 1 is: 192.168.43.0/26

The first usable address for subnet 1 is: 192.168.43.1

The last usable address in Subnet 1 is : 192.168.43.62

The broadcast address in Subnet 1 is : 192.168.43.63

Subnet-2

The network address for Subnet 2 is: 192.168.43.64/26

The first usable address for subnet 2 is: 192.168.43.65

The last usable address in Subnet 2 is : 192.168.43.126

The broadcast address in Subnet 2 is : 192.168.43.127

Subnet-3,4 and 5

In the subnet 3 , subnet 4, subnet 5 administrator want to connect 30 PC, 30 PC and 30 PC respectively. So Host requirement is less. So we have to create a network in which we can connet max 30 host.We can borrow 3 bits from the Host bit to create subnetwork.

IF i borrow 3 bit then the mask will be 11111111.11111111.11111111.11100000

Subnet = 2^N=2³=8 subnet
Block size (256 - subnet mask) = 256 - 224 = 32.
Valid subnets ( Count blocks from 0) = 0,32,64,128,160,192....256
Total hosts (2^H) =2⁵= 32
Valid hosts per subnet ( Total host - 2 ) = 32 - 2 = 30

Subnet-3

The network address for Subnet 3 is: 192.168.43.0/27

The first usable address for subnet 3 is: 192.168.43.1

The last usable address in Subnet 3 is : 192.168.43.30

The broadcast address in Subnet 3 is : 192.168.43.31

Subnet-4

The network address for Subnet 4 is: 192.168.43.32/27

The first usable address for subnet 4 is: 192.168.43.33

The last usable address in Subnet 4 is : 192.168.43.62

The broadcast address in Subnet 4 is : 192.168.43.63

Subnet-5

The network address for Subnet 5 is: 192.168.43.64/27

The first usable address for subnet 5 is: 192.168.43.65

The last usable address in Subnet 5 is : 192.168.43.94

The broadcast address in Subnet 5 is : 192.168.43.95

Subnet-6 and Subnet-7

In the subnet 6 and subnet 7 administrator want to connect 10 PC per subnet. So Host are less. So create a network in which we can connet 10 PCs.We can borrow 4 bits from the Host bit to create subnetwork.

IF i borrow 4 bit then the mask will be 11111111.11111111.11111111.11110000

Subnet = 2^N=2⁴=16 subnet
Block size (256 - subnet mask) = 256 - 240 = 16.
Valid subnets ( Count blocks from 0) = 0,16,32.......256.
Total hosts (2^H) =2⁴= 16
Valid hosts per subnet ( Total host - 2 ) = 16 - 2 = 14