Question

Given that the company "Slate Rock and Gravel" has business locations in Granitetown and Bedrock, suppose that the company has been assigned a large block of consecutive IP addresses starting with 192.172.0.0, and that it wishes to set up two subnets, one for each location, with each subnet having sufficient IP numbers assigned to cover the host computers at that location. The IP addresses must be assigned in this order: first to Granitetown, and then to Bedrock. Assume that there are 4,000 computers in Granitetown and 2,000 in Bedrock.
Your task is to assign IP address to the two locations. You must show your steps for each calculation.

Answer 1

Answer:

Check the explanation

Explanation:

As Granitetown requires 5000 IP address and the starting IP address is 192.172.0.0 , we have to perform subnetting using number of hosts required

Number of IP address required by Granitetown=4000

Number of hosts bits =

2n -2= 5000

n=13 (approximately)

Number of hosts bits are 13 therefore subnet prefix =(32-13)=19

We have used 32 because IP address is of 32-bit

Subnet mask corresponding to /19 = 255.255.224.0

Starting IP address (given)= 192.172.0.0

Now to calculate last IP address of the subnet, OR operation will be performed between First IP address of the subnet and !Netmask

First IP address of subnet ( in binary) =

11000000.10101100.00000000.00000000

OR

00000000.00000000.00011111.11111111 (!Netmask)

11000000.10101100.000 11111.11111111 (result)

Converting result into decimal gives IP address 192.172.31.255 which will be the broadcast IP address of the subnet therefore

Last usable IP address = 192.172.31.254

Now, bedrock requires 2000 IP address

Therefore number of host bits are =

2n -2 =2000

n=10 (approximately)

Number of hosts bits required are 10

Therefore Netmask prefix = (32 -10) =22

Now, using the next subnet for bedrock therefore the first IP address of subnet will be 192.172.32.0

Now, similarly calculating last IP address of subnet

First IP address in binary =

11000000.10101100.00100000.00000000

OR

00000000.00000000.00000011.11111111 (!Netmask)

11000000.10101100.00100011.11111111 (result)

Now, result in decimal = 192.172.35.255

Therefore ,

Broadcast address of subnet =192.172.35.255

Last usable IP address =192.172.35.254