Assignment 2 sums

10) Find the subnet mask, network ID, subnet address, broadcast ID and the host range from the following IP address.

A) 192.168.30.112/28, B) 172.160.16.224/20 C) 11.11.108.96/22.

Here’s how you can calculate the subnet mask, network ID, subnet address, broadcast ID, and host range for each of the given IP addresses:

A) 192.168.30.112/28

1. Subnet Mask:

   CIDR /28 means 28 bits for the network, so the subnet mask is 255.255.255.240.

2. Network ID:

   192.168.30.112 in binary is 11000000.10101000.00011110.01110000.

   Applying the /28 mask, the first 28 bits remain unchanged, and the remaining bits are set to 0, giving the Network ID as 192.168.30.112.

3. Subnet Address:

   Same as the network ID in this case: 192.168.30.112.

4. Broadcast Address:

   Set the host bits to 1, which means the last 4 bits are 1111. The broadcast address is 192.168.30.127.

5. Host Range:

   The first usable address is 192.168.30.113 (Network ID + 1).

   The last usable address is 192.168.30.126 (Broadcast Address - 1).

   So, the host range is 192.168.30.113 - 192.168.30.126.

B) 172.160.16.224/20

1. Subnet Mask:

   CIDR /20 means 20 bits for the network, so the subnet mask is 255.255.240.0.

2. Network ID:

   172.160.16.224 in binary is 10101100.10100000.00010000.11100000.

   Applying the /20 mask, the first 20 bits remain unchanged, and the remaining bits are set to 0, giving the Network ID as 172.160.16.0.

3. Subnet Address:

   Same as the network ID in this case: 172.160.16.0.

4. Broadcast Address:

   Set the host bits to 1, meaning the last 12 bits are 11111111.11111111. The broadcast address is 172.160.31.255.

5. Host Range:

   The first usable address is 172.160.16.1 (Network ID + 1).

   The last usable address is 172.160.31.254 (Broadcast Address - 1).

   So, the host range is 172.160.16.1 - 172.160.31.254.

C) 11.11.108.96/22

1. Subnet Mask:

   CIDR /22 means 22 bits for the network, so the subnet mask is 255.255.252.0.

2. Network ID:

   11.11.108.96 in binary is 00001011.00001011.01101100.01100000.

   Applying the /22 mask, the first 22 bits remain unchanged, and the remaining bits are set to 0, giving the Network ID as 11.11.108.0.

3. Subnet Address:

   Same as the network ID in this case: 11.11.108.0.

4. Broadcast Address:

   Set the host bits to 1, meaning the last 10 bits are 11111111.11111111. The broadcast address is 11.11.111.255.

5. Host Range:

   The first usable address is 11.11.108.1 (Network ID + 1).

   The last usable address is 11.11.111.254 (Broadcast Address - 1).

   So, the host range is 11.11.108.1 - 11.11.111.254.

These are the network details for each of the given IP addresses.

11) A company has three networks into three different cities, namely Ahmedabad Delhi and Chennai Ahmedabad has three branches. A has 120 PCs, B has 170 PCs and C has 60 PCs. Delhi branch has 29 PC and 14 PC respectively and Chennai branch has 14 ,6 and 4 PC respectively. Determine the IP address scheme using VLSM and submitting summaries of the design in tabular form. Use class C Private address for the Design.

To design an IP address scheme using VLSM (Variable Length Subnet Masking) for the given setup, we will:

1. Use a Class C private IP address range: 192.168.0.0/24.
2. Subdivide this address space to allocate IP addresses efficiently based on the number of PCs in each branch while reserving addresses for future growth.

Step 1: Calculate the Subnet Requirements

For each branch, calculate the required number of IP addresses (including network and broadcast addresses):

Step 2: Assign Subnets

Allocate the subnets sequentially, starting from 192.168.0.0.

Step 3: Summary in Tabular Form

This design ensures efficient allocation with no wastage of IP addresses and meets the requirements using VLSM.