Question

Scenario:

You own a chain of department stores. You currently have 19 stores. The stores are not all the same size - the smallest store employees only 5 people and the largest employees 38 people.

Problem:

What is the size (in bits) or the address space you will need to create your system?

• How many bits will you need for your NETWORK side?
• How many bits will you need for your HOST(NODE) side?

Assume you have been given the space you asked for, answer the following questions:

1. What would be a valid TCP/IP address range you could be given? (determine what class or part of class you would use and make up a VALID ip address in that range)
2. What is the address of the FIRST computer in the First network?
3. What is the address of the Last computer in the First Network?
4. What is the address of the 5th computer in the 3rd Network?

Requirements:

Answer 1

As per the scenario, there are 19 stores and the largest number of employees is of 38 people and smallest store employees only have 5 people.

/26 will be the bit size will be needed to create the system because:

/26: 11111111.11111111.11111111.11000000

So, number of bits left is the number of hosts in each subnet: 2^6 -2 = 62

2 bits will be needed for network side

6 bits will be needed for host side.

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It is better to take a CLASS C address for the entire setup. the address range will be 192.168.10.1 - 192.168.14.254

First Network address: 192.168.10.0

The address of the First computer in First network: 192.168.10.1

The address of the Last computer in first network: 192.168.10.38

3rd network address: 192.168.10.128

The address of the Fifth computer in the Third Network: 192.168.10.133