A big company has bought a subnet from one class B IP net. It is known that the

Question

A big company has bought a subnet from one class B IP net. It is known that the IP address is 125.110.22.0/23 (23 means the first 23 bits are fixed). There are major 4 departments in this company which requires 95,271,63 and 66. In the beginning, the engineer plans to do one IP address bind to one host case, but, he finds that it needs lots of time. Please find a solution for the engineer how to efficiently divide the available IP addresses into different subnets and satisfy the required number of hosts. 3.

Explanation / Answer

Solution :- First of all, We will define the configuration of class B IP net given below:-

(Hosts/Net Range:- 125.110.22.1 to 125.110.23.254)

From the above configuration we will derive IP address range for each department say D1,D2,D3,D4 given below:-

For department D1:-

Now we will choose first 95 addresses in between of 125.110.22.1 and 125.110.22.126 for Department D1, i.e. Range will be from 125.110.22.1 to 125.110.22.95 (Total required host=95).

Total Required Hosts for Department D1:- 95 Hosts

For department D2:-

We will choose from 125.110.22.96 to 125.110.22.126 (Total=30 Addresses) Now we will choose others 241 by:-

We will choose from 125.110.22.129 to 125.110.22.254 (Total=126 Hosts) and then

We will take remaining 115 from 125.110.23.1 to 125.110.23.115 ( Total Hosts=115)

So Total required Hosts for Department D2:=

125.110.22.96 to 125.110.22.126 (Total=30 Hosts)

125.110.22.129 to 125.110.22.254 (Total=126 Hosts)

125.110.23.1 to 125.110.23.115     (Total=115 Hosts)

Total= 271 Hosts As required by department D2.

For Department D3:- 125.110.23.116 to 125.110.23.126 (Total=11 Hosts)

Total Hosts for D3= 52+11=63

For D4:- We will choose from 125.110.23.182 to 125.110.23.248 = (Total Hosts=66)

For D4 Total Required Hosts = 66

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