Some might say in the world of networking any subnetting without VLSM (Variable Length Subnetting Masking) is way too easy, Have a look at this one.....

A Company is given Ip Address 145.72.0.0, they have 30 departments in their company and do not know and also do not want to assume which particular department will need more ip addresses or which department will need less ip addresses, so the clever Network Administrator initiates an idea that he would not use VLSM and devide the addresses in equal parts throughout the departments. How would you subnet it? Explain showing each subnet.

Answer:

Needed 30 EQUAL subnets

So, we need to borrow 5 bits from the host portion of the class B (128.0.0.0 - 192.255.255.255) network

but as we have to borrow 5 bits from network portion we will have two extra subnets.

how?
we borrow 1 bit we get two subnets 0 & 1
borrow 2 bits we get four subnets 00,01,10,11
3 gives 8 subnets 000,001,010,100,011,110,111,101
4 bits gives us 16
and 5 bits gives us 32

now back to our problem.

class B default mask: 255.255.0.0 = 11111111.11111111.00000000.00000000

after borrowing 5 bits from host portion (0 portion)

11111111.11111111.11111~000.00000000 = 255.255.248.0 = /21
* After "~" means start of host portion

starting subnet (0th subnet) = 145.72.0.0 - 145.72.7.255
1st subnet = 145.72.8.0 - 145.72.15.255
2nd subnet = 145.72.16.0 - 145.72.23.255
3rd subnet = 145.72.24.0 - 145.72.31.255
4th subnet = 145.72.32.0 - 145.72.39.255
5th subnet = 145.72.40.0 - 145.72.47.255
6th subnet = 145.72.48.0 - 145.72.55.255
7th subnet = 145.72.56..0 - 145.72.63.255
8th subnet = 145.72.64.0 - 145.72.71.255
9th subnet = 145.72.72.0 - 145.72..79.255
10th subnet = 145.72.80.0 - 145.72.87.255
11th subnet = 145.72.88.0 - 145.72.95.255
12th subnet = 145.72.96.0 - 145.72.103.255
13th subnet = 145.72.104.0 - 145.72.111.255
14th subnet = 145.72.112.0 - 145.72.119.255
15th subnet = 145.72.120.0 - 145.72.127.255
16th subnet = 145.72.128.0 - 145.72.135.255
17th subnet = 145.72.136.0 - 145.72.143.255
18th subnet = 145.72.144.0 - 145.72.151.255
19th subnet = 145.72.152.0 - 145.72.159.255
20th subnet = 145.72.160.0 - 145.72.167.255
21th subnet = 145.72.168.0 - 145.72.175.255
22nd subnet = 145.72.176.0 - 145.72.183.255
23rd subnet = 145.72.184.0 - 145.72.191.255
24th subnet = 145.72.192.0 - 145.72.199.255
25th subnet = 145.72.200.0 - 145.72.207.255
26th subnet = 145.72.208.0 - 145.72.217.255
27th subnet = 145.72.218.0 - 145.72.223.255
28th subnet = 145.72.224.0 - 145.72.231.255
29th subnet = 145.72.232.0 - 145.72.239.255
30th subnet = 145.72.240.0 - 145.72.247.255
31th subnet = 145.72.248.0 - 145.72.255.255

And they say this was easy......

How to determine network address, broadcast address and number of usable host addresses if you are given an ip address.

How to find network address and broadcast address of a given ip address?
also how to find a range of usable host addresses for a given ip address.

In our example we shall use
IP ADDRESS: 192.168.31.5
IP SUBNET MASK: /28 or 255.255.255.240



/28 or 255.255.255.240 = 11111111.11111111.11111111.11110000
192.168.31.5 = 11000000.10101000.00011111.00000101

11111111.11111111.11111111.1111~0000
11000000.10101000.00011111.0000~0101 *"~" means starting of host bits by the mask.

For a network address all the host bits have to be 0
therefore

11111111.11111111.11111111.1111~0000
11000000.10101000.00011111.0000~0000 = 192.168.31.0

For a broadcast address all host bits have to be 1
therefore

11111111.11111111.11111111.1111~0000
11000000.10101000.00011111.0000~1111 = 192.168.31.15

Now, if you want to know how many usable host address we have
remember we cannot use network address and the broadcast address for host addresses, So,
it is always
Square Of (n) - 2
where n is the number of host bits

in our example: n = 4
Square of 4 = 16;
16 - 2 = 14

therefore starting usable host address = 192.168.1.1
ending usable host address = 192.168.1.14

so .1 to .14 are the usable host addresses