subnetting

[nizamshaikh2504]

If you take the network 192.168.1.0/24 and subnet it to four /26 networks, what is the usable address range in the third subnet?

kindly explain me the answer in brief.

AND

please explain me the underlined text in simple manner.

THANKS !

[Mark Malone · Accepted solution]

The /24 is broken into blocks of 64 when split into /26s

Network addresses would be the 4 subnets below , so the useable addresses would be anything but the network address and the last address as its the broadcast

192.168.1.0 - 63 / 192.168.1.64 - 127 / 192.168.1.128 -191 /192.168.1.192 - 254

1- 63

65 -126

128 -191

193 - 254

[NF27]

what would be the subnet mask for each of these ranges?How would you calculate?Kindly help!

[nizamshaikh2504]

thanks for the answer ......

will you please explain me the highlighted text. i didn't get it.

[Mark Malone]

so you have 0-255 in /24 but if you break to /26 it starts 0 -63 which is 64 altogether 0 being the subnet and 63 the broadcast address , that's the first network always start from 0 when subnetting Class C /24

so you ask whats the 3rd useable address space of the /26s

1st you need to know what the first is

/26 in binary is 8+8+8+2  , first 3 8s or octets are used then you have 2 bits left for in the last octect that are used , when you break that down in binary its bit 64 as below

                    192 168 1 0

      11111111.11111111.11111111.                              1   1   1   1  1 1 1 1

       8 +             8 +           8 +  the last 2 in octet 4       128 64 32 16 8 4 2 1 = /26 in total bits

So 2 bits in is the block you start with that's how you get the 64 , how many 64s in 256 = 4

so 4 networks of /26

you want to know the 3rd so its not 0-63 / its not 64-127 its the next 3rd block available so its 192.168.1.127 - 191 that's your 3rd networks useable addresses after you break it down

This is a good website for interactively practising subnetting

https://subnettingpractice.com/