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Class B subnetting help

7 posts in this topic

Posted

Hi guys

Continuing my desire to teach myself how to subnet, I've now moved onto class b.

If you have time could you look at the workings in the link and tell me what you'd do next?

I'm stuck at this point. I'm not sure how to work out the address range due to it straying into the third octet from the left. Class c seemed a lot easier compared to this

Cheers

http://imageshack.com/a/img706/3166/b7by.jpg

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Posted

So as per your image, you've broken down the 32bit binary representation into three pieces. The first (16 bits) is a fixed value, representing the static portion of the address, the bit at the beginning that will not change. The second (four bits) are used to specify a particular subnet, in this case using four bits we have 16 of them, as requested in the text at the top of the image. The final piece (12 bits) is the host portion.

 

So, let's look at subnet #3 in your image (0011, 48).

 

First piece (static): 10101100 00010000

Second piece (subnet): 0011

Third piece (host): 0000 00000000 - 1111 11111111

 

The address of the network (subnet) is always the first: 10101100 00010000 00110000 00000000 (172.16.48.0)

The broadcast address is always the last: 10101100 00010000 00111111 11111111 (172.16.63.255)

Host addresses anything in between:  10101100 00010000 00110000 00000001 (172.16.48.1) - 10101100 00010000 00111111 11111110 (172.16.63.254)

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Posted

I've realised one of my mistakes.I've added up the subnets incorrectly right from the start. If I'm right in thinking, they should all be one value less. 16 should be 15,64 should be 63 etc.

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Posted

Hmm, or have I? I've just sat down again to give this another throw, and I'm doubting myself.

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Posted

Right ok, i think i'm there. Providing that the information in the pic below if correct...i had no reason to doubt myself, and to be honest, i've been over-complicating it in my head due to the addition octet which i have to play with.

 

What do you think to this? - http://imageshack.com/a/img716/8603/9rw1.png

 

Thanks to anyone who has a look and responds :)

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Posted

Hello,

Right ok, i think i'm there. Providing that the information in the pic below if correct...i had no reason to doubt myself, and to be honest, i've been over-complicating it in my head due to the addition octet which i have to play with.

 

What do you think to this? - http://imageshack.com/a/img716/8603/9rw1.png

 

Thanks to anyone who has a look and responds :)

http://trk.free.fr/ipcalc/

ip-calc.zip

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