Subnetting question

I came across this question in an online subnetting tutorial and was really confused on how to figure it out.

Given an IP address of 131.107.2.160 and a subnet mask of 255.255.255.192, to which subnet class does the host belong?

A. 131.107.2.32
B. 131.107.2.64
C. 131.107.2.96
D. 131.107.2.128
E. 131.107.2.192

How do you calculate this? What formula is used? Can you calculate using

(2^n-2) where n=the number of masked bits in subnet field. n is the number of 1s in the octet for the class that is available for subnetting

(2^y-2) where y=number of off bits in host field. y is bits available for subnetting - n

thanks

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dynamik

Since 2 of the 8 bits are used in the last octet for the subnets, you have 6 remaining for the hosts. Using that you can figure out the ranges.

131.107.2.0-63
131.107.2.64-127
131.107.2.128-191
131.107.2.192-255

Once you have those, it's obvious where it fits.

If you want to learn about the math behind it, research binary ANDing. I have an example here: http://www.techexams.net/forums/ccna-ccent/723-how-do-i-find-valid-host-addresses-subnet-mask.html#post272846 (Just be sure to read the next couple posts for the corrections!

)

phfaty

Thanks very much for the explanation and the link. This cleared it up for me.

DerekAustin26

I find it really simple to take the mask "192" and subtract it from 256.
256-192 = 64 (the difference will always be your block for the octect the mask is in)

so 0, 64, 128, & 192 will be the subnets

The block size is 64 so subtract 2 and you get your hosts.
62

Subnet 0 Hosts are 1-62 Broadcast 63
Subnet 64 Hosts are 65-126 Broadcast 127
Subnet 128 Hosts are 129-190 Broadcast 191

Subtracting the "mask" from 256 will always give you your block and once you find your block for the octet in question, you should be able to find everything else out from there.

Good luck

nevolved

I came across this question in an online subnetting tutorial and was really confused on how to figure it out.

Given an IP address of 131.107.2.160 and a subnet mask of 255.255.255.192, to which subnet class does the host belong?

A. 131.107.2.32
B. 131.107.2.64
C. 131.107.2.96
D. 131.107.2.128
E. 131.107.2.192

255.255.255.192(4th octet is interesting, ie: != 255)
so
256-192=64(Block size/increment size)

131.107.2.128 = subnet
usable addresses
131.107.2.129-190
131.107.2.191 = broadcast

block size explained:
so you have 64 with this problem, that means you can have the following subnets:

0,64,128,192

you start at 0 and add on the block size to find the subnets

131.107.2.0
131.107.2.64
131.107.2.128
131.107.2.192