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How many subnets can be gained by subnetting 172.17.32.0/23
into a /27 mask, and how many usable
host addresses will there be per subnet?
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Answer Posted / robert davenport
Dealing with a 172 IP places you into the Class B Network.
The default subnet mask is 255.255.0.0. However, with the
classless identifiers, or 'prefix length', of /23 and /27
your subnet masks will be as follows:
/23 255.255.254.0
/27 255.255.255.224
Which is broken down into binary as follows:
Default 11111111.11111111.00000000.00000000
/23 11111111.11111111.11111110.00000000
/27 11111111.11111111.11111111.11100000
If you'll notice there are 23 ones in the second subnet mask
and 27 in the third. Hence the /23 and /27. Pay attention
to the 'overlap' (area where the default reads 0 and the
prefix length reads a 1. In this case the last two octets)
For the /23 there are 7 and the /27 there are 11. Remember
these as they will come up later.
Note: There is a proper term for the 'overlap' but I don't
remember it
Beginning at the right of the binary version of the /23
subnet mask. Start with the number 1 and double it for every
'0' you have, stopping ON the first '1'. i.g.
512 256 128 64 32 16 8 4 2 1
/23 11111111.11111111.111111 1 0. 0 0 0 0 0 0 0 0
/27 11111111.11111111.111111 1 1. 1 1 1 0 0 0 0 0
As you can see for the /23 subnet mask, 512 is the TOTAL
number of hosts you can have per subnet. You'll have to
subtract 2 hosts for the network and broadcast addresses
leaving you with 510 USABLE hosts per subnet. So, both look
like this:
Prefix TOTAL USABLE
Length Subnets Subnets
/23 512 510
/27 32 30
Remember the 'overlap' areas from above? if not I'll give
them to you again:
/23 7
/27 11
This number will be used to find out the number of subnets
by using the following formula:
2^n
Where n is the the 'overlap'.
So...
2^7= 2x2x2x2x2x2x2=128 or
2x2=4x2=8x2=16x2=32x2=64x2=128
1 2 3 4 5 6 7
2^11= 2x2x2x2x2x2x2x2x2x2x2=2048 or
2x2=4x2=8x2=16x2=32x2=64x2=128x2=256x2=512x2=1028x2=2048
1 2 3 4 5 6 7 8 9 10 11
Like with the TOTAL hosts, we have to remove 2 from the
TOTAL subnets for network and broadcast ranges.
Now we know:
Prefix TOTAL USABLE
Length Subnets Subnets
/23 128 126
/27 2048 2046
Lets compile everything:
Prefix USABLE USABLE Total Network
Length Subnets Hosts HOSTS
/23 126 510 64,260
/27 2046 30 61,380
So, even though we gain 1920 subnets. We lose 480 hosts per
subnet. Which means we will lose a total potential of 2880
hosts.
| Is This Answer Correct ? | 2 Yes | 8 No |
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