As network administrator what is the subnet mask that allows 510 hosts given the IP address?

Answer 1

N - network bits

H - host bits

If you are following the first octet rule, this is a class B network with a subnet mask of 255.255.0.0 (N.N.H.H). We can subnet this by "borrowing" some bits from the host portion. There are 16 network bits and 16 host bits. There is a simple formula to calculate the proper subnet mask.

2 to the power of what equals at least 510(2^X)? We have a total of 16 host bits to borrow from. 2^1...2^2...2^3...Etc

2^9= 512 - 2 = 510 host addresses

We subtract two because the network and broadcast address are not usable addresses. As we can see we need at least 9 host bits to get 510 hosts per subnet.

Take 32 and subtract it from the host bits you need. So 32-9=23. Your subnet mask now has 23 network bits instead of 16.

In binary the original subnet mask would be 11111111.11111111.00000000.00000000.

In binary the new subnet mask is 11111111.11111111.11111110.00000000.

If you convert this into dotted decimal form you get 255.255.254.0.

TLDR: 172.30.0.0 - 172.30.1.255

255.255.254.0

Answer 2

The default mask 255.255.0.0 would do it. So would any subnet mask that had at least 10 host bits: 255.255.128.0 255.255.192.0 255.255.224.0 255.255.240.0 255.255.248.0 255.255.252.0 (1,022 hosts)

Answer 3

The idea is that you have enough bits in the host part (the zeroes at the end) to accomodate 510 hosts. That is, 9 or more bits at the right of the subnet mask should be zero. Like this:

11111111.11111111.11111110.00000000

Now just convert that to decimal.

Using the Windows Calculator (or doing it on paper, etc.) the complete subnet mask would be:

255.255.254.0

for your class B example. This would give you exactly 510 hosts. As suggested above, if you needed more than 510 you would add more zeroes at the end of the subnet mask.

Answer 4

As network administrator, what is the subnet mask that allows 510 hosts given the IP address 172.30.0.0?

Answer 5

255.255.255.0