# Subnetting and masks quick question

Just trying to see if I'm in the ballpark, doing my math and have a grasp of subnets correctly.
With a subnet range if you have a /21 or 255.255.248.0 mask would these ranges be correct?

10.50.16.1 - 10.50.23.254
10.50.40.1 - 10.50.47.254
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Commented:
10.50.0.1   - 10.50.7.254: Hosts (2046), 10.50.0.0: Network, 10.50.7.255: Broadcast
10.50.8.1   - 10.50.15.254: Hosts (2046), 10.50.8.0: Network, 10.50.15.255: Broadcast
10.50.16.1 - 10.50.23.254: Hosts (2046), 10.50.16.0: Network, 10.50.23.255: Broadcast
...
10.50.40.1 - 10.50.47.254: Hosts (2046), 10.50.40.0: Network, 10.50.47.255: Broadcast
...
10.50.248.1 - 10.50.255.254: Hosts (2046), 10.50.248.0: Network, 10.50.255.254: Broadcast

You're correct!
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Commented:
The following is a rather simple subnet calculation method:

With a CIDR mask of /21 we have 8+8+5=21 bits for subnets and 3+8=11 bits for host/net/broadcast addresses.

Given that the first two bytes (10.50 in your example) should remain the same we have 5 bits in the third byte left for 2*2*2*2*2=32 subnets and 11 bits in the third and fourth bytes for 2*2*2*2*2*2*2*2*2*2*2=2048 hosts/net addrs/broadcast addrs.

The remaining 3 bits in byte 3 give a stepping of 2*2*2=8 (or 256-248=8, if you prefer, or 256/32=8), so our subnets are 10.50.0.0, 10.50.8.0, 10.50.16.0 etc.

The lowest useable host address in a subnet is net address + 1, so our host addresses start at 10.50.0.1, 10.50.8.1, 10.50.16.1 etc.

The highest useable host address in a subnet is the last address in the range - 1 (or next network address - 2, if you prefer) ,so our host addresses end at 10.50.7.254, 10.50.15.254, 10.50.23.254 etc.

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Author Commented:
thank you! much appreciated, especially with the detailed answer!
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