Does anyone know how many host bits are needed to guarantee that a subnet could have 9 usable hosts?
I was thinking about around 4, but I'm not sure. Can some shed some light on this?
Does anyone know how many host bits are needed to guarantee that a subnet could have 9 usable hosts?
I was thinking about around 4, but I'm not sure. Can some shed some light on this?
Two subnet addresses (all-ones and all-zeros) can't be used to indicate a host, so with N bits you get up to (2**N - 2)
usable hosts. So, for 9 hosts, 4 bits is correct: it would do up to 14, but 3 bits would do only 6 hosts.
Here are networks that meet the requirement 192.168.1.0 /28
Mask:255.255.255.240 Host/Net - 14
Network Broadcast
192.168.1.0 192.168.1.15
192.168.1.16 192.168.1.31
192.168.1.32 192.168.1.47
192.168.1.48 192.168.1.63
192.168.1.64 192.168.1.79
192.168.1.80 192.168.1.95
192.168.1.96 192.168.1.111
192.168.1.112 192.168.1.127
192.168.1.128 192.168.1.143
192.168.1.144 192.168.1.159
192.168.1.160 192.168.1.175
192.168.1.176 192.168.1.191
192.168.1.192 192.168.1.207
192.168.1.208 192.168.1.223
192.168.1.224 192.168.1.239
192.168.1.240 192.168.1.255