03-24-2011, 09:29 PM
TurkFebruary
06-15-2011, 04:54 PM
b/c you need a max of 850 hosts. 255.255.252.0 leaves 10 host bits. 2^10 = 1024 hosts (1022 are usable)...2^9 would leave you with 510 usable hosts which obviously wouldn't work. Anything more than 10 host bits would be wasting subnet space in this scenario.
You may be looking at the R3 side and thinking 1450 hosts (650 + 800). But it's going to be two separate subnets because with each router port you'll have two separate broadcast domains.
You may be looking at the R3 side and thinking 1450 hosts (650 + 800). But it's going to be two separate subnets because with each router port you'll have two separate broadcast domains.
megathumpzilla
09-24-2011, 08:33 PM
Lowest 3rd octet will give the most networks. 255.255.0.0 would give the most, but it isn't an option. The only parameter is max networks for future growth. If they made 255.255.0.0 an option, you'd realize this and reread the question.
You got conned into trying to make the minimum # of networks to make it work. Different question than the one asked.
You got conned into trying to make the minimum # of networks to make it work. Different question than the one asked.