8 but only 6 are usableOriginally posted by ndmmxiaomayi:All 0s = calculate subnet address
All 1s = calculate broadcast address
How many do you think you have if you use 2^3?
You ask me, I ask who?Originally posted by StarPuppy:8 but only 6 are usable
but is the question stating it wants 8 usable?
i blur ma..dun understand the questionOriginally posted by ndmmxiaomayi:You ask me, I ask who?
You asked this question, isn't it?![]()
naturally..but if its asking for 8 valid subnets..the closest would be 14Originally posted by manyu882:hmm i'm starting to get this.. do u mean u want 8 valid IP addresses?
according to the calculator..
u only can have (these are class c)
1 sub nets each 254hosts
(notice the 2 ends of ip address.. the more subnet, the more hosts wasted)
2 subnets each 126 hosts
4 subnets each 62 hosts
8 subnets each 30 hosts
16subnet each 14 hosts
32 subnet each 6 hosts
if u want to make 8 its impossible... correct me if wrong. either u get 6 or 14