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Home > Archive > CCNA > July 2001 > Please Help!!
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| raydawg30 2001-07-26, 12:48 pm |
| OK, here's my outlook on the question. You need at least 500 subnets so your going to have to use 9 bits for your subnet, 2 raised to the 9th power minus 2 gives you 510 useable subnets. This gives you a subnet that would look like this 255.255.255.128, it looks almost like you should be doing a class C but because you need so many useable subnets your forced to borrow more bits than the 3rd octet has. Anyway, since you borrowed 9 bits for the subnet and you had a grand total of 16 bits to borrow from this leaves you with 16 - 9 = 7 bits to use for hosts, 2 raised to the 7th power minus 2 (for network and broadcast addys) should give you 126 available hosts per subnet. OK there it is in a nutshell, that is if I didn't goof on my math or something.
Good Luck
R. | |
| dmaftei 2001-07-26, 12:53 pm |
| 148.157.0.0 255.255.255.128 will give you 512 subnets, with 128 addresses per subnet (126 of which are usable for hosts). The subnets are:
148.157.0.0, 148.157.0.128,
148.157.1.0, 148.157.1.128,
148.157.2.0, 148.157.2.128,
...
148.157.254.0, 148.157.254.128,
148.157.255.0, 148.157.255.128 | |
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