It appears that a portion of the question may be missing.
My best answer would be - there are 32 bits total.
22 bits are for the network portion 2^22 = 4194304 subnets
10 bits are for the host portion 2^10 = 1024-2 or 1022 hosts in each
subnet.
This dosen't match any of your answers.
Mileage may vary based on the missing info. If you said it was a class B
address range and you had 22 bit mask, I would say:
Class B has 16 network bits. The mask is 22 bits, so we have 6 bits to play
with for networks.
2^6 is 64 networks. Beware that according to the Cisco Class meterials,
they still say the first and last subnet can not be used - so this leaves 62
usable subnets. We still have 10 host bits, so it is the same as above.
Beware of the termonolgy - In my example, we have 16 network bits, 6 subet
bits and 10 host bits. The network bits are determined by the address
class.
Ed
Edward Moss
CCNP + Voice, CCDP
Message Posted at:
http://www.groupstudy.com/form/read.php?f=7&i=4319&t=4310
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