Quote:
Originally Posted by scofiled83
Could u plz explain the solution?
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8 bits per octect (2^8 == 256; less 1 because we start count at 0 == 255)
(for simplicity) IP range: 0-255.0-255.0-255.0-255
your subnet mask of 255.255.255.240 borrows from the last octet which in binary is
11111111.11111111.11111111.11110000
since we know the first 3 octets are 255 we'll focus on the last only
time to get out your ol' binary conversion skills
[code]
d: 128 64 32 16 8 4 2 1
b: 1 1 1 1 0 0 0 0
[code]
128+64+32+16 = 240 (your subnet mask)
that leaves us with only 4 bits left (fields 8, 4, 2, and 1)
8+4+2+1=15
of those 15 possible, the first is the network and the last is the broadcast
we know that kenny has a .7 IP so he falls in the 0-15 network which allows us 1-14 IP for use.
since zathras has a .17 IP we know it is greater than the alotted 14 IPs in the first network (kenny's network) so it must be in another subnet. in this case it just happened to be the next in line. 16-31 (IPs for use 17-30).
I'm sure there are
much better explanations out there but that's my 5 minute explanation.