Answer / vipin.lathawal

209.165.201.30 / 255.255.255.224 will be the Last IP
address in This Subnet

Answer / guest

209.165.201.30

Answer / pankaj dubey

subnet mask=255.255.255.224, range=32
and subent=209.165.201.192-209.165.201.223
and valid host=209.165.201.193-209.165.201.222

Answer / ashraf .kalu

209.165.201.0/27
209.165.201.11100000
209.165.201.224
255.255.255.224
256-224=32
1rst subnet : 209.165.201.32
1rst valid ip : 209.165.201.33
lst valid ip : 209.165.201.62
Broad cast address : 209.165.201.63
next subnet : 209.165.201.64

Answer / anantharaman

209.165.201.254 is the last assignable host address. we cant
assign 209.165.201.255 coz it s the broadcast address. /27
means jump wud be 32.
N/W address Host add range Broadcast addr
209.16.201.0 209.16.201.1-209.16.201.30 209.16.201.31
209.16.201.32 209.16.201.33-209.16.201.62 209.16.201.63
209.16.201.64 209.16.201.65-209.16.201.94 209.16.201.95
209.16.201.96 209.16.201.97-209.16.201.126 209.16.201.127
209.16.201.128 209.16.201.129-209.16.201.158 209.16.201.159
209.16.201.160 209.16.201.161-209.16.201.190 209.16.201.191
209.16.201.192 209.16.201.193-209.16.201.222 209.16.201.223
209.16.201.224 209.16.201.225-209.16.201.254 209.16.201.255
here yo can see the last host addr is 209.16.201.254.

Answer / anantharaman

209.165.201.223/27 this ll be de last host address.. as
255.255.255.224 is the subnet mask, the first 3 portions
define the network portions and the last portion defines the
host portion so the possible host address ranges b'ween
1-224.. of this 1st address wud be the network address and
the last 1 be the broadcast address..so the available host
address ranges from 1-223,and the last host adddress would
be 209.165.201.223

Answer / anantharaman

excuse me vipin.. tat s not de ip address. yo ve given de
subnet mask .. read de question carefully vipin..

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