Sunday, February 11, 2007
IP Addressing: Two
Link To : IP Addressing: One
CIDR
Subnet Mask CIDR
---------------------------------------------
255.0.0.0 /8
255.128.0.0 /9
255.192.0.0 /10
255.224.0.0 /11
255.240.0.0 /12
255.248.0.0 /13
255.252.0.0 /14
255.254.0.0 /15
255.255.0.0 /16
255.255.128.0 /17
255.255.192.0 /18
255.255.224.0 /19
255.255.240.0 /20
255.255.248.0 /21
255.255.252.0 /22
255.255.254.0 /23
255.255.255.0 /24
255.255.255.128 /25
255.255.255.192 /26
255.255.255.224 /27
255.255.255.240 /28
255.255.255.248 /29
255.255.255.252 /30
Q. IP: 210.10.10.54
SubnetMask: 255.255.255.0/24
N/W Address ??
Ans: N/w address: 210.10.10.0
Q. 220.92.92.5/24
Ans:
11011100.01011100.01011100.00000101 (Binary of 220.92.92.5) AND
11111111.11111111.11111111.00000000 (Binary of 255.255.255.0)
-----------------------------------------------------------
11011100.01011100.01011100.00000000
i.e: 220.92.92.0 (N/W address)
Subnetting Class C
-----------------------
255.255.255.128 /25 1000 0000
255.255.255.192 /26 1100 0000
255.255.255.224 /27 1110 0000
255.255.255.240 /28 1111 0000
255.255.255.248 /29 1111 1000
255.255.255.252 /30 1111 1100
Ex1: (255.255.255.192/26)
Subnet the n/w address 192.168.10.0
Subnet Mask 255.255.255.192
Ans:
------
n/w address: 192.168.10.0
@How many subnets?
192 is 1100 0000, 2 bits on
i.e 2^2=4 subnets
@How many hosts per subnet?
6 bits off,
i.e (2^6-2)=62 hosts
@Valid subnets?
256-192=64
i.e subnets are 0,64,128,192
n/w' 'Broadcast 'Range
192.168.10.0 192.168.10.63 0-63
192.168.10.64 192.168.10.127 64-127
192.168.10.128 192.168.10.191 128-191
192.168.10.192 192.168.10.255 192-255
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