| |
https://www.exampointers.com/ccna/sub.php?ip=44.220.184.63&mask=30
https://www.exampointers.com/ccna/sub.php?ip=51.38.71.166&mask=31
https://www.exampointers.com/ccna/sub.php?ip=172.16.0.31&mask=30
https://www.exampointers.com/ccna/sub.php?ip=224.1.0.39&mask=26
https://www.exampointers.com/ipv4/44.220.184.63/30
https://www.exampointers.com/ipv4/51.38.71.166/31
https://www.exampointers.com/ipv4/172.16.0.31/30
https://www.exampointers.com/ipv4/224.1.0.39/26
/0 0.0.0.0 /1 128.0.0.0 /2 192.0.0.0 /3 224.0.0.0
/4 240.0.0.0 /5 248.0.0.0 /6 252.0.0.0 /7 254.0.0.0
/8 255.0.0.0 /9 255.128.0.0 /10 255.192.0.0 /11 255.224.0.0
/12 255.240.0.0 /13 255.248.0.0 /14 255.252.0.0 /15 255.254.0.0
/16 255.255.0.0 /17 255.255.128.0 /18 255.255.192.0 /19 255.255.224.0
/20 255.255.240.0 /21 255.255.248.0 /22 255.255.252.0 /23 255.255.254.0
/24 255.255.255.0 /25 255.255.255.128 /26 255.255.255.192 /27 255.255.255.224
/28 255.255.255.240 /29 255.255.255.248 /30 255.255.255.252 /31 255.255.255.254
/32 255.255.255.255
My method of working out subnetting for masks that are > /24
If we take the address 192.168.4.18/29
/29 is just the number of bits used for the subnet mask. So the subnet mask will
be 255.255.255.248.
If we use the last octet 248. 256-248 = 8, this means the last octet 0-255 is broken up into 32 groups of 8.
Network Address (18 taken from address above)
integer(18/8) = 2
* 8
= 16
192.168.4.16
Broadcast Address (18 taken from address above)
integer(18/8) = 2
+ 1
= 3
* 8
= 24
- 1
= 23
192.168.4.23
Inverse Mask
/0 255.255.255.255 /1 127.255.255.255 /2 63.255.255.255 /3 31.255.255.255
/4 15.255.255.255 /5 7.255.255.255 /6 3.255.255.255 /7 1.255.255.255
/8 0.255.255.255 /9 0.127.255.255 /10 0.63.255.255 /11 0.31.255.255
/12 0.15.255.255 /13 0.7.255.255 /14 0.3.255.255 /15 0.1.255.255
/16 0.0.255.255 /17 0.0.127.255 /18 0.0.63.255 /19 0.0.31.255
/20 0.0.15.255 /21 0.0.7.255 /22 0.0.3.255 /23 0.0.1.255
/24 0.0.0.255 /25 0.0.0.127 /26 0.0.0.63 /27 0.0.0.31
/28 0.0.0.15 /29 0.0.0.7 /30 0.0.0.3 /31 0.0.0.1
/32 0.0.0.0
|
