11-08-2019 02:51 AM - edited 11-08-2019 02:52 AM
We have the following IP's assigned to the VLAN interfaces
interface Vlan100
ip address 172.30.168.2 255.255.255.0
(Range will be 172.30.168.0 - 172.30.168.255)
interface Vlan200
ip address 172.30.169.2 255.255.255.128
(Range will be 172.30.169.0 - 172.30.169.127)
interface Vlan300
ip address 172.30.170.2 255.255.254.0
(Unable to calculate the range for this VLAN)
Could some one please explain me how would I calculate the range of VLAN300 in detail as I am just a beginner.
11-08-2019 03:02 AM
172.30.170.0 - 172.30.171.255
172.30.1010 110|0.0000 0000= 172.30.170.0/23
172.30.1010 110|1.1111 1111= 172.30.171.255/23
Regards
11-08-2019 03:03 AM
172.30.170.2 255.255.254.0 = 172.30.170.2/23
In binary 170 = 1010 1010
The mask [255.255.254.0] makes the last bit of 170 to be don't care/Host bit, so it could be 1010 1010 or 1010 1011. 1010 1011 = 171
The range of IPs for this vlan is 172.30.170.1 to 172.30.171.254.
HTH
11-08-2019 03:03 AM - edited 11-08-2019 03:06 AM
Hello
interface Vlan100
ip address 172.30.168.2 255.255.255.0
(Range will be 172.30.168.0 - 172.30.168.255)
interface Vlan200
ip address 172.30.169.2 255.255.255.128
(Range will be 172.30.169.0 - 172.30.169.127)
interface Vlan300
ip address 172.30.170.2 255.255.254.0
(Unable to calculate the range for this VLAN)
(range will be ip address 172.30.170.0 - 172.30.171.255 )
11111111.11111111.11111110.00000000
in this case you have used one bit to get more hosts and to know how many hosts you will have, look exemple below;
11111111.11111111.1111111 0. 0 0 0 0 0 0 0 0
128 64 32 16 8 4 2 1 = 256
512
if your fourth octect have 256 address available and you get more one, you will have + 256 host... in totally 512 address.
your network address will be: 172.30.170.0 255.255.254.0
11-08-2019 03:05 AM
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