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New Member

Subnetting - Valid Hosts

Hi all. A bit of newbie question this, but I can't get the answer straight in my head!

If I had a subnetted network as below, would I get say in subnets #127 & 128 the IP's marked with the arrows as valid host IP's to use?    

10.1.3.65/23255.255.254.0
Subnet #IDRangeBroadcast
12710.0.254.010.0.254.1 - 10.0.255.25410.0.255.25510.0.254.1
10.0.254.2
10.0.254.3
10.0.254.253
10.0.254.254
10.0.254.255
10.0.255.0◄◄
10.0.255.1
10.0.255.2
10.0.255.3
10.0.255.252
10.0.255.253
10.0.255.254
10.0.255.255Broadcast
12810.1.0.010.1.0.1 - 10.1.1.25410.1.1.25510.1.0.1
10.1.0.2
10.1.0.3
10.1.0.255
10.1.1.0◄◄
10.1.1.1
10.1.1.2
10.1.1.253
10.1.1.254
10.1.1.255Broadcast
2 REPLIES
Cisco Employee

Subnetting - Valid Hosts

Hello,

Yes indeed - however you need to make sure mask consistency across all devices within this subnet for it to work.

Nik

New Member

Subnetting - Valid Hosts

Hello Daran,

i asume you are using the following networks:

subnet 127 with network 10.0.254.0/23

subnet 128 with network 10.1.0.0/23

In a /23 network (or 255.255.254.0) you can change the full last byte and the last bit of the third octect so you can have 1 or 0 in X:

in subnet 127      00001010 00000000 1111111X XXXXXXXX

so range from      00001010 00000000 11111110 00000000   (network) 10.0.254.0

to                       00001010 00000000 11111111 11111111   (broadcast) 10.0.255.255

the rest of combinations are hosts so 2^9-2 = 510 hosts

in subnet 128      00001010 00000001 0000000X XXXXXXXX

so range from      00001010 00000001 00000000 00000000   (network) 10.1.0.0

to                       00001010 00000001 00000001 11111111   (broadcast) 10.1.1.255

the rest of combinations are hosts so 2^9-2 = 510 hosts

just make sure that your hosts use the same /23 mask.

If in doubt google for ipcalc.

My 2 cents,

Hernán Márquez

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