Broadcast addressing/Subnet mask

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Hi all

Maybe someone can answer me this question. A default class C address has a subnet mask of 255.255.255.0 and you can only have 254 hosts because a host address can not be all 0's or all 1's as it is a broadcast address e.g. 0 = (last octon in decimals) 00000000 and 255 (again last octon in decimals) 11111111.

Say you have an ip address of 131.109.158.x /20 (subnet mask 255.255.240.0) can you give a host ip address of 131.109.158.0 or .255 with this subnet? because when you break it down to network address and host address the host address isnt all 1's or all 0's

IP Address:
131.109.158.0
Subnetmask
255.255.240.0

IP Aaddress and Subnetmask in decimal:
100000011.01101101.10011110.00000000 = 131.109.158.0
11111111.11111111.11110000.00000000 = 255.255.240.0

100000011.01101101.1001<-- network / Host-->1110.00000000

you can see the the host address address isnt all 0's (in red) but it is in the last octon. Can the last octon be all 0's or all 1's if the host address is not on the subnet mask above (255.255.240.0)?

 

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The important part is the host portion of the address, not the last octet (not octon). You can have a host will all 0s or 1s in the last octet, but you can't have a host with all 0s or all 1s in the host portion of the address. Thus, 131.109.158.0/20 and 131.109.158.255/20 are valid host addresses on the 131.109.144.0/20 network. The network address (all 0s) is 131.109.144.0/20, and the broadcast address (all 1s) is 131.109.159.255/20. Everything in between is fair game for host address allocation.

 

Thanks Michael, what you've said backs up what I worked out as I had 144.0 to 159.255 which is great and I just wanted to know if you could have 0 or 255 on the last octet if the network portion isnt all 1's and 0's and from what you said you can so thanks for your reply.

 

Glad to help.