Router Config

[kuiperbelt]

If you have a network id of 172.88.10.0 (class c) and require 8 host IDs for each subnet, can the subnet mask 255.255.255.248 /29 be assigned?? Asking for a friend. TIA :-)

[ZebuSanctuary]

\\ Can I assign the subnet mask 255.255.255.248 /29 \\

In a word, No.

This a classic mistake and all down to assuming the 3 remaining bits (4th octet) are all available for host addressing.

The subnet mask above would only allow for 6 hosts. See below;

It is stated the netword address is class c, therefore default subnet mask 255.255.255.0 is applicable.

So we are only concerned with the 4th octet i.e., 0 0 0 0 0 0 0 0

Given your assumption '248' the 4th octet will read;

(binary) 1 1 1 1 1 0 0 0

This leaves 3 bits (denary zero through to denary 7) equal to 8 permuataions thus giving 8 available ip addresses.

However, 2 of those ip addresses **MUST be on hand for the network and broadcast addresses. This leaves only 6 host ID addresses.

**Should you omit the network and broadcast addresses from the router config it will never pass data.

To accommodate 8 hosts, the 4th bit in the 4th octet needs to be acquired. The 4th octet will now read 1 1 1 1 0 0 0 0

This gives 16 perms (0000 to 1111). Once again only 14 addresses can be assigned to the hosts, since Network and Broadcast addresses have to be catered for.

Conclusion;

Therefore subnet mask should read 255.255.255.240 /28 and NOT 255.255.255.248 /29

FYI;

First subnet address range 172.88.10.0 - 172.88.10.15 where 172.88.10.1 to 172.88.10.14 can be allocated to hosts. The remaining 2 addresses 172.88.10.0 and 172.88.10.15 are the network and broadcast addresses respectively.

Second subnet address range 172.88.10.16 - 172.88.10.31 where 16 and 31 are the designated network and broadcast addresses respectively.

Third subnet 172.88.10.32 through to 172.88.10.47 and so on... upto and including the 16th subnet.