Question : Learning TCP/IP Advice Required on Subnet Masking

Site A and Site B each have a lease line connected to a VPN Appliance and communicate with each other through a VPN tunnel.

Site A has a network with a subnet ID of 192.168.1.0.  It has a subnet mask of 255.255.255.192.  This should allow a maximum of 62 hosts. So, the range 192.168.1.1 to 192.168.1.62 is available to any host on the network, (server, printer, router etc).  The broadcast ID is 192.168.1.63 which obviously shouldnt be allocated to any host.


Site B has a network with a subnet ID of 192.168.2.0, (this is what I understand by subnetting).  The subnetmask is 255.255.255.192.  

1. Is it correct that both sites need to have the same subnet mask so they can communiate with each other.

2. Is my understanding of subnetting correct in the above example.

4. becausue each site is on its own subnet the traffic for 1.0 will stay in site A and the traffic for 2.0 will stay in site B unless Site A requests info from Site B and vice versa and the traffic travells across the VPN appliance.

3. The "Advanced IP Address calculator v1.1" says that IP 192.168.1.0 with a mask of 255.255.255.192 can have a maximum of 62 hosts which I agree with but a maximum of 2 subnets.   Its says the Subnets/Hosts information is :

ID            Range                  Broadcast
192.1681.64      192.168.2.65-192.168.2.126      192.168.2.127
192.168.2.128      192.168.2.129-129.168.2.1290      192.168.2.191

I dont understand this.  I thought my example of subnetting above was correct.  What does this mean in laymans terms please.




 

Answer : Learning TCP/IP Advice Required on Subnet Masking

its a bug.... I think.  Click the 'allow one subnet bit' check box, and then change the subnet bits to 1 and then back to 2 and you will see there are now 4 possible subnets possible.  

192.168.1.0       192.168.1.1 - 192.168.1.62       192.168.1.63
192.168.1.64     192.168.1.65 - 192.168.1.126    192.168.1.127
192.168.1.128   192.168.1.129 - 192.168.1.190   192.168.1.191
192.168.1.192   192.168.1.193 - 192.168.1.254   192.168.1.255

The calculator is a bit confusing.  If you change 255.255.255.192 into binary, you will see that it becomes:  

11111111.11111111.11111111.11000000

What this means is that any addresses with the same first 26 bits will be considered to be on the same network.  That means that 192.168.2.5, 192.168.1.61, and 192.168.1.66 are all on different networks.  So is 10.5.4.2

now if the mask was 255.255.255.0, then you are basically saying that only the first 24 bits determine the network.  This means that any IP addresses that don't have the first 3 octets identical are on different subnets.  So now 192.168.2.5 and 192.168.1.61 are on different networks, but 192.168.1.61 and 192.168.1.66 are on the same logical subnet.

make any sense?