Question

11. Consider a router with the IP address 192.168.1.254/26:

a) Indicate the subnet mask for the above address.

b) Demonstrate calculation of the network address (including binary and decimal values).

c) Write the relevant DHCP subnet configuration text for this network where the first 50 addresses will be dynamically allocated to hosts.

Answer 1

Network Address Block - 192.168.1.254/26

a) Subnet Mask for the above address -

SInce the subnet ID is 26 we will now assign all network bits to 1 and hosts to 0 so from the left we see

11111111.11111111.11111111.11000000

on converting them to decimal we see 255.255.255.192/26

So now the Subnet Mask - 255.255.255.192/26

b) From the above we see that -

Binary Form - 11111111.11111111.11111111.11000000

Decimal form - 255.255.255.192/26

Network Address Block - 192.168.1.254/26

c) Now looking at the host network representation using binary we find that 11111111.11111111.11111111.11000000, now the first host address binary digit we take from right which we have as 64 bit.

Now, this implies that we can have 64 hosts using this subnet

Now we calling the hosts we find

1st Subnet -

• Subnet Address - 192.168.1.192
• Host Range - 192.168.1.193 - 192.168.1.254
• Broadcast Address - 192.168.1.255

Only 1 subnet is possible which is 192.168.1.192/26

Thanks