Using classless addressing, an organization is granted the IP address block 14.1

Question

Using classless addressing, an organization is granted the IP address block 14.17.128.0/23. The network administrator wants to create 16 subnets. All subnets will have the same number of IP addresses

(a) Find the maximum number of physical hosts that can be connected in a subnet.

(b) Find the range of valid IP addresses that can be assigned to physical hosts in the very first subnet (with subnet ID 0) in dotted decimal IP address format.

(c) Find the broadcast address in the very first subnet (with subnet ID 0) in dotted decimal IP address format.

(d) Find the IP address assigned to the first physical host in the last subnet (with subnet ID 15) in dotted decimal IP address format.

Explanation / Answer

We are given that the administrator wants to create 16 subnets with same number of IP addresses.

The IP address given is 14.17.128.0/23 Now this is a class A address so we can find the number of subnets by = 2(given bits - default mask bits) = 2(23-8) = 215 .

We can create 215 subnets . So we will divide this subnets into 16 as we have to create 16 subnets.

Each subnet will have 211 sub-subnets.

a) The number of physical hosts per subnet would be = (2(32 - given bits) - 2) * 211 = (2(32-23) - 2)*211

= 10,44,480

b) The subnet address would be = AND of 14.17.128.0 with subnet mask 255.255.254.0 because we are given 23 bits .

subnet address = 14.17.128.0

The first valid IP address = subnet address + 1

= 14.17.128.1

The last valid IP address in first subnet = subnet address + 1044480

c) The last valid address in b) + 1 would give you the broadcast address.

d) proceed with the calculations made in a),b) and c.

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