Question

COMPUTER NETWORKS:

3.   Consider an organization that needs 400 IP addresses. The organization has been assigned two blocks Class C addresses. 200.128.64.0 – 200.128.64.255, and 200.128.65.0 – 200.128.65.255.
Having obtained these 512 addresses, the organization decided to subnet its address space with 3 bit subnetID.

e.   Determine the number of subnets possible in this organization. The organization plans to use zero and all-ones subnets.

Number of subnets possible: __________________________________________________

f.   Determine the number of hosts possible in each subnet.

Number of assignable hosts possible in each subnet: __________________________

g.   Consider host numbered 18 in subnet numbered 4. Determine its IP address.

IP address: _________________________________________________________________

h.   Determine the subnet mask to be set on hosts and routers within this organization.

Subnet Mask in dotted decimal notation: _______________________________________

i.   For subnet numbered 4, determine the network address.

Network address for subnet 4: ________________________________________________

j.   For subnet numbered 4, determine the broadcast address.   

Broadcast address for subnet 4: _______________________________________________

4.   The College has Class B address 137.21.0.0/16. When I used my desktop and connected to the campus network, the ipconfig command showed the following parameters:

IP Address . . . . . . . . . :   137.21.102.233
   Subnet Mask . . . . . . . . . :   255.255.255.0

When I connected a laptop, the ipconfig command showed the following parameters:

IP Address . . . . . . . . . :   137.21.152.226
   Subnet Mask . . . . . . . . . :   255.255.224.0

b.   Consider the laptop. How many bits are used in each of the following three fields?

netID: ______________       subnetID: __________   hostID: _________________

Given the above sizes of the various fields for the 137.21.0.0/16 address space, how many hosts (wireless devices) can be supported on this campus?

Maximum number of wireless devices supported on this campus: _______________

Answer 1

Allotted IPs are:
200.128.64.0 - 200.128.64.255 and
200.128.65.0 - 200.128.65.255.

e. 3 bits are being used for subnetting. Hence,
for first network there will be 2^3 = 8 subnets.
For second network there will be 2^3 = 8 subnets.
Total no of subnets  = 8+8 = 16.

f. After subnetting, no of host bits = 5
Hence, no of assignable hosts = 2^5 - 2 = 32-2 =30

g. I assume this question is about network 200.128.64.0.
Block size = 2^5 = 32.
Subnets will be:
200.128.64.0
200.128.64.32
200.128.64.64
200.128.64.96  <-- This is 4th subnet
.....
....

IP range for 4th subnet:
200.128.64.96 - 200.128.64.127
Host IPs in this subnet:
200.128.64.97 - 200.128.64.126
18th host will be:
200.128.64.(96+18) i.e. 200.128.64.114

h. Subnet mask will be 255.255.255.224

i. Network address for subnet no 4: 
200.128.64.96

j.Broadcast address for subnet no 4:
200.128.64.127


4.
Subnet mask for laptop is:
255.255.224.0
Hence, all bits of first 2 octets are part of network ID.
224 means first 3 bits of 3rd octet are part of network ID. 
Rest belong to host part along with 4th octet.
IP address is 137.21.152.226.
First 2 octets 137 and 21 are part of network address.
First 3 bits of 152 are also part of network address.
152 in binary is 100 11000.
To get the netowork ID make all host bits 0.
So, 3rd octet will become 10000000 i.e. 128 and 4th will be 0.
So subnet ID will be 137.21.128.0
Network ID is already given: 137.21.0
Host ID is 137.21.125.226.

Given network is a /16 network. So 16 bits are left for hosts.
Hence, no of hosts possible according to this network is 2^16 -2.

If I assume that for wireless hosts only above described subnet 
will be used then no of wireless hosts = 2^13 -2.