Assume that you have been assigned 19216811112928 1 How many

Assume that you have been assigned 192.168.111.129/28.

1. How many bits are borrowed to create the subnet field?_______

2. What is the maximum number of subnets that can be created with this number of bits?

3. How many bits can be used to create the host space?

4. What is the maximum number of host addresses available per subnet?

5. What is the subnet mask, in binary and decimal format?

6. Complete the following table and calculate the subnet that this address is on, and define all the other subnets (The range of host addresses on the subnet and the directed broadcast address on the subnet).

Subnet Number

Subnet Address

Range of Host Addresses

Direst Broadcast Address

0

1

2

3

4

5

6

7

…

Last subnet number

7. Answer the following:

                A. What subnet is 192.168.111.129 on?

                B. A junior network administrator is trying to assign 192.168.111.127 as a static IP address for a computer on the network but is getting an error message. Why?

                C. Can 192.168.111.39 be assigned as an IP address?

Solution

Answer:

We have given IP = 192.168.111.129/28

Now , 32 -28 = 4 , means 2^2

1) Two bits are borrowed to create subnet feild as you can see 32 -28 = 4 , 2^2 2 bits

2) 32 -28 = 4 subnets

3) 30 bits are choosen for host address space.

4) Since 4 subnets are there , therefore each subnet contains 64 hosts.

5) 32 - 28 = 4 , 2^2

Since we are choosing 2 bits , our subnet mask will be : 11111111.11111111.11111111.11

255.555.555.192