Question

1. Given an IP address of 220.84.9.123 and a subnet mask of 255.255.255.240

Write down the address of the network, show your calculation.

Write also the broadcast address

2. On a network with an IP address of 133.133.133.13 and a subnet mask of 255.255.248.0 what is the network ID?  Show your calculation.  Write also the broadcast address

2. On a Class C network, you have a subnet mask of 255.255.255.192.  

1. How many hosts per subnet do you have?
2. On how many subnets was divided this class C address?  Show your calculations.

4. Your company has leased a Class C network whose network ID is 222.91.37.0. You want to create 30 subnets within this network. Show how many bits, you would need to change, and the range of addresses in one of the subnets.

Answer 1

Answer1. (not specified which question to answer, As per the guidelines I am answering topmost question in the list)

Given IP address- 220.84.9.123 in binary ---> 11011100.01010100.00001001.01111011

Subnet Mask- 255.255.255.240 in binary ---> 11111111.11111111.11111111.11110000

Address of Network, (network id)-the address of a network is calculated as the bitwise AND operation of the IP address and the Subnet Mask.

Address of Network=

IP address- 1 1 0 1 1 1 0 0 . 0 1 0 1 0 1 0 0 . 0 0 0 0 1 0 0 1 . 0 1 1 1 1 0 1 1

subnet mask-   1 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1 . 1 1 1 1 1 1 1 1 . 1 1 1 1 0 0 0 0

Bitwise AND--------------------------------------------------------------------------------------------------------

   1 1 0 1 1 1 0 0 . 0 1 0 1 0 1 0 0 . 0 0 0 0 1 0 0 1 . 0 1 1 1 0 0 0 0 ==> 220.84.9.112

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Hence, Network Address is 220.84.9.112

Broadcast Address- Broadcast address is a special IP address used for broadcasting a message/data over the entire network. In broadcast address the host bits of the network are all 1.

Network Address is 220.84.9.112 ---> Class C address (192-223 class C range)

Class C has 8 host bits and 24 network bits in the binary notation.

Network address-- 1 1 0 1 1 1 0 0 . 0 1 0 1 0 1 0 0 . 0 0 0 0 1 0 0 1 . 0 1 1 1 0 0 0 0

last octet represent the host bits, so make them all 1 for broadcast address.

Broadcast Address-   1 1 0 1 1 1 0 0 . 0 1 0 1 0 1 0 0 . 0 0 0 0 1 0 0 1 . 1 1 1 1 1 1 1 1 ===> 220.84.9.255