Tutorial - IP Addressing

Tutorial - IP Addressing

QUESTION 1

model Answer

What is the default mask for class A,B and class C Networks?

The number of 1s in the default mask is 24 (class C).

Class A 255.0.0.0 11111111.00000000.00000000.00000000

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Class B 255.255.0.0 11111111.11111111.00000000.00000000

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Class C 255.255.255.0 11111111.11111111.11111111.00000000

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The company needs six subnets. This number 6 is not a power of 2. The next number that is a power of 2 is 8 (23 ). We need 3 more 1s in the subnet mask. The total number of 1s in the subnet mask is 27 (24 + 3). The total number of 0s is 5 (32 - 27). The mask is 11111111 11111111 11111111 11100000 or 255.255.255.224 The number of subnets is 8.

QUESTION 2

The number of addresses in each subnet is 25 (5 is the number of 0s) or 32.

Given the network address 133.22.0.0, find the class, the network id, and the range of the addresses

QUESTION 5

model Answer The 1st byte is between 128 and 191. Hence, Class B The block has a netid of 132.22. The addresses range from 133.22.0.0 to 133.23.255.255.

You currently use the default mask for your IP network 192.168.1.0. You need to subnet your network so that you have 30 additional networks, and 4 hosts per network. Is this possible, and what subnet mask should you use?

QUESTION 3

What is the subnetwork address if the destina- Yes it is possible tion address is 200.45.34.56 and the subnet mask we need to 8 = bb (bits borrowed) + bl (bits left) is 255.255.240.0? model Answer 11001000 00101101 00100010 00111000 and 11111111 11111111 11110000 00000000

4 ≤ 2bl − 2 hence bl = 3

11001000 00101101 00100000 00000000

11111111.11111111.11111111.1111000 255.255.255.248.

30 ≤ 2bb − 2 hence bb = 5 so the subnet mask is:

The subnetwork address is 200.45.32.0.

=

QUESTION 6

model Answer

QUESTION 4

We need to make a supernetwork out of 16 class C blocks. What is the supernet mask?

A company is granted the site address 201.70.64.0 (class C). The company needs six subnets. Design the subnets. Find the number of addresses of each subnet.

model Answer We need 16 blocks. For 16 blocks we need to change four 1s to 0s in the default mask. So the mask is 11111111 11111111 11110000 00000000

1

or

version starts with 0); three octets before colon identify host machine; final number identifies port number of email SMTP.

255.255.240.0

• 220.3.6.23:80 First three octets (decimal 220.3.6) identify network as class C (in range 192.n.n.n - 223.n.n.n or binary version starts with 110); octet before colon identifies host machine; final number identifies port number of WWW application (HTTP). (subject guide pp. 14, 22-3).

model Answer

QUESTION 7

A small organization is given a block with the beginning address and the prefix length 205.16.37.24/29 QUESTION 9 (in slash notation). What is the range of the block? model Answer The beginning address is 205.16.37.24.

(a) Explain the concept of Classless Inter-Domain Routing (CIDR).

To find the last address we keep the first 29 bits and change the last 3 bits to 1s.

(b) Given the following IP in CIDR format 192.168.12.0/23.Find the network, the host addresses, and the range of addresses.

Beginning: 11001111 00010000 00100101 00011000 Ending : 11001111 00010000 00100101 00011111 There are only 8 addresses in this block.

[ 10 Marks ]

We can find the range of addresses in Example 17 by another method. We can argue that the length of the suffix is 32 29 or 3. So there are 23 = 8 addresses in this block. If the first address is 205.16.37.24, the last address is 205.16.37.31 (24 + 7 = 31).

Model answer:

(a) CIDR is a classless addressing mode. it work prefix boundary to fall anywhere dress. the number bits identifying the pended to the IP. the remaining bits of with constitute the host.

QUESTION 8 In the TCP/IP model, which layer is responsible for logical addressing of host computers and routing messages? Say as much as you can about how the following sequences identify a particular network, host computer and application, assuming the classfull addressing system:

allows the netwithin the adnetwork is apthe IP address

(b) 192.168.12.0/23. mask 11111111 11111111 11111110 00000000 IP 11000001 10100111 0000100 00000000 network address 193.167.12.0 host address 0.0[2] the range of of host addresses is between 0.0.0.0 to 0.0.1.255

• 129.8.45.13:25 • 124.42.5.45:23 • 220.3.6.23:80 model answer: Logical addressing and routing: Internet or Network (not Network Access) layer. • 129.8.45.13:25 First two octets (decimal 129.8) identify network as class B (in range 128.n.n.n - 191.n.n.n or binary version starts with 10); two octets before colon identify host machine; final number identifies port number of email application (SMTP)). • 124.42.5.45:25 First octet (decimal 124) identifies network as class A (in range 1.n.n.n - 126.n.n.n or binary

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