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- : Internet Technology Suggestion 01 :-

**1. Difference between ARP and RARP. Ans:-

2. Identify the similarities and difference between OSI and TCP/IP model. Ans:-**

**3. Differentiate between Classful and Classless addressing with proper example. Ans:-

4. What is Autonomous system? Explain Inter and Intra Autonomous system with example. Ans:-** An autonomous system (AS) consists of several connected Internet Protocol routing prefixes. One or more network operators manage all these routing prefixes in place of a single administrative entity or organization. These IP prefixes have a clearly defined routing policy that states how the autonomous system will exchange the routing data with other nodes and similar systems. Each autonomous system is provided with a unique autonomous system number (ASN) to implement in the Border Gateway Protocol (BGP) routing. Their regional internet registries provide the ASN to the Local Internet Registries (LIRs) and end- user organizations. It will receive blocks of autonomous system numbers from the reassignment

6. A block of address is generated to a small organization. One of the address is 205.116.137.38/26.

• What is the first address in the block?
• What is the last address in the block?
• **Find the number of addresses in the block? Ans:-

1. First Address in the Block:** To find the first address in the block, you set all host bits to 0. In this case, the last octet of the IP address is used for hosts, and the remaining bits are part of the network. So: Binary representation of the given IP address: 11001101.01110100.10001001. Set the last 6 bits to 0 (since it's a /26 subnet): 11001101.01110100.10001001. Convert back to decimal: 205.116.137. So, the first address in the block is 205.116.137.32. 2. Last Address in the Block: To find the last address in the block, you set all host bits to 1: Binary representation of the given IP address: 11001101.01110100.10001001. Set the last 6 bits to 1: 11001101.01110100.10001001. Convert back to decimal: 205.116.137. So, the last address in the block is 205.116.137.63.

3. Number of Addresses in the Block: The number of addresses in a subnet can be calculated using the formula 2 ^ (32−subnet_mask), In this case : 2 ^(32−26) = 2 ^ 6 = 64 However, two addresses in the block are reserved for network and broadcast addresses, so the number of usable addresses for hosts is 64−2=62. Therefore, the block has a total of 62 addresses that can be assigned to hosts. 7. Explain with proper example of Default Mask and Subnet Mask. Identify the role of Sub Mask in network designing. Ans :- Subnet Mask :- A subnet mask is a 32-bit number created by setting the host bits to all 0s and setting network bits to all 1s. In this way, the subnet mask is separated the IP address into the host address and network address. The broadcast address is always assigned to the "255" address, and a network address is always assigned to the "0" address. Since the subnet mask is reserved for a special purpose, it cannot be assigned to the host. The subnetting process is the main goal of the subnet mask. Role of Subnet Mask :- Suppose we have a Class A network that means we have 16 million hosts in a network. The task we have to do is: - Maintenance of such a huge network - Security for the network – For example, we have 4 departments in a company and all of the 4 departments need not access the whole network. For this we need subnetting i.e., dividing a huge network into smaller network. Now every department will have their own network. In case of addressing without subnetting, the process of reaching an address is done by 3 steps - - Identification of the network - Identification of the host - Identification of the process In case of addressing with subnetting, the process of reaching an address is done by 4 steps –

Ans :- The CIDR notation "141.142.108.0/22" represents a block of IP addresses ranging from 141.142.108.0 to 141.142.111.255. This block contains 2^22 addresses. To divide this block as requested: Organization A (Half of the addresses): We need half of the addresses from the /22 block, which corresponds to a /23 block. The new block for Organization A is "141.142.108.0/23," which includes addresses from 141.142.108.0 to 141.142.109.255. Organization B (A quarter of the addresses): A quarter of the addresses from the /22 block corresponds to a /24 block. The new block for Organization B is "141.142.110.0/24," which includes addresses from 141.142.110.0 to 141.142.110.255. Remaining Addresses (ISP): The remaining addresses are from the range 141.142.111.0 to 141.142.111.255, which can be represented as "141.142.111.0/24." So, the valid allocation is as follows:

• Organization A: 141.142.108.0/
• Organization B: 141.142.110.0/
• ISP: 141.142.111.0/ Each of these allocations respects the requirements and ensures that the IP addresses are properly divided among the three entities. 10. Explain the working principle of TCP. Ans:- TCP (Transmission Control Protocol) is one of the main protocols of the Internet protocol suite. It lies between the Application and Network Layers which are used in providing reliable delivery services. It is a connection-oriented protocol for communications that helps in the exchange of messages between different devices over a network. The Internet Protocol (IP), which establishes the technique for sending data packets between computers, works with TCP. Working of TCP To make sure that each message reaches its target location intact, the TCP/IP model breaks down the data into small bundles and afterward reassembles the bundles into the original message on the opposite end. Sending the information in little bundles of information makes it simpler to maintain efficiency as opposed to sending everything in one go.

After a particular message is broken down into bundles, these bundles may travel along multiple routes if one route is jammed but the destination remains the same. For example, When a user requests a web page on the internet, somewhere in the world, the server processes that request and sends back an HTML Page to that user. The server makes use of a protocol called the HTTP Protocol. The HTTP then requests the TCP layer to set the required connection and send the HTML file. Now, the TCP breaks the data into small packets and forwards it toward the Internet Protocol (IP) layer. The packets are then sent to the destination through different routes. The TCP layer in the user’s system waits for the transmission to get finished and acknowledges once all packets have been received. Advantages

• It is a reliable protocol.
• It provides an error-checking mechanism as well as one for recovery.
• It gives flow control.
• It makes sure that the data reaches the proper destination in the exact order that it was sent. **11. Difference of TCP and UDP Ans :-

12. BOOTP Protocol Ans:-** The Bootstrap Protocol (BOOTP) is a legacy computer networking protocol used to automatically assign IP addresses to devices on a network. BOOTP was developed in the 1980s as a way to assign IP addresses to diskless workstations without the need for a DHCP server. BOOTP works by allowing a device to broadcast a request for an IP address on the network. The BOOTP server, which is typically a router or a dedicated BOOTP server, receives the request and

14. Describe BUS, MESH, RING, STAR Tropology and HDLC , RARP Ans:-

• The documents stored on the intranet are much more secure. Disadvantages of Intranet
• The expense of actualizing intranets is normally high.
• The staff of the company or organization require special training to know how to use the system.
• Data overloading.
• Although the intranet provides good security, but it still lacks in some places. 16. Justify “ In Go-Back-N-ARQ the size of sender window must be less” Ans:- Go-Back-N-ARQ :- Stop and wait ARQ mechanism does not utilize the resources at their best. When the acknowledgement is received, the sender sits idle and does nothing. In Go-Back-N ARQ method, both sender and receiver maintain a window. The sending-window size enables the sender to send multiple frames without receiving the acknowledgement of the previous ones. The receiving-window enables the receiver to receive multiple frames and acknowledge them. The receiver keeps track of incoming frame’s sequence number. When the sender sends all the frames in window, it checks up to what sequence number it has received positive acknowledgement. If all frames are positively acknowledged, the sender sends next set of frames. If sender finds that it has received NACK or has not receive any ACK for a particular frame, it retransmits all the frames after which it does not receive any positive ACK. Justification :- The above diagram is for justification
  • We choose m = 2, which means the size of the window can be 2m - 1, or 3. We can now show why the size of the send window must be less than 2m.
  • If the size of the window is 3 (less than 22) and all three acknowledgments are lost, the frame 0 timer expires and all three frames are resent.

• The receiver is now expecting frame 3, not frame 0, so the duplicate frame is correctly discarded. On the other hand, if the size of the window is 4 (equal to 22) and all acknowledgments are lost, the sender will send a duplicate of frame O.
• However, this time the window of the receiver expects to receive frame 0, so it accepts frame 0, not as a duplicate, but as the first frame in the next cycle. This is an error.

Important Questions :- 1, 2, 3, 5, 6, 7, 8, 9, 11, 12, 13, 14, 16