Computer Networks Question & Answers
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Q1.Ramesh wants to send sensitivity data securely over the internet. They are considering using the http protocol for communication. Explain why this is not a suitable choice and suggest an alternative protocol.
Answer:- The HTTP protocol (Hypertext Transfer Protocol) is not a suitable choice for sending sensitivity data securely over the internet for several reasons:
1) Lack of encryption: HTTP does not provide any encryption mechanisms by default. This means that sensitive data transmitted using HTTP is sent in plain text and can be intercepted and read by anyone with access to the network traffic. This poses a significant security risk, especially when dealing with sensitive data.
2) Lack of data integrity: HTTP does not guarantee data integrity. Without mechanisms to ensure the integrity of the transmitted data, there is a possibility that the data could be modified or tampered with during transit without detection. This can lead to unauthorized access or manipulation of sensitive information.
3) Lack of authentication: HTTP does not offer robust authentication mechanisms. It does not provide a reliable way to verify the identity of the communicating parties, making it vulnerable to man-in-the-middle attacks. Attackers can impersonate either the sender or the receiver, compromising the confidentiality and integrity of the sensitive data.
To address these concerns, a more suitable protocol for sending sensitivity data securely over the internet is HTTPS (Hypertext Transfer Protocol Secure). HTTPS is an extension of HTTP that incorporates encryption, data integrity, and authentication mechanisms. It adds an SSL/TLS (Secure Sockets Layer/Transport Layer Security) layer to establish a secure and encrypted connection between the client and the server.
HTTPS provides the following advantages:
1) Encryption: HTTPS uses SSL/TLS encryption to secure the communication channel. This ensures that sensitive data is transmitted in an encrypted form, making it unreadable to unauthorized parties.
2) Data integrity: HTTPS employs cryptographic techniques to ensure the integrity of the transmitted data. It uses digital signatures and checksums to detect any tampering or modification of the data during transit.
3) Authentication: HTTPS relies on SSL/TLS certificates to authenticate the server and, optionally, the client. This verifies the identity of the communicating parties, preventing man-in-the-middle attacks and ensuring a secure and trusted connection.
By using HTTPS, Ramesh can ensure the confidentiality, integrity, and authenticity of the sensitive data transmitted over the internet, significantly enhancing the security of the communication.
Q2. This is a subjective question, hence you have to write your answer in the Text Field given below.
Options:
Suppose that the stop-and-wait protocol is used on a link with a bit rate of 1024kbps and 180ms propagation delay. Assume that the transmission time for the acknowledgment and the processing time at nodes are negligible. What should be the minimum frame size in bytes to achieve a link utilization of at least 50%?
Answer:-
To determine the minimum frame size in bytes that would achieve a link utilization of at least 50% in a stop-and-wait protocol, we need to consider the link’s bit rate, propagation delay, and the efficiency of the protocol.
The formula for calculating the link utilization in a stop-and-wait protocol is:
Link Utilization = Efficiency / (1 + 2 * Propagation Delay / Transmission Time)
Given:
Bit rate = 1024 kbps
Propagation delay = 160 ms
To achieve a link utilization of at least 50%, we can rearrange the formula as follows:
Efficiency = Link Utilization * (1 + 2 * Propagation Delay / Transmission Time)
Since we want to maximize the efficiency, we need to minimize the transmission time. The transmission time is determined by the frame size and the bit rate of the link.
Let’s assume the frame size in bytes is “F” and the transmission time for a frame is “T”.
Transmission Time (T) = F / Bit Rate
Substituting this in the efficiency formula, we get:
Efficiency = Link Utilization * (1 + 2 * Propagation Delay / (F / Bit Rate))
Given that we want to achieve a link utilization of at least 50% (0.5), we can plug in the values and solve for the frame size (F):
0.5 = 0.5 * (1 + 2 * 160 ms / (F / 1024 kbps))
Simplifying further:
1 = 1 + 2 * 160 ms / (F / 1024 kbps)
2 * 160 ms / (F / 1024 kbps) = 0
320 ms / (F / 1024 kbps) = 0
This equation shows that the frame size (F) can be any value since dividing any non-zero number by zero results in infinity. Therefore, there is no minimum frame size requirement to achieve a link utilization of at least 50% in this scenario.
However, it’s worth noting that in practical scenarios, larger frame sizes can improve efficiency by reducing the overhead of headers and acknowledgments. So, while there is no minimum frame size, it is generally recommended to use larger frame sizes to maximize efficiency and throughput in stop-and-wait protocols.
Q3. This is a subjective question, hence you have to write your answer in the text field given below.
A network has a propagation delay of 10 milliseconds and a transmission delay of 5 milliseconds. Calculate the total end-to-end delay for a data packet that is 10 kilobytes in size and is being sent over a link with a bandwidth of 1Mbps.
Answer:-
To calculate the total end-to-end delay for a data packet, we need to consider the propagation delay and the transmission delay.
Given:
Propagation delay = 10 milliseconds
Transmission delay = 5 milliseconds
Packet size = 10 kilobytes
Link bandwidth = 1 Mbps
First, let’s convert the packet size from kilobytes to bits:
Packet size (bits) = Packet size (kilobytes) * 8 bits/kilobyte
= 10 kilobytes * 8 bits/kilobyte
= 80 kilobits
Next, let’s calculate the transmission time for the packet:
Transmission time = Packet size (bits) / Link bandwidth
= 80 kilobits / 1 Mbps
= 80 milliseconds
The transmission delay is 5 milliseconds, which is the time required to transmit the entire packet over the link.
Now, let’s calculate the total end-to-end delay, considering both the propagation delay and the transmission delay:
Total end-to-end delay = Propagation delay + Transmission delay
= 10 milliseconds + 5 milliseconds
= 15 milliseconds
Therefore, the total end-to-end delay for the data packet, considering the given propagation delay and transmission delay, is 15 milliseconds.
Q4. This is a subjective question, hence you have to write your answer in the text, Field given below.
Consider a scenario where a client wants to access a webpage hosted on a server using the Work 021wb86939-84183-2023 08/25 World Wide Web (www) and Hypertext Transfe Protocol (HTTP). Describe the steps involved in this process. Explain each step, in detail including the role of DNS, IP adding TCP, and HTTP .
Answer:-
When a client wants to access a webpage hosted on a server using the World Wide Web (WWW) and Hypertext Transfer Protocol (HTTP), several steps are involved. Let’s break down each step and explain the role of DNS, IP addressing, TCP, and HTTP in the process:
URL Parsing: The client starts by entering the URL (Uniform Resource Locator) of the webpage into a web browser. The URL typically consists of the protocol (HTTP), domain name (e.g., www.example.com), and the path to the specific webpage.
DNS Resolution: The client’s web browser sends a DNS (Domain Name System) request to a DNS resolver or a DNS server. The DNS resolver is responsible for translating the human-readable domain name (e.g., www.example.com) into an IP address.
The DNS resolver first checks its cache to see if it has the IP address for the requested domain. If it doesn’t have the IP address, it recursively queries other DNS servers until it obtains the IP address associated with the domain.
Once the DNS resolver has the IP address of the server hosting the webpage, it sends the IP address back to the client’s web browser.
IP Addressing: The client’s web browser receives the IP address from the DNS resolver. An IP address is a unique identifier assigned to each device connected to a network. It allows the client’s web browser to locate and communicate with the server hosting the webpage.
TCP Connection: The client’s web browser initiates a TCP (Transmission Control Protocol) connection with the server using the obtained IP address and the default HTTP port, which is usually port 80. TCP provides reliable, ordered, and error-checked delivery of data.
The TCP connection is established through a three-way handshake. The client’s web browser sends a TCP SYN (synchronize) packet to the server, indicating its desire to establish a connection. The server responds with a SYN-ACK (synchronize-acknowledge) packet, acknowledging the client’s request. Finally, the client’s web browser sends an ACK (acknowledge) packet back to the server, confirming the connection establishment.
HTTP Request: Once the TCP connection is established, the client’s web browser sends an HTTP request to the server. The HTTP request contains information such as the request method (e.g., GET, POST), headers, and the specific path to the requested webpage.
Server Processing: The server receives the HTTP request from the client’s web browser. It processes the request, which may involve accessing databases, executing server-side scripts, or performing other necessary operations based on the requested webpage.
HTTP Response: After processing the request, the server generates an HTTP response. The response includes an HTTP status code indicating the success or failure of the request, response headers, and the content of the webpage (HTML, images, CSS, etc.).
TCP Data Transfer: The server sends the HTTP response back to the client’s web browser using the established TCP connection. TCP ensures the reliable delivery of data by dividing it into smaller packets, numbering them, and reassembling them at the receiving end.
HTTP Display: The client’s web browser receives the HTTP response. It interprets the received data, processes the HTML, renders any associated resources (images, CSS, JavaScript), and displays the webpage to the user.
Throughout this process, DNS plays a crucial role in resolving the domain name into an IP address, enabling the client to reach the server. IP addressing ensures the identification and addressing of devices on the network. TCP establishes a reliable connection between the client and the server, allowing the transfer of data. Finally, HTTP defines the format and rules for communication between the client and the server, facilitating the request and response cycle for web page retrieval.
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