Key Concepts in Computer Networking and Internet Protocols

Posted on Dec 11, 2024 in Computers

Frequently Asked Questions

“The term for an internet connection is the amount of data that can be transferred in 1 second, measured in bits per second (bps):”
Answer: B. Throughput
“If (throughput of the server is greater than the throughput of the client ), then the average end-to-end throughput is:”
Answer: C. Min
“Identify the correct order in which the following steps take place in an interaction between a browser and a web server to establish a TCP connection. (Assume RTT = 2):”
Answer: A. 5, 1, 2, 3, 4
“Internet access by transmitting digital data over the wires of a local telephone network is provided by:”
Answer: D. Digital Subscriber Line (DSL)
“When creating datagram to send into UDP socket, must specify:”
Answer: C. Source IP-Port, Destination IP-Port
“The domain name system (DNS) is maintained by:”
Answer: C. A distributed database system
“The TCP protocol can deliver ___ out of order segments to the process:”
Answer: A. Some
“Which is not an application layer protocol?”
Answer: D. IP and ICMP
“DNS is an acronym for:”
Answer: A. Domain Name Service
“Application layer offers ___ service”
Answer: A) Process to process
“Internet works on?”
Answer: B) Packet switching
“What is the keyword used to define a domain alias in DNS?”
Answer: A) CName
“Which is a time-sensitive service?”
Answer: Internet telephony
“Consider different activities related to email: m1: Send an email from a mail client to a mail server, m2: Download an email from mailbox server to a mail client, m3: Checking email in a web browser:”
Answer: m1: SMTP, m2: IMAP, m3: HTTP
“Who determines source-destination paths taken by packets?”
Answer: Routing algorithm

Image Transmission Time Calculations

Problem Description:

An image has a size of 1200×1800 pixels (Full HD) with true color, which means that 2 Bytes per pixel are used for the color information. Calculate how long it takes to transmit the uncompressed image via different connection speeds:

56 kbps Modem connection
10 Mbps DSL connection
100 Mbps Ethernet connection
1 Gbps Ethernet connection

Given:

1 Byte = 8 Bits
1 kbps = 1,000 Bits per second
1 Mbps = 1,000,000 Bits per second
1 Gbps = 1,000,000,000 Bits per second

Calculations and Results:

Total Pixels = 1200 x 1800 = 2,160,000 pixels
Total Image Size = 2,160,000 pixels x 2 Bytes/pixel = 4,320,000 Bytes
Total Image Size in Bits = 4,320,000 Bytes x 8 = 34,560,000 Bits

Transmission Times:

(a) 56 kbps Modem:
- Time = 34,560,000 / 56,000 = 617.14 seconds
(b) 10 Mbps DSL:
- Time = 34,560,000 / 10,000,000 = 3.46 seconds
(c) 100 Mbps Ethernet:
- Time = 34,560,000 / 100,000,000 = 0.346 seconds
(d) 1 Gbps Ethernet:
- Time = 34,560,000 / 1,000,000,000 = 0.0346 seconds

Difference between HTTP and SMTP

HTTP (Hypertext Transfer Protocol):
- Used to transfer hypertext (web pages) between a client (browser) and a server.
- Stateless and request-response protocol.
- Operates on port 80 (by default).
SMTP (Simple Mail Transfer Protocol):
- Used to send emails between mail servers or from a client to a mail server.
- Primarily focuses on email delivery.
- Operates on port 25 (by default).

Three Major Components of E-Mail

User Agents (UA):
- Examples: Microsoft Outlook, Gmail, Thunderbird.
- Applications that allow users to send, read, and organize emails.
Mail Servers:
- Store and manage incoming and outgoing emails.
- Examples: SMTP server for outgoing mail, POP3/IMAP server for incoming mail.
Protocols:
- Rules used for sending and receiving emails.
- Examples: SMTP (sending), POP3, IMAP (receiving).

Difference between Application Layer and Network Layer

Application Layer:
- Topmost layer in the OSI model.
- Focuses on providing network services to end users (e.g., email, file transfer, web browsing).
- Examples of protocols: HTTP, SMTP, FTP.
Network Layer:
- Third layer in the OSI model.
- Responsible for data routing, packet forwarding, and logical addressing (IP addresses).
- Examples of protocols: IP, ICMP.

Difference between Forwarding and Routing

Forwarding:
- Process of moving packets from an input interface to the appropriate output interface within a router.
- A local, per-router process based on the forwarding table.
Routing:
- Process of determining the optimal path a packet should take across the network.
- Involves creating and updating the routing table using algorithms like Dijkstra or Bellman-Ford.

Difference between Encapsulation and Decapsulation

Encapsulation:
- Process of adding headers (and sometimes trailers) to data as it moves down the OSI layers.
- Happens at the sender’s side.
- Ensures data is properly formatted for transmission.
Decapsulation:
- Process of removing headers (and trailers) as data moves up the OSI layers.
- Happens at the receiver’s side.
- Ensures the original message is retrieved.

NASA Mars Communication Link

(a) Round Trip Time (RTT):
- Distance = 55,000,000 km = 55 x 10⁹ meters.
- Speed of light = 299,792,458 m/s.
- RTT = 2 x (Distance / Speed) = 2 x (55 x 10⁹ / 299,792,458) seconds
- RTT ≈ 366.18 seconds.
(b) Bandwidth-Delay Product:
- Bandwidth = 128 kbps = 128 x 10³ bps.
- Delay = RTT / 2 = 366.18 / 2 seconds = 183.09 seconds.
- Bandwidth-delay product = Bandwidth x Delay = 128 x 10³ x 183.09 bits.
- Bandwidth-delay product = 23,426,880 bits.
(d) 1 Gbps Ethernet Connection:
- Distance = 100 meters.
- RTT = 2 x (Distance / Speed of Light in Fiber (2/3 of c)).
- c = 299,792,458 m/s
- RTT = 1 microsecond.

Network Configuration Problem

Determine the bottleneck link:
- The bottleneck is the link with the lowest transmission rate, which is Link1 (2 Mbps).
Time to transfer an 8 Mb file:
- File size = 8 Mb = 8 x 10⁶ bits.
- Transfer rate = 2 Mbps = 2 x 10⁶ bits/second.
- Time = File Size / Transfer Rate = 8 x 10⁶ / 2 x 10⁶ = 4 seconds
Total delay with RTTs:
- RTT for the bottleneck link = 4 ms = 0.004 seconds.
- Considering HTTP request-response cycles:
- Assume 1 RTT for the request and an additional RTT for starting the download. Total RTT delay = 0.004 + 0.004 = 0.008 seconds.
- Total time = File transfer time + RTTs = 4 + 0.008 = 4.008 seconds.

Exercise 1: Network Layers, Protocols, and Units

Application Layer
- Protocols: HTTP, SMTP, FTP, DNS
- Protocol Data Unit (PDU): Messages
- Addressing: Domain names (e.g., www.example.com)
Transport Layer
- Protocols: TCP, UDP
- Protocol Data Unit (PDU): Segments
- Addressing: Port numbers (e.g., 80 for HTTP, 25 for SMTP)
Network Layer
- Protocols: IP (IPv4, IPv6), ICMP
- Protocol Data Unit (PDU): Packets
- Addressing: IP addresses (e.g., 192.168.1.1)
Link Layer
- Protocols: Ethernet, Wi-Fi, PPP
- Protocol Data Unit (PDU): Frames
- Addressing: MAC addresses
Physical Layer
- Protocols: IEEE 802.3 (Ethernet), IEEE 802.11 (Wi-Fi)
- Protocol Data Unit (PDU): Bits
- Addressing: None (transmission of raw data)

Exercise 2: Email Protocols and Mechanisms

(a) At point 6 in the diagram, what protocol is being used?
- The protocol at point 6 is HTTP or IMAP (depending on whether the user is retrieving the email from a web-based service or a mail client).
(b) At point 4 in the diagram, what protocol is being used?
- The protocol at point 4 is SMTP (Simple Mail Transfer Protocol).
(c) At point 2 in the diagram, what protocol is being used?
- The protocol at point 2 is SMTP.
(d) Does SMTP use TCP or UDP?
- SMTP uses TCP, typically on port 25, 465, or 587.
(e) Is SMTP a ‘push’ or ‘pull’ protocol? Is IMAP a ‘push’ or ‘pull’ protocol?
- SMTP is a push protocol because it actively sends (pushes) mail to the receiving server.
- IMAP is a pull protocol because it retrieves (pulls) emails from the server to the client.
(f) What is the difference between HTTP and SMTP?
- HTTP (Hypertext Transfer Protocol):
  - Used for transferring web pages and related data over the internet.
  - Works in a client-server model, primarily for browsing websites.
- SMTP (Simple Mail Transfer Protocol):
  - Used for transferring emails between mail servers and sending emails from clients to servers.
  - Specifically designed for email communication.
(g) Three major components of email:
1. User Agent (UA): Allows users to compose, send, and read emails (e.g., Outlook, Gmail).
2. Mail Transfer Agent (MTA): Transfers emails between servers using SMTP.
3. Mail Delivery Agent (MDA): Delivers email to the recipient’s mailbox (e.g., using IMAP or POP3).

Exercise 3: Packet Retransmission

(a) Calculate how many packets will be re-transmitted using Go-Back-N (GB3)?

Given:

Total packets: 10
Every 5th packet is lost.

In Go-Back-N, if a packet is lost, all subsequent packets in the window are retransmitted. Assuming a window size of 3 (GB3):

Packets 1–5 are sent. Packet 5 is lost, so packets 5, 6, and 7 will be retransmitted.
Packets 8–10 are sent after retransmission of packet 5. No further losses occur.

Answer:

Packets 5, 6, and 7 are retransmitted.
Total retransmissions = 3.

(c) Calculate how many packets will be re-transmitted using Selective Repeat (SR) with W_s = 3.

In Selective Repeat, only the lost packet is retransmitted. Given that only the 5th packet is lost:

Only Packet 5 will be retransmitted.