Here are the solutions for the given questions in a simple and easy-to-understand format.
Q1) Attempt any EIGHT of the following (out of TEN): [8×1=8]
a) Define Bandwidth.
Bandwidth is the maximum amount of data that can be transmitted over a communication channel in a given time, measured in bits per second (bps).
b) What is Throughput?
Throughput is the actual amount of data successfully transmitted over a network in a specific time period. It is always less than or equal to the bandwidth.
c) What is Jitter?
Jitter is the variation in the time delay of received data packets over a network. High jitter can lead to poor audio and video quality in real-time communication.
d) List the control access protocols.
Control access protocols manage how devices share a communication channel. Common protocols include:
- CSMA/CD (Carrier Sense Multiple Access with Collision Detection)
- CSMA/CA (Carrier Sense Multiple Access with Collision Avoidance)
- Token Passing
- Polling
e) Define Packetizing.
Packetizing is the process of breaking down data into smaller packets for efficient transmission over a network. Each packet carries part of the data along with necessary headers.
f) Write IPv6 address space.
IPv6 has a 128-bit address space, allowing for 2¹²⁸ unique IP addresses, which is significantly larger than IPv4.
g) List UDP services.
UDP (User Datagram Protocol) provides the following services:
- Connectionless Service: No need to establish a connection before sending data.
- Fast Data Transfer: Suitable for real-time applications.
- Minimal Overhead: No error checking or flow control, making it lightweight.
- Multiplexing: Uses port numbers to distinguish different applications.
h) Write the list of states for TCP.
TCP connection states include:
- LISTEN
- SYN-SENT
- SYN-RECEIVED
- ESTABLISHED
- FIN-WAIT-1
- FIN-WAIT-2
- CLOSE-WAIT
- CLOSING
- LAST-ACK
- TIME-WAIT
- CLOSED
i) What is Full Duplex Communication?
Full Duplex communication allows data to be sent and received simultaneously between two devices. Example: Telephone communication.
j) Write the registered ports.
Registered Ports range from 1024 to 49151 and are assigned to user processes or applications.
Q2) Attempt any FOUR of the following (out of FIVE): [4×2=8]
a) Write a note on BSS.
BSS (Basic Service Set) is the basic building block of a wireless LAN. It consists of:
- One Access Point (AP) and multiple wireless devices (stations).
- Devices communicate with the AP to access the network.
- Types of BSS:
- Independent BSS (IBSS): No AP, devices communicate directly (Ad-hoc network).
- Infrastructure BSS: Devices communicate through an AP.
b) What are the different layers in the TCP/IP reference model?
The TCP/IP model consists of four layers:
- Application Layer: Provides services to user applications (HTTP, FTP).
- Transport Layer: Ensures reliable or unreliable data transfer (TCP, UDP).
- Internet Layer: Handles logical addressing and routing (IP).
- Network Access Layer: Manages data transfer across physical networks.
c) Write the Net ID & Host ID of IP address: 117.149.29.4
- The IP address 117.149.29.4 belongs to Class A (range: 1.0.0.0 to 126.0.0.0).
- Net ID: 117 (First octet).
- Host ID: 149.29.4 (Remaining three octets).
d) What are the two sub-layers of the data link layer?
The Data Link Layer is divided into two sub-layers:
- LLC (Logical Link Control) Sub-layer:
- Responsible for error detection and flow control.
- MAC (Media Access Control) Sub-layer:
- Manages access to the physical transmission medium and data frame creation.
e) Write the different control bits or flags in the control field of TCP segment.
TCP control flags in the header are:
- URG (Urgent): Data is urgent.
- ACK (Acknowledgment): Acknowledgment field is valid.
- PSH (Push): Push data to the receiving application.
- RST (Reset): Resets the connection.
- SYN (Synchronize): Initiates a new connection.
- FIN (Finish): Indicates the end of data transmission.
This covers all the questions with straightforward explanations. Let me know if you need more details or examples!
Here are the solutions for the given questions in a clear and simple manner:
Q3) Attempt any TWO of the following (out of THREE): [2×4=8]
a) What is the propagation time for a 2.5 Kbyte message if the bandwidth of the network is 1 Gbps?
Given:
- Message size = 2.5 KB = 2.5 × 1024 × 8 = 20,480 bits
- Bandwidth = 1 Gbps = 1 × 10⁹ bits per second
- Distance = 12,000 km = 12,000,000 meters
- Speed of light in the medium = 2.4 × 10⁸ m/s
Propagation Time formula: Propagation Time=DistancePropagation Speed\text{Propagation Time} = \frac{\text{Distance}}{\text{Propagation Speed}} =12,000,0002.4×108=0.05 seconds= \frac{12,000,000}{2.4 \times 10^8} = 0.05 \text{ seconds}
Answer: The propagation time is 0.05 seconds.
b) Write some important design issues of the Data Link Layer.
- Framing:
- Dividing the data into frames for proper transmission.
- Error Control:
- Detecting and correcting errors in the transmitted data.
- Flow Control:
- Ensuring that the sender does not overwhelm the receiver with data.
- Access Control:
- Managing access to the shared communication medium.
- Addressing:
- Adding physical addresses (MAC addresses) to the frames for correct delivery.
c) What are the main properties of routing?
- Correctness:
- The algorithm must provide accurate routing decisions.
- Simplicity:
- The routing mechanism should be simple and easy to implement.
- Robustness:
- The algorithm should adapt to changes in the network, like link failures.
- Stability:
- The routing algorithm must prevent oscillations and ensure consistent routes.
- Fairness:
- All data packets should get equal chances to be transmitted.
- Optimality:
- The chosen path should be the shortest or fastest for data delivery.
Q4) Attempt any TWO of the following (out of THREE): [2×4=8]
a) Write the difference between TCP & UDP.
| Feature | TCP (Transmission Control Protocol) | UDP (User Datagram Protocol) |
|---|---|---|
| Connection | Connection-oriented | Connectionless |
| Reliability | Reliable (error checking and recovery) | Unreliable (no error checking) |
| Speed | Slower due to reliability features | Faster due to no reliability features |
| Data Order | Ensures data arrives in order | No guarantee of order |
| Overhead | High | Low |
| Use Cases | Email, Web browsing | Video streaming, Online gaming |
b) Explain sliding window in TCP.
- The Sliding Window Protocol is used for flow control in TCP.
- It allows multiple packets to be sent before needing an acknowledgment, improving efficiency.
- Key Points:
- Sender’s Window: Defines how many packets can be sent without receiving an acknowledgment.
- Receiver’s Window: Indicates how much buffer space is available to receive data.
- As packets are acknowledged, the window “slides” forward to allow more data to be sent.
- This ensures smooth and controlled data transfer.
c) Write the base header format of IPv6.
The IPv6 base header is 40 bytes long and includes the following fields:
- Version (4 bits): Specifies the IP version (IPv6 = 6).
- Traffic Class (8 bits): Identifies the packet’s priority.
- Flow Label (20 bits): Used to label packets for special handling.
- Payload Length (16 bits): Size of the data following the header.
- Next Header (8 bits): Identifies the type of the next header (like TCP, UDP).
- Hop Limit (8 bits): Similar to TTL, decreases with each hop.
- Source Address (128 bits): The IPv6 address of the sender.
- Destination Address (128 bits): The IPv6 address of the receiver.
Q5) Attempt any ONE of the following (out of TWO): [1×3=3]
a) Describe bus topology in detail.
- Bus topology is a network setup where all devices are connected to a single central cable (called the backbone).
- Data is sent in both directions, but only the intended recipient accepts the data.
Advantages:
- Simple and easy to set up.
- Cost-effective for small networks.
- Requires less cable compared to other topologies.
Disadvantages:
- If the backbone fails, the whole network is affected.
- Difficult to troubleshoot.
- Limited cable length and number of devices.
- Performance decreases with more devices.
b) Write the difference between IPv4 and IPv6.
| Feature | IPv4 | IPv6 |
|---|---|---|
| Address Length | 32 bits | 128 bits |
| Address Format | Decimal (e.g., 192.168.1.1) | Hexadecimal (e.g., 2001:0db8::1) |
| Address Space | Around 4.3 billion addresses | Virtually unlimited (2^128 addresses) |
| Header Size | 20-60 bytes | Fixed 40 bytes |
| Security | Optional (via IPSec) | Mandatory (built-in IPSec) |
| NAT (Network Address Translation) | Required due to limited addresses | Not required (sufficient addresses) |
| Broadcasting | Supported | Not supported (uses multicast instead) |
This detailed explanation covers all the questions. Let me know if you need more details or examples for any specific topic!