BIT254 · Exam intelligence
Network and Data Communications important questions
From 6 past TU papers: which questions keep coming back, how much they carry, and what is most likely to show up next. Every question links to a model answer.
Most likely in the next examStatistical
Ranked by how often a topic is asked, its marks weight, and whether it is due after skipping the 2082 paper. No guarantees; study the whole syllabus.
1asked 3xavg 7 marks · due (skipped 2082) · Time Division MultiplexingAnswerHideExplain the concept of TDMA with a neat diagram. [5]
Explain the concept of TDMA with a neat diagram. [5]
TDMA (Time Division Multiple Access) is a channel access method used in shared medium networks where multiple users share the same frequency channel by dividing the signal into different time slots. Each user is assigned a specific time ...
2asked 3xavg 5 marks · due (skipped 2082) · MAC address definitionAnswerHideWhat is MAC-address? The message sequence is 1011011 and generator polynomial $G(X) = x^3 + x^2 + 1$. Calculate the transmitted encoded frame [5]
What is MAC-address? The message sequence is 1011011 and generator polynomial $G(X) = x^3 + x^2 + 1$. Calculate the transmitted encoded frame [5]
MAC Address and CRC Calculation
MAC Address (2 marks)
A MAC (Media Access Control) address is a unique physical hardware address permanently assigned to a Network Interface Card (NIC) by its manufacturer. It operates at the Data Link Layer (Layer 2) of the OSI model and is used to uniquely identify a device on a local network.
Key features:
- It is a 48-bit (6-byte) address, normally written in hexadecimal.
- Example:
00:1A:2B:3C:4D:5E - The first 3 bytes (24 bits) form the OUI (Organizationally Unique Identifier) identifying the manufacturer.
- The last 3 bytes identify the specific device.
- Also called the physical / hardware address; usually burned into the NIC's ROM and globally unique.
CRC Calculation (3 marks)
Given data
- Message $M$ =
1011011 - Generator $G(x) = x^3 + x^2 + 1 \Rightarrow$ binary
1101 - Degree $r = 3$
Step 1: Append $r = 3$ zeros
$$ M' = 1011011\underbrace{000}_{3\text{ zeros}} = 1011011000 $$
Step 2: Modulo-2 (XOR) division by 1101
Dividend bits: 1 0 1 1 0 1 1 0 0 0
1101010 <-- quotient (not required)
┌───────────
1101│1011011000
1101
────
0110 1 bring down bit → 1101
1101
────
0000 11 bring down bits → 0011
(0011 < 1101) → q=0, bring down → 0111
(0111 < 1101) → q=0, bring down → 1110
1101
────
0011 0 bring down → 0110
(0110 < 1101) → q=0 (no more bits)
Remainder = 010
Working the XOR steps cleanly:
| Current bits | XOR with 1101 | Result |
|---|---|---|
| 1011 | 1101 | 0110 |
| bring 0 → 1100 | 1101 | 0001 |
| bring 1 → 0011 | - (0011<1101, q=0) | 0011 |
| bring 1 → 0111 | - (0111<1101, q=0) | 0111 |
| bring 0 → 1110 | 1101 | 0011 |
| bring 0 → 0110 | - (0110<1101, q=0) | 0110 |
| bring 0 → 0100 | - (last bit, 0100<1101) | 100 |
Let me track the final bits exactly. After processing all 10 dividend bits:
- Start:
1011→ XOR 1101 =0110 - +bit(0):
1100→ XOR 1101 =0001 - +bit(1):
0011→ < divisor, q=0 - +bit(1):
0111→ < divisor, q=0 - +bit(0):
1110→ XOR 1101 =0011 - +bit(0):
0110→ < divisor, q=0 - +bit(0):
1100→ XOR 1101 =0001
Final 3-bit remainder (CRC) = 001
Verification
Divide the 3 remaining bits carefully; the last register content after all 10 bits gives remainder = 001.
Check: transmitted frame $T = 1011011,001$. Dividing $1011011001$ by 1101 must give remainder 000:
1011011001
1101
────
0110 → 1100 → 0001 → 0011 → 0111 → 1110→0011 → 0110 → 1101 → 0000
$1101 \oplus 1101 = 0000$ ✓ remainder is 0, so the CRC is correct.
Step 3: Transmitted frame
$$ T = \text{Message} + \text{CRC} = 1011011 , \big| , 001 = \boxed{1011011001} $$
CRC (remainder) = 001
Transmitted encoded frame = 1011011001
3asked 4xavg 6 marks · Circuit switching advantages and disadvantagesAnswerHideDifferentiate between packet and circuit switched network.Explain the layers of OSI Reference Model in brief.[4+6]
Differentiate between packet and circuit switched network.Explain the layers of OSI Reference Model in brief.[4+6]
--- (a) Difference Between Packet Switched and Circuit Switched Network Feature Circuit Switched Network Packet Switched Network --------- Connection A dedicated physical path is established before communication begins No dedicated path;...
4asked 4xavg 6 marks · Attenuation distortion and noiseAnswerHideWhat are the major differences between noise, distortion and attenuation? [5]
What are the major differences between noise, distortion and attenuation? [5]
Differences Between Noise, Distortion, and Attenuation
Definitions and Key Differences
| Feature | Noise | Distortion | Attenuation |
|---|---|---|---|
| Definition | Unwanted random signals added to the original signal during transmission | Change in the shape/form of the signal due to different propagation speeds of signal components | Gradual loss of signal strength/energy as it travels through a medium |
| Cause | External interference (thermal, electromagnetic, crosstalk) | Different frequency components arriving at different times | Resistance of the transmission medium |
| Effect on Signal | Adds foreign/random components to the signal | Alters the waveform shape | Reduces the amplitude/power of the signal |
| Nature | Additive and random | Deterministic (related to medium properties) | Predictable and progressive |
| Reversibility | Difficult to reverse | Partially correctable with equalizers | Can be corrected using amplifiers or repeaters |
| Example | Static/hiss on a telephone line | Signal spreading in a coaxial cable | Signal weakening over long fiber optic cable |
Brief Explanations
1. Noise
- Noise is any unwanted energy that gets mixed with the transmitted signal.
- It is random and unpredictable.
- Types include: thermal noise, impulse noise, crosstalk, and intermodulation noise.
- It degrades the signal-to-noise ratio (SNR).
2. Distortion
- Distortion occurs when the shape of the signal changes during transmission.
- It is common in composite signals where each frequency component travels at a different speed.
- The signal arrives with components out of phase, altering the original waveform.
- It is a deterministic problem related to medium characteristics.
3. Attenuation
- Attenuation is the loss of signal energy as the signal propagates through a medium.
- It increases with distance and frequency.
- Measured in decibels (dB).
- It can be compensated by using amplifiers (analog) or repeaters (digital).
Summary
Noise adds unwanted signals, distortion changes the signal shape, and attenuation reduces the signal strength. All three are major impairments in data communication but differ in their cause, nature, and remedy.
5asked 2xavg 8 marks · due (skipped 2082) · Congestion control definitionAnswerHideWhat is Congestion Control? How can it be handled? Explain acknowledgement policy and discarding policy_[10]_
What is Congestion Control? How can it be handled? Explain acknowledgement policy and discarding policy_[10]_
Congestion Control
Definition
Congestion in a network occurs when the number of packets being transmitted through the network approaches the packet handling capacity of the network. Congestion control refers to the mechanisms and techniques used to control or prevent congestion so that network performance does not degrade.
Note: No specific reference notes were found for this topic; the answer below is based on standard networking curriculum (Forouzan / Tanenbaum) as taught in BSc CSIT.
Why Congestion Occurs
When too many packets are present in a part of the network, network performance degrades. This happens because:
- Routers have limited buffer (queue) space
- Links have limited bandwidth
- Processors in routers have limited processing speed
If the load exceeds capacity, packets are dropped, retransmissions increase, and the situation worsens -- leading to congestion collapse.
How Congestion Can Be Handled
Congestion control approaches are broadly classified as:
1. Open Loop Congestion Control (Prevention)
Policies are applied before congestion occurs -- at the design stage.
| Policy Point | Technique |
|---|---|
| Source | Retransmission policy, Windowing policy |
| Destination | Acknowledgement policy |
| Router/Switch | Discarding policy, Scheduling policy |
2. Closed Loop Congestion Control (Reaction)
Policies are applied after congestion is detected.
- Back pressure: A congested node sends a signal to the upstream node to slow down.
- Choke packet: A special packet is sent by the router to the source to reduce its transmission rate.
- Implicit signaling: Source detects congestion from delays or dropped packets (e.g., TCP timeout).
- Explicit signaling: Router explicitly marks packets to inform source/destination of congestion (e.g., ECN -- Explicit Congestion Notification).
Open Loop Congestion Control Policies (Detailed)
A. Acknowledgement Policy
The acknowledgement (ACK) policy at the receiver side can greatly affect congestion.
How it works:
- If the receiver sends an acknowledgement for every packet received, it encourages the sender to keep sending packets rapidly, which can contribute to congestion.
- By delaying acknowledgements, the receiver can slow down the sender and reduce network load.
Techniques:
| Technique | Description |
|---|---|
| Cumulative ACK | Receiver sends one ACK for multiple received packets instead of one ACK per packet. This reduces ACK traffic and slows the sender. |
| Delayed ACK | Receiver waits for a short time (e.g., up to 500 ms) before sending an ACK, hoping to piggyback it on a data packet or combine multiple ACKs. |
| Selective ACK (SACK) | Receiver acknowledges only specific packets, avoiding unnecessary retransmissions. |
Effect on Congestion:
- Fewer ACKs mean the sender's window advances more slowly.
- This reduces the rate at which new packets are injected into the network.
- Thus, acknowledgement policy directly helps in preventing congestion at the source.
B. Discarding Policy
The discarding policy is applied at routers/switches. When a router's buffer is full or near full, it must decide which packets to drop.
Approaches:
1. Tail Drop (Default / Simple)
- When the buffer is full, the router simply drops all incoming packets (from the tail of the queue).
- Problem: Can cause global synchronization -- many TCP connections reduce their window simultaneously, then increase together, causing oscillation.
2. Random Early Detection (RED)
- The router monitors its average queue length.
- When the queue length exceeds a minimum threshold (min_th), it starts randomly dropping packets with increasing probability.
- When the queue length exceeds a maximum threshold (max_th), all packets are dropped.
Queue Length:
0 -------- min_th -------- max_th -------- Max Buffer
| |
Start random Drop all
dropping packets
- Advantage: Prevents global synchronization; notifies individual connections early.
- Advantage: Maintains low average queue length and low delay.
3. Priority-Based Discarding
- Packets are assigned priority levels.
- When congestion occurs, lower priority packets are dropped first.
- Higher priority packets (e.g., real-time voice/video) are protected.
4. Weighted RED (WRED)
- Extension of RED where different traffic classes have different drop probabilities.
- Higher priority traffic has a lower drop probability.
Summary Table
| Policy | Applied At | Purpose |
|---|---|---|
| Acknowledgement Policy | Receiver (Destination) | Slow down sender by reducing ACK frequency |
| Discarding Policy | Router/Switch | Decide which packets to drop when buffer is full |
| Retransmission Policy | Sender (Source) | Avoid unnecessary retransmissions |
| Scheduling Policy | Router | Decide which packet to send next (e.g., FIFO, WFQ) |
Conclusion
Congestion control is essential for maintaining network performance and reliability. Open loop methods like acknowledgement policy and discarding policy prevent congestion proactively, while closed loop methods react to congestion after it is detected. Together, these mechanisms ensure fair and efficient use of network resources.
Most repeated questions
Topics asked at least twice, most-asked first.
asked 4xavg 6 marks · 2082, 2081, 2080, 0AnswerHideDifferentiate between packet and circuit switched network.Explain the layers of OSI Reference Model in brief.[4+6]
Differentiate between packet and circuit switched network.Explain the layers of OSI Reference Model in brief.[4+6]
--- (a) Difference Between Packet Switched and Circuit Switched Network Feature Circuit Switched Network Packet Switched Network --------- Connection A dedicated physical path is established before communication begins No dedicated path;...
asked 4xavg 6 marks · 2082, 2080, 2079, 0AnswerHideWhat are the major differences between noise, distortion and attenuation? [5]
What are the major differences between noise, distortion and attenuation? [5]
Differences Between Noise, Distortion, and Attenuation
Definitions and Key Differences
| Feature | Noise | Distortion | Attenuation |
|---|---|---|---|
| Definition | Unwanted random signals added to the original signal during transmission | Change in the shape/form of the signal due to different propagation speeds of signal components | Gradual loss of signal strength/energy as it travels through a medium |
| Cause | External interference (thermal, electromagnetic, crosstalk) | Different frequency components arriving at different times | Resistance of the transmission medium |
| Effect on Signal | Adds foreign/random components to the signal | Alters the waveform shape | Reduces the amplitude/power of the signal |
| Nature | Additive and random | Deterministic (related to medium properties) | Predictable and progressive |
| Reversibility | Difficult to reverse | Partially correctable with equalizers | Can be corrected using amplifiers or repeaters |
| Example | Static/hiss on a telephone line | Signal spreading in a coaxial cable | Signal weakening over long fiber optic cable |
Brief Explanations
1. Noise
- Noise is any unwanted energy that gets mixed with the transmitted signal.
- It is random and unpredictable.
- Types include: thermal noise, impulse noise, crosstalk, and intermodulation noise.
- It degrades the signal-to-noise ratio (SNR).
2. Distortion
- Distortion occurs when the shape of the signal changes during transmission.
- It is common in composite signals where each frequency component travels at a different speed.
- The signal arrives with components out of phase, altering the original waveform.
- It is a deterministic problem related to medium characteristics.
3. Attenuation
- Attenuation is the loss of signal energy as the signal propagates through a medium.
- It increases with distance and frequency.
- Measured in decibels (dB).
- It can be compensated by using amplifiers (analog) or repeaters (digital).
Summary
Noise adds unwanted signals, distortion changes the signal shape, and attenuation reduces the signal strength. All three are major impairments in data communication but differ in their cause, nature, and remedy.
asked 3xavg 7 marks · 2080, 2079, 0AnswerHideExplain the concept of TDMA with a neat diagram. [5]
Explain the concept of TDMA with a neat diagram. [5]
TDMA (Time Division Multiple Access) is a channel access method used in shared medium networks where multiple users share the same frequency channel by dividing the signal into different time slots. Each user is assigned a specific time ...
asked 3xavg 5 marks · 2080.2, 2079, 0AnswerHideWhat is MAC-address? The message sequence is 1011011 and generator polynomial $G(X) = x^3 + x^2 + 1$. Calculate the transmitted encoded frame [5]
What is MAC-address? The message sequence is 1011011 and generator polynomial $G(X) = x^3 + x^2 + 1$. Calculate the transmitted encoded frame [5]
MAC Address and CRC Calculation
MAC Address (2 marks)
A MAC (Media Access Control) address is a unique physical hardware address permanently assigned to a Network Interface Card (NIC) by its manufacturer. It operates at the Data Link Layer (Layer 2) of the OSI model and is used to uniquely identify a device on a local network.
Key features:
- It is a 48-bit (6-byte) address, normally written in hexadecimal.
- Example:
00:1A:2B:3C:4D:5E - The first 3 bytes (24 bits) form the OUI (Organizationally Unique Identifier) identifying the manufacturer.
- The last 3 bytes identify the specific device.
- Also called the physical / hardware address; usually burned into the NIC's ROM and globally unique.
CRC Calculation (3 marks)
Given data
- Message $M$ =
1011011 - Generator $G(x) = x^3 + x^2 + 1 \Rightarrow$ binary
1101 - Degree $r = 3$
Step 1: Append $r = 3$ zeros
$$ M' = 1011011\underbrace{000}_{3\text{ zeros}} = 1011011000 $$
Step 2: Modulo-2 (XOR) division by 1101
Dividend bits: 1 0 1 1 0 1 1 0 0 0
1101010 <-- quotient (not required)
┌───────────
1101│1011011000
1101
────
0110 1 bring down bit → 1101
1101
────
0000 11 bring down bits → 0011
(0011 < 1101) → q=0, bring down → 0111
(0111 < 1101) → q=0, bring down → 1110
1101
────
0011 0 bring down → 0110
(0110 < 1101) → q=0 (no more bits)
Remainder = 010
Working the XOR steps cleanly:
| Current bits | XOR with 1101 | Result |
|---|---|---|
| 1011 | 1101 | 0110 |
| bring 0 → 1100 | 1101 | 0001 |
| bring 1 → 0011 | - (0011<1101, q=0) | 0011 |
| bring 1 → 0111 | - (0111<1101, q=0) | 0111 |
| bring 0 → 1110 | 1101 | 0011 |
| bring 0 → 0110 | - (0110<1101, q=0) | 0110 |
| bring 0 → 0100 | - (last bit, 0100<1101) | 100 |
Let me track the final bits exactly. After processing all 10 dividend bits:
- Start:
1011→ XOR 1101 =0110 - +bit(0):
1100→ XOR 1101 =0001 - +bit(1):
0011→ < divisor, q=0 - +bit(1):
0111→ < divisor, q=0 - +bit(0):
1110→ XOR 1101 =0011 - +bit(0):
0110→ < divisor, q=0 - +bit(0):
1100→ XOR 1101 =0001
Final 3-bit remainder (CRC) = 001
Verification
Divide the 3 remaining bits carefully; the last register content after all 10 bits gives remainder = 001.
Check: transmitted frame $T = 1011011,001$. Dividing $1011011001$ by 1101 must give remainder 000:
1011011001
1101
────
0110 → 1100 → 0001 → 0011 → 0111 → 1110→0011 → 0110 → 1101 → 0000
$1101 \oplus 1101 = 0000$ ✓ remainder is 0, so the CRC is correct.
Step 3: Transmitted frame
$$ T = \text{Message} + \text{CRC} = 1011011 , \big| , 001 = \boxed{1011011001} $$
CRC (remainder) = 001
Transmitted encoded frame = 1011011001
asked 3xavg 7 marks · 2082, 2080, 2079AnswerHideExplain Link State Routing with suitable example. [5]
Explain Link State Routing with suitable example. [5]
Link State Routing is a dynamic routing algorithm where each router has complete knowledge of the entire network topology. Every router broadcasts information about its directly connected links (neighbors and link costs) to all other rou...
asked 3xavg 7 marks · 2082, 2080.2, 0AnswerHideBriefly explain ALOHA and Slotted ALOHA protocol with suitable diagram. [5]
Briefly explain ALOHA and Slotted ALOHA protocol with suitable diagram. [5]
Note: Reference notes were not available for this topic; the following answer is based on standard networking curriculum covered in TU BSc CSIT. --- ALOHA is one of the earliest random access protocols developed at the University of Hawa...
asked 3xavg 5 marks · 2082, 2081, 2080AnswerHideEncode the bit stream 111001011 with:
(i) Manchester
(ii) NRZ-I and
(iii) NRZ-L scheme. [5]
Encode the bit stream 111001011 with:
(i) Manchester
(ii) NRZ-I and
(iii) NRZ-L scheme. [5]
- Bit stream to encode: $1\ 1\ 1\ 0\ 0\ 1\ 0\ 1\ 1$ (9 bits) - Schemes required: (i) Manchester, (ii) NRZ-I, (iii) NRZ-L Note: These are standard line-coding schemes. I follow the conventions used in Forouzan (TU curriculum standard). An...
asked 3xavg 5 marks · 2082, 2081, 2080AnswerHideA bit stream 11011001 is transmitted using a standard CRC method. The generator polynomial is $x^3 + x - 1$. Show the actual bit string transmitted and show the error checking on the receiver side. [5]
A bit stream 11011001 is transmitted using a standard CRC method. The generator polynomial is $x^3 + x - 1$. Show the actual bit string transmitted and show the error checking on the receiver side. [5]
CRC (Cyclic Redundancy Check) - Worked Solution
Step 1 - EXTRACT: Given Data
- Message bit stream (M):
11011001(8 bits) - Generator polynomial: $x^3 + x - 1$
- In modulo-2 (GF(2)) arithmetic, $-1 \equiv +1$, so $x^3 + x - 1 = x^3 + x + 1$
- Generator (G):
1011(4 bits, degree $r = 3$)
Step 2 - SOLVE
Step 2.1: Append $r = 3$ zeros to the message
$$M' = 11011001 ,|, 000 = 11011001000$$
Step 2.2: Sender-side XOR division of 11011001000 by 1011
10000101 <-- quotient
______________
1011 ) 11011001000
1011
----
1101
1011
----
01100
1011
----
01110
1011
----
01011
1011
----
00000
000 <-- remainder (CRC)
Step trace (leading-bit alignment, XOR when leading bit = 1):
| Working segment | XOR with | Result | Bring down |
|---|---|---|---|
1101 | 1011 | 0110 | +1 → 1101 |
1101 | 1011 | 0110 | +0 → 1100 |
1100 | 1011 | 0111 | +0 → 1110 |
1110 | 1011 | 0101 | +1 → 1011 |
1011 | 1011 | 0000 | +0 → 0000 |
0000 | (leading 0, no XOR) | 000 | end |
Remainder (CRC) = 000
Step 2.3: Actual transmitted bit string
$$\text{Transmitted} = M ,|, \text{CRC} = 11011001 ,|, 000$$
$$\boxed{11011001000}$$
Step 2.4: Receiver-side error checking
Receiver divides the received frame 11011001000 by the same generator 1011:
1011 ) 11011001000
... (identical division) ...
000 <-- remainder
Remainder = 000
Step 2.5: Decision
| Remainder | Decision |
|---|---|
000 | No error - accept the frame |
| non-zero | Error detected - reject / retransmit |
Since the remainder is 000, the receiver concludes the frame was received without error.
Summary
| Parameter | Value |
|---|---|
| Message | 11011001 |
| Generator ($x^3+x+1$) | 1011 |
| Appended message | 11011001000 |
| CRC (remainder) | 000 |
| Transmitted frame | 11011001000 |
| Receiver remainder | 000 → No error |
asked 2xavg 8 marks · 2080, 0AnswerHideWhat is Congestion Control? How can it be handled? Explain acknowledgement policy and discarding policy_[10]_
What is Congestion Control? How can it be handled? Explain acknowledgement policy and discarding policy_[10]_
Congestion Control
Definition
Congestion in a network occurs when the number of packets being transmitted through the network approaches the packet handling capacity of the network. Congestion control refers to the mechanisms and techniques used to control or prevent congestion so that network performance does not degrade.
Note: No specific reference notes were found for this topic; the answer below is based on standard networking curriculum (Forouzan / Tanenbaum) as taught in BSc CSIT.
Why Congestion Occurs
When too many packets are present in a part of the network, network performance degrades. This happens because:
- Routers have limited buffer (queue) space
- Links have limited bandwidth
- Processors in routers have limited processing speed
If the load exceeds capacity, packets are dropped, retransmissions increase, and the situation worsens -- leading to congestion collapse.
How Congestion Can Be Handled
Congestion control approaches are broadly classified as:
1. Open Loop Congestion Control (Prevention)
Policies are applied before congestion occurs -- at the design stage.
| Policy Point | Technique |
|---|---|
| Source | Retransmission policy, Windowing policy |
| Destination | Acknowledgement policy |
| Router/Switch | Discarding policy, Scheduling policy |
2. Closed Loop Congestion Control (Reaction)
Policies are applied after congestion is detected.
- Back pressure: A congested node sends a signal to the upstream node to slow down.
- Choke packet: A special packet is sent by the router to the source to reduce its transmission rate.
- Implicit signaling: Source detects congestion from delays or dropped packets (e.g., TCP timeout).
- Explicit signaling: Router explicitly marks packets to inform source/destination of congestion (e.g., ECN -- Explicit Congestion Notification).
Open Loop Congestion Control Policies (Detailed)
A. Acknowledgement Policy
The acknowledgement (ACK) policy at the receiver side can greatly affect congestion.
How it works:
- If the receiver sends an acknowledgement for every packet received, it encourages the sender to keep sending packets rapidly, which can contribute to congestion.
- By delaying acknowledgements, the receiver can slow down the sender and reduce network load.
Techniques:
| Technique | Description |
|---|---|
| Cumulative ACK | Receiver sends one ACK for multiple received packets instead of one ACK per packet. This reduces ACK traffic and slows the sender. |
| Delayed ACK | Receiver waits for a short time (e.g., up to 500 ms) before sending an ACK, hoping to piggyback it on a data packet or combine multiple ACKs. |
| Selective ACK (SACK) | Receiver acknowledges only specific packets, avoiding unnecessary retransmissions. |
Effect on Congestion:
- Fewer ACKs mean the sender's window advances more slowly.
- This reduces the rate at which new packets are injected into the network.
- Thus, acknowledgement policy directly helps in preventing congestion at the source.
B. Discarding Policy
The discarding policy is applied at routers/switches. When a router's buffer is full or near full, it must decide which packets to drop.
Approaches:
1. Tail Drop (Default / Simple)
- When the buffer is full, the router simply drops all incoming packets (from the tail of the queue).
- Problem: Can cause global synchronization -- many TCP connections reduce their window simultaneously, then increase together, causing oscillation.
2. Random Early Detection (RED)
- The router monitors its average queue length.
- When the queue length exceeds a minimum threshold (min_th), it starts randomly dropping packets with increasing probability.
- When the queue length exceeds a maximum threshold (max_th), all packets are dropped.
Queue Length:
0 -------- min_th -------- max_th -------- Max Buffer
| |
Start random Drop all
dropping packets
- Advantage: Prevents global synchronization; notifies individual connections early.
- Advantage: Maintains low average queue length and low delay.
3. Priority-Based Discarding
- Packets are assigned priority levels.
- When congestion occurs, lower priority packets are dropped first.
- Higher priority packets (e.g., real-time voice/video) are protected.
4. Weighted RED (WRED)
- Extension of RED where different traffic classes have different drop probabilities.
- Higher priority traffic has a lower drop probability.
Summary Table
| Policy | Applied At | Purpose |
|---|---|---|
| Acknowledgement Policy | Receiver (Destination) | Slow down sender by reducing ACK frequency |
| Discarding Policy | Router/Switch | Decide which packets to drop when buffer is full |
| Retransmission Policy | Sender (Source) | Avoid unnecessary retransmissions |
| Scheduling Policy | Router | Decide which packet to send next (e.g., FIFO, WFQ) |
Conclusion
Congestion control is essential for maintaining network performance and reliability. Open loop methods like acknowledgement policy and discarding policy prevent congestion proactively, while closed loop methods react to congestion after it is detected. Together, these mechanisms ensure fair and efficient use of network resources.
asked 2xavg 8 marks · 2080, 0AnswerHideExplain leaky-bucket algorithm with an example [5]
Explain leaky-bucket algorithm with an example [5]
The leaky bucket algorithm is a traffic shaping and congestion control mechanism used in computer networks to control the rate at which data packets are sent into the network. It smooths out bursty traffic and converts it into a steady, ...
asked 2xavg 5 marks · 2080.2, 2080AnswerHideDifferentiate between reliable and unreliable protocol. Provide example of each. How does protocol check for errors? [5]
Differentiate between reliable and unreliable protocol. Provide example of each. How does protocol check for errors? [5]
A reliable protocol is a communication protocol that guarantees delivery of data from sender to receiver. It ensures that: - Data arrives without errors - Data arrives in order - No data is lost or duplicated - The sender receives an ack...
asked 2xavg 5 marks · 2080, 0AnswerHideExplain recursive resolution. What are its advantages? [5]
Explain recursive resolution. What are its advantages? [5]
Recursive resolution is a name resolution strategy used in the Domain Name System (DNS) where a DNS resolver takes full responsibility for resolving a query on behalf of the client. Instead of referring the client to another server, the ...
asked 2xavg 8 marks · 2082, 2080.2AnswerHideWhat are the services provided by Data Link Layer?A channel has a bandwidth of 5 MHz and a signal-to-noise ratio of 1000. Calculate the Shannon Capacity.[6+4]
What are the services provided by Data Link Layer?A channel has a bandwidth of 5 MHz and a signal-to-noise ratio of 1000. Calculate the Shannon Capacity.[6+4]
Data Link Layer Services & Shannon Capacity
Part A: Services Provided by Data Link Layer [6 Marks]
The Data Link Layer (Layer 2 of the OSI model) provides reliable transit of data across a physical link. Its main services are:
1. Framing
The bit stream received from the Network Layer is divided into manageable units called frames. A header and trailer are added to mark the start and end of each frame.
2. Physical (MAC) Addressing
A header containing the MAC addresses of the sender and receiver is added, enabling frame delivery between devices on the same network.
3. Flow Control
When the sender transmits faster than the receiver can absorb, the layer applies a flow-control mechanism to prevent receiver buffer overflow.
4. Error Control
Mechanisms to detect and retransmit damaged or lost frames are provided, along with prevention of duplicate frames. Techniques include CRC, parity, and checksums.
5. Access Control
On a shared medium, protocols decide which device controls the link at a given moment, avoiding collisions (e.g., CSMA/CD, Token passing).
6. Reliable (Node-to-Node) Delivery
Uses acknowledgements (ACK) and retransmission schemes such as ARQ to ensure error-free node-to-node delivery.
Part B: Shannon Capacity Calculation [4 Marks]
Given data
| Parameter | Value |
|---|---|
| Bandwidth $B$ | $5\ \text{MHz} = 5 \times 10^6\ \text{Hz}$ |
| Signal-to-Noise Ratio (SNR) | $1000$ |
Formula
$$C = B \times \log_2(1 + \text{SNR})$$
Step 1: Substitute
$$C = 5 \times 10^6 \times \log_2(1 + 1000) = 5 \times 10^6 \times \log_2(1001)$$
Step 2: Evaluate the logarithm
$$\log_2(1001) = \frac{\log_{10}(1001)}{\log_{10}(2)} = \frac{3.00043}{0.30103} \approx 9.9672$$
Step 3: Compute capacity
$$C = 5 \times 10^6 \times 9.9672 \approx 49.836 \times 10^6\ \text{bps}$$
$$\boxed{C \approx 49.84\ \text{Mbps} \approx 50\ \text{Mbps}}$$
Interpretation
This is the theoretical maximum data rate of the channel. No transmission scheme, regardless of the number of signal levels, can exceed this limit.
asked 2xavg 5 marks · 2082, 2081AnswerHideWhat is DHCP? Explain DNS name resolution process with suitable example. [1+4]
What is DHCP? Explain DNS name resolution process with suitable example. [1+4]
DHCP and DNS Name Resolution
What is DHCP? [1 Mark]
DHCP (Dynamic Host Configuration Protocol) is a network management protocol used to automatically assign IP addresses and other network configuration parameters (such as subnet mask, default gateway, and DNS server address) to devices on a network, so they can communicate on an IP network without manual configuration.
DNS Name Resolution Process [4 Marks]
DNS (Domain Name System) is a hierarchical, distributed naming system that translates human-readable domain names (e.g., www.example.com) into machine-readable IP addresses (e.g., 93.184.216.34).
Components Involved
| Component | Role |
|---|---|
| DNS Resolver | Client-side component that initiates the query |
| Root Name Server | Top of the DNS hierarchy; knows TLD servers |
| TLD Name Server | Handles top-level domains (.com, .org, .np, etc.) |
| Authoritative Name Server | Holds the actual DNS records for the domain |
| Local DNS Server | First server contacted; may have cached results |
Step-by-Step Name Resolution Process
Example: A user types www.example.com in a browser.
User's PC --> Local DNS --> Root Server --> TLD Server --> Authoritative Server
Step 1: Browser Cache Check
- The browser first checks its local cache to see if the IP for
www.example.comis already stored. - If not found, the query is passed to the OS resolver.
Step 2: Query to Local DNS Server (Recursive Resolver)
- The client sends a recursive query to the Local DNS Server (configured via DHCP or manually).
- The local DNS server checks its own cache.
- If not cached, it begins the resolution process on behalf of the client.
Step 3: Query to Root Name Server
- The local DNS server sends a query to one of the 13 Root Name Servers.
- The root server does not know the IP of
www.example.com, but it knows the address of the.comTLD Name Server. - Root server replies: "Ask the .com TLD server at IP X.X.X.X"
Step 4: Query to TLD Name Server
- The local DNS server queries the
.comTLD Name Server. - The TLD server does not know the exact IP, but knows the Authoritative Name Server for
example.com. - TLD server replies: "Ask the authoritative server for example.com at IP Y.Y.Y.Y"
Step 5: Query to Authoritative Name Server
- The local DNS server queries the Authoritative Name Server for
example.com. - This server has the actual A record (Address record) for
www.example.com. - It replies: "
www.example.comhas IP address93.184.216.34"
Step 6: Response to Client
- The local DNS server caches the result (for future queries) and sends the IP address back to the client.
- The browser can now connect to
93.184.216.34to load the website.
Diagram
Client
|
| (1) Query: www.example.com?
v
Local DNS Server
|
| (2) Query to Root Server
v
Root Name Server --> "Try .com TLD Server"
|
| (3) Query to TLD Server
v
.com TLD Server --> "Try example.com Authoritative Server"
|
| (4) Query to Authoritative Server
v
Authoritative Server --> "IP = 93.184.216.34"
|
| (5) Final Answer returned to Client
v
Client connects to 93.184.216.34
Types of DNS Queries
| Query Type | Description |
|---|---|
| Recursive Query | Client asks resolver to get the full answer |
| Iterative Query | Server returns the best answer it knows (referral) |
Note: The query from client to local DNS is typically recursive; queries from local DNS to other servers are typically iterative.
Summary
The DNS name resolution process converts www.example.com into an IP address through a hierarchical chain: Local DNS Server → Root Server → TLD Server → Authoritative Server, ultimately returning the IP address to the requesting client.
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