TCP flow control (sliding window) in Computer Networks - Mini Project: Build & Apply

Understanding TCP Flow Control with Sliding Window

📖 Scenario: You are learning how computers manage data transmission over the internet using TCP (Transmission Control Protocol). TCP uses a method called sliding window to control the flow of data packets between sender and receiver, ensuring that the receiver is not overwhelmed.

🎯 Goal: Build a simple step-by-step model that represents the TCP sliding window mechanism. You will create data structures to represent the window size, the packets sent, and the acknowledgement process.

📋 What You'll Learn

Create a list of data packets numbered 1 to 10

Define a window size variable to control how many packets can be sent before waiting for acknowledgement

Simulate sending packets within the window using a loop

Update the window to slide forward after receiving acknowledgements

💡 Why This Matters

🌍 Real World

TCP sliding window is used in internet communication to ensure reliable and efficient data transfer without overwhelming the receiver.

💼 Career

Understanding TCP flow control is important for network engineers, software developers working on network applications, and cybersecurity professionals.

1

Create the list of data packets

Create a list called packets containing integers from 1 to 10 representing data packets.

Computer Networks

# Create the list of packets from 1 to 10
# Your code here

Hint

Use the range function to create numbers from 1 to 10 and convert it to a list.

2

Define the window size

Create a variable called window_size and set it to 4 to represent the number of packets that can be sent before waiting for acknowledgement.

Computer Networks

packets = list(range(1, 11))
# Define the window size as 4
# Your code here

Hint

Set window_size to 4 as an integer.

3

Simulate sending packets within the window

Use a for loop with variable i to iterate over the first window_size packets in packets and create a list called sent_packets containing these packets.

Computer Networks

packets = list(range(1, 11))
window_size = 4
# Use a for loop to send packets within the window
# Your code here

Hint

Loop from 0 to window_size - 1 and add each packet to sent_packets.

4

Slide the window after acknowledgement

Create a variable called acknowledged and set it to 2 representing the number of packets acknowledged by the receiver. Then create a list called next_window containing the next set of packets after sliding the window by acknowledged positions.

Computer Networks

packets = list(range(1, 11))
window_size = 4
sent_packets = []
for i in range(window_size):
    sent_packets.append(packets[i])
# Set acknowledged to 2 and create next_window after sliding
# Your code here

Hint

Set acknowledged to 2. Then loop from acknowledged to acknowledged + window_size - 1 to get the next packets.

Practice

1. What is the main purpose of the sliding window in TCP flow control?

easy

A. To control the amount of data sent before receiving an acknowledgment

B. To encrypt data during transmission

C. To assign IP addresses to devices

D. To route packets between networks

Solution

1. Step 1: Understand the role of sliding window in TCP

   The sliding window limits how much data the sender can transmit before waiting for an acknowledgment from the receiver.

2. Step 2: Identify the purpose among options

   Only To control the amount of data sent before receiving an acknowledgment describes controlling data flow using acknowledgments, which matches the sliding window's purpose.

3. Final Answer:

   To control the amount of data sent before receiving an acknowledgment -> Option A

4. Quick Check:

   Sliding window controls data flow = To control the amount of data sent before receiving an acknowledgment [OK]

Hint: Sliding window limits unacknowledged data sent [OK]

Common Mistakes:

• Confusing flow control with encryption
• Mixing routing with flow control
• Thinking sliding window assigns IPs

2. Which of the following correctly describes the sliding window size in TCP?

easy

A. It is fixed and never changes during the connection

B. It is set by the IP address of the sender

C. It is always equal to the maximum segment size (MSS)

D. It increases or decreases based on acknowledgments received

Solution

1. Step 1: Recall how sliding window size behaves

   The window size changes dynamically as acknowledgments arrive, allowing the sender to send more data.

2. Step 2: Match behavior to options

   It increases or decreases based on acknowledgments received correctly states the window size adjusts based on acknowledgments, unlike fixed size or unrelated factors.

3. Final Answer:

   It increases or decreases based on acknowledgments received -> Option D

4. Quick Check:

   Window size changes with ACKs = It increases or decreases based on acknowledgments received [OK]

Hint: Window size changes with ACKs, not fixed [OK]

Common Mistakes:

• Assuming window size is fixed
• Confusing window size with MSS
• Linking window size to IP address

3. Consider a TCP connection with a window size of 4 segments. The sender has sent segments 1 to 4. The receiver acknowledges segment 2. Which segments can the sender send next?

medium

A. Segments 5, 6, and 7

B. Segments 3 and 4 again

C. Segments 5 and 6

D. No new segments until all are acknowledged

Solution

1. Step 1: Understand sliding window movement after ACK

   When segment 2 is acknowledged, the window slides forward, allowing sending of new segments beyond 4.

2. Step 2: Calculate which segments fit in the window

   Window size is 4, segments 1 and 2 acknowledged, so segments 3 and 4 are still unacknowledged. The sender can send segments 5 and 6 to fill the window.

3. Final Answer:

   Segments 5 and 6 -> Option C

4. Quick Check:

   ACK 2 slides window, send next 2 segments = Segments 5 and 6 [OK]

Hint: ACK moves window; send new segments within window size [OK]

Common Mistakes:

• Resending unacknowledged segments unnecessarily
• Sending more segments than window size allows
• Waiting for all segments to be acknowledged

4. A TCP sender has a window size of 3 segments and has sent segments 1, 2, and 3. The receiver sends an acknowledgment for segment 1 only. The sender mistakenly sends segment 4 and 5 immediately. What is the problem with this behavior?

medium

A. It violates the sliding window limit and may cause congestion

B. It improves throughput by sending more data

C. It is correct because ACK 1 means all previous segments are received

D. It causes the receiver to close the connection

Solution

1. Step 1: Analyze window size and ACK received

   The window size is 3, and only segment 1 is acknowledged, so segments 2 and 3 are still unacknowledged.

2. Step 2: Identify why sending segments 4 and 5 is wrong

   Sending segments 4 and 5 exceeds the allowed window size, risking congestion and data loss.

3. Final Answer:

   It violates the sliding window limit and may cause congestion -> Option A

4. Quick Check:

   Sending beyond window size causes issues = It violates the sliding window limit and may cause congestion [OK]

Hint: Don't send beyond window size before ACKs arrive [OK]

Common Mistakes:

• Assuming ACK 1 means all segments received
• Ignoring window size limits
• Thinking sending more data always helps

5. In a TCP connection, the sender's window size is 5 segments. The receiver's buffer can only hold 3 segments. How should the sliding window mechanism adapt to prevent buffer overflow?

hard

A. The sender should keep sending 5 segments regardless

B. The sender should reduce the window size to 3 segments

C. The receiver should increase its buffer size to 5 segments

D. The sender should stop sending any segments

Solution

1. Step 1: Understand receiver buffer limits

   The receiver can only hold 3 segments, so sending more than 3 risks overflow and data loss.

2. Step 2: Adjust sender window size accordingly

   The sliding window size must match or be less than the receiver's buffer to avoid overflow, so the sender should reduce its window size to 3.

3. Final Answer:

   The sender should reduce the window size to 3 segments -> Option B

4. Quick Check:

   Sender window ≤ receiver buffer to prevent overflow = The sender should reduce the window size to 3 segments [OK]

Hint: Match sender window to receiver buffer size [OK]

Common Mistakes:

• Ignoring receiver buffer size
• Assuming receiver buffer can always expand
• Stopping sending completely without reason