Flow control (stop-and-wait, sliding window) in Computer Networks - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

Understanding Stop-and-Wait Flow Control

In a stop-and-wait flow control protocol, what happens immediately after the sender transmits a data frame?

A The sender sends multiple frames at once and waits for acknowledgments.

B The sender immediately sends the next frame without waiting.

C The sender waits for an acknowledgment before sending the next frame.

D The sender closes the connection after sending one frame.

💡 Hint

Think about how the sender ensures the receiver got the data before continuing.

In stop-and-wait, the sender sends one frame and waits for an acknowledgment before sending the next. This prevents overwhelming the receiver.

📋 Factual

intermediate

Sliding Window Protocol Window Size

What is the main advantage of using a sliding window protocol with a window size greater than one?

A It allows multiple frames to be sent before waiting for acknowledgments, increasing throughput.

B It reduces the number of frames sent to one per connection.

C It guarantees zero packet loss in the network.

D It requires the sender to wait for each acknowledgment before sending any frame.

💡 Hint

Consider how sending multiple frames without waiting affects data flow.

A sliding window with size greater than one lets the sender send several frames before needing acknowledgments, improving efficiency and throughput.

🔍 Analysis

advanced

Effect of Window Size on Network Utilization

Given a network with a high round-trip time (RTT), which sliding window size will maximize utilization without causing buffer overflow at the receiver?

A A window size larger than the receiver's buffer size.

B A window size equal to the bandwidth-delay product of the network.

C A window size of 1, to avoid any congestion.

D A window size of zero, to stop all transmissions.

💡 Hint

Think about how the bandwidth-delay product relates to data in transit.

The bandwidth-delay product represents the amount of data that can fill the network pipe. Setting the window size to this value maximizes utilization without overflow.

❓ Comparison

advanced

Comparing Stop-and-Wait and Sliding Window Protocols

Which statement correctly compares stop-and-wait and sliding window flow control protocols?

A Stop-and-wait sends one frame at a time and waits for acknowledgment; sliding window sends multiple frames before waiting.

B Stop-and-wait uses a window size greater than one; sliding window uses a window size of one.

C Both protocols send multiple frames without waiting for acknowledgments.

D Sliding window sends one frame at a time; stop-and-wait sends multiple frames before waiting.

💡 Hint

Focus on how many frames each protocol sends before waiting.

Stop-and-wait sends one frame and waits for acknowledgment before sending the next. Sliding window allows multiple frames to be sent before waiting.

❓ Reasoning

expert

Impact of Lost Acknowledgments in Sliding Window

In a sliding window protocol, if an acknowledgment for a frame is lost but the frame was received correctly, what is the most likely outcome?

A The sender will ignore the lost acknowledgment and continue sending new frames.

B The sender will stop sending any further frames permanently.

C The receiver will discard all future frames until the acknowledgment is received.

D The sender will retransmit the frame, causing a duplicate at the receiver.

💡 Hint

Consider how the sender detects lost acknowledgments and what it does next.

If an acknowledgment is lost, the sender assumes the frame was not received and retransmits it. The receiver may get a duplicate but can detect and discard it.

Practice

1. Which flow control method sends one packet and waits for an acknowledgment before sending the next?

easy

A. Stop-and-Wait

B. Sliding Window

C. Multiplexing

D. Routing

Solution

1. Step 1: Understand Stop-and-Wait flow control

   Stop-and-Wait sends one packet and waits for an acknowledgment before sending the next packet.

2. Step 2: Compare with other methods

   Sliding Window sends multiple packets before waiting, while Multiplexing and Routing are unrelated to flow control.

3. Final Answer:

   Stop-and-Wait -> Option A

4. Quick Check:

   One packet at a time = Stop-and-Wait [OK]

Hint: One packet then wait means Stop-and-Wait [OK]

Common Mistakes:

• Confusing Sliding Window with Stop-and-Wait
• Thinking Multiplexing is a flow control method
• Mixing routing with flow control

2. Which of the following correctly describes the sliding window flow control method?

easy

A. Sends one packet and waits for acknowledgment

B. Sends multiple packets before waiting for acknowledgment

C. Sends packets without any acknowledgment

D. Sends packets only after a timeout

Solution

1. Step 1: Recall sliding window behavior

   Sliding Window sends multiple packets before waiting for acknowledgment to improve speed.

2. Step 2: Eliminate incorrect options

   Sends one packet and waits for acknowledgment describes Stop-and-Wait. The other options do not describe standard flow control methods.

3. Final Answer:

   Sends multiple packets before waiting for acknowledgment -> Option B

4. Quick Check:

   Multiple packets sent = Sliding Window [OK]

Hint: Multiple packets sent before wait means Sliding Window [OK]

Common Mistakes:

• Choosing Stop-and-Wait for sliding window question
• Thinking no acknowledgment is used
• Confusing timeout with acknowledgment wait

3. Consider a sliding window protocol with window size 4. If the sender has sent packets 1, 2, 3, and 4, but only packets 1 and 2 are acknowledged, which packets can the sender send next?

medium

A. Packets 5 and 6

B. Packets 3, 4, and 5

C. Packets 5, 6, 7, and 8

D. Packets 3, 4, 5, and 6

Solution

1. Step 1: Understand sliding window acknowledgments

   Packets 1 and 2 are acknowledged, so the window slides forward by 2 packets.

2. Step 2: Calculate new window range

   Window size is 4, so after acknowledging 1 and 2, sender can send packets 5 and 6 (next two packets).

3. Final Answer:

   Packets 5 and 6 -> Option A

4. Quick Check:

   Window slides by acknowledged packets = Packets 5 and 6 [OK]

Hint: Window slides by acknowledged packets count [OK]

Common Mistakes:

• Trying to resend unacknowledged packets
• Sending too many packets beyond window size
• Not sliding window after acknowledgments

4. A stop-and-wait protocol implementation is not receiving acknowledgments and keeps resending the same packet. What is the most likely cause?

medium

A. The sender is sending multiple packets at once

B. The sender window size is too large

C. The receiver is not sending acknowledgments

D. The network is using sliding window instead

Solution

1. Step 1: Analyze stop-and-wait behavior

   Stop-and-wait waits for acknowledgment before sending next packet; no acknowledgment causes resending.

2. Step 2: Identify cause of no acknowledgment

   If acknowledgments are missing, likely the receiver is not sending them or they are lost.

3. Final Answer:

   The receiver is not sending acknowledgments -> Option C

4. Quick Check:

   No acknowledgment = Resend same packet [OK]

Hint: No ack means receiver not responding [OK]

Common Mistakes:

• Blaming sender window size in stop-and-wait
• Confusing sliding window with stop-and-wait
• Assuming multiple packets sent in stop-and-wait

5. In a sliding window protocol with window size 5, the sender transmits packets 1 to 5. Packets 2 and 3 are lost, but packets 1, 4, and 5 are acknowledged. Which packets should the sender retransmit next to maintain correct flow control?

hard

A. Packets 6 and 7

B. Packets 2, 3, and 6

C. Packets 2, 3, 4, and 5

D. Packets 2 and 3 only

Solution

1. Step 1: Identify lost packets

   Packets 2 and 3 are lost, so they need retransmission.

2. Step 2: Understand acknowledgments impact

   Packets 1, 4, and 5 are acknowledged, so only lost packets 2 and 3 must be resent before sending new packets.

3. Final Answer:

   Packets 2 and 3 only -> Option D

4. Quick Check:

   Retransmit lost packets only = Packets 2 and 3 [OK]

Hint: Retransmit only lost packets, not acknowledged ones [OK]

Common Mistakes:

• Retransmitting already acknowledged packets
• Sending new packets before lost ones
• Confusing lost packets with acknowledged packets