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Chapter 2: Problem 22

Consider an ARQ protocol that uses only negative acknowledgments (NAKs), but no positive acknowledgments (ACKs). Describe what timeouts would need to be scheduled. Explain why an ACK-based protocol is usually preferred to a NAK- based protocol.

Short Answer

Expert verified
Timeouts are scheduled to detect lost packets and depend on RTT. ACK-based protocols are preferred due to their explicit confirmations, improving reliability and efficiency.

Step by step solution

01

- Understanding ARQ Protocols

An Automatic Repeat reQuest (ARQ) protocol is used for error control in data transmission. It ensures that the sender and receiver can communicate reliably even with occasional lost or corrupted packets.
02

- Explaining NAK-based ARQ

In a NAK-based ARQ protocol, the receiver sends a negative acknowledgment (NAK) when it detects an error or a missing packet. This requests the sender to retransmit the erroneous or missing packet.
03

- Scheduling Timeouts

Timeouts are needed to detect lost packets. If the sender does not receive a NAK within a certain time period, it assumes the packet has been received successfully. Scheduling a timeout for the retransmission depends on the network's round-trip time (RTT) and the expected processing delays.
04

- Timeout Calculation

Calculate the timeout interval using RTT estimation. For instance, timeout can be set to a default of two to three times the average RTT to ensure that any packet loss can be detected within a reasonable timeframe.
05

- Comparing ACK-based Protocols

In ACK-based protocols, the receiver sends positive acknowledgments for received packets. The sender will retransmit a packet if it does not receive an ACK within the timeout period. ACK-based protocols are preferred because they explicitly confirm successful receipt, making the detection of lost packets simpler and more reliable.
06

- Conclusion

While NAK-based protocols rely on detecting errors or losses to trigger retransmissions, ACK-based protocols continuously affirm the successful receipt, reducing ambiguity. Thus, ACK-based protocols can enhance reliability and efficiency in communications.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

NAK-based ARQ
Automatic Repeat reQuest (ARQ) protocols are vital in error control for data transmission. One type of ARQ protocol is the NAK-based ARQ. In this approach, the receiver sends a Negative Acknowledgment (NAK) whenever it detects an error or finds a missing packet. The NAK essentially tells the sender to retransmit the specific packet that had issues. This mechanism is reactive since retransmissions only happen after detecting a problem.
Using NAKs helps to control the data flow and ensures that even if a packet gets lost or corrupted, the communication remains intact by requesting a correction.
ACK-based ARQ
Another common ARQ protocol is the ACK-based ARQ. In this method, the receiver sends a Positive Acknowledgment (ACK) after successfully receiving and processing a packet. If the sender does not receive an ACK within a predetermined timeout period, it assumes the packet was lost and retransmits the packet.
ACK-based ARQ provides several benefits over NAK-based ARQ:
  • Guarantees confirmation for each packet received, reducing uncertainty.
  • Improves overall reliability of the data transmission process.
  • Helps quickly detect and correct errors.
This proactive confirmation method often leads to more efficient and reliable data communications.
Timeout Scheduling
Timeout scheduling is crucial for ensuring both NAK-based and ACK-based ARQ protocols function effectively. Timeouts help in detecting lost packets. Here's how you can approach it:
  • Estimate the network's Round-Trip Time (RTT). RTT is the time it takes for a signal to travel from the sender to the receiver and back again.
  • Set your timeout intervals based on this RTT to detect and correct packet loss efficiently.
For example, you might set a timeout to two to three times the average RTT. This approach provides a buffer to account for slight variations in network delays, ensuring timely detection of lost packets without causing unnecessary retransmissions.
Effective timeout scheduling enhances communication reliability, whether using NAK-based or ACK-based ARQ protocols.

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Most popular questions from this chapter

Suppose that we attempt to run the sliding window algorithm with SWS = RWS = 3 and with MaxSeqNum \(=5 .\) The Nth packet DATA[ \(N]\) thus actually contains \(N\) mod 5 in its sequence number field. Give an example in which the algorithm becomes confused; that is, a scenario in which the receiver expects DATA[5] and accepts DATA[0]-which has the same transmitted sequence number-in its stead. No packets may arrive out of order. Note this implies MaxSeqNum \(\geq 6\) is necessary as well as sufficient.

Consider an ARQ algorithm running over a \(20-\mathrm{km}\) point-to-point fiber link. (a) Compute the propagation delay for this link, assuming that the speed of light is \(2 \times 10^{8} \mathrm{~m} / \mathrm{s}\) in the fiber. (b) Suggest a suitable timeout value for the ARQ algorithm to use. (c) Why might it still be possible for the ARQ algorithm to time out and retransmit a frame, given this timeout value?

Suppose you are designing a sliding window protocol for a 1-Mbps point-to- point link to the moon, which has a one-way latency of \(1.25\) seconds. Assuming that each frame carries \(1 \mathrm{~KB}\) of data, what is the minimum number of bits you need for the sequence number?

Why is it important for protocols configured on top of the Ethernet to have a length field in their header, indicating how long the message is?

Suppose Ethernet physical addresses are chosen at random (using true random bits). (a) What is the probability that on a 1024-host network, two addresses will be the same? (b) What is the probability that the above event will occur on some one or more of \(2^{20}\) networks? (c) What is the probability that of the \(2^{30}\) hosts in all the networks of (b), some pair has the same address? Hint: The calculation for (a) and (c) is a variant of that used in solving the socalled Birthday Problem: Given \(N\) people, what is the probability that two of their birthdays (addresses) will be the same? The second person has probability \(1-\frac{1}{365}\) of having a different birthday from the first, the third has probability \(1-\frac{2}{365}\) of having a different birthday from the first two, and so on. The probability all birthdays are different is thus $$ \left(1-\frac{1}{365}\right) \times\left(1-\frac{2}{365}\right) \times \cdots \times\left(1-\frac{N-1}{365}\right) $$ which for smallish \(N\) is about $$ 1-\frac{1+2+\cdots+(N-1)}{365} $$
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