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In Section 3.5.4, we saw that TCP waits until it has received three duplicate ACKs before performing a fast retransmit. Why do you think the TCP designers chose not to perform a fast retransmit after the first duplicate ACK for a segment is received?

Short Answer

Expert verified
Three duplicate ACKs indicate likely packet loss, preventing unnecessary retransmissions caused by packet reordering.

Step by step solution

01

Understand the Scenario

In TCP (Transmission Control Protocol), data packets are sent and acknowledged by the receiver. The sender gets an acknowledgment (ACK) each time a packet is successfully received. If a segment of data is missing, the receiver sends a duplicate ACK for the last segment it received. The sender needs to decide when to retransmit any potentially lost segments.
02

Define Fast Retransmit

Fast retransmit is a mechanism that allows the sender to retransmit a packet presumed to be lost after receiving a certain number of duplicate ACKs, usually three. This helps recover lost data without waiting for a timeout.
03

Consider Immediate Duplicate ACK

If TCP initiated a fast retransmit upon receiving the first duplicate ACK, it might be too hasty. A single duplicate ACK does not necessarily indicate packet loss; it could result from minor out-of-order packet arrival, which is normal in network traffic.
04

Analyze Three Duplicate ACKs Strategy

By waiting for three duplicate ACKs, TCP distinguishes between actual packet loss and simple packet reordering. Three duplicate ACKs strongly suggest that a packet was lost, as it implies several packets arrived out of order due to the earlier packet not being received or acknowledged.
05

Discuss the Balance of Performance and Accuracy

The strategy of waiting for three duplicate ACKs balances the need to quickly recover from genuine packet loss while avoiding unnecessary retransmissions due to occasional out-of-order packet arrivals. This approach reduces network congestion and improves the performance of data transfers.

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Key Concepts

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

Understanding Duplicate ACK
In the Transmission Control Protocol (TCP), data is exchanged between sender and receiver packets. Each time a packet reaches its destination successfully, the receiver sends back an acknowledgment or ACK. When some packets go missing, the receiver sends what is called a duplicate ACK. This kind of ACK is specifically for the last packet that arrived correctly. It's the receiver's way of signaling, "Hey, I got this one, but I haven't seen the ones coming after it yet."

If a sender gets even just one duplicate ACK, it's not a clear sign of packet loss right away. It might just mean a packet arrived out of order, which is common in busy networks. Thus, immediately retransmitting a packet on the first duplicate could lead to unnecessary action if the packet was just delayed or reordered.
Decoding Packet Loss
Packet loss happens when one or more packets of data traveling across a network fail to reach their destination. In a reliable protocol like TCP, this loss has to be detected, and the missing packets must be resent. Packet loss might lead to missing parts of a conversation or an interrupted stream of data if not addressed properly.

Packet loss can occur due to various reasons:
- **Network Congestion**: Overloaded network paths can drop packets to reduce the load.
- **Faulty Hardware**: Old or malfunctioning routing devices can result in packet drops.
- **Signal Interference**: In wireless communications, interference from other devices can cause packet loss.

Fast retransmit is crucial as it triggers a resend of potentially lost packets sooner than waiting for a timeout, helping maintain the smooth flow of data.
What is Network Congestion?
Network congestion arises when a network segment experiences too much data traffic, clashing with its maximum capacity. Think of it like a busy highway in rush hour where every car moves at a crawl because the road can't handle the number of vehicles trying to use it.

This congestion can lead to packet loss as the network tries to manage the load, sometimes by dropping packets. TCP's role is to ensure data gets through despite these hiccups. It does so by managing how much data it sends at a given time, trying to operate just under the threshold where congestion becomes a problem.

This is why a mechanism like fast retransmit is critical, as it helps avoid exacerbating the congestion by quickly correcting packet losses without flooding the network with more data.
Navigating Packet Reordering
Packets can sometimes reach their destination in a different order from how they were sent. This is known as packet reordering and is normal in a dynamic network environment where packets might traverse different paths.

Unlike packet loss, packet reordering is not necessarily a problem but is something TCP must account for. Receiving packets out of order occasionally results in duplicate ACKs, as the receiver confirms receipt of the packets it has received while waiting for the others.

Therefore, TCP waits for three duplicate ACKs before fast retransmitting because this patience allows the protocol to distinguish between packets that are truly lost versus those that are simply arriving in the wrong order. This cautious approach helps maintain performance without unnecessary retransmissions.

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Most popular questions from this chapter

Consider the TCP procedure for estimating RTT. Suppose that \(\alpha=0.1\). Let SampleRTT \(_{1}\) be the most recent sample RTT, let SampleRTT \(_{2}\) be the next most recent sample RTT, and so on. a. For a given TCP connection, suppose four acknowledgments have been returned with corresponding sample RTTs: SampleRTT \(_{4}\), SampleRTT \(_{3}\), SampleRTT \(_{2}\), and SampleRTT \(_{1}\). Express EstimatedRTT in terms of the four sample RTTs. b. Generalize your formula for \(n\) sample RTTs. c. For the formula in part (b) let \(n\) approach infinity. Comment on why this averaging procedure is called an exponential moving average.

Consider sending a large file from a host to another over a TCP connection that has no loss. a. Suppose TCP uses AIMD for its congestion control without slow start. Assuming cwnd increases by 1 MSS every time a batch of ACKs is received and assuming approximately constant round-trip times, how long does it take for cwnd increase from 6 MSS to 12 MSS (assuming no loss events)? b. What is the average throughout (in terms of MSS and RTT) for this connection up through time \(=6\) RTT?

In this problem we investigate whether either UDP or TCP provides a degree of end-point authentication. a. Consider a server that receives a request within a UDP packet and responds to that request within a UDP packet (for example, as done by a DNS server). If a client with IP address \(\mathrm{X}\) spoofs its address with address Y, where will the server send its response? b. Suppose a server receives a SYN with IP source address Y, and after responding with a SYNACK, receives an ACK with IP source address Y with the correct acknowledgment number. Assuming the server chooses a random initial sequence number and there is no "man-in-the-middle," can the server be certain that the client is indeed at \(Y\) (and not at some other address \(\mathrm{X}\) that is spoofing \(\mathrm{Y})\) ?

Why is it that voice and video traffic is often sent over TCP rather than UDP in today's Internet? (Hint: The answer we are looking for has nothing to do with TCP's congestion-control mechanism.)

True or false? a. Host \(A\) is sending Host \(B\) a large file over a TCP connection. Assume Host B has no data to send Host A. Host B will not send acknowledgments to Host A because Host B cannot piggyback the acknowledgments on data. True or false? a. Host \(\mathrm{A}\) is sending Host \(\mathrm{B}\) a large file over a \(\mathrm{TCP}\) connection. Assume Host B has no data to send Host A. Host B will not send acknowledgments to Host A because Host B cannot piggyback the acknowledgments on data. b. The size of the TCP rwnd never changes throughout the duration of the connection. c. Suppose Host A is sending Host B a large file over a TCP connection. The number of unacknowledged bytes that A sends cannot exceed the size of the receive buffer. d. Suppose Host \(\mathrm{A}\) is sending a large file to Host \(\mathrm{B}\) over a TCP connection. If the sequence number for a segment of this connection is \(m\), then the sequence number for the subsequent segment will necessarily be \(m+1\). e. The TCP segment has a field in its header for rwnd. f. Suppose that the last SampleRTT in a TCP connection is equal to \(1 \mathrm{sec}\). The current value of TimeoutInterval for the connection will necessarily be \(\geq 1 \mathrm{sec}\). g. Suppose Host A sends one segment with sequence number 38 and 4 bytes of data over a TCP connection to Host B. In this same segment the acknowledgment number is necessarily 42 .

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