Question

Suppose a TCP connection has a window size of eight segments, an RTT of \(800 \mathrm{~ms}\), the sender sends segments at a regular rate of one every \(100 \mathrm{~ms}\), and the receiver sends ACKs back at the same rate without delay. A segment is lost, and the loss is detected by the fast retransmit algorithm on the receipt of the third duplicate \(\mathrm{ACK}\). At the point when the ACK of the retransmitted segment finally arrives, how much total time has the sender lost (compared to lossless transmission) if (a) the sender waits for the ACK from the retransmitted lost packet before sliding the window forward again? (b) the sender uses the continued arrival of each duplicate ACK as an indication it may slide the window forward one segment?

Short answer

a) 1100 ms, b) 300 ms.

Step-by-step solution

Understanding the Given Parameters

The window size is eight segments, the RTT (Round-Trip Time) is 800 ms, the sender sends a segment every 100 ms, and the fast retransmit algorithm detects a loss with the third duplicate ACK.

Calculating Time Until Loss Detection

Assume the loss happens at the 9th segment. The ACKs for the first 8 segments arrive in sequence. Once the 9th segment is lost, the sender receives three duplicate ACKs. Given that the sender sends segments every 100 ms, this detection happens in 300 ms (3 duplicate ACKs * 100 ms per ACK).

Calculating Retransmission Time for Lost Segment

After detecting the loss with the third duplicate ACK (at T+300 ms), the sender retransmits the lost segment. The RTT is 800 ms, so the ACK for the retransmitted segment will arrive after 800 ms.

Total Time Lost in Case (a)

In scenario (a), the sender waits for the ACK of the retransmitted segment before sliding the window. The total delay is the time for receiving duplicate ACKs (300 ms) plus the RTT for the retransmitted segment (800 ms), resulting in a total time loss of 1100 ms.

Handling Duplicate ACKs for Case (b)

For scenario (b), the sender uses each duplicate ACK as a signal to slide the window forward and transmit a new segment. The delay is only the 300 ms for receiving the three duplicate ACKs, as the sender does not wait for an ACK of the retransmitted segment before moving the window forward.

Final Time Lost in Case (b)

In scenario (b), the total time lost is just the 300 ms to detect the loss using the third duplicate ACK.

Key concepts

Round-Trip Time (RTT)

Round-Trip Time (RTT) is a critical concept in networking. It is the time it takes for a signal to travel from the sender to the receiver and back again. In our exercise, the RTT is given as 800 ms.
RTT includes the time taken for packet transmission and the propagation delays. Higher RTT can lead to increased latency, impacting the overall performance of network communication.
Understanding RTT helps in optimizing data flow and ensuring efficient packet delivery. Without an accurate measure of RTT, it can be difficult to manage the timing of data packets and acknowledgments effectively.

Fast Retransmit Algorithm

The fast retransmit algorithm is employed to quickly recover from packet loss. It triggers retransmission of a lost packet before the normal timeout period, based on duplicate ACKs received.
Here’s how it works:

• The sender monitors the arrival of ACKs for sent packets.
• If three duplicate ACKs for the same packet are received, the sender assumes the packet is lost.
• Instead of waiting for a timeout, the sender retransmits the lost packet immediately.

This rapid retransmission helps reduce the time lost due to packet loss, ensuring quicker data delivery and better network performance.

Duplicate ACKs

Duplicate ACKs are a key signal that something has gone wrong in transmission. When a packet is received out of order or a packet is lost, the receiver sends an acknowledgment (ACK) for the last in-sequence packet it received.
For example:

• If packets 1, 2, 3, and 4 are sent but packet 3 is lost, the receiver will keep acknowledging packet 2, resulting in duplicate ACKs for packet 2.

The sender uses these duplicate ACKs to detect packet loss. According to our exercise, upon receiving the third duplicate ACK, the fast retransmit algorithm is triggered to retransmit the lost packet without waiting for a timeout period.

Window Size

Window size in TCP determines how many segments can be sent without receiving an ACK. In our example, the window size is eight segments. This means the sender can send eight packets before needing to wait for an ACK to send more.
Various factors affect the optimal window size, including network bandwidth, RTT, and the sender/receiver capacities. A properly set window size ensures efficient data transfer and minimizes congestion in the network.
When facing packet loss, the sender can adjust the window size to control the flow of packets, thus accommodating the network conditions more effectively.

Packet Loss Detection

Detecting packet loss is essential for maintaining data integrity in TCP communication. It can be identified through various mechanisms, primarily using timeouts and duplicate ACKs.
In our exercise, packet loss is detected with three duplicate ACKs, which triggers the fast retransmit algorithm. Here’s a brief guide on these mechanisms:

• Timeouts: If an ACK does not arrive within the expected time (RTT), the sender assumes packet loss and retransmits the packet.
• Duplicate ACKs: Multiple duplicate ACKs for the same packet indicate packet loss, prompting immediate retransmission.

Fast packet loss detection minimizes data delay and ensures smoother communication despite network issues.