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Another interesting scenario is when the loss happens early in the data transfer. This is shown in the script below where the second segment is lost. We observe that by triggering transmissions of unacknowledged data, the :rfc:`3042` rule speeds up the recovery since a fast retransmit happens. This script is available from :download:`/exercises/packetdrill_scripts/slow-start-frr2.pkt`.
Our last scenario is when the first segment sent is lost. In this case, two round-trip-times are required to retransmit the missing segment and recover from the loss. This script is available from :download:`/exercises/packetdrill_scripts/slow-start-frr3.pkt`.
Open questions
Unless otherwise noted, we assume for the questions in this section that the following conditions hold.
the sender/receiver performs a single :manpage:`send(3)` of `x` bytes
the round-trip-time is fixed and does not change during the lifetime of the TCP connection. We assume a fixed value of 100 milliseconds for the round-trip-time and a fixed value of 200 milliseconds for the retransmission timer.
the delay required to transmit a single TCP segment containing MSS bytes is small and set to 1 milliseconds, independently of the MSS size
the transmission delay for a TCP acknowledgment is negligible
the initial value of the congestion window is one MSS-sized segment
the value of the duplicate acknowledgment threshold is fixed and set to 3
TCP always acknowledges each received segment
To understand the operation of the TCP congestion control, it is often useful to write time-sequence diagrams for different scenarios. The example below shows the operation of the TCP congestion control scheme in a very simple scenario. The initial congestion window (``cwnd``) is set to 1000 bytes and the receive window (``rwin``) advertised by the receiver (supposed constant for the entire connection) is set to 2000 bytes. The slow-start threshold (``ssthresh``) is set to 64000 bytes.
Can you explain why the sender only sends one segment first and then two successive segments (the delay between the two segments on the figure is due to graphical reasons) ?
Can you explain why the congestion window is increased after the reception of the first acknowledgment ?
How long does it take for the sender to deliver 3 KBytes to the receiver ?
Same question as above but now with a small variation. Recent TCP implementations use a large initial value for the congestion window. Draw the time-sequence diagram that corresponds to an initial value of 10000 bytes for this congestion window.
Same question as the first one, but consider that the MSS on the sender is set to 500 bytes. How does this modification affect the entire delay ?
Assuming that there are no losses and that there is no congestion in the network. If the sender writes `x` bytes on a newly established TCP connection, derive a formula that computes the minimum time required to deliver all these `x` bytes to the receiver. For the derivation of this formula, assume that `x` is a multiple of the maximum segment size and that the receive window and the slow-start threshold are larger than `x`.
In question 1, we assumed that the receiver acknowledged every segment received from the sender. In practice, many deployed TCP implementations use delayed acknowledgments. Assuming a delayed acknowledgment timer of 50 milliseconds, modify the time-sequence diagram below to reflect the impact of these delayed acknowledgment. Does their usage decreases or increased the transmission delay ?
Let us now explore the impact of congestion on the slow-start and congestion avoidance mechanisms. Consider the scenario below. For graphical reasons, it is not possible anymore to show information about the segments on the graph, but you can easily infer them.